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ISBN: 0-02-834177-5
Printed in the United States of America.
4 5 6 7 8 9 10 024 08 07 06 05 04

Chapter 1 Linear Relations and Functions
8.

Relations and Functions

1-1

Pages 8–9

Check for Understanding

1.
x

4
6
0
8
2

4

y

y
2
1
5

4
2
0

4
2

8 
4 O

2

4

8x

y

x

3. Determine whether a vertical line can be drawn
through the graph so that it passes through more
than one point on the graph. Since it does, the
graph does not represent a function.
4. Keisha is correct. Since a function can be
expressed as a set of ordered pairs, a function is
always a relation. However, in a function, there is
exactly one y-value for each x-value. Not all
relations have this constraint.
5. Table:
Graph:
x
1
2
3
4
5
6
7

y

3

2

1
0
1
2
3

x
1
2
3
4
5
6
7
8

y

5

5

5

5

5

5

5

5

y
12
8
4

4

O

2

4x


4

y

O

x

10. {
3, 0, 1, 2}; {
6, 0, 2, 4}; yes; Each member of the
domain is matched with exactly one member of
the range.
11. {
3, 3, 6}; {
6, 
2, 0, 4}; no; 6 is matched with
two members of the range.
12a. domain: all reals; range: all reals
12b. Yes; the graph passes vertical line test.
13. f(
3)  4(
3)3  (
3)2 
 5(
3)
 
108  9  15 or 
84
14. g(m  1)  2(m  1)2 
 4(m  1)  2
 2(m2  2m  1) 
 4m 
 4  2
 2m2  4m  2 
 4m 
 4  2
 2m2
15. x  1  0
x  
1
The domain excludes numbers less than 
1.
The domain is {xx  
1}.
16a. {(83, 240), (81, 220), (82, 245), (78, 200),
(83, 255), (73, 200), (80, 215), (77, 210), (78, 190),
(73, 180), (86, 300), (77, 220), (82, 260)}; {73, 77,
78, 80, 81, 82, 83, 86}; {180, 190, 200, 210, 215,
220, 240, 245, 255, 260, 300}

y

O

y

7

4

1
2
5
8
11
14
17

9.


4

2. Sample answer:

O

x

4

3

2

1
0
1
2
3
4

x

Equation: y  x 
4
6. {(
3, 4), (0, 0), (3,
4), (6, 
8)}; D  {
3, 0, 3, 6};
R  {
8, 
4, 0, 4}
7. {(
6, 1), (
4, 0), (
2, 
4), (1, 3), (4, 3)};
D  {
6, 
4, 
2, 1, 4}; R  {
4, 0, 1, 3}

1

Chapter 1

16b.

22. {(4, 0), (5, 1), (8, 0), (13, 1)};
D  {4, 5, 8, 13}; R  {0, 1}
23. {(
3, 
2), (
1, 1), (0, 0), (1, 1)};
D  {
3, 
1, 0, 1}; R  {
2, 0, 1}
24. {(
5, 5), (
3, 3), (
1, 1), (2, 
2), (4, 
4)};
D  {
5, 
3, 
1, 2, 4}; R  {
4, 
2, 1, 3, 5}
25. {(3, 
4), (3, 
2), (3, 0), (3, 1), (3, 3)}; D  {3};
R  {
4, 
2, 0, 1, 3}
y
26.
O
x
y

300
280
260
Weight 240
(lb)
220
200
180

O

70 72 74 76 78 80 82 84 86
Height (in.)

16c. No; a vertical line at x  77, x  78, x  82, or
x  83 would pass through two points.

Pages 10–12

Exercises

17. Table
x
1
2
3
4
5
6
7
8
9

y
24
18
12
6

O

2

4

6

10 x

8

x

4

3

2

1
0
1
2
3
4

y
O

y

11

10

9

8

7

6

Equation: y  x 
 5
19. Table:
y
4
5
6
7
8
9
10
11
12

y

1

2

3

4

5

6

x

5

4

3

2

1
0
1

x

y
x

O

y

y
5
4
3
2
1
0
1

x

O

29.
x
1
2
3
4
5

Graph:

y

y

y
0
3
6
9
12

x

O
30.

O

x
11
11

x

Equation: y  8  x
20. {(
5, 
5), (
3, 
3), (
1, 
1), (1, 1)};
D  {
5, 
3, 
1, 1}; R  {
5, 
3, 
1, 1}
21. {(
10, 0), (
5, 0), {0, 0), (5, 0)};
D  {
10, 
5, 0, 5}; R  {0}

Chapter 1

x
1
2
3
4
5
6
28.

Equation: y  3x
18. Table:
x

6

5

4

3

2

1


9

8

7

6

5

4

27.

Graph:
y
3
6
9
12
15
18
21
24
27


4

3

2

1
0
1

x

y
3

3

4
2

O
–2
–4

2

y

4

8

12 x

31.

x
4
4

y

y
2

2

O

3



5

51c.

x


2, 2

Number 5
Attending
(thousands) 4
3
2

O 12 16 20 24 28
Number Applied
(thousands)

52c. Yes; no member of the domain is paired with
more than one member of the range.

 1)


1

x

6

2
44. j(2a)  1 
 4(2a)3
 1 
 4(8a3)
 1 
 32a3
45. f(n 
 1)  2(n 
 1)2 
 (n 
 1)  9
 2(n2 
 2n  1) 
 n  1  9
 2n2 
 4n  2 
 n  1  9
 2n2 
 5n  12
b2

51b.

7

1

3


1

52a. {(13,264, 4184), (27,954, 4412), (21,484, 6366),
(23,117, 3912), (16,849, 2415), (19,563, 5982),
(17,284, 6949)}; {13,264, 16,849, 17,284, 19,563,
21,484, 23,117, 27,954}; {2415, 3912, 4184, 4412,
5982, 6366, 6949}
52b.


43. h(0.5)  
0.5


46. g(b2  1)  
5  (b2  1)

x

x

32. {4, 5, 6}; {4}; yes; Each x-value is paired with
exactly one y-value.
33. {1}; {
6, 
2, 0, 4}; no; The x-value 1 is paired with
more than one y-value.
34. {0, 1, 4); {
2, 
1, 0, 1, 2}; no; The x-values 1 and 4
are paired with more than one y-value.
35. {0, 2, 5}; {
8, 
2, 0, 2, 8}; no; The x-values 2 and 5
are paired with more than one y-value.
36. {
1.1, 
0.4, 
0.1}; {
2, 
1}; yes; Each x-value is
paired with exactly one y-value.
37. {
9, 2, 8, 9}; {
3, 0, 8}; yes; Each x-value is paired
with exactly one y-value.
38. domain: all reals; range: all reals; Not a function
because it fails the vertical line test.
39. domain: {
3, 
2, 
1, 1, 2, 3}; range: {
1, 1, 2, 3};
A function because each x-value is paired with
exactly one y-value.
40. domain: {x
8  x  8}; range: {y
8  y  8};
Not a function because it fails the vertical line
test.
41. f(3)  2(3)  3
 6  3 or 9
42. g(
2)  5(
2)2  3(
2) 
 2
 20 
 6 
 2 or 12

(b2

51a.

x
1

53. x  2m  1, so 2  m.

2 
 b2




6  b2 or 6  b2

47. f(5m)  (5m)2 
 13
 25m2 
 13
2
48. x 
 5  0
x2  5
x  
5
; x 
5

49. x2 
 9  0
x2  9

3  x  3; x  
3 or x  3
50. x2 
 7  0
x2  7


7
  x  
7
; x  

7
 or x

x
1
Substitute 2 for m
24m3  36m2  26m
x
1 3

in f(2m  1) to solve for f(x),

x
1 2

x
1

 242  362  262
x3 
 3x2  3x 
 1

x2 
 2x  1

x
1

 24 
 364  262
8

 3x3 
 9x2  9x 
 3  9x2 
 18x  9  13x 
 13
 3x3  4x 
 7
54a. t(500)  95 
 0.005(500)
 92.5°F
54b. t(750)  95 
 0.005(750)
 91.25°F
54c. t(1000)  95 
 0.005(1000)
 90°F
54d. t(5000)  95 
 0.005(5000)
 70°F


7


3

Chapter 1

54e. t(30,000)  95 
 0.005(30,000)
 
55°F
55a. d(0.05)  299,792,458(0.05)
 14,989,622.9 m
d(0.02)  299,792,458(0.2)
 59,958,491.6 m
d(1.4)  299,792,458(1.4)
 419,709,441.2 m
d(5.9)  299,792,458(5.9)
 1,768,775,502 m
55b. d(0.008)  299,792,458(0.08)
 23,983,396.64 m

3.

(1)(2)  1

56. P(4)  3  1
(2)(3)  1

P(5)  1  7
(3)(1)  1

4. Sample answer: The (sum/difference/product/
quotient) of the function values is the function
values of the (sum/difference/product/quotient)
of the functions.
5. Sample answer: For functions f(x) and g(x),
(f  g)(x)  f(x)  g(x); (f 
 g)(x)  f(x) 
 g(x);

4

P(6)  7  7
57. 72 
 (32  42)  49 
 (9  16)
 49 
 25 or 24
The correct choice is B.

f

1-2
Page 13

Composition of Functions

Page 17

Graphing Calculator Exploration

0

Check for Understanding

1. Sample answer: f(x)  2x 
 1 and g(x)  x  6;
Sample explanation: Factor 2x2  11x 
 6.
2. Iteration is composing a function on itself by
evaluating the function for a value and then
evaluating the function on that function value.
3. No; [f 
 g](x) is the function f(x) performed on g(x)
and [g 
 f ](x) is the function g(x) performed on f(x).
See students’ counter examples.
4. Sample answer: Composition of functions is
performing one function after another. An
everyday example is putting on socks and then
putting shoes on top of the socks. Buying an item
on sale is an example of when a composition of
functions is used in a real-world situation.
5. f(x)  g(x)  3x2  4x 
 5  2x  9
 3x2  6x  4
f(x) 
 g(x)  3x2  4x 
 5 
 (2x  9)
 3x2  2x 
 14
f(x) g(x)  (3x2  4x 
 5)(2x  9)
 6x3  35x2  26x 
 45

1.

2.

f

f(x)

g(x)  g(x)

3x2  4x 
 5

, x

2x  9

6. [f 
 g](x)  f(g(x))
 f(3  x)
 2(3  x)  5
 2x  11
[g 
 f ](x)  g(f(x))
 g(2x  5)
 3  (2x  5)
 2x  8

Chapter 1

f(x)


(f g)(x)  f(x) g(x); and g(x)  
g(x) , g(x)

4

9


2

7. [f 
 g](x)  f(g(x))
 f(x2 
 2x)
 2(x2 
 2x) 
 3
 2x2 
 4x 
 3
[g 
 f ](x)  g(f(x))
 g(2x 
 3)
 (2x 
 3)2 
 2(2x 
 3)
 (4x2 
 12x  9) 
 4x  6
 4x2 
 16x  15
8. Domain of f(x): x 1
Domain of g(x): all reals
g(x)  1
x31
x  
2
Domain of [f 
 g](x) is x 
2.
9. x1  f(x0)  f(2)
 2(2)  1 or 5
x2  f(x1)  f(5)
 2(5)  1 or 11
x3  f(x2)  f(11)
 2(11)  1 or 23
5, 11, 23
10a. [K 
 C](F)  K(C(F))
5
 K9(F 
 32)
5
 9(F 
 32)  273.15
5
10b. K(
40)  9(
40 
 32)  273.15
 
40  273.15 or 233.15

x

2

f(x) 
 g(x)  
x  1 
 (x 
 1)

f(x)  g(x)



13.

 
17.78  273.15 or 255.37
5

 0  273.15 or 273.15

14.

5

K(212)  9(212 
 32)  273.15
 100  273.15 or 373.15

Pages 17–19

Exercises

11. f(x)  g(x)  x2 
 2x  x  9
 x2 
 x  9
f(x) 
 g(x)  x2 
 2x 
 (x  9)
 x2 
 3x 
 9
f(x) g(x)  (x2 
 2x)(x  9)
 x3  7x2 
 18x
x2 
 2x


g(x)  
x9 , x
f

9

x

2

12. f(x)  g(x)  
x1  x 
 1






x3 
 x2  2x  1
 ,
x1
x
 (x2 
 1)

x1
x(x  1)(x 
 1)


x1
2
 x 
 x, x 
1

x


1

(x)  
x2 
 1
x
1


 x 
1 x2 
 1
x



1 or 1
x3  x2 
 x 
 1 , x
3
2

f(x)  g(x)  
x 
 7  x  5x
3
(x2  5x)(x 
 7)



x
7 
x
7
x3 
 7x2  5x2 
 35x
3


 x
7 
x
7
x3 
 2x2 
 35x  3
 
,x 7
x
7
3
2

f(x) 
 g(x)  
x 
 7 
 (x  5x)
(x2  5x)(x 
 7)
3



x
7
x
7 

3
x3 
 7x2  5x2 
 35x


 x
7 

x
7
x3 
 2x2 
 35x 
 3
 
 
,x 7
x
7
3
2

f(x) g(x)  
x 
 7 (x  5x)
2
3x  15x

7

x
7 , x
3

x
7
f
 (x)  
g
x2  5x
3
1
 

x 
 7 x2  5x
3
, x


5, 0, 7
x3 
 2x2 
 35x
2x

f(x)  g(x)  x  3  
x
5
(x  3)(x 
 5)
2x
  

x
5
x
5
x2 
 2x 
 15
2x
  

x
5
x
5
x2 
 15


5
x
5 , x
2x

f(x) 
 g(x)  x  3 
 
x
5
(x  3)(x 
 5)
2x
 
 

x
5
x
5
x2 
 2x 
 15
2x
 
 

x
5
x
5
x2 
 4x 
 15
, x

5
x
5
2x

f(x) g(x)  (x  3) 
 
x
5
2x2  6x


5
x
5 , x



5

 9(32 
 32)  273.15

x3  x2 
 x 
 1

x

x1

f

g

 
24.44  273.15 or 248.71

K(32)

x



5

 9(0 
 32)  273.15

(x2 
 1)(x  1)




x1 

x1

K(
12)  9(
12 
 32)  273.15
K(0)

x




x1 

x1









x3

g(x)  
2x

f

(x2 
 1)(x  1)
x
  
x1
x1
x
x3  x2 
 x 
 1
  
x1
x1
x3  x2 
 1
, x

1
x1

x
5

x
5

2x
x2 
 2x 
 15
, x
2x

x3


5

0 or 5

Chapter 1

15. [f 
 g](x)  f(g(x))
 f(x  4)
 (x  4)2 
 9
 x2  8x  16 
 9
 x2  8x  7
[g 
 f ](x)  g(f(x))
 g(x2 
 9)
 x2 
 9  4
 x2 
 5
16. [f 
 g](x)  f(g(x))
 f(x  6)

21. [f 
 g](x)  f (g(x))


 f 
x 
 1
1

1



x
1  1

x



x 
 1, x

1

[g 
 f ](x)  g(f(x))
 g(x  1)
1



x1
1
1

 x, x

1

 2(x  6) 
 7

0

22. Domain of f(x): all reals
Domain of g(x): all reals
Domain of [f 
 g](x): all reals
23. Domain of f(x): x 0
Domain of g(x): all reals
g(x)  0
7
x0
7x
Domain of [f 
 g](x) is x 7.
24. Domain of f(x): x  2
Domain of g(x): x 0
g(x)  2

1

 2x  3 
 7
1

 2x 
 4
[g 
 f ](x)  g(f(x))
1

 g(2x 
 7)
1

 2x 
 7  6
1

 2x 
 1
17. [f 
 g](x)  f(g(x))
 f(3x2)
 3x2 
 4
[g 
 f ](x)  g(f(x))
 g(x 
 4)
 3(x 
 4)2
 3(x2 
 8x  16)
 3x2 
 24x  48
18. [f 
 g](x)  f(g(x))
 f(5x2)
 (5x2)2 
 1
 25x4 
 1
[g 
 f ](x)  g(f(x))
 g(x2 
 1)
 5(x2 
 1)2
 5(x4 
 2x2  1)
 5x4 
 10x2  5
19. [f 
 g](x)  f(g(x))
 f(x3  x2  1)
 2(x3  x2  1)
 2x3  2x2  2
[g 
 f ](x)  g(f(x))
 g(2x)
 (2x)3  (2x)2  1
 8x3  4x2  1
20. [f 
 g](x)  f(g(x))
 f(x2  5x  6)
 1 x2  5x  6
 x2  5x  7
[g 
 f ](x)  g(f(x))
 g(1  x)
 (x  1)2  5(x  1)  6
 x2  2x  1  5x  5  6
 x2  7x  12

Chapter 1

x
1

1




x
1  x
1

1
x
4

2

1  8x
1

8

x
1

Domain of [f 
 g](x) is x  8, x
25. x1  f(x0)  f(2)
 9 
 2 or 7
x2  f(x1)  f(7)
 9 
 7 or 2
x3  f(x2)  f(2)
 9 
 2 or 7
7, 2, 7
26. x1  f(x0)  f(1)
 (1)2  1 or 2
x2  f(x1)  f(2)
 (2)2  1 or 5
x3  f(x2)  f(5)
 (5)2  1 or 26
2, 5, 26
27. x1  f(x0)  f(1)
 1(3 
 1) or 2
x2  f(x1)  f(2)
 2(3 
 2) or 2
x3  f(x2)  f(2)
 2(3 
 2) or 2
2, 2, 2

6

0.

34a. I  prt
 5000(0.08)(1)
 400
I  prt
 5400(0.08)(1)
 432
I  prt
 5832(0.08)(1)
 466.56
I  prt
 6298.56(0.08)(1)
 503.88
I  prt
 6802.44(0.08)(1)
 544.20
(year, interest): (1, $400), (2,$432), (3, $466.56),
(4, $503.88), (5, $544.20)
34b. {1, 2, 3, 4, 5}; {$400, $432, $466.56, $503.88,
$544.20}
34c. Yes; for each element of the domain there is
exactly one corresponding element of the range.
35. {(
1, 8), (0, 4), (2, 
6), (5, 
9)}; D  {
1, 0, 2, 5};
R  {
9, 
6, 4, 8}
36. D  {1, 2, 3, 4}; R  {5, 6, 7, 8}; Yes, every element
in the domain is paired with exactly one element
of the range.

28. $43.98  $38.59  $31.99  $114.56
Let x  the original price of the clothes, or
$114.56.
Let T(x)  1.0825x. (The cost with 8.25% tax rate)
Let S(x)  0.75x. (The cost with 25% discount)
The cost of clothes is [T 
 S](x).
[T 
 S](x)  T(S(x))
 T(0.75x)
 T(0.75(114.56))
 T(85.92)
 1.0825(85.92)
 93.0084
Yes; the total with the discount and tax is $93.01.
29. Yes; If f(x) and g(x) are both lines, they can be
represented as f(x)  m1x  b1 and g(x) 
m2x  b2. Then [f 
 g](x)  m1(m2x  b2)  b1
 m1m2x  m1b2  b1
Since m1 and m2 are constants, m1m2 is a
constant. Similarly, m1, b2, and b1 are constants,
so m1b2  b1 is a constant. Thus, [f 
 g](x) is a
linear function if f(x) and g(x) are both linear.
30a. Wn  Wp 
 Wf
 Fpd 
 Ff d
 d(Fp 
 Ff)
30b. Wn  d(Fp 
 Ff)
 50(95 
 55)
 2000 J
31a. h[f(x)], because you must subtract before
figuring the bonus
31b. h[f(x)]  h[f(400,000)]
 h(400,000 
 275,000)
 h(125,000)
 0.03(125,000)
 $3750
32. (f 
 g)(x)  f(g(x))
 f(1 
 x2)

(
4)3  5


37. g(
4)  
4(
4)

64  5




16

59

38.

x2(x2  1)



1  x2

 x2
 
(1 
 x2)  1
7p

1

1

33b. r(v)  0.84v

33a. v(p)  47

x

2

1
0
1
2
3

y
6
3
0
3
6
9

y

O

x

39. f(n 
 1)  2(n 
 1)2 
 (n 
 1)  9
 2(n2 
 2n  1) 
 n  1  9
 2n2 
 5n  12
The correct choice is C.

So, f(x)  
x  1 and f 2  
2  1  2.
1

11





16 or 3 16

33c. r(p)  r(v(p))
7p

 r47

7p

 0.8447
5.88p

1-3

14 7p


 4
7 or 1175

Graphing Linear Equations

147(4 23.18)


33d. r(423.18)  
1175

Page 23

 $52.94

Check for Understanding

1. m represents the slope of the graph and b
represents the y-intercept
2. 7; the line intercepts the x-axis at (7, 0)
3. Sample answer: Graph the y-intercept at (0, 2).
Then move down 4 units and right 1 unit to graph
a second point. Draw a line to connect the points.

147(2 25.64)


r(225.64)  
1175

 $28.23
147(7 97.05)


r(797.05)  
1175

 $99.72

7

Chapter 1

4. Sample answer: Both graphs are lines. Both lines
have a y-intercept of 8. The graph of y  5x  8
slopes upward as you move from left to right on
the graph and the graph of y  
5y  8 slopes
downward as you move from left to right on the
graph.
5. 3x 
 4(0)  2  0
3(0) 
 4y  2  0
3x  2  0

4y  2  0
3x  
2

4y  
2
2

1

9. 2x  6  0
1
x
2

y
12
8
4

1

x  
3

y  2


12 
8 
4 O

4

y

(
 23 ,0)

 
6

x  
12

x

10. Since m  0 and b  19, this function has no
x-intercept, and therefore no zeros.

(0, 12 )

y

x

O

24
16

6. x  2(0) 
 5  0
x
50
x5

0  2y 
 5  0
2y 
 5  0
2y  5

8

4 
2 O

5

y  2

2

4x

11a. (38.500, 173), (44.125, 188)

y

188 
 173


11b. m  
44.125 
 38.500

(0, 52 )

15



5.625 or about 2.667

11c. For each 1 centimeter increase in the length of a
man’s tibia, there is an 2.667-centimeter
increase in the man’s height.

x
O

(5, 0)

7. The y-intercept is 7. Graph (0, 7).
The slope is 1.

y

Pages 24–25

Exercises

12. The y-intercept is 
9. The slope is 4.

y

(1, 8)
(0, 7)

x

O

y  4x 
 9

O

x

8. The y-intercept is 5. Graph (0, 5).
The slope is 0.

13. The y-intercept is 3. The slope is 0.

y

y

y3

(0, 5)

O
O

Chapter 1

x

8

x

19. 2x  y  0
y  
2x
The y-intercept is 0. The slope is 
2.

14. 2x 
 3y  15  0

3y  
2x 
 15
2

y  3x  5

y

2

The y-intercept is 5. The slope is 3.

y
O

2x 
 3y  15  0

x

2x  y  0

2

20. The y-intercept is 
4. The slope is 3.

O x

y

15. x 
 4  0
x4
There is no slope. The x-intercept is 4.

O

x

y
y  23 x 
 4

x
40

x

O

21. The y-intercept is 150. The slope is 25.

y
y  25x  150
100

16. The y-intercept is 
1. The slope is 6.

50

y


6 
4 
2 O
50

y  6x 
 1

O

2x

22. 2x  5y  8
5y  
2x  8

x

2

8

y  
5x  5
8

2

The y-intercept is 5. The slope is 
5.

17. The y-intercept is 5. The slope is 
2.

y

y

2x  5y  8

y  5 
 2x

x
O
O

x
23. 3x 
 y  7

y  
3x  7
y  3x 
 7
The y-intercept is 
7. The slope is 3.

18. y  8  0
y  
8
The y-intercept is 
8. The slope is 0.

y
O

y

x

x

O
3x 
 y  7

y80

9

Chapter 1

24. 9x  5  0
9x  
5

33a. (1.0, 12.0), (10.0, 8.4)

f (x )

8.4 
 12.0


m
10.0 
 1.0

f (x )  9x  5

5

x  
9


3.6

 9 or 
0.4

The y-intercept is 5.


(
0.4)  0.4 ohms

(
 59 , 0)

12 
 v

25. 4x 
 12  0
4x  12
x3
The y-intercept is 
12.


33b. 
0.4  
1.0 
 25.0

x

O

12 
 v



0.4  

24

f (x )

9.6  12 
 v
v  2.4 volts

f (x )  4x 
 12

9
7

(3, 0)

2

26. 3x  1  0
3x  
1




7

x

O

(
 13 ,0)

The y-intercept is 1.

m

x



f (x )

1

14
(0, 0)

there is a 4-pascal increase in the pressure.

f (x )  14x

35c. 100 P

x

80
60

28. None; since m  0 and b  12, this function has
no x-intercepts, and therefore no zeros.

40

f (x )

20

f (x )  12
8

O

4

f (x )
f (x )  5x 
 8
2

8

5

The y-intercept is 
8.


2 O

4

6

8

( 85 , 0)
x

2

30. 5x 
 2  0
5x  2
2

x  5
3

10,000

31. The y-intercept is 3. The slope is 
2.
3


2x  3  0

20

y

 
3
x2

60

80 T

(0, 10,440)

8000

y

3

2x

40

36. No; the product of two positives is positive, so for
the product of the slopes to be 
1, one of the
slopes must be positive and the other must be
negative.
37a. 10,440 
 290t  0

290t  
10,440
t  36
The software has no monetary value after
36 months.
37b. 
290; For every 1-month change in the number
of months, there is a $290 decrease in the value
of the software.
v (t )
37c.

x

29. 5x 
 8  0
5x  8
x

100 – 90

126.85 – 86.85
10
1
 or 
40
4

35b. For each 1 degree increase in the temperature,

28

O

O


 32 x

3

6000
4000

(2, 0)

O

2000

x

(36, 0)

O

32. Sample answer: f(x)  5; f(x)  0

Chapter 1


8

35a. (86.85, 90), (126.85, 100)

O
27. 14x  0
x0
The slope is 14.

2


2(a  4)  
56

2a 
 8  
56

2a  
48
a  24

f (x )  3x  1

1

1
9



7  
a4

f (x )

x  
3

2



7  
a 
 (
4)


34. m  

4 
 3

10

8

16

24

32

t

38. A function with a slope of 0 has no zeros if its
y-intercept is not 0; a function with a slope of 0
has an infinite number of zeros if its y-intercept is
0; a function with any slope other than 0 has
exactly 1 zero.
39a. (56, 50), (76, 67.2)

1-3B Graphing Calculator Exploration:

Analyzing Families of Linear
Graphs

Page 26

67.2 
 50


m
76 
 56

1. See students’ graphs. All of the graphs are lines
with y-intercept at (0, 
2). Each line has a
different slope.
2. A line parallel to the ones graphed in the Example
and passing through (0, 
2).
3. See students’ sketches. Sample answer: The
graphs of lines with the same value of m are
parallel. The graphs of lines with the same value
for b have the same y-intercept.

17.2

 2
0 or 0.86
39b. 1805(0.86)  $1552.30
39c. 1 
 MPC  1 
 0.86
 0.14
39d. 1805(0.14)  $252.70
40. (f  g)(x)  2x  x2 
 4 or x2  2x 
 4
(f 
 g)(x)  2x 
 (x2 
 4)
 
x2  2x  4
41a. 1 
 0.12  0.88
d(p)  0.88p
41b. r(d)  d 
 100
41c. r(d(p))  r(0.88p)
 0.88p 
 100
41d. r(799.99)  0.88(799.99) 
 100
 603.9912 or about $603.99
r(999.99)  0.88(999.99) 
 100
 779.9912 or about $779.99
r(1499.99)  0.88(1499.99) 
 100
 1219.9912 or about $1219.99
42. [f 
 g](
3)  f(g(
3))
 f(
3 
 2)
 f(
5)
 (
5)2 
 4(
5)  5
 25  20  5 or 50
[g 
 f ](
3)  g( f(
3))
 g((
3)2 
 4(
3)  5)
 g(9  12  5)
 g(26)
 26 
 2 or 24
43. f(9)  4  6(9) 
 (9)3
 4  54 
 729 or 
671
44. No; the graph fails the vertical line test.
45.
x
y

3
14

2
13

1
12
{(
3, 14), (
2, 13), (
1, 12),
(0, 11)}, yes
0
11

1-4

Page 29 Check for Understanding
1. slope and y-intercept; slope and any point; two
points
2. Sample answer:
Use point-slope
Use slope-intercept
form:
form.
y 
 y1  m(x 
 x1)
y  mx  b
1

y 
 (
4)  4(x 
 3)
1

3

y  4  4x 
 4
x 
 4y 
 19  0

1


4  4(3)  b
19


4  b
Substitute the slope
and intercept into the
general form.
1

19

y  4x 
 4
Write in standard form.
x 
 4y 
 19  0
3. 55 represents the hourly rate and 49 represents
the fee for coming to the house.

3 
 0

1


4. m  
0
6

3

y  2x 
 3
1





6 or 2

5. Sample answer: When given the slope and the
y-intercept, use slope-intercept form. When given
the slope and a point, use point-slope form. When
given two points, find the slope, then use pointslope form.

46. Let s  sum.
s

4

Writing Linear Equations

1

6. y  mx  b → y  
4x 
 10

 15

7. y 
 2  4(x 
 3)
y 
 2  4x 
 12
y  4x 
 10

s  60
The correct choice is D.

9
2


8. m  
7
5
7

 2
7

y 
 2  2(x 
 5)
7

35

7

31

y 
 2  2x 
 2
y  2x 
 2
9. y 
 2  0(x 
 (
9))
y
20
y2

11

10a. y  5.9x  2

Chapter 1

10b. y  5.9(7)  2
 41.3  2 or 43.3 in.
10c. Sample answer: No; the grass could not support
it own weight if it grew that tall.

Pages 30–31

27b. Using sample answer from part a,
9

171

3

13. y  mx  b → y  
4x

3
9

14. y  mx  b → y  
12x 

9
5

4

y


20. x  
4
22. y 
 0  0.25(x 
 24)
y  0.25x 
 6
23.

y 
 (
4) 
y4
1
x
2

4

y 
 5  
9x  9

18. (
8, 0), (0, 5)

4
49
y  
9x  9
5
0  8(x 
 (
8))
5
y  8x  5

3

3x 
 2y 
 5  0

5

y  2x 
 2

O

y 
 1  0(x 
 8)
y
10
y1
21. x  0

82,805 
 70,583


m
1997–1999
12,222

 2 or 6111; $6111 billion
31b. The rate is the slope.
32. g[f(
2)]  g(f(
2))
 g((
2)3)
 g(
8)
 3(
8) or 
24
33. (f g)(x)  f(x) g(x)
 x3(x2 
 3x  7)
 x5 
 3x4  7x3

1

2(x 
 (
2))
1

2x 
 1

y50


3 
 0

x
, where g(x)
g(x)  
x 
 3x  7
3

f


24. m  
1 
 (
2)

3

3

34.

 
1
y 
 0  
1(x  2)
y  
x 
 2
x  y  
2

x

4

3

2

y
16
9
4

{(
4, 16), (
3, 9), (
2, 4)}, yes

x 
 7000


25a. t  2  
2000

35. x 

14,494 
 7000


25b. t  2  
2000

1

y

y1

y

C

y2  1

 y

A

y2  1

 2
y

27a. Sample answer: using (20, 28) and (27, 37),

The correct choice is A.

9


m
27 
 20

y 
 28  7(x 
 20)

9

y 
 28  7x 
 7

9

180

9

16

y  7x  7
Chapter 1

1

 y
 
2
 
2

y  
Bx 
 B; m  
B

 7

y2

y

y2  1

y

 2  3.747 or 5.747; about 5.7 weeks
26. Ax  By  C  0
By  
Ax 
 C

37 
 28

0

2

A

x

31a. (1995, 70,583), (1997, 82,805)

x  2y  10  0
x  2y  
10


3

lines would have the same slope and would share
a point, their equations would be the same. Thus,
they are the same line and all three points are
collinear.
y
30. 3x 
 2y 
 5  0

2y  
3x  5

4

4

19.

6
3



(
3, 3), and (1, 6) is 
1 
 (
3) or 4 . Since these two

y 
 5  
9(x 
 1)




9

3



is 

3 
 5 or 4 . The slope of the line through

1

2

16. x  12


17. m  

8 
 1

187

27c. Sample answer: The estimate is close but not
exact since only two points were used to write
the equation.
28a. See students’ work.
28b. Sample answer: Only two points were used to
make the prediction equation, so many points lie
off of the line.
29. Yes; the slope of the line through (5, 9) and (
3, 3)

Exercises

5
0

m
0 
 (
8)
5
 8
1
1

m

3 
 8
0



11 or 0

16

 7  7 or 7 or about 26.7 mpg

11. y  mx  b → y  5x 
 2
12. y 
 5  8(x 
 (
7))
y 
 5  8x  56
y  8x  61

15. y 
 5  6(x 
 4)
y 
 5  6x 
 24
y  6x 
 19

16

y  7(19)  7

12

1

2

Page 31

10b. y – 6,478,216  170,823.7(x – 1990)
y – 6,478,216  170,823.7x – 339,939,163
y  170,823.7x – 333,460,947

Mid-Chapter Quiz

1. {
2, 2, 4}, {
8, 
3, 3, 7}; No, 
2 in the domain is
paired with more than one element of the range.
3

3. g(n  2)  
2. f(4)  7 
 42
n2
1
 7 
 16 or 9
3


n1
4. Let x  original price of jacket
Let T(x)  1.055x. (The cost with 5.5% tax rate)
Let S(x)  0.67x. (The cost with 33% discount)
The cost of the jacket is [T 
 S](x).
[T 
 S](x)  T(S(x))
 T(0.67x)
 1.055(0.67x)
The amount paid was $49.95.
45.95  1.055(0.67x)
43.55  0.67x
65  x; $65
5. [f 
 g)(x)  f(g(x)
 f(x  1)

Pages 35–36



4

4 3


3; 4
4. All vertical lines have undefined slope and only
horizontal lines are perpendicular to them. The
slope of a horizontal line is 0.
5. none of these
6. perpendicular
7. y  x 
 6
x
y80
yx8
parallel
8. y  2x 
 8
4x 
 2y 
 16  0
y  2x 
 8
coinciding
9.
y 
 9  5(x 
 5)
y 
 9  5x 
 25
5x 
 y 
 16  0
10.
6x 
 5y  24

1


 g(
x 
 1)
1



x
1  1

x
1




x
1  x
1
x



x 
 1, x

1

y

6. 2x 
 4y  8

4y  
2x  8
1

y  2x 
 2

x

O

2x 
 4y  8

6

7. 3x  2y
3
x
2

19

y  
3x 
 3

[g 
 f ](x)  g(f(x))

1

Check for Understanding

1. If A, B, and C are the same or the ratios of the As
and the Bs and the Cs are proportional, then the
lines are coinciding. If A and B are the same and
C is different, or the ratios of the As and the Bs
are proportional, but the ratio of the Cs is not,
then the lines are parallel.
2. They have no slope.
3. 4x  3y  19  0

1

x1
1
1
, x
0
x



Writing Equations of Parallel and
Perpendicular Lines

1-5

24

y  5x 
 5

y

5

y 
 (
5)  
6(x 
 (
10))

y

5

25

y  5  
6x 
 3
6y  30  
5x 
 50
5x  6y  80  0

O
x

3x  2y

8
4


11. m of EF: m  
4
3
4

 1 or 4

8. 5x 
 3  0
5x  3

2
6


m of GH: m  
6
7

4




1 or 4

3

x  5
8
5

y 
 5  5(x 
 2)

3

3

2

 
3

6

3

Pages 36–37

19


5x  y 
 5  0

12. y  5x 
18

3x 
 5y  19  0
10a. (1990, 6,478,216), (2000, 8,186,453)



6
8


m of FG: m  
7
4

y 
 5  5x 
 5

 5

m

2

 
3

parallelogram

3


9. m  
7
2

2
4


m of EH: m  
6
3

2x  10y  10  0
1

y  
5x  1

8,186,453 
 6,478,216

2000 – 1990
1,708,237

10

Exercises

slopes are opposite reciprocals; perpendicular

or about 170,823.7

13

Chapter 1

13. y 
 7x  5  0
y 
 7x 
 9  0
y  7x 
 5
y  7x  9
same slopes, different y-intercepts; parallel
14. different slopes, not reciprocals; none of these
15. horizontal line, vertical line; perpendicular
16. y  4x 
 3
4.8x 
 1.2y  3.6
y  4x 
 3
same slopes, same y-intercepts; coinciding
17. 4x 
 6y  11
3x  2y  9
y

2
x
3




11

6

y

3

2x



4
4

y 
 8  5x  12
5y 
 40  4x  60
4x 
 5y  100  0
5

28b. perpendicular slope: 
4
5

y 
 8  
4(x 
 (
15))

9

2

5

y

y

5

9x



4y 
 32  
5x 
 75
5x  4y  43  0
29a. 8x 
 14y  3  0
kx 
 7y  10  0
8

4

7

29b.

same slopes, same y-intercepts; coinciding
20. y  4x 
 2  0

k

 7 → k  4

7

 
4
49

k  
4
30a. Sample answer: y 
 1  0, x 
 1  0
30b. Sample answer: x 
 7  0, x 
 9  0
31. altitude from A to BC:

5 
 (
5)


m of BC  
10 
 4
0

 6 or 0
m of altitude is undefined; x  7
altitude from B to AC:

4



23. m  

(
9) or 9


5 
 10

4

y 
 (
15)  9(x 
 12)
4

k

7


m of AC  
4
7

15

y  15  9x 
 3

16




3 or 5

9y  135  4x 
 48
4x 
 9y 
 183  0
24. y 
 (
11)  0(x 
 4)
y  11  0

m of altitude  
5

25.

1

1

y 
 (
5)  
5(x 
 10)
1

y  5  
5x  2
5y  25  
x  10
x  5y  15  0
altitude from C to AB:

1

y 
 (
3)  
5(x 
 0)
y3

1

5x

5y  15  
x
x  5y  15  0


5 
 10


m of AB  
10 
 7

6


15

 3 or 
5

1



26. m  

(
1) or 6; perpendicular slope is 
 6

1

m of altitude  5

1

y 
 (
2)  
6(x 
 7)
1

1

y 
 (
5)  5(x 
 4)

7

y  2  
6x  6
6y  12  
x  7
x  6y  5  0
27. x  12 is a vertical line; perpendicular slope is 0.
y 
 (
13)  0(x 
 6)
y  13  0

Chapter 1

k



m  

(
7) or 7

4k  
49

y  4x  1  0

y  
4x  2
y  
4x 
 1
same slopes, different y-intercepts; parallel
21. None of these; the slopes are not the same nor
opposite reciprocals.
22. y 
 (
8)  2(x 
 0)
y  8  2x
2x 
 y 
 8  0
4

k

4



m  

(
14) or 7

14

9

14

9



75

y 
 8  
4x 
 4

y  
3x  2
different slopes, not reciprocals; none of these
19. 5x  9y  14

4



m  

(
5) or 5

y 
 8  5(x 
 (
15))

3x  y  2

5

9x

4

4x 
 5y  10  0

Slopes are opposite reciprocals; perpendicular.
18. y  3x  2

5y 
 4x  10

28a.

1

4

y  5  5x 
 5
5y  25  x 
 4
x 
 5y 
 29  0
32. We are given y  m1x  b1 and y  m2x  b2 with
m1  m2 and b1 b2. Assume that the lines
intersect at point (x1, y1). Then y1  m1x1  b1
and y1  m2x1  b2. Substitute m1x1  b1 for y1 in
y1  m2x1  b2. Then m1x1  b1  m2x1  b2.
Since m1  m2, substitute m1 for m2. The result is
m1x1  b1  m1x1  b2. Subtract m1x1 from each
side to find b1  b2. However, this contradicts the
given information that b1 b2. Thus, the

14

36a. (40, 295), (80, 565)

assumption is incorrect and the lines do not share
any points.
33a. Let x  regular espressos.
Let y  large espressos.
216x  162y  783
248x  186y  914
y

4

3x



29

6

y

4

3x



565 
 295


m
80 
 40
270

 4
0 or 6.75
y 
 565  6.75(x 
 80)
y 
 565  6.75x 
 540
y  6.75x  25
36b. $6.75
36c. $25
37. 3x 
 2y 
 6  0

457

93

No; the lines that represent the situation do not
coincide.
33b. Let x  regular espressos.
Let y  large espressos.
216x  162y  783
344x  258y  1247
4

29

y  
3x  6

4

y

y

3x 
 2y 
 6  0

29

O

x

38. [g 
 h](x)  g(h(x))
 g(x2)
 x2 
 1
39. Sample answer: {(2, 4), (2, 
4), (1, 2), (1, 
2),
(0, 0)}; because the x-values 1 and 2 are paired
with more than one y-value.
40. 2x  y  12
x  2y  
6
y  
2x  12
x  2(
2x  12)  
6
x 
 4x  24  
6

3x  
30
x  10
2(10)  y  12
y  
8
2x  2y  2(10)  2(
8)
 20  (
16) or 4

1943.09 – 1939.20

m  
16 – 15
3.89

 1 or 3.89
y 
 1943.09  3.89(x 
 16)
y 
 1943.09  3.89x 
 62.24
y  3.89x  1880.85
(16, 1943.09), (17, 1976.76)
1976.76 – 1943.09

m 
17 – 16
33.67

 1 or 33.67
y 
 1976.76  33.67(x 
 17)
y 
 1976.76  33.67x 
 572.39
y  33.67x  1404.37
(17, 1976.76), (18, 1962.44)
1962.44 
 1976.76

18 
 17

14.32


or –14.32
1
y 
 1962.44  –14.32(x 
 18)
y 
 1962.44  –14.32x  257.76
y  –14.32  2220.2
(18, 1962.44), (19, 1940.47)
m


3

y  
3x  6

Yes; the lines that represent the situation
coincide.
34a. (15, 1939.20), (16, 1943.09)

m

3
x
2

1-6

Modeling Real-World Data with
Linear Functions

Pages 41–42

Check for Understanding

1. the rate of change
2. Choose two ordered pairs of data and find the
equation of the line that contains their graphs. Find
a median-fit line by separating the data into three
sets and using the medians to locate a line. Use a
graphing calculator to find a regression equation.
3. Sample answer: age of a car and its value
4a. Personal Consumption on Durable Goods

1940.47 
 1962.44

19 
 18
–21.97

 1 or –21.97
y 
 1940.47  
21.97(x 
 19)
y 
 1940.47  
21.97x  417.43
y  
21.97x  2357.9
34b. parallel lines or the same line; no
34c. y  3.89(22)  1880.85
 1966.43
y  33.67(22)  1404.37
 2145.11
y  
14.32(22)  2220.2
 1905.16
y  
21.97(22)  2357.9
 1874.56
No; the equations take only one pair of days into
account.
35. y 
 5  
2(x 
 1)
y 
 5  
2x  2
y  
2x  7

3500
3000
2500
2000
Dollars
1500
1000
500
0
1995 1997 1999 2001 2003
Year

15

Chapter 1

4b. Sample answer: using (1995, 2294) and
(2002, 3158)
m

7a.

3158 – 2294

2002 – 1995
864

 7 or 123.4
y 
 3158  123.4(x 
 2002)
y  123.4x 
 243,888.8
4c. y  132.8x 
 262,621.2; r  0.98
4d. y  132.8(2010) 
 262,621.2
 4306.8
$4306.80; yes, the correlation coefficient shows a
strong correlation.
Students per Computer
5a.

0
1995 1997 1999 2001 2003
Year

7b. Sample answer: using (1995, 23,255) and
(2002, 30,832)
m


25
15
10
5
0
’89

’91

’93

’95
Year

’97

’99

’01

5b. Sample answer: using (1997, 6.1) and (2001, 4.9)
m


30,832 – 23,255

2002 – 1995
7577
 or 1082.43
7

y 
 23,255  1082.43(x 
 1995)
y  1082.43x 
 2,136,192.85
7c. y  1164.11x 
 2,299,128.75; r  0.99
7d. y  1164.11(2010) – 2,299,128.75
 40,732.35
$40,732.35; yes, r shows a strong relationship.
Car Weight and Mileage
8a.

20
Average

Personal Income
35
30
25
Dollars
(thousands) 20
15

80

4.9 – 6.1

2001 – 1997
–1.2
 or 
0.3
4

60
Average
Mileage 40

y 
 6.1  
0.3(x 
 1997)
y  
0.3x  605.2
5c. y  
1.61x  3231.43; r  
0.97
5d.
1  
1.61x  3231.43

3230.43  
1.61x
2006.47  x
2006; yes, the number of students per computer is
decreasing steadily.

20
0
0

15 20 25 30 35
Weight (hundreds of pounds)

8b. Sample answer: using (17.5, 65.4) and
(35.0, 27.7)
27.7 
 65.4


m
35.0 
 17.5

37.7



17.5 or 
2.15

Pages 42–44

y 
 65.4  
2.15(x 
 17.5)
y  
2.15x  103.0
8c. y  
1.72x  87.59; r  
0.77
8d. y  
1.72(45.0)  87.59
y  10.19
10.19; no, r doesn’t show a particularly strong
relationship.
9a.
Acorn Size and Range

Exercises

6a.

All-Time NFL Coaching Victories
400
Wins 200
0
0

10

20

30

40

Years

6b. Sample answer: using (18, 170) and (40, 324)
m


30,000

324 – 170

40 – 18
154
 or 7
22

20,000
Range
(hundreds
of km2) 10,000

y 
 324  7(x 
 40)
y  7x  44
6c. y  7.57x  33.38; r  0.88
6d. y  7.57(16)  33.38
 154.5
155; yes, r is fairly close to 1. (Actual data is
159.)

Chapter 1

0
0

16

2 4 6 8 10 12
Acorn Size (cm3)

9b. Sample answer: using (0.3, 233) and (3.4, 7900)
m


12b. Sample answer: a medication that proves to help
delay the progress of a disease; because any
positive correlation is better than none or a
negative correlation.
12c. Sample answer: comparing a dosage of medicine
to the growth factor of cancer cells; because the
greater the dosage the fewer cells that are
produced.
13. Men’s Median Salary
Women’s Median Salary
LinReg
LinReg
y  ax  b
y  ax  b
a  885.2867133
a  625.041958
b  –1,742,768.136
b  –1,234,368.061
r  .9716662959
r  .9869509009
The rate of growth, which is the slope of the
graphs of the regression equations, for the women
is less than that of the men’s rate of growth. If
that trend continues, the men’s median salary will
always be more than the women’s.
14a. Let x  computers.
Let y  printers.
24x  40y  38,736
y  
0.6x  968.4
30x  50y  51,470
y  
0.6x  1029.4
No; the lines do not coincide.
14b. Let x  computers.
Let y  printers.
24x  40y  38,736
y  
0.6x  968.4
30x  50y  48,420
y  
0.6x  968.4
Yes; the lines coincide.
15.
y 
 (
4)  
6(x 
(
3))
y  4  
6x 
 18
6x  y  22  0
y
16a.

7900 
 233

3.4 
 0.3
7667
 or 2473.23
3.1

y 
 7900  2473.23(x 
 3.4)
y  2473.23x 
 508.97
9c. y  885.82  6973.14; r  0.38
9d. The correlation value does not show a strong or
moderate relationship.
Working Women
10a.
40
35
30
25
Percent in
20
Management
15
10
5
0
’86 ’88 ’90 ’92 ’94 ’96 ’98 ’00 ’02 ’04
Year

10b. Sample answer: using (1990, 26.2) and
(2003, 37.6)
37.6 – 26.2

m  
2003 – 1990
11.4

 1
3 or 0.88
y 
 26.2  0.88(x 
 1990)
y  0.88x 
 1725
10c. y  0.84x 
 1648.27; r  0.984
10d. y  0.84(2008) 
 1648.27 or 38.45
38.45%; yes, r is very close to 1.
11a.
World Population
7000
6000
5000
Millions 4000
of People
3000
2000
1000
0

25
20 y  0.82x  24
15
10
5
0

1000
Year

2000

O

11b. Sample answer: using (1, 200) and (2000, 6050)
m


1 2 3 4 5x

16b. $24 billion
16c. If the nation had no disposable income, personal
consumption expenditures would be $24 billion.
For each 1 billion increase in disposable income,
there is a 0.82 billion dollar increase in personal
consumption expenditures.
17. [f 
 g](x)  f(g(x))
 f(x  1)
 (x  1)3
 x3  3x2  3x  1
[g 
 f ](x)  g(f(x))
 g(x3)
 x3  1
18. Yes; each domain value is paired with exactly one
range value.

6050 – 200

2000 – 1
5850
 or 2.93
1999

y 
 200  2.93(x 
 1)
y  2.93x  197.07
11c. y  1.65x 
 289.00; r  0.56
11d. y  1.65(2010) 
 289.00
 3027.5
3028 million; no, the correlation value is not
showing a very strong relationship.
12a. Sample answer: the space shuttle; because
anything less than perfect could endanger the
lives of the astronauts.

17

Chapter 1

19. The y-intercept is 1.
The slope is 
3 (move down 3 and right 1).
The correct choice is C.

Piecewise Functions

1-7

Pages 48–49
1.
2.
3.
4.
5.

8.

Check for Understanding



x

1
0
1
2
3
4

y

f(x)
4
3
2
1
0
1

x

O

9. greatest integer function; h is hours, c(h) is the
50h if [[h]]  h
cost, c(h) 
50[[h  1]] if [[h]]  h

x if x  0
f(x) 

x if x  0
reals, even integers
x  2 if x  0
f(x)  x if 0  x  4
x 
 2 if x 4
Alex is correct because he is applying the
definition of a function.
y





x
0x1
1x2
2x3
3x4

y

f(x)
50
100
150
200

400
300
200
100

x
O

O
y

Pages 49–51
11.

x

3  x  
2

2  x  
1

1  x  0
0x1
1x2
2x3
3x4
4x5

10

y

x

12.

x
1
3
5
7
9

y

f(x)
4
2
0
2
4

x

O

Chapter 1

8

Exercises

O

f(x)
3
2
1
0

1

2

3

4

y

O

6

x

O
7.

4

10. long term lot:
2(6)  3(1)  12  3 or 15
shuttle facility:
4(3)  12
shuttle facility

x

6.

2

13.

x

18

x

2  x  
1

1  x  0
0x1
1x2
2x3

f(x)
0
1
2
3
4

y

O

x

14.

x

5

3

1.5
0
2

x

2  x  
1

1  x  0
0x1
1x2
2x3

x

x

f(x)

3

2

1
0
1

f(x)


1  x 

2

3

3

2

3

1

3

2

x

1

3

O
15.

18.

y

f(x)
7
3
0
3
7

x0

1

0x

1

3

0

1

3

x

2

3


1

2

3

x1


2

y
4
2

y
O

–1

1

x

–2
–4

O

x
19.

20.

y

y
16. ART y

O
x

O
21.

x

O
17.

x
0
1
2
3
4

f(x)
6
4
2
0
2

y

x

x

f(x)


5  x  
4


5


4  x  
3


2


3  x  
2


3


2  x  
1

1  x  0
1x2
2x3


1

2
2
1

3x4

2

3

4x5

1

2

5x6

2

5

2
1
2

y
2

O

x

1

x

O

1

2

19

Chapter 1

22.

x

5

3

1
0
1
3
5

26c.

y

f(x)
6
0
6
9
6
0
6

Shipping
(dollars)

O
x

O

23. Step; t is the time in hours, c(t) is
the cost in dollars,
 6 if t  1
2

1

  t  1
10
if
c(t)  
2
 16 if 1  t  2

 24 if 2  t  24

27. If n is any integer, then all ordered pairs (x, y)
where x and y are both in the interval [n, n  1)
are solutions.
28a. absolute value
28b. d(t)  65 
 t
28c.
d (t )
80

d(t)
24

60

16

40

8

20

O

2

4

6

8

10 12 14 16 18 20 22 24 t

2  37

2

c(w)

60

80

t

 19.5 heating degree days

5% if x  $10,000

29b. t(x)  7.5% if $10,000  x  $30,000

9.3% if x $30,000

0.8
0.6

29c.

0.4

y

0.2

10
O

2

4 w

Tax
Rate
(percent)

5

d(w)

f(x)
1
0
1
2

0

O

10

20
30
Income
(thousands of dollars)

x

29d. 9.5%
30. No; the functions are the same if x is positive.
However, if x is negative, the functions yield
different values. For example, [g 
 f ](1.5)  1
and [f 
 g](1.5)  1; [g 
 f ](
1.5)  2 and
[f 
 g](
1.5)  1.

w

26a. step
26b. v is the value of the order, s(v) is the shipping,
 3.50 if 0.00  v  25.00

 5.95 if 25.01  v  75.00
s(v)  
 7.95 if 75.01  v  125.00

 9.95 if 125.01  v

Chapter 1

40

29a. step

1.0

25. Absolute value; w is the weight in pounds, d(w) is
the discrepancy, d(w)  1 
 w
x
0
1
2
3

20

28d. d(63)  65 
 63 or 2
d(28)  65 
 28 or 37



f(x)
0.37
0.37
0.60
0.60
0.83
0.83

d (t )  |65 
 t |

O

24. Greatest integer; w is the weight in ounces, c(w) is
the cost in dollars,
0.37  0.23(w 
 1) if [[w]]  w
c(w) 
0.37  0.23[[w]] if [[w]]  w
x
0x1
1
1x2
2
2x3
3

25 50 75 100 125 150
Value of Order
(dollars)

20

31a.

Public Transport

Graphing Linear Inequalities

1-8

60
50

Pages 54–55

Percent who use 40
public transportation 30
20
10
0
0

5

10

15

20

25

30

35

Working Population
(hundreds of thousands)

31b. Sample answer: using (3,183,088, 53.4) and
(362,777, 3.3)
m


Check for Understanding

1. y  2x 
 6
2. Graph the lines 
3  2x  y and 2x  y  7. The
graph of 
3  2x  y is solid and the graph of
2x  y  7 is dashed. Test points to determine
which region(s) should be shaded. Then shade the
correct region(s).
3. Sample answer: The boundaries separate the
plane into regions. Each point in a region either
does or does not satisfy the inequality. Using a
test point allows you to determine whether all of
the points in a region satisfy the inequality.
4.
y

3.3 
 53.4

362,777 
 3,183,088

50.1
 or 0.0000178

2,820,311

y 
 53.4  0.0000178(x 
 3,183,088)
y  0.0000178x 
 3.26
31c. y  0.0000136x  4.55, r  0.68
31d. y  0.0000136(307,679)  4.55
y  8.73
8.73%; No, the actual value is 22%.
32.
y 
 2  2(x 
 4)
y 
 2  2x 
 8
2x 
 y 
 6  0
15 
 29

33a. (39, 29), (32, 15)
33b. m  
32 
 39

x

O
xy!4

5. ART


14

y
x

O




7 or 2
33c. The average number of points scored each
minute.
34. p(x)  (r 
 c)(x)
 (400x 
 0.2x2) 
 (0.1x  200)
 399.9x 
 0.2x2 
 200
35. Let x  the original price, or $59.99.
Let T(x)  1.065x. (The cost with 6.5% tax rate)
Let S(x)  0.75x. (The cost with 25% discount)
[T 
 S](x)  (T(S(x))
 (T(0.75x))
 (T(0.75(59.99))
 (T(44.9925)
 1.065(44.9925)
 $47.92
36. {
7, 
2, 0, 4, 9}; {
2, 0, 2, 3, 11}; Yes; no element
of the domain is paired with more than one
element of the range.
37. 5 612  10,883,911,680
5  612  2,176,782,341
So, 5  612 is not greater than 5 612.
The correct choice is A.

3x 
 y ^ 6

6.

y

7!xy^9

x

O
7.

y

y ! |x  3|

21

O

x

Chapter 1

8a. c(m)  45  0.4m
8b. c(m)

y

14.

70
60

c (m) ^ 45  0.4m

x

O

50

–4 ^ x 
 y ^ 5

40
30

15.

20

y

16.

10

O

20

40

60

x

100 m

80

O

8c. Sample answer: (0, 45), (10, 49), (20, 50)

Pages 55–56

O

x

Exercises

9.

17.

y

y

y 6 |x |  4

y<3

x

O

x

O
10.

18.

y

y

y ! |2x  3|

x

O
x–y>–5

11.

x

O
19.

y

y

8 ^ 2x  y ! 6

2x  4y 6 7

x

O

x

O
12.


y ! 2x 
 1

y
|x  3| ! y 
 1

y 6 2 x  19
5

O

20.

y

13.

y

5

x
O
O

Chapter 1

y

2 ^ x  2y ^ 4

y 6 |x |

x

22

x

21.

y

y

26b.

4.252

x60
y60

4.25
4.248

O

O

x

27a. 0.6(220 
 a)  r  0.9(220 
 a)
27b. r

y

22.

4
3
2
1 O

200

2 4 6 8 10x


10
8
6
4
2

2

3

4

0.6 (220 
 a) ^ r ^ 0.9 (220 
 a)

100

23a. 8x  10y  8(60)
8x  10y  480
23b. y

O

60
40

60

20

30

40

50

60

70 a

275
220
165
110
55

20
40

10

28a. step
28b. Let c(h) represent the cost for h hours.
55h if [[h]]  h
Then c(h) 
55[[h  1]] if [[h]]  h
28c. y

80

O 20

x

7.998 8 8.002

80

x

23c. Sample answer: (0, 48) (60, 0), (45, 6)
23d. Sample answer: using complex computer
programs and systems of inequalities.
24.
y

O

2 3 4
Hours

5

x

29a. 3x 
 y  10
y  3x 
 10
y 
 (
2)  3(x 
 0)
3x 
 y 
 2  0

|y| 6 x

x

O

1

1

29b. perpendicular slope: 
3
1

y 
 (
2)  
3(x 
 0)
1

y  2  
3x
3y  6  
x
x  3y  6  0

25a. points in the first and third quadrants
25b. If x and y satisfy the inequality, then either
x  0 and y  0 or x  0 and y  0. If x  0 and
y  0, then x  x and y  y. Thus,
x  y  x  y. Since x  y is positive,
x  y  x  y.
If x  0 and y  0, then x  
x and y  
y.
Then x  y  
x  (
y) or 
(x  y). Since
both x and y are negative, (x  y) is negative,
and x  y  
(x  y).
26a. 8 
 x 

1
;
500

1
44


 y 

7
4


30. m  
5
1

3

y 
 7  4(x 
 5)

3

3

1

y  4x  34

 4
31a. (0, 23), (16, 48);
48 
 23


m
16 
 0
25

 16 or 1.5625
31b. the average change in the temperature per hour

1

500

32.

95 
 94

8

59,049 
 6561

 8
52,488

 8 or 6561 or 94
The correct choice is E.

23

Chapter 1

20. (f  g)(x)  f(x)  g(x)
 4 
 x2  3x or 4  3x 
 x2
(f 
 g)(x)  f(x) 
 g(x)
 4 
 x2 
 (3x)
 4 
 3x 
 x2
(f g)(x)  f(x) g(x)
 (4 
 x2)(3x)
 12x 
 3x3

Chapter 1 Study Guide and Assessment
Page 57
1.
3.
5.
7.
9.

Understanding the Vocabulary
2.
4.
6.
8.
10.

c
d
i
h
e

f
g
a
j
b

 (x) 
f

g



Pages 58–60

15. g

 (x) 
f

g

2

1
5 3

  


2

5

3

or



6

5



f

g



 (x) 
f

g






4




x3 
 8x2  16x  4

x
4
4

x2 
 4x  
x 
 4,

(f 
 g)(x)  f(x) 
 g(x)
 x2 
 4x 




 (x2 
 4x)
x

2

 (x) 
f

g





 4x, x 
 4
f(x)

g(x)
x2 


x
4



4

x
4

x3 
 8x2  16x 
 4

x
4
4
2

x 
 4x 
 
x 
 4,

(f g)(x)  f(x) g(x)

f(x)



f(x)

g(x)
x2 
 1

x1
(x 
 1)(x  1)

x1


 x2 
 4x  
x
4


(x)  
g(x )
x2  4x
,
x
2

f(x)

g(x)
x2  7x  12

x4
(x  4)(x  3)

x4

 x  1, x 
 
1
23. (f  g)(x)  f(x)  g(x)

f(x)

g(x)
6x 
 4

2

 3x 
 2
19. (f  g)(x)  f(x)  g(x)
 x2  4x  x 
 2
 x2  5x 
 2
(f 
 g)(x)  f(x) 
 g(x)
 x2  4x 
 (x 
 2)
 x2  3x  2
(f g)(x)  f(x) g(x)
 (x2  4x)(x 
 2)
 x3  2x2 
 8x
f

g

0

 x  3, x 
 
4
22. (f  g)(x)  f(x)  g(x)
 x2 
 1  x  1
 x2  x
(f 
 g)(x)  f(x) 
 g(x)
 x2 
 1 
 (x  1)
 x2 
 x 
 2
(f g)(x)  f(x) g(x)
 (x2 
 1)(x  1)
 x 3  x2 
 x 
 1

16. k(4c)  (4c)2  2(4c) 
 4
 16c2  8c 
 4
17. f(m  1)  (m  1)2  3(m  1)
 m2  2m  1  3m  3
 m2  5m  4
18. (f  g)(x)  f(x)  g(x)
 6x 
 4  2
 6x 
 2
(f 
 g)(x)  f(x) 
 g(x)
 6x 
 4 
 (2)
 6x 
 6
(f g)(x)  f(x) g(x)
 (6x 
 4)(2)
 12x 
 8

 (x) 

x

21. (f  g)(x)  f(x)  g(x)
 x2  7x  12  x  4
 x2  8x  16
(f 
 g)(x)  f(x) 
 g(x)
 x2  7x  12 
 (x  4)
 x2  6x  8
(f g)(x)  f(x) g(x)
 (x2  7x  12)(x  4)
 x3  11x2  40x  48

Skills and Concepts

11. f(4)  5(4) 
 10
 20 
 10 or 10
12. g(2)  7 
 (2)2
 7 
 4 or 3
13. f(
3)  4(
3)2 
 4(
3)  9
 36  12  9 or 57
14. h(0.2)  6 
 2(0.2)3
 6 
 0.016 or 5.984
1

3

f(x)

g(x)
4 
 x2
,
3x

x

4



4

x
4

x3 
 8x2  16x
4x
 
or  , x 
 4
4
4

x
4

Chapter 1

24

24. [f 
 g](x)  f(g(x))
 f(2x)
 (2x)2 
 4
 4x2 
 4
[g 
 f ](x)  g(f(x))
 g(x2 
 4)
 2(x2 
 4)
 2x2 
 8
25. [f 
 g](x)  f(g(x))
 f(3x2)
 0.5(3x2)  5
 1.5x2  5
[g 
 f ](x)  g(f(x))
 g(0.5x  5)
 3(0.5x  5)2
 0.75x2  15x  75
26. [f 
 g](x)  f(g(x))
 f(3x)
 2(3x)2  6
 18x2  6
[g 
 f ](x)  g( f(x))
 g(2x2  6)
 3(2x2  6)
 6x2  18
27. [f 
 g](x)  f(g(x))
 f(x2 
 x  1)
 6  (x2 
 x  1)
 x2 
 x  7
[g 
 f ](x)  g(f(x))
 g(6  x)
 (6  x)2 
 (6  x)  1
 x2  11x  31
28. [f 
 g](x)  f(g(x))
 f(x  1)
 (x  1)2 
 5
 x2  2x 
 4
[g 
 f ](x)  g( f(x))
 g(x2 
 5)
 x2 
 5  1
 x2 
 4
29. [f 
 g](x)  f(g(x))
 f(2x2  10)
 3 
 (2x2  10)
 
2x2 
 7
[g 
 f ](x)  g(f(x))
 g(3 
 x)
 2(3 
 x)2  10
 2x2 
 12x  28
30. Domain of f(x): x  16
Domain of g(x): all reals
g(x)  16
5 
 x  16
x  
11
Domain of [f 
 g](x) is x  
11.

31. The y-intercept is 6. The slope is 3.
18

y

12
6

y  3x  6

O 6


6

12 x

32. The y-intercept is 8. The slope is 
5.
y
8

y = 8 – 5x

4
–4 O

x

4

33. y 
 15  0
y  15
The y-intercept is 15. The slope is 0.

y
20 y 
 15  0
10

10 O

10

x

34. 0  2x 
 y 
 7
y  2x 
 7
The y-intercept is 
7. The slope is 2.

y
O

x

y  2x 
 7

35. The y-intercept is 0. The slope is 2.

y
y  2x

x

O

36. The y-intercept is 
2. The slope is 
8.

y
O

x

y  
8x 
 2

25

Chapter 1

37. 7x  2y  
5
y

1

7

2x




The y-intercept is

50. y 
 (
8)  2(x 
 4)

5

2

1

5

2.

The slope is

y  8  2x 
 2

7

2.

2y  16  x 
 4
x 
 2y 
 20  0

y

1

4



51. m  

(
2) or 2, perpendicular slope is 
 2

7x  2y  
5

1

y 
 4  
2(x 
 1)

x

O

1

1

y 
 4  
2x  2
38. The y-intercept is 
6. The slope is
y

4 O 2 4 x

2y 
 8  
x  1
x  2y 
 9  0
52. x  
8 is a vertical line; perpendicular slope is 0.
y 
 (
6)  0(x 
 4)
y60
Overseas Visitors
53a.

1
.
4

y  1x 
 6

28

4

21

39. y  2x 
 3

40. y  
x  1

Visitors
14
(millions)

1

41. y 
 2  2(x 
 (
5))
1

5

y 
 2  2x  2
42.

7

1
9
y  2x  2
5
2

m
2 
 (
4)
3
1
 6 or 2
1
y 
 5  2(x 
 2)
1
y 
 5  2x 
 1
1
y  2x  4

0
1994 1996 1998 2000
Year

53b. Sample answer: using (1994, 18,458) and
(2000, 25,975)
m


43. (1, 0), (0, 
4)
m



4 
 0

0
1

4
 or 4

1

y 
 (
4)  4(x 
 0)
y  4  4x
y  4x 
 4
44. y  
1
46. y 
 0  0.1(x 
 1)
y  0.1x 
 0.1
47. y 
 1  1(x 
 1)
y
1x
1
x
y0

25,975 – 18,458

2000 – 1994
7517
 or 1252.8
6

y 
 25,975  1252.8(x 
 2000)
y  1252.8x 
 2,479,625
53c. y  1115.9x 
 2,205,568; r  0.9441275744
53d. y  1115.9(2008) 
 2,205,568
 35,159.2
35,159,200 visitors; Sample answer: This is a
good prediction, because the r value indicates a
strong relationship.
54. f (x )

45. y  0

1

y 
 6  3(x 
 (
1))

48.

1

O

1

y 
 6  3x  3
3y 
 18  x  1
x 
 3y  19  0
2

49. m  
1 or 
2
y 
 2  
2(x 
 (
3))
y 
 2  
2x 
 6
2x  y  4  0

Chapter 1

26

x

55.

y

60. ART

h (x )

x^5

x

O

O

x

61.

56.

x

2  x  
1

1  x  0
0x1
1x2
2x3

f(x)

1
0
1
2
3

y

x

O
xy^1

y

62.

f (x )

2y 
 x ^ 4

x

O

f (x )  x   1

x

O

63.

64.

y

y
y 
 3x 1 2

O
57.

x

2

1
0
1
2

O

y ^ |x |

g (x )

f(x)
8
4
0
4
8

x

65.

x

y
y ! |x| 
 2

g (x )  |4x |

O

x

O
x

58.

x

2

1
0
1
2

k (x)

f(x)
6
4
2
4
6

66.

y

O
O

y ! |x 
 2|

x

x

y

59.
y14

O

x

27

Chapter 1

Page 61
67a. d 

Applications and Problem Solving

2. Since this is a multiple-choice question, you can
try each choice. Choice A, 16, is not divisible by 12,
so eliminate it. Choice B, 24, is divisible by both 8
and 12. Choice C, 48, is also divisible by both 8
and 12. Choice D, 96, is also divisible by both 8
and 12. It cannot be determined from the
information given. The correct choice is E.
3. Write the mixed numbers as fractions.

1
(20)(1)2
2

 10
1

d  2(20)(2)2
 40
1

d  2(20)(3)2
 90
d

1

1

d  2(20)(5)2

1

43

3 
25

 250
10 m, 40 m, 90 m, 160 m, 250 m
67b. Yes; each element of the domain is paired with
exactly one element of the range.
68a. (1999, 500) and (2004, 636)


636 – 500

2004 – 1999
136
 or 27.2;
5

about $27.2 billion

Open-Ended Assessment

1. Possible answer: f(x)  4x 
 4, g(x)  x2;
[f 
 g](x)  f(g(x))  4(x2) 
 4  4x2 
 4
2a. No; Possible explanation: If the lines have the
same x-intercept, then they either intersect on
the x-axis or they are the same line. In either
case, they cannot be parallel.
2b. Yes; Possible explanation: If the lines have the
same x-intercept, they can intersect on the
x-axis. If they have slopes that are negative
reciprocals, then they are perpendicular.
 4 if x  4
3a. y  
2x 
 4 if x  4
x 
 1 if x  
1
3b. y  
3x  1 if x  
1

13

 3

1

5

13

5

 3

1

1

3

13

13




16  1
6  4  16  16  16 16

The correct choice is A.
7. Use your calculator. First find the total amount
per year by adding.
$12.90  $16.00  $18.00  $21.90  $68.80
Then find one half of this, which is the amount
paid in equal payments.
$68.80 2  $34.40
Then divide this amount by 4 to get each of 4
monthly payments.
$34.40 4  $8.60.
The correct choice is A.

SAT and ACT Practice

1. Prime factorization of a number is a product of
prime numbers that equals the number. Choices
A, B, and E contain numbers that are not prime.
Choice C does not equal 54. Choice D, 3 3 3
2, is the prime factorization of 54. The correct
choice is D.

Chapter 1

13

3

13

5

(2)
4  24  1
6

Chapter 1
SAT & ACT Preparation
Page 65

13

The correct choice is B.
4. Since this is a multiple-choice question, try each
choice to see if it answers the question. Start with
10, because it is easy to calculate with tens. If 10
adult tickets are sold, then 20 student tickets
must be sold. Check to see if the total sales
exceeds $90.
Students sales  Adult sales $90
20($2.00)  10($5.00)  40  50  $90
So 10 is too low a number for adult tickets. This
eliminates answer choices A, B, C, and D. Check
choice E. Eleven is the minimum number of adult
tickets.
19($2.00)  11($5.00)  38  55  $93
The correct choice is E.
5. Recall the definition of absolute value: the number
of units a number is from zero on the number line.
Simplify the expression by writing it without the
absolute value symbols.

7  7


7  
7


7

5
3
4  
7 
 5 
 12  
24
The correct choice is A.
6. Write each part of the expression without
exponents.
(
4)2  16

68b. y 
 500  27.2(x 
 1999)
y  27.2x 
 53,872.8
69. y  
0.284x  12.964; The correlation is
moderately negative, so the regression line is
somewhat representative of the data.

Page 61

3

25  5

Remember that dividing by a fraction is
equivalent to multiplying by its reciprocal

 160

m

13

43  3

1
(20)(4)2
2

28

10. First find the number of fish that are not tetras.

8. First combine the numbers inside the square root
symbol. Then find the square root of the result.

64
   
100
36
  10
The correct choice is A.
9. 60  2 2 3 5
 22 3 5
The number of distinct prime factors of 60 is 3.
The correct choice is C.

18(24) or 3 are tetras. 24 
 3 or 21 are not
2

tetras. Then 3 of these are guppies.

23(21)  14

The answer is 14.

29

Chapter 1

Chapter 2 Systems of Linear Equations and Inequalities
y

6.

Solving Systems of Equations in
Two Variables

2-1

no solution

y
x2

O

Page 69

Graphing Calculator Exploration

y
x5

1. (1, 480)
3

2. 3x 
 4y  320

→

5x  2y  340

y  4x 
 80
5

y  
2x  170

7. 7x  y  9
5x 
 y  15
12x
 24
x2
8. 3x  4y  
1
2(6x 
 2y)  2(3)

(76.923077, 
22.30769)
3. accurate to a maximum of 8 digits
5

4. 5x 
 7y  70

→

10x 
 14y  120

y  7x 
 10
5

60

y  7x 
 7

Inconsistent; error message occurs.
5. See students’ systems and graphs; any point in
TRACE mode will be the intersection point since
the two lines intersect everywhere.

7x  y  9
7(2)  y  9
y  
5
(2, 
5)
3x  4y  
1
12x 
 4y  6
15x  5

→

x

1

3

3x  4y  
1
33  4y  
1
1

Page 70

4y  
2

Check for Understanding

→

4y  16 
 6x



1

3



9. 30 3x 
 2y  30(
4)

2(5x 
 4y)  
2(14)

→

1

10x 
 45y  
120

10x  8y  
28

37y  
148
y4

5x 
4y  14
5x 
 4(4)  14
5x  30
x6
(6, 4)
10. Let b represent the number of baseball racks and
k represent the number of karate-belt racks.
b  6k
b  6k
3b  5k  46,000
 6(2000)
3(6k)  5k  46,000
 12,000
23k  46,000
k  2000
12,000 baseball, 2000 karate

y  
2x  3
3

y  
2x  4

Pages 71–72

Inconsistent; Sample answer: The graphs of the
equations are lines with slope

3

2,

Exercises

11. x  3y  18

but each


x  2y  7

equation has a different y-intercept. Therefore,
the graphs of the two equations do not intersect
and the system has no solution.
5.
(1, 3)
y

1

y  
3x  6

→

1

7

y  2x  2

consistent and independent
12. y  0.5x
y  0.5x
→
1
2y  x  4
y  2x  2
inconsistent
13. 
35x  40y  55

y 
 5x  2

(1, 3)

7x  8y 
 11

y 
 2x  5

Chapter 2

1

3

4. 2y  3x  6

O

3, 
2

1

y  
2

1. Sample answer:
4x  7x  21
y  2x 
 1
The substitution method is usually easier to use
whenever one or both of the equations are already
solved for one variable in terms of the other.
2. Sample answer: Madison might consider whether
the large down-payment would strap her
financially; if she wants to buy the car at the end
of the lease, then she might also consider which
lease would offer the best buyout.
3. Sample answer: consistent systems of equations
have at least one solution. A consistent,
independent system has exactly one solution; a
consistent, dependent system has an infinite
number of solutions. An inconsistent system has
no solution. See students’ work for examples and
solutions.

→

consistent and dependent

x

30

7

11

7

11

y  8x  8
y  8x  8

14.

y

15.

y
x
5

O

y 
 3

y

16.

O
y 
 x  2

(4, 3)

(4, 
3)
17.
y

(5, 0)

x

y 
 13 x  56

y 
 3x  10

x

O

(2, 4)

y


→


6x  20y  10
6x  21y  72
41y  82
y2

2x  7y  24
2x  7(2)  24
2x  10
x5
(5, 2)
26.
2x 
 y  
7
x  2y 
 8
2(2y 
 8) 
 y  
7
x  2(3) 
 8
3y  9
x  
2
y3
(
2, 3)
27. 3(2x  5y)  3(4)
6x  15y  12

2(3x  6y)  
2(5) → 
6x 
 12y  
10
3y 
2
2


2x  5y  4
y 3

x

O

x

(5, 0)

25. 2(
3x  10y  5)
3(2x  7y  24)

x
4

y 
45 x  4

13 x

2x  53  4
2

(2, 
4)

y

18.

19.
y
x2

no solution
y
(0, 3)

O

2

2x  3
1

28. 55x 
 6y  5(1)
3

x

y 
 12 x  2

O

(0, 2)

y 
 32 x  3

1
x
5

x

1

3, 3
1 2

x  3

y 
  14 x  3

5

 6y  1

→

5

3x 
 6y  5
1
5
x  y
5
6
16
x
5

 11
 16

x5
(0, 
2)
20. 3x 
 8y  10

→

16x 
 32y  75

21.

22.

23.

24.

1
x
5

5
 6y  11
1
5
(5)  y  11
5
6
5
y  10
6

(0, 3)
3

5

1

75

y  8x 
 2
y  2x 
 32

Consistent and independent; if each equation is
written in slope-intercept form, they have
different slopes, which means they will intersect
at some point.
3(5x 
 y)  3(16)
15x 
 3y  48
→
2x  3y  3
2x  3y  3
17x
 51
x3
5x 
 y  16
5(3) 
 y  16

y  1
y  
1
(3, 
1)
3x 
 5y  
8
3x 
 5y  
8

3(x  2y)  
3(1) →

3x 
 6y  
3

11y  
11
x  2y  1
y1
x  2(1)  1
x  
1
(
1, 1)
x  4.5  y
y6
x
x  4.5  6 
 x
y  6 
 5.25
2x  10.5
y  0.75
x  5.25
(5.25, 0.75)
5(2x  3y)  5(3)
10x  15y  15
→
12x 
 15y  
4
12x 
 15y  
4
22x
 11
1
2x  3y  3
x  2

y  12
(5, 12)
29. 7(4x  5y)  7(
8)
28x  35y  
56
5(3x 
 7y)  5(10) →
15x 
 35y  50
43x
 
6
6
4x  5y  
8
x  
4
3



320
64

y  
4
3


463 , 
6443 

30. 
2(3x 
 y)  
2(
9)

6x  2y  18
→
4x 
 2y  
8
4x 
 2y  
8

2x
 10
3x 
 y  
9
x  
5
3(
5) 
 y  
9

y  6
y  
6
(
5, 
6)
31. Sample answer: Elimination could be considered
easiest since the first equation multiplied by 2
added to the second equation eliminates b;
substitution could also be considered easiest since
the first equation can be written as a  b, making
substitution very easy.
a
b0
3a  2b  
15
ab
3(b)  2b  
15
5b  
15
b  
3
a
b0
a 
 (
3)  0
a  
3
(
3, 
3)
32a. B

1

3y  2
2



5y  
4
3

22  3y  3
y  3

6

4 
4
3  5y  
8

2, 3
1 2

31

Chapter 2

32b. S 
 4V  0
S  4V

S  V  30,000
4V  V  30,000
5V  30,000
V  6000

38. Let x represent the number of people in line
behind you. 200  x represents the number in
front of you. Let 
 represent the whole line.
200  x  1  x  


  3x
200  x  1  x  3x

  3x
201  x

  3(201)
603 people
 603
y
39.

S 
 4V  0
S 
 4(6000)  0
S 
 24,000  0
S  24,000
Spartans: 24,000; visitors: 6000
33a. Let b represent the base and 
 represent the leg.
Perimeter of first triangle: b  2
  20
Perimeter of second triangle: 6  b  
  20
b  2
  20
b  20 
 2

6  b  
  20
b  2
  20
6  20 
 2
  
  20
b  2(6)  20


  
6
b8

6
6, 6, 8; 6, 6, 8
33b. isosceles
34. y 
 (
3)  4(x 
 4)
y  4x 
 19

2x  7  y

40.

x

2

1
0
1
2

1

y 
 (
3)  
4(x 
 4)
1

y  
4x 
 2
35a. Let x represent the number of refills. Then,
x  1  number of drinks purchased.
C  2.95  0.50x
C  0.85  0.85x
C  2.95  0.50x
0.85  0.85x  2.95  0.50x
0.35x  2.1
x6
x17
C  2.95  0.50x
C  2.95  0.50(7) or 5.95
(7, 5.95)
35b. If you drink 7 servings of soft drink, the price for
each option is the same. If you drink fewer than
7 servings of soft drink during that week, the
disposable cup price is better. If you drink more
than 7 servings of soft drink, the refillable mug
price is better. See students’ choices.
35c. Over a year’s time, the refillable mug would be
more economical.
36a.
36b.
36c.

a
d


b
e
a
d
  , c
b
e
a
d
  , c
b
e

45.

f (x )

O
f (x) 
 2|x|  3


25


5



5

5

 
1

1
The correct choice is A.

2-2

Solving Systems of Equations in
Three Variables

f
f

Page 76

3.5
x
4
3.5
(516
4

Check for Understanding

1. Solving a system of three equations involves
eliminating one variable to form two systems of
two equations. Then solving is the same.

 264  y
 264  y

 y)  264  y

451.5  0.875y  264  y
y  $1500

Chapter 2

f(x)
1

1

3

1
1

41. y 
 6  2(x 
 0)
y  2x  6
42. $12,500
43. [f 
 g](x)  f(g(x))
 f(x  2)
 3(x  2) 
 5
 3x  1
44. {18}, {
3, 3}; no, because there are two range
values paired with a single domain value.

37. Let x represent the full incentive.
Let y represent the value of the computer.
x  516  y
3.5
x
4

x

O

32

x

1

2. The solution would be an equation in two
variables. Sample example: the system 2x  4y
 6z  12, x  2y  3z  6, and 3x 
 5y 
 6z 
27 has a solution of all values of x and y that
satisfy 5x 
 y  39.
2x  4y  6z  12

2(x  2y  3z)  
2(6)
3x 
 5y 
 6z  27

2x 
 4y 
 6z  
12
5x 
 y
 39
↓

2x 
 4y 
 6z  
12
2x  4y  6z  12
0
0
all reals
3. Sample answer: Use one equation to eliminate one
of the variables from the other two equations.
Then eliminate one of the remaining variables
from the resulting equations. Solve for a variable
and substitute to find the values of the other
variables.
4. 4(4x  2y  z)  4(7)
2(4x  2y  z)  2(7)
2x  2y 
 4z  
4
x  3y 
 2z  
8
↓
↓
16x  8y  4z  28
8x  4y  2z  14
2x  2y 
 4z  
4
x  3y 
 2z  
8
18x  10y
 24
9x  7y
 6
18x  10y  24

2(9x  7y)  
2(6) →


2.5(75)  
2.5(0.5a  v0  s0)
75  3.125a  2.5v0  s0
↓

187.5  
1.25a 
 2.5v0 
 2.5s0
75  3.125a  2.5v0 
s0

112.5  1.875a

 1.5s0

4(75)  
4(0.5a  v0  s0)
3  8a  4v0  s0
↓

300  
2a 
 4v0 
 4s0
3  8a  4v0  s0

297  6a

 3s0

2(
112.5)  
2(1.875a 
 1.5s0)

297  6a 
 3s0

18x  10y  24

18x 
 14y  
12

4y  12
y  
3
4x  2y  z  7
4(3)  2(
3)  z  7
z1

9x  7y  6
9x  7(
3)  6
x3
(3, 
3, 1)
5. 2(x 
 y 
 z)  2(7)

x  2y 
 3z  
12
↓
2x 
 2y 
 2z  14

x  2y 
 3z  
12
x

5z 
2

2(x 
 5z)  
2(2)
2x  4z  18

7. 75  2a(1)2  v0(1)  s0
 0.5a  v0  s0
1
75  2a(2.5)2  v0(2.5) s0
 3.125a  2.5v0  s0
1
3  2a(4)2  v0(4)  s0
 8a  4v0  s0

↓
225  
3.75a  3s0

297 
6a 
 3s0

72  2.25a

32  a

297  6a 
 3s0
3  8a  4v0  s0

297  6(
32) 
 3s0
3  8(
32)  4v0  s0
35  s0
56  v0
acceleration 
32 ft/s2, initial velocity: 56 ft/s,
initial height: 35 ft


x  2y 
 3z  
12
3x 
 2y  7z  30
2x
 4z  18

Pages 76–77


2x  10z  
4
2x  14z  18
14z  14
z1
x 
 5z  2
x
y
z7
x 
 5(1)  2
7
y
17
x7
y  
1
(7, 
1, 1)
6. 
2(2x 
 2y  3z)

2(2x 
 3y  7z)
 
2(6)
 
2(
1)
4x 
 3y  2z  0
4x 
 3y  2z  0
↓
↓

4x  4y 
 6z  
12

4x  6y 
 14z  2
4x 
 3y  2z 
0
4x 
 3y  2z  0
y 
 4z  
12
3y 
 12z  2

3(y 
 4z)  
3(
12)
3y 
 12z  2

→

→

Exercises

8. 3(5x  3y 
 z)  3(
11)
x  2y  3z  5
↓
15x  9y 
 3z  
33
x  2y  3z 
5
16x  11y
 
28

2(5x  3y 
 z)  
2(
11)
3x  2y 
 2z  
13
↓

10x 
 6y  2z  22
3x  2y 
 2z  
13

7x 
 4y

9
4(16x  11y)  4(
28)
64x  44y  
112
→
11(
7x 
 4y)  11(9)

77x 
 44y 
99

13x
 
 13
x1
16x  11y  
28
x  2y  3z  5
16(1)  11y  
28
1  2(
4)  3z  5
y  
4
z4
(1, 
4, 4)


3y  12z  36
3y 
 12z  2
0  38

no solution

33

Chapter 2

9. 7(
x  3y  2z)  7(16)
7x  5y  z  0
↓

7x  21y  14z  112
7x  5y  z  0
26y  15z  112

15(
3y  z)  
15(
2)
26y  15z  112

3y  z  
2

3(2)  z  
2
z4
(
2, 2, 4)
10. 2(2x  y 
 2z)  2(11)

x 
 2y  9z  13
↓
4x  2y 
 4z  22

x 
 2y  9z  13
3x
 5z  35

3(x 
 3z)  
3(7)
3x  5z  35

13. 3(x  y 
 z)  3(3)
2(x  y 
 z)  2(3)
4x 
 3y  2z  12

2x 
 2y  2z  5
↓
↓
3x  3y 
 3z  9
2x  2y 
 2z  6
4x 
 3y  2z  12

2x 
 2y  2z  5
7x
 z  21
0  11
no solution
14. 3(36x 
 15y  50z)  3(
10)

5(54x 
 5y  30z)  
5(
160)
↓
108x 
 45y  150z  
30

270x  25y 
 150z  800

162x 
 20y
 770


x  3y  2z  16
x 
 6y 
 2z  
18

3y  2z  
 2

45y 
 15z  30
26y  15z  112
71y
 142
y2
x 
 6y 
 z  
18
x 
 6(2) 
 4  
18
x  
2
→

81(2x  25y)  81(40)

162x 
 20y  770

162x  2025y  3240
→ 
162x 
 20y  770
2005y  4010
y2
36x 
 15y  50z  
10
36(
5) 
 15(2)  50z  
10
z4

→

2x  25y  40
2x  25(2)  40
x  
5
(
5, 2, 4)
15. 4(
x 
 3y  z)  4(54)
4x  2y 
 3z  
32
↓

4x 
 12y  4z  216
4x  2y 
 3z  
32

10y  3z  184


3x  9z  
21
3x  5z  35
14z  14
z1
2x  y 
 2z  11
2(10)  y 
 2(1)  11
y  
7

x 
 3z  7
x 
 3(1)  7
x  10
(10, 
7, 1)
11. 
3(x 
 3y 
 2z)  
3(
8) 
5(x 
 3y 
 2z) 

5(
8)
3x 
 5y  z  9
5x 
 6y  3z  15
↓
↓

3x  9y  6z  24

5x  15y  10z  40
3x 
 5y  z  9
5x 
 6y  3z  15
4y  7z  33
9y  13z  55
9(4y  7z)  9(33)
36y  63z  297
→

4(9y  13z)  
4(55)

36y 
 52z  
220
11z 
77
z7
4y  7z  33
x 
 3y 
 2z  
8
4y  7(7)  33
x 
 3(
4) 
 2(7)  
8
y  
4
x  
6
(
6, 
4, 7)
12. 8x  y 
 z  4
yz5
8x  y
9

1(8x  y)  
1(9)
11x  y  15
8x 
 z  4
8(2) 
 z  4
z  12
(2, 
7, 12)

Chapter 2

→

5(2y 8z)  5(78)

10y  z  184

→

2y  8z  78
2y  8(14)  78
y  
17
(11, 
17, 14)
16. 1.8x  1.2y 
 z  0.7
1.2y  z  0.7
1.8x  1.2y
 0

10y  40z  390

10y  z  184
41z  574
z  14

x 
3y  z  54

x 
 3(
17)  14  54
x  11

3(1.8x  1.2y)  3(0)
5.4x  3.6y  0
1.2(1.5x 
 3y)  1.2(3) → 1.8x 
 3.6y  3.6
7.2x
 3.6
x  0.5
1.5x 
 3y  3
1.8x 
 z  0.7
1.5(0.5) 
 3y  3
1.8(0.5) 
 z  0.7
y  
0.75
z  0.2
(0.5, 
0.75, 0.2)
17. y  x  2z
z  
1 
 2x
y  (y 
 14)  2z
7  
1 
 2x
7z

4  x
x  y 
 14

4  y 
 14
10  y
(
4, 10, 7)


8x 
 y  
9
11x  y  15
3x
 6
x2
yz5
y  12  5
y  
7

34

18.



  52(
12)

5 3
1
1
 x  y 
 z
2 4
6
3
1
2
5
x 
 y  z
8
3
6

20c. Sample answer: x  y  z  6;
2x 
 y 
 2z  8; x 
 2y 
 3z  2

 
8

1

21. 124  2a(1)2  v0(1)  s0

↓
15
x
8
1
x
8

5

1

5

272  2a(3)2  v0(3)  s0

5

82  2a(8)2  v0(8)  s0


 1
2 y 
 6 z  
30
2
y
3
1
y
4




2x 


1

 6z  
8
 
38





7 3
1
1

4 4x  6y 
 3z
3
7
5
x 
 y 
 z
16
12
8

124 
272 

7

4(
12)

7






5
y
8
11
y
12



1
2x 
 4y
11
9

8x 
 12y


1(124)  
12a  v0  s0

21

9

272  2a  3v0  s0

7


 1
2 z  
25

↓
1

124  
2a 
 v0 
 s0

 
4



9

16

9
(
38)
16

→

 
4

9
x
8
9

8x

9

 6
4y
11

 12y
203



192 y



272 
148 

171

8

 
4


2x 

1

3
x
4

 
38

2x 
 4(24)  
38

3
(
16)
4


1

1
y
6

9
a
2

 3v0  s0
4a  2v0


1(124)  
12a  v0  s0
82  32a  8v0  s0
↓
1

124  
2a 
 v0 
 s0
82  32a  8v0  s0

42  31.5a  7v0
1

203

8

y  24
1
y
4

 3v0  s0

1




1
6 x 
 24 y  12 z 
3
x
16
9

8x

↓
 v0  s 0

82  32a  8v0  s0

 
25

↓
7

21

1
a
2
9
a
2

1


 3z  
12
1

 6(24) 
 3z  
12

x  
16
z  12
(
16, 24, 12)
19. Let x represent amount in International Fund, y
represent amount in Fixed Assets Fund, and z
represent amount in company stock.

7(148)  7(4a  2v0)

2(
42)  
2(31.5a  7v0)
↓
1036  28a  14v0
84  
63a 
 14v0
1120  
35a

32  a
1
124  2a  v0  s0
148  4a  2v0

x  y  z  2000
x  2z
0.045x  0.026y 
 0.002z  58

148  4(
32)  2v0

x  y  z  2000
2z  y  z  2000
y  3z  2000
y  2000 
 3z

1

124  2(
32)  138  s0

138  v0
2  s0
(
32, 138, 2)
22a. Sample answer: A system has no solution when
you reach a contradiction, such as 1  0, as you
solve.
22b. Sample answer: A system has an infinite number
of solutions when you reduce the system to two
equivalent equations such as x  y  1 and
2x  2y  2.

0.045x  0.026y 
 0.002z  58
0.045(2z)  0.026y 
 0.002z  58
0.026y  0.088z  58
0.026y  0.088z  58
0.026(2000 
 3z)  0.088z  58
z  600
x  2z
x  y  z  2000
x  2(600)
1200  y  600  2000
x  1200
y  200
International Fund  $1200; Fixed Assets Fund 
$200; company stock  $600
20a. Sample answer: x  y  z  15;
2x  z  1; 2y 
 z  7
20b. Sample answer: 4x  y  z  12;

4x 
 y 
 z  
10; 5y 
 z  9

35

Chapter 2

23. x  yz  2
y  xz  2
z  xy  2
x  yz  2
y  xz  2

y

26.

D (3, 6)
(1, 3)
A

x
yz  2

y 

xz  
2
(x 
 y)  (yz 
 xz)  0
(x 
 y) 
 z(x 
 y)  0
(1 
 z)(x 
 y)  0
1 
 z  0 or x 
 y  0
z1
xy
y  xz  2
y

xz
 2
→
z  xy  2

z 

xy  
2
( y 
 z) 
 (xz 
 xy)  0
( y 
 z) 
 x(y 
 z)  0
(1 
 x)(y 
 z)  0
1 
 x  0 or y 
 z  0
x1
yz
If z  1 and x  1, 1  1y  2 and y  1.
If z  1 and y  z, x  1 1  2 and x  1.
If x  y and x  1, 1  1z  2 and z  1.
If x  y and y  z, x  y  z.
xx x2
x  x2  2
2
x x
20
(x  2)(x 
 1)  0
x  2  0 or x 
 1  0
x2  
2
x1
If x  
2, y  
2 and z  
2.
The answers are (1, 1, 1) and (
2, 
2, 
2).
24.
3x  4y  375
→
3x  4y  375

2(5x  2y)  
2(345)

10x 
 4y  
690

7x
 
315
x  45
5x  2y  345
5(45)  2y  345
y  60
(45, 60)
y
25.
→

C (6, 2)

O

B (2, 1)

x

AB: d  

(
1 

3)2 
(2 
 
(
1))2  5
2
(2 
 (

1)) 
 (6 

2)2  5
BC: d  

CD: d  

(6 
 2
)2  (3

 6)2  5
2
AD: d  

(6 
 3
)  (3

 (

1))2  5

1 
 3


AB: m  
2 
 (
1)

2 
 (
1)


B
C: m  

6
2

4

3

 
3

 4

AB  BC  CD  AD  5 units; ABCD is
4

3

rhombus. Slope of A
B
  
3 and slope of B
C
  4,
so 
AB
⊥B
C
. A rhombus with a right angle is a
square.
27a. (20, 3000), (60, 5000)
5000 
 3000


m
60 
 20
2000

 40 or 50
y 
 5000  50(x 
 60)
y  50x  2000
C(x)  50x  2000
27b. $2000; $50
27c.
c (x )
4

3
Cost
2
($1000)
1

c(x) 
 50x  2000

O 1 2 3 4 5 6 7 8 9x
Televisions Produced

x

28.

O
y  13 x  2

A
A  r2
2  s2
 (
2
 )2

2
s
 2
The correct choice is C.

2-3

s2

Modeling Real-World Data with
Matrices

Pages 82–83

Check for Understanding

1. Sample answer:
film
pain reliever blow dryer
(24 exp.)
(100 ct)
$4.03
$6.78
$18.98 
$4.21
$7.41
$20.49 
$3.97
$7.43
$32.25 

$7.08
$36.57
$63.71 








Atlanta
Los Angeles
Mexico City
Tokyo
2. 2 4
3. The sum of two matrices exists if the matrices
have the same dimensions.
Chapter 2

36

4. Anthony is correct. A third order matrix has 3
rows and 3 columns. This matrix has 4 rows and 3
columns.
5. 2y  x 
 3
2y  y  5 
 3
xy5
y2
xy5
x25
x7
(7, 2)
6. 18  4x 
 y
24  12y
24  12y
2y
18  4x 
 y
18  4x 
 2
5x
(5, 2)
7. 16  4x
16  4x
0y
4x
2x  8 
 y
(4, 0)
13
8. X  Z  4  (
1)

2  0 6  (
2)
3 4


2 4
9. impossible

1 
 4
3
1
10. Z 
 X 
0 
 (
2) 
2 
 6
2
 
5
2 
8
4(4) 4(1)
4(
2) 4(6)
16
4


8 24
12. impossible
4 1
13. YX  [0 
3]

2 6
 [0(4)  (
3)(
2) 0(1)  (
3)(6)]
or [6 
18]
14.
Budget
($ million)
soft-drink
40.1
package delivery
22.9
telecommunications
154.9

x  2y
x  2(2)
x4
(4, 2)
19. 27  3y
8  5x 
 3y

27  3y
9y

4x 
 3y  11
3(x  y)  1
xy1
2y1
y  
1
21. 2x  
10
y  3x

y  15
(
5, 
15)
22. 
12  6x
2y1
12y  10 
 x
23. x  y  0
y  y2
3  2y 
 x
6  4 
 2x
6  4 
 2x

1  x

Viewers
(million)
78.6 
21.9 
88.9 

24. x2  1  2
xy5
5
yx
y
42
y
42
y6






Exercises

15. y  2x 
 1
xy
5

25.

y  2(y 
 5) 
 1
y  11

xy
5
x  11 
 5
x6
(6, 11)
16. 9  x  2y
13  4x  1
9  x  2y
9  3  2y
3y
(3, 3)
17. 4x  15  x
5  2y
5  2y
2.5  y

y  2(2y) 
 6
y2

8  5x 
 3y
8  5x 
 3(9)
7x
(7, 9)
20. 4x 
 3y  11
xy1

11. 4X 

Pages 83–86

18. x  2y
y  2x 
 6

4x 
 3y  11
3x  3y  3
7x
 14
x2

(2, 
1)
2x  
10
x  
5


12  6x

2  x

y  3x
y  3(
5)
y  
15
2y1
1y
(
2, 1)

xy0

1  y  0
y1

(
1, 1)

xy5
x65
x  
1

(
1, 6)

6
3 x y 
 1  15
4
3z
6z 3x  y
3x 3y 
 3
15
6

12
9z
6z 3x  y
3x  15
12  6z
3y 
 3  6
9z  3x  y
3x  15
x5
(5, 3, 2)

13  4x  1
3x

→

12  6z
2z

3y 
 3  6
y3

4x  15  x
x5

(5, 2.5)

37

Chapter 2

26.


4

2 w  5 x 
 z  
16
3y
8
6 2x  8z

2w 
 10 
2x  2z

16

4


6y

16
6 2x  8z

2w 
 10  
16 
2w 
 10  
16 
6y  6

6y  6
w3
y  
1

2x  2z  
4

16  2x  8z

2x  2z  
4
2x  8z  
16
10z  
20
z  
2

5(5) 5(7)
5(
6) 5(1)
25 35


30
5
3
5
5 7
38. BA 

1 8

6 1
3(5)  5(
6)
3(7)  5(1)


1(5)  8(
6) 
1(7)  8(1)

15 26
or

53
1
39. impossible
4 
2
3
40. FC  
6 
1 0
5
0 
1
1
4 0
9
0
1

6(4)  (
1)(5)  0(9)

1(4)  4(5)  0(9)
37. 5A 


2x  2z  
4

2x  2(
2)  
4
x0
(3, 0, 
1, 
2)

53 75

6  (
1) 1  8
8 12


7
9
28. impossible
29. impossible
0  4 1  (2)
23
30. D  C  
2  5
3  0 0  (
1)
49
40

2  1
4 
1
5
 3
3 
1
13
4 
1
27. A  B 


6(
2)  (
1)(0)  0(0)
1(
2)  4(0)  0(0)

6(3)  (
1)(
1)  0(1)
1(3)  4(
1)  0(1)

29 12 
17

24 
2

1
0 1
2
8 
4
2

2 3
0
41. ED 
3
1 
5
4 4 
2
 8(0)  (
4)(
2)  2(4)
3(0)  1(
2)  (
5)(4)

31. B 
 A  B  (
A)
3 5  
5 
7


1 8
6 
1
3  (
5) 5  (
7)

2 
2

or

1  6 8  (
1)
5
7
32. C 
 D  C  (
D)
4 
2
3
0 
1 
2
0 
1 
 5
2 
3
0
9
0
1

4 
4
2
4  0 
2  (
1) 3 (
2)
52
0  (
3) 
1  0

9  (
4)
0  (
4)
12
4 
3
1
or 7 
3 
1
5 
4
3
33.
4(0) 4(1)
4(2)
4D  4(
2) 4(3)
4(0)
4(4) 4(4) 4(
2)
0
4
8
 
8 12
0
16 16 
8
34. 
2F  
2(
6) 
2(
1)

2(1)

2(4)
12
2 0


2 
8 0
35. F 
 E  F  (
E)

6 
1 0


1
4 0

14 3 
2


2 3
5
36. E 
 F  E  (
F)
8 
4
2


3
1 
5
14 
3
2

2 
3 
5

Chapter 2


42. AA 


43. FD 


8(1)  (
4)(3)  2(4)
3(1)  1(3)  (
5)(4)
8(2)  (
4)(0)  2(
2)
3(2)  1(0)  (
5)(
2)
16
4 12

22 
14 16
5 7
5 7

6 1

6 1
5(5)  7(
6)
5(7)  7(1)

6(5)  1(
6) 
6(7)  1(1)
42
or 
17

36 
41
0 1
2

6 
1 0

2 3
0
1
4 0
4 4 
2

6(0)  (
1)(
2)  0(4)
1(0)  4(
2)  0(4)


6(1)  (
1)(3)  0(4)
1(1)  4(3)  0(4)

6(2)  (
1)(0)  0(
2)
1(2)  4(0)  0(
2)
2

9 
12


8 13
2
8  2 
4  (
9) 2  (
12)
E  FD 
3  (
8)
1  13

5  2
10

13

10


5
14

3


2(0)

2(0)


8
4 
2

3 
1
5

6
1 0

1 
4 0

38

5 7
3 5

6 1

1 8
(
3)(5) 
3(7)
3 5

(
3)(
6) 
3(1)

1 8
3 5
 
15 
21
18

3

1 8
 
15(3)  (
21)(
1)
18(3)  (
3)(
1)

15(5)  (
21)(8)
18(5)  (
3)(8)

24

243

57
66
3
5
5 7
8 
4
2
45. (BA)E 

1 8

6 1
3
1 
5
3(5)  5(
6)
3(7)  5(1)


1(5)  8(
6) 
1(7)  8(1)
8 
4
2
3
1 
5
8 
4
2
 
15 26

53
1
3
1 
5
 
15(8)  26(3) 
15(
4)  26(1)

53(8)  1(3)

53(
4)  1(1)

15(2)  26(
5)

53(2)  1(
5)

42
86 
160


421 213 
111
46. F 
 2EC  F  (
2EC)
4 
2
3
2

2EC  
2 8 
4
5
0 
1
3
1 
5
9
0
1

2(8)

2(
4)

2(2)


2(3)

2(1) 
2(
5)
4 
2
3
5
0 
1
9
0
1
4 
2
3

16
8 
4

5
0 
1

6 
2 10
9
0
1

16(4)

8(5)

(
4)(9)


6(4)  (
2)(5)  10(9)

44. 
3AB  
3





2
4
3 
3
6
8 
4
5
4 
2

2
6
3(2)
3(4)
3 
3
6
3(8) 3(
4)
5
4 
2
3(
2)
3(6)
6
12
3 
3
6
24 
12
5
4 
2

6
18
6(3)  12(5)
6(
3)  12(4)
24(3)  (
12)(5) 24(
3)  (
12)(4)

6(3)  18(5)

6(
3)  18(4)
6(6)  12(
2)
24(6)  (
12)(
2)

6(6)  18(
2)
78
30
12
12 
120 168
72
90 
72

47. 3XY  3







5
48. 2K 
 3J  2 1 
7  (
3) 
4
3
2
1 
1

3(5)
 2(1) 2(
7)  (
3)(
4)
2(3)
2(2)
(
3)(1) 
3(
1)
 2 
14  12 
15
6
4

3
3
2

12

14

(
15)

6  (
3)
43
14 
29

3
7
49. Sample answer:
1996 2000 2006
18 to 24
8485 8526 8695 

 10,102 9316 9078 
25 to 34

35 to 44
8766 9036 8433 


45 to 54
6045 6921 7900 


55 to 64
2444 2741 3521 

65 and older 
2381 2440 2572 
18 24 19
26 31 24
16 24 17
22 28 21
50a.


6
6
6
12
9
7
12
2
4
17
4
6
18 24 19

26 
31 
24
 26 24 17  
22 
28 
21
6
6
6

12

9

7
12
2
4

17

4

6
18  (
26) 24  (
31) 19  (
24)
 16  (
22) 24  (
28) 17  (
21)
6  (
12)
6  (
9)
6  (
7)
12  (
17)
2  (
4)
4  (
6)

8 
7 
5

6 
4 
4
or

6 
3 
1

5 
2 
2


16(
2)  8(0)  (
4)(0)

6(
2)  (
2)(0)  10(0)

16(3)  8(
1)  (
4)(1)

6(3)  (
2)(
1)  10(1)
 
60 32 
60
56 12

6

6

(
60)

1  32 0  (
60)
F  (
2EC) 
1  56
4  12 0  (
6)

66
31

60

57 16

6

Classical
Jazz
Opera
Musicals

TV

8

6

6

5

Radio

7

4

3

2

Recording

5

4

1

2







50b. classical performances on TV

39








Chapter 2

51a.


2
3
a b

2
3

4 
5
c d
4 
5

2(a)  3(c)

2(b)  3(d)

2
3

4(a)  (
5)(c) 4(b)  (
5)(d)
4 
5

2a  3c  
2

2b  3d  3
4a 
 5c  4
4b 
 5d  
5

55.

2(
2a  3c)  2(
2) → 
4a  6c  
4
4a 
 5c  4
4a 
 5c  4
c 0
4a 
 5c  4
4a 
 5(0)  4
a1
2(
2b  3d)  2(3) → 
4b  6d  6
4b 
 5d  
5
4b 
 5d  
5
d 1
4b 
 5d  
5
4b 
 5(1)  
5
b0
51b. a matrix equal to the original matrix
52a. [42 59 21 18]
52b.
33.81 30.94 27.25 


15.06 13.25 8.75 

[42 59 21 18] 
54
54
46.44


52.06 44.69 34.38 



153  20z  10 2x  63  84  5
1
1
z  4
x  2
12, 13, 14
7
56. 4x 
 2y  7
y  2x 
 2
→
7

12x  6y  
21
y  2x 
 2
consistent and dependent
y
57.
1

1

1

6  3x  y  12

x

O

 4233.81  5915.06  21(54)  1852.06

58.

4230.94  5913.25  21(54)  1844.69
4227.25  598.75  2146.44  1834.38

 [4379.64 4019.65 3254.83]
July, $4379.64; Aug, $4019.65; Sep, $3254.83
53. The numbers in the first row are the triangular
numbers. If you look at the diagonals in the
matrix, the triangular numbers are the end
numbers. To find the diagonal that contains 2001,
find the smallest triangular number that is
greater than or equal to 2001. The formula for the
n(n  1)
n(n  1)
nth triangular number is 2. Solve 2 
2001. The solution is 63. So the 63rd entry in the
63(63  1)
first row is 2  2016. Since 2016 
 2001 
15, we must count 15 places backward along the
diagonal to locate 2001 in the matrix. This
movement takes us from the position (row,
column)  (1, 63) to (1  15, 63 
 15)  (16, 48).
54a.
A B C D
A  0 1 0
0

B  1 0 1
1


C
0 1 0
2


D  0 1 2
0


54b. No; since the matrix shows the number of nodes
and the numbers of edges between each pair of
nodes, only equivalent graphs will have the
same matrix.

Chapter 2

2x  6y  8z  5

2x  9y  12z  5
15y  20z  10
2(
2x  9y  12z)  2(5) → 
4x  18y  24z  10
4x  6y 
 4z  3
4x  6y 
 4z  3
24y  20z  13
15y  20z  10
→ 15y  20z  10

1(24y  20z)  
1(13)

24y 
 20z  
13

9y
 
3
1
y  3
15y  20z  10
2x  6y  8z  5

x

2

1
0
1
2

f(x)
8
5
2
5
8

f (x)

f (x) 
 |3x |  2
x

O

59. Sample answer: using (60, 83) and (10, 65),
65 
 83


m
10 
 60

18




50 or 0.36

y 
 65  0.36(x 
 10)
y  0.36x  61.4
7
4

3

y 
 7  4(x 
 5)


60. m  
5
1
3

3

61. f(x)  5x 
 3
0  5x 
 3
3

5

x

62. [f g](x)  f(x) g(x)

 5x(40x 
 10)
2

 16x2 
 4x
63. f(x)  4  6x 
 x3
f(14)  4  6(14) 
 (14)3
 
2656

40

1

y  4x  34

 4

64.

2x 
 3

x

2d. Let A  a11 a12 , B  b11 b12 ,
a21 a22
b21 b22

3
x

 2
2(2x 
 3)  x(3 
 x)
4x 
 6  3x 
 x2
x2  x 
 6  0
(x  3)(x 
 2)  0
x  3  0 or x 
 2  0
x  
3
x2
The correct choice is A.

Page 86

and C  c11 c12 .
c21 c22
(AB)C  a11b11  a12b21 a11b12  a12b22 c11 c12
a21b11  a22b21 a21b12  a22b22 c21 c22
 a11b11c11  a11b12c21  a12b21c11  a21b22c21
a21b11c11  a21b12c21  a22b21c11  a22b22c21
a11b11c12  a11b12c22  a12b21c12  a12b22c22
a21b11c12  a21b12c22  a22b21c12  a22b22c22

Graphing Calculator Exploration

A(BC)  a11 a12 b11c11  b12c21 b11c12  b12c22
a21 a22 b21c11  b22c21 b21c12  b22c22

1. All of the properties except for the Commutative
Property of Multiplication hold true. When
multiplying matrices, the order of the
multiplication produces different results.
However, in addition of matrices, order is not
important.
2a. Let A  a11 a12 and B  b11 b12 .
a21 a22
b21 b22
a
a
b
b
11
12
11
12
AB

a21 a22
b21 b22

a11b11c11  a11b12c21  a12b21c11  a21b22c21
 a b c a b c a b c a b c
21 11 11
21 12 21
22 21 11
22 22 21
a11b11c12  a11b12c22  a12b21c12  a12b22c22
a21b11c12  a21b12c22  a22b21c12  a22b22c22
Therefore (AB)C  A(BC).
3. All properties except the Commutative Property of
Multiplication will hold for square matrices. A
proof similar to the ones in Exercises 2a-2d can be
used to verify this conjecture.

 a11  b11 a12  b12
a21  b21 a22  b22


b11  a11
b21  a21

b12  a12
b22  a22



b11 b21  a11 a12
b21 b12
a21 a22

2-4A Transformation Matrices

2b. Let A  a11 a12 , B  b11 b12 ,
a21 a22
b21 b22
and C  c11
c21
a11  b11
(A  B)  C 
a21  b21


Page 87
1. The new figure is a 90° counterclockwise rotation
of LMN.
2. The new figure is an 180° counterclockwise
rotation of LMN.

c12
c22
a12  b12
 c11 c12
a22  b22
c21 c22

(a11  b11)  c11 (a12  b12)  c12
(a21  b21) c21 (a22  b22)  c22

 (b11  c11) a12  (b12  c12)
 a11
a21  (b21  c21) a22  (b22  c22)
 a11 a12  b11  c11 b12  c12
b21  c21 b22  c22
a21 a22
 A  (B  C)
2c. Let A  a11 a12 and B
a21 a23
a b  a12b21
AB  11 11
a21b11  a22b21
BA 
Thus, AB

3. The new figure is an 270° counterclockwise
rotation of LMN.

b11 b12 .
b21 b22
a11b12  a12b22
a21b12  a22b23

b11a11  b12a21 b11a12  b12a22
b21a11  b22a21 b21a12  b22a22

BA, since a12b21

b12a21.

41

Chapter 2

4. See students’ work for graphs. Multiplying a
vertex matrix by 0 
1 results in a vertex
1
0
matrix for a figure that is a 90° counterclockwise
rotation of the original figure. Multiplying a

1
0
vertex matrix by
results in a vertex
0 
1
matrix for a figure that is a 180° counterclockwise
rotation of the original figure. Multiplying a
0 1
vertex matrix by
results in a vertex

1 0
matrix for a figure that is a 270° counterclockwise
rotation of the original figure.

2 
2
 
2 2
1 1 
3 
3
A
(
2, 1), B
(2, 1), C
(2, 
3), D
(
2, 
3)
y
A

Pages 92–93

O

C x

D
D

7.

Check for Understanding

1. Translation, reflection, rotation, dilation;
translations do not affect the shape, size, or
orientation of figures; reflections and rotations do
not change the shape or size of figures; dilations
do not change the shape, but do change the size of
figures.
2. 90° counterclockwise  (360 
 90)° or 270°
clockwise; 180° counterclockwise  (360 
 180)°
or 180° clockwise; 270° counterclockwise 
(360 
 270)° or 90° clockwise.
3. Sample answer: the first row of the reflection
matrix affects the x-coordinates and the second
row affects the y-coordinates. A reflection over
the x-axis changes (x, y)  0 (x, 
y), so the first
row needs to be [1 0] so the x is unchanged
and the second row needs to be [0 
1] so the
y-coordinates are the opposite. Similar reasoning
can be used for a reflection over the y-axis, which
changes (x, y) to (
x, y) and a reflection over the line
y  x, which interchanges the values for x and y.
4a. 6
4b. 2
4c. 3
4d. 4

C


1 0

1 
4 
3
0
0 1
2
1 
2 
1
3
0
 1 4
2 1 
2 
1
A
(1, 2), B
(4, 1), C
(3, 
2), D
(0, 
1)

y
A

A

B

B

O
C

C

3 
1
1 
2 4 
2
2
4 
2

3 1 
1

P
(2, 
3), Q
(4, 1), R
(
2, 
1)
y
Q
P

O

Q

x

R
R

P

0

1 0
1
0
9. 
1

0 
1
0 1
0 
1
1
0

6 
3 
1

6 
3 
1

0 
1
4
2 
2

4 
2
2
L
(
6, 
4), M
(
3, 
2), N
(
1, 2)

y

y

L

J

x

D D

0 1
8.

1 0

0  
3 1.5
0
5. 1.5 
2 1
5 3 
2
7.5 4.5 
3
J
(
3, 7.5), K
(1.5, 4.5), L
(0, 
3)
J

B
B

A

Modeling Motion with Matrices

2-4


1 3
3 
1

1 
1 
1 
1

3 3 
1 
1

2 
2 
2 
2

6.

K

M N

K
N

x

O x

M

L
L

L

10b. x  3
y4

10a.
Landing
4N
3E
Ball

Chapter 2

42

Pages 93–96

0

3 0 3
0
14b. 3 
1 0 1

0 1 0 
1
0 3 0 
3

Exercises

1 1 5
3
3 15

1 4 1
3 12
3
A
(3, 3), B
(3, 12), C
(15, 3)

11. 3

0  
6 0 6
0
2 
3 0 3
0 3 0 
3
0 6 0 
6
A
(
3, 0), B
(0, 3), O
(3, 0), D
(0, 
3); A(
6, 0),
B(0, 6), C
(6, 0), D(0, 
6)

y

B
12
10
8
6 B
A
4
2 A
C

y B
B

C

2 4 6 8 10 12 14 x

3

12. 4

0 
5 
3

8
9
2

A

0


15

4

6

27

4

3



14c. The final results are the same image.
3 
3
3
3 
1 4
15. 
2 1
0 5 
1

2 
2 
2

2 3
W
(1, 
2), X
(4, 3), Y
(6, 
3)

X

y

6

3

X
X

Z
Z

x

O

W O


3 
2 1 4

6 
4 2 8
13. 2

0
2 3 2
0
4 6 4
P
(
6, 0), Q
(
4, 4), R
(2, 6), S
 (8,4)

y

W

16. 0 1 4 3  
2 
2 
2 
2
0 5 7 2

1 
1 
1 
1

2 
1 2 1


1
4 6 1
Q
(
2, 
1), P
(
1, 4), Q
(2, 6), R
(1, 1)

S
R
Q

S

y
x

O

P

x

Y

Y

R

Q

P

x

D

X

Y

C

D

3

2
1
1

24, 12

y

Y

C

C

D


9

4

X
(0, 6), Y

34, 64, Z

3

B

A

A

Q

Q
P

P

0

2 0 2
0
14a. 2 
1 0 1

0 1 0 
1
0 2 0 
2
0  
6 0 6
0
3 
2 0 2
0 2 0 
2
0 6 0 
6
A
(
2, 0), B
(0, 2), C
(2, 0), D
(0, 
2); A(
6, 0),
B(0, 6), C
(6, 0), D(0, 
6)

R
R
O

17.

y B

x

O


3 1 5
1
3 3 3 3
0 4 8 4


1 5 1 
3
4 4 4 4
5 9 5 1
C
(0, 5), D
(4, 9), E
(8, 5), F
(4, 1)

A

A A
D

B
B
C C
D

y

D

C
x

C

D

C

D

E

F

E

x

O
F

43

Chapter 2

4 0
2

6 
6 
6

2 
6 
4


1 3 
1

2 
2 
2

1
1 
3

18a.

22.

F
(
2, 
1), G
(
6, 1), H
(
4, 
3)
18a-b.
y

0 
1
1
2
3
1 2 1

1
0

1 
2 
1
1 2 3
L
(1, 1), M
(2, 2), N
(1, 3)

y
N

G

M

F
L

G
G

O L

F

H
F O

x

N
M

x
H

23.

H


2 
6 
4
1 1 1

1 
5 
3



1
1 
3
5 5 5
4
6
2
F (
1, 4), G(
5, 6), H(
3, 2)
18c. translation of 5 units left and 3 units up
1
0

1
0
2

1 0 2
19.

0 
1

2 
4 
3
2 4 3
A
(
1, 2), B
(0, 4), C
(2, 3)
18b.


1
0
0 4 4 0
0 
4 
4
0

0 
1
0 0 4 4
0
0 
4 
4
O
(0, 0), P
(
4, 0), Q
(
4, 
4), R
(0, 
4)

y
R

Q

P

O O

P

x

y
B

C

R

Q

A

24.

x

O
A




1 0
2 6
3 
1

2 
6 
3
1

0 1
4 2 
4 
2
4
2 
4 
2
D
(
2, 4), E
(
6, 2), F
(
3, 
4), G
(1, 
2)
D

y

V
E

S
T

x

U

S

Ox

G

G

V

F

F

b
 Rx-axis
d
a b
1 
2 
1 
1 
2 
1
c d
3 
1 
3

3
1
3
a  3b 
2a 
 b 
a 
 3b 
1 
2 
1
c  3d 
2c 
 d 
c 
 3d

3
1
3

y

a  3b  1

2a 
 b  
2 
a 
 3b  
1
c  3d  
3 
2c 
 d  1

c 
 3d  3
Thus, a  1, b  0, c  0, and d  
1. By
0 .
substitution, Rx-axis  1
0 
1

J
K
K
I

x
O

W

25a. Let a
c


1 
3 
1 2  
2
1
5 4
21. 0 1
1 0

2
1
5 4

1 
3 
1 2
H
(
2, 
1), I
(1, 
3), J
(5, 
1), K
(4, 2)

Chapter 2

T

W

O

H

y

U

D

E

H


1 
3 
5 
4 
2

2 
1 
2 
4 
4


2 
1 
2 
4 
4
1
3
5
4
2
S
(
2, 1), T
(
1, 3), U
(
2, 5), V
(
4, 4),
W
(
4, 2)

C
B

20.

0 1

1 0

J

I

44

0

6 
3 
1
6
3 1
26. 
1

0 
1
4
2 
2

4 
2 2
6
3 1  
2 
2 
2  4 1 
1

4 
2 2
5
5
5
1 3
7
J(4, 1), K(1, 3), L(
1, 7)

25b. Let a
c

b R
y-axis.
d
a b
1 
2 
1

1
2
1

c d
3 
1 
3
3 
1 
3
a  3b 
2a 
 b 
a 
 3b

1
2
1

c  3d 
2c 
 d 
c 
 3d
3 
1 
3

a  3b  
1 
2a 
 b  2

a 
 3b  1
c  3d  3

2c 
 d  
1 
c 
 3d  
3
Thus, a  
1, b  0, c  0, and d  1. By

1 0 .
substitution, Ry-axis 
0 1
25c. Let a b  Ry  x.
c d
a b
1 
2 
1  3 
1 
3
c d
3 
1 
3
1 
2 
1
a  3b 
2a 
 b 
a 
 3b  3 
1 
3
c  3d 
2c 
 d 
c 
 3d
1 
2 
1

L

y

J
K
L

K

J

x

O
L

K
J

a  3b  3 
2a 
 b  
1 
a 
 3b  
3
c  3d  1 
2c 
 d  
2 
c 
 3d  
1
Thus, a  0, b  1, c  1, and d  0. By
0 1
.
substitution, Ry  x 
1 0
a b
25d. Let
 Rot90
c d
a b
1 
2 
1

3
1
3

c d
3 
1 
3
1 
2 
1
a  3b 
2a 
 b 
a 
 3b

3
1
3

c  3d 
2c 
 d 
c 
 3d
1 
2 
1

0

6 
3 
1  
6 
3 
1
27. 1
0 
1
4
2 
2

4 
2
2

1 0

6 
3 
1 
6
3 1
0 1

4 
2
2

4 
2 2
J(6, 
4), K(3, 
2), L(1, 2)

y
J
K

a  3b  
3 
2a 
 b  1

a 
 3b  3
c  3d  1

2c 
 d  
2 
c 
 3d  
1
Thus, a  0, b  
1, c  1, and d  0. By
0 
1
substitution, Rot90 
1
0
25e. Let a b  Rot180.
c d
a b
1 
2 
1

1 2 1

c d
3 
1 
3

3 1 3
a  3b 
2a 
 b 
a 
 3b

1 2 1

c  3d 
2c 
 d 
c 
 3d

3 1 3

L L

x

O
K

K

L

J

28.

J


1 0

6 
3 
1
6 3
1

0 1
4
2 
2
4 2 
2
0 
1
6 3
1

4 
2 2

1
0
4 2 
2
6
3 1
J(
4, 6), K(
2, 3), L(2, 1)

y

a  3b  
1 
2a 
 b  2 
a 
 3b  1
c  3d  
3 
2c 
 d  1 
c 
 3d  3
Thus, a  
1, b  0, c  0, and d  
1. By

1
0 .
substitution, Rot180 
0 
1
25f. Let a b  Rot270.
c d

J
J

J
K

K

K
L

x

O

a b
1 
2 
1 
3 
1 
3
c d
3 
1 
3

1
2
1
a  3b 
2a 
 b 
a 
 3b 
3 
1 
3
c  3d 
2c 
 d 
c 
3d

1
2
1

L

L

29a. The bishop moves along a diagonal until it
encounters the edge of the board or another
piece. The line along which it moves changes
vertically and horizontally by 1 unit with
each square moved, so the translation
matrices are scalars. Sample matrices are
1 , c 
1 
1 , and
c 1 1 ,c 1
1 1

1 
1
1
1

a  3b  3

2a 
 b  
1 
a 
 3b  
3
c  3d  
1 
2c 
 d  2

c 
 3d  1
Thus, a  0, b  1, c  
1, and d  0. By
0 1 .
substitution, Rot270 

1 0

45

Chapter 2

37. Let x represent hardback books and y represent
paperback books.
4x  7y  5.75
3x  5y  4.25
3(4x  7y)  3(5.75)
12x  21y  17.25
→

4(3x  5y)  
4(4.25)

12x 
 20y  
17
y
0.25
4x  7y  5.75
4x  7(0.25)  5.75
x1
hardbacks $1, paperbacks $0.25
y
38. x  y  3
y  
x  3
(2, 4)
(3, 2)

c 
1 
1 , where c is the number of squares

1 
1
moved.
29b. The knight moves in combinations of 2 vertical-1
horizontal or 1 vertical-2 horizontal squares.
These can be either up or down, left or right.
Sample matrices are 1 1 ,
2 2
1
1 
1 
1 
1 
1 2 2
,
,
,
,

2 
2
2
2 
2 
2 1 1
2
2 , 
2 
2 , and 
2 
2 .

1 
1
1
1

1 
1
29c. The king can move 1 unit in any direction. The
matrices describing this are 1 1 ,
0 0

1 
1 , 0 0 ,
0
0 , 1 1 ,
1
1 ,
0
0 1 1 
1 
1 1 1 
1 
1

1 
1

1 
1
, and
.
1
1

1 
1

(4, 2)

O (0, 0)

xy
3

a b
.
c d
Dilation with scale factor 
1
a b

a 
b

1

c d

c 
d
Rotation of 180°

1
0 a b

a 
b

0 
1 c d

c 
d
The vertex matrices for the images of a dilation
with scale factor 
1 and a rotation of 180° are the
same, so the images are the same.
31. (0, 
125); (125, 0), (0, 125), (
125, 0)
32. Sample answer: There is no single matrix to
achieve this. You could reflect over the x-axis and
then translate 2(4) or 8 units upward.
33. See students’ work; the repeated dilations animate
the growth of something from small to large,
similar to a lens zooming into the origin.
2
34. 
1 1 4

1 2 0 
3
34a. Sample answer: the figure would be enlarged
disproportionally.
3 y0

1 1 4
2

3 3 12
6
34b.

0 62 B
1
2
0

3

2
4
0

6

30. Consider

4 B
2

A A

4

6

(3, 2)
y 
 1  4(x 
 2))
y 
 1  4x  8
4x 
 y  9  0
40. y 
 6  2(x 
 1)
y  2x  4
41. (f g)(x)  f(x) g(x)
 x3(x2 
 3x  7)
 x5 
 3x4  7x3
39.

g(x)  g(x)
f

f(x)

x3



x2 
 3x  7

42. 2x  y  12
2x  y  12
→

2(x  2y)  
2(
6)

2x 
 4y  12

3y  24
y  
8
2x  y  12
2x  (
8)  12
x  10
2x  2y  2(10)  2(
8)
4
The correct choice is B.

C

C

2 4 6 8 10 12 x
D

Page 96
1.

D

1
x
2

Mid-Chapter Quiz
1

 5y  17

3

y  
2x  9

y
3x  2y 
 18

34c. See students’ work; the figure appears as if
blown out of proportion.
3 8
1 5
31 85
35.



2 4

2 8

2  (
2) 4  8

36. x  y  z  1.8
y 
 z  
5.6
x  2y
 
3.8
x  2y  
3.8
Chapter 2

(4, 3)

4 13

4 12
x 
 2y  4.6
x  2y  
3.8
2x
 0.8
x  0.4
y 
 z  
5.6

1
2 x  5y 
 17

O

46

2


y  
1
0x  35

→

3x  2y  18



x

x

2. 4x  y  8
y  8 
 4x

6x 
 2y  
9
6x 
 2(8 
 4x)  
9
1
x  2

10. The result is the original figure. The original
figure is represented by a b c . The
d e f
reflection over the x-axis is found by
1
0
a b c 
a
b
c . The
0 
1
d e f

d 
e 
f
reflection of the image over the x-axis is found by
1
0
a
b
c  a b c . The matrix
0 
1

d 
e 
f
d e f
for the final image is the same as that of the
original figure.

4x  y  8

42  y  8
1

y6
2, 6
3. Let x represent trucks and y represent cars.
x  4y
6x  5y  29,000
x  4y
6(4y)  5y  29,000
x  4(1000)
y  1000
 4000
4000 trucks, 1000 cars
4. 2x  y  4z  13
3x 
 y 
 2z  
1
5x
 2z  12
1

2(3x 
 y 
 2z)  2(
1)
4x  2y  z  19

→


2(5x  2z)  
2(12)
10x 
 3z  17

→

5x  2z  12
5x  2(1)  12
x2
(2, 5, 1)
5. x  y  1
2x 
 y  
2
3x
 
1
1
x  
3
4

4x  y  z  8
  113  z  8
z8

1

3

6. y 
 3  x
y  2x
y  2x
y  2(3)
6

2-5

6x 
 2y 
 4z  
2
4x  2y  z  19
10x

 3z  17

Page 102

Determinants and Multiplicative
Inverses of Matrices
Check for Understanding

1. Sample answer: a matrix with a nonzero
determinant
2 0 is not a square
2. Sample answer: 3
4 
3 5
2 
5
matrix. 1 
1 also has no determinant.
0
9
1 0 0 0
0 1 0 0
3. Sample answer:
0 0 1 0
0 0 0 1


10x 
 4z  
24
10x 
 3z  17

7z  
7
z1
2x  y  4z  13
2(2)  y  4(1)  13
y5

xy1
1

3  y  1
1
y  13

4. Sample answer: The system has a solution if
ad 
 bc 0, since you can use the inverse of the
matrix a c to find the solution.
b d
4 
1  4(3) 
 (
2)(
1) or 10

2
3
12

26
6.
 12(32) 
 (
15)(
26) or 
6

15
32
4
1
0
7.
5 
15 
1

2
10
7

15 
1
5 
1
5 
15
4

1
0
10
7

2
7

2
10
 4(
95) 
 1(33)  0(20)
 
413
6 4 
1
3 3
0 3
0 3

4
 (
1)
8.
0 3
3 6
0 0

9 0

9 0

9 0
0
 6(0) 
 4(27)  (
1)(27)
 
135
9. 
2 3  
2(7) 
 5(3) or 
29
5 7
1
7 
3

2
9 
5 
2
10. 4 6  4(9) 
 6(6) or 0
6 9
does not exist
5.


13, 113, 8
y
3x
2x 
 3  x
x3

(3, 6)
5  8 
7  6
7. A  B  3  (
2)

1  5 0  (
9) 4  10
1 13 
1

4 
9 14
8. impossible
9. B 
 3A  B  (
3A)

3A  
3 3 5 
7

1 0
4

3(3)

3(5)

3(
7)


3(
1) 
3(0)

3(4)

9

15
21
or
3
0 
12

2
8
6

9 
15
21
B  (
3A) 

5 
9 10
3
0 
12
6  21
 
2  (
9) 8  (
15)
53

9  0 10  (
12)
 
11 
7 27
8 
9 
2

47

Chapter 2

4 
1 
2
2 1
0 1
0 2

 (
1)
 (
2)
20. 0
2
1 4
1 3
2 3
2 1
2
1
3
 4(5)  1(
2) 
 2(
4)
 26
2 
1
3
21. 3
0 
2
1 
3
0
0 
2
3 
2
3
0

 (
1)
3
2

3
0
1
0
1 
3
 2(
6)  1(2)  3(
9)
 
37
8 9 3
5 7
3 7
3 5

9
3
22.
3 5 7 8
2 4

1 4

1 2

1 2 4
 8(6) 
 9(19)  3(11)
 
90
4
6
7

2 
4
3 
4
3 
2
23. 3 
2 
4  4

6
7
1
1
1
1
1
1
1
1
1
 4(2) 
 6(7)  7(5)
1
25
36 15
24. 31 
12 
2
17
15
9

5
4
x  
3

3 
5
y

24
1

5 
4  
1 
5 
4
13
5
4
3
5
3
5

11.


3 
5


5 
4
3
5

1

1
3

5
4

3 
5

x
y
1 
5 
4
 
1
3
3
5


3

24

111
x
 
13
y
129

13



111 129

13, 1
3


6 
3
x
63

5
9
y
85
1 
9 3
1

9 3
 
3
9 
5 6

3

5 6

12.

6
5 
9


9 3

5 6

1


3
9

6 
3
5 
9

1 
9 3
x
 
3
9 
5 6
y
x
8

y

5

63
85

(8, 
5)
13. Let x represent the amount of metal with 55%
aluminum content, and let y represent the amount
of metal with 80% aluminum content.
x  y  20
0.55x  0.8y  0.7(x  y)
0.55x  0.8y  0.7x  0.7y
0.15x 
 0.1y  0
15x 
 10y  0
x  y  20
15x 
 10y  0
1
1
15
1


2
5

1

10

→

1
1
15 
10

 25 
12 
2 
 36 31 
2  15 31 
12
15
9
17
9
17
15
 25(
78) 
 36(313)  15(669)
 
3183
1.5 
3.6 2.3
25.
4.3
0.5 2.2

1.6
8.2 6.6
 1.5 0.5 2.2 
 (
36) 4.3 2.2 
8.2 6.6

1.6 6.6
4.3 0.5
23

1.6 8.2
 1.5(
14.74)  3.6(31.9)  2.3(36.06)
 175.668
0
1 
4
2 3
3 3
3
2

1
 (
4)
26. 3
2
3 0

3 4
8 4
8 
3
8 
3
4

x
20

y
0

1 
10 
1

10 
1
 
2
5 
15

15
1
1


10 
1

15
1

1
1
15 
10

x
y

1 
10 
1
 
2
5 
15
1

20
0

 0(17) 
 1(
12) 
 4(
25)
 112

x
8

y
12
8 kg of the metal with 55% aluminum and 12 kg of
the metal with 80% aluminum

Pages 102–105
14.
15.
16.
17.
18.
19.

2 
3  2(
2) 
 (
2)(
3) or 
10

2 
2
1 
2 3

1
0
2 2
2 0
 2(0) 
 1(0) or 0
28.
1 0
does not exist
29. 4 2  4(2) 
 1(2) or 6
1 2
1
2 
2

6 
1
4
6
7
30.
 6(7) 
 (
6)(7) or 84

6 7

27.

Exercises

3 4  3(5) 
 2(4) or 7
2 5

4 
1  
4(
1) 
 0(
1) or 4
0 
1
9 12  9(16) 
 12(12) or 0
12 16

2 3  
2(1) 
 (
2)(3) or 4

2 1
13
7  13(
8) 
 (
5)(7) or 
69

5 
8

6
5  
6(
8) 
 0(5) or 48
0 
8

Chapter 2

7 
7
6
6

4
6  
4(
12) 
 8(6) or 0
31.
8 
12
does not exist
1

84

48

32.

9 13  9(36) 
 27(13) or 
27
27 36
1
36 
13

2
7 
27
9
3

4


8


5

1

2

33.

1

2

1

1

34.
1
4 
1
1
2

4 
1
1
2

40.

1
x
2 1
 9
y

1 4
x
1

y
3

1
7

3
1

2

4
1
 
3
9
5


2
3
2
2
1 
1
1 
10
3

3
1
8

(1, 3)

1
9 
6
4
6
1
6

78 
4

1
1
5
3 
2
1 
2

1
7 
3

6
9 
6


9
4
6

1
x
6 6
 7
8 
4 9
y
x
0

y

2

1
5
3 
2

2 
5  
1
17

3
1

5
1

12

12

x
y 
z

x
26

y

41

2 
5

3
1



2
,
3

41.

1 
2 
5
x
26
 
1
7 
3
y
1 
41
x

9

y
7

1
5
3 
2

1
4
8
3 
3
1 
3

3
6 
3

4
8
3 
3

3 
8  
1
36

3
4

8
4

4
8
3 
3

1

1 
3 
8
x
 
3
6 
3
y
4

x
1
y  
9
z
x
1
y  
9
z

7
0

7

12
7

12

1
3 
5
5
4
1

37

4 5

5 3

x  1 4 5
37 
5 3
y
x  
3
y
3

2

3

1
4


6
5
3
x
9 
2 
1
y
3
1
1
z

1

2
1

9

12 
15
21
5
15
21 
33

1

1(
9)  (
2)(5)  1(
1)

12(
9)  (
15)(5)  21(
1)
15(
9)  21(5)  (
33)(
1)

2
12
3

2
x

9
y  4



z
3

x  
24
y

3
1
4 5
4 5
 3
7 
5 3

5 3

3 
5
5
4

1

3
2

3

x

9
y 
5
z

1

 
9

3 
5
5
4

38.




6
5
3
9 
2 
1
3
1
1

x
7

y
0

3 
8

3
4

x

y

172 , 172 

1,

1
1


3

4

3


1

2
1
1

9 
12 
15
21
15
21 
33

(
9, 7)
37.

1
x
1 
3
 
6
y

5
9

x

4
y 
0
z
1
3 
2
3
x
1
2
2
y

2
1 
1
z

4
1

10

4
1
 9 
3 3

3
0
5 1
8
1

4(
4)

1(0)

(
10)(1)
1
 9 
3(
4)  3(0)  (
3)(1)
5(
4)  1(0)  8(1)
6
1
 9
9

12

x
y
z
x
y
z

9 
6
x
12

4
6
y

12
1
6 6
6 6
 7
8 
4 9

4 9

(0, 
2)
36.

9 3
5 1

13, 
23

4 
1
x
1

1
2
y
7
2 1  1 2 1
9 
1 4

1 4

2 1

1 4

35.

1

 
6

x
1

y
1
1 
3

5
9

x

y

1

8
3

4


5

1

9

1
1 
3
9 3

5
9
5 1
1
1 
3

6

5
9

 42 
 5
8 or 1
3 1

9 3
5 1

39.

1


3

29, 
43, 
13
42. 
216
43. 30,143


24

3

(
3, 3)

49

Chapter 2

0.3
0.5
x
4.74

12 
6.5
y

1.2
1

6.5 
0.5  
1 
6.5 
0.5
7.95 
12
0.5

12
0.3
0.3

44.

0.3
12 
6.5
1 
6.5



7.95 
12


0.5
0.3

0.3
0.5
x
12 
65
y
1 
6.5 
0.5


 
 7.95

12
0.3
x
3.8

y
7.2

(3.8, 7.2)
45. 2(x 
 2y  z)  2(7)
6x  2y 
 2z  4

→

2(6x  2y 
 2z)  2(4)
4x  6y  4z  14

1
8 
2
16 10
1
10

112 
16

48. Let x represent the number of gallons of 10%
alcohol solution, and let y represent the number of
gallons of 25% alcohol solution.
x  y  12
0.10x  0.25y  0.15(x  y)
0.10x  0.25y  0.15x  0.15y

0.05x  0.10y  0
1
1
x  12
x  y  12
→

0.05
0.10
y
0

0.05x  0.10y  0
1
1
0.10 
1   0.10 
1
0.15 0.05
1
1
0.05
1
1

4.74

1.2


0.05

2x 
 4y  2z  14
6x  2y 
 2z  4
8x 
 2y
 18

1

0.15

8 
2
16 10

→ 12x  4y 
 4z  8
4x  6y  4z  14
16x  10y
 22

1
x
10 2


112 
16 8
y
x
2

y

1

18
22

1


A
1  
ad 
 bc

x 
 2y  z  7
2 
 2(
1)  z  7
z3
(2, 
1, 3)
46. Let x represent the number of cars produced in
the first year, and let y represent the number of
cars in the second year.
x  y  390,000
→
x  y  390,000
x  y  90,000
x 
 y  90,000
1
1
1 
1
1
1
1
1 
1
1


2


1 
1

1
1

(A
1)2 



Chapter 2


b 
a

d

ad 
 bc

c

ad 
 bc


b

ad 
 bc
a

ad 
 bc
2
d 
ab

1

2


 bd
a2  bc

bc 

ac 
 cd

1

a2d2 
 2abcd  b2d2



(1(4)  3(
3)  1(
12))

1

x
y

 2 
17 or 8.5
1

82 or 8.5 square units
51. Let x represent the cost of complete computer
systems, and let y represent the cost of printers.
day 1: 38x  53y  49,109
day 2: 22x  44y  31,614
day 3: 21x  26y  26,353
using day 1 and day 2:
38 53
x
49,109

22 44
y
31,614
1
44 
53  1 44 
53
506 
22
38 53

22
38
38

1
390,000
 
2 
1 
1

1
1
90,000
x
240,000

y
150,000
150,000 in the second year and 240,000 in the first year
47. Let A  a11 a12 and I  1 0 .
0 1
a21 a22
a22

d

c




1 
1  
1 
1 
1
2 
1

1
1
1



a11a22 
 a21a12
A
1  

a21


 a11a22 
 a21a12

a11a22 
 a21a12

 a11a22 

a21a12
AA
1   a21a22 
 a21a22

 a11a22 
 a21a12
1 0 I

0 1
Thus, AA
1  I.

x
y

Thus, (A2)
1  (A
1)2.
a b 1
1
50. A  2 c d 1
e f 1
1

3 1
1
 2 0
4 1
3
0 1
1
 2 1 4 1 
 (
3) 0 1  1 0 4
0 1
3 1
3 0

x
390,000

y
90,000

1
1
1 
1

1
1

0.05 0.10

1 0.10 
1
12


0.15 0.05
1
0
x
8

y
4
8 gal of 10% and 4 gal of 25%
49. Yes
A  a b . Does (A2)
1  (A
1)2
c d
2
2
A  a  bc ab  bd2
ac  cd bc  d
1
bc  d2 
ab 
 bd
(A2)
1  
a2d2 
 2abcd  b2d2
a2  bc

ac 
 cd

8 
2
x  18
16 10
y
22
1
10 2
10 2


 112

16 8

16 8
2
8

0.10

0.10 
1
0.05
1


a12

a11a22 
 a21a12 



a
11

a11a22 
 a21a12 



22

44

44 
53

22
38


a11a12  a12a11 

a11a22 
 a21a12 

1

506

a11a22 
 a21a12


a11a22 
 a21a12

44 
53

22
38
x  959
y
239
computer system: $959, printer: $239





38 53
22 44

x
y
1



506

50

49,109
31,614

60. [f 
 g](x)  f(g(x))
 f(x 
 1)
 (x 
 1)2  3(x 
 1)  2
 x2 
 2x  1  3x 
 3  2
 x2  x
[g 
 f ](x)  g(f(x))
 g(x2  3x  2)
 x2  3x  2 
 1
 x2  3x  1
61. No, more than one element of the range is paired
with the same element of the domain.
62. The radius of circle E is 3, so the circumference is
2(3) or about 18.85. The diagonal of the square
6
6
2
has length 6, so each side has length    
2

2
3

2. The perimeter of the square is 4(3

2) or
12

2  16.92.
The difference between the circumference of the
circle and the perimeter of the square is
approximately 18.85 
 16.92  1.93.
The correct choice is B.

52. Let x represent Jessi’s first test score, and let y
represent Jessi’s second test score.
x  y  179
x  y  179
→
yx7
x 
 y  
7
1
1
x  179
1 
1
y

7
1

1 
1  
1 
1 
1
2 
1
1
1

1
1
1
1 
1

1 
1 
1
x
179
 
2
y

1
1

7
x
86

y
93
first test: 86, second test: 93
53. 8 4 0 4  
3 
3 
3 
3
5 1 5 9
4
4
4
4
8

(
3)
4

(
3)
0

(
3)
4  (
3)

54
14
54
94
5
1

3
1

9 5
9 13
H
(5, 9), I
(1, 5), J
(
3, 9), K
(1, 13)
1


2

54.

3

4


1 
1

1
1

1
1
1 
1

8 
7 

4
0

21

6 
4

3

55. x 
 3y  2z  6
2(4x  y 
 z)  2(8)

0
→

Graphing Calculator Exploration:
2-5B Augmented Matrices and
Reduced Row-Echelon Form

x 
 3y  2z  6
8x  2y 
 2z  16
9x 
 y
 22

4(4x  y 
 z)  4(8)
→ 16x  4y 
 4z  32

7x 
 5y  4z  
10

7x 
 5y  4z  
10
9x 
 y
 22

1(9x 
 y)  
1(22) → 
9x  y  
22
9x 
 y  22
9x 
 y  22
0
0
infinitely many solutions
g (x)
56.
x
f(x)

6  x  
5
2
O

5  x  
4
0

4  x  
3

2

3  x  
2

4
g (x) 
 2x  5

2  x  
1

6

Page 106
2
1 
2 :
7
1 0 0 : 
1
1. 1 
2 
5 : 
1 , 0 1 0 :
5 ,
4
1
1 : 
1
0 0 1 : 
2
(
1, 5, 
2)
1
1 1 :
6
1 0 0 :
7
2. 2 
3 4 :
3 , 0 1 0 :
1 ,
4 
8 4 : 12
0 0 1 : 
2
(7, 1, 
2)
1
1
1
1 : 0 1 0 0 0 :
1
2
1 
1 
1 : 1 0 1 0 0 : 
1
3.
,

1 
1
1
1 : 0 0 0 1 0 :
2
0
2
1
0 : 0 0 0 0 1 : 
2
(1, 
1, 2, 
2)
4. Exercise 1: x  
1, y  5, z  
2; Exercise 2:
x  7, y  1, z  
2; Exercise 3: w  1, x  
1,
y  2, z  
2; They are the solutions for the
system.
5. The calculator would show the first part of the
number and follow it by ....

x

1

y 
 5  2(x 
 2)

57.

1

y 
 5  2x 
 1
2y 
 10  x 
 2
x 
 2y  8  0
3
5


m
2
1

58.


2

 1 or 
2
(y 
 5)  
2(x 
 1) or (y 
 3)  
2(x 
 2)
y 
 5  
2(x 
 1)
y 
 5  
2x  2
y  
2x  7
59a.
59b.

1

12
1

12

2-6

Solving Systems of Linear
Inequalities

Page 109–110

Check for Understanding

1a. the sum of twice the width and twice the height
1b. Sample answer: skis, fishing rods
2. Tomas is correct. There are functions in which the
coordinates of more than one vertex will yield the
same value for the function.

or approximately 0.0833
x

 18
18  12x
1.5  x; 1.5 ft

51

Chapter 2

8. Let x represent the number of greeting cards sold,
and let y represent the income in dollars.
x0
y0
y  0.45x  1500
y  1.70x

3. You might expect five vertices; however, if the
equations were dependent or if they did not
intersect to form the sides of a convex polygon,
there would be fewer vertices.
y
4.
x  2y 
 4

O

y y 
 1.70x
x

(3 13, 13)

6000
4000

xy
3

y 
 0.45x  1500

2000

y

5.
(1, 3)
(1, 0)

(1200, 2040)
2000 4000 6000 x

O
(0, 4)

at least 1200 cards
(7, 0.5)

O
(7, 0)

Pages 110–111

x

9.

Exercises

y

(
1, 0), (
1, 3), (0, 4), (7, 0.5), (7, 0)
6. y

y  x 
 1
(1,0)

(0, 2)

(4, 3)

x

O
O

(213, 13 )

x
yx
1

f(x, y)  4x  3y
f(0, 2)  4(0)  3(2) or 6
f(4, 3)  4(4)  3(3) or 25
1
1
1
1
1
f 23, 
3  423  3
3 or 83

y

10.
3

(0, 1)

25, 6
y
7.

( 23, 1)
y 
1

1
y 
3x1
1 O

(7, 7)

(3, 5)

y 
 3x  3

2

11.

(5, 3)

1

2

x

y
2x 5y 
 25

x
O
f(x, y)  3x 
 4y
f(3, 5)  3(3) 
 4(5) or 
11
f(7, 7)  3(7) 
 4(7) or 
7
f(5, 3)  3(5) 
 4(3) or 3
3, 
11

O

5x  7y 
 14

x

y 
 3x  2

12. yes, it is true for both inequalities:
1

y  3x  5
?

52

y  2x  1
?

5


2  2(3)  1


2  4 true


2  7 true


2 

Chapter 2

1
(3)
3

y

13.

y

18.
y 
 2x  2

y 
 0.5x  1

(0, 4)

(0, 2)

x2
2

( 25, 45 )

(25, 1 15 )

x

O

y 
 3x  2

f(x, y)  y 
 x
f(0, 0)  0 
 0 or 0
f(0, 4)  4 
 0 or 4
f(2, 0)  0 
 2 or 
2
4, 
2
4x  5y 
 10
19. y
y
6
(5, 6)

y
y
0

(0, 0)

x

O
(3, 3)
x3
0

y
0
y 
 2x  4

35, 315, 
25, 115, 135, 15
14.

(2, 0) x

(0, 0)
O

(0, 3)

(10, 6)

x
y

2x  5y 
 10

(0, 0), (0, 
3), (
3, 
3)
y
15.

(0, 2)

(1, 5)

O

f(x, y)  x  y
f(0, 2)  0  2 or 2
f(5, 6)  5  6 or 11
f(10, 6)  10  6 or 16
16, 2
y
20.

yx
7

5x  3y 
 20

x

(4, 0) (7, 0)

(2, 5), (7, 0), (4, 0), (1, 5)
16. f(x, y)  8x  y
f(0, 0)  8(0)  0 or 0
f(4, 0)  8(4)  0 or 32
f(3, 5)  8(3)  5 or 29
f(0, 5)  8(0)  5 or 5
32, 0
y
17.
x
5
2x  5y 
 10

x

O

(2, 5) y  5 
 0

(0, 4)

x
0
(0, 1)

xy
4
(3, 1)
y
1

O

x

f(x, y)  4x  2y  7
f(0, 1)  4(0)  2(1)  7 or 9
f(0, 4)  4(0)  2(4)  7 or 15
f(3, 1)  4(3)  2(1)  7 or 21
21, 9
y
21.

(5, 4)

(0, 2) (5, 2) y 
 2 x

O

y 
 4x  6
x  4y 
 7

f(x, y)  3x  y
f(0, 2)  3(0)  2 or 2
f(5, 4)  3(5)  4 or 19
f(5, 2)  3(5)  2 or 17
19, 2

(1, 2)
(2, 2) O
x  6y 
 10

(3, 1)

x

(4, 1)
2x  y 
 7

f(x, y)  2x 
 y
f(
2, 
 2)  2(
2) 
 (
2) or 0
f(
1, 2)  2(
1) 
 2 or 
4
f(3, 1)  2(3) 
 1 or 5
f(4, 
1)  2(4) 
 (
1) or 9
9, 
4

53

Chapter 2

22. y

2x  y  2 
 16

(3, 8)

(1, 8)

25b. f(x, y)  5x  6y

f 52, 0  552  6(0) or 272
1

y
8

1

1

f 62, 0  562  6(0) or 322
1

1

f 3, 3  53  63 or 863
29 19

(1, 4)

y
5x

1

y
2

x

y  2x 
 13
0  2x 
 13
1
62

1

1

1

6

23

1

1

1

max at 72, 82  882; min at 22, 2  242
1

1

1

1

1

26. x  y  200
2x  y  300
x0
y0

y
300
(0, 200)
200
x
0
100
(0, 0)

512, 0

2x  y 
 300
(100, 100)
x  y 
 200
(150, 0)



x

0

f(x, y)  $6.00x  $4.80y
f(0, 0)  $6.00(0)  $4.80(0) or 0
f(0, 200)  $6.00(0)  $4.80(200) or $960
f(100, 100)  $6.00(100)  $4.80(100) or $1080
f(150, 0)  $6.00(150)  $4.80(0) or $900
$1080
27a. Let x represent the Main Street site, and let y
represent the High Street site.
x  20
y  20
10x  20y  1200
y
x 
 20

y  16 
 x
2x 
 13  16 
 x
29
x  3
y  16 
 x

239 , 139 

2y  17
2(16 
 x)  17
1

x  72
2y  17
1

y  82

712, 812

(20, 50 )

2y  17

10x  20y 
 1200

1

y  82

y 
 20

y  3x  1
1
82  3x  1

(20, 20 ) (80, 20 )

O

22  x
22, 82
y  7 
 2x
3x  1  7 
 2x
6
x  5
y  7 
 2x
6
y  7 
 25
23
 5
65, 253 
3y  
2x  11
3(7 
 2x)  
2x  11
1
22  x
1

y  7 
 2x
1
y  7 
 222
2
Chapter 2

1

1

1
62,

19

3y  
2x  11
y  7 
 2x

1

O y 
 0 200 300 x

 3

y  7 
 2x
y  3x  1

1

f 22, 2  522  6(2) or 242

29

y  3x  1

1

f 5, 5  55  65 or 335

y  16 
 3

2y  17

1

6 23

1

y  16 
 x
2y  17

1

1

52  x

y  16 
 x
y  2x 
 13

19

f 22, 82  522  682 or 632

f(x, y)  
2x  y  5
f(1, 4)  
2(1)  4  5 or 7
f(1, 8)  
2(1)  8  5 or 11
f(3, 8)  
2(3)  8  5 or 7
f(6, 2)  
2(6)  2  5 or 
5
f(3, 2)  
2(3)  2  5 or 1
11, 
5
23. x  4, x  
4, y  4, y  
4
24. Sample answer: y  3, x  4, 4x  3y  12
25a. 3y  
2x  11
3y  
2x  11
y0
3(0)  
2x  11
y  2x 
 13
y0

29

f 72, 82  572  682 or 882

(6, 2)
(3, 2)

O

1

1

1

x

27b. f(x, y)  30x  40y
27c.
f(x, y)  30x  40y
f(20, 20)  30(20)  40(20) or 1400
f(20, 50)  30(20)  40(50) or 2600
f(80, 20)  30(80)  40(20) or 3200
80 ft2 at the Main St. site and 20 ft2 at the High
St. site
27d. Main Street: $1200 $10  120 ft2
120 30  3600 customers
High Street: $1200 20  60 ft2
60 40  240 customers
The maximum number of customers can be
reached by renting 120 ft2 at Main St.

212, 2
54

28a. 3 is $3 profit on each batch of garlic dressing and
2 is $2 profit on each batch of raspberry
dressing.
28b. 2x  3y  18
2x  y  10
x0
y0

2. Sample answer: In an infeasible problem, the
region defined by the constraints contains no
points. An unbounded region contains an infinite
number of points.
3. Sample answer: First define variables. Then write
the constraints as a system of inequalities. Graph
the system and find the coordinates of the vertices
of the polygon formed. Then write an expression to
be maximized or minimized. Finally, substitute
values from the coordinates of the vertices into the
expression and select the greatest or least result.

y
2x  y 
 10
(0, 6)

4.

(3, 4)
2x  3y 
 18

x
0
(0, 0)

(5, 0)

O

x

y
0

f(x, y)  3x  2y
f(0, 0)  3(0)  2(0) or 0
f(0, 6)  3(0)  2(6) or 12
f(3, 4)  3(3)  2(4) or 17
f(5, 0)  3(5)  2(0) or 15
3 batches garlic dressing, 4 batches raspberry
dressing

5a. 25x  50y  4200
5b. 3x  5y  480
5c.
y

160
140
120 (0, 84)
3x  5y 
 480
100
25x  50y 
 4200
80
60
(120, 24)
40
(160,0)
20 (0,0)
x

2 1  2(2) 
 (
3)(1) or 7

3 2
1
 2 
1
7 3
2
y
30.

29.

O

y  2x  8

80
60
40
20

31.

d

O


3
2
1
20

40

60

80

1 2 3 p

32. {16}, {
4, 4}; no, two y-values for one x-value
wxyz

4

33.

20

60

100 140

5d. P(x, y)  5x  8y
5e.
P(x, y)  5x  8y
P(0, 0)  5(0)  8(0) or 0
P(0, 84)  5(0)  8(84) or 672
P(120, 24)  5(120)  8(24) or 792
P(160, 0)  5(160)  8(0) or 800
160 small packages, 0 large packages
5f. $800
5g. No; if revenue is maximized, the company will
not deliver any large packages, and customers
with large packages to ship will probably choose
another carrier for all of their business.
6. Let x  the number of brochures.
Let y  the number of fliers.
3x  2y  600
y
x  50
x 
 50
300
y  150

x

O

y

8
7
6
5 0.5x  1.5y 
 7
4
3
2 3x  9y 
 2
1 O
x

2
2 4 6 8 10 12 14

 15

w  x  y  z  60

200
100

2-7

Linear Programming

Pages 115–116

O

(50, 225)
(100, 150)

y 
 150
(50, 150)
3x  2y 
 600
100 200 300 x

C(x, y)  8x  4y
C(50, 150)  8(50)  4(150) or 1000¢
C(50, 225)  8(50)  4(225) or 1300¢
C(100, 150)  8(100)  4(150) or 1400¢
50 brochures, 150 fliers

Check for Understanding

1. Sample answer: These inequalities are usually
included because in real life, you cannot make less
than 0 of something.

55

Chapter 2

11. y

7. Let x  the number of Explorers.
Let y  the number of Grande Expeditions.
x  y  375
y
400
2x  3y  450
x0
300
y0
x  y 
 375
x
0
200
(0, 150)
100
(0, 0)

O

O

(3, 0)

(225, 0)

f 4, 3  3  3(3) or 12
f(4, 3)  3  3(3) or 12
f(4, 0)  3  3(0) or 3
f(3, 0)  3  3(0) or 3
alternate optimal solutions
12a. Let g  the number of cups of Good Start food
and s  the number of cups of Sirius food.
0.84g  0.56s  1.54
12b. 0.21g  0.49s  0.56
12c. s
3

3 (0, 2.75)

2

y
0

5x  3y 
 15

x

56
48
40
32
24
16
8

O

0.21g  0.49 s 
 0.56
(1.5, 0.5)
(2.67, 0)
1

2

g

12d. C(g, s)  36g  22s
12e. C(g, s)  36g  22s
C(0, 2.75)  36(0)  22(2.75) or 60.5
C(1.5, 0.5)  36(1.5)  22(0.5) or 65
C(2.66, 0)  36(2.66)  22(0) or 95.76
0 cups of Good Start and 2.75 cups of Sirius
12f. 60.5¢
13a. Let d  the number of day-shift workers
and n  the number of night-shift workers.
d5
n6
d  n  14
13b. d

y
6

10.

0.81y  0.56 s 
 1.54

O

infeasible

y

x

f(x, y)  3  3y

x

100 200 300
y
0

Exercises

O

(4, 0)

4x  3y 
 12

P(x, y)  1x  1.50y
P(0, 0)  1(0)  1.50(0) or 0
P(0, 30)  1(0)  1.50(30) or 45
P(45, 0)  1(45)  1.50(0) or 45
alternate optimal solutions

9. y

y
3

2x  3y 
 450

1

Pages 116–118

3)

(4, 3)

R(x, y)  250x  350y
R(0, 0)  250(0)  350(0) or 0
R(0, 150)  250(0)  350(150) or 52,500
R(225, 0)  250(225)  350(0) or 56,250
225 Explorers, 0 Grande Expeditions
8. Let x  the number of loaves of light whole wheat.
Let y  the number of loaves of regular whole
wheat.
y
60
2x  3y  90
x  2y  80
40
x  2y 
 80
x0
(0, 30)
2x  3y 
 90
y0
20
x
0
(45, 0)
(0, 0)
x
20 40 60 80
O

20

x
4

( 34 ,

unbounded

(6, 8)

2x  y 
 48

d  n 
 14

d
5
x  2y 
 42

(9, 5)

x
n
6

8 16 24 32 40 48 56

O

n

13c. $5.50 4  $7.50 4  $52
$7.50 8  $60
C(n, d)  52d  60n
13d. C(n, d)  52d  60n
C(6, 8)  52(8)  60(6) or 776
C(9, 5)  52(5)  60(9) or 800
8 day-shift and 6 night-shift workers

Chapter 2

56

17. Let x  amount to deposit at First Bank.
Let y  amount to deposit at City Bank.
x  y  11,000
y
x 
 7500
0  x  7500
12,000
1000  y  7000
x  y 
 11,000

13e. $776
14a. Let x  the number of acres of corn.
Let y  the number of acres of soybeans.
x  y  180
y
x  40
200
x 
 40
y  20
x  2y
x  y 
 180
x 
 2y

100
(40, 20)

O

y 
 20
100

(120, 60)
(160, 20)

O

x
200

x 
 2y
(33.3, 16.7)
y 
 12
(24, 12) (38, 12)
20

O

40

60

x

C(x, y)  35,000x  18,000y
C(24,12)  35,000(24)  18,000(12) or
1,056,000
C(33.3, 16.7)  35,000(33.3)  18,000(16.7) or
1,466,100
C(38, 12)  35,000(38)  18,000(12) or
1,546,000
24 nurses, 12 nurse’s aides
19. Let x  units of snack-size bags.
Let y  units of family-size bags.
x  y  2400
y
x  600
x 
 600
2400
y  900

S(x, y)  10x  15y
S(0, 2)  10(0)  15(2) or 30
S(0, 6)  10(0)  15(6) or 90
S(10, 2)  10(10)  15(2) or 130
10 section I questions, 2 section II questions
16. Let x  the number of food containers.
Let y  the number of drink containers.
x  y  1200
y
x  300
1200
y  450
(300, 900)

O

4000 8000 12,000x
y 
 1000

40
x  y 
 20
20

8 (0, 6)
6x  15y 
 90
4
(10, 2)
x
(0, 2) y 
 2
O
4
8 12 16

x  y 
 1200
(750, 450)
x 
 300

(7500, 3500)
(7500, 1000)

I(x, y)  0.06x  0.065y
I(0, 1,000)  0.06(0)  0.065(1,000) or 65
I(0, 7000)  0.06(0)  0.065(7,000) or 455
I(4000, 7000)  0.06(4000)  0.065(7000) or 695
I(7500, 3500)  0.06(7500)  0.065(3500) or 677.5
I(7500, 1000)  0.06(7500)  0.065(1000) or 515
$4000 in First Bank, $7000 in City Bank
18. Let x  the number of nurses.
Let y  the number of nurse’s aides.
x  y  50
y
x  y  20
60
y  12
x  y 
 50
x  2y

P(x, y)  150x  250y
P(40, 20)  150(40)  250(20) or 11,000
P(120, 60)  150(120)  250(60) or 33,000
P(160, 20)  150(160)  250(20) or 29,000
120 acres of corn, 60 acres of soybeans
14b. $33,000
15. Let x  the questions from section I.
Let y  the questions from section II.
6x  15y  90
y
y2
12 x 
 0
x0

800
(300, 450)
400

(4000, 7000) y 
 7000

8000
(0, 7000)
4000
(0, 1000)

y 
 450

1600

400 800 1200 x

(600, 1800)
x  y 
 2400
(1500, 900)

800

P(x, y)  17.50x  20y
P(300, 450)  17.50(300)  20(450) or 14,250
P(300, 900)  17.50(300)  20(900) or 23,250
P(750, 450)  17.50(750)  20(450) or 22,125
$23,250

(600, 900)

O

y 
 900

800 1600 2400 x

P(x, y)  12x  18y
P(600, 900)  12(600)  18(900) or 23,400
P(600, 1800)  12(600)  18(1800) or 39,600
P(1500, 900)  12(1500)  18(900) or 34,200
600 units of snack-size, 1800 units of family-size

57

Chapter 2

20. Let x  batches of soap.
Let y  batches of shampoo.
12x  6y  48
y
20x  8y  76
(0, 8)
x0
y0

23b. Sample answer: Spend more than 30 hours per
week on these services.
y
24.
20

20x  8y 
 76

y 
 x  5 10

12x  6y 
 48

x
0

O


20 
10 O
10 20 x

10
(15, 10)
y  5 
 5

20
(15, 10)

(3, 2)
(3.8, 0)

(0, 0)

y
0

x

1

O

1

8x  3y 
 33
4
A (0, 1)

O

1

1

1

25. 4x  y  6
x  2y 
 12

(9, 0)

4x  y  6
4(2y 
 12)  y  6
8y 
 48  y  6
y6

x

4
8
12
(0, 0) y 
 0

x  12y 
 12
x  2(6) 
 12 or 0
26.

x

y


1

9

0

6

1

3

2

0

3

3

(0, 6)

y
y 
 3|x  2|

O

x

27. Sample answer: C  $13.65  $0.15(n 
 30);
C  $13.65  $0.15(n 
 30)
C  $13.65  $0.15(42 
 30)
 $15.45

x
1

Area of trapezoid ACDE  2(12)(13  1)
 84

28.

(40, 10)
(60, 0)
20

40 60
(25, 0)

3
x

 2

Chapter 2 Study Guide and Assessment

(25, 10)
20

2x 
 3

x

2(2x 
 3)  x(3 
 x)
4x 
 6  3x 
 x2
x2  x 
 6  0
(x  3)(x 
 2)  0
x  3  0 or x 
 2  0
x  
3
x2
The correct choice is A.

1

Area of ABF  2(10)(3)
 15
Area of shaded origin  84 
 15
 69 square units
23a. Let x  oil changes.
Let y  tune-ups.
y
x 
 25
x  25
60
0  y  10
30x  60y  30(60)
30x  60y 
 1800
40

Page 119 Understanding and Using the Vocabulary

x

1.
3.
5.
7.
9.

P(x, y)  12x  20y
P(25, 0)  12(25)  20(0) or 300
P(25, 10)  12(25)  20(10) or 500
P(40, 10)  12(40)  20(10) or 680
P(60, 0)  12(60)  20(0) or 720
$720
Chapter 2

1

minimum: 
22, maximum: 10

C (12, 9)

y 
 10
y 
 0O

1

f(15, 
10)  3(15) 
 2(
10) or 10

2x  3y 
 3
B (3, 3)
x 
 12
8
12
y
0

1

f(0, 5)  3(0) 
 2(5) or 
22

2x  6y 
 84
D (12, 10)

4

1

f(
15, 
10)  3(
15) 
 2(
10) or 0

P(x, y)  40x  40y
P(0, 0)  40(0)  40(0) or 0
P(0, 6)  40(0)  40(6) or 240
P(4, 5)  40(4)  40(5) or 360
P(9, 0)  40(9)  40(0) or 360
alternate optimal solutions
y E (0, 14)
22.
12
F (0, 11)
8

1

f(x, y)  3x 
 2y

3 batches of soap and 2 batches of shampoo
21. Let x  the number of small monitors.
Let y  the number of large monitors.
x  2y  16
y
xy9
xy
9
12
x  4y  24
x  2y 
 16
x0
x  4y 
 24
8
y0
(4, 5)
(0, 6)
4
x
0

y
x5
(0, 5)

58

translation
determinant
scalar multiplication
polygonal convex set
element

2.
4.
6.
8.
10.

added
inconsistent
equal matrices
reflections
multiplied

Pages 120–122

18. 
x  2y 
 6z  4
x  y  2z  3
3y 
 4z  7

Skills and Concepts

2y  
4x
2(
x 
 2)  
4x

2x 
 4  
4x
x2
(2, 
4)
12.
6y 
 x  0
6(x 
 5) 
 x  0
6x 
 30 
 x  0
x6
(6, 1)
13. 3y  x  
1
x  
1 
 3y
11.

y  
x 
 2
y  
2 
 2
y  
4

yx
5
y6
5
y1

2x  5y
2(
1 
 3y)  5y

2 
 6y  5y
19.

2


11  y
2x  5y

2x  5
11
2

5

x  
11


151, 
121

2y 
 15x  
4
y  6x  1
2(6x  1) 
 15x  
4
y  6(2)  1
12x  2 
 15x  
4
y  13
(2, 13)
x2
15. 5(3x 
 2y)  5(
1)
15x 
 10y  
5
→ 4x  10y  24
2(2x  5y)  2(12)
19x
 19
x1
2x  5y  12
2(1)  5y  12
y2
(1, 2)
16. x  5y  20.5
x  5y  20.5
→ 
x  3y  13.5
3y 
 x  13.5
8y  34
x  5y  20.5
y  4.25
x  5(4.25)  20.5
x  
0.75
(
0.75, 4.25)
17. 3(x 
 2y 
 3z)  3(2)
3(x 
 4y  3z)  3(14)

3x  5y  4z  0

3x  5y  4z  0
↓
↓
3x 
 6y 
 9z  6
3x 
 12y  9z  42

3x  5y  4z  0

3x  5y  4z  0

y 
 5z  6

 7y  13z  42

7(
y 
 5z)  
7(16)
7y  35z  
42
→

7y  13z  42

7y  13z  42
48z  0
z0

y 
 5z  6
x 
 2y 
 3z  2

y 
 5(0)  6
x 
 2(
6) 
 3(0)  2
y  
6
x  
10
(
10, 
6, 0)
14.

20.

21.

22.

23.

24.

25.
26.

59

2(
x  2y 
 6z)  2(4)
2x  3y 
 4z  5
↓

2x  4y 
 12z  18
2x  3y 
 4z  15
7y 
 16z  13

4(3y 
 4z)  
4(7) →

12y  16z  
28
7y 
 16z  13
7y 
 16z  13

5y
 
15
y3
3y 
 4z  7
x  y  2z  3
3(3) 
 4z  7
x  3  2(0.5)  3
z  0.5
x  
1
(
1, 3, 0.5)
x 
 2y  z  7
2(3x  y 
 z)  2(2)
3x  y 
 z  2
2x  3y  2z  7
4x 
 y
9
↓
6x  2y 
 2z  4
2x  3y  2z  7
8x  5y
 11
5(4x 
 y)  5(9)
→
20x 
 5y  45
8x  5y  11
8x  5y  11
28x
 56
x2
4x 
 y  9
x 
 2y  z  7
4(2) 
 y  9
2 
 2(
1)  z  7
y  
1
z3
(2, 
1, 3)
8  (
5)
A  B  7  (
3)
0  2 
4  (
2)
4
3

2 
6

3 
 7

5 
 8
B
A
2 
 0 
2 
 (
4)
 
10 
13
2
2
3(
3) 3(
5)
3B 
3(2) 3(
2)

9

15

6

6

4(2)

4C 

4(
5)
 
8
20
8

3 
5
AB  7
0 
4
2 
2
7(
3)

8(2)
7(
5)  8(
2)

0(
3)  (
4)(2) 0(2)  (
4)(
2)
or 
5 
51

8
8
impossible
4A  4B  4A  (
4B)
4(7)
4(8)

4(
3) 
4(
5)
4A 

4B 
4(0) 4(
4)

4(2) 
4(
2)
28
32
12
20


0 
16

8
8
28

12
32

20
4A  (
4B) 
0  (
8) 
16  8
 40 52

8 
8

Chapter 2

0

4 2
5

4 2 5
32. 1

0 
1
3 1 
3

3 1 3
0 
1

4 2 5 
3 
1 
3
1
0

3 1 3

4 
2
5
A
(3, 
4), B
(1, 2), C
(
3, 5)

27. impossible

4 2
5

3 
3 
3

7 
1
2
28.


3 1 
3

4 
4 
4

1 
3 
7
A
(
7, 
1), B
(
1, 
3), C
(2, 
7)

y

A

C

B

y

A
A

B

B
C

C
A

1
0

2 
1 0
1

2 
1
0 1

0 
1

3
2 4 
2
3 
2 
4 2
W
(
2, 3), X
(
1, 
2), Y
(0, 
4), Z
(1, 2)

33.
34.

y
Y

W
X

35.

Z

36.

x

O
X

Z

W
Y

30.


1
0
2
2 
1 
1

2 
2 1
1

0 
1
3 
5 
5
3

3
5 5 
3
D
(
2, 
3), E
(
2, 5), F
(1, 5), G
(1, 
3)

y

37.

F

E

D

G

38.
39.

x

O
D

G
F

40.

E

3 1 
1
1.5 0.5 
0.5


4 2
1

2
1
0.5
P
(6, 
8), Q
(2, 4), R
(
2, 2)

41.

31. 0.5

4

R

2

Q

y

x


2 O

2

4

2

4

6

43. 2 
5  2(1) 
 6(
5) or 32
6
1
1
1
5

32 
6 2
2 
4  2(2) 
 (
1)(
4) or 0
44.

1
2
no inverse

P


6

8

Chapter 2

5

4 
4 
4
5
5 


3 
3 
3
6
6
6

3 5  
3(7) 
 (
4)(5) or 
1

4 7
8 
4  8(3) 
 (
6)(
4) or 0

6
3
3 
1
4
5 
2
6
7
3 
4

2
6
5
6
5
3

 (
1)
4
3 
4
7 
4
7
 3(
10)  1(
62)  4(29)
 24
5
0 
4
7
3 
1
2 
2
6
3

1
5

 0 7 
1  (
4) 7

2
6
2
6
2
 5(16) 
 0(44) 
 4(
20)
 160
no, not a square matrix
3 8  3(5) 
 (
1)(8) or 23

1 5
1
 5 
8
23 1
3
5 2
 5(4) 
 10(2) or 0
10 4
no inverse

3
5  
3(
4) 
 1(5) or 7
1 
4
1
 
4 
5
7 
1 
3

42. 3 2  3(7) 
 5(2) or 11
5 7
1
7 
2

11 
5
3

Q

R

x

O

C

29.

B

x

O

P

60

1 1 1
3 3 3


2
3

3

2

2
5
x
1


1 
3
y
2
1

3 
5  
1 
3 
5
2
5
1
2
1
2

45.

50.


1 
1


3 
5

1
1
2

(13, 
5)

x

3 
5
 
1
y
1
2
x
13

y

5

1
2


6 4

4 
2
6
3

x  1 4 
2
24 6
y
3
x

1

y
0

3 2

6 4


3
6


3 5
x 
1

2 4
y

2
1
4 
5  
1 4 
5
2 2 
3

3 5
2 
3
1

4 
5
2 
3


3 5

2 4

1 4 
5
x
 
2
y
2 
3
x  
7
y

4

(
7, 
4)
48.

1
4.6 
2.7
2.9
8.8
1
8.8

48.31 
2.9

4.6 
2.7
2.9
8.8
1



48.31

x
y
8.8 2.7

2.9 4.6

1

2

8.4
74.61

Page 123

x
1

(25, 3)

10 20 30
(0, 0) (25, 0)

x

Applications and Problem Solving

2 5 5 5
2(5)  5(3)  5(1)
30
52. 8 2 3 3  8(5)  2(3)  3(1)  49
6 4 1 1
6(5)  4(3)  1(1)
43
Broadman 30; Girard 49; Niles 43

y 
 10  2x
(1, 5)

(22, 6)

m(x, y)  22x  42y
m(0, 0)  22(0)  42(0) or 0
m(0, 6)  22(0)  42(6) or 252
m(22, 6)  22(22)  42(6) or 746
m(25, 3)  22(25)  42(3) or 676
m(25, 0)  22(25)  42(0) or 550
22 gallons in the truck and 6 gallons in the
motorcycle

(5.7, 6.6)

y

x

y
0O

x
5.7

y
6.6
49.

12

10 (0, 6)
y
6

1
8.8 2.7
8.8 2.7


48.31 
2.9 4.6

2.9 4.6

2.7
4.6

8

20


2 4


2

4

f(x, y)  3x  2y  1
f(0, 4)  3(0)  2(4)  1 or 9
f(0, 9)  3(0)  2(9)  1 or 19
f(4, 7)  3(4)  2(7)  1 or 27
f(6, 5)  3(6)  2(5)  1 or 29
f(6, 4)  3(6)  2(4)  1 or 27
29, 9
51. Let x  gallons in the truck.
Let y  gallons in the motorcycle.
y
x  y  28
0  x  25
30 x  y 
 28 x 
 25
0y6

(
1, 0)
47.

x  y 
 11
12 (0, 9) x 
 6
(4, 7)
2y  x 
 18
8
(6, 5)
(0, 4)
4
(6, 4)
y
4

O

3 2
x  
3

6 4
y
6
1
4 
2  1 4 
2
24 6
3 2
6
3
3

46.

1

24

2
5

1 
3

y

yx
6
(4, 2)

x
O

(1, 2)
y 
 2

(6, 2)

f(x, y)  2x  3y
f(1, 
2)  2(1)  3(
2) or 
4
f(1, 5)  2(1)  3(5) or 17
f(4, 2)  2(4)  3(2) or 14
f(6, 
2)  2(6)  3(
2) or 6
17, 
4

61

Chapter 2

53. Let x  the shortest side.
Let y  the middle-length side.
Let z  the longest side.
x  y  z  83
x  y  z  83
z  3x
x  y  3x  83
1
z  2(x  y)  17
4x  y  83

Chapter 2 SAT & ACT Preparation
Page 125

1

z  2(x  y)  17

1

(1  2)(2  3)(3  4)  2(20  x)

1

3x  2(x  y)  17
5x 
 y  34

1
4
5 
1

1
4
1
5 
1
1


9


1 
1

5
4

1

(3)(5)(7)  2(20  x)
x  83
y
34

1

105  2(20  x)
210  20  x
x  190
The correct choice is D.
2. First convert the numbers to improper fractions.


1 
1  
1 
1 
1
9

5
4

5
4

4
1
5 
1

1 
1 
1
x
 
9
y

5
4

83
34

1

z  3x
z  3(13)
z  39
13 in., 31 in., 39 in.
54a. Let x  number of Voyagers.
Let y  number of Explorers.
y
5x  6y  240
3x  2y  120
60
5x  18y  540
x0
40
y0
(0, 30)
20
y
0

O

16

25

Express both fractions with a common
denominator. Then subtract.
1

1

16

25

64

75

53 
 64  3 
 4
 12 
 12
11

 
12
The correct choice is A.
3. You can solve this problem by writing algebraic
expressions.
Amount of root beer at start: x
Amount poured into each glass: y
Number of glasses: z
Total amount poured out: yz
Amount remaining: x 
 yz
The correct choice is D.
4. 2x 
2 
 1  5
2x 
2  6
x 
2  3

3x  2y 
 120
5x  6y 
 240
(18, 25)
(30, 15)
5x  18y 
 540
(40, 0)

20
(0, 0)

40

60

x

P(x, y)  2.40x  5.00y
P(0, 0)  240(0)  5.00(0) or 0
P(0, 30)  2.40(0)  5.00(30) or 150
P(18, 25)  2.40(18)  5.00(25) or 168.20
P(30, 15)  2.40(30)  5.00(15) or 147
P(40, 0)  2.40(40)  5.00(0) or 96
18 Voyagers and 25 Explorers
54b. $168.20

x 
2  3 or x 
2  
3
x  5 or xor x  
1
The correct choice is D.
5. The total amount charged is $113. Of that, $75 is
for the first 30 minutes. The rest (113 
 $75 
$38) is the cost of the additional minutes. At $2
per minutes, $38 represents 19 minutes. (19 $2
 $38). The plumber worked 30 minutes plus 19
minutes, for a total of 49 minutes.
The correct choice is C.

Open-Ended Assessment

1a. A(
2, 
2), B(1, 
2), C(2, 1), and D(
3, 0)
Sample answer: Two consecutive 90° rotations is
the same as one 180° rotation. An additional
180° rotation will return the image to its original
position.
1b. Two consecutive 90° rotations is the same as one
180° rotation.
2. No; such a coefficient matrix will not have an
inverse. Consider the matrix equation
2 4
x  12 . The coefficient 2 4 has a
4 8
y
24
4 8
determinant of 0, so it has no inverse.

Chapter 2

1

53 
 64  3 
 4

x
13

y
31

Page 123

SAT and ACT Practice

1. Translate the information from words into an
equation. Then solve the equation for x. Use the
correct order of operations.

62

8. Start by representing the relationships that are
given in the problem. Let P represent the number
of pennies; N the number of nickels; D the number
of dimes; and Q the number of quarters. He has
twice as many pennies as nickels.
P  2N
Similarly, N  2D and D  2Q. You know he has
at least one quarter. Since you need to find the
least amount of money he could have, he must
have exactly one quarter.
Since he has 1 quarter, he must have 2 dimes,
because D  2Q. Since he has 2 dimes, he must
have 4 nickels. Since he has 4 nickels, he must
have 8 pennies.
Now calculate the total amount of money.
1 quarter  $0.25
2 dimes  $0.20
4 nickels  $0.20
8 pennies  $0.08
The total amount is $0.73. The correct choice is D.

6. Start by simplifying the fraction expression on the
right side of the equation.
2x

5x
2x

5x

2

2

 5  5
4

 5

To finish solving the equation, treat it as a
proportion and write the cross products.
2x

5x

4

 5

5(2  x)  4(5  x)
10  5x  20  4x
x  10
The correct choice is E.
7. Notice that the question asks what must be true.
There are two ways to solve this problem. The
first is by choosing specific integers that meet the
criteria and finding their sums.
I. 2  3  5, 3  4  7
Choose consecutive integers where the first
one is even and where the first one is odd. In
either case, the result is odd. So statement I is
true. Eliminate answer choice B.
II. 2  3  4  9, 3  4  5  12
One sum is odd, and the other is even. So
statement II is not always true. Eliminate
answer choices B, C, and E.
III. 2  3  4  9, 3  4  5  12,
10  11  12  33
Statement III is true for these examples and
seems to be true in general. Eliminate answer
choice A.
Another method is to use algebra. Represent
consecutive integers by n and n  1. Represent
even integers by 2k, and odd integers by 2k  1.
I. n  (n  1)  2n  1
2n  1 is odd for any value of n. So statement
I is always true.
II. n  (n  1)  (n  2)  3n  3
If n is even, then 3n  3  3(2k)  3 is odd.
If n is odd, then 3n  3  3(2k  1)  3 
6k  3  3  6k  6 is even. So statement II
is not always true.
III. By the same reasoning as in II, the sum is a
multiple of 3, so statement III is always true.
n  (n  1)  (n  2)  3n  3  3(n  1)
The correct choice is D.

3

2

9. 
2 

 
3

2

3
2

9
4

3

 2

4

9

12

2


 1
8  3

The correct choice is C.
10. There are two products, CDs and tapes. You need
to find the number of tapes sold. You also have
information about the total sales and CD sales.
You might want to arrange the information in a
table. Let t be the number of tapes sold.
Price each

Number
Sold

Total
Sales

CDs

40

$480

Tapes

t

Total

$600

You can calculate the price of each CD. Since
40 CDs sold for $480, each CD must cost $12
($480 40  $12). You know that the price of a
CD is three times the price of a tape. So a CD
1
costs 3 of $12 or $4. You can calculate the total
sales of CDs by subtracting $480 from $600 to get
$120.
Price each

Number
Sold

Total
Sales

CDs

$12

40

$480

Tapes

$4

t

$120

Total

$600

Now you can find t using an equation.
4t  120
t  30
Thirty cassette tapes were sold. The answer is 30.

63

Chapter 2

Chapter 3 The Nature of Graphs
3-1
Page 133

Algebraically: Substituting (
x, 
y) into the
equation followed by substituting (
x, y) is the
same as substituting (x, 
y).
6. f(x)  x6  9x
f(
x)  (
x)6  9(
x)

f(x)  
(x6  9x)
f(
x)  x6 
 9x

f(x)  
x6 
 9x
no

Symmetry and Coordinate
Graphs
Graphing Calculator Exploration

1. f(
x)  f(x)
2. f(
x)  
f(x)
3. even; odd
4. f(x)  x8 
 3x4  2x2  2
f(
x)  (
x)8 
 3(
x)4  2(
x)2  2
 x8 
 3x4  2x2  2
 f(x)
f(x)  x7  4x5 
 x3
f(
x)  (
x)7  4(
x)5 
 (
x)3
 
x7 
 4x5  x3
 
(x7  4x5 
 x3)
 
f(x)
5. First find a few points of the graph in either the
first or fourth quadrants. For an even function, a
few other points of the graph are found by using
the same y-values as those points, but with opposite
x-coordinates. For an odd function, a few other
points are found by using the opposite of both the
x- and y-coordinates as those original points.
6. By setting the INDPNT menu option in TBLSET
to ASK instead of AUTO, you can then go to
TABLE and input x-values and determine their
corresponding y-values on the graph. By inputting
several sets of opposite pairs, you can observe
whether f(
x)  f(x), f(
x)  
f(x), or neither of
these relationships is apparent.

7.

1

f(x)  5x 
 x19
1

19

f(
x)  
5(
x) 
 (
x)
1


f(x)  
5x  x19
6a2  b 
 1
6a2  (
b) 
 1
6a2  
b 
 1 no
6(
a)2  b 
 1
6a2  b 
 1 yes
6(b)2  a 
 1
6b2  a 
 1 no
6(
b)2  (
a) 
 1
6b2  
a 
 1 no
y-axis
a3  b3  4
a3  (
b)3  4
a3 
 b3  4 no
(
a)3  b3  4

a3  b3  4 no
(b)3  (a)3  4
a3  b3  4 yes
3
(
b)  (
a)3  4

a3 
 b3  4 no
yx

y-axis
yx
y  
x

→

y-axis
yx
y  
x

Pages 133–134

Check for Understanding

1. The graph of y  
x2  12 is an even function.
The graph of xy  6 is an odd function. The graphs
of x  y2 
 4 and 17x2  16xy  17y2  225 are
neither.
2. The graph of an odd function is symmetric with
respect to the origin. Therefore, rotating the graph
180° will have no effect on its appearance. See
student’s work for example.
3a. Sample answer: y  0, x  0, y  x, y  
x
3b. infinitely many
3c. point symmetry about the origin
4. Substitute (a, b) into the equation. Substitute
(
b, 
a) into the equation. Check to see whether
both substitutions result in equivalent equations.
5. Alicia
Graphically: If a graph has origin symmetry, then
any portion of the graph in Quadrant I has an
image in Quadrant III. If the graph is then
symmetric with respect to the y-axis, the portion
in Quadrants I and II have reflections in
Quadrants II and IV, respectively. Therefore, any
piece in Quadrant I has a reflection in Quadrant
IV and the same is true for Quadrants II and III.
Therefore, the graph is symmetric with respect to
the x-axis.

Chapter 3

10.

y

(2.5, 3)
(4, 2)

(2.5, 3)

(1, 2)

(1, 2)

(4, 2)

O

x

2 
 x2
11. y  

x-axis
y-axis

b  

2 
 a2

b  

2 
 a2 no
b  

2 
 (

a)2
b  

2 
 a2 yes

→

y
2
(1,1)

1

(0√2)
(1,1)

(√2,0)

(√2,0)


2 
1 O

1

2

64

1

1

f(
x)  
5x  x19
yes
8. 6x2  y 
 1
→
x-axis

9. x3  y3  4
x-axis


f(x)  
5x 
 x19

1

2 x

12. y  x3
x-axis

→

b  a3

b  a3
b  a3 yes
b  (
a)3
b  
a3 no

y-axis

18.

y

19.

f(x)  
7x5  8x
f(
x)  
7(
x)5  8(
x)
f(
x)  7x5 
 8x

f(x)  
(
7x5  8x)

f(x)  7x5 
 8x
yes
1

f(x)  x 
 x100
1

 
 (
x)100
f(
x)  
(
x)

(1,1)

O

(1,1)

1

f(
x)  
x 
 x100

x


f(x)  
x 
 x100
1

1


f(x)  
x  x100

no
20. yes;
x2 
 1

g(x)  x
13. x-intercept:

x2

25

y2

9




x2

25


 9  1

1

g(
x) 

02


g(x) 

x2
  1
25
x2  25

36

25

36

25


 9  1


y2


 9  1
y2

9

11

25

y2 

99

25





y-axis

yx

99


3
11

3
11


y  
5

y  
x

y  
5

3
11

3
11




6, 
 
5 , 
6, 
 5 , 
6,
5

Pages 134–136



11

25

y2  25


3
11

(
x)2  x2

when x  
6

y2

y2

9

Replace x with 
x.

x2 
 1

x
x2 
 1

 x

g(
x)  
g(x)
21. xy  5
→
x-axis

x  5 (5, 0)
other points:
when x  6

(
x)2 
 1


g(
x)  
(
x)

22. x  y2  1
x-axis

Exercises

→

y-axis

f(x)  3x
f(
x)  3(
x)

f(x)  
(3x)
f(
x)  
3x

f(x)  
3x
yes
15. f(x)  x3 
 1
f(
x)  (
x)3 
 1

f(x)  
(x3 
 1)
3
f(
x)  
x 
 1

f(x)  
x3  1
no
16. f(x)  5x2  6x  9
f(
x)  5(
x)2  6(
x)  9
f(
x)  5x2 
 6x  9

f(x)  
(5x2  6x  9)

f(x)  
5x2 
 6x 
 9
no
14.

17.

yx
y  
x

23. y  
8x
x-axis
y-axis
yx
y  
x

1


f(x)  
4x7
1


f(
x)  
4(
x)7
1


f(
x)  

4x7

→

Determine the opposite
of the function.
ab  
5
a(
b)  
5

ab  
5
ab  5 no
(
a)b  
5

ab  
5
ab  5 no
(b)(a)  
5
ab  
5 yes
(
b)(
a)  
5
ab  
5 yes
y  x and y  
x
a  b2  1
a  (
b)2  1
a  b2  1 yes
(
a)  b2  1

a  b2  1 no
(b)  (a)2  1
b  a2  1 no
(
b)  (
a)2  1

b  a2  1 no
x-axis
b  
8a
(
b)  
8a

b  
8a no
b  
8(
a)
b  8a no
(a)  
8(b)
a  
8b no
(
a)  
8(
b)

a  8b no
none of these



f(x)  

4x7 
1

1



f(x)  

4x7

yes

65

Chapter 3

1

24. y  x2

→

28.

1

b  a2
1


b  a2

(2, 1)

no

1

b  a2

(4, 4)


(a)  
(b)2

(4, 4) y

1

a  b2

no


(
a)  
(
b)2
1


a  b2
→

y-axis
yx
y  
x
4x2

26. y2  9 
 4
x-axis

29.

1

y  
x

→

(2, 1)

a2  b2  4
a2  (
b)2  4
a2  b2  4
(
a)2  b2  4
a2  b2  4
(b)2  (a)2  4
a2  b2  4
(
b)2  (
a)2  4
a2  b2  4
all
4a2
b2  9 
 4
4a2
(
b)2  9 
 4

(2, 1)

O

x

yes
30. Sample answer:

yes

y
yes

(4, 4)
(1, 2)

yes

(2, 1)

O

x

(2, 3)

4a2

4(
a)2

→

31. y2  x2
x-axis

b2  9 
 4
4a2

b2  9 
 4 yes
4(b)2

b2  a2
(
b)2  a2
b2  a2
b2  (
a)2
b2  a2 yes; both

y-axis

(a)2  9 
 4

yx

(4, 4)

(1, 2) (1, 2)

no; y-axis

b2  9 
 4 yes
y-axis

x

(1, 2)

yes

1

yx

O

(2, 1)

1


b
(
a)2

y-axis

(4, 4)
(1, 2)

1

(
b)  a2

x-axis

25. x2  y2  4
x-axis

y

4b2

a2  9 
 4 no

y

4(
b)2

(
a)2  9 
 4

y  
x

4b2

a2  9 
 4 no
x-axis and y-axis
27.

x2



1
y2

x-axis

→

1

1

(
a)2 
a2 

yx

(b)2



b2 
a2 
y  
x

(
b)2 
b2 
a2 

1
b2
1
b2
1
(a)2
1
a2
1
b2
1
(
a)2
1
a2
1
b2

yes

32. x  
3y
x-axis

yes

→

a  
3b
a  
3(
b)
a  3b no
(
a)  
3b
a  
3b yes
y-axis

y-axis

y

yes

O
yes

x-axis, y-axis, y  x, y  
x

Chapter 3

x


a2  
(
b)2

a2  b2
y-axis

O

1

a2  b2

66

x

33. y2  3x  0
x-axis

→

37. y  x3 
 x
x-axis

b  a3 
 a
(
b)  a3 
 a
b  a3 
 a yes
y-axis
b  (
a)3 
 (
a)
b  
a3  a no
x-axis
The equation y  x3 
 x is symmetric about the
x-axis.
y

b2  3a  0
(
b)2  3a  0
b2  3a  0 yes
2
b  3(
a)  0
b2 
 3a  0 no
x-axis

y-axis

y

→

1

x

O

O


1

34. y  2x2
x-axis

→

b  2a2
(
b)  2a2
b  2a2 yes
b  2(
a)2
b  2a2
yes; both

y-axis

x

1


1

38a.

x2

8

y2

a2

8

→

 10  1

a2

8

x-axis

y

x

O

origin

(
b)2

 10  1

a2

8
(
a)2

8
a2

8
(
a)2

8

y-axis

b2

 1
0 1
b2

 1
0  1 yes
b2

 1
0 1
b2

 1
0  1 yes
(
b)2

 10  1
a2

8

b2

 1
0  1 yes

x- and y-axis symmetry


12 
 
8y2

35. x 
x-axis

→

y-axis



12 
 
8b2

a
a  

12 
 8(
b)
2
a  

12 
 
8b2
yes
(
a)  

12 
 
8b2
a  

12 
 
8b2
yes; both

38b.

y

x

O

y

x

O

36. y  xy
x-axis

), (
2, 
5
), (
2, 

5
)
38c. (2, 

5
39. Sample answer: y  0
40. Sample answer:
yx1
y  
x  1
y  
x 
 1
y  
2x  4
y  
2x 
 4
y  2x 
 4

→

y

b  ab
(
b)  a(
b)
b  
ab no
b  (
a)b
b  
ab no
neither

y-axis

yx
1
y  2x  4

2
1

2 
1 O

1

y

1

2

x


2
1
(1, 0)

O

(1, 0)

x


1

67

Chapter 3

41.

y2

12
(6)2

12

x2


 16  1

 16  1
x2

y  2x 
 2
7

y  2x 
 2

consistent and dependent
47.

x2


16  
2

O
y
x2

16 
 2

14




2 or 
7

y 
 2  
7(x 
 0)
y  
7x  2
49. [f 
 g](x)  f(g(x))
 f(x 
 6)
 
2(x 
 6)  11
 
2x  23
[g 
 f ](x)  g(f(x))
 g(
2x  11)
 (
2x  11) 
 6
 
2x  5
50. 753 757  7537
 7510
The correct choice is B.

1
y
x

x

43. Let x  number of bicycles.
Let y  number of tricycles.
3x  4y  450
y
5x  2y  400
5x  2y 
 400
150
x0
(0, 112.5)
y0

O

x


48. m  

2 
 0

y

100
x
0
50
(0, 0)

y
y
x

x2  32
x  4
2

(4
2
, 6) or (
4
2
, 6)
42. No; if an odd function has a y-intercept, then it
must be the origin. If it were not, say it were
(0, 1), then the graph would have to contain (
1, 0).
This would cause the relation to fail the vertical
line test and would therefore not be a function.
But, not all odd functions have a y-intercept.
1
Consider the graph of y  x.

(50, 75)
3x  4y 
 450

y
0
50 100 150 x
(80, 0)

3-2

P(x, y)  6x  4y
P(0, 0)  6(0)  4(0) or 0
P(0, 112.5)  6(0)  4(112.5) or 450
P(50, 75)  6(50)  4(75) or 600
P(80, 0)  6(80)  4(0) or 480
50 bicycles, 75 tricycles
8 5  4(8)  3(9) 4(5)  3(6)
44. 4 3
7 2
9 6
7(8)  2(9) 7(5)  2(6)
59 38

74 47
45. 3(2x  y  z)  3(0)
6x  3y  3z 
0
→
3x 
 2y 
 3z  
21
3x 
 2y 
 3z  
21
9x  y
 
21
3x 
 2y 
 3z  
21
4x  5y  3z  
2
7x  3y
 
23

3(9x  y)  
3(
21)

27x 
 3y  63
→
7x  3y  
23
7x  3y  
23

20x
 40
x  
2
9x  y  
21
2x  y  z  0
9(
2)  y  
21
2(
2)  (
3)  z  0
y  
3
z7
(
2, 
3, 7)

Chapter 3

→


12x  6y  
21

3 
 16  1

O

7

46. 4x 
 2y  7

x2

Page 142

Families of Graphs
Check for Understanding

1. y  (x  4)3 
 7
2. The graph of y  (x  3)2 is a translation of y  x2
three units to the left. The graph of y  x2  3 is a
translation of y  x2 three units up.
3. reflections and translations
4. When c 1, the graph of y  f(x) is compressed
horizontally by a factor of c.
When c  1, the graph of y  f(x) is unchanged.
When 0  c  1, the graph is expanded
1
horizontally by a factor of c.
3
5a. g(x)  
x  1
3
5b. h(x)  


x
1
3
5c. k(x)  

x21
6. The graph of g(x) is the graph of f(x) translated
left 4 units.
7. The graph of g(x) is the graph of f(x) compressed
1
horizontally by a factor of 3, and then reflected
over the x-axis.
8a. expanded horizontally by a factor of 5
8b. translated right 5 units and down 2 units
8c. expanded vertically by a factor of 3, translated
up 6 units

68

9a. translated up 3 units, portion of graph below
x-axis reflected over the x-axis
9b. reflected over the x-axis, compressed
1
horizontally by a factor of 2
9c. translated left 1 unit, compressed vertically by a
factor of 0.75
y
y
10.
11.

O

x

20a. reflected over the x-axis, compressed
horizontally by a factor of 0.6

20b. translated right 3 units, expanded vertically by a
factor of 4
1
20c. compressed vertically by a factor of 2, translated
down 5 units
21a. expanded horizontally by a factor of 5
21b. expanded vertically by a factor of 7, translated
down 0.4 units
21c. reflected across the x-axis, translated left 1 unit,
expanded vertically by a factor of 9
22a. translated left 2 units and down 5 units
22b. expanded horizontally by a factor of 1.25,
reflected over the x-axis
3
22c. compressed horizontally by a factor of 5,
translated up 2 units
23a. translated left 2 units, compressed vertically by
1
a factor of 3
23b. reflected over the y-axis, translated down 7 units
23c. expanded vertically by a factor of 2, translated
right 3 units and up 4 units
24a. expanded horizontally by a factor of 2
1
24b. compressed horizontally by a factor of 6,
translated 8 units up
24c. The portion of parent graph on the left of the yaxis is replaced by a reflection of the portion on
the right of the y-axis.
2
25a. compressed horizontally by a factor of 5,
translated down 3 units
25b. reflected over the y-axis, compressed vertically
by a factor of 0.75
25c. The portion of the parent graph on the left of the
y-axis is replaced by a reflection of the portion on
the right of the y-axis. The new image is then
translated 4 units right.

O
x

12a.

x
0x1
1x2
2x3
3x4
4x5

f(x)
50
100
150
200
250

$250
$200
$150
$100
$50
0

0

1

2 3 4
Time (h)

5

12b.
$250
$200
$150
$100
$50
0

26. y  x2
0

1

2

3 4 5
Time (h)

y

6

12c. $225

Pages 143–145

Exercises

13. The graph of g(x) is a translation of the graph of
f(x) up 6 units.
14. The graph of g(x) is the graph of f(x) compressed

O

x

3

vertically by a factor of 4.
15. The graph of g(x) is the graph of f(x) compressed

0.25


27. y  
x
4  3

1

16.
17.
18.
19.

horizontally by a factor of 5.
The graph of g(x) is a translation of f(x) right 5
units.
The graph of g(x) is the graph of f(x) expanded
vertically by a factor of 3.
The graph of g(x) is the graph of f(x) reflected over
the x-axis.
The graph of g(x) is the graph of f(x) reflected over
the x-axis, expanded horizontally by a factor of
2.5, and translated up 3 units.

28.

29.

y

O

y

x
O

69

x

Chapter 3

y

30.

31.

O

36a. 0

y

x
O

y

32.

y

33.

O

x

x

4


4 O

4

4

[
7.6, 7.6] scl:1 by [
5, 5] scl:1
36b. 0.66


8x


8

12

34.

y

8
4


8 
4 O

4

y 
 f (|x |)
4

[
7.6, 7.6] scl:1 by [
5, 5] scl:1
36c. 
0.25

x

y 
 f (x ) 
8

35a. 0

[
7.6, 7.6] scl:1 by [
5, 5] scl:1
37a. 0

[
7.6, 7.6] scl:1 by [
5, 5] scl:1
35b. 0.5

[
7.6, 7.6] scl:1 by [
5, 5] scl:1
37b. 
2.5

[
7.6, 7.6] scl:1 by [
5, 5] scl:1
35c. 
1.5

[
7.6, 7.6] scl:1 by [
5.5] scl:1

[
7.6, 7.6] scl:1 by [
5, 5] scl:1

Chapter 3

70

42a. (1) y  x2
(3) y  
x2
42b. (1)
y

37c. 0.6

(2) y  x3
(4) y  
x3
(2)
y

4

4 O

4

[
7.6, 7.6] scl:1 by [
5, 5] scl:1
38a. The graph would continually move left 2 units
and down 3 units.
38b. The graph would continually be reflected over
the x-axis and moved right 1 unit.

(3)

5
4
Price 3
(dollars) 2
1
0

1

2

3

4

5

Fare Units

41a.

y

1

A  2bh
1

8

 2(10)(5)

4

 25 units2

O

4

8

12

x

8x


8

O

y
8
4
8

12

46.

x
1

41c. The area of the triangle is A  2(10)(5) 
25 units2. Its area is the same as that of the
original triangle. The area of the triangle formed
by y  f(x  c) would be 25 units2.

47.
48.

y
8
4

O

4

8

12

8x


4

1

4


4 O

(30, 20)
(30, 0)
x  y 
 50

20
(0, 0)

41b. The area of the triangle is A  2(10)(10) or
50 units2. Its area is twice as large as that of the
original triangle. The area of the triangle formed
by y  c f(x) would be 25c units2.

O

4

y

(4)
4

42c. (1) y  (x 
 3)2 
 5
(2) y  (x 
 3)3 
 5
(3) y  
(x 
 3)2 
 5
(4) y  
(x 
 3)3 
 5
43a. reflection over the x-axis, reflection over the
y-axis, vertical translation, horizontal
compression or expansion, and vertical
expansion or compression
43b. horizontal translation
44. f(x)  x17 
 x15
f(
x)  (
x)17 
 (
x)15

f(x)  
(x17 
 x15)
f(
x)  
x17  x15

f(x)  
x17  x15
yes; f(
x)  
f(x)
45. Let x  number of preschoolers.
Let y  number of school-age children.
x  y  50
y
x  3(10)
60 (0, 50)
y  5(10)
y 
 50
x0
40
x 
 30
y0

40b.

0

8x


8


12

0.25[[x 
 1]]  1.50 if [[x]]  x

y
1.50
1.75
2.00
2.25
2.50

8x


8

0.25[[x]]  1.50 if [[x]]  x

4


4


4

b

x
0x1
1x2
2x3
3x4
4x5

4

y

4 O

39. The x-intercept will be 
a.
40a. y 


4 O

x

49.

40

60

x

I(x, y)  18x  6y
I(0, 0)  18(0)  6(0) or 0
I(0, 50)  18(0)  6(50) or 300
I(30, 20)  18(30)  6(20) or 660
I(30, 0)  18(30)  6(0) or 540
30 preschoolers and 20 school-age
0 
1
5 1 
2  4 
3
1
1
0

4 3 
1
5
1 
2
A
(4, 5), B
(
3, 1), C
(1, 
2)
x2  25
9y
12  2z
x  5
6z

5(6x  5y)  
5(
14) 
30x 
 25y  70
→
6(5x  2y)  6(
3)
30x  12y  
18

13y  52
y  
4
5x  2y  
3
5x  2(
4)  
3
x1
(1, 
4)
The graph implies a negative linear relationship.

50. 3x 
 4y  0

→

perpendicular slope:

71

20
y
0

3

y  4x

4

3

Chapter 3

y

7.

5

51. 5d 
 2p  500
→
p  2d 
 250

250
52. [f 
 g](x)  f(g(x))
 f(x2 
 6x  9)

O

2

 3(x2 
 6x  9) 
 2

8.

y

x

2

 3x2 
 4x  4

O

x

[g 
 f](x)  g(f(x))

 g3x 
 2

9.

2

y

 3x 
 2 
 63x 
 2  9
2

2

2

4

8

4

20

 9x2 
 3x  4 
 4x  12  9

O

 9x2 
 3x  25

x

50

53. If m  1; d  1 
 1 or 
49.
50

If m  10; d  10 
 10 or 5.

10. Case 1
x  6 4

(x  6) 4

x 
 6 4

x 10
x  
10
{xx  
10 or x 
2}
11. Case 1
3x 
 4  x

(3x 
 4)  x

3x  4  x

4x  
4
x1
{x 1  x  2}
12a. x 
 12  0.005
12b. Case 1
x 
 12  0.005

(x 
 12)  0.005

x  12  0.005

x  
11.995
x  11.995
12.005 cm, 11.995 cm

50

If m  50; d  50 
 50 or 49.
50


If m  100; d  100 
 
100 or 99.5
50


If m  1000; d  1000 
 
1000 or 999.95.

The correct choice is A.

3-3
Page 149

Graphs of Nonlinear Inequalities
Check for Understanding

1. A knowledge of transformations can help
determine the graph of the boundary of the
shaded region, y  5  
x.

2
2. When solving a one variable inequality
algebraically, you must consider the case where
the quantity inside the absolute value is nonnegative and the case where the quantity inside
the absolute value is negative.
3. Sample answer: Pick a point not on the boundary
of the inequality, and test to see whether it is a
solution to the inequality. If that point is a
solution, shade all points in that region. If it is not
a solution to the inequality, test a point on the
other side of the boundary and shade accordingly.
4. This inequality has no solution since the two
graphs do not intersect

Pages 150–151

17.

x

x2 
 6

y  x

9

y  
5x4  7x3  8
? 
5(
1)4  7(
1)3  8

3 

3  
4; yes
6. y  3x 
 4 
 1
? 3(0) 
 4 
 1
3
3  3; no

5.

Chapter 3

4
4

2

Case 2
3x 
 4  x
3x 
 4  x
2x  4
x2

Case 2
x 
 12  0.005
x 
 12  0.005
x  12.005

Exercises

13. y  x3 
 4x2  2
14. y  x 
 2  7
? (1)3 
 4(1)2  2
? 3 
 2  7
0
8
0  
1; no
8  8; no
15. y 
 
x
 11  1


1 ? 
 
(
2)


11  1

1 
2; yes
16. y  
0.2x2  9x 
 7
? 
0.2(10)2  9(10) 
 7
63 
63  63; no

y

O

Case 2
x  6
x6
x

(
6)2 
 6
? 



6


9  
5; yes
18. y  2x3 
 7
? 203 
 7
0 
0  
7; yes

72

19. y  
x  2
y  
x  2
? 
0
? 
1
0
2
4
2
0  2; yes
4  3; no
y  
x  2
y  
x  2
? 
1
? 

1
1
2
0
2
1  3; yes
0  i  2; no
y  
x  2
? 
1

1 
2

1  3; yes
(0, 0) (1, 1) and (1, 
1); if these points are in the
shaded region and the other points are not, then
the graph is correct.

y

20.

O

22.

x

y

O

32.

x

12
8
4

O
24.

4

8

12

O

x

y

y

25.

O
O
26.

x

x
y

O

27.

y

x

x

O
28.

x

y

29.

y

O

O

x

x

no solution
38. Case 1
2x  9 
 2x  0

(2x  9) 
 2x  0

2x 
 9 
 2x  0

4x  9
9
x  
4
all real numbers

x

73

x

x

33. Case 1
x  4 5

(x  4) 5

x 
 4 5

x 9
x  
9
{xx  
9 or x 1}
34. Case 1
3x  12  42

(3x  12)  42

3x 
 12  42

3x  54
x  
18
{xx  
18 or x  10}
35. Case 1
7 
 2x 
 8  3

(7 
 2x) 
 8  3

7  2x 
 8  3
2x  18
x9
{x
2  x  9}
36. Case 1
5 
 x  x

(5 
 x)  x

5  x  x

5  0; true
{xx 2.5}
37. Case 1
5x 
 8  0

(5x 
 8)  0

5x  8  0

5x  
8
8
x 5

y

23.

y

y

O

O

31.

O

y

21.

y

30.

Case 2
x  4
x4
x

5
5
1

Case 2
3x  12  42
3x  12  42
3x  30
x  10

Case 2
7 
 2x 
 8  3
7 
 2x 
 8  3

2x  4
x 
2

Case 2
5 
 x  x
5
xx

2x  
5
x  2.5
Case 2
5x 
 8  0
5x 
 8  0
5x  8
8
x 5

Case 2
2x  9 
 2x  0
2x  9 
 2x  0
9  0; true

Chapter 3

39. Case 1
2

3x  5  
8
2


3(
(x  5))  
8
2

3(
x 
 5)
2
10
x  
3
3
2
x
3

40.

41.

42.

43.

45b. The shaded region shows all points ( x, y) where
x represents the number of cookies sold and y
represents the possible profit made for a given
week.
46. The graph of g(x) is the graph of f(x) reflected over
the x-axis and expanded vertically by a factor of 2.

Case 2
2

3x  5  
8

 
8
 
8

2


3(x  5)  
8
2

10


3x 
 3  
8
2

14


3x  
3

1

34

1

→

47. y  
a4

 
3
x7
x  
17
{x
17  x  7}
x 
 37.5  1.2
Case 1
Case 2
x 
 37.5  1.2
x 
 37.5  1.2

(x 
 37.5)  1.2
x 
 37.5  1.2

x  37.5  1.2
x  38.7

x  
36.3
x  36.3
36.3  x  38.7
Case 1
Case 2
3x 
 7  x 
 1
3x 
 7  x 
 1
3(
(x 
 7))  x 
 1
3(x 
 7)  x 
 1
3(
x  7)  x 
 1
3x 
 21  x 
 1

3x  21  x 
 1
2x  20

4x  
22
x  10
x 5.5
{x5.5  x  10}
30 units2
The triangular region has vertices A(0, 10),
B(3, 4), and C(8, 14). The slope of side AB is 
2.
The slope of side AC is 0.5, therefore AB is
perpendicular to AC. The length of side AB is
3
5
. The length of side AC is 4
5
 the area of
the triangle is 0.5(3
5
)(4
5
) or 30.
0.10(90)  0.15(75)  0.20(76)
 0.40(80)  0.15(x)  80
0.15x  12.55
2
x  833

b  
a4
1

(
b)  
a4

x-axis

1


b  
a4
b

y-axis

b
yx

(a) 
a

y  
x

(
a) 

a 
y-axis

1
8 
3
4 
5

48.

49.

3

4


5

4

3  
1 
5
28 
4
8


7

0

8

4



3
(8)
4
3
(
4)
4

6

3


21

4

0

50.
x

2

1
0
1
2

3
8

3
(
7)
4
3
(0)
4

f(x)
11
8
5
8
11

44.

f (x)

O
51.
50
45
40
States 35
with 30
Teen 25
Courts 20
15
10
5
0

[
1, 8] scl:1 by [
1, 8] scl:1
44a. b  0
44b. none
44c. b  0 or b 4
44d. b  4
44e. 0  b  4
45a. P (x )

300
200
100

Chapter 3

0

1980 1990 2000 2010
Year

52. [f 
 g](4)  f(g(4))
 f(0.5(4) 
 1)
 f(1)
 5(1)  9
 14
[g 
 f ](4)  g(f(4))
 g(5(4)  9)
 g(29)
 0.5(29) 
 1
 13.5

400

O

no

1



(
a)4
1

a4 yes
1



(b)4
1



no
(b)4
1



 (
b)4
1



no
(b)4

900 1000 1100 1200 1300 1400 1500 x

74

x

4. y  x  1
x-axis

53. Student A  15
1

Student B  3(15)  15 or 20
Let x  number of years past.
20 
 x  2(15 
 x)
20 
 x  30 
 2x
x  10

→

y-axis
yx
y  
x

Page 151

y  x  1
→
f(
x)  
 x  1
f(
x)  x  1
y-axis
5a. translated down 2 units
5b. reflected over the x-axis, translated right 3 units
1
5c. compressed vertically by a factor of 4, translated
up 1 unit
6a. expanded vertically by a factor of 3
6b. expanded horizontally by a factor of 2 and
translated down 1 unit
6c. translated left 1 unit and up 4 units

Mid-Chapter Quiz

1. x2  y2 
 9  0
x-axis

→

a2  b2 
 9  0
a2  (
b)2 
 9  0
a2  b2 
 9  0 yes
y-axis
(
a)2  b2 
 9  0
a2  b2 
 9  0 yes
yx
(b)2  (a)2 
 9  0
a2  b2 
 9  0 yes
y  
x
(
b)2  (
a)2 
 9  0
a2  b2 
 9  0 yes
x2  y2 
 9  0 → f(x)  


x2 
9
2
f(
x)  

(
x)

 9 
f(x)  
 (


x2 
9)
2
f(
x)  


x 
9

f(x)   


x2 
9
yes
x-axis, y-axis, y  x, y  
x, origin
2. 5x2  6x 
 9  y →
5a2  6a 
 9  b
x-axis
5a2  6a 
 9  (
b)
5a2  6a 
 9  
b no
y-axis
5(
a)2  6(
a) 
 9  b
5a2 
 6a 
 9  b no
yx
5(b)2  6(b) 
 9  (a)
5b2  6b 
 9  a no
y  
x
5(
b)2  6(
b) 
 9  (
a)
5b2 
 6b 
 9  
a no
5x2  6x 
 9  y
→
f(x)  5x2  6x 
 9
f(
x)  5(
x)2  6(
x) 
 9
 5x2 
 6x 
 9

f(x)  
(5x2  6x 
 9)

f(x)  
5x2 
 6x  9 no
none of these
7

7.

x

9. Case 1
2x 
 7  15

(2x 
 7)  15

2x  7  15

2x  8
x 
4

4  x  11
10. x 
 64  3
Case 1
x 
 64  3

(x 
 64)  3

x  64  3

x  
61
x 61
61  x  67

7


a
(
b)
7

a  
b no
7

(
a)  b

y-axis

y

O

a  b

x-axis

8.

y

O

7

→

3. x  y

b  a  1
(
b)  a  1

b  a  1 no
b  (
a)  1
b  a  1 yes
(a)  (b)  1
a  b  1 no
(
a)  (
b)  1

a  b  1 no
f(x)  x  1

f(x)  
(x  1)

f(x)  
x 
 1 no

x

Case 2
2x 
 7  15
2x 
 7  15
2x  22
x  11

Case 2
x 
 64  3
x 
 64  3
x  67

7


a  b no
7

yx

(b)  (a)

3-4

7

a  b yes
y  
x
7

→

x  y
f(
x) 
f(
x) 

7

(
x)
7

x

7

(
b)  
(
a)
7
a  b yes
7
f(x)  x
7

f(x)  
 x
7

f(x)  
x yes

Inverse Functions and Relations

Pages 155–156

Check for Understanding

1. Sample answer: First, let y  f(x). Then
interchange x and y. Finally, solve the resulting
equation for y.
2. n is odd
3. Sample answer: f(x)  x2
4. Sample answer: If you draw a horizontal line
through the graph of the function and it intersects
the graph more than once, then the inverse is not
a function.

 

y  x, y  
x, origin

75

Chapter 3

5. She is wrong. The inverse is f 
1(x)  (x 
 3)2 
 2,
which is a function.
6.
f
1(x)
f(x)  x  1
x
f
1(x)
x
f(x)
3

2

2
3
2

1

1
2
1
0
0
1
2
1
1
2
3
2
2
3

f (x )

9.

f(x)  
3x  2
y  
3x  2
x  
3y  2
x 
 2  
3y
1

2

1

2

y  
3x  3
f
1(x)  
3x  3; f
1(x) is a function.
10.

1

f(x)  x3
1

y  x3
1

x  y3

f (x )

1

y3  x
y

f 1(x )

O

x

f
1(x) 

3

1

x

1

3

x

or

1


1
3

x ; f (x)

is a function.

f(x)  (x  2)2  6
y  (x  2)2  6
x  (y  2)2  6
x 
 6  (y  2)2

x


6y2
y  
2 
x

6


1
f (x)  
2  
x;

 6 f
1(x) is not a function.
12. Reflect the graph of y  x2 over the line y  x.
Then, translate the new graph 1 unit to the left
and up 3 units.
11.

7.

f(x)  x3  1
x
f(x)

2

7

1
0
0
1
1
2
2
9

x

7
0
1
2
9

f
1(x)
f
1(x)

2

1
0
1
2

y
f (x )

f (x )
f 1(x )

8.

x

O

x

O

13.
f(x)  
(x 
 3)2  1
x
f(x)
1

3
2
0
3
1
4
0
5

3

x

3
0
1
0

3

f
1(x)
f
1(x)
1
2
3
4
5

1

f(x)  2x 
 5
1

y  2x 
 5
1

x  2y 
 5
1

x  5  2y
y  2x  10
f
1(x)  2x  10
[f 
 f
1](x)  f(2x  10)
1
 2(2x  10) 
 5
x

f (x )

[f
1 
 f ](x)  f
12x 
 5
1

f 1(x )

 22x 
 5  10
x
Since [f 
 f
1](x)  [f
1 
 f ](x)  x, f and f
1 are
inverse functions.
14a. B(r)  1000(1  r)3
B  1000(1  r)3
1

O

x
f (x )

B

1000
B
3

1000

 (1  r)3
1r
3

r  
1 

Chapter 3

76


B


10

14b. r  
1 

18.

3


B


10
3

 
1 


1100


10

Pages 156–158
15.

or 0.0323; 3.23%

Exercises

f(x)  x  2
x
f(x)

2
4

1
3
0
2
1
3
2
4

f (x )

x
4
3
2
3
4

f (x )
10

10 O

O

19.

x

f(x)  2x
x
f(x)

2
4

1
2
0
0
1
2
2
4

x
4
2
0
2
4

f
1(x)
f
1(x)

2

1
0
1
2

f 1(x )

O

f 1(x )

f (x ) f 1(x )

x

f
1(x)
x
f
1(x)

10

2

3

1

2
0

1
1
6
2

f(x)  x3 
 2
x
f(x)

2

10

1

3
0

2
1

1
2
6

f (x )

20.

f (x )
f 1(x )

O

f(x)

2

1
0
1
2

f
1(x)
f
1(x)

2  x  
1

1  x  0
0x1
1x2
2x3

O f (x )
17.

x

10

f(x)  [x]
x

2  x  
1

1  x  0
0x1
1x2
2x3

x

2

1
0
1
2

f (x )
f (x )

f (x )


10

f 1(x )

f (x )

16.

f
1(x)
f
1(x)

2

1
0
1
2

f
1(x)
x
f
1(x)

42

2

11

1

10
0

9
1
22
2

f(x)  x5 
 10
x
f(x)

2

42

1

11
0

10
1

9
2
22

x

f(x)  3
x
f(x)

2
3

1
3
0
3
1
3
2
3

f (x )

x
3
3
3
3
3

f
1(x)
f
1(x)

2

1
0
1
2

f (x )

x
O

x
f 1(x )

77

Chapter 3

21.

f(x)  x2  2x  4
x
f(x)

3
7

2
4

1
3
0
4
1
7

x
7
4
3
4
7

24.

f
1(x)
f
1(x)

3

2

1
0
1

8

f (x )


8


4 O

f 1(x )
8x

4


4

f (x )
f (x )


8

22.

4

f(x)  
(x  2)2 
 5
x
f(x)

4

9

3

6

2

5

1

6
0

9

x

9

6

5

6

9

f
1(x)
f
1(x)

4

3

2

1
0


4 O

f 1(x )

O

25.

x

f(x)  2x  7
y  2x  7
x  2y  7
x 
 7  2y
x
7

y  2

x

x
7

f
1(x)  2; f
1(x) is a function.
26.

f (x )

f(x)  (x  1)2 
 4
x
f(x)

4
5

2

3

1

4
0

3
2
5

f (x )

4

f 1(x )


4

f (x )

23.

f
1(x)  
x

4
x
f
1(x)
8
2
5
1
4
0

f(x)  x2  4
x
f(x)

2
8

1
5
0
4
1
5
2
8

f (x )
4

f(x)  x2  4
y  x2  4
x  y2  4
x 
 4  y2
y  
x
;

 4 f
1(x)  
x


4

x
5

3

4

3
5

f
1(x)
f
1(x)

4

2

1
0
2

27.

f(x)  
x 
 2
y  
x 
 2
x  
y 
 2
y  
x 
 2
f
1(x)  
x 
 2; f
1(x) is a function.
1

f(x)  x
1

y  x
1

x  y
1

y  x
1

f
1(x)  x; f 
1(x) is a function.
28.

1

f(x)  
x2
1

f (x )

y  
x2

f 1(x )

1

x  
y2
1

y2  
x

O

1

y   
x
1

f
1(x)   
x; f 
1(x) is not a function.
29.

Chapter 3

78

f(x)  (x 
 3)2  7
y  (x 
 3)2  7
x  (y 
 3)2  7
x 
 7  (y 
 3)2

x


7y
3
y  3 
x

7

f
1(x)  3 
x
;

 7 f 
1(x) is not a function.

30.

31.

f(x)  x2 
 4x  3
y  x2 
 4x  3
x  y2 
 4y  3
x  1  y2 
 4y  4
x  1  (y 
 2)2

x

1y
2
y  2  
x
1
f
1(x)  2  
x;
 1 f 
1(x) is not a function.

35. Reflect the part of the graph of x2 that lies in the
first quadrant about y  x. Then, translate 5 units
to the left.

f (x )

x

O

1


f(x)  
x2
1


y
x2
1


x
y2

36. Reflect the graph of x2 about y  x. Then,
translate 2 units to the right and up 1 unit.
f (x )

1

y  2  x
1

y  x 
 2
1

f
1(x)  x 
 2; f 
1(x) is not a function.
32.

1


f(x)  
(x 
 1)2

y
x
( y 
 1)2 
y
1
y

x
O

1

(x 
 1)2
1

( y 
 1)2
1

x
1
 x
1
1 

x

37. Reflect the graph of x3 about y  x to obtain the
3
3
graph of 
x. Reflect the graph of 
x about the
x-axis. Then, translate 3 units to the left and
down 2 units.
f (x )

1
 ; f 
1(x) is not a function.

x
2

f(x)  

(x 
 2)3
2

y  

(x 
 2)3
2

x  

( y 
 2)3
2
( y 
 2)3  
x
2
y 
 2  
 3 x
2
y  2 
 3 x
2
f
1(x)  2 
 3 x; f 
1(x) is not a function.
3

g(x)  
x2  2x
3

y
x2  2x
3

x
y2  2y
3
y2  2y  x
3
y2  2y  1  x  1
3
(y  1)2  x  1
3
y  1   x  1
3
y  
1  x  1
3
g
1(x)  
1  x  1

f
1(x)  1 

33.

34.

O

x

38. Reflect the graph of x5 about y  x. Then,
translate 4 units to the right. Finally, stretch the
translated graph vertically by a factor of 2.
f (x )

O

79

x

Chapter 3

39.

2

1

2

1


42a. v  
2gh
v2  2gh

2

1

h

f(x)  
3x  6
y  
3x  6
x  
3y  6
1

h

2

x 
 6  
3y
3

1

3

1

h

y  
2x  4
[f 
 f
1](x)  f 
2x  4
1

 
3
2x  4  6
2

3

1

1

1

1

 x 
 6  6
x

x
0x1
1x2
2x3
3x4
4x5

[f
1 
 f ](x)  f
1
3x  6
2

1

 
2
3x  6  4
3

2

1

1

1

1

 x 
 4  4
x
Since [f 
 f
1](x)  [f
1 
 f ](x)  x, f and f
1 are
inverse functions.
40.
f(x)  (x 
 3)3  4
y  (x 
 3)3  4
x  (y 
 3)3  4
x 
 4  (y 
 3)3
3


x
4y
3
3
y  3  

x
4
3
f
1(x)  3  

x
4
3
[f 
 f
1](x)  f(3  
)
x
4
3
 [(3  
)
x 
 4 
 3]3  4
x
44
x
[f
1 
 f ](x)  f
1[(x 
 3)3  4]
3
 3  

[(x 
 
3)3  
4] 
 4
3x
3
x
Since [f 
 f
1](x)  [f
1 
 f ](x)  x, f and f
1 are
inverse functions.
41a.
d(x)  x 
 4
d
1(x)
x
d(x)
x
d
1(x)
6
2
2
6
5
1
1
5
4
0
0
4
3
1
1
3
2
2
2
2

C (x )
$0.80
$0.70
$0.60
$0.50
$0.40
$0.30
$0.20
$0.10
1 2 3 4 5 6 7 8x

44b. positive real numbers; positive multiples of 10
44c.
C
1(x)
0x1
1x2
2x3
3x4
4x5

x
$0.10
$0.20
$0.30
$0.40
$0.50

8
7
6
5
4
3
2
1

C 1(x )

O

x
20¢ 40¢ 60¢ 80¢

44d. positive multiples of 10; positive real numbers
44e. C
1(x) gives the possible lengths of phone calls
that cost x.
45. It must be translated up 6 units and 5 units to the
5
left; y  (x 
 6)2 
 5; y  6  
x.
46a.

1

KE  2mv2
2KE  mv2
2KE

m

 v2

v

46b. v  
v

2KE

m
2(15)

1

v  5.477; 5.5 m/sec
2KE

m

46c. There are always two velocities.
47a. Yes; if the encoded message is not unique, it may
not be decoded properly.
47b. The inverse of the encoding function must be a
function so that the encoded message may be
decoded.
47c.
C(x)  
2  
x
3
y  
2  
x
3
x  
2  
y
3
x  2  
y
3
(x  2)2  y  3
y  (x  2)2 
 3

1
C (x)  (x  2)2 
 3

x

41b. No; the graph of d(x) fails the horizontal line test.
41c. d
1(x) gives the numbers that are 4 units from x
on the number line. There are always two such
numbers, so d
1 associates two values with each
x-value. Hence, d
1(x) is not a function.

Chapter 3

C(x)
$0.10
$0.20
$0.30
$0.40
$0.50

O

d 1(x)

O

(75)2

h  64
h  87.89
Yes. The pump can propel
water to a height of about
88 ft.

v2

2g
v2

2(32)
v2

64

43a. Sample answer: y  
x
43b. The graph of the function must be symmetric
about the line y  x.
43c. Yes, because the line y  x is the axis of
symmetry and the reflection line.
44a.

f
1(x)  
2x  4
3

v2

42b. h  64

80

47d. C
1(x)  (x  2)2 
 3
C
1(1)  (1  2)2 
 3 or 6, F
C
1(2.899)  (2.899  2)2 
 3 or 21, U
C
1(2.123)  (2.123  2)2 
 3 or 14, N
C
1(0.449)  (0.449  2)2 
 3 or 3, C
C
1(2.796)  (2.796  2)2 
 3 or 20, T
C
1(1.464)  (1.464  2)2 
 3 or 9, I
C
1(2.243)  (2.243  2)2 
 3 or 15, O
C
1(2.123)  (2.123  2)2 
 3 or 14, N
C
1(2.690)  (2.690  2)2 
 3 or 19, S
C
1(0)  (0  2)2 
 3 or 1, A
C
1(2.583)  (2.583  2)2 
 3 or 18, R
C
1(0.828)  (0.828  2)2 
 3 or 5, E
C
1(1)  (1  2)2 
 3 or 6, F
C
1(2.899)  (2.899  2)2 
 3 or 21, U
C
1(2.123)  (2.123  2)2 
 3 or 14, N
FUNCTIONS ARE FUN
48. Case 1
Case 2
2x  4  6
2x  4  6

(2x  4)  6
2x  4  6

2x 
 4  6
2x  2

2x  10
x1
x  
5
{x
5  x  1}
49. both
50a. a  0, b  0, 4a  b  32, a  6b  54
50b.

53.

O
54.

Page 165

(8, 0)

1 
2

1
4

2.

a

4x  2y  10
xy6

4 2
x  10
1 1
y
6
1
1 
2   1 
2
2 
1

1
4
4
1
x
1 
2
 2
y

1
4
x

1

y
7

4 2
1 1

52.

9 
3


6
6

10
6



1
(
3)
2
1
(6)
2

1
(9)
2
1
(
6)
2
9

2


3

y 
 7  
1(x 
 0)

Check for Understanding

an

n

x→

p(x) →

positive
positive
positive
positive

even
even
odd
odd














an

n

x→

p(x) →

negative
negative
negative
negative

even
even
odd
odd
















3. Infinite discontinuity; f(x) →  as x → 
,
f(x) → 
 as x → .
4. f(x)  x2 is decreasing for x  0 and increasing for
x 0, g(x)  
x2 is increasing for x  0 and
decreasing for x  0. Reflecting a graph switches
the monotonicity. In other words, if f(x) is
increasing, the reflection will be decreasing and
vice versa.
5. No; y is undefined when x  
3.
6. No; f(x) approaches 6 as x approaches 
2 from the
left but f(x) approaches 
6 as x approaches 
2
from the right.

(
1, 7)
1

2

y 
 y1  m(x 
 x1)

1. Sample answer: The function approaches 1 as x
approaches 2 from the left, but the function
approaches 
4 as x approaches 2 from the right.
This means the function fails the second condition
in the continuity test.

a
0

1

2

neither

Continuity and End Behavior

3-5

4a  b 
 32

1
4 2
1 1

x

1
4; 4 4 
1;
2
7


5
0

5
 5 or 
1

y  
x  7
56. b  c  180
If 
PQ
 is perpendicular to Q
R
, then m∠PQR  90.
Since the angles of a triangle total 180,
a  d  90  180.
a  d  90
a  b  c  d  180  90 or 270
The correct choice is C.

(6, 8)

G(a, b)  a  b
G(0, 0)  0  0 or 0
G(0, 9)  0  9 or 9
G(6, 8)  6  8 or 14
G(8, 0)  8  0 or 8
14 gallons
51. 4x  2y  10
→
y6
x

1

4

55. m

b a  6b 
 54
(0, 9)

O (0, 0) b 
 0

y

3


2
3

81

Chapter 3

7. an: positive, n: odd
y →  as x → , y → 
 as x → 
.
8. an: negative, n: even
y → 
 as x → , y → 
 as x → 
.
9.

24.
x

10,000

1000

100

10
0
10
100
1000
10,000

f (x )

x
O

decreasing for x  
3; increasing for x
y
10.

1

y  x2


3

y
1 10
8
1 10
6
1 10
4
0.01
undefined
0.01
1 10
4
1 10
6
1 10
8

y → 0 as x → , y → 0 as x → 
.
25.

1

f(x)  
x3  2

O

x

decreasing for x  
1 and x

1  x  1
11a. t  4 11b. when t  4

Pages 166–168

x

10,000

1000

100

10
0
10
100
1000
10,000

1; increasing for
11c. 10 amps

Exercises

2
2.000000001
2.000001
2.001
undefined
1.999
1.999999
1.999999999
2

f(x) → 2 as x → , f(x) → 2 as x → 
.

12. Yes; the function is defined when x  1; the
function approaches 
3 as x approaches 1 from
both sides; and y  
3 when x  1.
13. No; the function is undefined when x  2.
14. Yes; the function is defined when x  
3; the
function approaches 0 as x approaches 
3 from
both sides; and f(
3)  0.
15. Yes; the function is defined when x  3; the
function approaches 1 (in fact is equal to 1) as x
approaches 3 from both sides; and y  1 when
x  3.
16. No; f(x) approaches 
7 as x approaches 
4 from
the left, but f(x) approaches 6 as x approaches 
4
from the right.
17. Yes; the function is defined when x  1; f(x)
approaches 3 as x approaches 1 from both sides;
and f(1)  3.
18. jump discontinuity
19. Sample answer: x  0; g(x) is undefined when
x  0.
20. an: positive, n: odd
y →  as x → , y → 
 as x → 
.
21. an: negative, n: even
y → 
 as x → , y → 
 as x → 
.
22. an: positive, n: even
y →  as x → , y →  as x → 
.
23. an: positive, n: even
y →  as x → , y →  as x → 
.

Chapter 3

f(x)

26.

[
6, 6] scl:1 by [
30, 30] scl:5
increasing for x  
3 and x 1; decreasing for

3  x  1
27.

[
7.6, 7.6] scl:1 by [
5, 5] scl:1
decreasing for all x

82

28.

33b. Since f is odd, its graph must be symmetric with
respect to the origin. Therefore, f is increasing
for 
2  x  0 and decreasing for x  
2. f must
have a jump discontinuity when x  
3 and
f(x) →  as x → 
.

f (x)

[
7.6, 7.6] scl:1 by [
8, 2] scl:1
decreasing for x  
1 and x 
1

O

29.

x

34a. polynomial
34b.
[
25, 25] scl:5 by [
25, 25] scl:5
increasing for x  
1 and x 5; decreasing for

1  x  2 and 2  x  5
30.

[
5, 80] scl:10 by [
500, 12000] scl:1000
0.5  t  39.5
34c. 0  t  0.5 and t 39.5
35a. 1954-1956, 1960-1961, 1962-1963, 1966-1968,
1973-1974, 1975-1976, 1977-1978, 1989-1991,
1995-1997
35b. 1956-1960, 1961-1962, 1963-1966, 1968-1973,
1974-1975, 1976-1977, 1978-1989, 1991-1995,
1997-2004
36a.

[
7.6, 7.6] scl:1 by [
1, 9] scl:1
decreasing for x  
2 and 0  x  2; increasing
for 
2  x  0 and x 2
31.

[
7.6, 7.6] scl:1 by [
1, 9] scl:1
3
3
decreasing for x  
2 and 0  x  2; increasing
3
3
for 
2  x  0 and x 2
32. As the denominator, r, gets larger, the value of
U(r) gets smaller. U(r) approaches 0.
33a. Since f is even, its graph must be symmetric
with respect to the y-axis. Therefore, f is
decreasing for 
2  x  0 and increasing for
x  
2. f must have a jump discontinuity when
x  
3 and f(x) → 
 as x → 
.

36b.
36c.

f (x)

O

36d.
36e.
37a.

x

83

[
1, 8] scl:1 by [
10, 1] scl:1
x4
Answers will vary.
The slope is positive. In an interval where a
function is increasing, for any two points on the
graph, the x- and y-coordinates of one point will
be greater than that of the other point, ensuring
that the slope of the line through the two points
will be positive.
See graph in 36a. x 4
The slope is negative; see students’ work.
The function has to be monotonic. If the function
were increasing on one interval and decreasing
on another interval, the function could not pass
the horizontal line test.

Chapter 3

37b. The inverse must be monotonic. If the inverse
were increasing on one interval and decreasing
on another interval, the inverse would fail the
horizontal line test. That would mean the
function fails the vertical line test, which is
impossible.

3-5B Gap Discontinuities
Page 170
1. {all real numbers xx

3}

38a.
40
35
Percent
30
with
25
Similar
20
Computer
15
Usage
10
5
0

0 2 4 6 8 10 12 14 16

[
10, 10] scl:1 by [
6, 50] scl:10
2. {all real numbers x
2  x  4}

38b. 0  x  1, 1  x  2, 2  x  4, 4  x  6,
6  x  8, x 8
39. For the function to be continuous at 2, bx  a and
x2  a must approach the same value as x
approaches 2 from the left and right, respectively.
Plugging in x  2 to find that common value gives
2b  a  4  a. Solving for b gives b  2. For the
function to be continuous at 
2, 

b

  and
x
bx  a must approach the same value as x
approaches 
2 from the left and right,
respectively. Plugging in x  
2 gives 

b

2
 
2b  a. We already know b  2, so the
equation becomes 0  
4  a. Hence, a  4.
40.
f(x)  (x  5)2
y  (x  5)2
x  (y  5)2

x
y5
y  
5  
x
f 
1(x)  
5  
x
41. The graph of g(x) is the graph of f(x) translated
left 2 units and down 4 units.
42. f (x, y)  x  2y
f(0, 0)  0  2(0) or 0
f(4, 0)  4  2(0) or 4
f(3, 5)  3  2(5) or 13
f(0, 5)  0  2(5) or 10
13, 0

4  5(2) 
 8(
4) or 42
43. 5
8
2
44a. c  47.5h  35

[
9.4, 9.4] scl:1 by [
6.2, 6.2] scl:1
3. {all real numbers xx  
3 or x  1}

[
18.8, 18.8] scl:1 by [
12.4, 12.4] scl:1
4. {all real numbers xx  
3 or x 
2}

44b. c  47.5h  35
1
c  47.5 24  35

[
4.7, 4.7] scl:1 by [
25, 25] scl:10
5. {all real numbers xx  
1 or x 1}

 

c  $141.875
f(x)  2x2 
 2x  8
f(
2)  2(
2)2 
 2(
2)  8
 8  4  8 or 20
46. The volume of the cube is x3.
The volume of the other box is
x(x 
 1)(x  1)  x(x2 
 1) or x3 
 x.
The difference between the volumes of the two
boxes is x3 
 (x3 
 x) or x.
The correct choice is A.
45.

Chapter 3

[
4.7, 4.7] scl:1 by [
3.1, 3.1] scl:1

84

6. {all real numbers xx  
6 or x


2}

12. Yes; sample justification: if f(x) is a polynomial
function, then the graph of y 

f(x)

(x 
 [[x]]  0.25)

is like the graph of f(x), but with an infinite
number of “interval bites” removed.
13. Yes; sample justification: the equation y 
x2(x  
2)  (2x 
 4)(x  4)

((x  
2) or (x  4))

is a possible equation for

the function described.
14a.
[
9.4, 9.4] scl:1 by [
6.2, 6.2] scl:1
7. {all real numbers xx  3 or x  4}

[
15, 15] scl:2 by [
10, 20] scl:2
14b.
[
9.4, 9.4] scl:1 by [
3, 9.4] scl:1
8. {all real numbers xx  
2 or 
1  x  1
or x  2}

[
9.1, 9.1] scl:1 by [
6, 6] scl:1

[
4.7, 4.7] scl:1 by [
2, 8] scl:1
9. {all real numbers xx  1 or 2  x  3 or x  4}

[
3, 6.4] scl:1 by [
2, 8] scl:1
10. {all real numbers xx  1 or x

3-6

Critical Points and Extrema

Page 176

Check for Understanding

1. Check values of the function at x-values very close
to the critical point. Be sure to check values on
both sides. If the function values change from
increasing to decreasing, the critical point is a
maximum. If the function values change from
decreasing to increasing, the critical point is a
minimum. If the function values continue to
increase or to decrease, the critical point is a point
of inflection.
2. rel. min.;
f(0.99)  
3.9997
f(1)  
4
f(1.01)  
3.9997
By testing points on either side of the critical
point, it is evident that the graph of the function
is decreasing as x approaches 1 from the left and
increasing as x moves away from 1 to the right.
Therefore, on the interval 0.99  x  1.01, (1, 4) is
a relative minimum.

2}

[
9.4, 9.4] scl:1 by [
6.2, 6.2] scl:1
11. Sample answer: y 
x2

((x  2) or ((x  5) and (x  7)) or x  8))

85

Chapter 3

12c. $58.80 per acre
12d. Rain or other bad weather could delay harvest
and/or destroy part of the crop.

3. Sample answer:

y
(4, 6)

Pages 177–179

(3, 1)

O

Exercises

13. abs. max.: (
4, 1)
14. abs. max.: (
1, 3); rel. min.: (0.5; 0.5);
rel. max.: (1.5, 2)
15. rel. max.: (
2, 7): abs. min.: (3, 
3)
16. rel. max.: (
6, 4), rel. min.: (
2, 
3)
17. abs. min.: (3, 
8); rel. max.: (5, 
2);
rel. min.: (8, 
5)
18. no extrema
19. abs. max.: (1.5, 
1.75)

x

(0, 4)

4. rel. min.: (
3, 
2); rel. max.: (1, 6)
5. rel. min.: (
1, 
3); rel. max.: (3, 3)
6. rel. max.: (0, 0); rel. min.: (2, 
16)

[
5, 5] scl:1 by [
8, 2] scl:1
20. rel. max.: (
1.53, 1.13);
rel. min.: (1.53, 
13.13)

[
4, 6] scl:1 by [
20, 20] scl:5
7. rel. min.: (
2.25, 
10.54)

[
6, 4] scl:1 by [
14, 6] scl:2
8. f(
1.1)  0.907
f(
1)  1
f(
0.9)  0.913
max.
9. f(
2.6)  
12.24
f(
2.5)  
12.25
f(
2.4)  
12.24
min.
10. f(
0.1)  
0.00199
f(0)  0
f(0.1)  0.00199
pt. of inflection
11. f(
0.1)  
0.97
f(0)  
1
f(0.1)  
0.97
min.
12a. P(x)  (120  10x)(0.48 
 0.03x)

[
5, 5] scl:1 by [
16, 4] scl:2
21. rel. max.: (
0.59, 0.07), rel. min.: (0.47, 
3.51)

[
5, 5] scl:1 by [
5, 5] scl:1
22. abs. min.: (
1.41, 
6), (1.41, 
6);
rel. max.: (0, 
2)

P (x )
80
70
60
Profit 50
(dollars) 40
30
20
10
0

x
0 2 4 6 8 10 12 14 16 18
Time (weeks)

[
5, 5] scl:1 by [
8, 2] scl:1

12b. 2 weeks
Chapter 3

86

23. rel. max.: (
1, 1); rel. min.: (0.25, 
3.25)

35b.

[
1, 12] scl:1 by [
200, 500] scl:100
2.37 cm by 2.37 cm
35c. See students’ work.
36a. P  sd 
 25d
 s(
200s  15,000) 
 25(
200s  15,000)
 
200s2  20,000s 
 375,000

[
5, 5] scl:1 by [
5, 5] scl:1
24. no extrema

[
5, 5] scl:1 by [
5, 5] scl:1
25. abs. min.: (
3.18, 
15.47); rel. min. (0.34, 
0.80);
rel. max.: (
0.91, 3.04)

[0, 100] scl:10 by [0, 130,000] scl:10,000
abs. max.: (50, 125,000)
$50
36b. Sample answer: The company’s competition
might offer a similar product at a lower cost.
37a. AM 2  MB2  AB2
AM 2  x2  22
AM  

x2  4
f(x)  5000(

x2  4)  3500(10 
 x)

[
5, 5] scl:1 by [
16, 5] scl:2
26. f(
0.1)  
0.001
f(0)  0
f(0.1)  0.001
pt of inflection
27. f(3.9)  5.99
f(4)  6
f(4.1)  5.99
max.
28. f(
2.6)  
19.48
f(
2.5)  
19.5
f(
2.4)  
19.48
min.
29. f(
0.1)  6.98
f(0)  7
f(0.1)  6.98
max.
30. f(1.9)  
3.96
f(2)  
4.82
f(2.1)  
3.96
min.
31. f(2.9)  
0.001
f(3)  0
f(3.1)  0.001
pt. of inflection
32. f(
2.1)  4.32
f(
2)  4.53
f(
1.9)  4.32
max.
33. f(0.57)  2.86

37b.

[
10, 20] scl:2 by [0, 60,000] scl:10,000
abs. min.: (1.96, 42,141.4)
1.96 km from point B
n
38. equations of the form y  xn or y  
x, where n is
odd
39.

[
3, 7] scl:1 by [
50, 10] scl:10
The particle is at rest when t  0.14 and when
t  3.52. Its positions at these times are
s(0.14)  
8.79 and s(3.52)  
47.51.

f3  2.85
f(0.77)  2.86
min.
34. The point of inflection is now at x  
6 and there
is now a minimum at x  
3.
35a. V(x)  2x(12.5 
 2x)(17 
 2x)
2

87

Chapter 3

47. Let x  number of 1-point free throws.
Let y  number of 2-point field goals.
Let z  number of 3-point field goals.
1x  2y  3z  32
x  y  z  17
y  0.50(18)
1x  2y  3z  32
→ 1x  2y  3z  32

1(x  y  z  17)

x 
 y 
 z  
17
y  2z  15
y  0.50(18)
y  2z  15
y9
9  2z  15
z3
x  y  z  17
x  9  3  17
x5
5 free throws, 9 2-point field
goals, 3 3-point field goals

40. If a cubicle has one critical point, then it must be
a point of inflection. If it were a relative maximum
or minimum, then the end behavior for a cubic
would not be satisfied. If a cubic has three critical
points, then one must be a maximum, another a
minimum, and the third a point of inflection.
41. No; the function is undefined when x  5.
42.

y

x

O

48. y  6  4
y  
2

43. Let x  units of notebook paper.
Let y  units of newsprint.
x  y  200
y
x  10
200 (10, 190)
y  80
x  y 
 200

y

O

x

(120, 80)
y 
 80

100
(10, 80)

x 
 10

O

2

49. 2x  3y  15 → y  
3x  5

x
100

200

3

6x  4y  16 → y  2x 
 4

P(x, y)  400x  350y
P(10, 80)  400(10)  350(80) or 32,000
P(10, 190)  400(10)  350(190) or 70,500
P(120, 80)  400(120)  350(80) or 76,000
120 units of notebook and 80 units of newsprint

2


3

A (3, 4)

B (2, 4)

x

O
D (3, 1)

C (2, 1)

T


3  x  2, 
1  y  4
3  1(5) 
 2(3) or 
1; yes
45. 1
2
5
4 
2
5
7
3(4) 3(
2) or

3(5)
3(7)

3
5
2B  2

4
3
2(
3)
2(5) or

2(
4)
2(3)
12

6 
3A  2B 
15
21
12  (
6)

15  (
8)
6
4

7
27

S

46. 3A  3

Chapter 3

12
15

 
1; perpendicular

50. A relation relates members of a set called the
domain to members of a set called the range. In a
function, the relation must be such that each
member of the domain is related to one and only
one member of the range. You can use the vertical
line test to determine whether a graph is the
graph of a function.
51. The area of PTX is equal to the area of RTY.
The area of STR is 25% of the area of rectangle
PQRS. The correct choice is D.
X
Q
P

y

44.

3

2


6
21


6 10

8
6

6 10

8
6

6  10
21  6

88

Y

R

3-7

x3

6. x  2, x  
1

Graphs of Rational Functions


y
(x 
 2)(x  1)
x3


y
x2 
 x 
 2

Pages 185–186
1.

Check for Understanding

f (x )
O

x3

x3

y


x2
x
2


 x3 
 x3
x3

y

1

1
2
1
 
 

 x3
x2
x

x

no horizontal
asymptotes
1


7. f(x)  
x1 
 2

8. The parent graph is translated 4 units right. The
vertical asymptote is now at x  4. The horizontal
asymptote, y  0, is unchanged.

1a. x  2, y  
6
1

y


1b. y  
x
2 
 6

8

2. Sample graphs:
Vertical Asymptote

4

Horizontal Asymptote

y

y


8


4 O

4

4

8x


8

O
x

O

x

9. The parent graph is translated 2 units left and
down 1 unit. The vertical asymptote is now at x 

2 and the horizontal asymptote is now y  
1.

y

Slant Asymptote

y
O

x

O
x

10.
x (x  1)


3. Sample answer: f(x)  
x1

4. False; sample explanation: if that x-value also
causes the numerator to be 0, there could be a
hole instead of a vertical asymptote.
5. x  5

3x  5
20

x 
 3

3x2 

4x  
5 → 3x  5  
x
3
2
3x 
 9x
5x  5
5x 
 15
20
y  3x  5

y

11.

x


f(x)  
x
5

12.

y

x


y
x
5
y(x 
 5)  x
xy 
 5y  x
xy 
 x  5y
x(y 
 1)  5y

O

x

O

x

5y


x
y 
 1; y  1

13a. P

O

89

V
Chapter 3

13b. P  0, V  0
13c. The pressure approaches 0.

Pages 186–188

(x  1)2

19. x  1


y
x2 
 1
x2  2x  1


y
x2 
 1
x2

x2

2x

14. x  
4


f(x)  
x4

1

y(x  4)  2x
xy  4y  2x
xy 
 2x  
4y
x(y 
 2)  
4y

4y

x
y 
 2; y  2

x3


y
(x 
 2)4

y

x2


y
x6
x2

x2

y

1

y

y 
6
1
  
x2
x

x5

y
y

0

1


22. f(x)  
x
2 
 3
1

23. f(x)  
x  1
24. The parent graph is translated 3 units up. The
vertical asymptote, x  0, is unchanged. The
horizontal asymptote is now y  3.

1

x2



y  
2x2
9x
5



x2 
 x2 
 x2
1

x

x3

x4

x4
8x3
24x2
32x
16





 x
4  x4 
 x4  x4
x4
1

x
 ; y 
8
24
32
16
1 
   

 x3  x4
x
x2
1

x
1

(2x  1)(x 
 5)
x
1


2x2 
 9x 
 5
x

x2

x3

x4 
 8x3  24x2 
 32x  16


21. f(x)  
x3  1

no horizontal
asymptote
16. x 

1

20. x  2

y 
6
x2

 x2
x2

1

2,

2

 x  x2

y
;y1
1
1 
 
x2

2x


y
x4

15. x  
6

1

2x

 x2  x2
y  
1
x2


 x2
x2

Exercises

y

1


 x2

y
5 ;y0
9
2 
  
 
x2
x
17. x  
1, x  
3

O

x
2


y
x2 
 4x  3
x

x2

x

2

 x2

2
y x
4x
3

 x2  x2
x2
1

x

25. The parent graph is translated 4 units right and
expanded vertically by a factor of 2. The vertical
asymptote is now x  4. The horizontal asymptote,
y  0, is unchanged.

2

 x2

y
4
3 ;y0
1    
x
x2
18. no vertical asymptote,

8

x2


y
x2  1

4

x2

O


x2


8

y 
x2
1

 x2
x2


4

4

4

8

1

y 
1 ;y1
1  
x2

Chapter 3

y

90

8x

x
1
1

x  4

x2  3
x
3
→
x
1
x4
2
x  4x

x 
 3

x 
 4
1
yx
1
31.
x3
4
x

x2  3
x
4
→
x  3 
 x
x2
3x 
 4
3x

4
yx3
32.
x
2
2

x2  1

x3 
 2
x2  x

4
→
x
2

x2  1
3
x
x

 2x2

4

 2x2

2

2
yx
2
30.

26. The parent graph is translated 3 units left. The
translated graph is then expanded vertically by a
factor of 2 and translated 1 unit down. The
vertical asymptote is now x  
3 and the
horizontal asymptote is now y  
1.

y
4

O

4

x

4

4

27. The parent graph is expanded vertically by a
factor of 3, reflected about the x-axis, and
translated 2 units up. The vertical asymptote,
x  0, is unchanged. The horizontal asymptote is
now y  2.

y

1
x
2

33.
2x 


8

3

x2 
 4
x1
x2

4

8


4 O

4

8x

4


8

→

3

 2x
5

2x  1
5
15

2x  4
11

4

1
x
2




5

4

1




11

4

2x 
 3

5

y  2x 
 4

34. No; the degree of the numerator is 2 more than
that of the denominator.
y
y
35.
36.

28. The parent graph is translated 3 units right. The
translated graph is expanded vertically by a factor
of 10 and then translated 3 units up. The vertical
asymptote is x  3 and the horizontal asymptote
is y  3.
12

5


 4

(2, 4)

x

O

y

O

x

8
4

y

37.
4 O
4

4

8

12

38.

y

x

8

(2, 0)

O

(2,

12

29. The parent graph is translated 5 units left. The
translated graph is expanded vertically by a factor
of 22 and then translated 4 units down. The
vertical asymptote is x  
5 and the horizontal
asymptote is y  
4.

14

)

x

39.

x

O

40.

y

y

y
8

O

x

4

16 
12 
8

x

O


4

O

4

(3, 0) x


4

8

12

16

91

Chapter 3

480  3t

48. abs. max.: (2, 1)


41a. C(t)  
40  t

480  3t


C(t) 
40  t

41b.

480  3t


10  
40  t

400  10t  480  3t
7t  80
t  11.43 L
42. Sample answer: The circuit melts or one of the
components burns up.
43. To get the proper x-intercepts, x 
 2 and x  3
should be factors of the numerator. The vertical
asymptote indicates that x 
 4 should be a factor
of the denominator. To get point discontinuity
at (
5, 0), make x  5 a factor of both the
numerator and denominator with a bigger
exponent in the numerator. Thus, a sample
answer is f(x) 
44a.

(x 
 2)(x  3)(x  5)2
 .
(x 
 4)(x  5)

V  x2 h

A(x)  4x h  2x2

2
A(x)  4xx
2   2x
120

120  x2 h
120

x2

[
1, 6] scl:1 by [
5, 2] scl:1
49. x2 
 9  y
y2 
 9  x
y2  x  9
y  
x
9
50. f(x, y)  y 
 x
f(0, 0)  0 
 0 or 0
f(4, 0)  0 
 4 or 
4
f(3, 5)  5 
 3 or 2
f(0, 5)  5 
 0 or 5
5; 
4
5  
4(
6)

4(5)
51. 
4 
6
8 
4

4(8) 
4(
4)
24 
20


32
16
52. Let x  price of film and y  price of sunscreen.
8x  2y  35.10
3x  y  14.30
8x  2y  35.10
y  14.30 
 3x
8x  2(14.30 
 3x)  35.10
2x  28.60  35.10
x  3.25
y  14.30 
 3x
y  14.30 
 3(3.25)
y  4.55
$3.25; $4.55
53. x  y  3
xy3
?
?
003
323
0  3 no
5  3 yes
xy3
xy3
?
?

4  2  3

2  4  3

2  3 no
2  3 no
(3, 2)

480

h

A(x)  x  2x2

44b. A(x )
320
300
280
260
240
220
200
180
160
140

O

2 4 6 8 10 12 14 1618 x

44c. The surface area approaches infinity.
45. If the degree of the denominator is larger than
that of the numerator, then y  0 will be a
horizontal asymptote. To make the graph intersect
the x-axis, the simplest numerator to use is x.
x

Thus, a sample answer is f(x)  
x2  1 .
46a. A vertical asymptote at r  0 and a horizontal
asymptote at F  0.
46b. The force of repulsion increases without bound
as the charges are moved closer and closer
together. The force of repulsion approaches 0 as
the charges are moved farther and farther apart.
47a.
47b.

1

55. [f 
 g](x)  f(g(x))
 f(2 
 x2)
 8(2 
 x2)
 16 
 8x2
[g 
 f ](x)  g(f(x))
 g(8x)
 2 
 (8x)2
 2 
 64x2

a2 
 9

a
3

x
m

2.9
5.9

2.99
5.99

3
—

3.01
6.01

3.1
6.1

The slope approaches 6.

Chapter 3

1


54. 15y 
 x  1 → y  1
5 x  15

92

56. Let x  the width of each card and y  the height
of each card. The rectangle has a base of 4x or 5y.
The rectangle has a height of x  y.
A  bh
180  4x(x  y)
4x  5y
180  4xx  5

y  5

180  5

y  5

4x

V  khg2
288  k(40)(1.5)2
3.2  k
V  3.2hg2
12b. V  3.2hg2
V  3.2(75)(2)2
V  960
50 960  48,000 m3
12a.

4x

36x2

4(5)

25  x2
y4
5x
Perimeter  2(4x)  2(x  y)
P  2(4 5)  2(5  4)
P  58 in.
The correct choice is B.

Pages 194–196
13.

14.

Direct, Inverse, and Joint Variation

3-8

Pages 193–194

15.

Check for Understanding

16.

1a. inverse
1b. neither
1c. direct
2. Sample answer:
Suppose y varies directly as xn.
Then y1  kx1n and y2  kx2n
y1  kx1n
y1

y2
y1

y2

kx1



kx
2



x1

x2

17.

y


54  k(9)2
2

3

7.

8.

y

k

19.

y  w
2

1

12

20.

k(3)(10)

y  4
2

21.

22.

24.

93

y  
48

1

6

k(20)

y  5
2
y  0.168

kxz

y  w

k(2)(
3)


y

4
y  14

kz2

y  x
2


6  3
3

k(9)2

y  6
3


2  k

y  
2
7

y  w

y  x
3

k(6)2

10. A varies jointly as 
 and w; 1
11. y varies inversely as x; 
3.

1

3
2
y  1
2 (
4) (
3)

y  z
2

kb2

y2

1

3 2
y  1
2x z

kx

23. a  c

1
x4; 7.

1


x
1


0.16


y  z2


3  4
2k

0.4(4)(20)

0.4  k

k

 6
2
0.3  k

0.4xz

3  2
2

y  kx3z2

9  k(
3)3(2)2

4
y  5
2
 
3x2
2
 
3(6)2

kxz

9. y varies directly as

y  0.484
y  1.21

y  0.5xz3
y  (0.5)(
8)(
3)3
y  108

y  w
2

r1


xy  k

1.21
(0.44)  k
0.484  k

xy  0.484 or y  0.484

y  
24

y  kxz3
16  k(4)(2)3
0.5  k

1 2

16  k
18.

Simplify.

y  kx2

r  164

1 2

3. The line does not go through the origin, therefore
its equation is not of the form y  kxn.
4a. Sample answer: The amount of money earned
varies directly with the number of hours worked.
4b. Sample answer: The distance traveled by a car
varies inversely as the amount of gas in the car.
4c. Sample answer: The volume of a cylinder varies
jointly as its height and the radius of its base.
5. xy  k
xy  12
4(3)  k
15y  12
6.

y  0.2x
y  0.2(6)
y  1.2
xy  
50
x(
40)  
50
x  1.25
y  15xz
y  15(0.4)(3)
y  18
x2y  36
32y  36
y4
r  16t2

4  k2

Division property of equality.

12  k

Exercises

y  kx
0.3  k(1.5)
0.2  k
xy  k
25(
2)  k

50  k
y  kxz
36  k(1.2)(2)
15  k
x2y  k
(2)2(9)  k
36  k
r  kt2

0.3x

0.3(14)

2xz

2(4)(
7)


2z2


2(
4)2
4

15b2

a  c
15b2

45  12

96  1
0

15  k
x2y  k
2
(4) (2)  k
32  k

8  b
yx2  32
8x2  32
x  2

Chapter 3

25. C varies directly as d; .
26. y varies directly as x;

4

10
11  G; 6.67

6.67

4
.
3

 (6.67

30. y varies inversely as the square root of x; 2.
31. A varies jointly as h and the quantity b1  b2; 0.5.
32. y varies directly as x and inversely as the square
1
of z; 3.
33. y varies directly as x2 and inversely as the cube of
z; 7.
34. y varies jointly as the product of the cube of x and
z and inversely as the square of w.
35a. Joint variation; to reduce torque one must either
reduce the distance or reduce the mass on the
end of the fulcrum. Thus, torque varies directly
as the mass and the distance from the fulcrum.
Since there is more than one quantity in direct
variation with the torque on the seesaw, the
variation is joint.
35b.
T1  km1d1 and T2  km2d2
T1  T 2
km1d1  km2d2
Substitution property
m1d1  m2d2
of equality
35c. m1d1  m2d2
75(3.3)  (125)d2
1.98  d2; 1.98 meters
36a. tr  k
36b.
tr  k
tr  36,000
45(800)  k
t(1000)  36,000
36,000  k
t  36 minutes
37. If y varies directly as x then there is a nonzero
constant k such that y  kx. Solving for x, we find
1
1
x  ky k is a nonzero constant, so x varies
directly as y.

10
11)

kL

1.68

1.07
1.68

k 2


10
2  
(0.001)2

10
8

k

16  62
576  k
39. a is doubled

10
8(3)


R
(0.003)2

R  1.78

10
3 

k

y

42.

8
6
4
2

O


8
6
4
2
2 4 6 8x

2

4

6

8

f(x)  (x 
 3)3  6
y  (x 
 3)3  6
x  (y 
 3)3  6
x 
 6  (y 
 3)3
3


x
6y
3
3
y  

x
63
3

1
f (x)  

x 
 6  3; f
1(x) is a function.

1 0
1
3 
1 
3

1 
3
1 3
44.

0 1
2 
2 
4
0
2 
2 
4 0
A
(
1, 2), B
(


2), C
(1, 
4), D
(3, 0)
43.

45. 4x 
 2y  7

12x  6y  
21

a2  b2  c2
(6)2  (25)2  c2
6.5  c

N m2


10
11 
kg2

(5.98 1024)(1.99 1030)

(1.50 1011)2


R
r2

41.

k

k

1022)

 3.53 1022 N
40d. 3.53 1022  (1.99 1020)x
178  x; about 178 times greater

38a. I  d2
I  d2

1024)(7.36

m1 m2

40c. F  G d
2

5

29. y varies inversely as the square of x; 4.

38b.

(5.98

1020  G 
(3.84 108)2

1.99

27. y varies jointly as x and the square of z; 3.
28. V varies directly as the cube of r;

m1 m2

F  G d
2

40b.

1
.
4

7

→

y  2x 
 2

→

y  2x 
 2

7

consistent
and dependent
y

576

I  d
2

46.

576


I
(6.5)2
I  13.6 lux

x

kb2

a  c
3

O

 

1
k 2b

2

a


3
12c

a

1
kb2
4

1
c3
8

18.6 
 23.2


47. m  
2000 
 1995

48.

kb2

a  2c
3
m1 m2

40a. F  Gd
2

Chapter 3

94


4.6
 5
144  42

or 
0.92

y 
 18.6  
0.92(x 
 2000)
y  
0.92x  1858.60

9 or 122 1
12 is divisible by 3, 4, 6, and 12.

a2 
 b  4 no
x-axis
a  
2b
a  
2(
b)
a  2b no
(
a)  
2b
a  2b no
(b)  
2(a)
b  
2a no
(
b)  
2(
a)
b  
2a no; none

The correct choice is D.
→

17. x  
2y
x-axis
y-axis
yx
y  
x
1

Chapter 3 Study Guide and Assessment
Page 197
1.
4.
7.
10.

2. continuous
5. maximum
8. monotonic

a2  b
1


a2  
(
b)

x-axis

Understanding and Using the
Vocabulary

even
decreasing
inverse
Joint

1

→

18. x2  y

1

a2  
b no

3. point
6. rational
9. slant

1

(
a)2  b

y-axis

1

a2  b yes
1

yx

(b)2  (a)
1

b2  a

Pages 198–200

11. f(
x)  
2(
x)
f(
x)  2x
12. f(
x)  (
x)2  2
f(
x)  x2  2
13. f(
x)  (
x)2 
 (
x)  3
f(
x)  x2  x  3

f(x)  
(x2 
 x  3)

f(x)  
x2  x 
 3 no
14. f(
x)  (
x)3 
 6(
x)  1
f(
x)  
x3  6x  1

f(x)  
(x3 
 6x  1)

f(x)  
x3  6x 
 1 no
15. xy  4
→
x-axis
y-axis
yx
y  
x

16. x  y2  4
x-axis
y-axis
yx
y  
x

y  
x

Skills and Concepts

→


f(x)  
(
2x)

f(x)  2x yes

f(x)  
(x2  2)

f(x)  
x2 
 2 no

19.
20.
21.
22.

(
b)2



no

1

(
a)
1

ay

b2 
no; y-axis
y
The graph of g(x) is a translation of the graph of
f(x) up 5 units.
The graph of g(x) is a translation of the graph of
f(x) left 2 units.
x graph of f(x) expanded
The graph of O
g(x) is the
vertically by a factor of 6.
O
x
The graph of g(x) is the graph of f(x) expanded
4

horizontallyy by a factor of 3 and translated
down
y
4 units.
23.

ab  4
a(
b)  4

ab  4 no
(
a)b  4

ab  4 no
(b)(a)  4
ab  4 yes
(
b)(
a)  4
ab  4 yes
y  x and y  
x
a  b2  4
a  (
b)2  4
a  b2  4 yes
(
a)  b2  4

a  b2  4 no
(b)  (a)2  4
a2  b  4 no
(
b)  (
a)2  4

24.

x

O

x

O

25.

27. Case 1
4x  5

(4x  5)

4x 
 5

95

26.

7
7
7

Case 2
4x  5
4x  5
4x

7
7
2

Chapter 3

f(x)
4x
 3x 
121
x  
3
x
f(x)
{xx  
3 or x 0.5}

2

7
28. Case 1

1

4
x 
 3  2  11
0

1

(x 
 3)  2  11
1
x  5 2 11

f (x )
x  6
x f (
6
x)
1
{x
6  x  12} f (x )

32.

x 0.5
f
1(x)
x
f
1(x)

7

2
Case 2

4

1
x 
 3  2  11

1
0
x 
 3  2  11
2
x1  12

f(x)  (x  1)2 
 4
x
f(x)

3
0

2

3

1

4
0

3
1
0
f 1(x )

f (x )

29.

f (x )

O

x

1

x

2

1
0
1
2

x
5.5
5.25
5
4.75
4.5

f(x)
5.5
5.25
5
4.75
4.5

f
1(x)
f
1(x)

2

1
0
1
2

f(x)  (x 
 2)3 
 8
y  (x 
 2)3 
 8
x  (y 
 2)3 
 8
x  8  (y 
 2)3
3


x8y
2
3
y  

x82
3

1
f (x)  

x  8  2; yes
34. f(x)  3(x  7)4
y  3(x  7)4
x  3(y  7)4

33.

f (x )
6
4
2

30.

f (x )


6
4 O 2 4 6

4 1

6 f (x )

x

3
x
 4 3

x

 (y  7)4
y7

y  
7 
f
1(x)  
7 

2

x

3

2

1

f(x)
2.3
2
1

f
1(x)
x
f
1(x)
2.3

3
2

2
1

1

1


1


1

0

—

7

1

2

1

2

7

1
2
3

5
4
3.7

5
4
3.7

1
2
3

f(x)  x  3


2

x

3
4 x
;
3

4

no

35. Yes; the function is defined when x  2; the
function approaches 6 as x approaches 2 from both
sides; and y  6 when x  2.
36. No; the function is undefined when x  
1.
37. Yes; the function is defined when x  1; the
function approaches 2 as x approaches 1 from both
sides; and y  2 when x  1.
38. an: negative, n: odd
y → 
 as x → , y →  as x → 
.
39. an: positive, n: odd
y →  as x → , y → 
 as x → 
.

1


2

40.

1

y  x2  1
x

1000

100

10
1
10
100
1000

y
f (x )

O
f 1(x )

x

O

f(x)  
4x  5

31.

x
0

3

4

3
0

f
1(x)
f
1(x)

3

2

1
0
1

x

y
1.000001
1.0001
1.01
2
1.01
1.0001
1.000001

y → 1 as x → , y → 1 as x → 
.
41. an: positive, n: odd
y →  as x → , y → 
 as x → 
.

Chapter 3

96

x

52. x  1

42.


y
x
1
x

x

y 
x
1
 
 
x
x
1

y 
1 ;
1 
 x

x2  1

53. x  
2

[
5, 5] scl:1 by [
20, 10] scl:5
decreasing for x  
2 and x 1;
increasing for 
2  x  1

y1


y
x2
x2
1


x2  x2

y 
x
2

43.



x2  x2
1
1  x
2

y 
1
2
  
x
x2

no horizontal asymptotes
(x 
 3)2

54. x  
3,

44.
45.
46.

47.


y
x2 
 9
x2 
 6x  9


y
x2 
 9

[
6, 6] scl:1 by [
5, 20] scl:5
decreasing for x  
3 and 0  x  3;
increasing for 
3  x  0 and x 3
abs. max.: (
2, 1)
rel. max.: (0, 4), rel. min.: (2, 0)
f(2.9)  0.029
f(3)  0
f(3.1)  0.031 min.
f(
0.1)  6.996
f(0)  7
f(0.1)  7.004 pt of inflection

9
x2
6x



x2 
 x2  x2

y  
x2
9


x2 
 x2

9
6
1 
 x  x
2

y  
;y1
9
1 
 x
2

x2
x2  2
x1
→
x

x2
2x
2x
1
56.
y  kxz
5  k(
4)(
2)
0.625  k
55.

1

48. f(x)  x  1
2

49. f(x)  
x
50. The parent graph is translated 2 units left and
expanded vertically by a factor of 3. The vertical
asymptote is now x  
3. The horizontal
asymptote, f(x)  0, is unchanged.

57.

f (x )

y
20 

k


x

k


49


140  k

x

O

58.

51. The parent graph is translated 3 units right and
then translated 2 units up. The vertical asymptote
is now x  3 and the horizontal asymptote is
f(x)  2.

O

97

y

140


x

10 

140


x

kx2

y  z

320x2

7.2  4

k(0.3)2

y  40

320  k

y8

320(1)2

Applications and Problem Solving

59. x 
 6.5  0.2;
Case 1
x 
 6.5  0.2

(x 
 6.5)  0.2

x  6.5  0.2

x  
6.3
x  6.3

x

yes; y  x  2
y  0.625xz
y  0.625(
6)(
3)
y  11.25


x  14
x  196

y  z

Page 201

f (x )

1

x  2  x

Case 2
x 
 6.5  0.2
x 
 6.5  0.2
x  6.7
6.3  x  6.7

Chapter 3

60a.

x
0x1
1x2
2x3
3x4
4x5
5x6
3.60
3.20
2.80
2.40
Cost 2.00
(dollars) 1.60
1.20
0.80
0.40
0

Page 201

C(x)
0.40
0.80
1.20
1.60
2.00
2.40

y

1b. Sample answer: y  x2

y

x

0123456789
Time (min)

O

1c. Sample answer: 
xy  1

y

C
1(x)
0x1
1x2
2x3
3x4
4x5
5x6

x
0.40
0.80
1.20
1.60
2.00
2.40

x
O

1d. Sample answer: xy  1

9
8
7
6
Time 5
(min) 4
3
2
1
0
0 0.40 1.20 2.00 2.80 3.60
Cost (dollars)

y

x

O

1e. Sample answer: y  x3

60d. positive multiples of $0.40;
positive real numbers
60e. C
1(x) gives the possible number of minutes
spent using the scanner that cost x dollars.
61a.

x

O

60b. positive real numbers;
positive multiples of $0.40
60c.

Open-Ended Assessment

1a. Sample answer: x  y2

y

h(t)

O

x

1
0.5

O

0.5

1

2. Sample answer: 
2(x 
 4)2  1

t

61b. 1.08 m

Chapter 3

98

3. Notice that 450 miles is the distance to
Grandmother’s house, not the round trip. This is a
multiple-step problem. First calculate the number
of gallons of gasoline used in each direction of the
trip.

3a. Sample answer:

y

miles

miles per gallon
450

25

(0, 0)

x

O

(2, 1)

3b. abs. min.: (2, 
3); rel. max.: (0, 0); rel. min.:
(
2, 
1)

Chapter 3 SAT & ACT Preparation

4.

y

SAT & ACT Practice

1. Always factor or simplify algebraic expressions
when possible. Notice that the numerator in the
problem is the difference of two squares, a2 
 b2.
Factor it.
y2 
 9

3y 
 9

 18 gallons

On the trip to Grandmother’s, the cost of gasoline
is 18 gallons $1.25 per gallon or $22.50.
On the trip back, the gasoline cost is 18 gallons
$1.50 per gallon or $27.00. The difference between
the costs is $4.50.
A faster way to find the cost difference is to reason
that each gallon cost $0.25 more on the trip back.
So the total amount more that was paid was
18 gallons $0.25 or $4.50. The correct choice is B.

(2, 3)

Page 203

 gallons

O

x

( y  3)(y 
 3)



3y 
 9

Factor the denominator. Both the numerator and
denominator contain the factor (y 
 3). Simplify
the fraction.
(y  3)(y 
 3)

3y 
 9

( y  3)(y 
 3)

The portion of the graph of f(x) which is shown
crosses the x-axis 3 times.
The correct choice is D.
5. Notice that the denominators are all powers of
ten. Carefully convert each fraction to a decimal.
Then add the three decimals.

y3

  

3( y 
 3)
3

The correct answer is E.
2. You need to find the statement that is not true.
Compare the given information with each answer
choice. Choice A looks like x  y  z, except for
the numbers. Multiply both sides of the equation
x  y  z by 2.
2(x  y)  2z or 2x  2y  2z
So choice A is true. For choice B, start with x  y
and subtract y from each side.
x
yy
y0
So choice B is true. For choice C, start with x  y
and subtract z from each side.
x
zy
z
So choice C is true. For choice D, substitute y for x
and x  y for z.
z
x  2
xy

900

10

90

9




100  1000  90  0.9  0.009  90.909

The correct choice is C. You could also use your
calculator on this problem.
6. Combine like terms.
(10x4 
 x2  2x 
 8) 
 (3x4  3x3  2x  9)
 (10x4 
 3x4)  (
3x3)  (
x2)  (2x 
 2x) 
(
8 
 9)
 7x4 
 3x3 
 x2  0 
 17
The correct choice is A.
7. One method of solving this problem is to “plug in”
a number in place of n. Choose a number that
when divided by 8, has a remainder of 5. For
example, choose 21.
21  2(8)  5
Then use this value for n in the answer choices.
Find the expression that has a remainder of 7.

2y

y  2  2y
So choice D is also true. For choice E, write each
side of the equation in terms of y.
z 
 y  (x  y) 
 x  y
2x  2y
y 2y
So choice E is not true. The correct choice is E.

n1

21  1

22

Choice A: 8  8  8  2R6
The remainder is 6.
n2

21  2

23

Choice B: 8  8  8  2R7
You could also reason that since n divided by 8 has
a remainder of 5, then (n  2) divided by 8 will
have a remainder of (5  2) or 7. The correct
choice is B.

99

Chapter 3

8.Simplify the expression inside the square root
symbol. Factor 100 from each term. Then factor
the trinomial.


100x2
 60
0x  9
00

x3





100(x2
 6x
 9)

x3



10

x2  6
x9

x3



(x  3
)(x  
3)
10


x3



10

(x  3
)2

x3



10(x  3)

x3

10. There are two equations and two variables, so this
is a system of equations. First simplify the
equations. Start with the first equation. Divide
both sides by 2.
4x  2y  24
2x  y  12
Now simplify the second equation. Multiply both
sides by 2x.
7y

2x

7y  7(2x)
7y  14x
Divide both sides by 7.
y  2x
You need to find the value of x. Substitute 2x for y
in the first equation.
2x  y  12
2x  (2x)  12
4x  12
x3
The answer is 3.

 10
The correct choice is B.
9. Since a  b  c, substitute a  b for c in
a 
 c  5. So, a 
 (a  b)  5. Then 
b  5 or
b  
5. Substitute 
5 for b in b 
 c  3. So,

5 
 c  3. Then 
c  8 or c  
8.
The correct choice is B.

Chapter 3

7

100

Chapter 4 Polynomial and Rational Functions
4-1

11. 2; x2 
 14x  49  0
(x 
 7)(x 
 7)  0
x
70
x7

Polynomial Functions

Pages 209–210

Check for Understanding

x
70
x7

f (x)

1. A zero is the value of the variable for which a
polynomial function in one variable equals zero. A
root is a solution of a polynomial equation in one
variable. When a polynomial function is the
related function to the polynomial equation, the
zeros of the function are the same as the roots of
the equation.
2. The ordered pair (x, 0) represents the points on
the x-axis. Therefore, the x-intercept of a graph of
a function represents the point where f(x)  0.
3. A complex number is any number in the form
a  bi, where a and b are real numbers and i is
the imaginary unit. In a pure imaginary number,
a  0 and b 0. Examples: 
2i, 3i;
Nonexamples: 5, 1  i
4.
y

60
40

(0, 49)

f (x ) 
 x 2  14x  49

20
(7, 0)

O

4

8

x

12

12. 3; a3  2a2 
 8a  0
a(a2  2a 
 8)  0
a(a  4)(a 
 2)  0
a0
a40
a  
4
30
20

a
20
a2

f (a)
f (a) 

a 2  2a 2  8a

10
(4, 0) (0, 0) (2, 0)

O

x


4


2 O

10

2

4a

t4 
 1  0
(t2 
 1)(t2  1)  0
(t 
 1)(t  1)(t2  1)  0
t
10
t10
t1
t  
1

13. 4;
5. 3; 1
6. 5; 8
7. no; f(x)  x3 
 5x2 
 3x 
 18
f(5)  (5)3 
 5(5)2 
 3(5) 
 18
f(5)  125 
 125 
 15 
 18
f(5)  
33
8. yes; f(x)  x3 
 5x2 
 3x 
 18
f(6)  (6)3 
 5(6)2 
 3(6) 
 18
f(6)  216 
 180 
 18 
 18
f(6)  0
9. (x 
 (
5))(x 
 7)  0
(x  5)(x 
 7)  0
x2 
 2x 
 35  0; even; 2
10. (x 
 6)(x 
 2i)(x 
 (
2i))(x 
 i) (x 
 (
i))  0
(x 
 6)(x 
 2i)(x  2i)(x 
 i)(x  i)  0
(x 
 6)(x2 
 4i2)(x2 
 i2 )  0
(x 
 6)(x2  4)(x2  1)  0
(x3 
 6x2  4x 
 24)(x2  1)  0
x5 
 6x4  5x3 
 30x2  4x 
 24  0;
odd; 1

t2  1  0
t2  
1
t  i

f (t )

(1, 0)

(1, 0)

O

t

f (t ) 
 t 4 1

14a. x2  r2  62
r2  36 
 x2
14b. V(x)  (36 
 x2)(2x)
V(x)  (36 
 x2)(2x)
V(x)  72x 
 2x3
14c. V(x)  72x 
 2x3
V(4)  72(4) 
 2(4)3
V(4)  502.65 units3

Pages 210–212

V(x)  Bh
V(x)  (36 
 x2)(2x)

Exercises

15. 4; 5
16. 7; 3
17. 3; 5
18. 5; 
25
19. 6; 
1
20. 2; 1
21. Yes; the coefficients are complex numbers and the
exponents of the variable are nonnegative
integers.

101

Chapter 4

37. (x 
 (
1))(x 
 1)(x 
 4)(x 
 (
4)(x 
 5)  0
(x  1)(x 
 1)(x 
 4)(x  4)(x 
 5)  0
(x2 
 1)(x2 
 16)(x 
 5)  0
(x4 
 17x2  16)(x 
 5)  0
x5 
 5x4 
 17x3  85x2  16x 
 80  0; odd; 5
38. (x 
 (
1))(x 
 1)(x 
 3)(x 
 (
3))  0
(x  1)(x 
 1)(x 
 3)(x  3)  0
(x2 
 1)(x2 
 9)  0
x4 
 10x2  9  0
39. 1; x  8  0
f (x)
x  
8
(0, 8)

1
a

22. No;   a1, which is a negative exponent.


 12a
23. yes; f(a) 
f(0)  (0)4 
 13(0)2  12(0)
f(0)  0
24. no; f(a)  a4 
 13a2  12a
f(
1)  (
1)4 
 13(
1)2  12(
1)
f(
1)  1 
 13 
 12
f(
1)  
24
25. yes; f(a)  a4 
 13a2  12a
f(1)  (1)4 
 13(1)2  12(1)
f(1)  1 
 13  12
f(1)  0
26. yes; f(a)  a4 
 13a2  12a
f(
4)  (
4)4 
 13(
4)2  12(
4)
f(
4)  256 
 208 
 48
f(
4)  0
27. no; f(a)  a4 
 13a2  12a
f(
3)  (
3)4 
 13(
3)2  12(
3)
f(
3)  81 
 117 
 36
f(
3)  
72
28. yes; f(a)  a4 
 13a2  12a
f(3)  (3)4 
 13(3)2  12(3)
f(3)  81 
 117  36
f(3)  0
29. f(b)  b4 
 3b2 
 2b  4
f(
2)  (
2)4 
 3(
2)2 
 2(
2)  4
f(
2)  16 
 12  4  4
f(
2)  12; no
30. f(x)  x4 
 4x3 
 x2  4x
f(
1)  (
1)4 
 4(
1)3 
 (
1)2  4(
1)
f(
1)  1  4 
 1 
 4
f(
1)  0; yes
31a. 3; 1
31b. 2; 2
31c. 4; 2
32. (x 
 (
 2))(x 
 3)  0
(x  2)(x 
 3)  0
x2 
 x 
 6  0; even; 2
33. (x 
 (
1))(x 
 1)(x 
 5)  0
(x  1)(x 
 1)(x 
 5)  0
(x2 
 1)(x 
 5)  0
x3 
 5x2 
 x  5  0; odd; 3
34.
(x 
 (
2))(x 
 (
0.5))(x 
 4)  0
(x  2)(x  0.5)(x 
 4)  0
(x2  2.5x  1)(x 
 4)  0
3
2
x 
 4x  2.5x2 
 10x  x 
 4  0
x3 
 1.5x2 
 9x 
 4  0
2x3 
 3x2 
 18x 
 8  0; odd; 3
35. (x 
 (
3))(x 
 (
2i))(x 
 2i)  0
(x  3)(x  2i)(x 
 2i)  0
(x  3)(x2 
 4i2)  0
(x  3)(x2  4)  0
3
x  3x2  4x  12  0; odd; 1
36. (x 
 (
5i))(x 
 (
i))(x 
 i)(x 
 5i)  0
(x  5i)(x  i)(x 
 i)(x 
 5i)  0
(x  5i)(x 
 5i)(x 
 i)(x  i)  0
(x2  25)(x2  1)  0
x4  26x2  25  0; even; 0
a4

Chapter 4

13a2

f (x) 
 x  8

O x

(8, 0)

40. 2;

a2 
 81  0
(a 
 9)(a  9)  0
a
90
a9

(9, 0) f (a)

20 
10 O

20

a90
a  
9

(9, 0)
10

20a

2

40 f (a) 
 a  81


60

80

(0, 81)

41. 2; b2  36  0
b2  
36
b  6i

80
60
f (b) 

b 2  36 40

f (b)

(0, 36)

20

4 
2 O

42. 3;

t3  2t2 
 4t 
 8  0
t2(t  2) 
 4(t  2)  0
(t  2)(t2 
 4)  0
(t  2)(t  2)(t 
 2)  0
t20
t20
t  
2
t  
2
4
(2, 0)


4


2 O

4
(0, 8)

12

102

f (t)
(2, 0)
2

4 t

f (t) 
 t 3 
2t 2  4t  8

2

4b

t
20
t2

n3 
 9n  0
n(n2 
 9)  0
n(n 
 3)(n  3)  0
n0
n
30
n3

43. 3;

20

47. 4;

n30
n  
3

4m4  17m2  4  0
(4m2  1)(m2  4)  0
4m  1  0
m2  4  0

1
m  

4

m   4
m  0.5i

f (n)

m  2i

f (m)

10
(3, 0)

4

(0, 0)


2 O

10

20

44. 3;

(3, 0)

(0, 4)

4n

2

f (m) 
 4m 4  17m 2  4

f (n ) 
 n 3 9n

m

O

6c3 
 3c2 
 45c  0
c(6c2 
 3c 
 45)  0
c(c 
 3)(6c  15)  0
c0
c
30
c3

48.
6c  15  0
15

c  
6

(u  1)(u2 
 1)  0
(u  1)(u  1)(u 
 1)  0
u10
u10
u  
1
u  
1

u
10
u1

f (u)

c  
2.5
100

f (c)

(1, 0)

O

(2.5, 0) 50
(0, 0)

4


2 O

50

100

45. 4;

(0, 1)

(3, 0)

2

u

f (u) 

(u  1)(u 2  1)

4c

f (c) 
 6c 3 
3c 2  45c

a 4  a2 
 2  0
(a2  2)(a2 
 1)  0
(a2  2)(a 
 1)(a  1)  0
a2  2  0
a
10
a2  
2
a1
a  
2
i

49a.

a10
a  
1

y

49b.

x

O

f (a)

(1, 0)

(1, 0)

49c.

y

49d.

y

x

O
y

(1, 0)

f (a) 
 O
a 4  a 2  2 (0, 2)

a

x4 
 10x2  9  0
(x2 
 9)(x2 
 1)  0
(x 
 3)(x  3)(x 
 1)(x  1)  0
x
30 x30 x
10
x3
x  
3
x1

46. 4;

20

x

O

49e.

y

O

x

49f. not possible

x10
x  
1

f (x)
O

x

10 (0, 9)
(3, 0)(1, 0) (1, 0) (3, 0)

4 
2 O

10

20

2

4x

f (x) 

x 4  10x 2  9

103

Chapter 4

50.

56.

(x 
 B)(x 
 C)  0
x2 
 Cx 
 Bx  BC  0
x2 
 (C  B)x  BC  0

C 
 B  B
from x2  Bx  C  0
BC  C
B1

C 
 1  1

C  2
C  
2
Sample answer: 1; 
2



[
5, 5] sc11 by [
2, 8] sc11
50a. 4
50b. 2; 
1, 1
50c. There are 4 real roots. However, there is a
double root at 
1 and a double root at 1.
51a. V(x)  99,000x3  55,000x2  65,000x
51b. r  0.15
x1r
x  1  0.15
x  1.15
V(x)  99,000x3  55,000x2  65,000x
V(1.15)  99,000(1.15)3  55,000(1.15)2
 65,000(1.15)
V(1.15)  150,566.625  72,737.5  74,750
V(1.15)  298,054.125; about $298,054.13
52. 1 and 3 are two of its zeros.
53a.

1

d(t)  2at2
1

x
2


57. y  
x(x  2)(x 
 2)
1

58a. Let x  the width. The length  2(52 
 2x) or
26 
 x. A(x)  x(26 
 x)
58b. A(x)  x(26 
 x)
A(x)  26x 
 x2

1

d(t)  2at2
1

d(30)  2(16.4)(30)2

d(60)  2(16.4)(60)2

d(30)  7380 ft

d(60)  29,520 ft

[
5, 30] sc15 by [
2, 200] sc120
x  13
26 
 x  26 
 13
 13
13 yd by 13 yd
59. The graph of y  2x3  1 is the graph of y  2x3
shifted 1 unit up.
60. (
6, 9)

1

d(t)  2at2
1

d(120)  2(16.4)(120)2
d(120)  118,080 ft
53b. It quadruples; (2t)2  4t2.
54. Let x  the width of the sidewalk.
The length of the pool would be 70 
 2x feet.
The width of the pool would be 50 
 2x feet.
A  
w
2400  (70 
 2x)(50 
 2x)
2400  3500 
 240x  4x2
0  4x2 
 240x  1100
0  x2 
 60x  275
0  (x 
 55)(x 
 5)
x 
 55  0
x
50
x  55
x5
Use x  5 since 55 is an unreasonable solution.
5 ft
55. Let x  the number of pizzas.
(160  16x)(16 
 0.40x)  4000

6.4x2  192x  2560  4000

6.4x2  192x 
 1440  0
x2 
 30x 
 225  0
(x 
 15)(x 
 15)  0
x 
 15  0
x 
 15  0
x  15
x  15
16 
 0.40x  16 
 0.40(15)
 $10

Chapter 4

61. 
15 5  
15(3) 
 (
9)(5)

9 3
 
45  45 or 0; no
2 
1
3 
9
2
62. AB 
3
4
5
7 
6
2(3)  (
1)(5) 2(
9)  (
1)(7)

3(3)  4(5)
3(
9)  4(7)
2(2)  (
1)(
6)
3(2)  4(
6)
1 
25
10
29
1 
18
63. x  4y  9
4y  
x  9


1

y

9

y  
4x  4

x

O
x  4y  9

64. Parallel; the lines have the same slope.

104

65. [f 
 g](x)  f(g(x))








Completing the Square
x2  4x 
 5  0
x2  4x  5
2
x  4x  4  5  4
(x  2)2  9
x  2  3
x23
x1
Quadratic Formula




1
f 2x  6
1
x  6 2 
 4
2
1
x2  6x  36
4
1
x2  6x  32
4


4

[g 
 f ](x)  g( f(x))
 g(x2 
 4)
1

x

 2(x2 
 4)  6
1


4  

42 
 4
(1)(
5)

2(1)

4  
36


 2x2 
 2  6

x  2

1

 2x2  4


4  6

x  2

66. The pictograph shows two more small car symbols
in the row for 1999 than it does for 2000. These
two small cars represent the 270 additional cars
that were sold in 1999 compared to 2000. Since
the two small cars represent 270 real cars, each

x  
2  3
x  
2  3
x1
See students’ work.
5. x2  8x 
 20  0
x2  8x  20
x2  8x  16  20  16
(x  4)2  36
x  4  6
x46
x2
6. 2a2  11a 
 21  0

270

small car symbol must represent  or 135 real
2
cars.
The correct choice is A.

4-2

Quadratic Equations

11

Pages 218–219

Check for Understanding

2 
 4(5)(7)


13  
(
13)

2(5)

3a. equals 0
3c. positive number
4. Graphing

or

21

11

21

121

21

11

121



a2  2a  1
6  2  16

a  1412  21869
11

17

a  4  4
11

17

11

a  4  4
3

7.

b2

17

a  4  
4

a  2

a  
7


 4ac 

 4(1)(36)
 0; 1 real
122

m  
12 



0

2(1)

12

m  
2 or 
6

13  
29


10

8. b2 
 4ac  (
6)2 
 4(1)(13)
 
16; 2 imaginary

3b. negative number

t


(
6)  

16


2(1)
6  4i

 2
 3  2i

x 2  4x  54

6

x  4  
6
x  
10

a2  2a  2

f (x ) 
 f (x )
(5, 0)

x  
2 
 3
x  
5

a2  2a 
 2  0

1. Add 4 to each side of the equation to get t2 
 6t
 4. Determine the value needed to make t2 
 6t
a perfect square trinomial. Add this value (9) to
each side. Take the square root of each side of the
equation and solve the two resulting equations.
t  3  
13

2. Quadratic Formula; Since the leading coefficient
does not equal 1 and the discriminant equals 185
which is not a perfect square, the Quadratic
Formula would be the best way to get an exact
answer. Completing the square can also be used,
but errors in arithmetic are more likely. A graph
will give only approximate solutions.
p

x  2  
3
x  
5

p2 
 6p  5  0
(p 
 5)(p 
 1)  0
p
50
p5
10. r2 
 4r  10  0
9.

(1, 0)


4 
2 O

4

2x


8

r
Factoring
x2  4x 
 5  0
(x  5)(x 
 1)  0
x50
x  
5

r

p
10
p1


(
4)  

(
4)2 

 4(1
)(10)

2(1)
4  

24


2
4  2i
6


2

r
r  2  i
6


x
10
x1

105

Chapter 4

11.

P  12I 
 0.02I 2
1600  12I 
 0.02I 2
0.02I 2 
 12I  1600  0
I 2 
 600I  80,000  0
(I 
 200)(I 
 400)  0
I 
 200  0
I 
 400  0
I  200 amps
I  400 amps

17. t2 
 3t 
 7  0
t2 
 3t  7
9

t

3 2

 2
3
t 
 2



9



p




z 
 1  
5
z  
4

m
m

21. b2 
 4ac  (
5)2 
 4(1)(9) or 
11; 2 imaginary
s

361

4
3

s

19

p 
 2  
2

d
d
x
x
x
24.

1

d2 
 4d  
8
1

9

9



d2 
 4d  6
4  
 8  64

d 
 382  614
3

p

1

3

1

d

1

2

3

16.

3g2

p

1

d 
 8  
8
d

1

4

25.


 12g  
4

g2

4


 4g  
3
4

k

8

(g 
 2)2  3
g
2
g


5  
97


2(2)


5  
97

4


26. 
7  i
5

2
6


3
2
6

2 
3

28. s 
s
s
s

Chapter 4


4  
112


2(3)

4  4
7


6


2  2
7

3
2k2  5k 

9
2k2  5k 
 9  0
b2 
 4ac  52 
 4(2)(
9) or 97; 2 real
k

g2 
 4g  4  
3  4

2  

140


2(4)
2  2i
35


8
1  i
35


4
3p2  4p 

8
3p2  4p 
 8  0
b2 
 4ac  42 
 4(3)(
8) or 112; 2 real
p

d 
 8  8
d 
 8  8

84  
0


2(36)
84
7
 or 
72
6

23. b2 
 4ac  (
2)2 
 4(4)(9) or 
140; 2 imaginary

1

3

5  

11


2(1)
5  i
11


2

22. b2 
 4ac  (
84)2 
 4(36)(49) or 0; 1 real

15. d2 
 4d  8  0
3



7  
121

2(6)

7  11

12
3 1

2, 3

m

p  11
p  
8
14. x2 
 10x  21  0
x2 
 10x  
21
x2 
 10x  25  
21  25
(x 
 5)2  4
x 
 5  2
x
52
x 
 5  
2
x7
x3
3

1

19. b2 
 4ac  (6)2 
 4(4)(25)
 
364
2 imaginary; the discriminant is negative.
20. b2 
 4ac  72 
 4(6)(
3) or 121; 2 real

9





37

2

18. 
2  4

p2 
 3p  4  88  4
3 2
p 
 2
3
19
  
2
2


37


 
2
3

Exercises

12. z2 
 2z 
 24  0
z2 
 2z  24
z2 
 2z  1  24  1
(z 
 1)2  25
z 
 1  5
z
15
z6
13. p2 
 3p 
 88  0
p2 
 3p  88

37

 4

t  2 
1 2

Pages 219–221

9

t2 
 3t  4  7  4

106


b  

b2 
 4ac


2a
5  

(
5)2 

 4(3)(9)


2(3)
5  

83


6
5  i
83


6

27. 5  2i

29. x2 
 3x 
 28  0
(x 
 7)(x  4)  0
x
70
x7
30. 4w2  19w 
 5  0
(4w 
 1)(w 
 5)  0
4w 
 1  0
4w  1
1
w  4
31.

4r2 
 r  5
4r2 
 r 
 5  0
(4r 
 5)(r  1)  0
4r 
 5  0
4r  5
5
r  4

32. p 
p
p

b2 
 4ac  0
82 
 4(1)(c)  0
64 
 4c  0

4c  
64
c 16
36a. A  bh
A  12(16)
A  192
1
(12 
 2x)(16 
 2x)  2(192)
35.

x40
x  
4

w50
w  
5

(12 
 2x)(16 
 2x)  96
36b. (12 
 2x)(16 
 2x)  96
192 
 56x  4x2  96
4x2 
 56x  96  0
x2 
 14x  24  0

r10
r  
1

f (x)


b  

b2 
 4ac


2a

2  

22 
 4
(1)(8)

2(1)

2  

28


2

2  2i
7


2

20
10

O

p
p  
1  i
7


x

2 
 4(
2
6
  

(2
6
 )
1)(
2)

2(1)
2
6
  
32


2

x

2
6
  4
2


2

33. x 


20

A
A
A


b  

b2 
 4
ac

2a

0.05  

0.052 

 4(0.01)(

18)


2(0.01)

0.05  
0.722
5


0.02
5


0.05  
0.722

or
0.02

A  40
40 years old
34c.

x

A

f (x ) 

x 2  14x  24

1

37a. d(t)  v0t 
 2gt2

d (t)

1

d(t)  5t 
 2(32)t2

t

O

d(t)  5t 
 16t2
d (t) 
 5t  16t 2

37b. 0 and about 0.3
37c. The x-intercepts indicate when the woman is at
the same height as the beginning of the jump.
37d. d(t)  5t 
 16t2

50  5t 
 16t2
37e.

50  5t 
 16t2
16t2 
 5t 
 50  0


0.05 
 
0.722
5


0.02

A  
45

P

150
125
100 P 

2
75 0.01A  0.05A  107
50
25

O

20

36c. roots: 2, 12
12 
 2x  12 
 2(2) or 8
16 
 2x  16 
 2(2) or 12
8 ft by 12 ft
12 
 2x  12 
 2(12) or 
12
16 
 2x  16 
 2(12) or 
8
∅

  2
2

x  
6
34a. P  0.01A2  0.05A  107
P  0.01(25)2  0.05(25)  107
P  6.25  1.25  107
P  114.5 mm Hg
34b.
P  0.01A2  0.05A  107
125  0.01A2  0.05A  107
0  0.01A2  0.05A 
 18
A

10


10

t
t
t
t

25 50 75 100 A


b 

b2 
 4
ac

2a
2
5  

(
5) 

 4(1
6)(
5
0)

2(16)
5  
3225


32
5  
3225


32

t  1.93 s
about 1.93 s

As a woman gets older, the normal systolic
pressure increases.

107

t

5 
 
3225


32

t  
1.62

Chapter 4

ax2  bx  c  0

38.

b

c

x2  ax  a  0
b

2

2

x2  ax  2a  
a  2a
b

c

b

2

2

2

2

2

b

2a



b2 
 4ac


  
2a
b

x  
2a 
x



b2 
 4
ac

2a

b2 
 4
ac

b  


2a

18a2  3a 
 1  0
(3a  1)(6a 
 1)  0
3a  1  0
3a  
1

39. 2;

6a 
 1  0
6a  1

1

1

a  
3

a  6

f (a)
f (a) 

18a 2  3a  1

a

O
40.

x

2

1
0
1
2

y

y
0

1

2

1
0

x

O
y  |x |  2

f(x)  (x 
 9)2
y  (x 
 9)2
x  (y 
 9)2

x
y
9
y  
x  9
f1(x)  
x  9
42. 3x  4y  375

2(5x  2y)  
2(345)
41.

3x  4y  375
3(45)  4y  375
y  60
43. m 

→

3x  4y  375

10x 
 4y  
690

7x
 
315
x  45

(45, 60)

619 – 595

2.8 – 2.4
24


m
0.4

m  60

60 

619 – x

2.8 – 3.2

619 
 x  
24
x  $643

44. 3y  8x  12
3y  
8x  12
8
y  3x  4; 
3
45. x2  x 
 20  (x  5)(x 
 4)
The correct choice is A.
Chapter 4

Page 226

Check for Understanding

1. The Remainder Theorem states that if a
polynomial P(x) is divided by x 
 r, the remainder
is P(r). If a division problem has a remainder of 0,
then the divisor is a factor of the dividend. This
leads to the Factor Theorem which states that the
binomial x 
 r is a factor if and only if P(r)  0.
2. (x3 
 4x2 
 7x  8) (x 
 5); x2  x 
 2; 
2
3. The degree of a polynomial is one more than the
degree of its depressed polynomial.
4. Isabel; if f(
3)  0, then (x 
 (
3)) or (x  3) is a
factor.
5. 2 1 
1 4
6. 
5 1 1 
17
15
2 2

5
20 
15
1
1 6
1 
4
3
0
x  1, R6
x2 
 4x  3
7. f(x)  x2  2x 
 15
f(3)  (3)2  2(3) 
 15
 9  6 
 15 or 0; yes
8. f(x)  x4  x2  2
f(3)  (3)4  (3)2  2
 81  9  2 or 92; no
9. f(x)  x3 
 5x2 
 x  5
1 1 
5 
1
5
f(1)  (1)3 
 5(1)2 
 1  5
1 
4 
5
 1 
 5 
 1  5 or 0
1 
4 
5
0
x 
 1 is a factor
x2 
 4x 
 5  (x 
 5)(x  1)
(x 
 5), (x  1), (x 
 1)
10. f(x)  x3 
 6x2  11x 
 6
f(1)  (1)3 
 6(1)2  11(1) 
 6
 1 
 6  11 
 6 or 0
1 1 
6
11 
6
1 
5
6
1 
5
6 0
x 
 1 is a factor
x2 
 5x  6  (x 
 2)(x 
 3)
(x 
 1), (x 
 2), (x 
 3)
11. 
1 1
6k2
0 
7
k
k  
4

1
1
6
1 
1 
6
6k
12a. 12
12b. 12
12c. 11
12d. f(x)  x7  x9  x12 
 2x2
 x12  x9  x7 
 2x2
 x(x11  x8  x6 
 2x)
 x2(x10  x7  x5 
 2)
x, x2, x11  x8  x6 
 2x, or x10  x7  x5 
 2
13. h  r  4
V  r2h
V  r2(r  4)
5  r2(r  4)
5  r3  4r2
0  r3  4r2 
 5
0  (r3  4r2 
 5)
1 1 4 0 
5
1 5
5
1 5 5
0
r
10
hr4
r  1 in.
h  1  4 or 5 in.

x  2ba  
ac  4ba
2

4ac  b

x  2ba  
4a
x

The Remainder and Factor
Theorems

c

x2  ax  
a
b

4-3

108

Pages 226–228
14. 
7 1

20

7
13

91

91
0


9
27
3 
18
1 
6
9
x2 
 6x  9, R 
1

15. 3 1


5
2
8
4 
4 
8
1 
1 
2
0
x2 
 x 
 2  (x 
 2)(x  1)
(x 
 4), (x 
 2), (x  1)
2 1 
2 
4
8
2
0 
8
1
0 
4
0
x2 
 4  (x 
 2)(x  2)
(x 
 2), (x 
 2), (x  2)
1 1
4 
1 
4
1
5
4
1
5
4
0
x2  5x  4  (x  1)(x  4)
(x 
 1), (x  1), (x  4)

1 1
3
3
1

1 
2 
1
1
2
1
0
x2  2x  1  (x  1)(x  1)
(x  1), (x  1), (x  1)
2 1
0 
9
0
24
0 
16
2
4 
10 
20
8
16
1
2 
5 
10
4
8
0
x5  2x4 
 5x3 
 10x2  4x  8
2 1 2 
5 
10
4
8
2
8
6 
8 
8
1 4
3

4 
4
0
x4  4x3  3x2 
 4x 
 4
2 1 4
3

4

4
2
12
30 
52
1 6
15 
26 
56
2 times

1 1
2 
1 
2

1 
1
2
1
1 
2
0
x2  x 
 2  (x  2)(x 
 1)
1 time; 
2, 1
f(x)  2x3 
 x2  x  k
f(1)  2(1)3 
 (1)2  1  k
02
11k

2  k
f(x)  x3 
 kx2  2x 
 4
f(2)  (2)3 
 k(2)2  2(2) 
 4
0  8 
 4k  4 
 4
0  
4k  8
2k
f(x)  x3  18x2  kx  4
f(
2)  (
2)3  18(
2)2  k(
2)  4
0  
8  72 
 2k  4
0  
2k  68
34  k
f(x)  x3  4x2 
 kx  1
f(
1)  (
1)3  4(
1)2 
 k(
1)  1
0  
1  4  k  1
0k4

4  k

31. 4 1

Exercises

28
27

1

1
x  13
0 
1
16. 2 1 1 0
2 6 12
24
1 3 6 12
23
x3  3x2  6x  12, R23
0 
8 0
16
17. 
2 1

2
4 8 
16
1 
2 
4 8
0
x3 
 2x2 
 4x  8
5

4 
2
18. 
1 3 
2

3
5 
10
14
3 
5 10 
14
12
3x3 
 5x2  10x 
 14, R12
19. 1 2 0 
2 
3
2
2
0
2 2
0 
3
2x2  2x, R 
3
20. f(x)  x2 
 2
21. f(x)  x5  32
f(1)  (1)2 
 2
f(
2)  (
2)5  32
 1 
 2 or 
1; no
 
32  32 or 0;
yes
22. f(x)  x4 
 6x2  8
f(
2
)  (
2
)4 
 6(
2
 )2  8
 4 
 12  8 or 0; yes
23. f(x)  x3 
 x  6
f(2)  (2)3 
 2  6
 8 
 2  6 or 12; no
24. f(x)  4x3  4x2  2x  3
f(1)  4(1)3  4(1)2  2(1)  3
 4  4  2  3 or 13; no
25. f(x)  2x3 
 3x2  x
f(1)  2(1)3 
 3(1)2  1
 2 
 3  1 or 0; yes
26a-d.
r
1
3

2

8
1
1
4
2

6

1
1
2

4

4
2
1
5
8
8

2
1
1

4
0
d

32.

33.

34.

35.

36.

37.

38.

27. (
6
)4 
 36  36 
 36 or 0
28. 
1 1
7 
1 
7

1 
6
7
1
6 
7
0
x2  6x 
 7  (x 
 1)(x  7)
(x 
 1)(x  1)(x  7)
29. 2 1 1 
4 
4
2
6
4
1 3
2
0
x2  3x  2  (x  1)(x  2)
(x 
 2), (x  1), (x  2)
30. 1 1 
1 
49
49
1
0 
49
1
0 
49
0
x2 
 49  (x 
 7)(x  7)
(x 
 1), (x 
 7), (x  7)

39.

40.

109

Chapter 4

1

44b.

41. d(t)  v0t  2at2
1
25  4t  2(0.4)t2
0  0.2t2  4t 
 25
5 0.2
4 
25
1
25
0.2
5  0
t
50
t5s
42. 1 1 1 
7
a
b
1
2

5

5  a
1 2 
5 
5  a 
5  a  b

V (x )

20O
15O
10O

V (x ) 

2x 3  38x 2  180x

50
4
8
12 x
O
3
2
44c. V(x)  2x 
 38x  180x
224  2x3 
 38x2  180x
44d. 224  2x3 
 38x2  180x
0  2x3 
 38x2  180x  224
0  x3 
 19x2  90x  112
2 1 
19
90
112
2 
34 
112
1 
17
56
0
2 in.
45. P(3  4i)  0 and P(3 
 4i)  0 implies that these
are both roots of ax2  bx  c. Since this
polynomial is of degree 2 it has only these two
roots.
x  3  4i
x 
 3  4i
(x 
 3)2  
16
2
x 
 6x  9 16
x2 
 6x  25  0
a  1, b  
6, c  25
46. r2  5r 
 8  0
r2  5r  8
25
25
2
r  5r  4  8  4


2 1

2 
5 
5  a

5  a  b

2
0
10

10 
 2a
1
0 
5
5  a 
15 
 a  b

5  a  b  0

5  a  b  0

15 
 a  b  0

5  a  10  0

20
 2b  0
a50
2b  20
a  
5
b  10
43a. V(x)  (3 
 x)(4 
 x)(5 
 x)
V(x)  (12 
 7x  x2)(5 
 x)
V(x)  
x3  12x2 
 47x  60
43b. V (x )
V (x ) 

x 3  12x 2  47x  60

O

25O

x

2

r  52

43c. V  
 w h
V  3 4 5 or 60
3
3
V  (60)
5
5

5

57

 4


57


r  2   
2
5


57


r  
 2  2

 36
V(x)  
x3  12x2 
 47x  60
36  
x3  12x2 
 47x  60
43d. 36  
x3  12x2 
 47x  60
0  
x3  12x2 
 47x  24

47a. f(x)  x4 
 4x3 
 x2  4x
f(2)  (2)4 
 4(2)3 
 (2)2  4(2)
f(2)  16 
 32 
 4  8 or 
12; no
47b. f(0)  (0)4 
 4(0)3 
 (0)2  4(0)
f(0)  0 
 0 
 0  0 or 0; yes
47c. f(
2)  (
2)4 
 4(
2)3 
 (
2)2  4(
2)
f(
2)  16  32 
 4 
 8 or 36; no
47d. f(4)  (4)4 
 4(4)3 
 (4)2  4(4)
f(4)  256 
 256 
 16  16 or 0; yes
48. f(x)  x5 
 32

[
3, 10] sc11 by [
200, 100] sc125
about 0.60 ft
1

44a. 
  2(20 
 2x) or 10 
 x
w  18 
 2x
h x
V(x)  (10 
 x)(18 
 2x)(x)
V(x)  (180 
 38x  2x2)(x)
V(x)  2x3 
 38x2  180x

Chapter 4

(0, 
32); point of inflection
[
4, 4] sc11 by [
50, 10] sc110
49. wider than parent graph and moved 1 unit left

110

50. Let x  number of 100 foot units of Pipe A and y 
number of 100 foot units of Pipe B.
4x  6y  48
2x  2y 
 18
y
2x  2y  18
2x  y 
 16
(0, 8)
2x  y  16
(3, 6) 4x  6y 
 48
x0
y0

4-4
Page 232

O (0, 0) y 
 0 (8, 0) x
P(x, y)  34x  40y
P(0, 0)  34(0)  40(0) or 0
P(0, 8)  34(0)  40(8) or 320
P(3, 6)  34(3)  40(6) or 342
P(7, 2)  34(7)  40(2) or 318
P(8, 0)  34(8)  40(0) or 272
3 
 100 foot units of A, or 300 ft of A
6 
 100 foot units of B, or 600 ft of B
51. 4x  2y  3z  6
4x  2y  3z  6
2x  7y  3z
→ 2x  7y 
 3z  0

3x 
 9y  13  
2z

3x 
 9y  2z  
13
4x  2y  3z  6
2x  7y 
 3z  0
6x  9y
6
2(2x  7y 
 3z)  2(0)
3(
3x 
 9y  2z)  3(
13)
↓
4x  14y 
 6z  0

9x 
 27y  6z  
39

5x 
 13y
 
39
5(6x  9y)  5(6)
6(
5x 
 13y)  6(
39)
↓
30x  45y  30

30x 
 78y  
234

33y  
204
68
y  11
6x  9y  6

6x  9

68

11

Page 233

6
546

6x  
1
1
91

x  
11

Check for Understanding

1. possible values of p: 1, 2, 3, 6
possible values of q: 1
possible rational roots: 1, 2, 3, 6
2. If the leading coefficient is 1, then q must equal 1.
p
p
Therefore, q becomes 1 or p, and p is defined as a
factor of an.
3. Sample answer: f(x)  x3 
 x2  x 
 3;
f(
x)  (
x)3 
 (
x)2  (
x) 
 3
f(
x)  
x3 
 x2 
 x 
 3; 0
3 or 1 possible positive zeros and no possible
negative zeros
4. Sample answer: You can factor the polynomial,
graph the function, complete the square, or use
the Quadratic Formula if it is a second-degree
function, or use the Factor Theorem and the
Rational Root Theorem. I would factor the
polynomial if it can be factored easily. If not and it
is a second-degree function, I would use the
Quadratic Formula. Otherwise, I would graph the
function on a graphing utility and use the
Rational Root Theorem to find the exact zeros.

4x  2y  3z  6

91
4(
11)

Graphing Calculator Exploration

1. 3; 1, 
1, 
2
2. 2; 
1, 2
3. (1) 1 positive;
f(
x)  (
x)4  4(
x)3  3(
x)2 
 4(
x) 
 4
f(
x)  x4 
 4x3  3x2  4x 
 4; 3 or 1
(2) 1 positive; f(
x)  (
x)3 
 3(
x) 
 2
f(
x)  
x3  3x 
 2; 2 or 0
4. In the first function, there are 2 negative zeros,
but according to Descartes’ Rule of Signs, there
should be 3 or 1 negative zeros. This is because
the 
2 is a double zero. In the second function,
there is one negative zero, but according to
Descartes’ Rule of Signs, there should be 2 or 0
zeros. This is because 
1 is a double root.
5. One number represents two zeros of the function.

(7, 2)

x
0

The Rational Root Theorem

 211  3z  6
68

294

3z  11
98

z  11


9111 , 6181, 9118 

5.


7 
 2 2  6

52. M2, 2 or (
4.5, 4)

2 
 2 6 
 3

N2, 2 or (
2, 1.5)
4 
 1.5


slope of 
MN


4.5 
 (
2) or 
1
2 
 (
3)


slope of 
RI  

7 
 (
2) or 
1

p
1,
q

2

r
1

1
2

2

1
1
1
1
1


4

3

5

2

6

1

2
6

3
13

2
0

4

4

24

rational root1

RI.
Since the slopes are the same, M
N
L
53. a b
a b
a b
ac  bc
ac bc
a
c b
c
I. true
II. true
III. false
The correct choice is D.

111

Chapter 4

6. p: 1, 3
q: 1, 2
P
:
q

Pages 234–235
1

10.

3

1, 3, 2, 2
r
1

1

2
2
2

3
5
1


8

3

9

3
0
12

1

2

2

4


6

0


2

2

2


6


1.5

3

3

2
2

9

3

19
1

60
0

1

p

q:

Exercises

1, 2, 3, 6
r
1
2

1
1
1

11.

p
:
q

1, 2, 3, 6, 9, 18

1

r

1

2

7. 2 or 0; f(
x)  8(
x)3 
 6(
x)2 
 23(
x)  6
f(
x)  
8x3 
 6x2  23x  6; 1
1

3

1

3

1

3

1, 2, 3, 6, 2, 2, 4, 4, 8, 8
r
1
2


6
2
10

8
8
8


23

21

3

8x2  10x 
 3  0
(4x 
 1)(2x  3)  0
4x 
 1  0
4x  1
1
x  4
1 1

12, 4,

6

15
0

12.

2x  3  0
2x  
3
3
1
x  
2 or 
12

7
8
6
4
2

7
15
1

5

3

18
14
0


5

4

9
5


7

2

5
1


2
0

1, 2

r
2

2

1
1
1
1
1

1
4
9

1
1

2
0

x3 
 4x2  5x 
 2

4

2

1
1

x2 
 2x  1  0
(x 
 1)(x 
 1)  0
x  1, x  1
rational roots: 1, 2


15
0

16
0
0

13.

p
:
q

1, 2, 4, 5, 10, 20
r
1

1
2


5, 
3, 1
9. r2  152 
 x2


5

4

6

3

1
1
1
1


4

8
2

10

20
12
18
0

x2 
 3x 
 10  0
(x 
 5)(x  2)  0
x  5, x  
2
rational roots: 
2, 2, 5
14. p: 1, 3
q: 1, 2

1

V  3r2h
1

1152  3(152 
 x2)(15  x)
3456  (152 
 x2)(15  x)
3456  3375  225x 
 15x2 
 x3
3
2
x  15x 
 225x  81  0
Possible rational roots: 1, 9, 81
f(x)  x3  15x2 
 225x  81  0
f(1)  
128
f(
1)  320
f(9)  0
f(
9)  2592
f(81)  611,712
f(
81)  
414,720
x represents 9 cm.

Chapter 4

p
:
q

r
1

1, 3, 5, 15
r
1

1

3

5


2

3

4

1
1
1

x2 
 4x  9  0
does not factor
rational root: 
2

8. 1; f(
x)  (
x)3  7(
x)2  7(
x) 
 15
f(
x)  
x3  7x2 
 7x 
 15; 2 or 0
p
:
q


6

8
0

x2  4x  3  0
(x  3)(x  1)  0
x  
3, x  
1
rational roots: 
3, 
1, 2

rational roots: 
3, 2, 1

p
:
q


5

2
3

2
3
4

p
:
q

1

r

2


1

0


6

3

1

2

2

0

0


6

0

2x3 
 6  0
x3  3
3
x  
3

1
rational root: 2

112

3

1, 3, 2, 2

15. p: 1
q: 1, 2, 3, 6
p
:
q

1

1

20. 2 or 0 positive
f(
x)  
10x3 
 17x2  7x  2
1 negative

1

1, 2, 3, 6
r

6

35


1


7


1

1

2

6

38

18

2

0


17


7

2

10


22

4

0

6

38

18

2

6

36

6

0

rational zeros: 
2, 5, 2

1 1

1


3

21. 2 or 0 positive
f(
x)  x4 
 2x3 
 9x2  2x  8
2 or 0 negative

6x2  36x  6  0
x2  6x  1  0
does not factor
1 1
rational roots: 
3, 2

r
1

16. 4; 3 or 1;
f(
x)  (
x)4 
 2(
x)3  7(
x)  4(
x) 
 15
f(
x)  x4  2x3 
 7x 
 4x 
 15; 1 negative
1 positive
17. f(
x)  
x3  7x 
 6
0 or 2 negative
r
1

1

1
1
1

0
1

1


7

6

6

1
1

3
6


10
8

1
1


9

6

2
3


2

8

8
0

x3  3x2 
 6x 
 8
r

1

1
1

3
2


6

8


8
0

x2  2x 
 8  0
(x  4)(x 
 2)  0
x  
4, x  2
rational zeros: 
4, 
1, 1, 2
22. 2 or 0 positive
f(
x)  x4 
 5x2  4
2 or 0 negative


6

12
0

x2 
 x 
 6  0
(x 
 3)(x  2)  0
x  3, x  
2
rational zeros: 
2, 
1, 3
18. 1 positive
f(
x)  
x3 
 2x2  8
1 negative
f(x)  x3 
 2x2 
 8x
0  x(x2 
 2x 
 8)
 x(x 
 4)(x  2)
x  0, x  4, x  
2
rational zeros: 
2, 0, 4
19. 1 positive
f(
x)  
x3  3x2  10x 
 24
2 or 0 negative
r
3

10

1

2

10x2 
 22x  4  0
5x2 
 11x  2  0
(5x 
 1)(x 
 2)  0
1
x  5, x  2

6x3  38x2  18x  2
r

r

r
1

1
1


5

4

0
1

0

4

4
0

x3  x2 
 4x 
 4
r

1

1
1

1
0


4

4

x2 
 4  0
(x 
 2)(x  2)  0
x  2, x  
2
rational zeros: 
2, 
1, 1, 2
23a. f(x)  (x 
 2)(x  2)(x  1)2
0  (x 
 2)(x  2)(x  1)2
x
20
x20
x2
x  
2

24
14


4
0

(x  1)2  0
x10
x  
1

23b. f(x)  (x 
 2)(x  2)(x  1)2
f(x)  (x2 
 4)(x2  2x  1)
f(x)  x4  2x3 
 3x2 
 8x 
 4
23c. 1 positive
f(
x)  x4 
 2x3 
 3x2  8x 
 4
3 or 1 negative
23d. There are 2 negative zeros, but according to
Descartes’ Rule of Signs, there should be 3 or 1.
This is because 
1 is actually a zero twice.
24a. Let 
  the length.
w

4
h  2
 
 1
V(
)  
 w h
V(
)  
(
 
 4)(2
 
 1)
V(
)  (
2 
 4
)(2
 
 1)
V(
)  2
3 
 9
2  4


x2  6x  8  0
(x  4)(x  2)  0
x  
4, x  
2
rational zeros: 
4, 
2, 3

113

Chapter 4

24b. V(
)  2
3 
 9
2  4

2208  2
3 
 9
2  4

24c. 2208  2
3 
 9
2  4

0  2
3 
 9
2  4
 
 2208
r
12

2
2


9
15


2208
0

4
184

w

4
h  2
 
 1
w  12 
 4 or 8
h  2(12) 
 1 or 23
12 in. 8 in. 23 in.
Sample answer: x4  x3  x2  x  3  0
Sample answer: x3 
 x2 
 2  0
Sample answer: x3 
 x  0
Let 
  the length.
h

9
1
V(
)  3Bh

Page 235

1
1

V(
)  3
3 
 3
2
1

26b. V(
)  3
3 
 3
2

25

2

x  52

1

26c. 6300  3
3 
 3
2

5

0
630


5  25


5  25


18,900
0

0

10,000

b

39  
441


2(6)
39  21

b  1
2
39  21

b  1
2

39 
 21

b  1
2

or

3

b5

b  2

5. 
2 1

3 
2 
8

2 
2
8
1
1 
4
0
x2  x 
 4
6. 4 1 
4 2 
6
4 0
8
1
0 2
2
2; no
6
7. 1 1 
2 
5
1 
1 
6
1 
1 
6
0
x2 
 x 
 6  (x 
 3)(x  2)
(x 
 3)(x 
 1)(x  2)

1,000,000
0

x  100 ft
28. The graphs are reflections of each other over the
x-axis. The zeros are the same.
29. 
7 1 
1 
56

7
56
1 
8
0
x
8
30. b2 
 4ac  62 
 4(4)(25)
 
364; 2 imaginary
31. (x 
 1)(x 
 (
1))(x 
 2)(x 
 (
2))  0
(x 
 1)(x  1)(x 
 2)(x  2)  0
(x2 
 1)(x2 
 4)  0
x4 
 5x2  4  0
32. y  4.3x 
 8424.3
y  4.3(2008) 
 8424.3
y  $210.10

Chapter 4

x  2

x  10
x  
15
2
2
4. b 
 4ac  (
39) 
 4(6)(45)
 441; 2 real roots

h

9
h  30 
 9 or 21
base: 30 in. by 30 in., height: 21 in.
27. d  0.0000008x2(200 
 x)
0.8  0.0000008x2(200 
 x)
0  0.00016x2 
 0.0000008x3 
 0.8
0  8x3 
 1600x2 
 8,000,000
0  x3 
 200x2 
 1,000,000

200

100


5 
 25

x  2


  30

1
1

25

x  2

0  
3 
 9
2 
 18,900

r
100

625

 4

x  2   2

1

0  3
3 
 3
2 
 6300

9
21

25

x2  5x  4  150  4

1

6300  3
3 
 3
2

1
1

Mid-Chapter Quiz

1. (x 
 1)(x 
 (
1))(x 
 2i)(x 
 (
2i))  0
(x 
 1)(x  1)(x 
 2i)(x  2i)  0
(x2 
 1)(x2  4)  0
x4  3x2 
 4  0
3
2
2. 3; x 
 11x  30x  0
x(x2 
 11x  30)  0
x(x 
 6)(x 
 5)  0
x0
x
60
x
50
x6
x5
3.
x2  5x  150

V(
)  3(
2)(
 
 9)

r
30

3
x

 2

2(2x 
 3)  x(3 
 x)
4x 
 6  3x 
 x2
x2  x 
 6  0
(x  3)(x 
 2)  0
x30
x
20
x  
3
x2
The correct choice is A.


  12

25a.
25b.
25c.
26a.

2x 
 3

x

33.

8.

p
:
q

1, 3
r

3

1
1

6
3

x2  3x  1  0
does not factor
rational root: 
3

114

10
1

3
0

4. Nikki; the sign changes between 
2 and 
1.
5. r
1

4

2

2
1

6
10

1
1

5
3
0
1

4

2
1
1

3

5
2
1

2

6
3
1

1

5
4
1
0

2
5
1
1
3
4 and 5, 
1 and 0
6. r
1

3

2
4

2
1

5
8

12

1
1

4
2
2
0
1

3

2
4
1
1

2

4
0←
2
1

1

4

4
3
1
0

2

2
4
1
1
2
12

2 and 
1, at 1, 3 and 4
2

4
0

3
7. r
0
2

4
0

3
1
2

2

2

5
2
2
0
0

3
3
2
2
6
15
approximate zero: 2.3
1
3
2
8. r

2
1
1
0

1
1
2
0
zeros: 
2, 
1
9. Sample answer:
r
1
0
0

8
2
1
1
1
1

7

5
2
1
2
4
0
2
upper bound: 2
f(
x)  x4  8x  2
r
1
0
0
8
2
0
1
0
0
8
2
lower bound: 0
10. Sample answer:
r
1
0
1
0

3
1
1
1
2
2

1
2
1
2
5
10
17
upper bound: 2
f(
x)  x4  x2 
 3
r
1
0
1
0

3
1
1
1
2
2

1
2
1
2
5
10
17
lower bound: 
2
11a. Let x  amount of increase.
V(x)  (25  x)(30  x)(5  x)
V(x)  (750  55x  x2)(5  x)
V(x)  x3  60x2  1025x  3750

9. 1 positive
F(
x)  x4 
 4x3  3x2  4x 
 4
3 or 1 negative
r
1

1
1

4
5


4
4

3
8


4
0

x3  5x2  8x  4  0
r

1

1
1

5
4

8
4

4
0

x2  4x  4  0
(x  2)(x  2)  0
x  
2, x  
2
rational zeros: 
2, 
1, 1
10. Let r  radius.
hr6
1
V  3r2h
1

27  3r2(r  6)
1

0  3r3  2r2 
 27
0  r3  6r2 
 81
r
3

1
1

6
9


81
0

0
27

r3

hr6
h  3  6 or 9

r  3 cm, h  9 cm

Locating Zeros of a Polynomial
Function

4-5

Pages 239–240

Check for Understanding

1. If the function is negative for one value and
positive for another value, the function must cross
the x-axis in at least one point between the two
values.
f (x )
(b, f (b ))

f (b )

O

a

b

x

f (a )
(a, f (a))

2. Use synthetic division to find the values of the
polynomial function for consecutive integers.
When the values of the function change from
positive to negative or from negative to positive,
there is a zero between the integers.
3. Use synthetic division to find the values of the
polynomial function for consecutive integers. An
integer that produces no sign change in the
quotient and the remainder is an upper bound.
To find a lower bound of a function, find an upper
bound for the function of 
x. The lower bound is
the negative of the upper bound for the function
of 
x.

115

Chapter 4

11b. V  
 w h
1.5V  1.5(3750)
V  25(30)(5)
1.5V  5625
V  3750
V(x)  x3  60x2  1025x  3750
5625  x3  60x2  1025x  3750
11c. 5625  x3  60x2  1025x  3750
0  x3  60x2  1025x 
 1875
r
1
60
1025

1875
1
1
61
1086

789
2
1
62
1149
423
x  1.7
25  x  25  1.7
30  x  30  1.7
 26.7
 31.7
5  x  5  1.7
 6.7
about 26.7 cm by 31.7 cm by 6.7 cm

Pages 240–242

17. r
2
0
1

3
3

3
2

6
19

60
183

2
2

4
9

21
45

1
2

2
3

6
9
0
2
0
1

3
3
1
2
2
3
0
3
2
2
4
9
16
35
no real zeros
18. r
6
24

54

3

6
6

12
18

111

5
6

6

24
117
yes; f(
6)  
111, f(
5)  117
19–25. Use the TABLE feature of a graphing
calculator.
19. 
0.7, 0.7
20. 
2.6, 
0.4
21. 
2.5
22. 
0.4, 3.4
23. 
1, 1
24. 
1.3, 0.9, 7.4
25. 
1.24
26. Sample answers:
r
3

2
5

1
1
3
1
6
5
upper bound: 1
f(
x)  
3x3 
 2x2 
 5x 
 1
r

3

2

5

1
0

3

2

5

1
lower bound: 0
27. Sample answers:
r
1

1

1
1
1
0

1
2
1
1
1
upper bound: 2
f(
x)  x2  x 
 1
r
1
1

1
1
1
2
1
lower bound: 
1
28. Sample answers:
r
1

6
2
6

13
1
1

5

3
3

10
2
1

4

6

6

25
3
1

3

7

15

58
4
1

2

6

18

85
5
1

1

3

9

58
6
1
0
2
18
95
upper bound: 6
f(
x)  x4  6x3  2x2 
 6x 
 13
r
1
6
2

6

13
1
1
7
9
3

10
2
1
8
18
30
47
lower bound: 
2

Exercises

12. r
1
0
0

2

1
1

1
1

3
0
1
0
0

2
1
1
1
1

1
2
1
2
4
6
3
1
3
9
25
1 and 2
13. r
2

5
1

1
2

7
8
0
2

5
1
1
2

3

2
2
2

1

1
3
2
1
4
0 and 1, 2 and 3
1

2
0
1

2
14. r

2
1

4
8

15
28

1
1

3
3

2
0
0
1

2
0
1

2
1
1

1

1
0

1
2
1
0
0
1
0
at 
1, at 2
15. r
1
0

8
0
10

3
1

3
1

3
19

2
1

2

4
8

6

1
1

1

7
7
3
0
1
0

8
0
10
1
1
1

7

7
3
2
1
2

4

8

6
3
1
3
1
3
19

3 and 
2, 
2 and 
1, 1 and 2, 2 and 3
1
0

3
1
16. r

2
1

2
1

1

1
1

1

2
3
0
1
0

3
1
1
1
1

2

1
2
1
2
1
3

2 and 
1, 0 and 1, 1 and 2

Chapter 4

116

32f. 
1.4, 3.4 (Use TABLE feature of a graphing
calculator.)
33a. 1890: P(0)  
0.78(0)4  133(0)3 
 7500(0)2
 147,500(0)  1,440,000
 1,440,000
1910: P(20)  
0.78(20)4  133(20)3 
 7500(20)2
 147,500(20)  1,440,000
 2,329,200
1930: P(40)  
0.78(40)4  133(40)3 
 7500(40)2
 147,500(40)  1,440,000
 1,855,200
1950: P(60)  
0.78(60)4  133(60)3 
 7500(60)2
 147,500(60)  1,440,000
 1,909,200
1970: P(80)  
0.78(80)4  133(80)3 
 7500(80)2
 147,500(80)  1,440,000
 1,387,200
The model is fairly close, although it is less
accurate at for 1950 and 1970.
33b. 1980 
 1890  90
P(90)  
0.78(90)4  133(90)3 
 7500(90)2
 147,500(90)  1,440,000
P(90)  
253,800
33c. The population becomes 0.
33d. No; there are still many people living in
Manhattan.
34. Sample answer:
f(x)  (x 
 
2
)(x  
2
)(x  1)
f(x)  (x2 
 2)(x  1)
f(x)  x3  x2 
 2x 
 2; 
2
, 
1

29. Sample answers:
r
1
5

3

20
1
1
6
3

17
2
1
7
11
2
upper bound: 2
f(
x)  
x3  5x2  3x 
 20
r

1
5
3

20
1

1
4
7

13
2

1
3
9

2
3

1
2
9
7
4

1
1
7
8
5

1
0
3

5
6

1

1

3

38
lower bound: 
6
30. Sample answers:
r
1

3

2
3

5
1
1

2

4

1

6
2
1

1

4

5

15
3
1
0

2

3

14
4
1
1
2
11
39
upper bound: 4
f(
x)  x4  3x3 
 2x2 
 3x 
 5
r
1
3

2

3

5
1
1
4
2

1

6
2
1
5
8
13
21
lower bound: 
2
31. Sample answers:
r
1
5

3
20
0

15
1
1
6
3
23
23
8
upper bound: 1
f(
x)  
x5  5x4  3x3  20x2 
 15
r

1
5
3
20
0

15
1

1
4
7
27
27
8
2

1
3
9
38
76
137
3

1
2
9
47
141
408
4

1
1
7
48
192
753
5

1
0
3
35
175
860
6

1

1

3
2
12
57
7

1

2

11

57

399

2808
lower bound: 
7
32a. 4
32b. 1, 5
32c. 3 or 1; f(
x)  x4  3x3 
 2x2 
 3x 
 5
1 negative real zero
r
1

3

2
3

5
32d.
5
1
2
8
43
210
4
1
1
2
11
39
3
1
0

2

3

14
2
1

1

4

5

15
1
1

2

4

1

6
0
1

3

2
3

5

1
1

4
2
1

6

2
1

5
8

13
21

3
1

6
16

45
130

2 and 
1, 3 and 4
32e. Sample answers:
upper bound: 4 (See table in 32d.)
f(
x)  x4  3x3 
 2x2 
 3x 
 5
r
1
3

2

3

5
1
1
4
2

1

6
2
1
5
8
13
21
lower bound: 
2

f (x )
f (x ) 
 x 3 
x 2  2x  2

x

O

35a. 37.44  60x3  60x2  60x
35b. f(x)  60x3  60x2  60x 
 37.44
35c.

f (x )

f (x ) 
 4O
60x 3  60x 2  60x  37.44
2O

2 
1 O

20

1

2x


40
1

about 2
35d. 0.4 (Use TABLE feature of a graphing
calculator.)
36. Sample answer: f(x)  x2 
 1

117

Chapter 4

37a.

f (x )
100,000

f (x ) 

0.125x 5  3.125x 4  4000

42.

7 9
 7(6) 
 3(9) or 15
3 6

3  8 
2  4

43. (x, y)  2, 2

80,000

 (2.5, 1)
44. x 
 2y 
 4  0

60,000

1

40,000
2O,000

O

45.
4

8

x

Page 247

4.


b  

b2 
 4
ac

2a
4  

(
4)2 

 4(4.9)(
1
750)


2(4.9)
4  
34,31
6


9.8
4  
34,31
6

4 
 
34,31
6


t  
9.8
9.8

41.

2

x(x 
 2)  2  2(x 
 2)  2
x2 
 2x  2  2x 
 4  2
x2 
 4x  4  0
(x 
 2)(x 
 2)  0
x
20
x
20
x2
x2
If you solve the equation, you will get x  2.
However, if x  2, the denominators will equal 0.

4.9t2 
 4t 
 1750  0

t  19.3
about 19.3 s

2



x
x
2  2  x
2

2
2


x  
x 
 2 (x 
 2)  2  x 
 2 (x 
 2)

1

t

Check for Understanding

1. Multiply by the LCD, 6(b 
 2). Then, solve the
resulting equation.
2. If a possible solution causes a denominator to
equal 0, it is not a solution of the equation.
3. Decomposing a fraction means to find two
fractions whose sum or difference equals the
original fraction.


1750  4t 
 2(9.8)t2

t

Rational Equations and Partial
Fractions

4-6

1

t

y

C
D
B
y  x cannot be true.
The correct choice is B.

d(t)  v0t 
 2gt2

40.

A

12 16 20 24 x

37b. f(0)  
0.125(0)5  3.125(0)4  4000
f(0)  4000 deer
37c. 1920 
 1905  15
f(15)  
0.125(15)5  3.125(15)4  4000
f(15)  67,281.25
about 67,281 deer
37d. in 1930
38a. 81.58  6x4  18x3  24x2  18x
38b. 81.58  6x4  18x3  24x2  18x
0  6x4  18x3  24x2  18x 
 81.58
about 1.1 (Use TABLE feature of a graphing
calculator.)
38c. x  rate  1
1.1  rate  1
0.1  rate
about 10%
39. 2 or 0; f(
x)  
2x3 
 5x2  28x  15
1 negative zero
r
2

5

28
15

3
2

11
5
0
2x2 
 11x  5  0
(2x  1)(x 
 5)  0
x  0.5
x5
rational zeros: 
3, 0.5, 5

t

1

y  2x 
 2; 2; 
2

5.

5

b 
 b  4

b 
 5bb  (4)b
b2 
 5  4b
b2 
 4b 
 5  0
(b 
 5)(b  1)  0
b
50
b5

t  
18.5

y
6.

9

b5

b10
b  
1
3



b
3

9
3



b  5 (b  5)(b 
 3)   b 
 3 (b  5)(b 
 3)

O

Chapter 4

y 
 x 4x 1

9(b 
 3)  3(b  5)
9b 
 27  3b  15
6b 
 42  0
b7

x

118

7.
t4


t 

3
t4
  
t
4
t
3
 (t)(t 
 4)
t
4




16

Pages 247–250



t2 
 4t


16


 
t2 
 4t (t)(t 
 4)

12.



(t  4)(t 
 4)  3(t)  
16
t2 
 16  3t  
16
t2  3t  0
t(t  3)  0
t0
t30
t  
3
But t 0, so t  
3.
3p 
 1


p2 
 1
3p 
 1


p2 
 1

8.

1

1

16
;
x

exclude: 0

1

But m

1

16

1

6

16 false

1

16

4

1

5

2y  3(y  2)  
y2
5y  6  
y2
y2  5y  6  0
( y  3)( y  2)  0
y30
y20
y  
3
y  
2
But y 
2, so y  
3.

10  (2n 
 5)(n  1)  (2n  5)(n 
 1)
2n2 
 3n  5  2n2  3n 
 5

6n  
10
5

n  3

3

5

7
 6
7
 6

false

11b.
3

3a



2(a  1)

4(a 
 1)7a 
 3(a  1)5  6(a 
 1)3a
28a2 
 28a 
 15a 
 15  18a2 
 18a
10a2 
 25a 
 15  0
2a2 
 5a 
 3  0
(2a  1)(a 
 3)  0
2a  1  0
a
30

true

 57.14

3 60  20

3x

3a



2a  2

7a
5



3(a  1) 
 4(a 
 1) (12)(a 
 1)(a  1) 
3a


2(a  1) (12)(a 
 1)(a  1)

Solution: a  1, a  31
3 60  20

3x

7a
5
 
 
3a  3
4a 
 4
7a
5
 
 
3(a  1)
4(a 
 1)

17.

7

7
5



36 
 1  6
1
7
1  7  6
48
49
  
42
42

3

b  1  b  1  3(b  2)
2b  2  3b  6

4  b

7

3

1




b2  b1

1
1
3




b  2  b  2 (b  2)(b  1)   b  1 (b  2)(b  1)


2  6 true

6

1

b2

16.

7

5

2
1

2n  5

2



Test a  
1: 1  

1 
 1  6

11a.

2n 
 5




n
1  n1

10
2n 
 5
2n  5




n 
 1  n 
 1 (n 
 1)(n  1)   n  1 (n 
 1)(n  1)

6(a 
 1)  30  7(a 
 1)
6a 
 6  30  7a 
 7
31  a

Test a  36: 1

10


n2 
 1

15.



1
a 
 1  6 ; exclude: 1

Test a  2: 1 


y

4 true

Solution: x  0, x
10.

3


 y  
y2

2

y
3




y  2  y (y)(y  2)   y  2 (y)(y  2)


16 true

4

m  34  0
m  
34

0, so m  
34.
2

y2

14.

16

(
1)


Test x  
1: 5  
(
1)

54

m 
 34



2m2

m2 
 34m  2m2
0  m2  34m
0  m(m  34)
m0

5x  1  16
5x  15
x3

Test x  4: 5  4

t
20
t2

2

2

Test x  1: 5  1

 (0)t

m 
 34
2

m12m2  
2m 2m

B




p1  p
1




p1  p
1

9. 5  x

1

m

13.

3p 
 1

3p 
 1  A(p 
 1)  B(p  1)
Let p  1.
3(1) 
 1  A(1 
 1)  B(1  1)
2  2B
1B
Let p  
1.
3(
1) 
 1  A(
1 
 1)  B(
1  1)

4  
2A
2A
3p 
 1


p2 
 1



0

0
t2 
 8t  12  0
(t 
 6)(t 
 2)  0
t
60
t6



(p  1)(p 
 1)
A

Exercises

12
  t 
 8
t
12
  t 
 8 t
t
12  t2 
 8t

1

a  
2

 57.14

a3

60  20  57.14(3  x)
200  171.42  57.14x
0.50  x; 0.50 h

119

Chapter 4

1

a



1
1
a  a
1

18.

1(1 
 a)(a 
 1)  

1

1
a



a

a
1

23.

(1 
 a)(a 
 1)

a 
 1 
 a2  a  a 
 1  a(1 
 a)

a2  2a 
 1  
a2  2a 
 1
00
all reals except 1
19.



2q(2q 
 3) 
 2q(2q  3)  (2q  3)(2q 
 3)
4q2 
 6q 
 4q2 
 6q  4q2 
 9
0  4q2  12q 
 9

q




6m 
 9
1
  
3m
3m
6m 
 9
 (12m)
3m

31m 



24.

3m 
 3


4m

3m 
 3
(12m)
 
4m


4
7
3
    
x
1
2
x
x1

4
 (x 
 1)(2 
 x)(x
x
1





25.
7
3
  
 1)  
2
x
x1

x
2

5m 
 4


(m  2)(m 
 2)

 A  B
m2

m2


4y


3y2 
 4y  1

4y


3y2 
 4y  1




m
2


4y

(3y 
 1)(y 
 1)
A
B
  
3y 
 1
y
1

2A

2

 2  


4y


3y2 
 4y  1

26.

9 
 9x


x2 
 9
9 
 9x


x2 
 9




3y 
 1

y
1

9 
 9x

(x  3)(x 
 3)
A
B
  
(x  3)
(x 
 3)

Let x  3.
9 
 9(3)  A(3 
 3)  B(3  3)

18  6B

3  B
Let x  
3.
9 
 9(
3)  A(
3 
 3)  B(
3  3)
36  
6A

6  A

6

3
9 
 9x

    
2

1  
1
(2)(
1

 48)


2 2
1  
145


4

x3
x 
9

6

3
, 
x3 x
3

27a. a(a 
 6)

Chapter 4

m
2


43  
23A

22b. 
1, 2

(n  1)(n 
 2)  (n 
 2)(n  6)  4(n  1)
n2 
 n 
 2  n2  4n 
 12  4n  4
2n2 
 n 
 18  0


x


413  A13 
 1  B313 
 1

n6
4
(n  1)(n 
 2)  (n  1)(n 
 2)
1  
n1
n
2

n

x
2


4y  A(y 
 1)  B(3y 
 1)
Let y  1.

4(1)  A(1 
 1)  B(3(1) 
 1)

4  2B

2  B
Let y  13.

n6
4
  
1
n1
n
2

22c.

5m 
 4


m2 
 4
5m 
 4


m2 
 4

m 
4

(x 
 1)(2 
 x)(x  1)

4(2 
 x)(x  1)  7(x 
 1)(x  1)
 3(x 
 1)(2 
 x)

4(
x2  x  2)  7(x2 
 1)
 3(
x2  3x 
 2)
4x2 
 4x 
 8  4x2  9x 
 13
5  13x
5
  x
13

22a. (n  1)(n 
 2)

x

5m 
 4  A(m 
 2)  B(m  2)
Let m  2.
5(2) 
 4  A(2 
 2)  B(2  2)
6  4B
1.5  B
Let m  
2.
5(
2) 
 4  A(
2 
 2)  B(
2  2)

14  
4A
3.5  A
3.5
1.5
5m 
 4

    
2

4  4(6m 
 9)  3(3m 
 3)
4  24m 
 36  9m 
 9
15m  23
m  2135
21.

x(x 
 2)

 A  B

x 
 2x


12  
144


9)
4(4)(


2 4

12  
288


8

12  12
2


8


3  3
2

2

20.

x
6



x 
 6  A(x 
 2)  B(x)
Let x  2.
2 
 6  A(2 
 2)  B(2)

4  2B

2  B
Let x  0.
0 
 6  A(0 
 2)  B(0)

6  
2A
3A
x
6

2



 3  
2

2q
2q
 
   1
2q  3
2q 
 3
2q
2q
 
  (2q  3)(2q 
 3)  1(2q  3)(2q 
 3)
2q  3
2q 
 3



x
6


x2 
 2x
x
6


x2 
 2x

120

x
3

a
2

a

27b.



a
2

a

a
4


a
6

(a)(a 
 6)  

a
4

a
6

30.

(a)(a 
 6)

(a 
 2)(a 
 6)  (a 
 4)(a)
a2 
 8a  12  a2 
 4a
12  4a
3a

1 
 2


1

1
2

Test a  1: 
1


1
4
2

Test a  4: 
4

Test a  7:

1

2
7
2

7
5

7


1 
 4



1 
 6
5
3  7 false
1
4


1
6
3
 5 true
4
4


4
6

Test x 
Test x 
Test x 

 0 false
7
4


7
6

Test x 

 3 true

Solution: 0  a  3, 6  a
28.

2

w

3

29
;
w

Test w  1:

2

1

3

31.
29


1

9
  0 true
40
(
2)2 
 16

2: 
0
(
2 
 5)(
2  1)

12
  0 false
7
0 
 16


0
0: 
02 
 4(0) 
 5

16
  0 true

5
4.52 
 16
4.5: 
0
(4.5 
 5)(4.5  1)
4.25
  0 false

2.75
2
6 
 16
0
6: 
62 
 24 
 5
20
  0 true
7

1

4a
1

4a




5

8a
5

8a



1
;
2
1

2

7

4

29

1

Solution: w  0, w
29.

(x 
 3)(x 
 4)

(x 
 5)(x 
 6)2



Test a  
1: 
4(
1)  8(
1)

1

2

7

1

2

1

29

10

Test x 

Test x 

Test x 

5


8
Test a  1:

true

9

 0; exclude 5, 6

Test a  2:

(x 
 3)(x 
 4)  0
x
30
x
40
x3
x4
(0 
 3)(0 
 4)
Test x  0: 2  0
Test x 

exclude: 0

a

5 29 false
2
29
3

Test w  10: 
(10)
10
32

10

5

2  5  4a


29 true

1

 0; exclude 5, 
1

Solution: x  
4, 
1  x  4, x

exclude: 0

2  3w  29
w9
2
3
Test w  
1: 

1

0

0
x2  16
x  4
(
5)2 
 16
Test x  
5: 
0
(
5 
 5)(
5  1)

27c. 0, 6
27d. Test a  
1:

x2 
 16


x2 
 4x 
 5
x2 
 16

(x 
 5)(x  1)
x2 
 16

Solution: 0

5

8(1)

1

2

7

8
1
5
  
4(2)
8(2)
7

16
7
 a  4

1

2
1

2
1

2

1

4(1)



false

true

false

(0 
 5)(0 
 6)
12
  0 true

180
(3.5 
 3)(3.5 
 4)
3.5: 2  0
(3.5 
 5)(3.5 
 6)

0.25
  0 false

9.375
(4.5 
 3)(4.5 
 4)
4.5: 2  0
(4.5 
 5)(4.5 
 6)
0.75
  0 true

1.125
(5.5 
 3)(5.5 
 4)
5.5: 2  0
(5.5 
 5)(5.5 
 6)
3.75
  0 false
0.125
6.5 
 3)(6.5 
 4)
6.5:( 2  0
(6.5 
 5)(6.5 
 6)
8.75
  0 false
0.375

Solution: x  3, 4  x  5

121

Chapter 4

1

2b  1
1

2b  1

32.

1



b1



1

b1



8
;
15
8

15

1

exclude: 
2, 
1

15(b  1)  15(2b  1)  8(2b  1)(b  1)
45b  30  16b2  24b  8
0  16b2 
 21b 
 22
0  (16b  11)(b 
 2)
16b  11  0
b
20
11
b2
b  
16
1

1

8

15
8

15



Test b  
2: 
2(
2)  1  
2  1
4


3
1

1



Test b  
0.8: 
2(
0.8)  1  (
0.8)  1
10

3
1

2(
0.6)  1

1


Test b  
0.6:
(
0.6)  1
5

2
1
1
8



Test b  0: 
2(0)  1  0  1
15
8
2 1
5 true
1
1
8






Test b  3: 2(3)  1  3  1
15
11
8

 false
28
15
1
11
Solution: 
1  b  
16, 
2  b  2

33.

7

y1


6  2

14

0.30 false

62

0.30

8 0.30 true
Solution: x  
5 or x 5

true

36a.

false

1

8

1

1

 d  3
2
i

1

8

36b.

1

1

 d  3
2
i

18(32di)  d1  312 (32di)
i

4di  32  di
3di  32
2
di  103 cm
x

1



37. Sample answer: 
x
3  x2

5(x 
 3)  2x
5x 
 15  2x
3x  15
x  5 tons

7

39a.

7 true

1

1

1

 2r  r  2
0
1

10

1

1

1

 2r  r  2
0

110 (20r)  21r  1r  210 (20r)

7 false

2r  10  20  r
r  30
2r  2(30) or 60; 60 ohms, 30 ohms
40. Let x  the number of quiz questions to be
answered.

4x  x  105
2

20  5x2  52x
5x2 
 52x  20  0
(5x 
 2)(x 
 10)  0
5x 
 2  0
2
x  5

1

10

39b.

7

Solution: 
1  y  0
34. Let x  the number.
1


0.4

Test x  6: 
6
5

7 false

7

Test y  1:

0.30

38. Let x  capacity of larger truck.
5
x
  
2
x
3


Test y  
0.5: 

0.5  1
7

11
7

2

0.30 true

02

0
5

Test x  0:
false
8

15
8

15
8

15
8

15

0.30

0.36

7


7

 0.30; exclude 5


Test x  
6: 

6 
 5

7  7(y  1)
1y1
0y
Test y  
2:

0.30

x  2  0.30(x 
 5)
x  2  0.30x 
 1.5
0.7x  
3.5
x  
5

7; exclude 
1

7


2  1

x2

x
5
x2

x
5

35.

11  x

20  x

 0.70

11  x  0.70(20  x)
11  x  14  0.70x
0.3x  3
x  10 questions
41. Let x  the speed of the wind.

x 
 10  0
x  10

1062

200  x

738



200 
 x

1062(200 
 x)  738(200  x)
212,400 
 1062x  147,600  738x
64,800  1800x
36  x; 36 mph

Chapter 4

122

42.

48. 
5

1
1
1
    
b
c
a
1
1
   (a)(b)(c)
b
a





 c(a)(b)(c)
1

43a.

1

x
1

x

a

 2y  z
1 1

1

1

1

1

x
1

x

43b.


 23
0  45 
1

1

1

1

1


 6
0  90

 (360x)  
1

x

1

60



(360x)


3x  12
x  
4

? 2(6)  3

3  6
5

3  2 false
no
52. y2  121x2 → b2  121a2
52a. (
b)2  121a2
52b. b2  121(
a)2
b2  121a2 yes
b2  121a2 yes
2
2
52c. (a)  121(b)
52d. (
a)2  121(
b)2
a2  121b2 no
a2  121b2 no
53a. Let x  short answer questions and y 
essay questions.
y
20
x  y  20
2x  12y  60
16
x0
x  y 
 20
y0
12

15,000 m

20x  15,000
x  750 gallons
750 $1.20  $900
x $1.20  $900 
 $200
1
x  5833 gallons
Let y  number of miles per gallon.
15,000 m
ym
  
1
g
5833 g

8 2x  12y 
 60
(0, 5)
4
x
0

1

5833y  15,000
y  25.7; about 25.7 mpg
d

T  s
2

2x  3  0
3
x  
2

2x  3

 x 
g

45.

0
25
25

51. y  x

1

90

360  6x  4x
360  10x
36  x
44. Let x  number of gallons of gasoline.
20 m

g


30
25

5

50. 3x 
 12  0
3x  12
3x  12
x4


 23
0  45 
1

0

5

5

1
no
49. 2; 12x2  8x 
 15  0
(6x 
 5)(2x  3)  0
6x 
 5  0
5
x  6

bc  ac  ab
bc  ab 
 ac
bc  a(b 
 c)
bc

b
c

1

O

26

26



103  
s5  s
5

32(s  5)(s 
 5)  26(3)(s 
 5)  26(3)(s  5)
32s2 
 800  78s 
 390  78s  390
2
32s 
 156s 
 800  0
8s2 
 39s 
 200  0
(8s  25)(s 
 8)  0
8s  25  0
s
80
25

12

x
16 20
(20, 0)

S(x, y)  5x  15y
S(0, 0)  5(0)  15(0) or 0
S(0, 5)  5(0)  15(5) or 75
S(18, 2)  5(18)  15(2) or 120
S(20, 0)  5(20)  15(0) or 100
18 short answer and 2 essay for a score of
120 points

26
26


1032(3)(s  5)(s 
 5)  
s  5  s 
 5 (3)(s  5)(s 
 5)

s  
8

4
8
(0, 0) y 
 0

(18, 2)

s8

8 mph
46.

3x 
 5y

5y

3x

5y

 5y 
 5y
 11 
 1
 10

47.

r

3

2

1
0
1
2

1
1
1
1
1
1
1

2

1
0
1
2
3
4


3
0

3

4

3
0
5


5

5
1

1

5

5
5


3 and 
2, 
2 and 
1, 1 and 2

123

Chapter 4

53b. Let x  short answer questions and y 
essay questions.
y
20
x  y  20
x  y 
 20
2x  12y  120
16
x0
2x  12y 
 120
y0
12

4-7

Pages 254–255

(12, 8)

x
O

4
8
(0, 0) y 
 0

12

16 20
(20, 0)

S(x, y)  5x  15y
S(0, 0)  5(0)  15(0) or 0
S(0, 10)  5(0)  15(10) or 150
S(12, 8)  5(12)  15(8) or 180
S(20, 0)  5(20)  15(0) or 100
12 short answer and 8 essay for a score of 180
points
1
1
3
5 x
54.
1
1 
3

5
1(3)  1(
3)
1(5)  1(
5)  x
1(3)  1(
3)
1(5)  1(
5)
0
0
x
0
0
55.
y 
 y1  m(x 
 x1)
y 
 1  2(x 
 (
3))
y 
 1  2x  6
2x 
 y  7  0
3000 
 5000


56a. m  
20 
 60

56b. $2000; $50

m  50
y 
 3000  50(x 
 20)
y  50x  2000
C(x)  50x  2000
56c.

C (x )
$4000
$3000

C (x ) 
 50x

Cost

 2000

62 
 4  22  10
7

$2000

0

2
4
6
8
Televisions Produced

x

1

57. A of JKL  2(9)(7) or 31.5
1

A of small triangle  2(5)(3) or 7.5
A of shaded region  31.5 
 7.5 or 24
The answer is 24.

Chapter 4

7


21
4

   
7
0
 1

17
  
17

8. 
a
 4  
a



37

a

 4  7 
 
a



3
a 
 4  49 
 14
a


3a
3

42  
14
a

3

1764  196(a 
 3)
9a
3
12  a
Check:

a
 4  
a



37

12
   
12
4
3

7

16
  
9
7
437

$1000
0

Check for Understanding

1. To solve the equation, you need to get rid of the
radical by squaring both sides of the equation. If
the radical is not isolated first, a radical will
remain in the equation.
2. The process of raising to a power sometimes
creates a new equation with more solutions than
the original equation. These extra or extraneous
solutions do not solve the original equation.
3. When solving an equation with one radical, you
isolate the radical on one side and then square
each side. When there is more than one radical
expression in an equation, you isolate one of the
radicals and then square each side. Then you
isolate the other radical and square each side. In
both cases, once you have eliminated all radical
signs, you solve for the variable.
4. 
1

 4  2
t
Check:

1

 4  2
t
3
1 
 4t  4
1 
 4
4  2

4t  3
3


1
32
t  
4
22
3
3
5. 

x  4  12  3
Check: 

x  4  12  3
3
3


x  4  
9


733

 4  12  3
3
x  4  
729


729
  12  3
x  
733

9  12  3
33
6. 5  
x

42
Check: 5  
x

42

x

 4  
3

5  
13
4

2
x
49
5  
9
2
x  13
53 2
no real solution
7. 
6x

 4  
2x

0
 1
6x 
 4  2x  10
4x  14
x  3.5
Check:

6x

 4  
2x

0
 1

(0, 10)

8
x
0
4

Radical Equations and
Inequalities

124

16. 4

3m2 

15  4


3m2 

15  1
3m2 
 15  1
3m2  16

9. 
5x

48
5x  4  0
5x  4  64
5x  
4
5x  60
x  
0.8
x  12
Test x  
1: 
5(
1)

48


1
  8 meaningless
Test x  0: 
5(0)


48

4
  8 true
Test x  13: 
5(13)


48

69
  8 false
Solution: 
0.8  x  12
10. 3  
4a

   10
5
4a 
 5  0

4a

7
5
4a  5
4a 
 5  49
a  1.25
4a  54
a  13.5
Test a  0: 3  
4(0)



5  10
3  

5
  10 meaningless
Test a  2: 3  
4(2)



5  10
4  
3
  10 true
Test a  14: 3  
4(14)



5  10
3  
51
  10 false
Solution: 1.25  a  13.5
11a. v  

v02  
64h

16

m2  3
4

m  3
3

Check:

4

Check:

14. 
8n


12
5

Check: 
x
85

17
8
5

25
5
55
3
Check: 

y
74
3

71

74
3

64
4
44
Check: 
8n
5


12


8n

3
5

84 
 5 
 1  2

3

13. 

y
74
y 
 7  64
y  71

8n 
 5  9
8n  14
7
n  4

4
4
1
44
3m2 

15  4
4

2

 
 15  4
4 3
3
3
4

17.

18.

Exercises

12. 
x
85

x
 8  25
x  17

2

 
 15  4
4 33
3

90  

102 
64h
90  
100


64h
11b. 90  
100


64h Check: 90  
100


64h
8100  100  64h
90  
100

5)
64(12
8000  64h
90  
8100

1125  h; 125 ft
90  90 

Pages 255–257

4

3m2 

15  4

19.

7

5

14


12
3
12
22

20.


15. 
x
 16  
x  4
x  16  x  8
x  16
0  8
x
0  
x
0x
Check: 
x
 16  
x  4


0
 1  
0
6
4

16
4
44

125

4
4
1
44

9u

   
7u
4
20


9u 
 4  7u 
 20
2u  
16
u  
8
Check:

9u

   
7u
4
20



9(
8)


 4  
7(
8)


 20


76
  

76

no real solution
3

5
6u 
   2  
3
3

5
6u 
   
5
6u 
 5  
125
6u  
120
u  
20
3
Check:

5
6u 
   2  
3
3

6(
20
)

 5  2  
3
3


125
  2  
3

5  2  
3

3  
3 


4m2 

3m 
2 
 2m 
 5  0


4m2 

3m 
2  2m  5
4m2 
 3m  2  4m2  20m  25

23  23m

1  m
Check:


4m2 

3m 
2 
 2m 
 5  0
2 
 3(


4(
1)

1) 
2 
 2(
1) 
 5  0

9
2
50
3
30
00

k
 9 
 
k

  
3


k

 9  
3
  
k

k  9  3  2
3k
k
6  2
3k

36  4(3k)
36  12k
3k
Check: 
k
 9 
 
k

  
3


3

 9 
 
3
  
3


12
 
 
3
  
3

2
3
 
 
3
  
3


3
  
3


Chapter 4

26. 
2x
 1  
2x


65

2x
 1  5 
 
2x


6
2x  1  25 
 10
2x

 6  2x  6

30  
10
2x
6

3  
2x

6
9  2x  6
3  2x


a
1
 2 
 1  
a
2
 1
a  21 
 2
a
1
 2  1  a  12

2
a
 2  
10
1
4(a  21)  100
a  21  25
a4
Check: 
a
 2 
 1  
a
1
2
 1

4
1
 2 
 1  
4
2
 1

25
 
 1  
16

5
14
44
22. 
3x
 4 
 
2x



73

3x
 4  3  
2x



7
3x  4  9  6
2x


 7  2x 
 7
x  2  6
2x

7

x2  4x  4  36(2x 
 7)
x2 
 68x  256  0
(x 
 4)(x 
 64)  0
x
40
x 
 64  0
x4
x  64
Check:

3x
 4 
 
2x



73

3(4)


4 
 
2(4)



73

16
 
 
1
3
4
13
33
Check:

3x
 4 
 
2x



73

3(64)


4 
 
2(64)



73

196
 
 
121
3
14 
 11  3
33
3
23. 2
1
7b 
  
 4  0
3
2
1
7b 
   4
8(7b 
 1)  64
7b 
 1  8
7b  9
21.

3

2

Check:

3

3

2
1
7b 
  
 4  0
2 77 
 1 
 4  0
9

3


40
2
8
4
40
00
4
24. 
3t

20
Check:
4


3t
2

4


3t

20
4

3t  16
t
25.

16

3

33 
 2  0
16

4


20

16
2
20
00


x
 2 
 7  
x
9
x  2 
 14
x
 2  49  x  9

14
x
 2  
42
196(x  2)  1764
x29
x7
Check: 
x
 2 
 7  
x
9

7

 2 
 7  
7

9
3
74

4 4
no real solution

Chapter 4

22  6  5
3

  
9
5

4
235
55
27.

3x
 1  
x
0
 11 
 1
3x  10  x  11 
 2
x
 11  1
2x 
 2  
2
x
 11

x  1  
x
 11
x2 
 2x  1  x  11
x2 
 3x 
 10  0
(x 
 5)(x  2)  0
x
50
x20
x5
x  
2
Check: 
3x
 1  
x
0
 11 
 1

3(5)


10  
5
1
 1 
 1

25
  
16

1
54
1
5 3
Check:

3x
 1  
x
0
 11 
 1

3(
2)

 10  

2
11

1

4
  
9

1
23
1
22
Solution: x  
2
28a. 
3t

 1  t  6
4

3t

 1  6 
 t
4
3t 
 14  36 
 12t  t2
0  t2 
 15t  50
0  (t 
 5)(t 
 10)
t
50
t 
 10  0
t5
t  10
Check: 
3t

 1  t  6
4

3(5)



14  5  6

1
56
156
66
Check:

3t

 1  t  6
4

3(10)



14  10  6

16
  10  6
4  10  6
14 6
10
28b. 5

9

3



65

2x
 1  
2x
22  1 

b  7
Check:

x

126

29. 
2x


75
2x 
 7  25
2x  32
x  16
2x 
 7  0
2x  7
7

x  2

Test x  0: 
2(0)



75


7
5
meaningless
Test x  4: 
2(4)



75

8

75

15
false
Test x  17: 
2(17)



75

27
5
true

34. 
m

 2  
3m


4
m  2  3m  4

2m  2
m  
1
m20
m  
2
3m  4  0
3m  
4
4

m  
3
Test m  
3: 

3
2
   
3(
3)

4


1
  

5
 meaningless
Test m  
1.6: 

1.6


2  
3(
1.
6)

4

0.4
  

0.8

meaningless
Test m  
1.2: 

1.2


2  
3(
1.
2)

4

0.8
  
0.4
 false
Test m  0: 
0
 2  
3(0)



4

2
  
4
 true
Solution: m  
1
35. 
2c

5 7
Test c  0: 
2(0)




57
2c 
 5 49


5
7
2c 54
meaningless
c 27
Test c  5: 
2(5)



57

5
7
2c 
 5  0
false
2c  5
Test
c

28:

2(28)



5

7
c  2.5

51
7
true
Solution: c 27

Solution: x  16
30. 
b
46

b  4  36
b  32
b40
b  
4

Test b  
5: 

5
6
4


1
6
meaningless
Test b  0: 
0
46


4
6
26
true
Test b  33: 
33
6
4

37
6
false
Solution: 
4  b  32
Test a  0: 
0

54

31. 
a

54



5
4
a 
 5  16
meaningless
a  21
Test a  6: 
6

54

a
50

1
4
a5
14
true
Test a  22: 
22

4
5

17
4
false
Solution: 5  a  21
Test x  0: 
2(0)



56
32. 
2x

56



5
6
2x 
 5  36
meaningless
2x  41
Test x  5: 
2(5)



56
x  20.5

5
6
2x 
 5  0
true
2x  5
Test x  22: 
2(22)



56
x  2.5

39
6
false
Solution: 2.5  x  20.5
4
4
Test y  0: 

5(0) 

92
33. 

5y 
 9  2
4


9
2
5y 
 9  16
meaningless
5y  25
4
Test y  2: 

5(0) 

92
y5
4

1
2
5y 
 9  0
true
5y  9
4
Test y  6: 

y  1.8
5(6) 

92
4

21
2
false
Solution: 1.8  y  5

36a. t 

37.

2s

g

3

2(7.2)

g

3

14.4

g

3

36b.

9

14.4

g
14.4

g

9g  14.4
g  1.6 m/s2
3


x

 5  

x
3
3
x 
 5  

(x 
 3
)2
3
2
x 
 5  

x 
 6
x9
(x 
 5)3  x2 
 6x  9
x3 
 15x2  75x 
 125  x2 
 6x  9
x3 
 16x2  81x 
 134  0
Use a graphing calculator to find the zero.

[
2, 10] sc11 by [
10, 10] sc11
about 7.88
38a. s  
30fd

s  
30(0.6
)(25)

s  
450

s  21.2 mph
38b.
s  
30fd

35  
30(0.6
)d

1225  18d
68.06  d; about 68 ft

127

Chapter 4

38c. No; it is not a linear function.



g

39a. T  2




g

39b. t  2

1

9.8

T  2

44.

T2

22
4
2

  2




g

 2




g


4g

r
1

r

1

x

g
x

g

Ta

Tb

x

 g



ra 3

r
b

p

r  141,433,433.8; about 141,433,434 mi
41. 
2x

9
ab


2x

9ab

2x  9  0, so a  b  0
no real solution when a  b  0
tc

T  2 
t  (
200)

2

t
c


2 

t 
 200


t  416

2
2
t 
 832t  173,056

4

t2




 p2

2

t2 
 400t  40,000
 
4
t2  400t  50,000

4
t2  400t  50,000

2500


 832t  173,056 
123,056  1232t
99.88  t; about 99.88 psi
a2

2a  1

43.

a

3

1



 3  
4a  2 ; exclude: 
 2

a2
3
a




2a  1 (6)(2a  1)   3  2(2a  1) (6)(2a  1)

6(a  2)  a(2)(2a  1)  3(3)
6a  12  4a2  2a  9
0  4a2 
 4a 
 3
0  (2a 
 3)(2a  1)
2a 
 3  0
2a  1  0
3

a  2

6
5

11
6

6
0

1056

w

p 
 1056
w

O

10Ow


10O

4

3

y  
7x  7

1

a  
2

7

perpendicular slope: 4

3

2

7

y 
 5  4(x 
 2)
7

3

y  4x  2
51. A  r2

A  r2

1 2

2

 


1

4



 (1)2

1

4

 1 

 1
5

4

The correct choice is C.
Chapter 4

p
10O

46b. x- and y-axes
46c. It increases.
46d. It is halved.
0
3
47. 4 
1 6 2 
2  4(0)  (
1)(2)  6(5)
4
0 2
4(0) 
0(2)  2(5)
5
1
4(3)(
1)(
2)  6(1)
4(3)  0(
2)  2(1)
28
20

10 14
48. 
4(a  b  c)  
4(6) → 
4a 
 4b 
 4c  
24
2a 
 3b  4c  3
2a 
 3b  4c  3

2a 
 7b
 
21

4(a  b  c)  
4(6) → 
4a 
 4b 
 4c  
24
4a 
 8b  4c  12
4a 
 8b  4c  12

12b
 
12
b1

2a 
 7b  
21
abc6

2a 
 7(1)  
21
71c6
a7
c  
2
(7, 1, 
2)
49. y  
3.54x  7125.4
y  
3.54(2010)  7125.4
y  10 students
50. 7y  4x 
 3  0

t 
 (
200)
  502

2
t  200 2
  2500

2




6
0


10O

108  2

108 
 2 

1
1

v



42.


5
6

5
11

46a. p  w

225
67,200,000 3
 

687
rb
50,625
3.03 1023
  
471,969
rb 3
50,625rb3  1.43 1029
rb3  2.83 1024



5
6

x2  5x  6  0
(x  3)(x  2)  0
x30
x20
x  
3
x  
2

3, 
2, 
1, 1
45a. point discontinuity
45b. jump discontinuity
45c. infinite discontinuity

4
  x
It must be multiplied by 4.
40.

1
1

x3  6x2  11x  6  0

x

g



6, 3, 2, 1

1

8.9

T  2.01 s
T  2.11 s
39c. Let x  the new length of the pendulum.



g

p
:
q

128

4-8

12. f(x)  
1.25x  5
13. f(x)  8x2 
 3x 
 9
14. Sample answer:
f(x)  1.03x4 
 5.16x3  6.08x2  0.23x  0.94
15. Sample answer:
f(x)  0.09x3 
 2.70x2  24.63x 
 65.21
16. Sample answer:
f(x)  4.05x4 
 0.09x3  6.69x2 
 222.03x 
2697.74
17. Sample answer:
f(x)  
0.02x3  8.79x2  3.35x  27.43
18a. Sample answer: f(x)  1.99x2 
 1.74x  2.76
18b. Sample answer:
f(x)  
0.96x3  0.56x2  0.36x  4.05
18c. Sample answer: Cubic; the value of r2 for the
cubic function is closer to 1.
19a. Sample answer: f(x)  0.126x  22.732
19b. Sample answer:
2010 
 1900  110
f(x)  0.126x  22.732
f(110)  0.126(110)  22.732
f(110)  36.592
37
19c. Sample answer:
2025 
 1900  125
f(x)  0.126x  22.732
f(125)  0.126(125)  22.732
f(125)  38.482
38
20. Sample answer:

Modeling Real-World Data with
Polynomial Functions

Pages 261-262

Check for Understanding
y

1a. Sample answer:

O

x

y

1b. Sample answer:

O

x

y

1c. Sample answer:

O

x

2. You need to recognize the general shape so that
you can tell the graphing calculator which type of
polynomial function to use as a model.
3. Sample answer: If companies use less packaging
materials, consumers keep items longer, and old
buildings are restored instead of demolished, the
amount of waste will decrease more rapidly. If
consumers buy more products, companies package
items in larger containers, and many old buildings
are destroyed, the amount of waste will increase
instead of decrease.
4. quartic
5. Sample answer:
f(x)  1.98x4  2.95x3 
 5.91x2  0.22x  4.89
6. Sample answer: f(x)  3.007x2  0.001x 
 7.896
7a. Sample answer: f(x)  0.48x  58.0
7b. Sample answer: 2010 
 1950  60
f(x)  0.48x  58.0
 0.48(60)  58.0
 86.8%
7c. Sample answer: f(x)  0.48x  58.0
89  0.48x  58.0
64  x
1950  64  2014

Pages 262-264
8. cubic
10. linear

x
f(x)

1
1

2
3

3
6

4
3

5
6
7
8
9
–13 –49 –112 –209 –347

21a. Sample answer:
f(x)  0.008x4 
 0.138x3  0.621x2  0.097x
 18.961
21b. Sample answer: 1994 
 1992  2
f(x)  0.008x4 
 0.138x3  0.621x2  0.097x
 18.961
f(2)  0.008(2)4 
 0.138(2)3  0.621(2)2 
0.097(2)  18.961
f(2)  20.663
about 21%
22. A sixth-degree polynomial; there are 5 changes in
direction.
23a. Sample answer:
f(x)  0.109x2 
 0.001x  48.696

Exercises
9. quadratic
11. quadratic

129

Chapter 4

27.

23b. Sample answer:
f(x)  0.109x2 
 0.001x  48.696
100  0.109x2 
 0.001x  48.696
0  0.109x2 
 0.001x 
 51.304

r

1
0
1

2
2
2
2


1

3

1
1

0
3
0
1

1

2
1
2


2
0

2
0


1, 1
28a. Let x  number of weeks.
P  (120  10x)(0.48 
 0.03x)
P  57.6  1.2x 
 0.3x2

23c.

24a.
24b.

24c.

[
5, 25] sc11 by [
50, 10] sc15
root: (21.7, 0)
1985  22  2007
Sample answer: 1998 
 1985  13
f(x)  0.109x2 
 0.001x  48.696
f(13)  0.109(13)2 
 0.001(13)  48.696
f(13)  67.104
No; according to the model, there should have
been an attendance of only about 67 million.
Since the actual attendance was much higher
than the projected number, it is likely that the
race to break the homerun record increased the
attendance.
Sample answer:
f(x)  
0.033x3  1.471x2 
 1.368x  5.563
Sample answer: 1996 
 1990  6
f(x)  
0.033x3  1.471x2 
 1.368x  5.563
f(6)  
0.033(6)3  1.471(6)2 
 1.368(6)  5.563
f(6)  43.183
about 43.18 million
Sample answer:
f(x)  
0.033x3  1.471x2 
 1.368x  5.563
200  
0.033x3  1.471x2 
 1.368x  5.563
0  
0.033x3  1.471x2 
 1.368x 
 194.437

[
20, 20] sc12 by [
40, 60] sc15
maximum: (2, 58.8)
2 weeks
28b. $58.80 per tree
29. x 
 0.10x  0.90x
0.90x  0.10(0.90x)  0.90x  0.09x
 0.99x
The correct choice is B.

4-8B Fitting a Polynomial Function to a

Set of Points

Page 266
1. y  7x3 
 4x2 
 17x  15
2. y  7x3 
 4x2 
 17x  15; yes
3. Sample answer: y  
5x6 
 2x5  40x4 
 2x3 
x2  8x 
 4
4. Infinitely many; suppose that you are given a set
of n points in a coordinate plane, no two of which
are on the same vertical line. You can pick an
infinite number of other points with different xcoordinates. You could find polynomial functions
that went through the original n points and any
number of the other points.
5. There is no problem with using L10 with list L1
for the example. However, if you are using a
different list which happens to have 0 as one of its
elements, using L10 will result in an error
message, since 00 is undefined.

[
5, 25] sc15 by [
300, 50] sc150
root: (14.8, 0)
14  1990  2004
about 2004
25. 5 
 
b
20

Check: 5 
 
b

20
5  
b
2

5 
 
23
2
0
25  b  2
5 
 
25
0
23  b
5
50
00
26.

6

p3

Chapter 4 Study Guide and Assessment

p



p
3  1

6
p



p  3  p 
 3 (p  3)(p 
 3)  1(p  3)(p 
 3)

Page 267

6(p 
 3)  p(p  3)  (p  3)(p 
 3)
6p 
 18  p2  3p  p2 
 9
9p  9
p1

Chapter 4

Understanding and Using the
Vocabulary

1. Quadratic Formula
3. zero
5. polynomial function

130

2. Integral Root Theorem
4. Factor Theorem
6. lower bound

7. Extraneous
9. complex numbers

20. b2 
 4ac  42 
 4(1)(4)
 0; 1 real

8. complex roots
10. quadratic equation

a

Pages 268-270

a

Skills and Concepts

a  
2
21. b2 
 4ac  (
1)2 
 4(5)(10)
 
199; 2 imaginary

11. no; f(a)  a3 
 3a2 
 3a 
 4
f(0)  (0)3 
 3(0)2 
 3(0) 
 4
f(0)  
4
12. yes; f(a)  a3 
 3a2 
 3a 
 4
f(4)  (4)3 
 3(4)2 
 3(4) 
 4
f(4)  0
13. no; f(a)  a3 
 3a2 
 3a 
 4
f(
2)  (
2)3 
 3(
2)2 
 3(
2) 
 4
f(
2)  
18
14. f(t)  t4 
 2t2 
 3t  1
f(
3)  (
3)4 
 2(
3)2 
 3(
3)  1
f(
3)  73
no
15. 3; x3  2x2 
 3x  0
x(x2  2x 
 3)  0
x(x  3)(x 
 1)  0
x0
x30
x
10
x  
3
x1

4

1

7

25. f(x)  x4 
 10x2  9
f(3)  (3)4 
 10(3)2  9
 81 
 90  9 or 0; yes
p
:
q

1, 2

r
1

x2 
 x 
 2  0
(x 
 2)(x  1)  0
x
20
x2
rational roots: 
1, 1, 2

x

27.

79

x  4
7
9

x  4

x  4

x4

x  
2

1

p
:
q

1

1
1


2

1


1

2

2
0

x10
x  
1

r

1

0


1

1


1


1
1
rational root: 1
28. p: 1, 2, 4
q: 1, 2

1
1


1
1

0
0

1
1


2
0

2
2
2


2
0
2


2

2
2


4

6
0

p
:
q

17. b2 
 4ac  (
10)2 
 4(3)(5)
 40; 2 real

m

1 3

 
8  2  1 or 4; no

7  
81


2(2)

m

1  i
199


10

1

16. b2 
 4ac  (
7)2 
 4(2)(
4)
 81; 2 real

m

r

f 
2  4
2 
 7
2  1

f (x) 
 x 3  2x 2  3x

79

1  

199


2(5)

f(x)  x3 
 x2 
 10x 
 8
f(
2)  (
2)3 
 (
2)2 
 10(
2) 
 8
 
8 
 4  20 
 8 or 0; yes
23. f(x)  2x3 
 5x2  7x  1
f(5)  2(5)3 
 5(5)2  7(5)  1
 250 
 125  35  1 or 161; no
24.
f(x)  4x3 
 7x  1

26.

x

r

22.

f (x)

O


4  
0


2(1)
4

2

1

1, 2, 4; 2

r
1
2

2x2  2x  2  0
x2  x  1  0
does not factor

10  
40


2(3)
10  2 
10


6
5  
10


3

rational root: 2

18. b2 
 4ac  (
1)2 
 4(1)(6)
 
23; 2 imaginary
1  

23


2(1)

1  i
23
x  
2
b2 
 4ac  32

x

19.


 4(
2)(8)
 73; 2 real

y


3  
73


2(2)

y

3  
73


4

131

Chapter 4

29. p: 1, 3
q: 1, 2
p
:
q

33.
1

1

2
1

2
3

2
3


 2

3

2
2
2
2
2

3
5
1
9

3


6

1

7
21
3


11

12

4
52

20


3

15
1
153
57

2

4


4


13


2

2

2


7


2

15

2

6


3


2

3

4
105



 4

0


6


2

0

2

rational root:
p
:
q

r
1

r

1

19

31

3

2

1

0


7

1

12


4

1
2

1
1

1
2


6

3


5

5

7
2

3
0

1
1


3

3

2
0


5
1

1
1
1
1
1


3
1

2
13
13

0

2

1
4

4

2
0

1

1
3
53

51

2

4

8

1, 2, 4, 8, 3, 3, 3,  3
r
1

3
3


2
8

7
10

x20
x  
2

1


4

1

8

2

4

0

8

0

4x2  8  0
x2  2  0
does not factor
1
rational root: 
4
Chapter 4


34
8


56
0

2


11

12

9

2


12

18

0

1
1

0
2


13

9

1
1

2
5


9
6

x2  5x  6  0
(x  3)(x  2)  0
x30
x20
x  
3
x  
2
rational zeros: 
3, 
2, 2, 3

1

1, 2, 4, 2
4


1
6

1
1

1

2

r
3

32. p: 1, 2
q: 1, 2, 4

r

5
0

0

18

x3  2x2 
 9x 
 18  0

4

1

1
0

36. 2 or 0 positive
f(
x)  x4 
 13x2  36
2 or 0 negative


8
0

rational roots: 
2, 
3, 1

p
:
q

5
5


5
0

r

r
2

3x2  10x  8  0
(3x  4)(x  2)  0
3x  4  0
4
x  
3

0
5

2x2 
 12x  18  0
x2 
 6x  9  0
(x 
 3)(x 
 3)  0
x
30
x
30
x3
x3
1
rational zeros: 
2, 3

rational roots: 
2, 2
31. p: 1, 2, 4, 8
q: 1, 3
1

4
5

x2  6x  8  0
(x  4)(x  2)  0
x40
x20
x  
4
x  
2
rational zeros: 
4, 
2, 7
35. 2 or 0 positive
f(
x)  
2x3 
 11x2 
 12x  9
1 negative

x3 
 3x  1  0
r

2

1
4

4

1
1

r
7

r

0
1

x2  5  0
does not factor
rational roots: 
1, 1
34. 1 positive
f(
x)  
x3 
 x2  34x 
 56
2 or 0 negative

1, 2, 4

r

2

1
1

x3  x2  5x  5  0

x4  2x3 
 3x2 
 5x  2  0

p
:
q

1, 5

1, 3, 2, 2
r
1

1
3

3

30.

p
:
q

132


18
0

36
0

37.

r

2

1
0

3
3
3
3

0

6

3
0

0
12
3
0

r
0
1
2
3
4

2

2x  x  3
x3


4

4

3

2

1
0

1
1
1
1
1
1

r

1
0
1
2
3
4


3

4

3

2

1
0
1

r

2

1
0
1

1
1
1
1
1

m

r

2

1
0
1
2

47.

3 
 2y  5
y  
1
3

5



Test y  
2: 

2 
 2  
2
7

5


2  
2 true
3

5



Test y  
0.5: 

0.5 
 2  
0.5


1

3

2

1
0

0
6
2
0
0

1

11

1
1
1


8  
10 false
3
5
Test y  1: 1 
 2  1
1  5 true
Solution: y  
1, y 0
2

x1

48.


11
3

8

11

6
7

2

7

3
1
5
9

3

3
11
3

3
17

x0

1


1

x 
 1 ; exclude 
1, 1


9

9

9

9

9

25
34
25
16
7

2

1

4


2  3

24

58

24

8

10

Test x  
0.5:

6
64
6

2

14

2


0.5  1

true
1


1


0.5  1

4  
1 false
2
1
  1 
 
Test x  0.5: 
0.5  1
0.5 
 1
4

3

Test x  2:
Test x  4:

6

n 
 n  5  0

n 
 n6  5(n)  0(n)
 5n 
 6  0
(n  6)(n 
 1)  0
n60
n  
6

2(x 
 1)  (x  1)(x 
 1) 
 (x  1)
2x 
 2  x2 
 x 
 2
0  x2 
 3x
0  x(x 
 3)
x
30
x3



Test x  
2: 

2  1  1 
 
2 
 1

0 and 1
43. Use the TABLE feature of a graphing calculator.

4.9, 
1.8, 2.2
44.

5


 2  y; exclude: 0

2
1



x  1 (x  1)(x 
 1)  1 
 x 
 1 (x  1)(x 
 1)

4
4
4
4
4
4

r

1
0
1
2

3

y

3y 
 2y  5yy


2 and 
1, 0 and 1, 1 and 2
42.

8  

(
8)2 

 4(1)(3)

2(1)
8  

52

2


m  4  
13


3
1

3

5

5

3
1


1 and 0
41.

1

6(m  1)(m 
 1)
5(m  1)(m 
 1)  (2m)(3)(m 
 1) 
 2(m  1)
5m2 
 5  6m2 
 6m 
 2m 
 2
0  m2 
 8m  3

m

1
1
1
1
1
1
1

2m




2m  2 
 3m 
 3

2m
1


566(m  1)(m 
 1)  
2(m  1) 
 3(m 
 1) 

2
2

1

2

1
2


1 and 0, 3 and 4
40.

5

6

46.

0 and 1, 3 and 4
39.

x3



2x2

x3
 (2x2)
1x(2x2)  
2x 


1 and 0
38.

1

x

45.

1

23

2
1

2

21
2

3
2

41
2

5

 3 true
1


1

2
1

 0 false
1


1

4
1
2

 3 true

Solution: x  
1, 0  x  1, x 3
49. 5 
 
x
20
Check: 5 
 
x
20
5  
x
2
5 
 
23
2
0
25  x  2
5 
 
25
0
23  x
5
50
00

n2

n
10
n1

133

Chapter 4

3

3

50. 
1
4a 
   8  5
Check: 
1
4a 
   8  5
3
3

1
4a 
   
3

4(
6.
5)


185
3
4a 
 1  
27


27
85
4a  
26

3  8  5
a  
6.5
55
51.
3  
x
 8  
x
 35
9  6
x
 8  x  8  x  35
6
x
 8  18

x
83
x89
x1
Check: 3  
x
 8  
x
 35
3  
1

 8  
1
5
 3
3  
9
  
36

336
66
52. 
x

57
x
5 0
x 
 5  49
x 5
x  54
Test x  0: 
0

57



5
  7 meaningless
Test x  10: 
10

7
5

5
  7 true
Test x  60: 
60

7
5

55
  7 false
Solution: 5  x  54
53. 4  
2a
6
7
2a  7 0

2a
2
7
2a 
7
2a  7  4
a 
3.5
2a  
3
a  
1.5
Test a  
5: 4  
2(
5)

76
4  

3
  6 meaningless
Test a  
2: 4  
2(
2)

76
4  
3
  6 false
Test a  0: 4  
2(0)


76
4  
7
  6 true
Solution: a  
1.5
54. cubic
55. f(x)  2x2 
 x  3

Page 271

58a.

g (x ) 

0.006x 4 
0.140x 3  0.053x 2
100  1.79x

O 10

x

20

58b. g(x)  
0.006x4  0.140x3 
 0.053x2  1.79x
 x(
0.006x3  0.140x2 
 0.053x  1.79)
 x(x3 
 23.3
x2  8.83
x 
 298.3
)
r
1

23.333
8.833

298.333
1
1

22.333

13.503

311.836
5
1

18.333

82.835

712.508
23.5
1
0.167
12.758
0
rational zeros: 0, about 23.5



g

59. T  2
1.6  2
0.25 
0.06 




9.8




9.8



9.8

0.64  
; about 0.64 m

Page 271

Open-Ended Assessment
x

2



1. Sample answer: 
x  3  2x  1

x
2



x  3 (x  3)(2x  1)   2x  1 (x  3)(2x  1)

x(2x  1)  2(x  3)
2x2  x  2x  6
2
2x 
 x 
 6  0
(2x  3)(x 
 2)  0
2x  3  0
x
20
3
x  
2
x2
2a. Sample answer: x 
 4  
x

2
2b. Sample answer: x 
 4  
x

2
(x 
 4)2  x 
 2
x2 
 8x  16  x 
 2
x2 
 9x 
 18  0
(x 
 6)(x 
 3)  0
x
60
x
30
x6
x3
Check: x 
 4  
x

2
x 
 4  
x

2
6 
 4  
6


2
3 
 4  
3


2
2  
4


1  
1

22

1 1
The solution is 6. Since 
1 1, 3 is an
extraneous root.
3a. Sample answer:

Applications and Problem Solving

56. Let x  width of window.
Let x  6  height of window.
A  
w
315  x(x  6)
315  x2  6x
0  x2  6x 
 315
0  (x  21)(x 
 15)
x  21  0
x 
 15  0
x  
21
x  15
Since distance cannot be negative, x  15 and
x  6  21. the window should be 15 in. by 21 in.
57. Let x  width.
Let x  6  length.
(x  12)(x  6) 
 (x  6)(x)  288
x2  18x  72 
 x2 
 6x  288
12x  72  288
x  18
x  6  24
18 ft by 24 ft
Chapter 4

g (x )

200

x
f(x)


3

12


2
0


1
2


0.5
1.125

x
f(x)

0
0

0.5

0.625

1
0

2
8

3b. Sample answer: f(x)  x3  x2 
 2x
3c. Sample answer: 
2, 0, 1

134

Chapter 4 SAT & ACT Preparation
Page 273

4.You may want to draw a diagram.

SAT and ACT Practice

s

1. There are two ways to solve this problem. You can
use the distance formula or you can sketch a
graph.
d  

(x2 
 
x1)2 
(y2 

y1)2
2
 

(1 
 (

2) 
(3 
 
(
1))2
 

32  42
 
9
6
 1
 
25
 or 5
y

s6
Use the formula for the perimeter of a rectangle,
where 
 represents the length and w represents
the width.
2
  2w  P
Replace w with s. Replace 
 with s  6.
2(s  6)  2s  60
The correct choice is E.
5. First find the slope of the given line. Write the
equation in the form y  mx  b.
3x 
 6y  12
6y  3x 
 12
1
1
y  2x 
 2
The slope is 2.

(1, 3)
4

O

x

(2, 1) 3

So the slope of the line perpendicular to this line
is the negative reciprocal of this slope. The slope
of the perpendicular line is 
2. The correct choice
is A.
6. Be sure to notice the small piece of given
information: x is an integer. You need to find the
number, written in scientific notation, that could
be x3. This means that the cube root of the
number is an integer.
Take the cube root of each of the answer choices
and see which one is an integer. You can use your
calculator or do the calculations by hand. Notice
that 2.7 is one-tenth of 27, which is 33.
2.7 1013  27 1012
3


27 
1012  3 104 or 30,000. 30,000 is an
integer.
When you try the same calculation with each of
the other answer choices, the resulting power of
10 has a fractional exponent. So the number
cannot be an integer. The correct choice is C.
7. This is a system of equations, but you do not need
to solve for x or y. You need to find the value of
6x  6y.
Notice that the first equation contains 5y and the
second contains 
1y. If you subtract the second
from the first, you have 6y. Similarly, subtraction
of the x values gives a result of 6x. Use the same
strategy that you would for solving a system.
Subtract the second equation from the first.
10x  5y  14

4x  5y  
2
6x  6y  12
The correct choice is C.

When you sketch the points and draw a right
triangle as shown above, you can see that this is a
3-4-5 right triangle. Using the Pythagorean
Theorem, you can calculate that the length of the
hypotenuse is 5.
52  42  32 The correct choice is C.
2. Points on the graph of f(x) are of the form (x, f(x)).
To move the entire graph of f(x) up 2 units,
2 must be added to each of the second coordinates.
Points on the translated graph are of the form
(x, f(x)  2). The function which represents the
translation of the graph up 2 units is f(x)  2.
The correct choice is E.
3. You need to find both the x- and y-coordinates of
point C. Use the properties of a parallelogram.
First find the y-coordinate. Since opposite sides of
a parallelogram are parallel and side AD is on the
x-axis, point C must have the same y-coordinate
as point B. So the y-coordinate is b. This means
you can eliminate answer choices A and B
Now find the x-coordinate. Since opposite sides of
a parallelogram have equal length and side AD
has length d, side BC must also have length d.
Point B is a units from the y-axis, so point C must
be a  d units from the y-axis. The x-coordinate of
point C is a  d. So point C has coordinates
(a  d, b). The correct choice is E.

135

Chapter 4

(x 
 a2  x2 
 2ax  a2
 (x2  a2) 
 2ax
So the quantity in Column A equals the quantity
in Column B plus the sum of the squares of x and
a. Since neither x nor a equal 0, their squares
must be greater than 0. So the quantity in
Column A is always greater than the quantity in
Column B. The correct choice is A.
10. Since the problem does not include a figure, draw
one. Label the four points.

8. You can solve this problem using the midpoint
formula or by sketching a graph.
The midpoint formula:
x1  x2 y1  y2

3  (
4) 5  3


1 8

2, 2  2, 2  2, 2  
2, 4
1

y

(

 12,

4)

(3, 5)

(4, 3)

O

x

3

2

E

F

8

G

One method of solving this problem is to “plug-in”
numbers for the segment lengths. Since
5
EG  3 EF, let EF  3. Then EG  5. This means
that FG must equal 2, since EF  FG  EG.

The correct choice is B.
9. The expression x2 
 2ax  a2 is a perfect square
trinomial and can be factored as (x 
 a)2. The
square of a real quantity is never negative.
The correct choice is A.

HF  5FG  5(2)  10
HG  HF 
 FG  10 
 2  8
EF

HG

3

 8

The answer is .375 or 3/8.

Chapter 4

H

136

Chapter 5 The Trigonometric Functions
Pages 281–283

Angles and Degree Measure

5-1

Pages 280–281

Check for Understanding

1. If an angle has a positive measure, the rotation is
in a counterclockwise direction. If an angle has a
negative measure, the rotation is in a clockwise
direction.
45

26


2. Add 29, 60, and 
3600 .

3. 270°  360k° where k is an integer
4.
y

x

O

1260°
5. 34.95°  34°  (0.95 60)

 34°  57

34° 57

6. 
72.775°  
(72°  (0.775 60)
)
 
(72°  46.5
)
 
(72°  46
  (0.5 60))
 
(72°  46
  30)

72° 46
 30



24. 23° 14
 30  23°  14

60
   30 3600 
 23.242°
1°

1°

1°

13.

453

360

 1.26

1°

3600



26. 233° 25
 15  233°  25

60
   15 3600 
 233.421°
1°

1°

1
(360°)  15°
24
1 1
  360°

60 24
1 1
1
   360°
60 60 24

1°



28. 
405° 16
 18  
405°  16

60
   18 3600 
1°

1°

 
405.272°



29. 1002° 30
 30  1002°  30

60
   30 3600 
 1002.508°
1°

1°

3 
360°  
1080° 31. 2 360°  720°
1.5 360°  540°
33. 7.5 (
360°)  
2700°
2.25 360°  810° 35. 5.75 (
360°)  
2070°
4 360°  1440°
30°  360k°; Sample answers:
30°  360k°  30°  360(1)° or 390°
30°  360k°  30°  360(
1)° or 
330°
38. 
45°  360k°; Sample answers:

45°  360k°  
45°  360(1)° or 315°

45°  360k°  
45°  360(
1)° or 
405°
39. 113°  360k°; Sample answers:
113°  360k°  113°  360(1)° or 473°
113°  360k°  113°  360(
1)° or 
247°
30.
32.
34.
36.
37.


14. 

360  
2.22


2.22  2  
0.22

0.22 360°  
78°
360° 
 78°  282°; IV

15.  
 180°  227° 
 180°
 47°
16. 360° 
 210°  150°
180° 
   180° 
 150°
 30°
17.

1°



27. 173° 24
 35  173°  24

60
   35 3600 
 173.410°

798

  360(1)°  453°
  360  453°
  93°; II

1°

 
14.089°

8. 29° 6
 6  29°  6
   6

 29.102°
9. 2 (
360°)  
720°
10. 4.5 360°  1620°
11. 22°  360k°; Sample answers:
22°  360k°  22°  360(1)° or 382°
22°  360k°  22°  360(
1)° or 
338°
12. 
170°  360k°; Sample answers:

170°  360k°  
170°  360(1)° or 190°

170°  360k°  
170°  360(
1)° or 
530°
1°

60


1°



25. 
14° 5
 20  
14°  5

60
   20 3600 



7. 
128° 30
 45
  
128°  30

60
   45 3600 
 
128.513°
1°

Exercises

18. 
16.75°  
(16°  (0.75 60)
)
 
(16°  45
)

16° 45

19. 168.35°  168°  (0.35 60)

 168°  21

168° 21

20. 
183.47°  
(183°  (0.47 60)
)
 
(183°  28.2
)
 
(183°  28
  (0.2 60)
 
(183°  28
  12)

183° 28
 12
21. 286.88°  286°  (0.88 60)

 286°  52.8

 286°  52
  (0.8 60)
 286°  52
  48
286° 52
 48
22. 27.465°  27°  (0.465 60)

 27°  27.9

 27°  27
  (0.9 60)
 27°  27
  54
27° 27
 54
23. 246.876°  246°  (0.876 60)

 246°  52.56

 246°  52
  (0.56 60)
 246°  52
 
 33.6
246° 52
 33.6

   0.25°, or 0.25(60
)  15

     0.0042°,

or 0.0042(60)(60)  15

137

Chapter 5

40. 217°  360k; Sample answers:
217°  360k°  217°  360(1)° or 577°
217°  360k°  217°  360(
1)° or 
143°
41. 
199°  360k°; Sample answers:

199°  360k°  
199°  360(1)° or 161°

199°  360k°  
199°  360(
1)° or 
559°
42. 
305°  360k°; Sample answers:

305°  360k°  
305°  360(1)° or 55°

305°  360k°  
305°  360(
1)° or 
665°
43. 310°  360k°  310°  360(0)° or 310°
44. 60°  360k°  60°  360(2)° or 780°
60°  360k°  60°  360(
3) or 
1020°
45.

400°

360°

 1.11

46.

  360(1)°  400°
  360°  400°
  40°; I
47.

940°

360°

 2.61

48.


624°

360°

1059°

360°

36,000,000 or
1.08
63.

52.
53.
54.
55.

57.

1275°

360°

  360(2)°  1059°
  720°  1059°
  339°; IV

Chapter 5

1

2



360°

rotation

60 seconds

minute

60 minutes

hour



1 rotation

70 minutes

360°
 
revolution
1
 2 minute

1°

1°

60 minutes

hour

24 hours

day



day

1 revolution


San Antonio: 
1 hour

20.6 rotations

24 hours

day

24 revolutions



day

24 
 20.6  3.4 revolutions
about 3.4 revolutions
66b. Sydney:
20.6 rotations

day

7 days

week

360°

rotation

24 revolutions


San Antonio: 
day

 51,840°
7 days

week

360°

revolution



60,480°
60,480° 
 51,840°  8640°
67a. Use graphing calculator to find cubic regression.
Sample answer: f(x)  
0.00055x3  0.0797x2

 3.7242x  76.2147
67b. 2010 
 1950  60
f(x)  
0.00055x3 
 0.0797x2 
 3.7242x
 76.2147
f(60)  
0.00055(60)3  0.0797(60)2

 3.7242(60)  76.2147
 20.8827
Sample answer: about 20.9%

 32,400°/second
360°

revolution

1°

66a. Sydney:



1,944,000°/minute

34,200°

60 seconds

minute



81° 45
 34.4  81°  45

60
   34.4 3600 
 81.760°

 2.90

30 seconds

 22,320° second

1°

 3.54

95 revolutions

minute

360°

rotation



65b. 24° 33
 32  24°  33

60
   32 3600 
 24.559°

60. 90k, where k is an integer
61.

62 rotations

second

107 degrees

64. 25°  120k°, where k is an integer
65a. 44.4499°  44°  (0.4499 60)

 44°  26.994

 44°  26
  (0.994 60)
 44°  26
  59.64
44° 26
 59.64
68.2616°  68°  (0.2616 60)

 68°  15.696

 68°  15
  (0.696 60)
 68°  15
  41.76
68° 15
 41.76

 2.94

 
2.75

90 revolutions
360°
  
second
revolution
90 revolutions 60 seconds
  
second
1 minute

360°

rotation

62 rotations
60 seconds 60 minutes
360°
   
second
minute
hour
rotation
hours


24 day  1,928,448,000°/day

  360(2)°  1045°
  720°  1045°
  325°
360° 
   360° 
 325° or 35°
58. 20°
180° 
   180° 
 20° or 160°
180°    180°  20° or 200°
360° 
   360° 
 20° or 340°
59.

107 to 3.6



80,352,000°/hour

 
0.78

  360(3)°  1275°
  1080°  1275°
  195°; III
360° 
   360° 
 327° or 33°
180° 
   180° 
 148° or 32°
563° 
 360°  203°
 
 180°  203° 
 180° or 23°

420°  360°  
60°
56. 360° 
 197°  163°
360° 
 60°  300°
180° 
 163°  17°
360° 
 300°  60°
1045°

360°

360°

revolution
3.6 107

62 rotations

second

62 rotations

second


2.75  2  
0.75

0.75 360°  
269°
360° 
 269°  91°; II
51.



107

1,339,200°/minute

 
1.73


989°

360°

360°

revolution

100,000 revolutions

minute


1.73  1  
0.73

0.73 360°  
264°
360° 
 264°  96°; II
50.

30,000 revolutions

minute

10,800,000 or 1.08

  360(
1)°  
280°
 
 360°  
280°
  80°; I

  360(2)°  940°
  720°  940°
  220°; III
49.


280°

360°

62.

34,200°/minute

 17,100°

138

3

68. 
5
6n   
 15  
10
3

5
6n    5
6n  5  125
6n  120
n  20
3
Check: 
5
6n   
 15  
10
3

6(20)


5 
 15  
10
3

125
 
 15  
10
5 
 15  
10

10  
10 
x3

x2
x3

x2
x3

x2

69.

(x  2)(x  3)

(3, 5)
(1, 5)
(0, 3)

77. [f g](x)  f(g(x))
 f(x 
 0.3x)
 (x 
 0.3x) 
 0.2(x 
 0.3x)
 x 
 0.3x 
 0.2x  0.06x
 0.56x
78. m∠EOD  180° 
 m∠EOA 
 m∠BOD
 180° 
 85° 
 15°
 80°
m∠OED  m∠EDO
1
m∠OED  2(180° 
 m∠EOD)

3


2

x2  5x  6
3


2

(x  2)(x  3)

  (x  2)(x  3)(2) 

3

(x  2)(x  3)
(x  2)(x  3) 

1

 2(180° 
 80°)
 50°
m∠ECA  180° 
 m∠EOC 
 m∠OED
 180° 
 (80°  15°) 
 50°
 35°
The correct choice is D.

5-2

52

y

20

y 
 xx 11

26
10

2

48

2. R1 

x

R1 
R2 
R2 

f (x)

R3 
R3 
f (x) 
|(x  1)2  2|

O

Graphing Calculator Exploration

1. Sample answers:

O

74.

Trigonometric Ratios in Right
Triangles

Page 284

 r2

4.91  r2
about 4.91
73. x  1  0
x 
1
Point discontinuity

x

O

(x  3)(x  3)  (x  2)(x  3)(2) 
 3
x2  6x  9  2x2  10x  12 
 3
0  x2  4x
0  x(x  4)
x  0 or x  4  0
x  
4
70. 2 1 1 0 8
1
2 4
24
1 2 12 25
25
71. (x 
 (
5))(x 
 (
6))(x 
 10)  0
(x  5)(x  6)(x 
 10)  0
(x2  11x  30)(x 
 10)  0
x3  x2 
 80x 
 300  0
72.
r1t1  r2t2
18(
3)  r2(
11)
18(
3)


11

y

76.

3. R1 
R1 

x

R2 
R2 

decreasing for x  
1, increasing for x 
1
75. expanded vertically by a factor of 3, translated
down 2 units

R3 
R3 

5

13
15

39
12

13
36

39
5

12
15

36
12

13
36

39
5

13
15

39
12

5
36

15

24

or about 0.3846
or about 0.3846
or about 0.9231
or about 0.9231
or about 0.4167
or about 0.4167
or about 0.9231
or about 0.9231
or about 0.3846
or about 0.3846
or 2.4
or 2.4

4. Each ratio has the same value for all 22.6° angles.
5. yes
6. Yes; the triangles are similar.

139

Chapter 5

Pages 287–288

11. (AC)2  (CB)2  (AB)2
82  52  (AB)2
89  (AB)2

89
  AB
side opposite

sin A  
hypotenuse

Check for Understanding

1. The side opposite the acute angle of a right
triangle is the side that is not part of either side of
the angle. The side adjacent to the acute angle is
the side of the triangle that is part of the side of
the angle, but is not the hypotenuse.
2. cosecant; secant; cotangent
a

b

c
,
a

c
,
b

a

tan A 

b

a

tan A 

3. sin A  c, cos A  c, tan A  b,
csc A 

sec A 

cot A 

side opposite

side adjacent

6.

15
514

15
17
 or 
cos T  

514

514

514

side opposite

tan T  
side adjacent
15

tan T  1
7
1
1



csc v  sin v
7. tan v  
cot v

1
5

csc v  
2 or 2

12


cos T  
hypotenuse

sin T 

or

17
514


514

or

1

cot v 

cot P 

2
91


6

1


csc v 

or 3

1

3

1

cos v 

1

5

9

9

1

1

1


18. tan v  
cot v


19. sec v  
cos v

1


tan v  
0.75 or about 1.3333

side opposite

side adjacent


sin R  
hypotenuse


91


3


cos R  
hypotenuse


I
t

Io

sec R 

I

sec R   or 

sin R 

 It
o

0.5I0  It

Exercises

10. (AC)2  (CB)2  (AB)2
802  602  (AB)2
10,000  (AB)2
100  AB
side opposite

sin A  
hypotenuse
60

3



sin A  
100 or 5

side adjacent


cos A  
hypotenuse
80

4



cos A  
100 or 5

side opposite


tan A  
side adjacent
60

3


tan A  8
0 or 4

Chapter 5

1


sec v  
0.125 or 8

20. (RT )2  (TS)2  (RS)2
142  (TS)2  482
(TS)2  2108
TS  
2108
 or 2
527



2


2

Pages 288–290

7

or 3


sin v  
2.5 or 0.4

or 5

tan R 

2

4

3

7


17. sin v  
csc v

cos 45° 

or

1

1


16. cos v  
sec v

2
527


14




91


10

1

tan R 

or

or


15. csc v  
sin v


14. cot v  
tan v


I
t
Io

cos v

20

2
91

I
 It
o


4
91

40

3
91


91



527

2
527
 or 
48
24
side opposite

side adjacent

sec P 
9.

10
91


91

cos A 

13. tangent

cot P 

csc P 

12

4
91


tan A 

2
91


91

 or 
20
10
hypotenuse

side opposite
20
10
 or 
6
3
side adjacent

side opposite

csc P 

8
89


89

or

side opposite

8. (PS)2  (QS)2  (QP)2
(PS)2  62  202
(PS)2  364
PS  
364
 or 2
91

side opposite
side adjacent


sin P  
cos
P
hypotenuse
hypotenuse
cos P 

8


89



tan A  
side adjacent

1

6
3

sin P  2
0 or 10
side opposite

tan P  
side adjacent
3
91

6

tan P  
2
91
 or 91
hypotenuse

sec P  
side adjacent

3

sin A  40 or 1
0


tan v  
1.5 or about 0.6667


5

cos A 

12. (AC)2 
(AC)2  122  402
(AC)2  1456
AC  
1456
 or 4
91

side opposite
side adjacent


sin A  
cos
A

hypotenuse
hypotenuse

4. sin A  cos B, csc A  sec B, tan A  cot B
5. (TV)2  (VU)2  (TU )2
172  152  (TU )2
514  (TU )2

514
  TU

sin T  
hypotenuse

5
5
89

 or 

89

89
side opposite

side adjacent
5

8
(BC)2  (AB)2

sin A 

side adjacent


cos A  
hypotenuse

140

14

7

cos R   or 
csc R 

48
24
hypotenuse

side opposite

csc R 

48

2
527


hypotenuse

side adjacent

cot R 

48

24

14

7

cot R 

side adjacent

side opposite
14
7
527

 or 
527
2
527


or


527


7

or

24
527


527

21. (ST)2  (TR)2  (SR)2
382  (TR)2  402
(TR)2  156
TR  
156
 or 2
39

side opposite
side adjacent


sin R  
cos
R
hypotenuse
hypotenuse
38

19

sin R  40 or 20
side opposite


tan R  
side adjacent

tan R 
sec R 
sec R 

38
19
39

 or 
39
2
39

hypotenuse

side adjacent

40
20
39
 or 
2
39

39

cos R 
csc R 
csc R 
cot R 
cot R 

side opposite


28. sin R  
hypotenuse
3

sin R  7
a2  b2  c2
32  b2  72
b2  40
b  
40
 or 2
10

side adjacent

cos R  
hypotenuse

2
39


39

 or 
40
20
hypotenuse

side opposite
40
20
 or 
38
19
side adjacent

side opposite


39

2
39
 or 
38
19

cos R 
csc R 

side opposite


cos R  
hypotenuse


154


7

 or 
2
22

44
side opposite

side adjacent

9
22

9
cos R  
or 
2
22

44
hypotenuse

csc R  
side opposite


sin R  
hypotenuse

sin R 
tan R 
tan R 
sec R 
sec R 


7


9
hypotenuse

side adjacent
2
22


9

23. cot (90° 
 v)  tan v
cot (90° 
 v)  1.3
24b. 0.186524036
24d. 1.37638192
25.
v
sin
cos
v
sin
cos

72°
0.951
0.309
82°
0.990
0.139

74°
0.961
0.276
84°
0.995
0.105

25a. 1
26.

cot R 
cot R 

14°
0.242
0.970
0.249

12°
0.208
0.978
0.213

v
sin
cos
tan

8°
0.139
0.990
0.141

6°
0.105
0.995
0.105

4°
0.070
0.998
0.070

2°
0.035
0.999
0.035

v2


tan 13°  
9.8(15.5)

35.07  v2
5.9  v
about 5.9 m/s
v2

29d. increase

v2

9.8(15.5)
v2

40.70 
6.4  v
about 6.4 m/s

side opposite

side adjacent


30. sin v  
hypotenuse

80°
0.985
0.174


cos v  
hypotenuse

side opposite

hypotenuse
——
side adjacent

hypotenuse

sin v

cos v



sin v

cos v



hypotenuse

sin v

cos v



hypotenuse

sin v

cos v
sin v

cos v



side adjacent

side opposite
side opposite

side adjacent

hypotenuse
hypotenuse

side adjacent

side opposite

 tan v

31a. ∠  90° 
 L 
 23.5°

(N  10)360

365
(172  10)360


cos
365

cos

∠  90° 
 26° 
 23.5° (
0.99997)
∠  87.5°
(N  10)360

∠  90° 
 L 
 23.5° cos 
365

10°
0.174
0.985
0.176

∠  90° 
 26° 
 23.5°

cos

∠  90° 
 26° 
 23.5°
∠  40.5°

1

31b. ∠  90° 
 L 
 23.5°
∠  90° 
 64° 
 23.5°

(355  10)360

365

(N  10)360

365
(172  10)360


cos
365

cos

n

∠  90° 
 64° 
 23.5° 
0.99997
∠  49.5°
(N  10)360

∠  90° 
 L 
 23.5° cos 
365

n

∠  90° 
 64° 
 23.5°

]
cos[
365

1.5103  n

∠  90° 
 64° 
 23.5°
∠  2.5°

1

26a. 0
26c. 0
sin v
i
sin vr
sin 45°

sin 27° 55



7
10

20

v2

∠  90° 
 26° 
 23.5°
16°
0.276
0.961
0.287

or

29b. tan v  gr

tan v  gr
tan 15° 

25b. 0
18°
0.309
0.951
0.325

27.

v2

tan v  gr

29c.

88°
0.999
0.035

v
sin
cos
tan

7

2
10


29.53  v2
5.4  v
about 5.4 m/s

24c. 35.34015106

86°
0.998
0.070

sec R 

v2

2
154

2
22

 or 

7

7
side adjacent

side opposite
9
7

9
 or 

7

7

78°
0.978
0.208

2
10


3

sec R 

3
10

3
 or 
2
10

20
hypotenuse

side adjacent

tan R 


tan 11°  
9.8(15.5)

24a. 0.7963540136

76°
0.970
0.242

cot R 
29a.

side adjacent

csc R 

cot R 

7

3
side adjacent

side opposite

csc R 

22. (ST )2  (TR)2  (SR)2
(
7
)2 
 92  (SR)2
88  (SR)2

88
  SR; 
88
 or 2
22


2
10


7
hypotenuse

side opposite

side opposite


tan R  
side adjacent

26b. 1

141

(355  10)360

Chapter 5

31c. 87.5° 
 40.5°  47°
49.5° 
 25°  47°
neither

2.

y

sin(B 
 A)


32. x  t
c os A 
sin(60° 
 41°)

cos 41°

x  10

x

O



x  10(0.4314)
x  4.31; about 4.31 cm
33. 88.37° 88°  (0.37 60)

 88°  22.2

 88°  22
  (0.2 60)
 88°  22
  12
88° 22
 12
34. positive: 1
f(
x)  x4 
 2x3  6x 
 1
negative: 3 or 1
35.
35a. 23 employees
35b. $1076

As v goes from 0° to 90°, the y-coordinate
increases. As v goes from 90° to 180°, the
y-coordinate decreases.
x

cos v


3. cot v  y  
sin v

4.

1

1

y

1x

O
1

Function
sin  or cos 
cos  or sec 
tan  or cot 

[
10, 50] scl:10 by [
10, 1200] scl:100
7 
3
5
0 
1
4 
1
4 0
36. 4

 (
3)
5
0 
1  7
2
0
8
0
8 2
8
2
0
 7(2) 
 (
3)(8)  5(8)
 78

y

1

or

7.

 1

3


2 , 2

sin 30°  y
sin 30° 

1

2

y

tan 30° 

1

2


3


2

tan 30° 

1


3


tan 30° 


3


3

1

y 
 3  
2(x 
 6)
y

6

1

1

12  2(2x)(3x)

3x  3(2) or 6

12  3x2
4  x2
2x

a2  b2  c2
42  62  c2
52  c2

52
  c; 
52
 or 2
13


1

sec 30°  x

The correct choice is C.

5-3
Page 296

Trigonometric Functions on the
Unit Circle
Check for Understanding

1. Terminal side of a 180° angle in standard position
is the negative x-axis which intersects the unit
1
1
circle at (
1, 0). Since csc v  y, csc 180°  0
which is undefined.

Chapter 5

cos 30°  x
cos 30° 


3


2

1

csc 30°  y
csc 30° 

1

1

2

csc 30°  2

2x  2(2) or 4

38. A  2bh

1

2

tan 30°  x

y 
 y1  m(x 
 x1)
1

2x

1

6. (0, 
1); sec(
90°)  x or 0; undefined

3
5

m

0


5. (
1, 0); tan 180°  x or 

1 ; 0


37. m  
6
2

2

4

Quadrant
II
III











I




142

x

cot 30°  y


3

2

1

2

sec 30° 

1


cot 30° 

sec 30° 

2


3


cot 30°  
3


sec 30° 

2
3


3



3

2

IV






x

8. terminal side — Quadrant III
reference angle: 225° 
 180°  45°

12. cos v  r
1

cos v  
2


2



2


 
2 ,
 2 

sin 225°  y

cos 225°  x



2

sin 225°  
 
2
y



2

2



2


2

csc 225° 
csc 225° 

1

y

1

sin v  r

1


2



2
2



2


sin v 
sec v 
sec v 

x

cot 225° 

cos v  r

x

tan v  x

sin v  5

4

cos v  5

3

tan v  3

csc v  y

r

sec v  x

r

cot v  y

5

sec v  3

5

cot v  4

or 

3


csc v 

cot v 
or 
2

cot v 

2


3

2
3


3

x

y

1


3



3


3

Exercises

4

15. (1, 0); tan 360°  x or 1; 0

y

x

0


1

x

16. (
1, 0); cot(
180°)  y or 0; undefined

3

1

1


17. (0, 
1); csc 270°  y or 

1 ; 
1

18. (0, 1); cos(
270°)  x or 0
1

1


19. (
1, 0); sec 180°  x or 

1 ; 
1
y

x

cos v  r
v
v

y

11. tan v  x
tan v  
1
x  1, y  
1
y

sin v  r

1


2


2

sin v  
 
2
r
csc v  y


2
csc v  
or 

2


1
x
cot v  y
1

cot v  

1 or 
1

sin v 

r

x
2


1

14. (0, 1); sin 90°  y or 1

10. r 
r  

(
6)2 
 62
r  
72
 or 6
2

6

cos
6
2



2
sin v  
cos
2
r
csc v  y

6
2
csc v  

6 or 
2
x
cot v  y

6
cot v  6 or 
1


3



1

y



x2  y2

sin v 

tan v 

Pages 296–298

y

y



3

2

13. C  2r cos L
C  2r cos L
C  2(3960) cos 0°
C  2(3960) cos 90°
C  24,881.41
C0
The circumference goes from about 24,881 miles to
0 miles.

cot 225°  1

sin v  r

r

csc v  y

cot v 


2



2


2



2

sec 225°  

2

9. r  

x2  y2
r  

32  42
t  
25
 or 5

sin v  r

y

tan v  x

csc v 

cot 225°  y

1


2


2
2



2


csc v  4

3  y2

3
y
Quadrant II, so y  
3



2


csc 225°  

2


sec 225°  x

sec 225° 

x  
1, r  2

csc 225°  y

tan 225°  1

sec 225° 

22  (
1)2  y2

cos 225°  
 
2

tan 225°  x
tan 225° 

r 2  x2  y2

20. Sample answers: 0°, 180°

tan v  x


6

6
2


2



2

sec v 
sec v 

6

22.


tan v  

6 or 
1

r

x
6
2



6



 
2


2
, 
2
2



sin 45°  y

cos 45°  x

sin 45° 

cos 45° 

tan 45° 

or 

2


tan 45° 

r2  12  (
1)2
r2  2
r  
2

x
cos v  r
cos v 
sec v 
sec v 

1


2


2


2
r

x

2


1



2

2
y

x

2


2


2


2

csc 45° 
or 1

csc 45° 
csc 45° 


2


2
1

y
1


2


2
2


2



csc 45°  
2

r 2  x2  y2

cos v 

21. undefined

sec 45° 

1

x

sec 45° 

1


2


2
2


2


sec 45° 

x

cot 45°  y
cot 45° 


2


2


2


2

or 1

sec 45°  
2


or 
2


143

Chapter 5

23. terminal side — Quadrant II
reference angle: 180° 
 150°  30°

25. terminal side — Quadrant III
reference angle: 210° 
 180°  30°
1

3



 
2 , 
2


3
 1


 
2 , 2

sin 150°  y

cos 150°  x

1
sin 150°  2
y
tan 150°  x

cos 150° 

1

2

tan 150°  
3


2
1

tan 150°  
 
3

3

tan 150°  
 
3
1
sec 150°  x
1
sec 150°  

3



2

csc150° 


3



2
1

csc 150°  y
1

1

2

x

cot 150°  y

2


3


2

1

2

1



2

tan 315° 

y

x

tan 315° 



2


2


2


2

or 
1

cos 315° 


2


2

csc 315° 

1

y

csc 315° 

1


2



2
2



2


csc 315° 

cot 210° 


3


2


 1


2

cot 210°  
3


1

cot 315°  y

1


2


2
2


2


cot 315° 

y

x


2


2


2



2

cos 330°  x
1

sin 330°  
2

csc 315°  

2

sec 315°  x

x

cot 210°  y


2
3


3

sin 330°  y
cos 315°  x

sin 315°  
 
2

csc 210°  
2

1


3


2 , 
2


2



2


2 ,
 2 

sin 315°  y

1

1

2

26. terminal side — Quadrant IV
reference angle: 360° 
 330°  30°

24. terminal side — Quadrant IV
reference angle: 360° 
 315°  45°

or 
1

sec 315°  
2


Chapter 5

csc 210° 

sec 210° 

2
3


sec 315° 

1

2


3



2
1
tan 210°  

3


3


tan 210°  3
1
sec 210°  x
1
sec 210°  

3


2


3


csc 210°  y

sec 210°  
 

3


sec 150°  
 
3

sec 315° 

cos 210°  
 
2

2

cot 150°  

3


sec 150°  
 

3


cos 210°  x

1
sin 210°  
2
y
tan 210°  x

tan 210° 

csc 150°  2

cot 150° 

sin 210°  y

144

cos 330° 

tan 330°  x

csc 330° 

1


2

tan 330°  
3

2
1
tan 330°  
 

3


3

tan 330°  
 
3
1
sec 330°  x
1
sec 330°  

3


2
2
sec 330°  

3

2
3


sec 330°  3

csc 330° 


3


2
1

y
1

1


2

csc 330°  
2
x

cot 330°  y
cot 330° 


3


2

1


2

cot 330°  

3


31. r  

x2  y2
r  

(
6)2 
 62
r  
72
 or 6
2


27. terminal side — Quadrant I
reference angle: 420° 
 360°  60°

3

12, 
2 

sin 420°  y

cos 420°  x

sin 420° 

cos 420°  2

tan 420° 
tan 420° 


3


2
y

x

1

1

csc 420° 
csc 420° 

tan 420°  
3

sec 420° 
sec 420° 

sin v 

csc 420°  y


3


2

1

2

csc 420° 

1

x
1

1

2

cot 420° 
cot 420° 

sec 420°  2

cot 420° 
cot 420° 

sin v 

1



3

2
2


3


csc v 
csc v 


6

6
2


6


tan v  

6 or 
1


2


cos v  
 
2
x

r

sec v  x
sec v 

cot v  y

6

6
2



6

cot v  6

or 

2


or 
1

cos v  r

y

0

cos v  2 or 1

r

2

sec v  2 or 1

x

tan v  x

2

tan v  2 or 0

sec v  x

r

cot v  y

2

cot v  0

0

x

2

undefined
33. r  

x2  y2
r  

12  (

8)2
r  
65


undefined

y

sin v  r
or 
1

or 2
34.

y

x

tan v  x

1


65


65


65

tan v  1 or 
8

cos v  r


8

cos

65

8
65

sin v  
 
cos
65
r


csc v  y


65

65

csc v  
or 
 

8
8
x
cot v  y
1
1


cot v  

8 or 
 8
r  

x2 
 y2

sin v 


3
 1


2 , 2

v
v


8

r

sec v  x
sec v 


65


1

or 
65


r  

52  (

3)2
r  
34


r  

(
4)2 
 (
3
)2
r  
25
 or 5

y

sin v 

y

r

cos v 

sin v 

3

5
r

y
5

3

cos v  
5

csc v 

tan v  x

y

csc v  0

29. terminal side — Quadrant 1
reference angle: 390° 
 360°  30°

csc v 


2


2
r

y

6
2

6

csc v  y

x

30.

cos v 

sin v  2 or 0

cot (
45°)  y

1
csc 390°  y
1
csc 390°  
1

2
r  

x2  y2

6

6
2


y

x

sin v  r


2


2



2 ,
 2 

cot (
45°) 

cos v  r

or 
2

32. r  

x2  y2
r  

22  02
r  
4
 or 2

2
3


3
x

y
1

2


3


2
1


3



3

3

28. terminal side — Quadrant IV
reference angle: 45°


2



2



2


2

y

sin v  r

x

r
4

r

sec v  x
5

sec v  
4

sin v  r

y
tan v  x

3
3


tan v  

4 or 4
x
cot v  y

4
4


cot v  

3 or 3


3

cos v

34

3
34

sin v  
 
cos v
34
r
csc v  y

34

34



csc v  

3 or 
 3
x
cot v  y
5
5


cot v  

3 or 
 3
r  

x2  y2

sin v 

35.

y

x

tan v  x

5


34

5
34


34

tan v  5 or 
5

cos v  r



r

sec v 

y

cos v  r

15

cos v  17 or 
17

r

sec v  x

17



sec v  

8 or 
 8

sin v  1
7
csc v  y
csc v  1
5

145

tan v  x
8

r

17


34


5

y

x


8

3

sec v  x

r  

(
8)2 
 152
r  
289
 or 17
sin v  r


3

15

15



tan v  

8 or 
 8
x

cot v  y
17


8

8

cot v  15 or 
1
5

Chapter 5

36. r  

x2  y2
r  

52  (

6)2
r  
61


r  

x2  y2
r  

(
5)2 
 62
r  
61


y

sin v  r


6


61

6
61


sin v  
 
61

6


61


sin v  r

sin v 


6
61

61

sin v 
sin v 

The sine of one angle is the negative of the sine of
the other angle.
37. If sin v  0, y must be negative, so the terminal
side is located in Quadrant III or IV
38. cos v 
cos v 

r2

12

1
3



x2



sec v 
sec v 

y2

y


5


5

5

y

sin v 

5



tan v  

12 or 12

r

sin v 

r

csc v  y

sec v  x
13

13

csc v 

13

13



csc v  

5 or 
 5



sec v  

12 or 
 12

csc v 

r

cot v  y

12

csc v 

12



cot v  

5 or 5
r

39. csc v  y

sin v 
sin v 

y

r
1

2

x

cos v  r
cos v 




3

2

or


3



2

tan v 

y

x

tan v 

1



3


3



 3

tan v 
sec v 

r

x

cot v 

sec v 

2



3

2
3



 3
y

r
1

5

cot v 

sec v 
40. sin v 
sin v 

y  
1, r  5

x

cos v  r
cos v 

2
6


5

sec v 

r

x

sec v 

5

2
6

5
6


12

Chapter 5

52

tan v 





 (
1)2

sin v 
sin v 
csc v 

tan v 

y2

csc v 

or 

2


cot v 
cot v 

1



2


2



 2

r2  (
1)2  (
1)2
r2  2
r  
2

y

x

cos v  r


1


2



2


 2
r

y

2

 or

1

cos v 
cos v 

tan v  x


1


2


2



 2


1


tan v  

1 or 1

r

sec v  x


2


sec v 


2



1

or 

2


r 2  x2  y2

y

22  x2  (
1)2
3  x2

3
x
Quadrant III, so x  

3


tan v  x
tan v 

r

5


1



2


1

x

r

tan v 


1



3


3


3

45. g sin v cos v  0
sin v  0 or cos v  0
v  0°
v  90°
46a. k is an even integer.

or 
5

x

cot v  y
cot v 

1


3


3


3

cot v  y

44. csc v  y

csc v  y

2
6



1

cos v 

csc v  
2
r  2, y  
1

x2


1

2
6


6



12

cos v 

r 2  x2  y2

y

or 

3




y



2



3



6


3
r

y

3




2



6


 2

sin v  r

csc v 

x2

tan v  x

tan v  x

cot v  1 (Quadrant III)
x  
1, y  
1

24  x2
2
6
x
Quadrant IV, so x  2
6


tan v 

sec v 

r2

x

y


3


1

y

x

cos v  r

43. cot v  y

22  x2  12
3  x2

3
x
Quadrant II, so x  

3



5


2

(
3
)2  12  y2
2  y2

2
y
Quadrant IV, so y  

2


x

r 2  x2  y2

csc v  2
r  2, y  1


5


csc v 

r 2  x2  y2

sin v  r

y

tan v  x

sin v  13 or 
1
3

1


5


5

cos v  
5
x
cot v  y
1
cot v  2

sec v  
3

r  
3
, x  1

25 
5  y
Quadrant III, so y  
5

r

csc v  y

cos v 

r

y2

sin v  r

2


5

2
5


5
r

x

5

 or
1

42. sec v  x

132  (
12)2  y2

x  
12, r  13

r2  12  22
r2  5
r  
5

x
cos v  r

y

sin v 

x

r

r 2  x2  y2

tan v  2
y  2, x  1

y

sin v  r
sin v 

y

41. tan v  x

47. cos v 

or 
2
6


cos v  
1

cos v  1
v  0°

146

46b. k is an odd integer.

I
t
Io

It  Io

48. Let x  
1. y  
3(
1)
y3
r2  x 2  y 2
r2  (
1)2  (3)2
r2  10
r  
10

sin v 

y

r

3
sin v  

10

3
10

sin v  
10
r
csc v  y


10
csc v  
3
x
cot v  y

1
cot v  3 or

x
cos v  r

1
cos v  

10


10

cos v  
 
10
r
sec v  x

10

sec v  

1 or

tan v 
tan v 

y

x
3


1

56.


10


C

sin v 
sin v 
51.


840

360

  9n 
 212  
4
2

n  
3

34, 
23, 12
2x 
 4y  7
?
2(9) 
 4(3)  7
6  7; yes
2x 
 4y  7
?
2(2) 
 4(
2)  7
12  7; no

2x 
 4y  7
?
2(
1) 
 4(2)  7

10  7; yes

1

59. absolute value; f(x)  22 
 x

  360(
2)°  
840°

60. A of square 
 A of circle  A
s2 
 r2  A
2
2 
 (1)2  A
0.86  A
The correct choice is C.

 
 720°  
840°
  
120°

5-4


b  

b2 
 4
ac

2a

9  

(
9)2 

 4(4
)(5)

2(4)
9  
1

x 
8
91
9
1
x  8 or x  8
10
8
x  8 or 1.25 x  8 or 1
y
y  kx
k  x
9

y  (
0.6)(21)
k

15

k  
0.6

4

1

x

54.

 
 11p  23

4m  9n 
 2p  
4

3

4

p  2

58.

360° 
 120°  240°; III
52. 5 
 
b
20

5  
b
2

25  b  2
23  b
53. x 

38

3

4

1

csc v
1

7

5
5

7

 
2.33

2
3

1
 2(2) 
 (
3)1
2
7

1
2

38m 
 11p  23

4

50. sin v 

5x


15 
10 
5 O

2

3

Since AC  36, AD  18.
18  4  22 ft
49c. Refer to 49b for diagram and reasoning.
Since AC  30, AD  15.
15 4  19 ft
1
r
2

2

m  4

4

49d.

4

57. 3(8m 
 3n 
 4p)  3(6)
24m 
 9n 
 12p  18
→ 4m  9n 
 2p  
4
4m  9n 
 2p  
4
28m 
 14p  14
4(8m 
 3n 
 4p)  4(6)
32m 
 12n 
 16p  24
6m  12n  5p  
1 → 6m  12n  5p  
1
38m

 11p  23
11(28m 
 14p)  11(14)

14(38m 
 11p)  
14(23)
↓
308m 
 154p  154

532m  154p  
322

224m
 
168

36

D

2

3
1

7

60˚

A

f (x) 
 x 2  16

f (x)


6

ABC is equilateral.
m∠BCA  60°
m∠ACD  m∠BCA  90°
m∠ACD  60°  90°
m∠ACD  30°

300˚

6


4

1

B

f(x)  x2 
 16
y  x2 
 16
x  y2 
 16
x  16  y2

x

 16  y

or 
3


3

49a. 4  2(36)  76 ft
49b.

36

55.

Applying Trigonometric Functions

Pages 301–302

Check for Understanding

1a. cos or sec
1b. tan or cot
1c. sin or csc
A
2. Sample answer: Find a.
10

C

y  
12.6

147

a

38˚

B

Chapter 5

3. ∠DCB; ∠ABC; the measures are equal; if parallel
lines are cut by a transversal, the alternate
interior angles are congruent.
4. Sample answer: If you know the angle of elevation
of the sun at noon on a particular day, you can
measure the length of the shadow of the building
at noon on that day. The height of the building
equals the length of the shadow times the tangent
of the angle of elevation of the sun.
5.

a

tan A  b

10

tan 49° 13
  —a—
a tan 49° 13
  10
10
—
a—
tan 49° 13

a  8.6
a

—
sin 16° 55
  —
13.7

b

a

13.7 sin 16° 55
  a
4.0  a

18


tan 76°  
113

sin 26°  c
c sin 26°  18

52.1  a

c

cos B 

7.

cos 16° 45
 

22.3

cos 47° 18
  —c—

8b. Let x 

of the base.

cos 55° 30
 

x

sin 55° 30
  10

c  32.9

10 sin 55° 30
  x
8.2  x
about 8.2 cm
8c.

x

10

10 cos 55° 30
  x
5.66  x
base  2x
base  2(5.66)
base  11.3 cm
9.

c cos 47° 18
  22.3
22.3
—
c—
cos 47° 18


8a. Let x  altitude.

13 cos 16° 45
  a
12.4  a
1

2

a

cos B  —c—

18.

18

sin 26°

c  41.1
a

c
a
——
13

a

sin A  —c—

17.

sin B  c

6.

13 tan 76°  a

b

tan B  —a—

16.

h

1
A  —2—bh
1
A  —2—(11.3)(8.2)
A  46.7 cm2

n

cos 30°  —12—

sin 30°  —12—

19.

12 sin 30°  h
6h

12 cos 30°  n
10.4  n

6

sin 45°  —p—

m tan 45°  6

p sin 45°  6

6

tan 45°  m
m

6

tan 45°

6


p
sin 45°

m6
20a.

p  8.5

10.8

cos 36°  —x—
x cos 36°  10.8

175

tan 13° 15
  —x—

10.8

—
x—
cos 36°

x tan 13° 15
  175
175

x
tan 13° 15


x  13.3 cm

x  743.2 ft

tan 36° 

20b.

1
 s
2

10.8
1

Pages 302–304
10.

Exercises
a

tan A  —b—

11.

a

cos 67°  —16—

6 tan 37°  a
4.5  a
b

sin B  —c—
sin 62° 

a

cos B  —c—
a

tan 37°  6

12.

10.8 tan 36°  2 s
2 10.8 tan 36°  s
15.7  s
about 15.7 cm
20c. P  5s
P  5(15.7)
P  78.5 cm

16 cos 67°  a
6.3  a
13.

b
——
24

a

sin A  —c—
sin 29° 

24 sin 62°  b
21.2  b

4.6
——
c

21a.

cos 42° 30
 

1

c sin 29°  4.6
4.6
—
c—
sin 29°

x cos 42° 30
  —2—(14.6)

c  9.5
14.

a

cos B  —c—
17.3

15.

c cos 77°  17.3

a tan 61°  33.2
—
a—
tan 61°

c  76.9

a  18.4

tan 42° 30
 

21b.

33.2

—
c—
cos 77°

1
——(14.6)
2

x

1
(14.6)
2

tan 42° 30
  x
6.7  x

about 6.7 m

Chapter 5

1
(14.6)
2

cos 42° 30


x  9.9 m

33.2

tan 61°  —a—

17.3

x

b

tan B  —a—

cos 77°  —c—

1
(14.6)
2

x

148

1

1

21c. A  —2—bh

28. Let M represent the point of intersection of the
altitude and E
F
. Since GEF is isosceles, the
altitude bisects 
EF
. EMG is a right triangle.
a
a
—
Therefore, sin v  —s— or s sin v  a and tan v  —
0.5b
or 0.5b tan v  a.
29. Latasha:
Markisha:

22a. r  —2—(6.4) or 3.2

1

a

A  —2—(14.6)(6.7)

—
cos 30°  —
3.2

3.2 cos 30°  a
2.771281292  a
about 2.8 cm
22b. Let x  side of hexagon. 22c. P  6s
1
P  6(3.2)
x
2
P  19.2 cm

sin 30°  3.2
A  48.8 m2

x

1

2 3.2 sin 30°  x
3.2  x; 32 cm
1

22d. A  —2—pa
1

y tan 47° 30
  x

A  —2—(19.2)(2.771281292)

y

A  26.6 cm2
195.8
sin 10° 21
 36  —x—
x sin 10° 21
 36  195.8
195.8
—
x—
sin 10° 21
 36
1

V  —3— area of base height

tan  
1
——s
2

x

1
s
2

1

1

V  —6— s3 tan 

tan   x



3

—12—, 
2 

84 ft
8 ft
25b. 84 
 8  76
76

x  87.8 ft

sec 120° 

csc 120° 

csc 120° 
cot 120° 

1

1


2

cot 120° 

sec 120°  
2

x tan 6°  3900
3900
——
tan 6°

32.

x  37,106.0 ft
3900

sin 6°  —x—
x sin 6°  3900
3900

—
x—
sin 6°

sin P 

Barge:

tan 20°  —x—

tan 12° 30
  —x—

tan P 

x tan 20°  208

x tan 12° 30
  208

tan P 

208

208
——
tan 20°

x  571.5
938.2 
 571.5  366.8 ft; no

x

cot 120° 



72  22  (PQ)2
53  (PQ)2

53
  PQ
P
—
—
sin P  r
(PR)2

sin P 

x  37,310.4 ft

x

1

csc 120°  y

1

3900

208

1

sec 120°  —x—

tan 6°  —x—

27. Yacht:


3


2
y

x

3


2

1


2

tan 120°  

3


x  43.9 ft

26b.

cos 120°  
—2—

tan 120° 

76

x sin 60°  76
76
—
x—
sin 60°

x tan 60°  76
76
—
x—
tan 60°

cos 120°  x

sin 120° 

sin 60°  —x—

25c.

tan 60°  —x—

sin 120°  y
tan 120° 

60˚

x

40  x

—
—
tan 54° 54


40 tan 47° 30


25a.

26a.

40  x

—
y—
tan 54° 54


x  ———
tan 54° 54
 
 tan 47° 30

x  131.7 ft
31. terminal side — Quadrant II
reference angle: 180° 
 120°  60°

V  —3— (s2)—2— s tan 
1

y tan 54° 54
  40  x

x
——
tan 47° 30

x

tan 47° 30


tan 54° 54
(x)  tan 47° 30
(40  x)
x tan 54° 54
  40 tan 47° 30
 
x tan 47° 30

x(tan 54° 54
 
 tan 47° 30
)  40 tan 47° 30


x  1088.8 ft
24. height:

—
sin 42°  —
225

250 sin 35°  x
225 sin 42°  x
143.4  x
150.6  x
1506 
 143.4  7.2
Markisha’s; about 7.2 ft
30. Let x  the height of the building.
Let y  the distance between the buildings.
x
40  x
tan 47° 30
  —y—
tan 54° 54
  y

32 sin 30°  —2—x

23.

x

—
sin 35°  —
250

(RQ)2

2


53

2
53


53
P
——
q
2
——
7

1


3


2
2
3


3
x
——
y
1


2



3

2

3



3

(PQ)2

q

cos P  —r—
cos P 
cos P 

7


53

7
53


53



33. 43° 15
 35  43°  15°
60
   35 3600 
 43.260°
1°

208
——
tan 12° 30


x  938.2

149

1°

Chapter 5

4. r  

x2  y2
r  

22  (

5)2
r  
29


y

34.

y

sin v  —r—

y  |x  2|

sin v 

x

O

36.

m miles
——
h hours


5


29

5
29


sin v  
 
29

tan v  x

cos v 

2


29


tan v  —2— or 
2

cos v 

2
29


29

r

35. Let x  the cost of notebooks and y  the cost of
pencils.
3x  2y  5.89
4x  y  6.20
3x  2y  5.80
→
3x  2y 
5.80

2(4x  y)  
2(6.20)

8x 
 2y  
12.40

5x
 
6.60
4x  y  6.20
x  $1.32
4(1.32)  y  6.20
y  $0.92


5

5

r

csc v  —y—

sec v  x


29

29


 or 
 
5

5
x
cot v  —y—
2
2
—

cot v  —

5 or 
 5
550
tan 27.8°  —x—

csc v 

5.

y

x

cos v  r

sec v 


29


2

x tan 27.8°  550
550


x
tan 27.8°

mx

x  1043.2 ft

x hours  —h— miles

The correct choice is E.

Page 304

1. 34.605 °  34°  (0.605 60)

 34°  36.3

 34°  36
  (0.3 60)
 34°  36
  18
34° 36
 18
2.


400°
——
360°

Pages 308–309


1.11  1  
0.11

0.11 360°  
40°
360° 
 40°  320°; IV
3. (GI )2  (IH )2  (GH )2
(GI )2  102  122
(GI )2  44
GI  
44
 or 2
11


5. 60°, 300°

cos G 
csc G 

10

2
11

5
11


tan G  11
hypotenuse
—
sec G  —
side adjacent
12
sec G  
2
11


csc G 

tan G 

Chapter 5

1

7.

side adjacent

side opposite



11

2
11
 or 
12
6
hypotenuse
——
side opposite
12
6
 or ——
10
5

8.

side adjacent
2
11


10

or

6. 150°, 330°


3

sin sin
1 
2


3


3

Let A  sin
1 
2 . Then sin A  2 .


3


3
sin sin
1 
 
2
2
3
3
Let A  cos
1 —5—. Then cos A  —5—
r 2  x2  y2









52  32  y2
16  y2
4y
4
3
4
tan A  —3—; tan cos
1 —5—  —3—

—
cot G  —
side opposite

cot G 

30˚

—
not —
cos .

10

5

1b. angle

4. Marta; they need to find the inverse of the cosine,

—
cos G  —
hypotenuse

—
tan G  —
side adjacent

sec G 

60˚

side opposite


sin G  —1—
2 or 6

Check for Understanding

1a. linear
2. They are complementary.
3. Sample answer:

 
1.11

—
sin G  —
hypotenuse

Solving Right Triangles

5-5

Mid-Chapter Quiz


11


5

6
11


11

150

r

r

9. tan R  —s—

4

10. cos S  —t—

7

cos arccos —5—  —5—

12

tan R  —1—
0

4

cos S  —20—
7

12

R  tan
1 —1—
0
R  35.0°
11. A  78°  90°
A  12°
Find b.

S  cos
1 —20—

2

tan tan
1 —3—  —3—
2

Find c.

—
sec A  —
cosA

41

sec A 

cos 78°  —c—

41 tan 78°  b

c cos 78°  41

sec A 

41

192.9  b

—
c—
cos 78°

2

A  12°, b  192.9, c  197.2
a2  b2  c2
Find B.

tan B 


562
c

tan B 

1

—
csc A  —
sin A

b
——
a
21
——
11

1

csc A  —1— or 1
csc (arcsin 1)  1
21

5

r 2  x2  y2
132  52  y2
144  y2
12  y

Find A.
A  62.35402464  90
A  27.64597536
c  23.7, A  27.6°, B  62.4°
13. 3.2°  B  90°
B  58°
Find a.
Find b.

——
tan A  —5—; tancos
1 —1—
3  5
12

2

cos A  —c—

a

cos 32°  —1—
3

b



21
;
5
n
 —m—
15
 —9—

cos A 
28. tan N

1280

—
14a. tan x  —
2100

tan N
1280

—
x  tan
1 —
2100

cos sin
1 —5— 
2

1280

tan 38°  —x—

m
8

sin M  —1—
4
M  sin
1 —14—

N  59.0°

M  34.8°

cos M 

1280
——
tan 38°

n
——
p
22
——
30

n

31. tan N  —m—
18.8

—
tan N  —
14.3
22

M  cos
1 —30—

x  1638.3 ft

M  42.8°
m
32. cos N  —p—

1280

tan 65°  —x—
x tan 65°  1280

32.5

7.2

—
sin M  —
54.7
7.2

x  596.9 ft

Exercises

—
M  sin
1 —
54.7

N  65.1°

M  36.5°

18
——
24

24

tan B  1
8
18

151

32.5

—
N  cos
1 —
17.1

34. tan A 
90°
16. 120°, 300°
30°, 330°
18. 90°, 270°
225°, 315°
20. 135°, 315°
Sample answers: 30°, 150°, 390°, 510°

18.8

—
N  tan
1 —
14.3

N  52.7°
m
33. sin M  —p—

—
cos N  —
17.1

1280

—
x—
tan 65°

8

N  tan
1 —9—
30. cos M 

x tan 38°  1280


21


5

29. sin M  —p—
15

x  31.4°

15.
17.
19.
21.

2

r 2  x2  y2
52  x2  22
21  x2

21
x

b

13 sin 32°  a
13 cos 32°  b
6.9  a
11.0  b
B  58°, a  6.9, b  11.0

Pages 309–312

12

5

27. Let A  sin
1 —5—. Then sin A  —5—

a

sin 32°  —13—

5

——
26. Let A  cos
1 —1—
3 . Then cos A  13

B  62.4°

sin A  —c—

5

25. Let A  arcsin 1. Then sin A  1.

B  tan
1 —11—

23.7  c

1

2

5
5
——
2

sec cos
1 —5—  —2—

c  197.2

112  212  c2

2

1

a

b

tan 78°  —4—
1

14c.

2

2

cos B  —c—

x

2

24. Let A cos
1 —5—. Then cos A  —5—.

b

14b.

4

23. Let A  tan
1 —3—. Then tan A  —3—.

S  53.1°

tan B  —a—

12.

4

22. Let A  arccos —5—. Then cos A  —5—.

24

A  tan
1 —24—

B  tan
1 —18—

A  36.9°

B  53.1°

Chapter 5

42. A  33°  90°
A  57°

1

35. —2—(14)  7
vertex angle  2m∠B

base angles:
tan A 

8
——
7

b

8

b

B  tan
1 —8—

A  48.8°

15.2 sin 33°  b
8.3  b
43. 14°  B  90°
B  76°

B  41.2°
2m∠B  82.4°

about 48.8°, 48.8°, and 82.4°
36. a2  b2  c2
212  b2  302
b  
459

b  21.4
44.427004  B  90
B  45.6°
37. 35°  B  90°
B  55°

9.8 sin 14°  a
2.4  a
647

—
v  sin
1 —
1020

v  39.4°
44b.

c cos 35°  8
8
—
c—
cos 35°

647

—
x—
tan 39.4°

x  788.5 ft
45a. Since the sine function is the side opposite
divided by the hypotenuse, the sine cannot be
greater than 1.
45b. Since the secant function is the hypotenuse
divided by the side opposite, the secant cannot
be between 1 and 
1.
45c. Since cosine function is the side adjacent divided
by the hypotenuse, the cosine cannot be less
than 
1.
46. 10 
 6  4

b

sin B  —c—
12.5

tan 47°  —a—

sin 47°  —c—

a tan 47°  12.5

c sin 47°  12.5

a  11.7
a2  b2  c2

12.5

—
c—
sim 47°

c  17.1
a

tan A  —b—
3.8

3.82  4.22  c2

—
tan A  —
4.2

4

tan v  —1—
5

3.8

—
A  tan
1 —
4.2


32.08
c
5.7  c
42.13759477  B  90
B  47.9°

4

v  tan
1 —1—
5

A  42.1°

v  14.9°
8

a2  b2  c2
sin B 
a2  3.72  9.52
sin B 
a  
76.56

3.7
—
a  8.7
B  sin
1 —
9.5
A  22.92175446  90
B  22.9°
A  67.1°
41. 51.5°  B  90°
B  38.5°
tan A 
tan 51.5° 

a
——
b
13.3
——
b

b tan 51.5°  13.3
13.3

Chapter 5

sin A 
sin 51.5° 

8

c  17.0

v  4.6°

v  2.9°

45
——
2200
45

v  1.2°
49.

65 miles
——
hour

tan v 
tan v 
v

13.3

b  10.6

—
v  tan
1 —
100

—
v  tan
1 —
2200

c sin 51.5°  13.3
—
c—
sin 51.5°

5

—
v  tan
1 —
100

48. tan v 

a
——
c
13.3
——
c

—
b—
tan 51.5°

5

—
47b. tan v  —
100

—
47a. tan v  —
100

b
——
c
3.7
——
9.5

40.

647

tan 39.4°  —x—
x tan 39.4°  647

c  9.8

12.5
——
tan 47°

9.8 cos 14°  b
9.5  b

647

8

12.5

b

—
cos 14°  —
9.8

—
44a. sin v  —
1020

cos 35°  —c—

a

a

A  44.4°

a

b

b

cos A  —c—

—
sin 14°  —
9.8

21

A  sin
1 —30—

b

tan B  —a—

15.2 cos 33°  a
12.7  a

a

cos A  —c—

38. A  47°  90°
A  43°

a

—
cos 33°  —
15.2

sin A  —c—

21

sin A  —30—

a

8 tan 35°  a
5.6  a

39.

a

sin A  c

tan A  —b—
tan 35°  —8—

cos B  —c—

—
sin 33°  —
15.2

7

A  tan
1 —7—

a

sin B  —c—

7

tan B  —8—

5280 feet
1 hour
—— ——
mile
3600 seconds
v2
——
gr
95.32
——
32 (1200)
95.32
—
tan
1 —
32(1200)

v  13.3°

152

feet

—
 95.3 —
second

50.

sin v
——i
sin vr
sin 60°
——
sin vr
sin 60°
——
2.42

 2.42
y-axis

 sin vr

0.3579  sin vr
sin
1 0.3579  vr
21.0°  vr
51. Draw the altitude from Y to XZ. Call the point of
intersection W.
m∠X  m∠XYW  90°
30°  m∠XYW  90°
m∠XYW  60°
In XYW:
XW
WY
cos 30°  —1—
sin 30°  —16—
6
16 cos 30°  XW
16 sin 30°  WY
13.9  XW
8  WY
In ZYW:
8

yx
y  
x

y-axis
56. 
1 0
0 1

—
tan 19.5°  —
WZ
8

Z  sin
1 —2—
4

8

—
WZ  —
tan 19.5°

WZ  22.6
cos m∠WYZ 

8

m∠WYZ  cos
1 —2—
4
m∠WYZ  70.5°
Y  m∠XYW  m∠WYZ
y  XW  WZ
Y  60°  70.5°
y  13.9  22.6
Y  130.5°
y  36.5
52. baseball stadium:
football stadium:
1000

x tan 63°  1000
1000
—
x—
tan 63°

y tan 18°  1000
1000
—
y—
tan 18°

x  509.5
distance  x  y
distance  509.5  3077.7
distance  3587.2 ft
53. (FD)2  (DE)2  (FE)2
72  (DE)2  152
(DE)2  176
DE  
176
 or 4
11

side opposite

—
sin F  —
hypotenuse

tan F 
tan F 
sec F 
sec F 

4
11


15
side opposite
——
side adjacent
4
11


7
hypotenuse
——
side adjacent
15

7

22.2 
 42.5


20.3

m  —7—
0 or 
0.29

y  3077.7

y 
 22.2  
0.29(x 
 1950)
y  
0.29x  587.7
59. 2x  5y 
 10  0
5y  
2x  10
2

y  
—5—x  2
2


—5—; 2
side adjacent

csc F 
csc F 
cot F 
cot F 


5 
5 
3 
1 
2

3
4
6
3 
2


58. m  
1950 
 1880

—
cos F  —
hypotenuse

cos F 

no

4  (
2) 
3  2
2  (
2)
 8  (
5) 
2  1
01
9  (
7)
6  2 
3  (
2)
2 
1
0
 3 
1
1
2
8 
5

1000

tan 18°  —y—

no

(5, 
3), (5, 4), (3, 6), (1, 3), (2, 
2)
4 
3
2

2 2 
2
57. 8 
2
0  
5 1
1
9
6 
3

7 2 
2

8
——
24

tan 63°  —x—

yes


1(
2)  0(
2)
0(
2)  1(
2)
5
5 3 1
2


3 4 6 3 
2

WZ tan 19.5°  8

Z  19.5°

no

 
1(
5)  0(
3) 
1(
5)  0(4)
0(
5)  1(
3)
0(
5)  1(4)

1(
3)  0(6) 
1(
1)  0(3)
0(
3)  1(6)
0(
1)  1(3)

8

sin Z  —2—
4

sin F 

y3 
 x2  2
(
y)3 
 x2  2

y3 
 x2  2;
y 3 
 x2  2
3
y 
 (
x)2  2
y3 
 x2  2;
y 3 
 x2  2
3
(x) 
 (y)2  2
x3 
 y2  2;
y3 
 x2  2
(
x)3 
 (
y)2  2

x3 
 y2  2;

55. x-axis

n

60.

7

15
hypotenuse
——
side opposite
15
15
11

 or 
4
11

44
side adjacent
——
side opposite
7
7
11

 or 
4
11

44

a
——
ac

ab  cd

c

—
——
b—
a  c d  10  a  c  10

The correct choice is A.

The Law of Sines

5-6
Page 316

54. Use TABLE feature of a graphing calculator.

0.3, 1.4, 4.3

1.

x
——
sin 30°
x

1

2

Check for Understanding



2x 

153

x
3


sin 60°



2x

sin 90°


x
3



3

2



2x

1

2x



2x

60˚

x

2x

x 
3

30˚

Chapter 5

Pages 316–318

2. Sample answer:

A
b

C

b
——
sin B
b
——
sin 70°

c

35˚

40˚
10

B

1

a
——
sin A
a
——
sin 30°

1

—
K—
2ab sin X

b
——
sin B
b
——
sin 59°

14

—
—
sin 81°
14 sin 40°

a
——
sin A
a
——
sin 25°

8.6

8.6 sin 55.9°

—
b—
sin 27.3°

a

—
—
sin A

8.6 sin 96.8°

—
c—
sin 27.3°

8.2

—
—
sin 93.9°

a
——
sin 39
 51


19.3

—
—
sin 64° 45

19.3 sin 39° 15


—
a—
sin 64° 45


78°

sin A sin C

sin 22° sin 53°

—
K  —2—(14)2 —
sin 105°

b
——
sin 51° 30


K  30.4 units2
10. Let d  the distance from the fan to the pitcher’s
mound.
v  180° 
 (24° 12
  5° 42
) or 150° 6


c  18.11843058

b
——
sin B
b
——
sin 50°

c

—
—
sin C
12

—
—
sin 65°
12 sin 50°

—
b—
sin 65°

c
——
sin 61.3°

8.2

—
—
sin 93.9°
8.2 sin 61.3°

—
c—
sin 93.9°

b
——
sin 76°

19.3

—
—
sin 64° 45

19.3 sin 76°

—
b—
sin 64° 45


125

—
—
sin 91° 10

125 sin 51° 31


—
b—
sin 91° 10


b  97.8
18. A  180° 
 (29° 34
  23° 48
) or 126° 38

a
b
——  ——
sin A
sin B

60.5

—
—
sin 5° 42


a
——
sin 126° 38


60.5 sin 150° 6


—
d—
sin 5° 42


11

—
—
sin 29 ° 34

11 sin 126° 38


—
a—
sin 29° 34


d  303.7 ft

a  17.9

Chapter 5

12 sin 120°

a  13.50118124
b  20.7049599
B  76°, a  13.5, b  20.7
17. C  180° 
 (37° 20
  51° 30
) or 91° 10

a
c
——  ——
sin B
sin C


K  —2— b2 
sin B

d
——
sin 150° 6


12

—
—
sin 35°

b  3.447490503
c  7.209293255
A  93.9°, b  3.4, c  7.2
16. B  180° 
 (39° 15
  64° 45
) or 76°
a
c
b
c
——  ——
——  ——
sin A
sin C
sin B
sin C

17

—
—
sin 63° 50


K
9. C  180° 
 (22°  105°) or 53°
1

b

—
—
sin B

—
c—
sin 35°

8.2 sin 24.8°

c  18.7

1

12

—
b—
sin 93.9°

a

17 sin 98° 15


K

30 sin 100°

c
——
sin C
c
——
sin 120°

—
—
sin 35°

b
——
sin 24.8°

—
c—
sin 63° 50


8. K 

30

—
—
sin 50


a  12
b  10.14283828
C  65°, a  12, b  10.1
15. A  180° 
 (24.8°  61.3°) or 93.9°
b
a
c
a
——  ——
——  ——
sin B
sin A
sin C
sin A

8.6

—
—
sin 27.3°

—
—
sin A

1
—— bc sin A
2
1
——(14)(12) sin
2
82.2 units2

b

—
—
sin B

a
c
——  ——
sin A
sin C
a
12
——  ——
sin 65°
sin 65°
12 sin 65°
—
a—
sin 65°

b  15.52671055
c  18.61879792
C  96.8°, b  15.5, c  18.6
7. A  180° 
 (17° 55
  98° 15
) or 63° 50

c
——
sin C
c
——
sin 98° 15


c

—
—
sin C

—
b—
sin 50°

a  8.84174945
C  120°, a  8.8, c  18.1
14. C  180° 
 (65°  50°) or 65°

14

—
—
sin 81°

c
——
sin C
c
——
sin 96.8°

—
—
sin 27.3°

30

—
—
sin 50°

12 sin 25°

a  9.111200533
b  12.14992798
C  81°, a  9.1, b  12.1
6. C  180° 
 (27.3°  55.9°) or 96.8°
a

b
——
sin B
b
——
sin 100°

—
a—
sin 35°

c

—
—
sin C

—
b—
sin 81°

—
—
sin A

c

—
—
sin C

a  19.58110934
b  38.56725658
A  30°, a  19.6, b  38.6
13. C  180° 
 (25°  35°) or 120°

14 sin 59°

—
a—
sin 81°

b
——
sin B
b
——
sin 55.9°

20 sin 70°

30 sin 30°

K  ab sin X
4. Both; if the measures of two angles and a nonincluded side are known or if the measures of two
angles and the included side are known, the
triangle is unqiue.
5. C  180° 
 (40°  59°) or 81°
c

20

—
—
sin 40°

—
c—
sin 40°

—
a—
sin 50°

1

K  —2—ab sin X  2ab sin X

—
—
sin C

20

—
—
sin 40°

a

—
—
sin A

b  29.238044
c  29.238044
B  70°, b  29.2, c  29.2
12. A  180° 
 (100°  50°) or 30°

k  —2—ab sin X

1

c
——
sin C
c
——
sin 70°

20 sin 70°

1

—
K—
2ab sin Z

a

—
—
sin A

—
b—
sin 40°

3. Area of WXYZ  Area of triangle ZWY  Area of
triangle XYW.
m∠X  m∠Z
triangle ZWY:
triangle XYW:

a
——
sin A
a
——
sin 40°

Exercises

11. B  180° 
 (40°  70°) or 70°

154

28c. P  112.7  72.7  80
P  265.4 ft

1

19. K  —2— bc sin A
K

1
——(14)(9)
2

sin 28°

m

1

r
s
n sin N
——  ——. Thus sin M  —— and sin R
sin R
sin S
n
r sin S
——. Since ∠M  ∠R, sin M  sin R and
s
m sin N
r sin S
——  ——. However, ∠N  ∠S and
n
s
m
r
m
n
sin N  sin S, so —n—  —s— and —r—  —s—. Similar

sin B sin C

—
K  —2—a2 —
sin A
1

sin 37° sin 84°

—
K  —2—(5)2 —
sin 59°

K  8.7 units2
21. C  180° 
 (15°  113°) or 52°
1

MNP  RST.
30. 360° 5  72°

sin 15° s in 52°

—
K  —2—(7)2 —
sin 113°

K  5.4 units2
22. K 
K
K

triangle:

1
——bc sin A
2
1
——(146.2)(209.3) sin
2
13,533.9 units2

62.2°

1
1

K  —2—(12.7)(5.8) sin 42.8°
K  25.0 units2
24. B  180° 
 (53.8°  65.4°) or 60.8°

4 sin 62° 30


—
x—
sin 97° 15


x  3.6 mi
31b. v  180° 
 (20° 15
  62° 30
) or 97° 15

Let y  the distance from the balloon to the
football field.
4
4
——  ——
sin 20° 15

sin 97° 15


sin B sin C

—
K  —2—a2 —
sin A

K

1
sin 60.8° sin 65.4°
— (19.2)2 ———
2
sin 53.8°
181.3 units2

K
25. K  ab sin X (formula from Exercise 3)
K  (14)(20) sin 57°
K  234.8 cm2
26.
Area of pentagon 
5 Area of triangle
360° 5  72°
9

K

72˚

1
——(9)(9)
2

K  38.51778891
27.

5

4 sin 20° 15


—
y—
sin 97° 15


y  1.4 mi
32. 180° 
 30°  150°
v  180° 
 (26.8°  150°) or 3.2°
Let x  the length of the track.
x
——
sin 26.8°

9

sin 72°

45˚

1

K  —2—(300)(300)sin 72°

K  42,797.54323
pentagon: 5K  5(42,797.54323)
5K  213,987.7 ft
31a. v  180° 
 (20° 15
  62° 30
) or 97° 15

Let x  the distance from the balloon to the
soccer fields.
x
4
——  ——
sin 62° 30

sin 97° 15


23. K  —2—ac sin B

1



proportions can be derived for p and t. Therefore,

sin A sin C

—
K  —2— b2 —
sin B
1

n

—
——
29. Applying the Law of Sines, —
sin M  sin N and

K  29.6 units2
20. A  180° 
 (37°  84°) or 59°

100

—
—
sin 3.2°
100 sin 26.8°

—
x—
sin 3.2°

x  807.7 ft
33a. Let x  the distance of the second part of the
flight.
v  180° 
 (13°  160°) or 7°

5K  5(38.51778891)
5K  192.6 in2
Area of octagon 
8 Area of triangle
360° 8  45°

x
——
sin 13°

80

—
—
sin 7°
80 sin 13°

—
x—
sin 7°

x  147.6670329
distance of flight  80  147.7 or about 227.7 mi
33b. Let y  the distance of a direct flight.

5

y
——
sin 160°

80

—
—
sin 7°
80 sin 160°

—
y—
sin 7°
1

K  —2—(5)(5) sin 45°

K  8.838834765
8K  70.7
28a. 180° 
 (95°  40°)  45°
28b.

x
——
sin 95°

y  224.5 mi

8K  8(8.838834765)

80

—
—
sin 45°
80 sin 95°

—
x—
sin 45°

x  112.7065642
about 112.7 ft and 72.7 ft

y
——
sin 40°

ft2
80

—
—
sin 45°
80 sin 40°

—
y—
sin 45°

y  72.72311643

155

Chapter 5

x

34.

55˚

y

90° 
 63°  27°
180° 
 (55°  63°)  62°
Let x  the vertical
distance.
Let y  the length of the
overhang.

63˚
62˚

13 ft

sin v 

cos v 
13

y
——
sin 63°

—
—
sin 63°
13 sin 27°
——
sin 63°

35b.

a
——
sin A
a
——
b
a
——
sin A
a
——
c

a
——
c
a
——
c


35


6


1


35


35


tan v  
 35
x
cot v  —y—

35

cot v  

1

r

sec v 

y  6.684288563

sec v 

6


35

6
35


35

y

sin A

—
—
sin B
c

—
—
sin C
sin A

—
—
sin C

8

sin A
sin A

sin C

—
——

 —c—  —
sin C 
 sin C



sin A 
 sin C
——
sin C
a
c

O

sin A 
 sin C

—
35c. From Exercise 34b, —c—  —
sin C
sin A 
 sin C
sin C
—  ——.
or —
a
c
c
a
c
——  ——
sin A
sin C
a
sin A
——  ——
c
sin C
a
sin A
——  1  ——  1
c
sin C
sin A
a
c
sin C
——  ——  ——  ——
sin C
c
c
sin C
ac
sin A  sin C
——  ——
c
sin C
sin C
sin A  sin C
——  ——
c
ac
sin A 
 sin C
—
Therefore, —
a
c
ac
sin A  sin C
—
——
or —
a 
 c  sin A 
 sin C .
a
——
sin A
a
——
b
a
——  1
b
a
b
——  ——
b
b
ab
——
b
b
——
ab

or 

35


sin A  sin C

—
—
ac

b

—
—
sin B
sin A

—
—
sin B
sin A

—
—
sin B  1
sin B

sin A

(2, 11)

(8, 7)
y
4

(2, 4)

(8, 4)

4

12

x
16

M(x, y)  100x  250y
M(2, 4)  100(2)  250(4) or 1200
M(2, 11)  100(2)  250(11) or 2950
M(4, 11)  100(4)  250(11) or 3150
M(8, 7)  100(8)  250(7) or 2550
M(8, 4)  100(8)  250(4) or 1800
4 standard carts, 11 deluxe carts
40. 4x  y  2z  0
3x  4y 
 2z  20
7x  5y 
 2z  20
3(3x  4y 
 2z)  3(20)
2(
2x  5y  3z)  2(14)
↓
9x  12y 
 6z  60

4x  10y  6z  28
5x  22y
 88

5(7x  5y)  
5(20) → 
35x 
 25y  
100
7(5x  22y)  7(88)
35x  154y  616
129y  516
y4
7x  5y  20
4x  y  2z  0
7x  5(4)  20
4(0)  4  2z  0
x0
z  
2
(0, 4, 
2)
41. 
6  3x 
 y
3x 
 y  12
y  3x  6
y  3x 
 12

—
——
—
sin B  sin B
sin A  sin B

—
—
sin B
sin B

—
—
sin A  sin B
45

36. tan v  —20—
45

v  tan
1 —20—

y

v  66.0°

6  3x  y  12

O

Chapter 5

6

—
csc v  —

1 or 
6

38. 83°  360k°
39. Let x  standard carts and let y  deluxe carts.
2x8
y x
2 x
8
16
4  y  11
x  y 
 15
x  y  15
(4, 11)
y 
 11
12

b

—
—
sin B

c

r

csc v  —y—

tan v 

—

1—
sin C 
 1

a
c
——
c

35d.

tan v  —x—

sec v  —x—

6.6 sin 63°
——
sin 62°

35  x2

35
x

Quadrant IV, so x  
35

x

66

—
—
sin 62°

y

x  6.623830843
about 6.7 ft
35a.

62  x2  (
1)2

cos v  —r—

63˚

x

r 2  x2  y2

1

—6—

y  
1, r  6

27˚

x
——
sin 27°

y

37. sin v  —r—

156

x

5. Since 44°  90°, consider Case I.
a sin B  23 sin 44°
a sin B  23 (0.6947)
a sin B  15.97714252
12  16.0; 0 solutions
6. Since 17°  90°, consider Case I.
a sin C  10 sin 17°
 2.923717047
2.9  10  11; 1 solution

42. Area of one face of the small cube  12 or 1 in2.
Surface area of the small cube  6 1 or 6 in2.
Area of one face of large cube  22 or 4 in2.
Surface area of large cube  6 4 or 24 in2.
Surface area of all small cubes  8 6 or 48 in2.
The difference in surface areas is 48 in2 
 24 in2
or 24 in2.
The correct choice is A.

Page 319

c

sin C
11

sin 17°

History of Mathematics

1. See students’ work; the sum is greater than 180°.
In spherical geometry, the sum of the angles of a
triangle can exceed 180°.
2. See students’ work. Sample answer: Postulate 4
states that all right angles are equal to one
another.
3. See students’ work.

Page 323

10 sin 17°

A  sin
1 11
10 sin 17°

A  15.41404614
B  180° 
 (15.4°  17°) or about 147.6°
c

sin C
11

sin 17°

b  20.16738057
A  15.4°, B  147.6°, b  20.2
7. Since 140°  90°, consider Case II.
3  10; no solutions
8. Since 38°  90°, consider Case I.
b sin A  10 sin 38°
b sin A  6.156614753
6.2  8  10; 2 solutions
a

sin A
8

sin 38°

Check for Understanding

30˚

A
6

sin 30°

10



sin B 
B

93.6˚

C

A

10

sin B
10 sin 30°

6
10 sin 30°
sin
1 6





10



sin B
10 sin 38°

B  sin
1 8
10 sin 38°

B  50.31590502
180° 
   180° 
 50.3° or 129.7°
Solution 1
C  180° 
 (50.3°  38°) or 91.7°

B
6

b



sin B

sin B  8

1. A triangle cannot exist if m∠A  90° and
a  b sin A or if m∠A  90° and a  b.
2.
B
123.6˚ 6
30˚
26.4˚
10

b



sin 147.6°
11 sin 147.6°

Graphing Calculator Exploration

56.4˚

b



sin B


b
sin 17°

1. B  44.1°, C  23.9°, c  1.8
2. B  52.7°, C  76.3°, b  41.0; B  25.3°,
C  103.7°, b  22.0
3. The answers are slightly different.
4. Answers will vary if rounded numbers are used to
find some values.

Page 324

10



sin A

sin A  11

The Ambiguous Case for the Law
of Sines

5-7

a



sin A

C

a

sin A
8

sin 38°

c



sin C
c



sin 91.7°
8 sin 91.7°

180° 
   180° 
 56.4°
 123.6°
C  180° 
 (30°  123.6°)
 26.4°


c
sin 38°

c  12.98843472
Solution 2
C  180° 
 (129.7°  38°) or 12.3°

B  56.44269024
C  180° 
 (30°  56.4°)
 93.6°
3. Step 1: Determine that there is one solution for
the triangle.
Step 2: Use the Law of Sines to solve for B.
Step 3: Subtract the sum of 120 and B from 180 to
find C.
Step 4: Use the Law of Sines to solve for c.
4. Since 113°  90°, consider Case II.
15  8; 1 solution

a

sin A
8

sin 38°

c



sin C
c



sin 12.3°
8 sin 12.3°


c
sin 38°

c  2.768149638
B  50.3°, C  91.7°, c  13.0; B  129.7°,
C  12.3°, c  2.8

157

Chapter 5

13. Since 61°  90°, consider Case I.
a sin B  12 sin 61°
a sin B  10.49543649
8  10.5; 0 solutions
14. two angles are given; 1 solution
15. Since 100°  90°, consider Case II.
15  18; 0 solutions
16. Since 37°  90°, consider Case I.
a sin B  32 sin 37°
a sin B  19.25808074
27 19.3; 2 solutions
17. Since 65°  90°, consider Case I.
b sin A  57 sin 65°
b sin A  51.65954386
55 51.7; 2 solutions
18. Since 150°  90°, consider Case II.
6  8; no solution
19. Since 58°  90°, consider Case I.
b sin A  29 sin 58°
b sin A  24.59339479
26 24.6; 2 solutions

9. Since 130°  90°, consider Case II.
17 5; 1 solution
c

sin C
17

sin 130°

b



sin B
5



sin B
5 sin 130°

sin B  17
B  sin
117
5 sin 130°

B  13.02094264
A  180° 
 (13.0  130°) or 37.0°
c

sin C
17

sin 130°

a



sin A
a



sin 37.0°
17 sin 37.0°


a
sin 130°

a  13.35543321
A  37.0°, B  13.0°, a  13.4
10a.

45 ft

70 ft

a

sin A
26

sin 58°

10˚

90° 
 10°  80°
180° 
 80°  100°

10b.

b



sin B
29



sin B
29 sin 58°

sin B  2
6
B  sin
12

6
29 sin 58°

45

70
100˚
80˚

70

sin 100°

B  71.06720496
180° 
   180° 
 71.1° or 108.9°
Solution 1
C  180° 
 (58°  71.1°) or 50.9°

x˚
10˚



sin x 
x

a

sin A
26

sin 58°

45

sin x
45 sin 100°

70
45 sin 100°
sin
1 7
0



26 sin 50.9°

c  23.80359004
Solution 2
C  180° 
 (58°  108.9)° or 13.1°

x  39.3°
v  180° 
 (100° 
 39.3°)
or about 40.7°
45

c



sin 50.9°


c
sin 58°



10c.

c



sin C

a

sin A
26

sin 58°

70

c



sin C
c



sin 13.1°
26 sin 13.1°

100˚
80˚
y

sin 40.7°

39.3˚
y


c
sin 58°

c  6.931727606
B  71.1°, C  50.9°, c  23.8; B  108.9°,
C  13.1°, c  6.9

10˚

70



sin 100°
70 sin 40.7°


y
sin 100°

y  46.4 ft

Pages 324–326

Exercises

11. Since 57°  90°, consider Case I.
b sin A  19 sin 57°
b sin A  15.93474079
11  15.9; 0 solutions
12. Since 30°  90°, consider Case I.
c sin A  26 sin 30°
c sin A  13
13  13; 1 solution.
Chapter 5

158

24. Since 36°  90°, consider Case I.
c sin B  30 sin 36°
c sin B  17.63355757
19 17.6; 2 solutions

20. Since 30°  90°, consider Case I.
b sin A  8 sin 30°
b sin A  4
4  4; 1 solution
a

sin A
4

sin 30°

b

b

sin B
19

sin 36°



sin B
8



sin B
8 sin 30°

sin C  1
9

B  sin
14

C  sin
11

9

8 sin 30°

30 sin 36°

B  90
C  180° 
 (30° 
 90°) or 60°

C  68.1377773
180° 
   180° 
 68.1° or 111.9°
Solution 1
A  180° 
 (36°  68.1°) or 75.9°

c



sin C
c



sin 60°

C

b

sin B
19

sin 36°

4 sin 60°

sin 30°

C  6.92820323
B  90°; C  60°, c  6.9
21. Since 70°  90°, consider Case I.
a sin C  25 sin 70°
a sin C  23.49231552
24 23.5; 2 solutions
c

sin C
24

sin 70°




sin A 
A

a

sin A
25

sin A
25 sin 70°

24
25 sin 70°
sin
1 2
4



a  31.34565276
Solution 2
A  180° 
 (36°  111.9°) or 32.1°
b

sin B
19

sin 36°

a  17.1953669
A  75.9°, C  68.1°, a  31.3; A  32.1°,
C  111.9°, a  17.2
25. Since 107.2°  90°, consider Case II.
17.2 12.2; 1 solution



a

sin A
17.2

sin 107.2°


C  sin
1
17.2
12.2 sin 107.2°

b  13.46081025
Solution 2
B  180° 
 (70°  101.8°) or 8.2°

C  42.65491459
B  180° 
 (107.2°  42.7°) or 30.1°
a

sin A
17.2

sin 107.2°

b

sin B
b

sin 8.2°
24 sin 8.2°

sin 70°

a

a



sin 40°
20 sin 40°


a
sin 80°

b



sin B
b



sin 30.1°
17.2 sin 30.1°


b
sin 107.2°

b  9.042067456
B  30.1°, C  42.7°, b  9.0
26. Since 76°  90°, consider Case I.
b sin A  20 sin 76°
b sin A  19.40591453
5  19.4; no solution

b  3.640196918
A  78.2°, B  31.8°, b  13.5; A  101.8°,
B  8.2°, b  3.6
22. C  180° 
 (40°  60°) or 80°


sin A

12.2



sin C


sin C  
17.2

24 sin 31.8°

c

sin C
20

sin 80°

c



sin C

12.2 sin 107.2°

b



sin 31.8°

b

a



sin 32.1°
19 sin 32.1°

b



a



sin A


a
sin 36



sin B



a



sin 75.9°
19 sin 75.9°


b
sin 70°

c

sin C
24

sin 70°

a



sin A


a
sin 36°

A  78.1941432
180° 
   180° 
 79.2° or 101.8°
Solution 1
B  180° 
 (70°  79.2°) or 31.8°
c

sin C
24

sin 70°

30



sin C
30 sin 36°

sin B  4

a

sin A
4

sin 30°

c



sin C

c

sin C
20°

sin 80°

b



sin B
b



sin 60°
20 sin 60°


b
sin 80°

a  13.05407289
b  17.58770483
C  80°, a  13.1, b  17.6
23. Since 90°  90°; consider Case II.
12  14; no solution

159

Chapter 5

27. Since 47°  90°, consider Case I.
16  10; 1 solution
c

sin C
16

sin 47°

a

a

sin A
15

sin 29°

10

sin B  1
5

31.



sin A

A

B  sin
11

5
20 sin 29°

10 sin 47°

16
10 sin 47°

1

sin
16



B  40.27168721
180° 
   180° 
 40.3° or 139.7°
Solution 1
C  180° 
 (29° 
 40.3°) or 110.7°



A  27.19987995
B  180° 
 (47° 
 27.2) or 105.8°
c

sin C
16

sin 47°

a

sin A
15

sin 29°

b



sin B
b



sin 105.8°

c  28.93721187
Perimeter  a  b  c
 15  20  28.9 or about 63.9 units
Solution 2
C  180° 
 (29° 
 139.7°) or 11.3°
a

sin A
15

sin 29°

c



sin C



sin C 
C






a

c  6.047576406
Perimeter  a  b  c
 15  20  6.0 or about 41.0 units


32.

15 sin 55°

A  70.93970395
180° 
   180° 
 70.9° or 109.1°
Solution 1
C  180° 
 (70.9°  55°) or 54.1°
b
c
  
sin B
sin C
13

sin 55°

a

c



sin 54.1°
13 sin 54.1°



sin 26.7°


c
sin 55°

42 sin 26.7°

c  12.8489656
Perimeter  a  b  c
 15  13  12.8 or about 40.8
Solution 2
c  180° 
 (109.1°  55°) or 15.9°
b
c
  
sin B
sin C

a  29.33237132
A  73.3°, C  66.7°, a  62.6; A  26.7°,
C  113.3°, a  29.3
29. Since 125.3°  90°, consider Case II.
32  40; no solution

13

sin 55°

19.3



sin x

13 sin 15.9°

c  4.35832749
Perimeter  a  b  c
 15  13  4.4 or about 32.4
A  70.9°, B  55°, C  54.1°


x  sin
1
21.7
19.3 sin 57.4°

x  48.52786934
v  180 
 (57.4°  48.5°) or 74.1°
y



sin 74.1°
21.7 sin 74.1°


y 
sin 57.4°

c



sin 15.9°


c
sin 55°

19.3 sin 57.4°


sin x  
21.7

19.3 cm

74.1˚

21.7 cm

y  24.76922417
48.5˚
57.4˚
24.8 cm

Chapter 5

15



sin A

A  sin
113

a

21.7

sin 57.4°

a



sin A

15 sin 55°



sin A

21.7

sin 57.4°

b

sin B
13

sin 55°

sin A  1
3

a

sin A
a

sin 73.3°
42 sin 73.3°

sin 40°


a
sin 40°

30.

c



sin 11.3°
15 sin 11.3°

a  62.58450564
Solution 2
A  180° 
 (40°  113.3°) or 26.7°
b

sin B
42

sin 40°

c



sin C


c
sin 29 °

C  66.67417652
180° 
   180° 
 66.7° or 113.3°
Solution 1
A  180° 
 (40°  66.7°) or 73.3°
b

sin B
42

sin 40°

c



sin 110.7°


c
sin 29 °

b  21,0506609
A  27.2°, B  105.8°, b  21.1
28. Since 40°  90°, consider Case I.
c sin B  60 sin 40°
c sin B  38.56725658
42 38.6; 2 solutions
60

sin C
60 sin 40°

42
60 sin 40°
sin
1 4
2

c



sin C

15 sin 110.7°

16 sin 105.8°


b
sin 47°

b

sin B
42

sin 40°

20



sin B
20 sin 29°



sin A

sin A 

b



sin B

160

37. Distance from satellite to center of earth is 3960 
1240 or 5200 miles.
angle across from 5200 mi side  45°  90° or 135°
5200
3960
  
sin 135°
sin x

33. side opposite 37°  15  18 or 33
side between v and 37°  15  22 or 37
Let x  the measure of the third angle.
33

sin 37°

37



sin x

3960 sin 135°

37 sin 37°


sin x  
5200

sin x  3
3

x  sin
133
x  42.43569405
v  180 
 (37°  42.4°) or about 100.6°
34a. a  b sin A
34b. a  b sin A
a  14 sin 30°
a  14 sin 30°
a7
a  7 or a  14
34c. a b sin A
a 14 sin 30°
a 7 and a  14
7  a  14


x  sin
1
5200

37 sin 37°

35.

184.5

sin 59°

3960 sin 135°

x  32.58083835
v  180° 
 (135°  32.6°) or about 21.4°
21.4°
 (2 hours)  0.0689953425 hours or about 4.1
360°
minutes
38. P turns 20(360°) or 7200° every second which
equals 72° every 0.01 second.
PQ

sin O
15

sin 72°

140

sin Q  15
Q  sin
11

5

140 sin 59°

5 sin 72°


sin x  
184.5

x  sin
1
184.5 
140 sin 59°

Q  18.48273235
m∠P  180° 
 (72°  18.5°) or about 89.5°

x  40.57365664
v  180° 
 (59°  40.6°) or about 80.4°
90° 
 80.4°  9.6°
36a. 12°  90° and 316 450 sin 12°; 2 solutions

QO

sin P
QO

sin 89.5°

QO  15.77133282
QO 
 5  15.8 
 5 or about 10.8 cm
39a.
b  c sin B
12  17 sin B


v  sin
1
316
450 sin 12°

v  17.22211674
180° 
   180° 
 17.2° or 162.8°
turn angle  180° 
 162.8° or 17.2°
about 17.2° east of north
36b. v  180° 
 (162.8°  12°) or about 5.2°
x

12

17
12
sin
1 1
7

12

17
12
sin
1 1
7

x  138.3094714
d  rt
138.3  23t
6.013455278  t; about 6 hr
36c. 180° 
 20°  160°
180° 
 (160°  12°)  8°
200
c
  
sin 8°
sin 160°

Port

12

17
12
sin
1 1
7

20˚
160˚

 sin B
B

sin B
B

44.90087216 B
B  44.9°
40. Area of rhombus  2(Area of triangle)
triangle:

c
200 mi.

B

44.90087216  B
B  44.9°
39c.
b c sin B
12 17 sin B

200 sin 160°

c  491.5032301
Since 491.5 450, the ship
will not reach port.

 sin B

44.90087216  B
B 44.9°
39b.
b  c sin B
12  17 sin B

316

sin 12°
316 sin 5.2°

sin 12°


c
sin 8°

15



sin 72°
15 sin 89.5°

450 sin 12°



PQ



sin O


QO  
sin 72°

450



sin v


sin v  
316

x

sin 5.2°

5



sin Q
5 sin 72°



sin x

316

sin 12°

OP



sin Q

12˚

1

K  2bc sin A
1

K  2(24)(24) sin 32°
K  152.6167481
rhombus:
A  2(152.6) or about 305.2 in2

161

Chapter 5

4. Sample answers:

75

41. tan 22°  x

1

42. 3; 2

x
x
x
x

c

C

53˚
a

10

sin 53°  c
10


c
sin 53°

B

c  12.52135658
b

tan B  a
10

tan 53°  a

2  

188


8
2  2i
47


8
1  i
47


4

3x

x
1

10


a
tan 53°

x
1

3x

  
1
x
1
x
1




9x
3x




x
1


3 
x
1

5x 
 2y  9
5x 
 2(3x 
 1)  9
5x 
 6x  2  9
x  
7
(
7, 
22)
45. 
2x  5y  7
2

4x 
 1

x
1

9x

x
1

a  7.535540501
A  37°, a  7.5, c  12.5
C  180° 
 (5.5°  45°) or 80°
A
b
a
  
sin B
sin A
4x 
 1

 9
x

x

b

55˚

b

sin 45°

c
45˚

C

10


b
sin 55°

B

b  8.6321799
c
a
  
sin C
sin A
c
10
  
sin 80°
sin 55°
10 sin 80°

c
sin 55°

7

c  12.0222828
C  80°, b  8.6, c  12.0
c2  a2  b2 
 2ab cos C
c2  102  82 
 2(10)(8) cos 50°
c2  61.15398245
c  7.820101179

y  5x  5
5

perpendicular slope: 
2
5

y 
 4  
2(x 
 (
6))
5

y 
 4  
2x 
 15
2y 
 8  
5x 
 30
5x  2y  
22
46. Perimeter of XYZ  4  8  9 or 21

A
8

1

length of A
B
  3 of perimeter
1

 3(21) or 7

C

c

a
c
  
sin A
sin C
10
7.8
  
sin A
sin 50°
10 sin 50°

sin A  
7.8
10 sin 50°

B A  sin
1 
7.8



50˚
10

The Law of Cosines

322 
 382 
 462


2(38)(46)

322 
 382 
 462
cos
1 

2(38)(46)



Check for Understanding

1. The Law of Cosines is needed to solve a triangle if
the measures of all three sides or the measures of
two sides and the included angle are given.
2. Sample answer: 1 in., 2 in., 4 in.
3. If the included angle measures 90°, the equation
becomes c2  a2  b2 
 2ab cos C. Since
cos 90°  0, c2  a2  b2 
 2ab(0) or c2  a2  b2.

 cos A

A

43.49782861  A
a

sin A
32

sin 43.5°

b



sin B
38



sin B
38 sin 43.5°

sin B  3
2
B  sin
132
38 sin 43.5°

B  54.8
C  180° 
 (43.5°  54.8°) or 81.7°
A  43.5°, B  54.8°, C  81.7°

Chapter 5



A  78.4024367
B  180° 
 (78.4°  50°) or 51.6°
A  78.4°, B  51.6°, c  7.8
5.
a2  b2  c2 
 2bc cos A
322  382  462 
 2(38)(46) cos A

The answer is 7.

Pages 330–331

10



sin 55°
10 sin 45°

y  3x 
 1
y  3(
7) 
 1
y  
22

44.

5-8

10

2  

(
2)2 

 4(4)(12)


2(4)

43. no



A  180° 
 (90°  53°) or 37°
b
sin B  c

A

75


x
tan 22°
x  185.6 m
4

4
13

6
2

1
6
4

2
12  0
4x2 
 2x  12  0

162

6. c2  a2  b2 
 2ab cos C
c2  252  302 
 2(25)(30) cos 160°
c2  2934.538931
c  54.1713848

Pages 331–332

c
a
  
sin C
sin A
54.2
25
  
sin 160°
sin A
25 sin 160°

sin A  
54.2
25 sin 160°

A  sin
1 
54.2



7.8

sin 51°



7 sin 51°




B  sin
1
7.8 
7 sin 51°

B  44.22186872
C  180° 
 (51°  44.2°) or 84.8°
B  44.2°, C  84.8°, a  7.8
12.
c2  a2  b2 
 2ab cos C
72  52  62 
 2(5)(6) cos C
72 
 52 
 62


2(5)(6)
72 
 52 
 62

cos
1 

2(5)(6)

 cos v



v

 cos C

 C

78.46304097  C

81.0  v
about 81.0°
1
8. s  2(2  7  8)  8.5
K  
8.5(8.
5
)(8.5

 2


7)(8.5

 8)
 6.4 units2
1
9. s  2(25  13  17)  27.5
K  
27.5(2
7.5


7.5
25)(2

7.5
13)(2


17)
 102.3 units2
10.
a2  b2  c2
652  652  c2
65 ft
8450  c2
91.92388155  c
c
65 ft

7



sin B


sin B  
7.8

A  9.1
B  180° 
 (9.1°  160°) or 10.9°
A  9.1°, B  10.9°, c  54.2
7. The angle with greatest measure is across from
the longest side.
212  182  142 
 2(18)(14) cos v
212 
 182 
 142


2(18)(14)
212 
 182 
 142

cos
1 

2(18)(14)

Exercises

11. a2  b2  c2 
 2bc cos A
a2  72  102 
 2(7)(10) cos 51°
a2  60.89514525
a  7.803534151
a
b
  
sin A
sin B

a

sin A
5

sin A

c



sin C
7



sin 78.5°
5 sin 78.5°

sin A  7
A  sin
17
5 sin 78.5°

A  44.42268919
B  180° 
 (44.4°  78.5°) or 57.1°
A  44.4°, B  57.1°, C  78.5°
13.
c2  a2  b2 
 2ab cos C
72  42  52 
 2(4)(5) cos C
72 
 42 
 52


2(4)(5)






 cos C


cos
1

2(4)(5)   C
72



42

52

101.536959  C
a
c
  
sin A
sin C
4
7
  
sin A
sin 101.5°
4 sin 101.5°
sin A  7
4 sin 101.5°
A  sin
1 7


 
 2bc cos v
652  652  91.92 
 2(65)(91.9) cos v
a2

652 
 652 
 91.92


2(65)(91.9)

b2

c2

 cos v

652 
 652 
 91.92

cos
1 
v

2(65)(91.9)



45.01488334  v



A  34.05282227
B  180° 
 (34.1°  101.5°) or 44.4°
A  34.1°, B  44.4°, C  101.5°
14. b2  a2  c2 
 2ac cos B
b2  162  122 
 2(16)(12) cos 63°
b2  225.6676481
b  15.02223845

c
50 ft
65 ft

a

sin A
16

sin A

45.0˚

c2  a2  b2 
 2ab cos C
c2  652  502 
 2(65)(50) cos 45.0°
c2  2128.805922
c  46.1 ft

b



sin B
15.0



sin 63°
16 sin 63°


sin A  
15.0

A  sin
1
15.0 
16 sin 63°

A  71.62084388
C  180° 
 (71.6°  63°) or 45.4°
A  71.6°, C  45.4°, b  15.0

163

Chapter 5

b2  a2  c2 
 2ac cos B
13.72  11.42  12.22 

2(11.4)(12.2) cos B

15.

13.72 
 11.42 
 12.22


2(11.4)(12.2)
13.72 
 11.42 
 12.22
cos
1 

2(11.4)(12.2)



a

sin A
11.4

sin A

b

1

23. s  2(174  138  188)  250
76
12
12
6
K  
250
 11,486.3 units2

 cos B

1

24. s  2(11.5  13.7  12.2)  18.7

B

8.7


(18.7
11.5)

7)(18.
13.7
2.2)

 1
K  
187(1
 66.1 units2
25a. d2  302  482 
 2(30)(48) cos 120°
d2  4644
d  68.1 in.
25b. Area of parallelogram  2(Area of triangle)

70.8801474  B



sin B
13.7



sin 70.9°
11.4 sin 70.9°


sin A  
13.7

1

K  2(30)(48) sin 120°


A  sin
1
13.7
11.4 sin 70.9°

K  623.5382907
2K  2(623.5382907) or about 1247.1 in2

A  51.84180107
C  180° 
 (51.8°  70.9°) or 57.3°
A  51.8°, B  70.9°, C  57.3°
16. c2  a2  b2 
 2ab cos C
c2  21.52  132 
 2(21.5)(13) cos 79.3°
c2  527.462362
c  22.96654876
a
c
  
sin A
sin C
21.5
23.0
  
sin A
sin 79.3°
21.5 sin 79.3°

sin A  
23.0
21.5 sin 79.3°

A  sin
1 
23.0



1

26a. s  2(15  15  24.6)  27.3
7.3


7.3
15)(2

7.3
15)(2


24.6)
K  
27.3(2
 105.6
Area of rhombus  2(105.6)  211.2 cm2
26b.
24.62  152  152 
 2(15)(15) cos v
24.62 
 152 
 152


2(15)(15)

14.92 
 23.82 
 36.92


2(23.8)(36.9)



 cos v




152

74 ft

v

x

38 ft



13.75878964  v
about 13.8°
18. d12  402  602 
 2(40)(60) cos 132°
d12  8411.826911
d1  91.71601229
180° 
 132°  48°
d22  402  602 
 2(40)(60) cos 48°
d22  1988.173089
d2  44.58893461
about 91.7 cm and 44.6 cm

88 ft

382  742  882 
 2(74)(88) cos v

382 
 742 
 882


2(74)(88)






 cos v


cos
1

2(74)(88)   v
382

742

882

25.28734695  v
side opposite


sin v  
hypotenuse
x

sin 25.3°  74

1

19. s  2(4  6  8)  9

31.60970664  x
about 31.6 ft
29a. x2  1002  2202 
 2(100)(220) cos 10°
x2  15,068.45887
x  122.7536511
about 122.8 mi
29b. (100  122.7536511) 
 220  2.7536511
about 2.8 mi


4)(9 

6)(9 

8)
K  
9(9


 11.6 units2
1

20. s  2(17  13  19)  24.5
4.5


4.5
17)(2

4.5
13)(2


19)
K  
24.5(2
 107.8 units2
1

21. s  2(20  30  40)  45
)(45

 20

5
30)(40)

 4
K  
45(45
 290.5 units2
1

22. s  2(33  51  42)  63
)(63

 33

3
51)(62)

 4
K  
63(63
 690.1 units2

Chapter 5

152

110.1695875  v
180° 
 110.2°  69.8°
about 110.2°, 69.8°, 110.2°, 69.8°
27. The angle opposite the missing side  45°.
x2  4002  902 
 2(400)(90) cos 45°
x2  117,188.3118
x  342.3 ft
28.



A  66.90667662
B  180° 
 (66.9° 
 79.3°) or 33.8°
A  66.9°, B  33.8°, c  23.0
17.
14.92  23.82  36.92 

2(23.8)(36.9) cos v

14.92 
 23.82 
 36.92
cos
1 

2(23.8)(36.9)




 cos v

cos
1 
v

2(15)(15)
24.62

164

30.

side opposite

202 ft
82.5˚
I
201.5 ft

x

y

II

125 ft


34. tan v  
side adjacent

180.25 ft

570


tan v  
700
570


v  tan
1 
700

III
75˚

v  39.2°
35. 
775°  2(360°)  
55°
reference angle  
55° or 55°
36. 3 1

7

k
6
3

12

36 
 3k

30 
 3k
1

4

12 
 k

30 
 3k  0
k  
10

158 ft

1

I: K  2(201.5)(202) sin 82.5°
K  20,177.3901
II: x2  201.52  2022 
 2(201.5)(202) cos 82.5°
x2  70,780.6348
x  266.046302
y2  1582  180.252 
 2(158)(180.25) cos 75°
y2  42.711.98851
y  206.6687894
206.72  266.02  1252 

2(266.0)(125) cos v
206.72 
 266.02 
 1252


2(266.0)(125)

5t 
 t


37. m  
5t 
 2t
4t

4

3

23x6

m  3t or 3

2yx
2

38.

 cos v

 y
3
8x6

 y
3
The correct choice is A.

206.72 
 266.02 
 1252

cos
1 
v

2(266.0)(125)
48.93361962  v
K

1
(266.0)(125)
2

5-8B Graphing Calculator Exploration:

sin 48.9°

Solving Triangles

K  12,536.58384
III: K 

1
(180.25)(158)
2

sin 75°

Page 334

K  13,754.54228
Area of pentagon  I  II  III
 20,177.4  12,536.6  13,754.5
 46,468.5 ft
31. I:
242  352  402 
 2(35)(40) cos v
242 
 352 
 402


2(35)(40)
242 
 352 
 402


2(35)(40)

cos
1 

1. AB  12.1, B  25.5°, C  119.5°
2.

C

 cos v

v

242 
 302 
 202


2(30)(20)
242 
 302 
 202

cos
1 

2(30)(20)



40
51˚
O A (0, 0)

 cos v

v

x
50

B (50, 0)

51˚
O A (0, 0)

x
50

B (50, 0)

Find y using A
C
.
y  (tan 51°)x
Find y using 
BC
.
40
50
  
sin 51°
sin C

52.89099505  v
the player 30 ft and 20 ft from the posts
20,000

32a. sin 6°  x

50 sin 51°

sin C  4
0

20,000


x
sin 6°

C  sin
140
50 sin 51°

x  191,335.4 ft
15,000

C  76.27180414
B  80 
 51 
 76.27180414
 52.72819586
y

tan(180 
 52.72819586)  
x 
 50

32b. sin 3°  y
y

C

40

36.56185036  v
242  302  202 
 2(30)(20) cos v

II:

y

y

15,000

sin 3°

y  286,609.8 ft
32c. 6° 
 3°  3°
d2  191,335.42  286,609.82 

2(191,335.4)(286,609.8) cos 3°
d2  9,227,519,077
d  96,060.0 ft
33. Since 63.2°  90°, consider Case I.
b sin A  18 sin 63.2°
b sin A  16.06654473
17 16.1; 2 solutions

(x 
 50) tan(127.2718041)  y
Set the two values of y equal to each other.
(tan 51°)x  (x 
 50), tan 127.2718041°
(tan 51°)x  x(tan 127.2718041° 

50(tan 127.2718041°)
x


50(tan 127.2718041°)

tan 51° 
 tan 127.271804°

x  25.77612538
y  (tan 51°) (25.77612538)  31.83086394

165

Chapter 5

C could also equal 180 
 76.27180414 or
103.728195°
B  180 
 51 
 103.7281959°
 25.2718041
y

tan (180 
 25.2718041)  
x 
 50

15.

16.

17.


50(tan 154.7281959°)

tan 51° 
 tan 154.7281959°

18.

Pages 336–338

2.
4.
6.
8.
10.

19.

860°

360°

20.

false; arcosine
false; adjacent to
false; coterminal
false; Law of Cosines
true

  360(2)°  860°
  720°  860°
  140°; II

Chapter 5

14.

1146°

3 60°

1072°

360°

 2.98

654°

360°

 1.82


832°

360°

 
2.31

  360(
2)°  
832°
 
 720°  
832°
  
112°
360° 
 112°  248°; III
21. 
284° has terminal side in first quadrant.
360° 
 284°  76°
22.

592°

360°

 1.64

  360(1)°  592°
  360°  592°
  232°
terminal side in third quadrant
232° 
 180°  52°
23. (BC)2  (AC)2  (AB)2
152  92  (AB)2
306  (AB)2

306
  AB
3
34
  AB

Skills and Concepts

 2.39

 
0.83

  360(1)°  654°
  360°  654°
  294°; IV

11. 57.15°  57°  (0.15 60)

 57°  9

57° 9

12. 
17.125°  
(17°  (0.125 60)
)
 
(17°  7.5
)
 
(17°  7
  (0.5 60))
 
(17°  7
  30)

17° 7
 30
13.


300°

360°

  360(2)°  1072°
  720°  1072°
  352°; IV

Understanding and Using the
Vocabulary

false; depression
true
true
true
false; terminal side

 2.77

  360(
1)°  
300°
 
 360°  
300°
  60°; I

Chapter 5 Study Guide and Assessment

1.
3.
5.
7.
9.

998°

360°

  360(2)°  998°
  720°  998°
  278°; IV

x  13.82829048
y  (tan 51°) (13.82828048)
 17.07651659
B  52.7°, C  76.3°, b  40.9; B  25.3°,
C  103.7°, b  220
3. Law of Cosines
4. Sample answer: put vertex A at the origin and
vertex C at (3, 0).

Page 335

 
0.43

  360(
1)°  
156°
 
 360°  
156°
  204°; III

(x 
 50) tan 154.7281959°  y
Set the two values of y equal to each other.
(tan 51°) x  (x 
 50)tan 154.7281959°
(tan 51°) x  x(tan 154.7281959°) 
 50(tan
154.7281959°)
x


156°

360°

opposite side

 3.18


cos A  
hypotenuse

sin A 

cos A 

tan A 

  360(3)°  1146°
  1080°  1146°
  66°; I

tan A 

166

adjacent side


sin A  
hypotenuse

5
34

15
 or 
34
3
34

side opposite

side adjacent
15
5
 or 
9
3

9

3
34


or

3
34


34

29. r  

x2  y2
r  

82  (

2)2
r  
68
 or 2
17


24. (PM)2  (PN)2  (MN)2
82  122  (MN)2
208  (MN)2

208
  MN
4
13
  MN

y

sin v  x

opposite side

sin M 
tan M 

3
13

12
 or 
13
4
13

side opposite

side adjacent
12

cos M 
csc M 

3

tan M  8 or 2

csc M 

hypotenuse

cot M 


sec M  
side adjacent

4
13


13
 or 
2
8
(PN)2  (MN)2

sec M 


2

cos v 
2
17


17

sin v  
 
cos v 
17
r
csc v  y
2
17

csc v  


2 or 

17
x
cot v  y
8

cot v  

2 or 
4
2
x  y2
30. r  


sin v 

adjacent side


cos M  
hypotenuse


sin M  
hypotenuse


2
13
8
 or 
13
4
13

hypotenuse

side opposite


4
13

13
 or 
12
3
side adjacent

side opposite
2

8


cot M  1
2 or 3

25. (MP)2 
(MP)2  102  122
(MP)2  44
 or 2
11

MP  
44
opposite side

26.

10
5

sin M  1
2 or 6
side opposite

tan M  
side adjacent
10
5
11

tan M  
or 
11
2
11

hypotenuse

sec M  
side adjacent
12
6
11

sec M  
or 
11
2
11

1

sec v  
cos v
1

cos v  
secv



11

2
11

cos M  
12 or 6
hypotenuse

csc M  
side opposite
12
6

csc M  1
0 or 5
side adjacent

cot M  
side opposite

cot M 

28.


2

sin v  
2
r
csc v  y
3
2

csc v  
3 or
x
cot v  y
3
cot v  3 or 1
x2  y2
r  


cos v 
sec v 

2
11


10

or


11


5


2


tan v 
tan v 

y

x
3

3

y

sin v 
csc v 
csc v 

12

13
r

y
13

12

sec v 
sec v 

x

or

or

13

5

tan v 
cot v 
cot v 

12


5
x

y

5

12

cot v  0
undefined

x

tan v  x

4


41


4
41

41
r

x

41


4

tan v  4

cos v  r

5


41


5
41

sin v  41
r
csc v  y


41
csc v  
5
r  

x2  y2

cos v 
cos v 
sec v 
sec v 

y

5

x

cot v  y
4

cot v  5


9


106



106

 or

9
x
cot v  y

5
5


cot v  

9 or 9
r  

x2  y2

y

tan v  x


5


106


cos v 


9

9



tan v  

5 or 5


5
106

cos v  
 
106
r

sec v  x

106




9

sec v 


106



5


106


or 
 
5

r  

(
4)2 
 42
r  
32
 or 4
2


tan v  x
5

1
3


2

2

csc v  y

y

x

0


tan v  

2 or 0

cot v  y

r

cos v  r
cos v 


sec v  

2 or 
1


9
106

33.


5

13
r

x
13


5

2

csc v 


17


4

y

r

sin v  
 
106
or 
2


or

tan v  x

sin v  r

sin v 

r  

(
5)2 
 122
r  
169
 or 13
sin v  r

2
17


8

r  

(
5)2 
 (
9
)2
r  
106

y
x
sin v  r
cos v  r


2


2
r

x

sec v 

sec v 


2

y

or 1

1

r

undefined
31. r  

x2  y2
r  

42  52
r  
41


32.

3
2


3


2

sec v  x

sec v  x

csc v  0


5

cos v 

tan v  8 or 
4

r

csc v  y

1
5

cos v  
7 or 7

sin v 

8

2
17

4
17


17

cos v  2 or 
1

sin v 

3

3
2


y

0

sin v  2 or 0

adjacent side


cos M  
hypotenuse

3

3
2


tan v  x

r  

(
2)2 
 02
r  
4
 or 2
y
x
sin v  r
cos v  r


sin M  
hypotenuse

27. r  

x2  y2
r  

32  32
r  
18
 or 3
2

y
x
sin v  r
cos v  r

x

cos v  y

or

y

cos v  r

4

4
2


cos v 

sin v  r

12

5

sin v 
sin v 

5

or 
1
2

csc v 
csc v 


2


2
r

y

4
2

4


csc v  
2

167

x

y

tan v  x


4

4
2


4


tan v  

4 or 
1



2

cos v  
 
2
r

sec v  x
sec v 

4
2



4

x

cot v  y

4

cot v  4 or 
1

sec v  

2

Chapter 5

34. r  

x2  y2
r  

52  02
r  
25
 or 5
y
sin v  r
0

csc v 

tan v  x

y

5

tan v  5 or 0

cos v  r

sin v  5 or 0
csc v 

x

0

cos v  5 or 1

r

y
5

0

sec v 
sec v 

r

x
5

5

cot v 

or 1

cos v 

undefined

x

r

r2



x2

3

8

82



(
3)2

x  
3, r  8
y

sin v  r

55


8
r
csc v  y
8

csc v  

55

8
55

csc v  55
y
tan v  x

sin v 

36.

sin v 
sin v 
sec v 
sec v 
37.

sin 42° 



tan v 
sec v 
sec v 


55


55

 or 
 

3
3
r

x
8
8
 or 

3

3

cot v 
cot v 
cot v 

cos v 
cos v 
cos v 

x
——
csc
r

1
 csc


10


10

1
0

x2

v
v

b
——
sin B
b
——
sin 70°

x

y

3



55
3
55



 55

15 sin 42°  b
10.0  b

cot v 
38.

b

tan B  —a—
tan 67° 

r
——
x


10


3

a
——
sin A
a
——
sin 74°

or

or

sin A 
sin 38° 

a

—
—
sin A
84

—
—
sin 52°
84 sin 58°

—
c—
sin 52°

c

—
—
sin C
8

—
—
sin 49°

b
——
sin B
b
——
sin 57°

sin B sin C

1

sin 96° sin 64°

—
K  —2—(19)2 —
sin 20°

a
——
c
24
——
c

K  471.7 units2
48. C  180° 
 (56°  78°) or 46°
1

c sin 38°  24
24
—
c—
sin 38°

sin A sin C

—
K  —2—b2 —
sin B
1

sin 56° sin 46°

—
K  —2—(24)2 —
sin 78°

K  175.6 units2
1

49. K  —2— bc sin A
1

K  —2—(65.5)(89.4) sin 58.2°
K  2488.4 units2
1

50. K  —2— ac sin B
1

K  —2—(18.4)(6.7) sin 22.6°

41. 180°
42.
A  49°  90°
A  41°

K  23.7 units2
a

cos B  —c—
16

cos 49°  —c—
c cos 49°  16
16
—
c—
cos 49°
c  24.4

A  41°, b  18.4, c  24.4

Chapter 5

84

—
—
sin 52°

c
——
sin C
c
——
sin 58°

—
K  —2—a2 —
sin A

1

3

a  10.2

b

a

—
—
sin A

c

—
—
sin C
8

—
—
sin 49°
8 sin 57°

—
b—
sin 49°

a  10.1891739
b  8.889995197
A  74°, a  10.2, b  8.9
47. B  180° 
 (20°  64°) or 96°
1


1


3

b

8 sin 74°

a tan 67°  24
24

a
tan 67°

tan 49°  —1—
6
16 tan 49°  b
18.4  b

cos 64°  —2—
8

—
a—
sin 49°


10


3

40. 30°, 210°

b

a

b  100.1689124
c  90.39983243
A  52°, b  100.2, c  90.4
46. A  180° 
 (57°  49°) or 74°

y2

24
——
a

tan B  —a—

b

84 sin 70°

c  39.0
39.

cos A  —c—

—
b—
sin 52°

x



10


a

28 sin 64°  a
28 cos 64°  b
25.2  a
12.3  b
B  26°, a  25.2, b  12.3
45. A  180° 
 (70°  58°) or 52°

cot v  —y—

b
——
c
b
——
15

15

A  cos
1 —20—

sin 64°  —28—

 
r2  (
1)2  (
3)2
r2  10
r  
10

r2

15

cos A  —20—

sin A  —c—

y2

55 

55
y
Quadrant 11, so y  
55

y
tan v  x

y
——
r

3



10
3
10


10
r
——
x


10
 or

1

sin B 

y2

y2

tan v  3; Quadrant III
y  
3, x  
1
sin v 



b

cos A  —c—

A  41.4°
41.40962211°  B  90°
B  48.6°
a  13.2, A  41.4°, B  48.6°
44. 64°  B  90°
B  26°

x

y
5

0

cot v 

undefined
35. cos v 

43. a2  b2  c2
a2  152  202
a  
175

a  13.2

168

B  113.7°, C  37.3°, b  22.7;
B  8.3°, C  142.7°, b  3.6
54. Since 45°  90°, consider Case I.
83 79; 1 solution

51. Since 38.7°  90°, consider Case I.
c sin A  203 sin 38.7°
c sin A  126.9242592
172 126.9; 2 solutions
a
——
sin A
172

sin 38.7°

c

a
——
sin A
83
——
sin 45°

—
—
sin C
203

—
—
sin C

203 sin 38.7°
—
sin C  —
172
203 sin 38.7°
—
C  sin
1 —
172



79 sin 45°



B  sin
1 —8—

3
79 sin 45°

B  42.30130394
C  180° 
 (45°  42.3°) or 92.7°
a
c
——  ——
sin A
sin C
83
——
sin 45°

a

—
—
sin A

83 sin 92.7°

c  117.2495453
B  42.3°, C  92.7°, c  117.2
55. a2  b2  c2 
 2bc cos A
a2  402  452 
 2(40)(45) cos 51°
a2  1359.446592
a  36.87067388

172 sin 93.7°

b  274.5059341
Solution 2
B  (180°
 (38.7°  132.4°) or 8.9°
b
a
  
sin B
sin A
b
172
——  ——
sin 8.9°
sin 38.7°
172 sin 8.9°
—
b—
sin 38.7°

a
——
sin A
36.9
——
sin 51°

sin

—
B  sin
1 —
36.9 
40 sin 51°

B  57.39811237
C  180° 
 (51°  57.4) or 71.6°
a  36.9, B  57.4°, C  71.6°
56. b2  a2  c2 
 2ac cos B
b2  512  612 
 2(51)(61) cos 19°
b2  438.9834226
b  20.95193124
b
a
——  ——
sin B
sin A
21.0
51
——  ——
sin 19°
sin A
51 sin 19°
—
sin A  —
21.0
51 sin 19°
—
A  sin
1 —
21.0

c







202 
 112 
 132
——

2(11)(13)
202 
 112 
 132
—
cos
1 —

2(11)(13)

b

—
—
sin B



b



b
sin 113.7°
12 sin 113.7°
——
sin 29 °

 cos C

C

112.6198649  C
a
c
——  ——
sin A
sin C
11
20
——  ——
sin A
sin 112.6°
11 sin 112.6°
sin A  —20—
11 sin 112.6°
A  sin
1 —20—

b  22.6647614
Solution 2
B  180° 
 (29°  142.7°) or 8.3°
a
——
sin A
12
——
sin 29°



A  52.4178316
C  180° 
 (52.4°  19°) or 108.6°
b  21.0, A  52.4°, C  108.6°
57.
c2  a2  b2 
 2ab cos C
202  112  132 
 2(11)(13) cos C

C  37.30170167
180 
   180° 
 37.3° or 142.7°
Solution 1
B  180° 
 (29°  37.3°) or 113.7°
a
——
sin A
12
sin 29°

40

—
—
sin B
40 sin 51°

—
—
sin C



b

—
—
sin B

—
sin B  —
36.9

b  42.34881128
B  93.7°, C  47.6°, b  274.5; B  8.9°,
C  132.4°, b  42.3
52. Since 57°  90°, consider Case I.
b sin A  19 sin 57°
b sin A  15.93474074
12  15.9; no solution
53. Since 29°  90°, consider Case I.
c sin A  15 sin 29°
c sin A  7.272144304
12 7.3; 2 solutions
15
sin C
15 sin 29°
C  —1—
2
15 sin 29°

1
——
C  sin
12

c

—
—
sin 92.7°

—
c—
sin 45°

172

—
—
sin 38.7°

—
b—
sin 38.7°

a
——
sin A
12
sin 29°

79

—
—
sin B

sin B  —8—
3

C  47.55552829
180° 
   180° 
 47.6° or 132.4°
Solution 1
B  180° 
 (38.7°  47.6°) or 93.7°
b
——
sin B
b
——
sin 937°

b

—
—
sin B

b

—
—
sin B



b

—
—
sin 8.3°



A  30.51023741
B  180° 
 (30.5°  112.6°) or 36.9°
A  30.5, B  36.9°, C  112.6°

12 sin 8.3°

—
b—
sin 29°

b  3.573829815

169

Chapter 5

y2  1
y  1
Since y is a length, use only the positive root.
Another method is to use the Triangle Inequality
Theorem. The hypotenuse must be shorter than
the sum of the lengths of the other two sides.
5y  3  4
5y  7
Which of the answer choices make this inequality
true?
5(1)  5  7
5(2)  10 7
The correct choice is A.
2. If you recall the general form of the equation of a
circle, you can immediately see that this equation
represents a circle with its center at the origin.
(x 
 h)2  (y 
 k)2  r2
If you don’t recall the equation, you can try to
eliminate some of the answer choices. Since the
equation contains squared variables, it cannot
represent a straight line. Eliminate choice D.
Similarly, eliminate choice E. Since both the x and
y variables are squared, it cannot represent a
parabola. Eliminate choice C. The choices
remaining are circle and ellipse. This is a good
time to make an educated guess, since you have a
50% chance of guessing correctly. It represents a
circle. The correct choice is A.
3. Use factoring and the associative property.
999 111  3 3 n2
(9 111) 111  3 3 n2,
3 3 (111)2  3 3 n2.
So n must equal 111. The correct choice is C.
4.
A

58. b2  a2  c2 
 2ac cos B
b2  422  6.52 
 2(42)(6.5) cos 24°
b2  1307.45418
b  36.15873588
b
c
——  ——
sin B
sin C
36.2
6.5
——  ——
sin 24°
sin C
6.5 sin 24°
—
sin C  —
36.2
6.5 sin 24°
—
C  sin
1 —
36.2





C  4.192989407
A  180° 
 (24°  4.2°) or 151.8°
b  36.2, A  151.8°, C  4.2°

Page 339

Applications and Problem Solving
8

59a. sin v  —1—
2
8

v  sin
1 —1—
2
v  41.8°
x

cos v  —12—

59b.

x

cos 41.8°  —12—
12 cos 41.8°  x
8.94427191  x
about 8.9 ft
60a. x2  4.52  8.22 
 2(4.5)(8.2) cos 32°
x2  24.9040505
x  5.0 mi
60b.

8.2
——
sin v

5.0

—
—
sin 32°
8.2 sin 32°

—
sin v  —
5.0
—
v  sin
1 —
5.0
8.2 sin 32°

v  60.54476292
180 
 v  180 
 60.5 or about 119.5°

45˚

D

Page 339
1.

435.86  ab
Sample answer: about 40 cm and 10.9 cm
2a. Sample answer: a  10, b  24, A  30°;
10  24, 10  24 sin 30°
2b. Sample answer: b  18; 10  18, 10 18 sin 30°

Chapter 5 SAT & ACT Preparation
SAT and ACT Practice

1. There are several ways to solve this problem. Use
the Pythagorean Theorem on the large triangle.
(2y  3y)2  42  32
(5y)2  16  9
25y2  25
25y2
——
25

Chapter 5

7

7

B
7
C
Since ABC is an equilateral triangle and one side
is 7 units long, each side is 7 units long. so AC  7.
A
D
 is the hypotenuse of right triangle ACD. One
leg is 7 units long. One angle is 45°, so the other
angle must also be 45°. A 45°
45°
90° triangle is
a special right triangle. Its hypotenuse is 
2

times the length of a leg. (The SAT includes this
triangle in the Reference Information at the
beginning of the mathematics sections.) The
hypotenuse is 7
2
. The correct choice is B.

Open-Ended Assessment

1
K  —2— ab sin C
1
125  —2—ab sin 35°

Page 341

45˚

25

 —25—

170

8. Factor the polynomial in the numerator of the
fraction. Simplify the fraction. Solve for x.

5. You need to find the fraction’s range of values,
from the minimum to the maximum. The
minimum value of the fraction occurs when a is as
small as possible and b is as large as possible.
Since the smallest value of a must be slightly
greater than 4, and the largest value of b must be
slightly less than 9, this minimum value of the
4
fraction must be larger than —9—. The maximum
value of the fraction occurs when a is as large as
possible and b is as small as possible. This
7
maximum must be smaller than —7— or 1. The
correct choice is A.
6. Start by making a sketch of the situation.

x2  7x  12
——  5
x4
(x  3)(x  4)
——  5
x4

x35
x2
The correct choice is B.
9.

O

3

C 30
T

9000

A

T
 is a radius of the circle.
T is on the circle, so O
The length of O
T
 is 3. Since 

TA is tangent to the
circle, OTA is a right angle, and OTA is a right
triangle. In particular, OTA is a 30°-60°-90° right
triangle. In a 30°-60°-90° right triangle, the length
of the hypotenuse is 2 times the length of the
shorter leg.

1:00 P.M.
9000
36,000
9000
10:00 A.M.
9000

By 1:00 P.M. the pool is three-fourths full. Three
fourths of 36,000 gallons is 27,000 gallons. The
pool contained 9,000 gallons at the start. So
27,000 
 9,000 or 18,000 gallons were added in
3 hours. The constant rate of flow is 18,000
gallons 3 hours or 6,000 gallons per hour. To
fill the remaining 9,000 gallons at this same rate
will take 9,000 gallons 6,000 gallons per hour
or 1.5 hours. One and a half hours from 1:00 P.M.,
is 2:30 P.M. The correct choice is C.
7. There are two right triangles in the figure. You
need to find the length of one leg of the larger
triangle, but you don’t know the length of the
other leg. Use the Pythagorean Theorem twice—
once for each triangle. Let y represent the length
of side AC.
In the smaller right triangle,
y2  42  62
y2  16  36
x2  52
You do not need to solve for y.
In the larger triangle,
102  x2  y2
100  x2  52
x2  48
x  
48

x  4
3

The correct choice is B.

OA  2(OT )
OC  CA  2(OT )
3  CA  2(3) or 6
CA  3
The correct choice is B.
10. Draw a diagram from the information given in the
problem. Drawing a valid diagram is the most
difficult part of solving this problem. Your
diagram could be different from the one below and
still be valid.

C
5

A

4
4

X

B

5

D
Since two segments bisect each other, you know
the length of each half of the segment. Notice that
B
D
 is a side of a right triangle. It is a 3-4-5 right
triangle. So BD  3. The answer is 3.

171

Chapter 5

Chapter 6 Graphs of Trigonometric Functions
Pages 348–351

Angles and Radian Measure

6-1

16. 135°  135°

Exercises


180°

3

Pages 347–348
1.



180°

5

5

 
2


180°

20. 
75°  
75°
5

22.

7

12

7

 12

125

 18
11

3

3. Divide 10 by 8.
4. Let R  2r. For the circle with radius R, s
  Rv or
2rv which is 2(rv). Thus, s
  2s. For the circle
1
1
with radius R, A
  2R2v or 2(2r)2v which is

23.

180°



25. 
3.5 
3.5

1
(4r2)v
2

or 42r2v. Thus, A
  4A.


180°

4

3

2

9. reference angle:
sin

3

4


2


tan



3

11

6

s  15

3

4


  or


;
4

Quadrant 2

1

11

6


  or

5
;
6

7

32.
Quadrant 3

2

1

A  2.1 units2



180°

1

14

3

2

is coterminal with 3
2



14


tan 3  

3
19

3

A

5

33. 
6 is coterminal with 6
5



reference angle:  
 6 or 6; Quadrant 2

3


19

cos 
6  
2
34. s  rv

2

3

s  14



180°

36. 150°  150°

1
r2v
2
1
3
(62) 
2
10

5

 6

 

s  rv

A  17.0 units2

5

s  146
s  36.7 cm



 6

38. s  rv

s  rv

s  1.46
s  0.7 m

35. s  rv



s  29.3 cm

 —10—
A

15. 30°  30°



reference angle:  
 3 or 3; Quadrant 2



180°

14. 54°  54°

A  2(1.42)3

7

sin 6  
2

s  rv
77

s  15
180 
s  20.2 in.

13. A  2r2v



31. reference angle: 6 
  or 6; Quadrant 3

77



5



2

5



180

s  39.3 in.



cos 4  
2

12. 77°  77°
5

6

7

30. reference angle: 4 
  or 4; Quadrant 3

180°



 
3

11. s  rv

 1002.7°


3


7

 2

10. reference angle:



tan 6  3

 270°
 100.3°

5

180°



29. reference angle: 6 
  or 6; Quadrant 3

180°



8. 
1.75  
1.75

27. 17.5 17.5

sin 3  
2

19

3

 
200.5°
180°




3


5



180°

 6

 2



180°



28. reference angle: 2 
 3 or 3; Quadrant 4

6. 570°  570°

 3





6.2

 
29.0°

1

5. 240°  240°

7.

26.

180°



 3
 660°

 974.0°





6.2

11

180°



 105°
24. 17  17



180°

21. 1250°  1250°

 
12

x

1

5

s  1412



180°

s  18.3 cm
37. 282°  282°
47

s  rv

47

s  143
0 

s  68.9 cm
39. 320°  320°

3

s  1411
s  12.0 cm

172



180°

 30

16

 9
s  rv

16

s  149
s  78.2 cm

Chapter 6



180°

19. 
450°  
450°

 3

3

4

O

 6

18. 300°  300°



2. 90°; —4—

y

7

 4

Check for Understanding



180°

17. 210°  210°



180°

1

78°  78°

1

 30

40. r  2d



180°

53b.

13

r  2(22)
r  11

s  rv

s  rv

5

4

70.7  r
18.00360716  r


180°

42. 60°  60°

11

s  rv



s  2

d  36.0 m

s  11.5 in.

A

A  65.4

44. 90°  90°

 



units2

A
A



180°

45. A 
A

46. A 

 

A

A  9.6 units2


——
180°

47. 225°  225°

A
A

5
——
4

A

 



s  rv

s  rv

s  3960—6—
0
s  207.3 mi
57. 84.5°  84.5°





52a.

1

58a. r  2d
3

r

52c. s  rv
s  3(3.4)
s  10.2 m

53a. 225°  225°

s  rv

5

4

1.5 rotations  1.5 2 radians
 3 radians

 

1

2

1
22

r  1.25
s  rv
s  1.25(3)
s  11.8 ft

mm2

58b.

s  rv

3.6  3.6

1

42  1.25v

180°



 206.3°

3.6  v


3

59a. v  2 
 2 or 2

180°



1

A  2r2v

 194.8°

3.4  v; 3.4 radians

s  rv
4
s  0.679

s  1.03 mi
s  0.94 mi
1.03  1.46  0.94  1.8  5.23 mi

1

52b. 3.4  3.4

4

 9

s  rv
169

s  0.70
360 

48.38  r; about 48.4 mm
1
51a.
A  2r2v
51b. s  rv
1 2
s  12.2(0.2)


15  2 r (0.2)
s  2.4 in.
150  r2
12.247  r
about 12.2 in.
1
A  2r2v
1
15.3  2(32)v



180°

80°  80°

169

50b. A  2r2v
A  2757.8



180°



360

A  15 ft2
1

114  r



180°


——
180°

A  2(48.42)4
3

4

s  967.6 mi



1

3
——
4

s  rv
7
s  396090

 60

A  2(52)(1.2)


180°

7



180°

 —9—
0

3°  3°

49b. A  2r2v

50a. 135°  135°

14°  14°

s  760.3 mi
56c. 34° 
 31°  3°

1

49a. s  rv
6  r(1.2)
5  r; 5 ft

56b. 45° 
 31°  14°



180°

 



A  70.7 units2

 26.3°

s  rv
11

s  3960
180 

1
r2v
2
4
1
(12.52) ——
2
7
140.2 units2

180°



0.5  0.5

11

41
 —90—
1
A  —2—r2v
1
41
A  2(7.32) 90
A  38.1 units2

1
——r2v
2
1
5
(62) 
2
4

1.44  v; about 1.4 radians



180

 

48. 82°  82°

11.5  8v

11°  11°



2
1
r2v
2
1

(222) 
2
2

A  380.1 units2
1
r2v
2

1
(72) ——
2
8



10.5  22.9v
0.46  v; about 0.5
56a. 45° 
 34°  11°



13.56  r; about 13.6 cm
43. A 

s  rv

s  rv

55.

14.2  r3

1
r2v
2
5
1
(102) ——
2
12

11

6

d  2(18.0)
s  rv

 3



180°

 6

s  15.0 in.
d  2r



 143.2°

54. 330°  330°

13

180°



2.5  2.5

5  2v
2.5  v

s  113
0 

41.

s  rv

3

A  2(152)—2—
1

A  530.1 ft2



180°

59b.

1

A  2r2v
3

750  2r22
1

318.3098862  r2
17.84124116  r; about 17.8 ft

5

s  24
s  7.9 ft

173

Chapter 6

60. 3.5 km  350,000 cm
s  rv
350,000  32v
10,937.5  v; 10,937.5 radians
61. Area of segment  Area of sector 
 Area of triangle

yx

y  
x

1

1

A  2r2a 
 —2—r r sin a

all
70. 4x 
 2y  3z  
6
5x 
 4y 
 3z  
75
9x 
 6y
 
81
2(4x 
 2y  3z)  2(
6)
→
3(3x  3y 
 2z)  3(2)

1

A  2r2(a 
 sin a)
1

62. s  2(6  8  12)
 13
K  
s(s


a)(s


b)(s 

c)
K  
13(13
13

 6)(8)(13

 


12)
K  
455

K  21.3 in2
63. Since 152°  90°, consider Case II.
10.2  12, so there is no solution.
64. C  180° 
 38° 
 27°  115°
560

sin 115°

5(9x 
 6y)  5(
81)
6(17x  5y)  6(
6)

x


sin 38°  
280.52

x  172.7 yd
65. I, III
66a. Find a quadratic regression line using a
graphing calculator. Sample answer: y  102x2

 505x  18,430
66b. 2020 
 1970  50
y  102x2 
 505x  18,430
y  102(50)2 
 505(50)  18,430
y  248,180
Sample answer: about 248,180
67.
r
1

3

2
6
10
1
1

2

4
2
12
2
1

1

4

2
6
3
1
0

2
0
10
4
1
1
2
14
66

3
4
5


2
2
8


6

4
10

Linear and Angular Velocity

Page 355

Check for Understanding

2.

5 rev

1 min

2 radians

1 rev

1 min

60 s

3. Linear velocity is the movement along the arc
with respect to time while angular velocity is the
change in the angle with respect to time.
4. Both individuals would have the same change in
angle during the same amount of time. However,
an individual on the outside of the carousel would
travel farther than an individual on the inside
during the same amount of time.
v
5. Since angular velocity is —t—, the radius has no
effect on the angular velocity. Let R  2r. For
v
v
a circle with radius R, v
  Rt or (2r)t which is

10
6
30

Sample answers: 4; 
2
68. 
2 1
6
12
12

2

8

8
1
4
4
4
No; there is a remainder of 4.
69. x2  y2  16 → a2  b2  16
x-axis
a2  b2  16
a2  (
b)2  16
a2  b2  16; yes
y-axis
a2  b2  16
(
a)2  b2  16
a2  b2  16; yes

Chapter 6

6-2

1.

f(
x)  (
x)4 
 3(
x3 
 2(
x)2  6(
x)  10
f(
x)  x4  3x3 
 2x2 
 6x  10
1
1
1

8x 
 4y  6z  
12
9x  9y 
 6z 
6
17x  5y
 
6
45x 
 30y  
405
→
102x  30y  
36
147x
 
441
x  
3
4x 
 2y  3z  
6
4(
3) 
 2(9)  3z  
6
z8

9x 
 6y  
81
9(
3) 
 6y  
81
y9
(
3, 9, 8)
71. b
72. Since q  0, 
q 0. Given that p 0,
p 
 q  p q and p q 0. So the
expression p 
 q is nonnegative.
The correct choice is B.

a



sin 27°

a  280.52

r
1
2

a2  b2  16
(b)2  (a)2  16
a2  b2  16; yes
a2  b2  16
(
b)2  (
a)2  16
a2  b2  16; yes

2r t. Thus v
  2v.
v

2  11.6 or about 36.4 radians
2  1420  or about 4461.1 radians
2  6.4
9. 700 2  1400

6. 5.8
7. 710
8. 3.2
v

q  t

q  7

6.4

q  15

q  2.9 radians/s

q  293.2 radians/min

q  t

174

v

1400

10. v  rq
11. v  rq
v  12(36)
v  7(5)
v  432 in./s
v  110.0 m/min
12a. r  3960  22,300 or 26,260 mi
s  rv
s  26,260(2)
s  164,996.4 mi
v
12b. v  r t
2
v  26,26024
v  6874.9 mph

1

35a. In 1 second, the second hand moves 6
0 (360°)
or 6°.


180°

6°  6°

v  rv

v  3030
v  3.1 mm/s
35b. In 1 second, the minute hand moves

 

1
1
 
60 60

(360°) or 0.1°.

0.1°  0.1°

Pages 355–358
13.
14.
15.
16.
17.
18.
19.

3 2  6 or about 18.8 radians
2.7 2  5.4 or about 17.0 radians
13.2 2  26.4 or about 82.9 radians
15.4 2  30.8 or about 96.8 radians
60.7 2  121.4 or about 381.4 radians
3900 2  7800 or about 24,504.4 radians
1.8 2  3.6
20. 3.5 2  7
v

q  t

3.6

q  3

v

1

v  404
5

v

2

v  5.6 ft/s

27.

85 radians

1 second

60 seconds

1 minute

28. v  rq
v  8(16.6)
v  132.8 cm/s
30. v  rq
v  1.8(6.1)
v  34.5 m/min
32. v  rq
v  39(805.6)
v  31,418.4 in./min
60 seconds

1 minute


120°  120° 
180°
2
 3
v
q  t
120°

1 second

2

3

1
2

3

36b. v 

2

t  31 s
37a. 3 2  6 radians
v

1 revolution

2 radians

1 6

v  7.1 ft/s
v

37c. 7.1 
 3.1  4 ft/s

37b. v  r t

 52.4
radians/s

6

60

3.1  r



9.87  r; about 9.9 ft

 811.7 rpm

38a. 35°  35°

29. v  rq
v  4(27.4)
v  109.6 ft/s
31. v  rq
v  17(75.3)
v  4021.6 in./s
33. v  rq
v  88.9(64.5)
v  18,014.0 mm/min
1 revolution

360°

1 minute  60 seconds

v
r t

v  22—2—60

0.1 radian/s
2 radians
——
1 revolution

v
r t

8  40t

q  28.5 radians/min

26.

q

v  r t

r  2(80) or 40

v
q  t
245.2
q  27

1 revolution
2 radians

 
50 seconds
1 revolution
500 revolutions
1 minute
——
——
1 minute
60 seconds

q

1

36a. r  2d

q  9.4 radians/s
24. 122.6 2  245.2

q  39.3 radians/min

34b.

v  0.003 mm/s

56.8

v

t
200

16

0.008


or 
180

v  rv
0.008

v  18
180 

q  1
9

q  9.0 radians/s
23. 100 2  200

34a.



180°

v

34.4

25.

0.008°  0.008°

q  t

q  1
2

q

 610  (360°) or about 0.008°.

1 1
 
12 60

q  7.3 radians/min
22. 28.4 2  56.8

q  t

q

35c. In 1 second, the hour hand moves

7

q  1.3 radians/s
21. 17.2 2  34.4

0.1


or 
180

v  0.05 mm/s

v

q  9



180°

v  rv
0.1

v  27
180 

Exercises

q  t



or 30



180°

7

 36
v

lighter child: q  t
q

7

36

1

2

q  1.2 radians/s
v

heavier child: q  t

 20 rpm

q

7

36

1

2

q  1.2 radians/s

v  rq
2
v  53
v  10.5 in./s

175

Chapter 6

38b. lighter child: v  rq
v  9(1.2)
v  11.0 ft/s
heavier child: v  rq
v  6(1.2)
v  7.3 ft/s
39a. 3 miles  190,080 inches

42c. 3960  500  4460; C  2(4460) or
28023.00647
t  28,023.00647 17,000 or 1.648412145
v

q  t
2


q
1.65

1

r  2d
r

q  3.8
Its angular velocity is between 3.8 radians/h and
4.1 radians/h.
43a. B clockwise; C counterclockwise

s  rv

1
(30)
2

190,080  15v

r  15

43b. vA  rAtA
v

12,672  v

vA  3.01
120

1 revolution
  2017 revolutions
2
2 radians
2.75 revolutions
60 seconds
60 minutes




1 revolution
second
1 minute
1 hour

12,672
39b.

vA  360
The linear velocity of each of the three rollers is
the same.

 19,800 radians/hour
v  rq
v  15(19,800)
v  933,053.0181
933,053.0181 inches


vB  rBtB

1 mile


40a. Mercury:

 14.7 mph

v

v  10.9 km/h
Earth:

v  6.5 km/h
Mars:

v

v
r t

v
2

23.935

v  6356

 12
1

A  2r2v

v  3375

1

A  47.5 cm2



1

r  2d

45.

v  1668.5 km/h
v  861.2 km/h
40b. The linear velocity of Earth is about twice that
of Mars.
41a. v  vm cos qt

1

r  2(7.3)
r  3.65



x



v  360°

41b. v  4 cost
0  4 cos t
0  cos t

t  2
1

t  2 or 0.5 s

sin v 
or

A
A



46. 35°20
55  35°  20

60
   55 3600 
1°

2

 35.349°
47. 10  
k

58


k

 5  
2


k

54

k9

q  4.1 radians/h

t

tC


2


—— (17,000)
2

 (17,000)
2

2

 2r

speed
17,000

 2r
r

176

1°

Check: 10  
k


58
10  
9


58
10  
4
8
10  2  8
12 8

no real solution
48. (x 
 (
4))(x 
 3i)(x 
 (
3i))  0
(x  4)(x 
 3i)(x  3i)  0
(x  4)(x2  9)  0
3
x  4x2  9x  36  0

4250  r
4250 
 3960  290; about 290 mi
Chapter 6

y  2.952912029

1
bh
2
1
(2.15)(2.95)
2

A  3.16761261
Area of pentagon  10(3.17) or about 31.68 cm2

17,000


q
1.54

2

t

y


cos 36°  
3.65

x  2.145416171

3

t  2 or 1.5 s

v

v

y

cos v  r


sin 36°  
3.65

q  t

42b. q  t

10 or 36°
x

r
x

3

t  2

42a. 3960  200  4160 miles
C  2r
t  C speed
C  2(4160)
t  26,138.05088
C  26138.05088
t  1.537532405


——
2


 y

r



v  4 cos t

4

7

A  2(7.22)12

v
r t
2

24.623





180°

7

2


v  6052
5832.5 

v
C

1

75  vC
75 rpm

44. 105°  105°

v

2

360  4.8

180  vB
180 rpm

v  r t


v  2440
1407.6 

v

v
B

1

360  2.0

Venus:

v  r t

vC  rCtC

v

y

49.
x3

y

maximum value of the cosine function occurs
when x  n, where n is an even integer, and its
minimum value occurs when x  n, where n is
an odd integer.
5. yes; 4
6. 0
7. 1

1

x

O

8.

3

2

y

9.
0
5

1


50. m  

6 
 8

m


5


14

or

5

14

O

y 
 y1  m(x 
 x1)
y
0
y

5
(x 
 (
6))
14
5
15
x  
14
7

y 
 cos x

5

y

10.

  2b

3
b
4

P
P
2P

7

3
b
2
7
b
2

y 
 sin x

1

 2b

4

b


3

O


2

x

The correct choice is D.

1

Graphing Sine and
Cosine Functions

6-3

7 x

6


1

51. P  2a  2b
P  2

 2n, where n is an integer

Page 363

11. Neither; the period is not 2.
12. April (month 4):

y  49  28 sin 6(t 
 4)

Check for Understanding

y  49  28 sin

1. Sample answer:


 4)

y  49
October (month 10):

y  49  28 sin 6(t 
 4)

y
O


(4
6

x

y  49  28 sin


(10
6


 4)

y  49
The average temperatures are the same.

period: 6
3  5

2. Sample answers: 
2, 2, 2

Pages 363–366

3. cos x  cos(x  2)
4. y

yes; 6
14. no
15. yes; 20
no
18. no
19. 1
0
22. 1
23. 
1
sin   cos   0  (
1)
 
1
26. sin 2 
 cos 2  0 
 1
 
1
27.   2n, where n is an integer

y 
 sin x

1

O




1

2

Exercises

13.
17.
21.
25.
3

4

5 x

y 
 cos x

28.

Both functions are periodic functions with the
period of 2. The domain of both functions is the
set of real numbers, and the range of both
functions is the set of real numbers between 
1
and 1, inclusive. The x-intercepts of the sine
function are located at n, but the x-intercepts of

the cosine function are located at 2  n, where n
is an integer. The y-intercept of the sine function
is 0, but the y-intercept of the cosine function is 1.
The maximum value of the sine function occurs

when x  2  2n and its minimum value occurs

29.



2


2

16. no
20. 0
24. 
1

 2n, where n is an integer
 n, where n is an integer

30. v  2n, where n is an integer

y

31.
y 
 sin x


5


4

1


3

O

x


1

3

when x  2  2n, where n is an integer. The

177

Chapter 6

32.

y
1


43b. csc v  
sin v

1

1



1  
sin v

8

1

—
1—
sin v

y 
 cos x

O

10 x

9

sin v  1

sin v  
1


2

3


2

 2n, where n

33.
y 
 cos x


5


4


3

y


44a. sec v  
cos v

1


1
cos v

O

1


44b. sec v  
cos v

1

1



1  
cos v

1

cos v  1
cos v  
1
2n, where n
  2n, where n
is an integer
is an integer
44c. sec v is undefined when cos v  0.
  n, where n is an integer
2

x


1

45.

y
1

y 
 sin x

O

5

x

6


1



[0, 2] sc1—2— by [
2, 2] sc11
3 7

x  4, 4

y

35.
y 
 cos x


3

46.

1

O


2

x


1

36.

 2n, where n

is an integer
is an integer
43c. csc v is undefined when sin v  0.
n, where n is an integer


1

34.

1


43a. csc v  
sin v

y



[0, 2] sc1—2— by [
2, 2] sc11

y 
 sin x

1

 5

0  x  4, 4  x  2
47.

O

4

5

x


1

37. y  cos x; the maximum value of 1 occurs when
x  4, the minimum value of 
1 occurs when
7 9
11
x  5, and the x-intercepts are 2, 2, and 2.
38. Neither; the graph does not cross the x-axis.



[0, 2] sc12 by [
2, 2] sc11
none

39. y  sin x; the maximum value of 1 occurs when
11
x  
2, the minimum value of 
1 occurs
13
when x  
2, and the x-intercepts are 
7,

6, and 
5.

48.



40. Sample answer: a shift of 2 to the left


41. x  2  n, where n is an integer
42. x  n, where n is an integer



[0, 2] sc12 by [
2, 2] sc11


3

x  0, 2  x  , 2  x  2

Chapter 6

178

54a. v  3.5 cos t

49.

k
——
m

v  3.5 cos 0.9




19.6
——
1.99

v  
3.3 cm
v  3.5 cos t

k

m

v  3.5 cos 1.7



0  3.5 cos t
0  cos t



1.570796327 

v  3.5 cos t

54c.

 5

x  4, 4
(t 
 4)
6
(7 
 4)
6

1  cos t
cos1 1  t

y  74
January (month 1):

2  t

(t 
 4)
6
(1 
 4)
6

55a.


4

n

 2, where n is an integer
55c. 
1

55b. 1
55e.

55d. 

y
1


2

O




y 
 cos 2x



2 x


1

y 
 2 sin x

1

O

1



19.6
——
1.99
19.6
——
1.99

y
2




19.6
——
1.99



2.00206591  t; about 2.0 s

y  12
74 
 12  62; it is twice the coefficient.
51b. Using answers from 51a., 74  12  86; it is
twice the constant term.
52a. n, where n is an integer
52b. 2
52c. 
2
52d. 2


2



k
——
m
19.6
——
1.99

3.5  3.5 cos t

51a. July (month 7):

52e.



19.6
——
1.99
19.6
—
t —
1.99
19.6
—
t —
1.99

0.5005164776  t; about 0.5 s



[0, 2] sc12 by [
2, 2] sc11

y  43  31 sin



k
——
m
19.6
——
1.99

cos1 0 

y  43  31 sin



v  3.5 cos t

54b.

50.

y  43  31 sin

19.6
——
1.99

v  2.0 cm

[0, 2] sc12 by [
2, 2] sc11

x  0, 2, 2

y  43  31 sin





56a. P  500  200 sin [0.4(t 
 2)]
P  500  200 sin [0.4(0 
 2)] or about 357
pumas
D  1500  400 sin (0.4t)
D  1500  400 sin (0.4(0)) or 1500 deer
56b. P  500  200 sin [0.4(t 
 2)]
P  500  200 sin [0.4(10 
 2)] or about 488
pumas
D  1500  400 sin (0.4t)
D  1500  400 sin (0.4(10)) or about 1197 deer
56c. P  500  200 sin [0.4(t 
 2)]
P  500  200 sin [0.4(25 
 2)] or about 545
pumas
D  1500  400 sin (0.4t)
D  1500  400 sin (0.4(25)) or about 1282 deer

2 x


2

52f. It expands the graph vertically.
53a. P  100  20 sin 2t
P  100  20 sin 2(0) or 100
P  100  20 sin 2(0.25) or 120
P  100  20 sin 2(0.5) or 100
P  100  20 sin 2(0.75) or 80
P  100  20 sin 2(1) or 100
53b. 0.25 s
53c. 0.75 s

57.

500 revo
lutions

1 minute

1 min
ute


60 seconds

58. 
1.5  
1.5

2 rad
ians


1 revolution

 52.4 radians
per second

18
0°



 
85.9°
59. 45°, 135°

179

Chapter 6

2
x
x2 
4
    

x2
2
x
x2 
 4
2
  
1(x  2)(x 

x2

60.


1(x  2)(x 
 2)



68. Perimeter of square RSVW
 RS  SV  VW  WR
 5  5  5  5 or 20
Perimeter of rectangle RTUW
 RT  TU  UW  WR
 (5  2)  5  (5  2)  5
 24
24 
 20  4
The correct choice is B.


2)
2 
 x
x


 (
1)(x  2)(x 
 2)
x2 
 4 
x2  4


1(x 
 2)(2)  (x  2)(x)  (
1)(x2  4)

2x  4  x2  2x 
 x2 
 4
x2
But, x 2, so there is no solution.
61. 1 positive real zero
f(
x)  
2x3  3x2  11x 
 6
2 or 0 negative real zeros
2 2 3 
11 
6
4
14
6
2 7
3  0
2x2  7x  3  0
(2x  1)(x  3)  0
2x  1  0
or
x30
1

x  
2

3,

1

2,

62. 1

1

1
12; no

n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

2

9
18
3 
6

6  12

2
1
3

g (x ) 


O

x2
x2  x

x

vertical:
x2  x  0
x(x  1)  0
x  0 or x  1  0
x  
1
horizontal:
y1

64. reflected over the x-axis, expanded vertically by a
factor of 3

2
4 
1
65.
1 
1
0

3
4
5
 
2 
1 0 
 4 1 0  (
1) 1 
1
4 5

3 5

3
4
 
2(
5) 
 4(5) 
 1(1)
 
11
3
2 1
66. 
1 0
0 1
2 
4 6

1(3)  0(2) 
1(2)  0(
4) 
1(1)  0(6)
0(3)  1(2)
0(2)  1(
4)
0(1)  1(6)

3

2

1

2 
4
6
A
(
3, 2), B
(
2, 
4), C
(
1, 6)


3

67. x  2y

y

2
x
3

ART

y

O

Chapter 6

History of Mathematics

1.

x  
3

g (x )

63.

Page 367

n 2  n3
2
12
36
80
150
252
392
576
810
1100
1452
1872
2366
2940
3600
4352
5202
6156
7220
8400
9702
11,132
12,696
14,400
16,250
18,252
20,412
22,736
25,230
27,900

[0, 30] sc15 by [0, 30,000]
sc15000

The graph is not a straight line. It curves upward,
increasing more rapidly as the value of n
increases.
2. See students’ work.

x 
 23 y

x

180

Amplitude and Period of Sine and
Cosine Functions

6-4
Page 369

7.

2

4



 2

y

y 
 sin 4 

1

Graphing Calculator Exploration

1.

O




2

3
2




1
2

8. 10  10; 2  

y

2. The graph is shrunk horizontally.
3. The graph of f(x)  sin kx for k  0 is the graph of
f(x)  sin kx reflected over the y-axis.

y 
 10 sin 2

8
4

O

Pages 372–373

1. Sample answer: y  5 sin 2v
2. The graphs are a reflection of each other over the
v-axis.

y

2

y 
 3 cos 2

3
2
1

or 4

D: period  2
C has the greatest period.
4. Period and frequency are reciprocals of each other.

O




1

2

3

y
y 
 3 cos 

O




2

10. 0.5  0.5;

y 
 cos 3

2

2

3

2

2

1

6

y





3

 12

y 
 0.5 sin

1


6

y 
 cos 



O
All three graphs are periodic and curve above and
below the x-axis. The amplitude of y  3 cos v is 3,
while the amplitude of y  cos v and y  cos 3v is
2
1. The period of y  cos 3v is 3, while the period
of y  cos v and y  3 cos v is 2.

2

4

6

8

10


1
1

1

11. 
5  5;

6. 
2.5  2.5

2

1

4

 8

y

y

y 
  15 cos 4

0.2

y 
 2.5 cos 
2

O



2

B: period  5

5.

3

9. 3  3; 2  

2

2

1

2

2


8

3. A: period  2 or 

C: period 




4

Check for Understanding




2

3

4

O




2

2

4

6

8


0.2

12. A  0.8
A  0.8

2

k


2

k   or 2

y  0.8 sin 2v

181

Chapter 6

2

k

13. A  7
A  7



 3

k

y  7 sin 6v

2

k

14. A  1.5

2



3

20.

 5

O

2

2

y  1.5 cos 5v
2

k

3

y

3
4



21.


v
3
2

k

16. A  0.25
A  0.25
y  0.25 sin (588

2

1

4

y

1



294

y 
 cos 4

1

k  588
t)



Exercises

2

4

6

8


1

17. 2  2
y
2



3

 8

O

Pages 373–377

2


1

k  6 or 3

cos



6
2

3

A  4

y 
 cos 2

1

k  5 or 5

15. A  4



y

or 6

2

A  1.5

2

2

y 
 2 sin 

22.

2

6



 3

y
O



2

3

y 
 sin 6

1




2

O

3




2




1

3

18. 
4  4

y

2

y 
  34 cos 

23. 5  5; 1  2

0.5

y

O

4

y 
 5 cos 


2

3




0.5

O



y


1



2

y


2

3



2

y 
 2 cos 0.5 




2

O

2

Chapter 6

4


24. 
2  2; 
0.5  4

y 
 1.5 sin 

1

O

3


4

19. 1.5  1.5
2

2

182



2

3

4

2 2

2

2


30. 3  3; 
0.5  4

25. 
5  5; 9

y

y
y


0.4

2
5


4

O

y 
 3 cos 0.5

sin 9

2




2

O



2

3

4

5 

4

5 

4

5


2


0.4

1 2

1

2

31. 
3  3; 3


26. 8  8; 
0.5  4

y

y
y 
 8 sin 0.5

8

1
1

y 
  3 cos 3

4

O



2

3

4

5

O





2

3


4

1


8

27. 
3  3;

2



2

4

32.

1

3

1

 3;

2

1

3

y

y
4



sin 2 

y 
 3

1


2

O

2



3
2

2

5 
2

O

2

3

2

 3;



2


3


1


4

28.



1

y 
 3 sin 3

2


2

 6

2

3

7

14

33. 
4  4;

 3

y

y


0.8

2
3

2

1

2

 4

y

3


cos 7 



y 
 4 sin 2

4
2

0.4



O

0.4


2



3
2

2

5
2

O

3


2



2

4

5 


4


0.8
2

34. 
2.5  2.5;

29. 3  3; 2  

x

y 
 3 sin 2

2

1

5

 10

y


2

y 
 2.5 cos 5

2

O

3



2

3

4

5 

O


2



2

3

4

5 


2

2

1



35. 0.5  0.5; 
698  349

183

Chapter 6

2

k

36. A  0.4
A  0.4

k
2

k

37. A  35.7
A  35.7
y  35.7 sin 8v
A
y

1

4
1
4

1
4



1

5



1

v

y  
1.5 sin 2

 0.75

A  3.8
y  3.8 sin(240

k  3
2

k

A  15
y  15 cos (72

5

 4
8

A  4.5

k  5

1



120

k  240
t)
2

k

54. A  15

8

y  0.34 sin 3v
40. A  4.5

2

k

53. A  3.8

8

A  0.34

 4

k  2

sin 6v

39. A  0.34

2

k

52. sine curve A  
1.5



3

 2

k1

y  
3 cos v

k6
2

k

2

k

A  
3

51. cosine curve



4



k2

y  0.5 sin 2v

k8
2

k

2

k

50. sine curve A  0.5

v

5

y  0.4 sin

38. A 

 10

1

 36

k  72
t)

55.

8

y  4.5 sin 5v
2

k

41. A  16

 30


A  16

k  15


y  16 sin 15v
2

k

42. A  5
A  5
y  5 cos v

2

k

5

5

A   8
y

2

k

y  7.5 cos

A 

y  0.5 cos



2

k
2

3

 5

57c.

10
v
3

47. A  17.9

37d.

 2

y
y 
 2  sin 

 16

k  8

2


v
8

48. A  1.5

2

k



 2

A  1.5
k4
y  1.5 sin 4v, y  1.5 cos 4v
49. cosine curve A  2

2

k

O



2

3

A  0.2
y  0.2 sin (524

1

v

y  2 cos 2

184

4
2

k

58a. A  0.2

 4

k  2

Chapter 6

2

1



A  17.9
y  17.9 cos

y  
1.5 sin 4(12)

y  
1.1 ft
y  0 ft
57a. Maximum value of sin v  1.
Maximum value of 2  sin v  2  1 or 3
57b. Minimum value of sin v  
1
Minimum value of 2  sin v  2  (
1) or 1

20

3

k  3
2

k



y  
1.5 sin 4(3)

10

A  5



56c. y  
1.5 sin 4t

56b. y  
1.5 sin 4t

 0.3

k

2

cos



y  
1.5 sin 4t

1

3

20
v
3

46. A  5

3

2



2

k

A  0.5

8

A  1.5; down first, so A  
1.5

v

3

45. A  0.5

2

k



 6

k

A 
 (
A)

2

k  4

cos 14v

A  7.5

y

56a. A 



 7

k  14

44. A  7.5

2
5

All the graphs have the same shape, but have
been translated vertically.

k1

43. A  8
5
8

 2

5 
1



262

k  524
t)

2

k

1

58b. A  2(0.2)
A  0.1
y  0.1 sin (1048

59a. y  A cos t



2

k

65.

k  262

9.8

6

0.6525352667 



9.8

6

0.458022743 



y
y 
 cos(  
)



15.1


A  tan1 
19.5

2




g



9.8

4

3




9.8



9.8

4.17  
; about 4.17 m
67. b2 
 4ac  52 
 4(3)(10)
 
95
2 imaginary roots
68a. Let x  the number of Model 28 cards and let
y  the number of Model 74 cards.
y
30x  15y  240
40
20x  30y  360
12x  10y  480
32
12x  10y 
 480
x0
x
0
24
y0

 2

O

15.1


tan A  
19.5

4.1  2

 n, where n is an integer

1

a

tan A  b

T  2

66.

60b. 
1
60d.



c2

A  37.75273111
B  180° 
 (90°  37.8°) or 52.2°
c  24.7, A  37.8°, B  52.2°

9.8

6

y  
1.2
about 1.2 m to the left

2

1



b2

24.66292764  c

y  1.5 cos 7.9

60c.

a2

15.12  19.52  c2

t)


9.8
y  1.5 cos t 6
9.8
y  1.5 cos 4 6



2

73


s  9
180 

s  11.5 in.

y  0.6
about 0.6 m to the right

60a.

s  rv

73

180

1



131



59c. y  1.5 cos t



180°

t)

g




y  1.5 cos t
59b.

64. 73°  73°

k  1048

58c. A  2(0.2)
A  0.4
y  0.4 sin (262

1



524

16 30x  15y 
 240
(3, 10)
(0, 12)
20x  30y 
 360
8
(8, 0) y 
 0
(0, 0)

5 


1

O
61a. y  1.5 cos t
y  1.5 cos t

k

m

2

k



18.5

0.4



61b. y  1.5 cos t
y  1.5 cos t

k

m

18.5

0.6

2

k



y  1.5 cos 5.6t
61c. y  1.5 cos t
y  1.5 cos t

k

m

k  0.9 s/cycle
1



 5.6

k  1.1 s/cycle
frequency:
2

k



18.5

0.8

y  1.5 cos 4.8t



1

1.1

0
69. 1
0 
1

 0.9 hertz

q

24

32

x
40


2
1
3 
3

1 
1 
4 
2

1(1)  0(
1)
 1(
2)  0(
1)
0(
2)  (
1)(
1) 0(1)  (
1)(
1)
1(3)  0(
4)
1(
3)  0(
2)
0(3)  (
1)(
4) 0(
3)  (
1)(
2)
 
2 1 3 
3
1 1 4
2
(
2, 1), (1, 1), (3, 4), (
3, 2)

 4.8

k  1.3 s/cycle
1


frequency: 
1.3  0.8 hertz

61d. It increases.
61e. It decreases.
62. 0
63. 84 2  168 radians
q

16

P(x, y)  100x  60y
P(0, 0)  100(0)  60(0) or 0
P(0, 12)  100(0)  60(12) or 720
P(3, 10)  100(3)  60(10) or 900
P(0, 8)  100(0)  60(8) or 480
3 of Model 28, 10 of Model 74
68b. $900

 6.8


frequency: 
0.9  1.1 hertz

y  1.5 cos 6.8t

8

70.

g (x )

v

t
168

6

q  88.0 radians/s

O

185

x

Chapter 6

71. y  14.7x  140.1
y  14.7(20)  140.1
y  $434.10
72.

x2
(
4)2
(
3)2
(
2)2

x

4

3

2

Translations of Sine
and Cosine Functions

6-5
Page 378

y
16
9
4

Graphing Calculator Exploration

1.

{(
4, 16), (
3, 9), (
2, 4)}; yes
73.

1

radius  2(10) or 5

A  s2

A  r2
A  (5)2 or 25
4(25)  100

100  s2
10  s

2. The graph shifts farther to the left.
3. The graph shifts farther to the right.

The correct choice is C.

Page 377
1.

5

6

Mid-Chapter Quiz

5

Page 383

180°



 6
 150°
1

s  rv

2. r  2d
r

1
(0.5)
2

5

s  0.253

or 0.25

s  1.3 m
1

3. A  2r2v
2

A  2(82)5
1

A  40.2 ft2
4. 7.8 2  15.6 or about 49.0 radians
5. 8.6 2  17.2
6. v  rq
v
v  3(8)
q  t
v
 75.4 meters/s
17.2
q  7

5. Jamal;
6.

q  7.7 radians/s
7. 1
8.

Check for Understanding

1. Both graphs are the sine curve. The graph of y 
sin x  1 has a vertical shift of 1 unit upward,
while the graph of y  sin (x  1) has a horizontal
shift of 1 unit to the left.
2. sine function
3a. increase A
3b. decrease h
3c. increase k
3d. increase c
4. Graph y  sin x and y  cos x, and find the sum of
their ordinates.
c

k



2





2

—


6

or 3

y




y 
 3 cos (  2 )

2

y
y 
 cos x

1

O



2

3

4

5 


2

O

9 x

8

7

7. 3; y  3
A  1;


1

9. 
7  7;

2

1

3

2

2

2

y



y 
 7 cos 3

O

4

O

2

4

6

8

10 12 


8

10. A  5
A  5
y  5 sin 6v
Chapter 6

y 
 sin 2  3

4

 6

8


4



y

2

k



 3

k6

186



2

3

4

5 

2



13c.

8. 2  2; 2  ; 
2; 
5

y
O




2

80

5 

4

3

2

P 
 30 sin 2
t  100

P
120

y 
 2 sin (2  
)  5

P 
 30 sin 2
t  100

40


4

O

1


6



1

1

9. 
2  2;

2

1

2

Pages 383–386


4

 4; 
 
 2; 3
1

14.


2

c

k



1

5

6

t

Exercises

2

or 2; A  1; 1  2

y

y

y 
 sin(  2
)

1




y 
 3  2 cos ( 2  4 )

4





2

1

4

3

2

O

2



2

3

4

5 

4

5 


1

O



5 

4

3

2



c

2

k

10. A  20

1

A  20
k  2
y  20 sin 2v  100

2

k

A  0.6

11.

 
2.13
c


  
2.13

 12.4

O

h7

16.

2.13





y  0.6 cos 
6.2 v  6.2 
7

12.

sin x 
 cos x
sin 0 
 cos 0



2

sin 2 
 cos 2

1



sin  
 cos 

1

3

2

3

2

sin

2

sin 2 
 cos 2

y

c

k

3





2

—
1

4

2
or 
2; A  2; 
1  8

4

2

y

1




 cos

2

y

x
0






1

2.13


c
6.2

y 
 sin(2  
)

1


k
6.2

c

k


6.2

y

h  100



A  0.6

2

15. 
k  
 2; A  1; 2  

c0

3

2

1


O

2

3

4 5

6 7 


1


1

y 
 2 cos ( 4  
2 )


2


1

1

y 
 sin x  cos x

1

17. 2; y  2
2
A  1; 
1  4

1


2

O



y

x

2

2


1



1

y 
 sin 2  2
1

13a.

130  70

2

13b. A 

 100; P  100

130 
 70

2

A  30
A  30

2

k

O
1



2

3

4

5 


1

k  2
P  30 sin 2t  100

187

Chapter 6

2

18. 
4; y  
4
A  5;

2

1

 2

y

y

30

O




2



24. 5  5; 3; 
3; 20

2

4

3

5 

y 
 20  5 cos (3  
)

20


4

10


6

O


8

y 
 5 cos   4

A  1;


2


2



y

y

O

y 
 7  cos 2

8




4

O
20.





4

2

21. 3  3;

2

1

y

2

4

3

5 

2

4
  16; 
5
26. 10  10; 
1  8; 
 1

4

or 2; 
3



2

y



 2; 
1  2; 0



2

4

3


15

5 

2 
 (
6)



3

A  4
A  4; 4



y 
 6 sin (



3

28. A  7

29. A  50


2

4

3

k

c


2 

h  
7


3

2

3

2

c  
3

2

2

k

A  50

5 

3

 4

k

c




8  2

3

8

3

h  
25
4

c  
3

4

y  50 sin 3v 
 3 
 25
8


3

3

A  4

1

3

y

y


2

3

2

k

3

30. A  4




12
2

23. 1  1; 
1  6; 
   4; 
2

4

3
4

5 

y 
 2  sin (

c


1
0 

k  10

c  
10

2

k



c

 2

2

k

32. A  5
4

A  5


4

y

188

4
5



 6

c



c  
4
7

5

h7

c  



12  3

k  12

cos (12v 
 4) 




4  4

A  3.5
k4
y  3.5 cos (4v 
 )  7

)

1

h  4

1

4




3  12



 5

sin (10v 
 10)  4

31. A  3.5


2

Chapter 6

 3

2


4

O

2

k

y  7 sin 3v 
 3 
 7

2



4; 2 to the left or 
2;
down 2, or 
2

A  7

) 2

6
4

O



27. A  2

2


2

5 

4

3

2


10


22. 6  6; 1  2; 
 
1  
3; 2

y




5


2

8



O




y 
 3 cos (  2 )


4

y 
 10 sin ( 4  4
)  5

5

2

O



1

y 
 4 cos 4  3

4
2

5 

4

3

2


2

6

c

k

5 

4

3

2

0
1
1 2

25. 4  4; 
1  4; 
 1  0; 
3

19. 7; y  7

2

2



7

h  5

2

k

33. A  100

c
 45 
 
2  0
2

A  100

k  45

38.

h  
110


45

2

y  100 cos 45v 
 110
34. A 

1 
 (
3)

2

2

k

A  2

 4

k

h  
1

1

2

y  
2 cos 2 
 1
v

3.5 
 (2.5)

2

k

A  0.5
A  0.5; 0.5
y  0.5 sin 2v  3
36.

x
0

1



0

3

2


1

2

0

y
8

c




2  0

k2

c0

sin 2x
0

sin x  sin 2x
0

0.71



2

1

1.71

1



0

1



0

2

0

0

3

2


1

3

0


1

h3

2

0

4

0

0

y
y 
 sin x  sin 2x

2
1

sin x
0



2

2x
0



4


2

A  2; 
2

35. A  2

sin x
0

x
0

c0

sin x  x
0


1

 1  2.57


2



2

O

  3.14
3

2

39.


 1  3.71



y
4

2  6.28

O

4

40.


x
0

cos x
1

3

2

sin x
0

x

3

y

y 
 2 sin x

y 
 3 cos x

y 
 cos 2x  cos 3x

2

x

cos x 
 sin x
1



2

0

1


1




1

0


1

3

2

0


1

1

2

1

0

1

O

2


2



3
2

y 
 cos 3x

2

y 
 cos 2x

41a. 2000  1000  3000
2000 
 1000  1000
41b. 10,000  5000  15,000
10,000 
 5000  5000

y

O

2

4

O


1




2

2

1

y 
 2 sin x  3 cos x

x
y 
 x  sin x

4

2

x

2

6

37.

2

41c.
y 
 cos x  sin x


2

x


2

[0, 24] sc11 by [0, 16,000] sc11000
41d. months number 3 and 15
41e. months number 0, 12, 24
41f. When the sheep population is at a maximum,
the wolf population is on the increase because of
the maximum availability of food. The upswing
in wolf population leads to a maximum later.

189

Chapter 6

y

42.

3


f(x)  
x
1

49.

y
x

3

3


y
x
1

2

3


x
y
1



x(y 
 1)  3

x
O






2

3

3

y 
 1  x

4

3


2

y  x  1

y 
 cos x

y 
 x cos x

1 1
3
5 X
1 1 
3 
5
1(3)  1(
3) 1(5)  1(
5)  X
1(3)  1(
3) 1(5)  1(
5)
0 0
X
0 0
51. 7(3x  5y)  7(4)
21x  35y  28
→
14x 
 35y  21
14x 
 35y  21
35x
 49
x  1.4
3x  5y  4
3(1.4)  5y  4
y  
0.04
(1.4, 
0.04)
50.


3

43a. 46 
 42  4 ft
1

t  21  4

1
(42)
2

t  25

43b. r  2d
r

r  21
43c.

3 revolutions

60 seconds

1 revolution



x seconds

x  20 s
2

k

43d. A  21
A  21; 21
h  25  21 sin
43e.

 20

k

t

10

h  25

52.



10

x  4

6  4

4  4

2  4
0  4

x

6

4

2
0

t

10
t

10

h  25  21 sin 


46  25  21 sin  
t
1  sin 10
t
sin
1  sin 10
5  t; 5 s

y

53. 3x 
 y  7  0
y  3x  7
slope: 3

t

43f. h  25  21 sin 1
0

 10

h  25  21 sin 1
0 
0



2


 2

There is a 4

or

75

cos x

45c. y  cos x2

45d. y  sin 
x




6-6

1

294

k  588
47. v  rq
v  7(19.2)
v  134.4 cm/s
48. asymptote: x  2
x
3


y
x
2

y(x 
 2)  x 
 3
yx 
 2y  x 
 3

2y 
 3  x 
 yx

2y 
 3  x(1 
 y)

2y 
 3

1
y

42

1

3

 1050 cubic feet

Check for Understanding

1. any function that can be written as a sine function
or a cosine function
2. Both data that can be modeled with a polynomial
function and data that can be modeled with a
sinusoidal function have fluctuations. However,
data that can be modeled with a sinusoidal
function repeat themselves periodically, and data
that can be modeled with a polynomial function
do not.
3. Sample answers: the amount of daylight, the
average monthly temperatures, the height of a
seat on a Ferris wheel

y
x3

y
 x2

O

Modeling Real-World Data
with Sinusoidal Functions

Pages 390–391

y  0.25 sin 588t

x

x

asymptote: y  1

Chapter 6

y 
 y1  m(x 
 x1)
y 
 (
2)  3(x 
 3)
y  2  3x 
 9
3x 
 y 
 11  0

phase difference.
45b. y  x

46.

x

1050 7.48  7854 gal
The correct answer is 7854.



4

x
45a. y  
sin
2

k

O

1

44. 
k  
2 or 0
c

k

y  |x  4|

54. 4 inches  3 foot

h  25 ft
c

y
2
0
2
4

190



4a. y  
5 cos6t

5

8d. h  
3 cos 3t  3.5

4b. 5 units above
equilibrium



y  
5 cos 6 0

5

h  
3 cos 3(25)  3.5

y  
5
5 units below equilibrium

t
6


6

4c. y  
5 cos 
y  
5 cos 



h  2 units



R  1200  300 sin 2 0

7

R  1200

y  
4.33
about 4.33 units above equilibrium
140 
 80



1

6a. A 

P  30 sin 2t  110

66° 
 41°

2

6b. h 

A  12.5°
6c. 12 months



H  232
9c. R: 1200  300  1500
H: 250  25  275
no

h  110

k  2



H  250  25 sin 2 0 
 4

140  80

A  30



9b. H  250  25 sin 2t 
 4

h  2

5. A  2
2

k



9a. R  1200  300 sin 2t



R  1200  300 sin 2t

9d.

66°  41°

2



1500  1200  300 sin 2t

h  53.5°



300  300 sin 2t


2

k

6d. A  12.5

 12

h  53.5



sin1 1  2t


k  6

cos 6t  c

cos 6 1 

cos 6  c


41  
12.5


12.5  
12.5


1  cos 6  c
y  
12.5

1  sin 2t
sin11  t
2

  53.5
c  53.5


1t
January 1, 1971
9e. 250 
 25  225










25  25 sin 2t 
 4



cos1 1 
 6  c






1  sin 2t 
 4


0.5  c



sin1 
 1  4  t



2






0.5  t
2
2
July 1, 1969; k  


6e. y  
12.5 cos 6t 
 0.5  53.5

y  
12.5 cos 6(2) 
 0.5  53.5

y  42.82517529
Sample answer: About 42.8°; it is somewhat
close to the actual average.

t 
 0.5 
6

(10) 
 0.5
6





sin1 
 1  2t 
 4



Sample answer: y  
12.5 cos 6t 
 0.5  53.5

y  
12.5 cos 



225  250  25 sin 2t 
 4



cos1 1  6  c

6f. y  
12.5 cos 



H  250  25 sin 2t 
 4


2

k4
next minimum: July 1, 1973
9f. See students’ work.

53.5

  53.5

2

k

4

10. A  2

y  53.20504268
Sample answer: About 53.2°; it is close to the
actual average.

 10


A2

k  5


y  2 cos 5t

c
2
11. h  4.25; A  3.55; k  12.40; 
 
  
4.68

Pages 391–394

Exercises

k

7a. 0.5
7b.

2

k




6.2



6.2

2.34


c
3.1

2.34



y  3.55 sin 
6.2 t  3.1   4.24

 660

c


2

1


k
330
1
7c. 
1  330 hertz

12a. h  47.5; A  23.5; k  12; 
   4
6




330



k  6

2

c  
3

2

y  23.5 sin 6t 
 3  47.5

8a. 3.5  
3  6.5 units
8b. 3.5 
 3  0.5 units
2
2
6

8c. k  
5 or 5

3

191

Chapter 6



2

12b. y  23.5 sin 6t 
 3  47.5





c

14. 
k  
 1 or 

2





3

y  23.5 sin 6 3 
 3  47.5

increase shift by 2;   2  2

y  35.75
about 35.8°


k  2

t
6


6

12c. y  23.5 sin 
y  23.5 sin 

y  67.9°

81° 
 73°

3

Sample answer: y  3 cos x 
 2  5
13.25 
 1.88

2

k

 12

13.25  1.88

15b. h  2

15a. A  2

A  5.685 ft
h  7.565 ft
15c. 4:53 P.M. 
 4:30 A.M.  12:23 or about 12.4 h

h  77°

13d. A  4

3

3

81°  73°

A  4°
13c. 12 months

c

c  
2

13b. h  2

13a. A  2

3


1  2

2

3

  47.5
2
8 
 3  47.5




c

2

k

15d. A  5.685

h  77

k



k  6
73  
4 cos 

1  c  77




13.25  5.685 sin 
6.2 4.5  c
4.5


5.685  5.685 sin 
6.2  c



6

1  cos   c

cos1 1 
cos1 1 




6




h  5.685 sin 
6.2 t  c  7.565


4  
4 cos   c


6

h  7.565



6.2

4:30 A.M.  4.5 hrs



y  
4 cos 6t  c  77


6


6

 12.4

4.5


1  sin 
6.2  c

c

4.5


sin1 1  
6.2  c

c

4.5


sin1 1 
 
6.2  c


0.5235987756  c

Sample answer: y  
4 cos 6t 
 0.5  77


0.7093918895  c


Sample answer: h  5.685 sin 
6.2 t 
 0.71 
7.565



13e. y  
4 cos 6t 
 0.5  77


15e. 7:30 P.M.  19.5 hrs

y  
4 cos 6 8 
 0.5  77




h  5.685 sin 
6.2 t 
 0.71  7.565

y  80.41594391
Sample answer: About 80.4°; it is very close to
the actual average.




h  5.685 sin 
6.2 19.5 
 0.71  7.565

h  8.993306129
Sample answer: about 8.99 ft
16a. Table at bottom of page.



13f. y  
4 cos 6t 
 0.5  77


y  
4 cos 6 5 
 0.5  77

y  79.08118409
Sample answer: About 79.1°; it is close to the
actual average.

Month
January
February
March
April
May
June
July
August
September
October
November
December
Chapter 6

Sunrise
A.M.
7:19
6:56
6:16
5:25
4:44
4:24
4:33
5:01
5:31
6:01
6:36
7:08

A.M. Time
in Decimals
7.317
6.933
6.267
5.416
4.733
4.4
4.55
5.017
5.517
6.017
6.6
7.133

Sunset
P.M.
4:47
5:24
5:57
6:29
7:01
7:26
7:28
7:01
6:14
5:24
4:43
4:28

192

P.M. Time
in Decimals
16.783
17.4
17.95
18.483
19.017
19.433
19.467
19.017
18.233
17.4
16.717
16.467

Daylight Hours
(P.M.-A.M.)
9.47 h
10.47 h
11.68 h
13.07 h
14.28 h
15.03 h
14.92 h
14 h
12.72 h
11.38 h
10.12 h
9.33 h

15.03 
 9.33

15.03  9.33

16b. A  2

24. 402  322  202 
 2(32)(20) cos v

16c. h  2

A  2.85 h
16d. 12 months

402 
 322 
 202


cos v  

2(32)(20)

h  12.18 h

402 
 322 
 202

2

k

16e. A  2.85

 12

k


v  cos1

2(32)(20) 

h  12.18

v  97.9°
180° 
 97.9°  82.1°
about 97.9°, 82.1°, 97.9°, 82.1°



6


y  
2.85 cos 6t  c  12.18


6


6

9.47  
2.85 cos 

25.

1  c  12.18


2.71  
2.85 cos   c


cos1 0.950877193  6  c


cos1 0.950877193 
 6  c

0.2088597251  c

Sample answer: y  
2.85 cos 6t 
 0.21  12.18


y  70.5  19.5 sin 6t  c

A

m
4



51  70.5  19.5 sin 6 1  c

26. 
2



6


19.5  19.5 sin   c


sin1 
 1  6  c


sin1 
 1 
 6  c

18a.

B

B


1

3





m4  m
4  m4

2


1

6

2k  8
4k 
 14

2k  7  4k 
 20

k

4
k
4

f (x ) 
 2 x  1  5


2.094395102  c
Sample answer: about 
2.09
1 minute

60 seconds
7
t
15
7
t
15
7
 4
15

A




m
4  m4

2
4k 
 20  0
k5
f (x )
27.




1  sin 6  c

14 revolutions

1 minute

2m  16



(m 
 4)(m  4)

2m  16  A(m  4)  B(m 
 4)
Let m  
4.
2(
4)  16  A(
4  4)  B(
4 
 4)
8  
8B

1  B
Let m  4.
2(4)  16  A(4  4)  B(4 
 4)
24  8A
3A



0.950877193  cos 6  c

17. 70.5 
 19.5  51

2m  16


m2 
 16
2m  16

(m 
 4)(m  4)

x

O

2 radians

1 revolution



7

15

rad/s

y  
3.5 cos 
18b.


y  
3.5 cos  
y  
3.5 cos 


increasing: x 
1; decreasing: x  
1
28. The correct choice is E.

y  
3.197409102
about (4, 
3.20)
120 
 (
120)

2

k

19. A  2
A  120

 60

k

t
30

VR  120 sin 

6-7



30



Page 400

20. See students’ work.
2



y 
 3 cos(2  
)  5

8

Check for Understanding

1. Sample answers: 
, , 2
2. The asymptotes of y  tan v and y  sec v are the
same. The period of y  tan v is  and the period
of y  sec v is 2.

21. 
3  3; 2  ; 
2; 5

y

Graphing Other Trigonometric
Functions



3

3. Sample answers: 2, 
2

6
4

4. 0
5. 1
6. n, where n is an odd integer

2

7.

O



2

3

4



4

 n, where n is an integer

5 

22. 2n where n is an integer
23. 800°



180°

40

 9

193

Chapter 6

8.



1



4

1

 ;



24.



4

25.

y

26.




y 
 tan (  4 )

4

27.

2

2




2

2

 n, where n is an integer



4  n, where n is an integer
3
  n, where n is an integer
4

n, where n is an odd integer
2

28. n, where n is an integer

O




2

2


29.


4

9.



4



1

 ;




2


1

y



y 
 cot(  2 )

8



 ; 
2; h  
1

4

y 
 sec(2  
)  1

y


2

4







O




4



2


8

2

2



 2



O


2

2


30.


4

2

1

3

 6; no phase shift; no vertical shift

y
10. k:

2

k

 3

k

c:

2

3

c


2

3





3

4

h: h  
4
2

c  
9


2

2

y  csc 3v 
 9 
 4
2



11. k: k  2

c: 


1

k  2

c

1

2



 
4




31.

2

1

12c.

y



8

12b. F  f sec v
F  715.4 sec v

6
4

 2; no phase shift, no vertical shift
800


2




 2;




4


1

2

F 
 715.4 sec 
400



O

32.

2

y 
 csc   5

2

y


2



1

2

O




y



12d. 715.4 N
12e. The tension becomes greater.



23.

 2n, where n is an integer

Chapter 6





2

4

Exercises

0
14. 0
undefined
16. 
1

1
18. undefined
undefined
20. 0
n, where n is an integer
n, where n is an even integer
3

2




y 
 tan ( 2  4 )  1

2

2

13.
15.
17.
19.
21.
22.



2

 2; h  1

4

Pages 400–403

2

 2; no phase shift; h  5

y  cot 2v  8

2

1




2

4



12a. f  ma
f  73(9.8)
f  715.4 N



O

h: h  0

c  8
1



y 
 sec 3

2

194

O



2



33.

2

2



2

 ; 
2; h  
3

O







2

2

k


6

8

34.

 6;



6


1

3







2;

2



O







2

2





40



2



d
40


6


40



d  10 tan 40(3)

y 
 sec 

d  2.4 ft from the center


y 
 cos 

O

44c. d  10 tan 40t


2 




2

d  10 tan 40(15)


4

d  24.1 ft from the center
y

45.

4


2, 
, 0, , 2
c



c: 
 
1 0

36. k: k  2


2

1

k  —2—

2

h: h  
6

2

c0






37. k: k  2

c



c



—4—

c: 
2  8

k2

h: h  7

2



c

38. k: k  

c: 
2  
4

k2

c

h: h  
10



2

2



2

c

c: 
 
2 

39. k: k  3


3

2

k  3

c

c:

1

v

F  2(68.6) sec 2
v

F  34.3 sec 2
46c.

2

3

2

1

5

v

46b. F  2f sec 2

h: h  
1

y  csc 3v 
 3 
 1
 5

1

46a. f  m 9.8
f  7 9.8
f  68.6 N



y  sec 2v  2 
 10

2

1

2

 4

F

c


1

5

 


40

h: h  12



F 
 34.3 sec 2



20

c  5


y  cot 5  5  12
v



The graph of y  csc v has no range values
between 
1 and 1, while the graphs of y  3 csc v
and y  
3 csc v have no range values between

3 and 3. The graphs of y  3 csc v and y 

3 csc v are reflections of each other.



y  cot 2v 
 4  7

k

O


2

4

v

y  tan2 
 6

40. k:

t

20

44b. d  10 tan 40 t

2



k




d 
 10 tan 40 t

20

O 10

20 
10

20

4

2

h: h  7

7

no phase shift
no vertical shift

y



c



2


4

35.






 40


8


2

c

c: 
2  4

y  tan 2v 


y 
 sec ( 3  6 )  2

2



k2
44a.




2

c  3

43. k: k  2

h  
2

y

2

2

3
2

y 
 csc(2  
)  3

h: h  
8


3

y  sec 3v  3 
 8


12

2

1

3

c

c: 
 
2  


42. k: k  3


4

h: h  
5

c  3

y  csc (6v  3) 
 5



2



c

c: 
6  
2

k6

y

2



41. k: k  3

O

195

 

Chapter 6

52. C  180° 
 (62°31
  75°18
) or 42°11


46d. 34.3 N
46e. The tension becomes greater.
47a. 220 A
47b.
47c.

2

60

a

sin A
57.3

sin 62°31


1

 3
0 s




6


60

57.3 sin 75°18


b  62.47505783

1



360



6



a

sin A
57.3

sin 62°31




I  220 sin 60 60 
 6
I  
110 A

49b. h 

c  43.37198044
C  42°11
, b  62.5, c  43.4

3.99  0.55

2

53a.

A  1.72 ft
h  2.27 ft
49c. 12:19 P.M. 
 12:03 A.M.  12:16 or about 12.3 hr
2

k

49d. A  1.72

 12.3

k

h  2.27

2

12.3

12:03  0.05 hr since midnight

73˚

2
 t  c
12.3
2
 0.05
12.3
0.1
  c
12.3

h  1.72 sin 

3.99  1.72 sin 
1.72  1.72 sin 
0.1


1  sin 
12.3  c



  2.27
 c  2.27

4m
x

54.

0.1


sin1 1 
 
12.3  c

72



b2



42



sin A 



c2

7


65
7
65


sin A  65
a
tan A  b
7
tan A  4

sin A  

b

4


65
4
65


65

cos A  

 1.55  2.27

cos A 

12  1.55  2.27

55.

h  3.964014939
Sample answer: 3.96 ft
 4

x2 
 4
——  0
x2 
 3x 
 10
(x 
 2)(x  2)
——  0
(x 
 5)(x  2)

Test 
3:

zeros: 2, 
2

(
3)2 
 4

(
3)2 
 3(
3) 
 10

y
2

y 
 2 cos


2




O

1

5

5


2

1

Test 0:




2

Test 3:


2

51. s  rv

Test 6:


s  183


4


02 
 3(0) 
 10

4


10
32 
 4


32 
 3(3) 
 10
5


10
2
6 
4


62 
 3(6) 
 10
32

8
02

excluded values: 5, 
2
0
 0 false

0
 0 false
0
 0 true
0
 0 false


2  x  5

s  6 cm

6


56. k  
0.5

k  12

Chapter 6

y  13.7 m
a

c

c2

cos A  c

49e. noon  12 hrs since midnight

2

1

2

a2


y
cos 73°


65
c

1.545254923  c
2

Sample answer: h  1.72 sin 
12.3 t  1.55  2.27

h  1.72 sin 

4

x  4 tan 73°
x  13.1 m

0.1

h  1.72 sin 

4

53c. cos 73°  y

53b. tan 73°  4


sin1 1  
12.3  c

2
t
12.3
2

12.3

c



sin 42°11

57.3 sin 42°11




3.99 
 0.55

2

c



sin C


c
sin 62°31


48. y  
1 tan v  2

50. 2  2;

b



sin 75°18



b
sin 62°31


47d. I  220 sin 60t 


49a. A 

b



sin B

196

t  kr
10  12r
r  0.83

57. 3x  2y  8
y

3

2x

y

y  2x  1

2y  
x  4
y

1
x
2


2

O

Trigonometric Inverses and Their
Graphs

6-8

4

Page 410

x

Check for Understanding

1. y  sin1 x is the inverse relation of y  sin x, y 
1
1

(sin x)1 is the function y  
sin x , and y  sin(x )
1

is the function y  sin x.
58. y  17.98x  35.47; 0.88
59. A: impossible to tell
B: 6(150)  10(90)
900  900; true
C: impossible to tell
D: 150  30  2(90)
150  210; false
E: 3(90)  30  2(150)
270  330; false
The correct choice is B.

2. For every y value there are more than one x value.
The graph of y  cos1 x fails the vertical line
test.
3. The domain of y  Sin x is the set of real numbers


between 
2 and 2, inclusive, while the domain of
y  sin x is the set of all real numbers. The range
of both functions is the set of all real numbers
between 
1 and 1, inclusive.
4. Restricted domains are denoted with a capital
letter.
5. Akikta; there are 2 range values for each domain
value between 0 and 2. The principal values are
between 0 and , inclusive.
6.
y  Arcsin x
x  Arcsin y
Sin x  y or y  Sin x

6-7B Sound Beats
Page 404
1.


2

y

y
1

y 
 Sin x

y 
 Arcsin x

O


1



1 x




2

the third graph

x


1


y  Cos x  2

7.

2.


2

O


2



x  Cos y  2


Cos1 x  y  2


y  Cos
1 x 
 2

y


2

1


y 
 Cos (x  2 )

3.
4.
5.

6.

7.
8.

Sample answer: The graph seems to stay above
the x-axis for an interval of x values, and then
stay below the x-axis for another interval of x
values.
0.38623583
no

1.78043; yes; the value for f(x) is negative and
corresponds to a point not graphed by the
calculator.
Sample answer: As you move 1 pixel to the left or
right of any pixel on the screen, the x-value for the
adjacent pixel decreases or increases by almost 7.
Thus, the “find” behavior of the function cannot be
observed from the graph unless you change the
interval of numbers for the x-axis.
See students’ work.
Yes; no, they only provide plausible visual
evidence.




2

O

1

8. Let v  Arctan 1.
Tan v  1

v  4


2

y



y 
 Cos1x  2

x


1

O

1 x




2

9. Let v  Tan1 1.
Tan v  1

v  4
cos (Tan1 1)  cos v

 cos 4

2


 2

197

Chapter 6

16.

10. Let v  Cos1 2.

2



2


Cos v  2

y  arctan x
x  arctan y
tan x  y or y  tan x



y

v  —4—
cos Cos1 



2

2






2


v 
 2


 
 
4
2


 —4—

 cos 

 cos 
 cos




 y 
 arctan x

O


2

11. true
12. false; sample answer: x  1; when x  1,
Cos1(
1)  , 
Cos1(1)  0
13a. C  2r
13b. C  40,212 cos v
C  2(6400)
C  40,212 km
13c.
C  40,212 cos v
3593  40,212 cos v
cos1 

2



x

2





2


 2

3593

40,212
3593

40,212

y
y 
 tan x

O

1

2


2

1

Cos x  y or y  2 Cos x

y

y



1

y 
 12 Cos x

 cos v

O


1

 x

O

y 
 Arccos 2x

1.48  v; about 1.48 radians
13d. C  40,212 cos v
C  40,212 cos 0
C  40,212 km

x

y  Arccos 2x
x  Arccos 2y
Cos x  2y

17.

v



1 x


1



y  2  Arcsin x

18.



x  2  Arcsin y

Pages 410–412
14.



x 
 —2—  Arcsin y

Exercises

y  arccos x
x  arccos y
cos x  y or y  cos x



y
2

y

y

1


1

y 
 cos x



y 
 arccos x



Sin x 
 2  y or y  Sin x 
 2



O

2

x

y

y



1


2

 2

1 x

 x


2

O

 Arcsin x

O

y 
 Sin (x  
2 )


1

x

y  tan 2

19.

y


1

15.

O

1 x

x  tan 2


1

y

tan1 x  2

y  Sin x
x  Sin y
Sin1 x  y or y  Sin1 x

2 tan 1 x  y; y  2 tan1 x

y

y


2

1

y 
 Sin x



1

Chapter 6


2

x




O


1

x

y 
 tan 2

2

y 
 Sin1 x

O


2

y

y

O

2

1 x




2

198

y 
 2 tan1 x




x

O


2



2

x



y  Tan x 
 2

20.

1

30. Let a  Sin1 1 and   Cos1 2.



x  Tan y 
 2

1

Sin a  1

Cos   2





Tan1 x  y 
 2



a  2

  3

sin Sin1 1 
 Cos1 2  sin (a 
 )



1

Tan1 x  2  y



21. y



 sin 6

y y 
 Arccot x

4



y 
 Cot x

2

1

 2


2

2

O

2

31. No; there is no angle with the sine of 2.
32. false; sample answer: x  2; when x  2,
Cos1(cos 2)  Cos1 1, or 0, not 2.
33. true
34. false; sample answer: x  
1; when x  
1,
Arccos (
1)   and Arccos (
 (
1))  0.
35. true
36. true

 x

4 
2 O

4 x

2


4

22. Let v  Sin1 0.
Sin v  0
v0

23. Let v  Arccos 0.
Cos v  0

v  —2—


3


25. If y  tan 4, then


3


Tan v  3
v

y  1.



6

tan 

4

Sin1 

26. If y

sin 

then y 

 Sin1 1

O


——.
4



2

2



2

0  sin 6t 
 3


2

sin1 0  6t 
 3


2

3

1

2

Tan a  1

Sin   1



4

  2



cos (Tan1 1 
 Sin1 1)  cos (a 
 )




 cos 4 
 2
 cos 
4



41.


2

 2
1
 Sin1 2.

42.



5

3

 6t



4

2



 6t

 n, where n is an integer
I  I0 cos2 v
1  8 cos2 v

1

Sin   2


  6

cos1

cos Cos1 0  Sin12  cos (a  )
1





or   6t 
 3

4t
10  t
April and October
40.
P  VI Cos v
7.3  122(0.62) Cos v
0.0965097832  Cos v
Cos1 0.0965097832  v
1.47  v; about 1.47 radians

28. Let a  Tan1 1 and   Sin1 1.





0  23.5 sin 6t 
 3

0  6t 
 3

a  2

2

54.5  54.5  23.5 sin 6t 
 3



2

 2

Cos a  0



y  54.5  23.5 sin 6t 
 3

39.

)  cos y
cos (Tan1 
3

 cos 3

29. Let a  Cos1 0 and 

x


——
2

1

27. If y  Tan1 
3
, then y  3.

a



  Sin1 y

  sin (2y)

 sin 2 4
 sin



y 
 tan(Tan1x )




2
 Cos1 2,

2

2 Cos1 2



37. false; sample answer: x  2; when x  2, cos1 2
is undefined.
y
38.



24. Let v  Tan1 3.



 sin 2 
 3



No; the inverse is y  Tan1 x  2.

1

8
1

8
1

8

 cos2 v
 cos v
v

1.21  v; about 1.21 radians
43a. 6:18  12:24  18:42 or 6:42 P.M.
43b. 12.4 h



 cos 2  6
2

 cos 3

7.05 
 (
0.30)

43c. A  2

1

 
2

A  3.675 ft

199

Chapter 6

2

k

43d. A  3.675

7.05  (
0.30)

 12.4



k
6.2

y

48.

h  2

1

y 
 cos x

h  3.375

6:18  6.3 h


y  3.675 sin 
6.2 t  c  3.375


6.2
6.3

6.2

7.05  3.675 sin 

3.675  3.675 sin 
6.3

6.2

1  sin 

sin1
sin1

1

6.3

6.2

1


6.3

6.2

 c


11

6.3  c  3.375

 c

49.

30

c

x

30

sin 25°







sin 
6.2 t 
 1.62

 t
6.2

2.625  3.675 sin 




sin1 






 1.62

30

sin 65°


 1.62

1

a  2(180° 
 50°) or 65°
30

y



sin 50°
30 sin 50°


y
sin 65°


 1.62

y  25.4 units
50. 210° 
 180°  30°
51. p: 1, 2, 3, 6
q: 1, 2

4.767243867  t
0.767243867 60  46.03463204;
Sample answer: about 4:46 A.M.

p
——:
q

52.

y

1

3

1, 2, 3, 6, 2, 2

g (x )
1

y 
 sin (Tan1 x ) 1

g (x ) 
 x  2  3

O

2

v  2(25°) or 50°

30

y
30

  1.62
  1.62  t

44.



30

6  3.375  3.675

2.625

sin 1 
3.675
6.2
2
.
6
2
5
 sin1 

3.675

x



sin 130°

x  54.4 units




—
—
6.2 t



6.2 t



30

30 sin 130°


sin 
6.2 t 
 1.62

 sin 




x
sin 25°

y  3.375  3.675


 t
6.2

v  25°
a  180° 
 (25°  25°)
or 130°

30

25˚
30


1.621467176  c
Sample answer:


y  3.375  3.675 sin 
6.2 t 
 1.62

2.625

3.675
2.625

3.675

x


1

c

43e.


9 O


10

x

2

O

x


1

45a. v 
v

decreasing for x  2 and x 2
53. [f 
 g](x)  f(g(x))
 f(3x)
 (3x)3 
 1
 27x3 
 1
[g 
 f ](x)  g(f(x))
 g(x3 
 1)
 3(x3 
 1)
 3x3 
 3
54. D  4, F  6, G  7, H  8
value: (4  6  7  8)4  (25)4 or 100
The correct choice is D.

D
d
cos1 2
c
6
4
1


cos
2(10)

v  1.47 radians
45b. L  D  (d 
 D)v  2C sin v
L  (6)  (4 
 6)1.47  2(10) sin 1.47
L  35.81 in.
46. n, where n is an integer
47. A  5

2

k

A  5

 3

k

2
——
3

c



2  


3

h  
8

2

c  3

2

y  5 sin 3v  3 
 8
2

Chapter 6

200

2

Chapter 6 Study Guide and Assessment
Page 413
1.
3.
5.
7.
9.

y

Understanding and Using the
Vocabulary
2.
4.
6.
8.
10.

radian
the same
angle
radian
sunusoidal

1

y 
 0.5 sin 4

angular
amplitude
phase
frequency
domain

O



11. 60°  60°




180°

12. 
75°  
75°



180°

13. 240°  240°

14.

5

6

4

 3
15.

7

4







180°

180°



 
315°
17. s  rv
3
s  154
s  35.3 cm

16. 2.4  2.4

O

q

q  2.3 radians/s
27. 15.4 2  30.8
q
q

q
q

q  6.5 radians/s
29. 
1
30. 0





c

 2


4  
2

A  4
k4
y  4 sin (4v  8) 
 1
2

k

37. A  0.5

2

k

3

38. A  —4—
3

A  4

h  
1

c  8



c



c

2

3


2  3

A  0.5
k2
2
y  0.5 sin 2v 
 3  3


h3

c

 4


8  0

k8

c0

h5

3

y  4 cos 8v  5
120 
 80

2

k

39. A  2
A  20
y  20 sin 2t  100
130 
 100

40. A  2

v

t
7.2

2

120  80

1

h  2

k  2
2
——
k

h  100
130  100

1

h  2

A  15
k  2
y  15 sin 2t  115

h  115

2

41. period: 1 or 2, no phase shift, no vertical shift

q  11.3 radians/min
28. 50 2  100

v

t
30.8

15

2

k

36. A  4

s  19.6 cm
20. s  rv

s  155
s  9.4 cm

q

6


1

5

q

2

180°



19. 150°  150°
5
 6
s  rv
5
s  156
s  39.3 cm
21. 5 2  10 or about 31.4 radians
22. 3.8 2  7.6 or about 23.9 radians
23. 50.4 2  100.8 or about 316.7 radians
24. 350 2  700 or about 2199.1 radians
25. 1.8 2  3.6
26. 3.6 2  7.2
q

4

 4

y 
  13 cos 2

s  1512

v

t
3.6

5



1

 137.5°


18. 75°  75° 
180°
5


 12
s  rv



180°

3

y

5
 
12
5
180°


6


 150°

7

4

2

1

2

1

35. 
3  3;

Skills and Concepts



3

2


1
1

Pages 414–416



34. 0.5  0.5; 4  2

y
1

v

t
100

12

y 
 13 csc 

q  26.2 radians/min
31. 1
32. 0

O



2

3

4 

2

33. 4  4; 2  

y


1

y 
 4 cos 2

4
2

O



2

3 


2

4

201

Chapter 6

42.


;
3



2


3

Page 417



 
6; no vertical shift

Applications and Problem Solving


50a. A  11.5

2k  12

y

k

8

y 
 2 tan ( 3  
2 )

4

O



h  64


6



6



c  
2



y  11.5 sin 6t 
 2  64
50b. April: month 4

2 



c



 3





y  11.5 sin 6t 
 2  64


4





y  11.5 sin 6 4 
 2  64


8

y  69.75; about 69.8°
50c. July: month 7

43. vertical shift: 4





y  11.5 sin 6t 
 2  64

y





y  11.5 sin 6 7 
 2  64

6
4

y  74.0°

y 
 sec   4

F


B
IL sin v

51.

2

0.2

O



2


0.04  
5.0(1) sin v



3


2

0.04(5.0(1) sin v)  0.2
0.2


sin v  
0.04(5.0)(1)

44. vertical shift: 
2

sin v  1

v  2

y
2

O



2



3

Page 417


2

1.

y 
 tan   2

A
26.2 


6

Open-Ended Assessment
1
r2v
2
1
r2v
2
2

Sample answer: r  5 in., v  3
46. Let v  Sin1 1.
Sin v  1

v  2

45. Let v  Arctan 
1.
Tan v  
1

v  
4

2a. Sample answer: If the graph does not cross the
y-axis at 1, the graph has been translated. The
first graph has not been translated and the
second graph has been translated.



47. If y  tan 4, then y  1.

y



Cos1tan 4  Cos1y

 Cos1 1
Let v  Cos1 1.
Cos v  1
v0

3


1

O



2

x



2

x



48. If y  Sin1 2, then y  —3—.


1

sinSin1 2  sin y

3




y

 sin —3—

1


3


 2
1

49. Let a  Arctan 
3
 and   Arcsin 2.
Tan a  
3

a

1



3

cos (Arctan 
3
  Arcsin

O

Sin   2

1

2



6


1

 cos (a  )




 cos 3  6
 cos

2b. Sample answer: If the equation does not have
the form y  A cos kv, the graph has been
translated. The graph of y  2 cos 2v has not
been translated. The graph of y  2 cos (2v  )

 3 has been translated vertically and
horizontally.



2

0

Chapter 6

202

Chapter 6 SAT & ACT Preparation
Page 419

If either of the two factors equals 0, then the
statement is true. Set each factor equal to 0 and
solve for x.
x
40
or
x20
x4
x  
2
The solutions of the equation are 4 and 
2. To
find the sum of the solutions, add 4  
2  2. The
correct choice is D.
6. You may want to label the triangle with opposite,
adjacent, and hypotenuse.

SAT and ACT Practice

1. Since there is no diagram, draw one. Sketch a
right triangle and mark the information given.

4



A

Notice that this is one of the “special” right
triangles. Its sides are 3-4-5. So the hypotenuse is
5. The sine is opposite over hypotenuse (SOH).




C

5
hypotenuse

4

sin v  5
The correct choice is B.
2. Let x be the smaller integer. The numbers are two
consecutive odd integers. So, the larger integer is 2
more than the first integer. Represent the larger
integer by x  2. Write an equation that says that
the sum of these two integers is 56. Then solve for x.
x  (x  2)  56
2x  2  56
2x  54
x  27
Be sure to read the question carefully. It asks for
the value of the larger integer. The smaller integer
is 27 and the larger integer is 29.
The correct choice is C.
3. Factor the numerator.
a2 
 b2  (a  b)(a 
 b)
(sin v  cos v)(sin v 
 cos v)

sin v 
 cos v

opposite
3

adjacent

3

B

To find cos v, you need to know the length of the
adjacent side. Notice that the hypotenuse is 5 and
one side is 3, so this is a 3-4-5 right triangle. The
adjacent side is 4 units.
Use the ratio for cos v.
adjacent

4



cos v  
hypotenuse  5

The correct choice is C.
7. Look at the powers of the variables in the
equation. There is an x2 term, an x term, and a y
term, but no y2 term. It cannot represent a line,
because of the x2 term. It cannot represent a circle
or an ellipse or a hyperbola because there is no y2
term. So, it must represent a parabola.
The general form of the equation of a parabola is
y  a(x 
 h)2  k. The correct choice is A.
8. Factor each of the numerators and determine if
the resulting expression could be an integer, that
is, the numerator is a multiple of the denominator.

 sin v  cos v

The correct choice is B.
4. First find the coordinates of point B. Notice that
there are two right triangles. One has a
hypotenuse of length 15 and a side of length 12.
This is a 3-4-5 right triangle. The coordinates of
point B are (9, 12).
Since point A has coordinates (0, 0), each point on
side AB must have coordinates in the ratio of 9 to
12 or 3 to 4.
The only point among the answer choices that has
this ratio of coordinates is (6, 8).
A slightly different way of solving this problem is
to write the equation of the line containing points
A and B.

I
II
III

16n  16

n1
16n  16

16n
16n2  n

16n

16(n  1)



n  1  16; an integer
16(n  1)

n1

  ; not an integer

16n
n
n(16n  1)

16n  1

  ; not an integer

16n
16

Only expression I is an integer.
The correct choice is A.
1

9. Since x 1, 1 
 x  0. So x1 
 x  
x

1.
x
1

Since x 1, x x 
 1 1. So 
 1.
x


1
x
The correct choice is D.

12

y  9 x
Then test each point to see whether it makes the
equation a true statement.
You could also plot each point on the figure and
see which point seems to lie on the line segment.
The correct choice is E.
5. Factor the polynomial on the left side of the
equation.
x2 
 2x 
 8  0
(x 
 4)(x  2)  0

203

Chapter 6

Add the two equations. m∠CAD  m∠BAD 
m∠ACD  m∠BCD  160, so two of the angles in
ABC have the combined measure of 160°.
Therefore, the third angle in this triangle, ∠B,
must measure 20°. The correct answer is 20.

10. Notice that the triangles are not necessarily
isosceles. In ADC, the sum of the angles is 180°,
so m∠CAD  m∠ACD  80. Since segment AD
bisects ∠BAC, m∠BAD  m∠CAD. Similarly,
m∠BAC  m∠ACD. So, m∠BAD  m∠BCD  80.

Chapter 6

204

Chapter 7 Trigonometric Identities and Equations
12.

Basic Trigonometric Identities

7-1
Page 427

1. Sample answer: x 
 45°
2. Pythagorean identities are derived by applying
the Pythagorean Theorem to a right triangle. The
opposite angle identities are so named because A
is the opposite of A.
3.

1







 cot v,
14.

sin A

cos A
sin A


cos A

1

3

2

tan v 

tan v 


10.

v

cos2

 
15. cos x csc x tan x 
 cos x
sin x  cos x 
1


16. cos x cot x  sin x 
 cos x 
sin x   sin x
cos x

cos2

cos2 x  sin2 x




sin x
1




sin x


 csc x
17.

B


F csc v

I


BI
 
 F csc v
BI



F

csc v

F 
 BI

csc v 
1

1



5
2
2


5

2
5
5

F 
 BI
sin v

Pages 427–430

v
1

 cos2 v 
 1


2


2



2  2  
 1

24

1

2

2
6

cos v 
 5

1

19. Sample answer: 45°

2
6

Quadrant III, so 5

sec v

tan v
sec 45°

tan 45°


2

1

11. tan2 v  1 
 sec2 v

472  1 
 sec2 v
 1 
 sec2 v

65

49

65


7

Exercises

18. Sample answer: 45°
sin v cos v 
 cot v
sin 45° cos 45° 
 cot 45°

cos2 v 
 25

16

49

x




sin x  sin x

152  cos2 v 
 1
1

25

sin x


1

1

tan v 


1

sin v
—
cos v

sin v
1
sin v
  
sin v cos v
1

cos v


 sec v


9. tan v 
 
cot v

3

sin2








 tan A
5. Rosalinda is correct; there may be other values for
which the equation is not true.
6. Sample answer: v 
 0°
sin v  cos v 
 tan v
sin 0°  cos 0° 
 tan 0°
01
0
10
7. Sample answer: x 
 45°
sec2 x  csc2 x 
 1
sec2 45°  csc2 45° 
 1
(
2
)2  (
2
)2 
 1
22
1
41

sec v 


csc v

cot v





8. sec v 
 
cos v
1
2
sec v 
 


1

sin (360°  30°)
1

sin 30°


 csc 30°

sin (A)








csc (330°) 
 
sin (330°)


4. tan(A) 
 
cos (A)




2
 
3


 cos 3
13. 330° 
 360°  30°

Check for Understanding

1
1
cos v

 
tan v 
 
cot v , cot v 
 tan v , sin v
1  cot2 v 
 csc2 v

7



 
 2
  
3
3
7



cos 3 
 cos


 sin v

 sin 45°


2


 2


2


2
  2


 sec2 v

 sec v

65


Quadrant IV, so 
7

205

Chapter 7

29. 1  cot2 v 
 csc2 v

20. Sample answer: 30°
sec2

x1


sec2 30°  1 




2

3 2

3

2

cos x

csc x
cos 30°

csc 30°


11


1  cot2 v 
 
3 

1  cot2 v 
 191


3


2



1
 
2

12

9

1




3

4

1

3



3

4



cot2 v 
 29

2


cot v 
  
3

2

Quadrant II, so  
3

30. tan2 v  1 
 sec2 v

2

tan2 v  1 
 54

21. Sample answer: 30°
sin x  cos x 
 1
sin 30°  cos 30° 
 1
1

2

tan2 v  1 
 2156
tan2 v 
 19
6
tan v 
 34


3


 2 
 1

1  
3

 
2

1

Quadrant II, so 34

22. Sample answer: 0°
sin y tan y 
 cos y
sin 0° tan 0° 
 cos 0°
00 
1
0 1
23. Sample answer: 45°
tan2 A  cot2 A 
 1
tan2 45°  cot2 45° 
 1
11
1
21
24. Sample answer: 0

31. sin2 v  cos2 v 
 1
2

13

1

9

cos2 v 
 89
2
2

cos v 
  
3
2
2

Quadrant III, so cos v 
 
3
sin v

tan v 
 
cos v













1



cos v  2  cos v  cos 2

3
tan v 
 —
2
2
3

cos 0  2  cos 0  cos 2





1
2
2


tan v 
 




cos 2  cos 0  cos 2

25. csc v 


2

26. cot v 


1
csc v 
 
2

23

1

tan v

4

9

1
cot v 
 

3



5

4

5

cot v 
 

3
4
3

cot v 
 
3

 

cos v 

cos v 


 cos2 v 
 1
15


15



cos v 
 


15
Quadrant I, so 

sec2 v
sec v

1

cos v 
 
sec v

 cos2 v 
 1

cos2 v 
 1
6

 1 
 sec2 v


13


Quadrant III, so sec v 
 
3

27. sin2 v  cos2 v 
 1
1 2

4
1

16


2


4

 1 
 sec2 v

13
 

9


13
3 



4

csc v 
 2

or

32. tan2 v  1 
 sec2 v

010
01
1

sin v

 cos2 v 
 1
 cos2 v 
 1

4

4

3
13

13

or  

1

33. cos v 
 
sec v

sin2 v  cos2 v 
 1

1
cos v 
 
7
5

sin2 v  57 
 1

cos v 


28. sin2 v  cos2 v 
 1

1



13

3
3


13


57

2 2

sin2 v  3 
 1
2
6

Quadrant III, so  
7

sin2 v  49 
 1
sin2 v 
 59

5
sin v 
  
3
Quadrant II, so

Chapter 7


5

3

206

2

sin2 v  2459 
 1
sin2 v 
 2449
2
6

sin v 
  
7

1

34. sec v 
 
cos v

tan2 v  1 
 sec2 v
tan2 v  1 
 82
tan2 v  1 
 64
tan2 v 
 63
tan v 
  3
7


1
sec v 
 
1

8

sec v 
 8
Quadrant IV, so 3
7


40.

19


5





 2(2
)  5
19


sin 5
19


 —
tan 
19

5
cos 5

1

35. 1  cot2 v 
 csc2 v


sin v 
 
csc v

2
43

1
sin v 
 
53

sin 5

 —


cos 5

sin v 
 35


 tan 5

1




 csc2 v

6
1  19

 csc2 v
25

9
 53








 csc2 v

41.


 csc v

Quadrant IV, so 53
36. 1  cot2 v 
 csc2 v
1  (8)2 
 csc2 v
1  64 
 csc2 v
65 
 csc2 v

65
 
 csc v
Quadrant IV, so 
65


10

——
3





 3
  3
1
10

csc —3— 
 
10

sin 3
1


 sin 3
  




3

1

 


sin 3







 csc 3

1

37. sec v 
 
cos v

42. 1290° 
 7(180°)  30°

sin2 v  cos2 v 
 1

3 2

sin2 v  
4
sin2 v  13
6
sin2 v



1
sec v 
 

3



4

4
3
 or 

sec v 
 
3


3
4



1


sec (1290°) 
 
cos (1290°)


1





1




13


 1
6

1

cos (7(180°)  30°)
1

cos 30°


 sec 30°
43. 660° 
 2(360°)  60°


13


sin v 
 
4

13


cos (660°)

Quadrant II, so 4


cot (660°) 
 
sin (660°)

sin v

tan v 
 
cos v






13





4
tan v 
 —

3



4

cos (2(360°)  60°)

sin (2(360°)  60°)
cos 60°

sin 60°


 cot 60°



13

3



39

3

tan v 
  or 
sec2 A  tan2 A

2sin2 A  2cos2 A




sin 2(2
)  5
——




cos 2(2
)  5




44.


39
 2
4
3 2

 
3
3




3 2
13 2
2 
 2 
4
4























 csc x

48

9

9
 39

 
21136  213
6




sec x
——
tan x

1

cos x
—
sin x

cos x
1

sin x

45.

cot v
——
cos v

cos v

sin v


 
cos v
1




sin v

9

9

32

16


 csc v

46.


 12

sin (v  
)
——
cos (v  
)

sin v

—

—
cos v


 tan v
47. (sin x  cos x)2  (sin x  cos x)2

 sin2 x  2sin x cos x  cos2 x  sin2 x
 2sin x cos x  cos2 x

 2sin2 x  2 cos2 x

 2(sin2 x  cos2 x)

2

38. 390° 
 360°  30°
sin 390° 
 sin (360°  30°)

 sin 30°
27

3

 
 3
  
39. 
8
8

27

3

 
 cos 3
  
cos 
8
8


 cos 38


207

Chapter 7

— ——
48. sin x cos x sec x cot x 
 sin x cos x—
cos x  sin x 
cos x

1


 cos x

eAs 
 W sec v

—
——
49. cos x tan x  sin x cot x 
 cos x—
cos x   sin x sin x 
cos x

sin x

eAs
 

sec v


 sin x  cos x

W

W 
 eAs cos v
56b. W 
 eAs cos v
W 
 0.80(0.75)(1000) cos 40°
W  459.6266659
459.63 W
57. FN  mg cos v 
 0
FN 
 mg cos v
mg sin v  mkFN 
 0
mg sin v  mk(mg cos v) 
 0
mk(mg cos v) 
 mg sin v

—
——
50. (1  cos v)(csc v  cot v) 
 (1  cos v)—
sin v  sin v 
1

W sec v

e 
 A
s

56a.

cos v

1  cos v

—

 (1  cos v)—
sin v 
1  cos2 v

—

—
sin v
sin2 v




sin v


 sin v
51. 1  cot2 v  cos2 v  cos2 v cot2 v

 1  cot2 v  cos2 v(1  cot2 v)

 csc2 v  cos2 v(csc2 v)

 csc2 v(1  cos2 v)

 csc2 v(sin2 v)

mg sin v


mk 
 
mg cos v
sin v


mk 
 
cos v

1

mk 
 tan v

—(sin2 v)

—
sin2 v


1
52.

58.

sin x
sin x
  
1  cos x
1  cos x
sin x  sin x cos x

 
1  cos2 x



sin x  sin x cos x

1  cos2 x




2 sin x




1  cos2 x

h

2 sin x

a




sin2 x
2




sin x


 2csc x
53. cos4 a  2cos2 a sin2 a  sin4 a 
 (cos2 a  sin2 a)2

 12 or 1
2
54.
I 
 I0 cos v
0 
 I0 cos2 v
0 
 cos2 v
0 
 cos v
cos1 0 
 v
90° 
 v
55. Let (x, y) be the point where the terminal side of
A intersects the unit circle when A is in standard
position. When A is reflected about the x-axis to
obtain A, the y-coordinate is multiplied by 1,
but the x-coordinate is unchanged. So,
sin (A) 
 y 
  sin A and
cos (A) 
 x 
 cos A.

v


360°

2n




180°
,
n

a

2

a
a


tan v 
  , so h 
 
2 tan v 
 2 cot v.
h

The area of the isosceles triangle is 2(a)2 cot n
1

a

180°


 4 cot n. There are n such triangles, so
a2

180°

A 
 4na2 cot n.
1

180°

y

59.

A

B

C

E

x




F D

O

y
sin v 
 EF and cos v 
 OF since the circle is a unit

(x, y)

CD

CD



circle. tan v 
 
OD 
 1 
 CD.

A

O

CO

CO


A



sec v 
 
OD 
 1 
 CO. EOF  OBA, so

x

OF

EF

BA

BA

cos v

EO

(x, 
y)

OB

1

OB





Also by similar triangles, 
EF 
 OA , or EF 
 1 .
1

1

OB




Then csc v 
 
sin v 
 EF 
 1 
 OB.

2


60. Cos1  
2  
 135°

Chapter 7

OF







OA 
 1 
 BA. Then cot v 
 sin v 
 EF 
 BA.

208

61.

42

y
1

y  cos (x



6


68. m 
 
4  5

)




2

9

or

y  y1 
 m(x  x1)
2
9

2

y  4 
 9(x  (4))
2

28

y 
 9x  9

x
O

2
3

6

7
6

5
3

69. m∠BCD 
 40°
1
40 
 m (BC)

13
6

2

80 
 m (BC)
1
m∠BAC 
 2mBC

1

1




62. 2(3° 30
) 
 7°

m∠BAC 
 40°
The correct choice is C.


7° 
 7°  
180°



s 
 rv

m∠BAC 
 2(80)

7


180

7



s 
 20
180 

s  2.44 cm
63. B 
 180°  (90°  20°) or 70°
a

b

sin A 
 c

cos A 
 c

a

cos 20° 
 3
5

35 sin 20° 
 a
35 cos 20° 
 b
11.97070502  a
32.88924173  b
a 
 12.0, B 
 70°, b 
 32.9
64. 2 2 1 8 4

4
1
0
4

2 5
2  0
2x2  5x  2 
 0
(2x  1)(x  2) 
 0
2x  1 
 0 or x  2 
 0
1

x 
 2
65.

2,

1
2,

2x2

 7x  4 
 0

Verifying Trigonometric Identities

Page 433

Graphing Calculator Exploration

1. yes
2. no
3. no
4. No; it is impossible to look at every window since
there are an infinite number. The only way an
identity can be proven is by showing algebraically
that the general case is true.

b

sin 20° 
 35

7-2

5.

x 
 2

2




[2
, 2
] sc12 by [2, 2] sc11
sin x

7

x2  2x  2 
 0
7

x2  2x 
 2
7

49

Pages 433–434

49

x2  2x  16 
 2  16

x  742 
 8116
7

9

x  4 
 4
7

Check for Understanding

1. Answers will vary.
2. Sample answer: Squaring each side can turn two
unequal quantities into equal quantities. For
example, 1  1, but (1)2 
 12.
3. Sample answer: They are the trigonometric
functions with which most people are most
familiar.
4. Answers will vary.

9

x 
 4  4
x 
 0.5 or 4
66. continuous
67. 4(x  y  2z) 
 4(3)
4x  4y  8z 
 12
4x  y  z 
 0
→ 4x  y  z 
 0
3y  9z 
 12
x  y  2z 
 3
x  5y  4z 
 11
4y  2z 
 14
4(3y  9z) 
 4(12)
→ 12y  36z 
 48
3(4y  2z) 
 3(14)
12y  6z 
 42
30z 
 90
z 
 3
3y  9z

 12
x  y  2z 
 3
3y  9(3) 
 12
x  (5)  2(3) 
 3
y 
 5
x 
2
(2, 5, 3)

cot x


5. cos x 
 
csc x

cos x 

cos x 


cos x

sin x
—
1

sin x
cos x

1

cos x 
 cos x

209

Chapter 7

6.

1

tan x  sec x

Pages 434–436

cos x



sin x  1

13. tan A 


1
cos x


sin x
1


sin x  1
  
cos x
cos x
1
cos x


sin x  1 
 
sin x  1

cos x
cos x

sin x  1

tan A 

tan A 


cos x




sin x  1

tan A 
 tan A
14. cos v 
 sin v cot v

1


7. csc v  cot v 
 
csc v  cot v
1

csc v  cot v

cos v


cos v 
 sin v 
sin v

 
csc v  cot v 
 
csc v  cot v  csc v  cot v
csc v  cot v

cos v 
 cos v


csc v  cot v 
 
csc2 v  cot2 v

csc v  cot v 

csc v  cot v 


1  sin x

csc v  cot v

(1  cot2 v)  cot2 v
csc v  cot v

1


15. sec x  tan x 
 
cos x
1

sin v tan v 

sin v tan v 

sin v tan v 

sin v tan v 


sec x  tan x 
 sec x  tan x

sin x

cos x

cos x

sin A


 sec x

1

cos x


 sec x

sin x

cos A

cos x

sin x

sin x

cos x

cos x

 
 
sec x csc x 
 
cos x  sin x  sin x  cos x

1  2 sin2 A cot A 
 1  2 sin2 A cot A

sin2 x

cos2 x



sec x csc x 
 
cos x sin x  sin x cos x

1

4

sin2 x  cos2 x


sec x csc x 
 
cos x sin x

sec x

1


sec x csc x 
 
cos x sin x
1

1

 
sec x csc x 
 
cos x  sin x

sec x csc x 
 sec x csc x

1


 4

2 sin2 v  1


18. sin v  cos v 
 
sin v  cos v

1

sin x 
 4

sin v  cos v 


11. Sample answer: cos x 
 1
cot x  sin x 
 cos x cot x
cos x

sin x

sin2 v  cos2 v

sin v  cos v 


cos x


 sin x 
 cos x 
sin x

I cos v

R2




R2 csc v

I cos v

R2


I
sin v

 —
1

R2 
sin v

I cos v

R2


I
sin v
sin v


 —

sin v
1
2


R sin v

I cos v

R2


 R
2

2 sin2 v  (sin2 v  cos2 v)

sin v  cos v


sin v  cos v 
 
sin v  cos v

(sin v  cos v)(sin v  cos v)

sin v  cos v

sin v  cos v 
 sin v  cos v

cos x  sin2 x 
 cos2 x
cos2 x  sin2 x 
 cos x
1 
 cos x
cos x 
 1

2  sec A csc A


19. (sin A  cos A)2 
 
sec A csc A
sec A csc A

2



(sin A  cos A)2 
 
sec A csc A  sec A csc A
1

1

 
(sin A  cos A)2 
 2
sec A  csc A  1

I cot v

(sin A  cos A)2 
 2 cos A sin A  1
(sin A  cos A)2 
 2 cos A sin A  sin2 A  cos2 A
(sin A  cos A)2 
 (sin A  cos A)2

cos v

cos v

Chapter 7

cos x  sin x

cos x(sin x  cos x)



sec x csc x 
 
cos x  sin x

2

1  2 sin2 A 
sin A 
 1  2 sin A cot A

12.


1
cos x

 sec x
 
sin x  cos x

sec x 
 sec x
17. sec x csc x 
 tan x  cot x

2

1  2 sin A cos A 
sin A 
 1  2 sin A cot A

sin x

cos x
—
1

cos x


 sec x

 
 sec x
sin x  cos x

sin v tan v 
 sin v tan v
9.
(sin A  cos A)2 
 1  2 sin2 A cot A
sin2 A  2 sin A cos A  cos2 A 
 1  2 sin2 A cot A
1  2 sin A cos A 
 1  2 sin2 A cot A

1
tan x 
 4
tan x
1
 
 
sec x
4

1  tan x

sin x  cos x
sin x

1
cos x

16.

1
  cos v
cos v
1
cos2 v
  
cos v
cos v
1  cos2 v

cos v
sin2 v

cos v
sin v

sin v 
cos v

10. Sample answer: sin x 


sin x



sec x  tan x 
 
cos x  cos x

csc v  cot v 
 csc v  cot v
8. sin v tan v 
 sec v  cos v
sin v tan v 


Exercises

sec A

csc A
1

cos A
—
1

sin A
sin A

cos A

I cos v

210

20.

(sin v  1)(tan v  sec v) 
 cos v
sin v tan v  tan v  sin v sec v  sec v 
 cos v
sin v

sin v

1

cos x

cos x
sin x

sin x  cos x 
 
cos x
sin x 


1
1
sin x
cos x

1





sin v 
cos v  cos v  sin v cos v  cos v 
  cos v
sin2 v  sin v  sin v  1

cos v
sin2 v  1

cos v
cos2 v

cos v




1  sin y

cos y




1  sin y

cos y

cos y(1  sin y)

1  sin2 y




1  sin y

cos y

cos y(1  sin y)

cos2 y




cos y

1  sin y

cos y

1  sin y

cos y

cos y

1  sin y

21.
cos y

1  sin y



1  sin y

1  sin y


 cos v

cos x
sin x
cos x
sin x



sin x  cos x 
 
cos x  sin x
sin x  cos x 




1  cos x
1  sin x


 cos v

cos2 x


 cos v

cos v 
 cos v

cos2 x

sin2 x  cos2 x


sin x  cos x 
 
sin x  cos x
(sin x  cos x)(sin x  cos x)

sin x  cos x 
 
sin x  cos x
sin x  cos x 
 sin x  cos x
28. sin v  cos v  tan v sin v 
 sec v  cos v tan v
sin v


sin v  cos v  
cos v sin v 
 sec v  cos v tan v
sin2 v


sin v  cos v  
cos v 
 sec v  cos v tan v
cos2 v

sin2 v



sin v  
cos v  cos v 
 sec v  cos v tan v
cos2 v  sin2 v

 
 sec v  cos v tan v
sin v  
cos v
1

cot2 x


sin v  
cos v 
 sec v  cos v tan v

csc2 x  1

sin v  sec v 
 sec v  cos v tan v


23. csc x  1 
 
csc x  1


csc x  1 
 
csc x  1

cos v


sin v 
cos v  sec v 
 sec v  cos v tan v

(csc x  1)(csc x  1)

csc x  1

sin v


cos v 
cos v  sec v 
 sec v  cos v tan v

csc x  1 
 csc x  1
24. cos B cot B 
 csc B  sin B
cos B cot B 

cos B cot B 

cos B cot B 

cos B cot B 

cos B cot B 


cos v tan v  sec v 
 sec v  cos v tan v
sec v  cos v tan v 
 sec v  cos v tan v
29. Sample answer: sec x 
 
2


1
  sin B
sin B
sin2 B
1
  
sin B
sin B
1  sin2 B

sin B
cos2 B

sin B
cos B

cos B 
sin B

csc x

cot x
1

sin v
—
cos v

sin v
1

cos x

cos B cot B 
 cos B cot B
25. sin v cos v tan v  cos2 v 
 1
sin v

sin2

v

cos2

v 
1
1 
1

1  cos x

1

cos x

x

cos x  sin x

cos x
——
sin x  cos x

sin x
sin x

cos x

1  cos x

 


 2
sin x  sin x  sin2 x 
 1  cos x
1  2 cos x  cos2 x

sin2 x
(1  cos x)2

1  cos2 x
(1  cos x)2

(1  cos x)(1  cos x)
1  cos x

1  cos x


 
2



2

—— 
 2
cos x

1
sin x

1  cos x


csc2 x  2 csc x cot x  cot2 x 
 
1  cos x
cos2


 
2


1  tan x

1  cot x
sin x

1
cos x


(csc x  cot x)2 
 
1  cos x

26.


 
2


sec x 
 
2

30. Sample answer: tan x 
 2

2

sin v cos v 
cos v  cos v 
 1

1

sin2 x

sin2 x



sin x  cos x 
 
sin x  cos x  sin x  cos x

22. cos v cos (v)  sin v sin (v) 
 1
cos v cos v  sin v(sin v) 
 1
cos2 v  sin2 v 
 1
1
1

csc x  1 


sin2 x



sin x  cos x 
 
cos x  sin x  sin x  cos x

1  sin y




sin x



27. sin x  cos x 
 
1  tan x  1  cot x

1  cos x




1  cos x
1  cos x




1  cos x


2

2

tan x 
 2
31. Sample answer: cos x 
 0

1  cos x




1  cos x

1

cot x

1  cos x




1  cos x

sec x



csc x 
 cos x
1

cos x
—
1

sin x
sin x


cos x

tan x 


 cos x

tan x


 cos x

tan x  tan x 
 cos x
0 
 cos x

211

Chapter 7

38. yes

1

32. Sample answer: sin x 
 2
1  cos x
sin x
  
sin x
1  cos x
1  2 cos x  cos2 x
sin2 x
  
sin x(1  cos x)
sin x(1  cos x)
1  2 cos x  cos2 x  sin2 x

sin x(1  cos x)
2  2 cos x

sin x(1  cos x)
2(1  cos x)

sin x(1  cos x)
2

sin x


4

4

4

4




4

[2
, 2
] sc12 by [4, 4] sc11


4

39. no

2 
 4 sin x
1

2


 sin x

33. Sample answer: sin x 
 1
cos2 x  2 sin x  2 
 0
1  sin2 x  2 sin x  2 
 0
0 
 sin2 x  2 sin x  1
0 
 (sin x  1)2
0 
 sin x  1
sin x 
 1
34. Sample answer: cot x 
 1
csc x 
 sin x tan x  cos x




[2
, 2
] sc12 by [4, 4] sc11
40a. P 
 I 02 R sin2 2
 ft
P 
 I 02 R(1  cos2 2pft)
40b. P 
 I 02 R sin2 2
ft

sin x

I 02 R


csc x 
 sin x 
cos x  cos x
sin2

x

cos2


P

csc2 2
ft

x



csc x 
 
cos x  cos x

x

41. f(x) 
 


1  4
x2

1


csc x 
 
cos x
1

sin x
cos x

sin x

1

 tan v
2
f(x) 
 

1




cos x

tan v
1
4


cot x 
 1
35.

2

1

2


1

tan3 v  1
  sec2 v
tan v  1
(tan v  1)(tan2 v  tan v  1)
  (tan2 v  1)
tan v  1
tan2 v  tan v  1  tan2 v  1

1

2

tan v

f(x) 
 

1  ta
n2 v

1
0

1
0
1
0
tan v  1 
 0
tan v 
 1

1

 tan v
2

f(x) 
 

sec2 v
1

2

tan v
f(x) 
 
sec v

1


cot v 
 
tan v
1

cot v 
 1

sin v

1

2



cos v
f(x) 
 
1

cot v 
 1
36. no


cos v

1

f(x) 
 2 sin v
sin a


42. sin a 
 sin a sin c ⇒ sin a 
 
sin c
cos b


cos b 
 
sin a ⇒ cos b 
 sin a cos b
cos c


cos c 
 cos a cos b ⇒ cos b 
 
cos a

Then cos b 
 sin a cos b




[2 
, 2 
] sc12 by [2, 8] sc11

sin a

cos c

sin a

cos c

 


sin c  cos a

37. yes

 


cos a  sin c


 tan a cot c
43. y 


gv2


2v02 cos2 v

x sin v



cos v

gv2

 sec2 v  x tan v
y

2v 2
0

[2 
, 2 
]

Chapter 7



sc12

g x2

 (1  tan2 v)  x tan v
y 
 
2v 2

by [4, 4] sc11

0

212

51. Let x 
 the number of shirts and y 
 the number
of pants.
y
100
x  1.5y 100
2.5x  2y  180
2.5x  2y 180
(0,
65)
80
1.5x  3y 195
x  1.5y  100
x 0
60
(40, 40)
y 0
1.5x  3y  195

44. We find the area of ABTP by subtracting the area
of OAP from the area of OBT.
1
OB
2

1

1

1

 BT  2OA  AP 
 2  1  tan v  2 cos v sin v


 2
cos v  cos v sin v
1 sin v



 2 sin v
cos v  cos v
1

1




 2 sin v
cos v  cos v 
1










1

2
1

2
1

2
1

2

cos2 v

1

sin v

40
x0
20
(0, 0)



1 cos2 v

cos v
sin2 v

cos v

sin v



sin v

a sin b




45. By the Law of Sines, 
sin b 
 sin a , so b 
 sin a .
Then
1

A 
 2ab sin 

A 
 2a
sin a  sin 
1

a sin b

a2 sin

sin 


A

2 sin a

A

A

46.

53.

sin b sin 

2 sin (180°  (b  ))
a2 sin b sin 

2 sin (b  )
a2

tan x  cos x  sin x tan x

sec x  tan x

sin x

cos x

sin x  cos2 x  sin2 x

cos x
1  sin x

cos x
sin x  1
cos x
  
cos x
1  sin x

15


ab

ba

 


ab  ba

ab

ab

Sum and Difference Identities
Check for Understanding

1. Find a counterexample, such as x 
 30° and
y 
 60°.
2. Find the cosine, sine, or tangent, respectively, of
the sum or difference, then take the reciprocal.
3. The opposite side for 90°  A is the adjacent side
for A, so the right-triangle ratio for sin (90°  A) is
the same as that for cos A.

2 
 45°
c 
 90°

180°



 1
6  



 168.75°

ba

ba

Pages 441–442

c


 180°

A 
 2
k
2
y 
 2sin (2x  90°)
15


16

ab

ab

7-3


1

48.

80 100


 1
The correct choice is D.

sin x


 cos x  sin x 
cos x

 
1
sin x
  
cos x
cos x



360°

k

40 60
y0

 


a  b  1(a  b)




47. A 
 2

20

P(x, y) 
 5x  4.5y
P(0, 0) 
 5(0)  4.5(0) or 0
P(0, 65) 
 5(0)  4.5(65) or 292.50
P(40, 40) 
 5(40)  4.5(40) or 380
P(72, 0) 
 5(72)  4.5(0) or 360
40 shirts, 40 pants
52. {16}, {4, 4}; no, 16 is paired with two elements of
the range

tan v sin2 v
a

x

O

2


cos v sin v

b

(72, 0)

60


168.75° 
 168°  0.75°  1
°

 168°  45

168° 45

3
49. 

3y  1  2 
 0
3


3y  1 
 2
3y  1 
 8
y
3
50. x  1 
 0
x 
 1
f(x) 


90˚ 
 A

3

Check: 

3y  1  2 
 0
3


3(3) 
12 
0
3

8
2 
0
22 
0

A
1


4. cot (a  b) 
 
tan (a  b)

1

 
tan a  tan b

1  tan a tan b

3x

x1

1  tan a tan b

3x


y 

x1

y(x  1)
yx  y
y
y
y

3y




tan a  tan b


 3x

 3x

 3x  yx

 x(3  y)

1

1

 
1
cot a  cot b
cot a cot b


 

1
1
cot a cot b
  
cot a
cot b
cot a cot b  1




cot a  cot b


x

3y
0
y
3

213

Chapter 7

10.

5. cos 165° 
 cos (45°  120°)

 cos 45° cos 120°  sin 45° sin 120°





2

2

 2 
1



2

2




3


2


2
  
6


4




tan 3  4




tan 3  tan 4








6. tan 12 





sin v  2
—— 
 cot v


cos v  2






(sin v)  0  (cos v)  1

(cos v)  0  (sin v)  1
cos v


sin v



 2  
3
7. 795° 
 2(360°)  75°
sec 795° 
 sec 75°
cos 75° 
 cos (30°  45°)

 cos 30° cos 45°  sin 30° sin 45°




sec 795° 





3

2
  
2
2

6
  
2


4
4


6
  
2


1

 2 




sin v cos 2  cos v sin 2

 cot v
———




cos v cos 2  sin v sin 2


3
1

1  
3
1
4  2
3


2







tan v  2 
 cot v

11.


 ——




1  tan 3 tan 4



sin (90°  A) 
 cos A
sin 90° cos A  cos 90° sin A 
 cos A
1  cos A  0  sin A 
 cos A
cos A 
 cos A

cot v 
 cot v
12. sin (x  y) 



2


2

sin (x  y) 
 ——
1
1
  
sin x cos y
sin y

 
1
sin x  cos y
sin x cos y

sin (x  y) 
 ——

sin x cos y
1
1
  
sin x cos y


 
6
  
2





2
1





2
1





65

65



81 or 9





15

15



16 or 4

4

9

sin (x  y) 
 sin (x  y)
13. sin (nq0t  90°) 
 sin nq0t cos 90°  cos nq0t sin 90°

 sin nq0t  0  cos nq0t  1

 cos nq0t

1

4

sin (x  y) 
 sin x cos y  cos x sin y

65


sin x cos y  cos x sin y

1

sin (x  y) 


cos y 
 

1  si
n2 y


15



 cot v

1  cot x tan y

csc x sec y
cos x sin y
 
1
sin x  cos y

cos x

8. cos x 
 

1  si
n2 x


 cot v

 

 9 
4    9  4 
4

1

Pages 442–445

4
15
  
65





36

cos x 
 

1  si
n2 x

1


9. csc x 
 
sin x
1

5

3




sin x




3




sin x 
 5



1




3
 2
5




1265 or 45




or




sin y

16. cos


tan y 
 
cos y

 
1

5 2

13




144
12
 or 

169
13

12

13
—
5

13

12

or 5

tan x  tan y


tan (x  y) 
 
1  tan x tan y
3

4

17. sin

12

5



 ——
3 12
1  45





Chapter 7




3


2


3

2



1 
2
     
2
2
2
2
  
2


6

 
4
7





 
 cos   
12
4
3









 cos 4 cos 3  sin 4 sin 3

2

2



3

1




 
2  2  2  2

2
  
6


 
4






 
 sin   
12
3
4











 sin 3 cos 4  cos 3 sin 4



2


3

2
1

 

 
2  2  2  2

6
  
2


 
4

 




3

4

sin y 
 

1  co
s2 y




2

2


1
    
2
2
2

2
  
6


4

15. sin 165° 
 sin (120°  45°)

 sin 120° cos 45°  cos 120° sin 45°

sin x


tan x 
 
cos x
3

5

4

5

Exercises

14. cos 105° 
 cos (45°  60°)

 cos 45° cos 60°  sin 45° sin 60°

63

20
—
4
5
63
16

214











18. tan 195° 
 tan (45°  150°)



25.

tan 45°  tan 150°

1  tan 45° tan 150°


3

1  
3

 

3
1  1
3



113
17
 
 4(2)  
12
12
113
17
cot 12 
 cot 1
2
17
7




tan 12 
 tan 6  4


tan 6  tan 4





 ——
p

1  tan 6 tan 4

3  
3

3
—
3  
3

3


3

12  63

 or 2  3

6



12 
 cos 4  3




 cos 4 cos 3  sin 4


3

2
2
1




 
2  2  2  2

  1
3
——



3
1  3  1




19. cos 















p

sin 3
cot

2
  6


4
















2




225
15
 or 

289
17












1

3





3

9
3



25 or 5

3

8

17






8

17
—
15

17
8

15

tan (x  y) 






2





 2  2  2  2
  6

2




4




4

csc 2 
 




225
15
 or 

289
17
sin x


2

sin y 
 
1  co
s2 y

2



1


 sin 6 cos 4  cos 6 sin 4

5

4


tan x 
 
cos x




2

4

28. cos x 
 
1  si
n2 x


 2
  6


1

4

5

24









2



1


 25

4






16
4



25 or 5

6
  2




sin y 
 
1  co
s2 y

3 2

5


 
5
5  
5
5

sec 1275° 
 


1225
35




1369 or 37

12

8

cos (x  y) 
 cos x cos y  sin x sin y

  2

6


 sin 
6  4

35

27. sin x 
 
1  co
s2 x



 
4

24. sin

12

37

621


 2  2  2  2

5

12

2
1
 






629

1

2

1

1




15

23. 1275° 
 3(360°)  195°
sec 1275° 
 sec 195°
cos 195° 
 cos (150°  45°)

 cos 150° cos 45°  sin 150° sin 45°

3

cos y 
 
1  si
n2 y

8 2

17


 

 
1
7 
 37   
 17 
 37 

22. 735° 
 2(360°)  15°
sin 735° 
 sin 15°
sin 15° 
 sin (45°  30°)

 sin 45° cos 30°  cos 45° sin 30°

2

2

1




3
2

sin (x  y) 
 sin x cos y  cos x sin y



 3

2
  
2
2
  2

6


4


 3
2

26. sin x 
 
1  co
s2 x

tan 45°  tan 120°

1  tan 45° tan 120°
1  (3
)

 
1  1 (3
)
1  3



 1  3
4  2
3
 or 2  3



2
23

5
 
 tan   
12
4
3

5
tan 4  tan 3

 ——

5
1  tan 4 tan 3
1  (3
)

 
1  1 (3
)
4  2
3




or 2  3

2







113

12

33


3
—
3  
3

3


 2  3


20. tan 165° 
 tan (45°  120°)

21. tan





sin y

2



1
3

5

16
4



25 or 5


tan y 
 
cos y




tan x  tan y

1  tan x tan y
4
8
  
3
15




4

5
—
3

5
4

3


8
4
1  15  3

12
1
5
—
77

45
36
77

  6

2


 6
  2


215

Chapter 7

29. sec x 
 
tan2 x
1




cos x 

tan x 

5

3

cos y 
 
1  si
n2 y

2

  1

5

3





34
34
 or 

3
9




34
3
3

 or 
34
34

sin x

cos x

1

2



1
1

3

2
2

89 or 
3

5

3
3

5




3
34


34
5
34


34











1

4

5




2
1







 or
16

25





 





5
5

13

5

13

144
12
 or 

169
13

 

 
5
1
3   
 5 
 13 
5

4



sin y 






1


sec (x  y) 
 
cos (x  y)

3

2
2

3

1

 —
56

55


1  co
s2 y

65


 56




1

32. cos a 
 
1  si
n2a






5
5

 or 
9
3
sin y

cos y


5

3
—
2

3

12


 65

2 2

3

tan y 


3

56

1






sin b 
 
1  co
s2b

2



1
1

5




24
2
6



25 or 5




2



1
2

7

45
3
5



49 or 7

sin (a  b) 
 sin a cos b  cos a sin b

5


or 2

2
6

3
5


 
5
7  
5
7
1

tan x  tan y

2


2  630


tan (x  y) 
 
1  tan x tan y
5

2
1




cos(x  y) 
 cos x cos y  sin x sin y

cos y 
 
sec y
1

 —

6

5
5

6


 
35


5

33. sin x 
 
1  co
s2 x

  
2
6

 ——

5 5
1  
6
2




169
13
 or 

25
5

3

5




5
34 1

 
3
34

8
4
66
53




102  102
1217
  534


102


30. tan x 
 
cot x
1

 —







12

5

sin y 
 
1  co
s2 y
3
34 2
2

 
34
3




2

  1



cos y 
 
sec y
1
—

 13

cos (x  y) 
 cos x cos y  sin x sin y






1

cos x 
 
1  si
n2 x

sin x

 —

sin x 


sec y 
 
tan2 y
1


31. sin x 
 
csc x
1

 —







10  65

12
——
12  5
5

12

sin y 
 
1  co
s2 y

2



1
1

3




2
2

89 or 
3




2



1
3

4


7
7



16 or 4

cos (x  y) 
 cos x cos y  sin x sin y
2
2


7


 
3
4  
3
4
1


10  6 5




12  5 5


3

3  2 
14




12


270  1225


 1
9



cos 
2  x 
 sin x

34.




cos 2 cos x  sin 2 sin x 
 sin x
0  cos x  1  sin x 
 sin x
sin x 
 sin x
35.
cos (60°  A) 
 sin (30°  A)
cos 60° cos A  sin 60° sin A 
 sin 30° cos A 
cos 30° sin A
1

2

36.

Chapter 7

216


3

1


3

cos A  2 sin A 
 2 cos A  2 sin A

sin (A  ) 
 sin A
sin A cos   cos A sin  
 sin A
(sin A)(1)  (cos A)(0) 
 sin A
sin A 
 sin A

37.



43. VL 
 I0qL cos 
qt  2

cos (180°  x) 
 cos x
cos 180° cos x  sin 180° sin x 
 cos x
1  cos x  0  sin x 
  cos x
cos x 
 cos x



VL 
 I0qL(cos qt  0  sin qt  1)
VL 
 I0qL(sin qt)
VL 
 I0qL sin qt

1  tan x


38. tan (x  45°) 
 
1  tan x
tan x  tan 45°

1  tan x tan 45°
tan x  1

1  (tan x)(1)
1  tan x

1  tan x

1  tan x




1  tan x

sin 2(a  b)
44. n 
 
b
sin 2
1

1  tan x




1  tan x
1  tan x




1  tan x

sin 2(a  60°)
n 
 
60°
sin 2
1

tan A  tan B


39. sin (A  B) 
 
sec A sec B
sin A

sin B

 + 
cos A
cos B
sin (A  B) 
 ——
1
1
  
cos A cos B
sin A

cos A

sin 
2  30°
n 
 
sin 30°
a

sin B


+
cos B
cos A cos B
——

sin (A  B) 
 1

1
cos A cos B
  
cos A cos B

sin (A  B) 


a


2

sin A cos B  cos A sin B

1

n 
 2



3

2

a

 
cos 2  2
a

1

a

45. The given expression is the expanded form of the




sine of the difference of 3  A and 3  A. We have

cos (A  B) 
 ——
1
1
  
cos A cos B





sin 
3  A  
3  A 
 sin (2A)

 sin 2A

sin B

 
1
cos A  cos B
cos A cos B

cos (A  B) 
 ——

1
1
cos A cos B
  
cos A cos B

cos (A  B) 




a
sin2

n 
 3
 sin 2  cos2

1  tan A tan B

sec A sec B
sin A sin B
 
1
cos A  cos B

sinA

a

sin 2 cos 30°  cos 2 sin 30°
n 
 
1

sin (A  B) 
 sin (A  B)
40. cos (A  B) 




VL 
 I0qL
cos qt cos 2  sin qt sin 2

46a.

f(x  h)  f(x)

h






cos A cos B  sin A sin B

1

46b.

cos (A  B) 
 cos (A  B)

sin (x  h)  sin x

h
sin x cos h  cos x sin h  sin x

h

y y 
 sin x cos 0.1  cos x sin 0.1  sin x
0.1

1

sec A sec B


41. sec (A  B) 
 
1  tan A tan B
1

cos A

1

1

cos A

1

cos B

0.5



cos B
sec (A  B) 
 ——
sin A sin B
 
1
cos A  cos B

O


cos A cos B

sec (A  B) 
 ——
sin A sin B  cos A cos B




1  cos A  cos B
sec (A  B) 

sec (A  B) 


1 2 3 4 5 6 7 8 x

0.5
1

1

cos A cos B  sin A sin B
1

cos (A  B)

46c. cos x
sin (a  b)


47. tan (a  b) 
 
cos (a  b)

sec (A  B) 
 sec (A  B)
42.
sin (x  y) sin (x  y) 
 sin2 x  sin2 y
(sin x cos y  cos x sin y)(sin x cos y  cos x sin y)

 sin2 x  sin2 y
2
2
(sin x cos y)  (cos x sin y) 
 sin2 x  sin2 y
sin2 x cos2 y  cos2 x sin2 y 
 sin2 x  sin2 y
sin2 x cos2 y  sin2 x sin2 y  sin2 x sin2 y
 cos2 x sin2 y 
 sin2 x  sin2 y
2
2
2
sin x(cos y  sin y)  sin2 y(sin2 x  cos2 x)

 sin2 x  sin2 y
(sin2 x)(1)  (sin2 y)(1) 
 sin2 x  sin2 y
sin2 x  sin2 y 
 sin2 x  sin2 y

tan (a  b) 


sin a cos b  cos a sin b

cos a cos b  sin a sin b
sin a cos b
cos a sin b
  
cos a cos b
cos a cos b

tan (a  b) 
 ———
cos a cos b
sin a sin b
  
cos a cos b
cos a cos b
tan a  tan b


tan (a  b) 
 
1  tan a tan b

Replace b with b to find tan(a  b).
tan (a  (b)) 

tan (a  b) 


tan a  tan (b)

1  tan a tan (b)
tan a  tan b

1  tan a tan b

48a. Answers will vary.

217

Chapter 7

48b. tan A  tan B  tan C 
 tan A tan B tan C
tan A  tan B  tan (180°  (A  B))

 tan A tan B tan(180°  (A  B))
tan A  tan B 


tan A  tan B 



56.

1

s 
 rv

A 
 2r2 v

18 
 r(2.9)

tan 180°  tan (A  B)

1  tan 180° tan (A  B)
tan 180°  tan (A  B)
tan A tan B 1
 tan 180° tan (A  B)
0  tan (A  B)

1  0  tan (A  B)
0  tan (A  B)
tan A tan B 1
 0  tan (A  B)

A

6.2 r; 6.2 ft
A
57. c2 
 702  1302  2(70)(130) cos 130°
c2 33498.7345
c 183 miles
58. 120°  90°, consider Case 2.
4  12, 0 solutions

tan A  tan B  tan (A  B)

 tan A tan B tan (A  B)

59.


1  tan A tan B


(tan A  tan B)  
1  tan A tan B  tan (A  B)

 tan A tan B (A  B)
tan (A  B)(1  tan A tan B)  tan (A  B)

 tan A tan B (A  B)
(1  tan A tan B  1) tan (A  B)

 tan A tan B (A  B)
tan A tan B tan (A  B) 
 tan A tan B (A  B)

35 ft

1  cos2 x

37˚12′

2
2

49. sec2 x 
 
1  sin2 x  csc x  cot x






x

6˚40′

1  cos2 x

2
2

sec2 x 
 
cos2 x  1  cot x  cot x

v 
 37° 12
  6° 40
 or 30° 32

a 
 90°  6° 40
 or 96° 40

b 
 180°  (30° 32
  96° 40
) or 52° 48


cos2 x

1



sec2 x 
 
cos2 x  cos2 x  1

sec2 x 
 sec2 x  1  1
sec2 x 
 sec 2 x

35

sin 30° 32

sin v

50. sin2 v  cos2 v 
 1

51. Arctan 3




x

sin 30° 32


1
8
—


3
7
8
1



37

7




21

63
cos2 v 
 64
3
7
cos v 
 8
3
7
Quadrant III, so  8

60.


3

1

61. Case 1
x  1  4
(x  1)  4
x  1  4
x  5
x 5



3


 2
52. k, where k is an integer
86  50

2

4

53. A 
 2
A 
 18 
 68



h 
 2


t  c
2

  1 
2

  c
2

y 
 18 sin 


50 
 18 sin 


18 
 18 sin 



2

1 
 sin 
  c

sin1

86  50


 2



  68
c  68


 68



c


c


y 
 18sin 
2t    68
360

Case 2
x  1  4
x14
x3
{xx

8

30°

54. 8 
 8; 1 
 360; 1 
 30°

The correct choice is A.

55. sin (540°) 
 sin (360°  180°)

0

Chapter 7

x 
 1

5 or x  3}

1 2
62.

 1(6)  3(2)
3 6

 6  6 or 12
63. fg(4) 
 f(g(4))

 f(5(4)  1)

 f(21)

 3(21)2  4

 1319
gf(4) 
 g(f(4))

 g(3(4)2  4)

 g(44)

 5(44)  1

 221
(8)62
8 62
64. (8)62 862 
 6

 
 
 (1)62 
 1
2

( 1) 
 2  c
3

 
  
2
2

x 54.87 ft
4x3  3x2  x 
 0
x(4x2  3x  1) 
 0
x(4x  1)(x  1) 
 0
x 
 0 or 4x  1 
 0 or x  1 
 0
x 
 4



3

sin (Arctan 3
) 
 sin

x




sin 52° 48

35 sin 52° 48



tan v 
 
cos v


182  cos2 v 
 1

1
(6.2)2(2.9)
2
55.7 ft2

218

8

Page 445
1. csc v 


1




2

7
7

2

1

v1


v


cot2

v


cot v 


sec2

cos v 


v

4 2

1

25

9
5

3

csc2




sec2

v

sec2

v

v




7 2

2
49

4
45

4
5
3

2






2



1
2

3


5

tan y 
 
sec2 y
1

  
3
4

 
5
1  
4

3
5










5.

6.

v
v

sec2 v
2
2
sec v
csc v


sec2 v  sec2 v
1

sin2 v
—
1
1

cos2 v
cos2 v

sin2 v  1
cot2 v  1


80  413

7-3B Reduction Identities


 csc2 v

Page 447


 csc2 v

1. sin, cos, sin
2. cot, tan, cot
3. tan, cot, tan
4. csc, sec, csc
5. sec, csc, sec
6a. (1) cos, sin, cos
(2) sin, cos, sin
(3) cot, tan, cot
(4) tan, cot, tan
(5) csc, sec, csc
(6) sec, csc, sec
6b. Sample answer: If a row for sin a were placed
above Exercises 1-5, the entries for Exercise 6a
could be obtained by interchanging the first and
third columns and leaving the middle column
alone.
7a. (1) cos, sin, cos
(2) sin, cos, sin
(3) cot, tan, cot
(4) tan, cot, tan
(5) csc, sec, csc
(6) sec, csc, sec
7b. Sample answer: The entries in the rows for cos a
and sec a are unchanged. All other entries are
multiplied by 1.
8a. Sample answer: They can be used to reduce
trigonometric functions of large positive or
negative angles to those of angles in the first
quadrant.
8b. Sample answer: sum or difference identities


 csc2 v

 csc2 v

1

sin x

1





cos x  sin x 
 2  cos x  cos x  sin x

1
21
1
1

1  cot a tan b


7. tan (a  b) 
 
cot a  tan b
1


1
tan a  tan b
——
tan (a  b) 

1
  tan b
tan a
1


1
tan a  tan b
tan a

tan (a  b) 
 ——

1
tan a
  tan b
tan a
tan a  tan b


tan (a  b) 
 
1  tan a tan b

tan (a  b) 
 tan(a  b)
8. cos 75° 
 cos(30°  45°)

 cos 30° cos 45°  sin 30° sin 45°

3


2


80  413


1


1
1
1


 csc2 v
csc2 v 
 csc2 v
cot x sec x sin x 
 2  tan x cos x csc x

cos x

sin x


5  43

4

4  5
3

4
5  43


4  53


 or 


59
59

1



1  cot2 x 
 1

sec2


 
22 1

 3


tan x  tan y


 cos 4

csc2

3


tan (x  y) 
 
1  tan x tan y

3
5



1
1


sec2 x  csc2 x
cos2 x  sin2 x

2

10. tan x 
 4

1

sec v


 5  4

1

1  tan2 x

7
7




16 or 4

35  6


1

 
5
3



3

4




12

cos 4 
 cos 
5  4
4.


7

2



1


 
3
4  
3
4


 sec v

19




cos (x  y) 
 cos x cos y  sin x sin y

5

19

4




59 or 35

Quadrant II, so 3
3.

cos y 
 
1  si
n2 y

5


3  1 
 sec2 v
16

9

v


cot2

5
3

2

Quadrant 1, so
2.

cot2

1  cot2 v 
 


1

 —

tan2

9. cos x 
 
1  si
n2 x

Mid-Chapter Quiz
1

sin v


2


 
2
2  
2
2
1

  2

6




4

219

Chapter 7

Double-Angle and Half-Angle
Identities

7-4
Page 453

2 2

5


 

Check for Understanding

cos2 v 
 25
21


cos v 
 
5

4
21




25

cos 2v 
 cos2 v  sin2 v
 2
21

1  cos


tan 2v 
 
1  tan2 v

21
2


  ,


2

21 

 ——
21 2
2

1  

21 

1  cos 2 2

2

3c. I, II, III or IV





4 2

3


 

0 
 2(1)
0 2

Sample answer: v 
 2
5. Both answers are correct. She obtained two
different representations of the same number. One
way to verify this is to evaluate each expression
with a calculator. To verify it algebraically, square
each answer and then simplify. The same result is
obtained in each case. Since each of the original
answers is positive, and they have the same
square, the original answers are the same
number.

cos v

21
4


or 
17

sin2 v  cos2 v 
 1
3 2

sin2 v  
5 
 1

 1 
 sec2 v

25
 

9
5
3 

1



sec v

16

sin2 v 
 2
5

sec2 v
sec v (Quadrant III)


3

sin 2v 
 2 sin v cos v

 2
5
5
4

3

24


 25
cos 2v 
 cos2 v  sin2 v
3 2

4 2


 
5  
5


 sin 
2

7


 2
5




4

1  cos

2

2 tan v


tan 2v 
 
1  tan2 v

(Quadrant I)

2
3

 —
4 2
1  
3
4



2
1  2

2



2  2




2




330°

7. tan 165° 
 tan 2
1  cos 330°



1  co
s 330°

(Quadrant II)

8

3
—
7
9

24

or 7
2


10. tan 2v 
 
cot v  tan v


3


1  3
—

3
1  3

2

tan v

 
tan 2v 
 
cot v  tan v  tan v

tan 2v 



 (2  3
)

 3
2

tan 2v 


2 tan v

cot v tan v  tan2 v
2 tan v

1  tan2 v

tan 2v 
 tan 2v

220

4

sin v 
 5

(Quadrant III)

1
3

 
5 or 5



4




4
21


21
—
17

21

9. tan2 v  1 
 sec2 v



Chapter 7

2 2

2 tan v


2
.





21

2

sin  
 2 sin 2




(Quadrant I)


 2
5

5 

sin 2
2 
 2 sin 2




2
21


or 
21

17

a



2

5
—
21



5
2

21


sin 2v 
 2 sin v cos v


 25

3a. III or IV
3b. I or II
4.
sin 2v 
 2 sin v

6. sin





 sin2 v

a



8






 

5   
5

Letting v 
 2 yields sin 2 

a

 cos2 v 
 1


 sin2 v

 
 sin v
or sin 2 



tan v 
 
cos v

21

1. If you are only given the value of cos v, then
cos 2v 
 2 cos2 v  1 is the best identity to use.
If you are only given the value of sin v, then
cos 2v 
 1  2 sin2 v is the best identity to use. If
you are given the values of both cos v and sin v,
then cos 2v 
 cos2 v  sin2 v is just as good as the
other two.
2.
cos 2 
 1  2 sin2 v
cos 2v  1 
 2 sin2 v
cos 2v  1

2
1  cos 2v

2
1 cos 2v

2

sin v

8. sin2 v  cos2 v 
 1

sec A  sin A

1


11. 1  2 sin 2A 
 
sec A

1

1

2

 sin A
sin 2A 
 ——

1

1

12.

1

cos A

 sin A
——
1

cos A

cos A



cos A




 2 sin A cos A
sin 2A






2  3

2
—

2  3

2

(2  3
)(2  3
)

(2  3
)(2  3
)

(2  3
 )2

43


 2  3

3

17. sin 8 
 sin

cos 2v
v1
cos 2v  1 
 2 cos2 v

3

4

2



 
—

3




1

cos 2v  2 
 cos2 v

1  cos 4
——
2

(Quadrant I)


2

P 
 I02 R sin2 qt
P 
 I02 R (1  cos2 qt)




P 
 I02 R
1 
2 cos 2qt  2
1

1




P 
 I02 R
2  2 cos 2qt
1





 

sin x

 2

 2 cos2

1

(Quadrant I)

1  
2


3
1  
2




sin x

2

1

2


1  cos 6
——
5
1  cos 6

3


 

13.


 tan 
2



sin 2A 
 1  sin A cos A

1
1  sin 2A 
 1  2
1
1  sin 2A 
 1  2
x
x
sin x
sin 2 cos 2 
 2
x
x
2 sin 2 cos 2
sin x
—— 
 

2
2
x
sin 2 2
sin x
— 


2
2

5

6

5

1

cos A

1  2 sin 2A 

1

2
1

2
1

2

16. tan

1

cos A

5

12

1

1

P 
 2 I02 R  2 I02 R cos 2qt

7

18. cos 12 


1  2
2


  2

2

2
7

6

cos 2

7π

Pages 454–455






Exercises
30°

14. cos 15° 
 cos 2



1  cos 30°



2



15. sin 75° 






45°

19. tan 22.5° 
 tan 2







1  cos 45°


1  cos
45°






(Quadrant I)


2

(Quadrant I)



3




1

2


  2





3

2



2  3


1  2
—
2

2  3


2
150°
sin 2
1  cos 150°

2

(Quadrant II)


 





(Quadrant I)


3




1  cos 6
——
2

1  
2

2


2  
3


2




1  2
—

2
1  2



2  2

2
—
2  
2

2



 

 



(2  2
)(2  2
)

(2  2
)(2  2
)
 )2
(2  2

42





2  2

2

2
22

2


 2
1

221

Chapter 7

v

20. tan 2 









23. tan2 v  1 
 sec2 v
(2)2  1 
 sec2 v
5 
 sec2 v
 
 sec v
5
cos v 
 sec v

1  cos v



1  cos
v


1
1  4

1
1  
4
3

4
—
5

4






v




 2
5

20

sin2 v 
 25

sin v 


2
5

sin v


tan v 
 
cos v


1

v


sin2

9

25
3

5

(Quadrant I)




3

5
—
4

5




3

4

sin v 
 5
2
5

4


 5
cos 2v 
 cos2 v  sin2 v
 2
5

4

3


 5

24

2 tan v


tan 2v 
 
1  tan2 v

cos 2v 
 cos2 v  sin 2 v


 

22.

sin2

2(2)




1  (2)2


 

4 2
3 2

 5  5
7

 2
5
2 tan v

2v 
 
1  tan2 v



v


 

1
 2
3



4

1


24. cos v 
 
sec v





 

 

cos2

v
1

cos2

v
1



sin2

7

tan v 




2
2

cos v 
 3




sin v

cos v
1

3
—
2
2

3
1

22


sin2 v 
 1
6

7

sin v 
 4
tan v 





2

or 4

(Quadrant I)




sin 2v 
 2 sin v cos v



cos 2v 





tan 2v 





7


 2
4
4
7
3

cos 2v 
 cos2 v  sin2 v

1 2

3 2

1

2



 1
6 or 8

2 tan v

1  tan2 v

2 tan v


tan 2v 
 
1  tan2 v


 


 

2

2




37 


 

7
 2
1  

3 


 

 2
1 
4

Chapter 7

 2
7


 
4  
4

7

9


2

2 
4




3


 8

v  sin2 v

2
2 2


2

2
—
14

16

(Quadrant II)

sin v

cos v

7

4
—
3
4

7
3

sin 2v 
 2 sin v cos v


 
3  
3



v  cos2 v 
 1
3 2

8

4
2

9
cos2

1

4
3
3
4

sin2 v  
4 
 1

cos2 v 
 9

1 2
2
 
3
3

4





3 or 3

3
2 4

3 2
1  4
3

2
24
— or 
7
7

16


 2


2
5 2


 
5  
5


 25





5


 2
5
5


 2
5
5

tan

(Quadrant II)

sin 2v 
 2 sin v cos v

sin 2v 
 2 sin v cos v
3

or 5

sin2 v  
5 
 1

21. sin2 v  cos2 v 
 1
sin2


5

1

5


sin2 v  cos2 v 
 1

15


35 or 
5
4
 2
5

(Quadrant II)

4
2

or 7




222

2
7

3
—
2

9

or 37


1

25. 1  cot2 v 
 csc2 v
1  
2 
 csc2 v
3 2

sin

13
 

4
13


2 

1

v

csc v






2 tan v


tan v 
 
cot v






csc2 v
csc v


tan 2v 
 
1  tan2 v

1

3

2
2

3

2

1

13


 
2
2

13






2
21 2

1  

21 




4
21


21
——
17

21

(Quadrant III)
sin2 v  cos2 v 
 1


213 2
2



13   cos v 
 1

cos2

2
13


or 
13

v


cos v 


sin2

 2
2
3

a


3
13


7

sin a 
 3

12

tan 2a 


cos 2v 
 cos2 v  sin2 v

tan















 

12

5


252  cos2 v 
 1

2

sin v






30.

2
21


or 
21

1

sec v csc v 
 2 sec v csc v

(cos A  sin A)(cos A  sin A)

cos A  sin A

cos A  sin A 
 cos A  sin A
(sin v  cos v)2  1 
 sin 2v
2
sin v  2 sin v cos v  cos2 v  1 
 sin 2v
2 sin v cos v  1  1 
 sin 2v
2 sin v cos v 
 sin 2v
sin 2v 
 sin 2v


31. cos x  1 
 
2(cos x  1)


21


 2
5

5 

2 cos2 x  1  1


cos x  1 
 
2(cos x  1)

4
21

2 cos2 x  2



 
25


cos x  1 
 
2(cos x  1)

2(cos2 x  1)

cos 2v 
 cos2 v  sin2 v



1

csc v sec v 
 2 sec v csc v

cos 2x  1

sin 2v 
 2 sin v cos v



1


 2 sec v csc v

cos A  sin A 


tan v 
 
cos v





1


 2 sec v csc v


cos A  sin A 
 
cos A  sin A

(Quadrant IV)


21

 2
5
17

25

1


 2 sec v csc v

cos2 A  sin2 A

21


2

1

sin 2v
1

2 sin v cos v
1
1
  
sin v cos v

cos 2A

cos v 
 
5

2
5
—
21



5
2

21


2
14

or 
5


29. cos A  sin A 
 
cos A  sin A

21

cos2 v 
 25


 5



1

2
1

2

sin2 v  cos2 v 
 1


14

5

 
2

1

1

2

1



csc 2v 
 2 sec v csc v

28.


26. sin v 
 
csc v
1

 —
5
2

(Quadrant II)


 
14

2
1  

2

2 2
1  3

or


14

7





 
2
 or  2

14



2


4

3

5

9




7


3


2

3

2 tan a

1  tan2 a
2

213
 2
13


 





1
7


 1
3






sin a


tan a 
 
cos a

sin2 a 
 9





 2

13 
 13 

313
 2

 13
5

 1
3
2 tan v

2v 
 
1  tan2 v
2
2 3

4
21


or 
17

27. sin2 a  cos2 a 
 1

117

16 9
3
13


13

(Quadrant III)
sin 2v 
 2 sin v cos v
2
13



221



21 





cos x  1 
 
2(cos x  1)



2 2
5

cos x  1 


2(cos x  1)(cos x  1)

2(cos x  1)

cos x  1 
 cos x  1
cos2 v  sin2 v


32. sec 2v 
 
cos2 v  sin2 v
1


sec 2v 
 
cos 2v

sec 2v 
 sec 2v
A

sin A


33. tan 2 
 
1  cos A

223

Chapter 7

1


 
A
1  cos 2
2
A
sin 2 2

A

L
tan 45°  tan 2

39a. tan
45°  2 
 
L
1  tan 45° tan 2
L
1  tan 2

tan 2 
 


 
L
1  1  tan 2

A
A
2 sin  cos 

A

2
2
tan 2 
 
A
A

tan 2 


tan

A

2




A

tan 2 


1  2 cos2 2  1
A
A
2 sin 2 cos 2

A
2 cos2 2
A
sin 2

A
cos 2
A
tan 2




39b.

1  cos L

1
1  cos L

1  cos L

1 
1  cos L










sin 3x 
 3 sin x  4 sin3 x
sin(2x  x) 
 3 sin x  4 sin3 x
sin 2x cos x  cos 2x sin x 
 3 sin x  4 sin3 x
2 sin x cos2 x  (1  2 sin2 x) sin x 
 3 sin x  4 sin3 x

34.

1  cos L

1
1  cos L

1  cos L

1 
1  cos L




1  cos 60˚
1  
1  cos 60˚

1  cos 60˚

1 
1  cos 60˚
1
1  2

1
1
1  
2




1

2 sin x(1  sin2 x)  (1  2 sin2 x) sin x 
 3 sin x  4 sin3 x

2 sin x  2 sin3 x  sin x  2 sin3 x 
 3 sin x  4 sin3 x
3 sin x  4 sin3 x 
 3 sin x  4 sin3 x
35.
cos 3x 
 4 cos3 x  3 cos x
cos(2x  x) 
 4 cos3 x  3 cos x
2
(2 cos x  1)cos x  2 sin2 x cos x 
 4 cos3 x  3 cos x




2 cos3 x  cos x  2 cos x  2 cos3 x 
 4 cos3 x  3 cos x
4 cos3 x  3 cos x 
 4 cos3 x  3 cos x
v2
 sin2 2v
2g



40.

sin2 2v


36. 


sin2 v
v2

R

R


21

4 sin2 v cos2 v


3

3

tan a 

trigonometry,

1
tan2v





3


3

(1  3
) tan a 
 3  3

PA

BA

cos v sin(v  a)

g cos2 a
2v2 cos v sin(v  45°)

g cos2 45°
2v2 cos v(sin v cos 45°  cos v sin 45°)

g cos2 45°


2
2


2v2 cos v (sin v) 
2  (cos v) 2

tan a 


3

3  
3

1  3


tan a 


9  3


3 33



 


6  53

tan a 
 
3



 






2


3





1


2


 2  2  2  2
  6

2




4


 



sec 12 



2
 2
2
2 v cos v(sin v  cos v)

R 
 
1
g  2




2
v2 (2 cos v sin v  2 cos2 v)

g

1



cos 
12
1

2  
6



4
4
2  4
6

 or 6


  2

4

v2 
2

R 
 g(2 cos v sin v  (2 cos2 v  1) 1)

Chapter 7




 cos 3 cos 4  sin 3 sin 4


2
 2
g 
2

v2 
2
(sin
g



41. cos 12 
 cos 
3  4

R 
 

R



 3  3
 tan a

2v2




R



3

 tan a 
 3  3
tan a  3


 4 cos2 v
37. ∠PBD is an inscribed angle that subtends the
same arc as the central angle ∠POD, so m∠PBD

38. R 


or 2  3


tan (a  30°) 
 7

(2 sin v cos v)2




sin2 v






3  3


3

3  3


3

12  6 3

6

tan a  tan 30°

1  tan a tan 30°




sin2 v






tan (a  30°) 
 3

 sin2 v
2g

1
v. By right triangle
2
PA
sin v
  
 .
1  OA
1  cos v

1
1  
2

1
1  2

1
1  
3

1
1  3




(2 cos2 x  1)cos x  2(1  cos2 x)cos x 
 4 cos3 x  3 cos x




42. Sample answer:
sin(
)2  cos(
)2 
 sin   cos 

 0  (1)

 1
1

2v  cos 2v  1)

224

17

10

43. s 
 rv
17 
 10  v
17

10

17

180°



 1
0  

7-5

Solving Trigonometric Equations

Page 458

Graphing Calculator Exploration

97.4°


v

44. Let x 
 the distance from A to the point beneath
the mountain peak.
h

tan 21°10
 
 
570  x
h 
 (570  x) tan 21°10

h
tan 36°40
 
 x
h 
 x tan 36°40

(570  x) tan 21°10
 
 x tan 36°40

570 tan 21°10
 
 x tan 36°40
  x tan 21°10

570 tan 21°10
 
 x(tan 36°40
  tan 21°10
)
570 tan 21°10


tan 36°40
  tan 21°10



x

617.7646751

x

1.

2.

h

tan 36°40
 
 x
tan 36°40


h

617.8

h 460 ft
45. (x  (3))(x  0.5)(x  6)(x  2) 
 0
(x  3)(x  0.5)(x  6)(x  2) 
 0
(x2  2.5x  1.5)(x2  8x  12) 
 0
x4  5.5x3  9.5x2  42x  18 
 0
2x4  11x3  19x2  84x  36 
 0
46.

y 
 2x  5
x 
 2y  5
x  5 
 2y
x5

2

3. Exercise 1: (1.1071, 0.8944), (4.2487, 0.8944)
Exercise 2: (5.2872, 0.5437), (0.9960, 0.5437)
4. The x-coordinates are the solutions of the
equations. Substitute the x-coordinates and see
that the two sides of the equation are equal.

y

5.

y 
 2x  5


y

O

x
y


x5
2



47. x  2y 
 11
x 
 11  2y

x  2y 
 11
x  2(2) 
 11
x 
7
48. ab 
 3
3
b 
 a
a2

[0, 2] sc14 by [3, 3] sc11
5a. The x-intercepts of the graph are the solutions of
the equation sin x 
 2 cos x. They are the same.
5b. y 
 tan 0.5x  cos x or y 
 cos x  tan 0.5x

3x  5y 
 11
3(11  2y)  5y 
 11
33  6y  5y 
 11
11y 
 22
y
2

Page 459

(7, 2)

(a  b)2 
 64
 2ab  b2 
 64

a2  2a
a  
a 
 64
3

Check for Understanding

1. A trigonometric identity is an equation that is
true for all values of the variable for which each
side of the equation is defined. A trigonometric
equation that is not an identity is only true for
certain values of the variable.
2. All trigonometric functions are periodic. Adding
the least common multiple of the periods of the
functions that appear to any solution to the
equation will always produce another solution.
3. 45°  360x° and 135°  360x°, where x is any
integer

3 2

a2  6  
a 
 64
3 2

a2  
a 
 70
3 2

a2  b2 
 70
The correct answer is 70.

225

Chapter 7

12. tan2 x  2 tan x  1 
 0
(tan x  1)(tan x  1) 
 0
tan x  1 
 0
tan x 
 1

4. Each type of equation may require adding,
subtracting, multiplying, or dividing each side by
the same number. Quadratic and trigonometric
equations can often be solved by factoring. Linear
and quadratic equations do not require identities.
All linear and quadratic equations can be solved
algebraically, whereas some trigonometric
equations require a graphing calculator. A linear
equation has at most one solution. A quadratic
equation has at most two solutions. A
trigonometric equation usually has infinitely
many solutions unless the values of the variable
are restricted.
5. 2 sin x  1 
 0
6. 2 cos x  3

0
2 sin x 
 1


2 cos x 
 3

12

cos x 
 2

sin x 


sin x 





cos x

sin x

cos x 


cos2 x  3 cos x 
 2
cos2 x  3 cos x  2 
 0
(cos x  1)(cos x  2) 
 0
cos x  1 
 0
or cos x  2 
 0
cos x 
 1
cos x 
 2
x 
 (2k  1)
no solutions
14.
sin 2x  cos x 
 0
2 sin x cos x  cos x 
 0
cos x (2 sin x  1) 
 0
cos x 
 0
or 2 sin x  1 
 0
13.


3

x 
 30°
7. sin x cot x 


3

x 
 4  k



x 
 30°


3

2

3

2

3

2



5

or x 
 6  2k

11

x 
 6 or x 
 6
11.

x


 k

0
1

cos v

2

1

2

1

2

3

4

4

v 3
16.
W 
 Fd cos v
1500 
 100  20 cos v
0.75 
 cos v
v 41.41°

Pages 459–461

Exercises

17. 2
 sin x  1 
 0
2
 sin x 
 1

18. 2 cos x  1 
 0
2 cos x 
 1

1


2

2

1

cos x 
 2

sin x 
 
2
sin x 

19.

no solutions

cos x 


15. 2 cos v  1
2 cos v

cos v 
 2 at 3 and 3

sin2 2x  cos2 x 
 0
1  cos2 2x  cos2 x 
 0
1  (2 cos2 x  1)2  cos2 x 
 0
1  (4 cos4 x  4 cos2 x  1)  cos2 x 
 0
4 cos4 x  5 cos2 x 
 0
cos2 x(4 cos2 x  5) 
 0
cos2 x 
 0
or
4 cos2 x  5 
 0
5
cos x 
 0
cos2 x 
 4


2

sin x 
 2
x 
 6  2k

x 
 30° or x 
 330°
8.
cos 2x 
 sin2 x  2
2 cos2 x  1 
 (1  cos2 x)  2
2 cos2 x  1 
 cos2 x  1
3 cos2 x 
 0
cos2 x 
 0
cos x 
 0
x 
 90° or x 
 270°
9. 3 tan2 x  1 
 0
3 tan2 x 
 1
1
tan2 x 
 3

3
tan x 
 3

5
7
11
x 
 6 or x 
 6 or x 
 6 or x 
 6
10.
2 sin2 x 
 5 sin x  3
2
2 sin x  5 sin x  3 
 0
(2 sin x  1)(sin x  3) 
 0
2 sin x  1 
 0
or
sin x  3 
 0
sin x 
 12
sin x 
 3
7

1

x 
 2  k

x 
 120°

x 
 45°
sin 2x  1 
 0
2 sin x cos x  1 
 0
sin2 x cos2 x 
 14
1

sin2 x (1  sin2 x) 
 4
1

sin2 x  sin4 x  4 
 0
1

sin4 x  sin2 x  4 
 0


sin2 x  12
sin2 x  12 
 0
1

sin2 x  2 
 0


5

2

1

sin2 x 
 2

no solutions

1



2

sin x 
 
or 2
2
x 
 45°

Chapter 7

226

tan 2x  3

0
tan 2x 
 3


20.

2 tan x

1  tan2 x

28.


 3



2

1
3

3


3

tan x 

21.


2


2

 cos x  2  sin x  2 
 2

2
 cos x 
 2

cos x 
 1
x 
 0°
29. 2 sin v cos v  3
 sin v 
 0
sin v (2 cos v  3
) 
 0
sin v 
 0
or 2 cos v + 3

0
v 
 0° or v 
 180°
2 cos v 
 3


tan x 
 3

x 
 60°

x 
 30°
cos2 x 
 cos x
cos2 x  cos x 
 0
cos x(cos x  1) 
 0
cos x 
 0
or
x 
 90°


3

cos v 
 2
v 
 150°
or v 
 210°

cos x  1 
 0
cos x 
 1
x 
 0°
2
22.
sin x 
 1  cos x
sin x 
 1  1  sin2 x
sin2 x  sin x  2 
 0
(sin x  1)(sin x  2) 
 0
sin x  1 
 0
or
sin x  2 
 0
sin x 
 1
sin x 
 2
x 
 90°
no solution
23. 2
 cos x  1 
 0
2
 cos x 
 1

30. (2 sin x  1)(2 cos2 x  1) 
 0
2 sin x  1 
 0
or 2 cos2 x  1 
 0
2 sin x 
 1
2 cos2 x 
 1
1

1

cos2 x 
 2

sin x 
 2


x 
 6

cos x 


5


2

2



or x 
 6

3

x 
 4 or x 
 4
5

7

or x 
 4 or x 
 4
31.

2


cos x 
 2
x 
 135° or x 
 225°
1
24. cos x tan x 
 2
sin x


2

cos x  2  sin x  2

2 tan x 
 3
 (1  tan2 x)
  3
 tan2 x
2 tan x 
 3
2
3
 tan x  2 tan x  3

0
(3
 tan x  1)(tan x  3
) 
 0
3
 tan x  1 
 0
tan x  3

0
tan x 
 

cos (x  45°)  cos (x  45°) 
 2

cos x cos 45°  sin x sin 45°
 cos x cos 45°  sin x sin 45° 
 2


4 sin2 x  1 
 4 sin x
4 sin2 x  4 sin x  1 
 0
(2 sin x  1)(2 sin x  1) 
 0
2 sin x  1 
 0
2 sin x 
 1
1

sin x 
 2
7

11

x 
 6 or x 
 6

1



cos x 
cos x 
 2

32. 2
 tan x 
 2 sin x

1

sin x 
 2

sin x



2
cos x 
 2 sin x

x 
 30° or x 
 150°
25. sin x tan x  sin x 
 0
sin x (tan x  1) 
 0
sin x 
 0
or
tan x  1 
 0
x 
 0° or x 
 180°
tan x 
 1
x 
 45° or x 
 225°
2
26. 2 cos x  3 cos x  2 
 0
(2 cos x  1)(cos x  2) 
 0
2 cos x  1 
 0
or
cos x  2 
 0
2 cos x 
 1
cos x 
 2
1
cos x 
 2
no solution

 
 2 cos x
2

2

2


 cos x


7

x 
 4 or x 
 4
2
 tan x 
 2 sin x would also be true if both tan x
sin x

and sin x equal 0. Since tan x 
 
cos x , tan x equals
0 when sin x 
 0. Therefore x can also equal 0 and
.


7

0, 4, , 4
33.

x 
 60° or x 
 300°
27.
sin 2x 
 sin x
2 sin x cos x 
 –sin x
2 sin x cos x  sin x 
 0
sin x (2 cos x  1) 
 0
sin x 
 0
or 2 cos x  1 
 0
x 
 0° or x 
 180°
2 cos x 
 1

sin x 
 cos 2x  1
sin x 
 1  2 sin2 x  1
2
2 sin x  sin x 
 0
sin x(2 sin x  1) 
 0
sin x 
 0
or
2 sin x  1 
 0
x 
 0 or x 
 
2 sin x 
 1
1

sin x 
 2
7

x 
 6 or

1

cos x 
 2

11

x 
 6

x 
 120°
or x 
 240°

227

Chapter 7

34.

cot2 x  csc x 
 1
csc2 x  1  csc x 
 1
csc2 x  csc x  2 
 0
(csc x  2)(csc x  1) 
 0
csc x  2 
 0
or
csc x  1 
 0
csc x 
 2
or
csc x 
 1
1

sin x 
 2
x

35.

or x 


5

6

3

x 
 2

sin x  cos x 
 0
sin x 
 cos x
sin2 x 
 cos2 x
sin2 x  cos2 x 
 0
sin2 x  1  sin2 x 
 0
2 sin2 x  1 
 0
1

sin2 x 
 2
sin x 


1

1


2

or

36.

v

37. sin x 

x

38.

or v 


1

sin x cos x 
 2
1

sin2 x cos2 x 
 4
1

sin2 x(1  sin2 x) 
 4
1

sin2 x  sin4 x) 
 4
1

sin4 x  sin2 x  4 
 0


sin2 x  12
sin2 x  12 
 0

11

6

1

sin2 x  2 
 0
1

2 k

or

sin2 x 
 2

11

x 
 6  2k

sin x 



sin x  2(1  sin2 x) 
 1
2 sin2 x  sin x  1 
 0
(2 sin x  1)(sin x  1) 
 0
2 sin x  1 
 0
or
sin x  1 
 0
2 sin x 
 1
sin x 
 1
x


1

2


6

x

 2k or x 


5

6

3

2

44.


3


3

2 cos2 x  1 
 2
2  3


2 cos2 x 
 2

 2k

2  3


cos2 x 
 4

 2k


2  
3

2

cos x 


 tan x
3 tan2 x 
 3
2
3 tan x  3
 tan x 
 0
tan x(3 tan x  3
) 
 0
tan x 
 0
or
3 tan x  3

0
x 
 k
3 tan x 
 3




11

x 
 12  k or x 
 12  k
sin4 x  1 
 0
(sin2 x  1)(sin2 x  1) 
 0
sin2 x  1 
 0
or
sin2 x  1 
 0
2
sin x 
 1
sin2 x 
 1
sin x 
 1
no solutions

x 
 2  k
46. sec2 x  2 sec x 
 0
sec x(sec x  2) 
 0
sec x 
 0
or
sec x  2 
 0
45.


3


x 
 6  k
2(1  sin2 x) 
 3 sin x
2  2 sin2 x 
 3 sin x
2
2 sin x  3 sin x  2 
 0
(2 sin x  1)(sin x  2) 
 0
2 sin x  1 
 0
or
sin x  2 
 0
2 sin x 
 1
sin x 
 2
1
sin x 
 2
no solution

1

cos x 
 0

sec x 
 2

no solution

cos x 
 2

1

2

x 
 3  2k or



4

x 
 6  2k or

x 
 3  2k

5

x 
 6  2k
Chapter 7


3

cos2 x  sin2 x 
 2
cos2 x  (1  cos2 x) 
 2

tan x 
 3
40.


2

2

x 
 4  k

sin x

2

cos x 
cos x  2 cos x 
 1

39.

5

x 
 3  2k

43.

cos x tan x  2 cos2 x 
 1

sin x 




x 
 3  2k or

no solution

1  3 sin v 
 cos 2v
1  3 sin v 
 1  2 sin2 v
2
2 sin v  3 sin v  2 
 0
(2 sin v  1)(sin v  2) 
 0
2 sin v  1 
 0
or sin v  2 
 0
2 sin v 
 1
sin v 
 2
1
sin v 
 2
no solution
1
2
7
 
6

cos x 
 2

cos x 
 2

3

4

7
.
4

7

6

1

sec x 
 2


2

2

sin x and cos x must be opposites, so x 

or x 



 cos x  sin x

(cos x  sin x)(cos x  sin x) 
 1
cos2 x  sin2 x 
 1
cos2 x  (1  cos2 x) 
 1
2 cos2 x  1 
 1
2 cos2 x 
 2
cos2 x 
 1
cos x 
 1
x 
 k
42.
2 tan2 x  3 sec x 
 0
2(sec2 x  1)  3 sec x 
 0
(2 sec x  1)(sec x  2) 
 0
2 sec2 x  3 sec x  2 
 0
2 sec x  1 
 0
or
sec x  2 
 0
2 sec x 
 1
sec x 
 2

sin x 
 1



6

1

cos x  sin x

41.

228

47.

48.

sin x  cos x 
 1
sin2 x  2 sin x cos x  cos2 x 
 1
sin2 x  2 sin x cos x  1  sin2 x 
 1
2 sin x cos x 
 0
sin x cos x 
 0
sin2 x cos2 x 
 0
sin2 x (1  sin2 x) 
 0
2
sin x 
 0
or
1  sin2 x 
 0
sin x 
 0
sin2 x 
 1
x 
 2k
sin x 
 1

x 
 2  2k

58a.

1.00 sin 35°


sin r 
 
2.42

sin r 0.2370150563
r 13.71°
58b. Measure the angles of incidence and refraction
to determine the index of refraction. If the index
is 2.42, the diamond is genuine.
59.
D 
 0.5 sin (6.5 x) sin (2500t)
0.01 
 0.5 sin (6.5(0.5)) sin (2500t)
0.02 
 sin 3.25 sin 2500t
0.1848511958 sin 2500t
0.1859549654 2500t
The first positive angle with sine equivalent to
sin (0.1859549654) is   0.1859549654 or
3.326477773.

2 sin x  csc x 
 3
2 sin2 x  1 
 3 sin x
2
2 sin x  3 sin x  1 
 0
(2 sin x  1)(sin x  1) 
 0
2 sin x  1 
 0
or
sin x  1 
 0
2 sin x 
 1
sin x 
 1



1

x 
 2  2k

sin x 
 2


x 
 6  2k or
x

49.

50.

5

6

 2k



4

3

4

0v

or

1

sin(bx  c) 
 2
360°

The period of the function sin(bx  c) is b, so
the given interval consists of
1

61.


 b periods.

 xy 
 34
3
cos v sin v
17




 sin
v
cos v 4 22

cos v  4 sin v
17



 33 sin
v  4 cos v 22

3 cos v  4 sin v 
 17

3 sin v  4 cos v 
 22

↓
9 cos v  12 sin v 
 317

16 cos v  12 sin v 
 82

25 cos v

 82
  317


2
53. 0, 1.8955

5.5  107


sin v 
 
0.003

  317

82


cos v 
 
25

sin v
v

0.0001833333333
0.01°
56.
sin 2x sin x
2 sin x cos x sin x
2 sin x cos x – sin x 0
sin x(2 cos x  1) 0
The product on the left side of the inequality is

5
equal to 0 when x is 0, 3, , or 3. For the product
to be negative, one factor must be positive and the

5
other negative. This occurs if 3 x  or 3 x
2.
57.

360°

360°

b

The equation sin (bx  c) 
 2 has two solutions
per period, so the total number of solutions is 2b.

3

v

0.0013 s
a

sin v 
 2 at 4 and 4
52. 0.4636, 3.6052
54. 0.3218, 3.4633

55. sin v 
 D

t

a sin(bx  c) 
 2


2

2



3.326477773

2500
a

0
1


2

t

60. a sin(bx  c)  d 
 d  2


3
cos v  2

5
7
3
cos v 
 2 at 6 and 6
5
7
  v  
6
6
1


cos v  2  0
1
cos v  2
1

5
cos v 
 2 at 3 and 3

5
0  v 3 or 3 v 2

 sin v  1
51. 2
2
 sin v
sin v

n1 sin i 
 n2 sin r
1.00 sin 35° 
 2.42 sin r

360  v

v
341.32°

18.68020037

v2

R 
 g sin 2v
152


20 
 
9.8 sin 2v

0.8711111111 sin 2v
2v 60.5880156
or
v 30.29°

2v
v

119.4119844
59.71°

229

Chapter 7

135°







0 3
2
2
4
2
1 2
1 
0
x2  2x  1 
 0
(x  1)(x  1) 
 0
x1
0
x 
 1
(x  2)(x  1(x  1)


cot v 
 
tan v

1 – co s 135°



1  co s 135°

135°

tan 2 


67. 2


1

62. cot 67.5° 
 cot 2

(Quadrant 1)













2
1  
2

2

1  
2


2  2


2

2 – 2


2






(2  
2)



4 
2

(2  2
)(2  2
)

(2  2
)(2  2
)
2

cot 67.5 


22


2

1

2  2



2




2


2  2





2
 (2  2
)

(2  2
)(2  2
)




22
2

42

[5, 5] sc11 by [2, 8] sc11
max: (1, 7), min: (1, 3)
69. 3x  4 
 16
6 
 2y
x
4
y
3
(4, 3)
70. x  y  z 
 1
xyz
1
2x  y  3z 
 5
x  y  z 
 11
3x
 4z 
 6
2x

 12
x
6
3x  4z 
 6
x  y  z 
 11
3(6)  4z 
 6
6  y  (3) 
 11
4z 
 12
y
2
z 
 3
(6, 2, 3)


 2
1
63.


2


 5

2


 5

71.


2

sin x 
 5


2

Sample answer: sin x 
 5
64. A 


2
,
3

2

y

x
7
5
3
1
1

g(x)
4
2
0
2
4

2
3

g (x)

g (x ) 
 |x  3|

O
2
3

y 
 cos
1

72. A 
 2bh

O

90˚

1

A 
 2(6)(1)

180˚ 270˚ 360˚

A
3
The correct choice is C.

 23

65.

x1
0
x 
 1

68.







tan x

sec x
sin x

cos x
—
1

cos x

1

45 miles

hour

5280 ft

12 inches

1 hour

    
 792 in/sec

mile 
ft
3600 sec
v

v 
 r t
v

792 
 7 1
792
 
 
7
792
 radians
7

2

18 rps

66. undefined

Chapter 7

230

x

Page 462

History of Mathematics

4.

1. x2 
 52  52  2(5)(5) cos 10°
x 0.87
x2 
 52  52  2(5)(5) cos 20°
x 1.74
x2 
 52  52  2(5)(5) cos 30°
x 2.59
x2 
 52  52  2(5)(5) cos 40°
x 3.42
x2 
 52  52  2(5)(5) cos 50°
x 4.23
x2 
 52  52  2(5)(5) cos 60°
x
5
x2 
 52  52  2(5)(5) cos 70°
x 5.74
x2 
 52  52 2(5)(5) cos 80°
x 6.43
x2 
 52  52  2(5)(5) cos 90°
x 7.07
Angle
Measure
10°
20°
30°
40°
50°
60°
70°
80°
90°

5.

Slope-Intercept Form: y 
 mx  b, displays
slope and y-intercept
Point-Slope Form: y  y1 
 m(x  x1),
displays slope and a point on the line
Standard Form: Ax  by  C 
 0, displays
no information
Normal Form: x cos f  y sin f  p 
 0,
displays length of the normal and the angle
the normal makes with the x-axis
See students’ work for sample problems.
x cos f  y sin f  p 
 0
x cos 30°  y sin 30°  10 
 0

3
x
2

6.

1

 2y  10 
 0

x  y  20 
 0
3
x cos f  y sin f  p 
 0
x cos 150°  y sin 150°  3

0

3

1


0
2x  2y  3

Length of
Chord (cm)
0.87
1.74
2.59
3.42
4.23
5.00
5.74
6.43
7.07

x  y  23

0
3
x cos f  y sin f  p 
 0

7.

7

7

x cos 4  y sin 4  52

0

2
x
2

8.





2



 2 y  52

0

x  2
y  102

0
2
x  y  10 
 0
4x  3y 
 10

A2  
B2 
 
42  32 or 5
4
x
5

4x  3y  10 
 0

3

10




5 y  5 
 0
4

3

5x  5y  2 
 0
3

4

sin f 
 5, cos f 
 5, p 
 2; Quadrant III

7-6
Page 467

tan f 


Normal Form of a Linear Equation
9.

Check for Understanding

1. Normal means perpendicular
2. Compute cos 30° and sin 30°. Use these as the
coefficients of x and y, respectively, in the normal

3

1

sin f

1
2


10
10






10 y  5 
 0
10
10
3






10 , cos f 
 10 ,
y  

0

10
10

tan f 

f

231

3

or 4

f
y 
 3x  2
3x  y  2 
 0

A2  
B2 
 
32  12 or 10

3

x
10

3
10

x
10

form. The normal form is 2x  2y  10 
 0.
3. The statement is true. The given line is tangent to
the circle centered at the origin with radius p.

3
5

4
5

10



10
—
3
10


10

10


p 
 
5 ; Quadrant I

1

or 3

18°

Chapter 7

10.

2
x  2
y 
 6
2
x  2
y  6 
 0

A2  
B2 
 
2
2 
(2

)2 or 2


6
2
 2y  2 
 0

2
 2y  3 
 0


2
2
sin f 
 2, cos f 
 2,
2

2
tan f 
 — or 1

2

2

17.


3

1

2x  2y  5 
 0


2
x
2

2
x
2

18.

x  y  10 
 0
3
x cos f  y sin f  p 
 0
4

4

x cos 3  y sin 3  5 
 0

p 
 3; Quadrant IV


3

1

2x  2y  5 
 0
y  10 
 0
x  3
x cos f  y sin f  p 
 0

19.

3

x cos 300°  y sin 300°  2 
 0

f 315°
11a. 3x  4y 
 8
y


x cos f  y sin f  p 
 0
x cos 210°  y sin 210°  5 
 0

1
x
2

y

3
x
4

2


3

y  3 
 0
x  3
x cos f  y sin f  p 
 0

20.

11

O

x


3
x
2

4

p


5

12

12

Exercises
22.


3

 2y  15 
 0





sin

 12 
 0

x  2
y  24 
 0
2
14.
x cos f  y sin f  p 
 0
x cos 135°  y sin 135°  32

0

2
2x




2
y
2

x cos

5

6

 y sin


3
2x



5

6
1
y
2

 32

0

x cos

 y sin



2

or 1

3x  4y 
 15

23.

3x  4y  15 
 0

A2  
B2 
 
32  (
4)2 or 5

 23

0
 23

0

3
4
15
x  y   

5
5
5
3
4
x  y  3 

5
5
4
sin f 
 5, cos

2
0

0x  1y  2 
 0
y2
0

tan f 


f
Chapter 7


2

2
—

2

2


2

p 
 2; Quadrant I

f 
 45°

x  y  43

0
3
16. x cos f  y sin f  p 
 0


2

1
1




2
2
 2y  2 
 0


2
2
f 
 2, cos f 
 2,

 
y  

0
2
2

tan f 


x  2
y  62

0
2
xy6
0
x cos f  y sin f  p 
 0

15.

f 247°
xy
1
xy1
0

A2  
B2 
 
12  12 or 2

1

x

2

2
x
2

x cos 4  y sin 4  12 
 0

2
y
2

5

–1
12
3
tan f 
 — or 5
5
–1
3

y  30 
 0
x  3
x cos f  y sin v  p 
 0

2
x
2

12


sinf 
 1
3 , cos f 
  13 , p 
 5; Quadrant III

x cos f  y sin f  p 
 0
x cos 60°  y sin 60°  15 
 0



65


1
3 x  13 y  5 
 0

or 1.6 miles

1
x
2

13.

12




13 y  13 
 0

8

Pages 467–469
12.

5
x
13

 5y  5 
 0
8

5

1

 2y  43

0

x  y  83

0
3
21. 
A2  
B2 
 
52  1
22 or 13

3x  4y 
 8
3x  4y  8 
 0

A2  
B2 
 
32  (
4)2 or 5
3
x
5

11

x cos 6  y sin 6  43

0

3x  4y 
 8

11b.

3

 2y  2 
 0

232

4
5
—
3

5

307°

0
0
3

f 
 5, p 
 3; Quadrant IV
4

or 3

24.

y 
 2x – 4
2x  y  4 
 0

A2  
B2 
 
(2)2 
 12 or 5

2

4



5
4
5



 5 y  5 
 0

5
2
5
5, cos f 
 5, p

5
–5
1
—
or 2
2
5

5

x

3

sin f 

tan f 




3

1

5
4


 5; Quadrant IV

1

f


3

1

2

f


3

30.

3

4

8

x cos f  y sin f  p 
 0
3
x
5

4

 5y  10 
 0

3x  4y  50 
 0

1

y  2 
 4(x  20)

31.

1


A2  
B2 
 
(4)2 
 42 or 42
; p 
 42


2

4

y  2 
 4x  5

4


2


 or , sin f 
  or 
cos f 
 
2
2
4 2
4 2



x  4y  28 
 0

x cos f  y sin f  p 
 0


A2  
B2 
 
(1)2 
 42 or 17



2


2


0
2x  2y  42
xy8
0
32.
22
x 
 y  18
22
x  y  18 
 0

A2  
B2 
 
2(2)2 
(1)2 
 9

3


2817


p

17 ; Quadrant II

2
1
18
2
x  y   

3
3
3
18
p 
 3 
 6 units

tan f 
 — or 4
17


–
17
f

40°




cos f 
 1
0 or 5 , sin f 
 10 or 5

210°

1
4
28
x  
y  

0

17

7
17

1

4
2817
17
17







 17 x  17 y  17 
 0
4
17
17




sin f 
 
17 , cos f 
  17 ,
4
17


17


12061


 
61 ; Quadrant I


A2  
B2 
 
62  82 or 10; p 
 10
6

1

–2

3
tan f 
 — or 3

3
–2
27.


0

6
5
124

x  
y  

0
61
61
61




6
5
12061
61
61






0
61
61 x  61 y 
61
61
5
6
, cos f 
 

sin f 
 
61
61 , p
5
61


61
5
—
tan f 
 6
61 or 6


61

sin f 
 2, cos f 
 2, p 
 1; Quadrant III

f


1

6x  5y  124 
 0

A2  
B2 
 
62  52 or 61


2x  2y  2 
 0
1

108°

x
y
  
20
24
x
y
    1
20
24

29.


0

2x  2y  1 
 0


610

p 
 
5 ; Quadrant II

tan f 
 —
or 3
10


–
10

x3
0
sin f 
 0, sin f 
 1, p 
 3
0
tan f 
 1 or 0
f 
 0°
26.
3
x  y 
 2
x  y  2 
 0
3

A2  
B2 
 
(3

)2  (
1)2 or 2

3

y4
0

1
3
12

x  
y  

0
10
10

10



10
3
10
610







 10 x  10 y  5 
 0
3
10
10




sin f 
 
10 , cos f 
  10 ,
3
10


10

f 333°
25.
x
3
x3
0

A2  
B2 
 
12  02 or 1
1
x
1


y4

x  3y  12 
 0

A2  
B2 
 
12  (
3)2 or 10


1

x  y   
 0

 5
 5
 5


5
2
x
5

x

3

28.

104°

33a.

0

y

45˚

O

x

1.25 ft

233

Chapter 7

33b. p 
 1.25, f 
 45°
x cos (45°)  y sin (45°)  1.25 
 0

2
x
2

y

36a.


2

 2y  1.25 
 0

x  2
y  2.5 
 0
2

2

y

34a.

1

x

O
The angles of the quadrilateral are 180°  a,
90°, f2  f1, and 90°. Then 180°  a  90° 
f2  f1  90° 
 360°, which simplifies to
f2 
 f1  a. If the lines intersect so that a is an
interior angle of the quadrilateral, the equation
works out to be f2 
 180°  f1  a.
36b. tan f2 
 tan(f1  a)





x

O


tan f  tan a

f and the supplement of v are complementary
angles of a right triangle, so f  180°  v 
 90°.
Simplifying this equation gives v 
 f  90°.
34b. tan v. The slope of a line is the tangent of the
angle the line makes with the positive x-axis
34c. Since the normal line is perpendicular to 
, the
slope of the normal line is the negative
1

reciprocal of the slope of 
. That is, 
tan v 

cot v.
34d. The slope of 
 is the negative reciprocal of the
1

slope of the normal, or 
tan f 
 cot f.

1



1  tan f tan a
1

If the lines intersect so that a is an interior
angle of the quadrilateral, the equation works
tan f1  tan a

out to be tan f2 
 
1  tan f1 tan a .
37.
5x  y 
 15
5x  y  15 
 0

A2  
B2 
 
52  (
1)2 or 26

5
1
15

x  
y  

26

6

26
2
6

26
5
1526
2

x  
y  

26
26
26

35b.

5x  2y 
 20
5x  2y  20 
 0

52  (
2)2 
 25
4
  or 29


Quadrant I

5
2
20

x  
y  



9
29
29
2

29
29
5
2
2029

x  
y  

29
29
29

f 67°
f  90° 
 67°  90° or 157°
x cos 157°  y sin 157°  3 
 0
12
1
3x



5
y
13


1526


26

3
0

x

5
y
 13


2029

36


 5  
29


0

2029



 0, p 
 
29

13.85564879

13.85564879  500 6927.824395; $6927.82
38. 2 cos2 x  7 cos x  4 
 0
(2 cos x  1)(cos x  4) 
 0
2 cos x  1 
 0
or
cos x  4 
 0
2 cos x 
 1
cos x 
 4

y
12
13


1526



 0, p 
 
26

3x  4y 
 36
3x  4y – 36 
 0

A2  B2 
 
32  42 or 5
3
4
36
36
x  y   
 0, p 
 
5
5
5
5

A2  
B2 
 
52  1
22 or 13
35a. 
5
12
39
x  y   
 0
13
13
13
5
12
x  y  3 
 0
13
13
12
5

sin f 
 1
3 , cos f 
 13 ;
12

13
12
tan f 
 —
5 or 5

13


0

3
0

1

cos x 
 2
x




3

no solution
or x 


5

3

39. sin x 
 
1  co
s2 x

O

x

35c. See students’ work.
35d. The line with normal form x cos f  y sin f 
p 
 0 makes an angle of f with the positive
x-axis and has a normal of length p. The graph
of Armando’s equation is a line whose normal
makes an angle of f  d with the x-axis and also
has length p. Therefore, the graph of Armando’s
equation is the graph of the original line rotated
d° counterclockwise about the origin. Armando is
correct. See students’ graphs.
Chapter 7




1
1  
6






36
35





36 or 6

sin y 
 
1  co
s2 y
2
 
3
1


5
5



 
9 or 3

2




sin(x  y) 
 sin x cos y  cos x sin y
 2
35


5


 

 

6 
 3   
 6 
 3 

  5

235




18

234

1

2

2



40. A 
 1, 4 
 2 or 90°

y
1

1
1 2
4 3

34

y 
 sin 4


 



O

45˚







1 3 2
2

 5
4 1
1
0
1 2
x



15
4 3
y
0
1 3 2
1 3 2
1 2
x


5 4 1  4 3  y 
 5 4 1  15
1 30
x

 5
y
15
x
6


y
3
(6, 3)
46. The value of 2a  b cannot be determined from
the given information. The correct choice is E.

45.

90˚

1
d

41. r 
 2
13.4

   
  
  
  
  

r 
 2 or 6.7
x2 
 6.72  6.72  2(6.7)(6.7) cos 26°20

x2 9.316604344
x 3.05 cm
x

x5
x

x5

42.

17

1

17

1




25  x2 
 x  5

Page 474




x2  25 
 x  5

x



(x  5)(x  5)

(x  5)(x  5)  
 (x  5)(x  5)
 x  5 
1

x(x  5)  17 
 x  5
x2  5x  17 
 x  5
x2  4x  12 
 0
(x  6)(x  2) 
 0
x  6 
 0 or x  2 
 0
x 
 6
x
2
43. original box: V 
 
wh

462

 48
new box:
V 
 
wh
1.5(48) 
 (4  x)(6  x)(2  x)
72 
 x3  12x2  44x  48
0 
 x3  12x2  44x  24
x
0.4
0.5

P

Q
4. The formula is valid in either case. Examples will
vary. For a vertical line, x 
 a, the formula
subtracts a from the x-coordinate of the point. For
a horizontal line, y 
 b, the formula subtracts b
from the y-coordinate of the point.
5. 2x  3y 
 2 → 2x  3y  2 
 0

V(x)
4.416
1.125

V(0.5) is closer to zero, so x 
 0.5.
4  x 
 4  0.5 or 4.5
6  x 
 6  0.5 or 6.5
2  x 
 2  0.5 or 2.5
4.5 in. by 6.5 in. by 2.5 in.

y

44.

x
2

d


Ax1  By1  C


A2  
B2

d


2(1)  (3)(2)  2


22  (
3)2
2

13
2

 or 

d

13
 13

(2, 6)



6. 6x  y 
 3 → 6x  y  3 
 0

xy
8
(2, 3)

Check for Understanding

1. The distance from a point to a line is the distance
from that point to the closest point on the line.
2. The sign should be chosen opposite the sign of C
where Ax  By  C 
 0 is the standard form of
the equation of the line.
3. In the figure, P and Q are any points on the lines.
The right triangles are congruent by AAS. The
corresponding congruent sides of the triangles
show that the same distance is always obtained
between the two lines.


(x  5)(x  5)

x  5 
17

Distance From a Point to a Line

7-7

(5, 3)
y
3

d


x

d


O

Ax1  By1  C


A2  
B2
6(2)  (1)(3)  3


62  (
1)2
12


1237


 or 
d

37

37

f(x, y) 
 3x  y  4
f(2, 3) 
 3(2)  3  4 or 7
f(2, 6) 
 3(2)  6  4 or 4
f(5, 3) 
 3(5)  3  4 or 16
16, 4

235

Chapter 7

7. 3x  5y 
 1
When x 
 2, y 
 1. Use (2, 1).
3x  5y 
 3 → 3x  5y  3 
 0
Ax1  By1  C


A2  
B2
3(2)  (5)(1)  3
d 
 

32  (
5)2
2
34
4

d 
  or 
17

34
2
34


17
1
y 
 3x  3
Use (0, 3).
1
y 
 3x  7 → x  3y  21
Ax1  By1  C
d 
 

A2  
B2
1(0)  3(3)  21
d 
 

12  32
30
d 
  or 310


10

2

14. y 
 4  3x → 2x  3y  12 
 0
d


d


8.

310


6x1  8y1  5


62  82

9. d1 


d


25

13

2513


13

15. y 
 2x  5 → 2x  y  5 
 0

16.

2x1  3y1  4


22  (
3)2
6x1  8y1  5
2x1 3y1  4
 
 
10
13





17.

2x1 3y1 
 4


13


613
x  813
y  513
 
 20x  30y  40
(20  613
)x  (813
  30)y  40  513

0
10. (2000, 0)

d

d


11. d 

d

d


10



10

d

d


3

5(2000)  (3)(0)  0


52  (
3)2
10,000
 or about 1715

34

d

ft

d

d


Exercises

d

d


d




Ax1  By1  C


A2  
B2
4(3)  (5)(0)  (6)


42  (
5)2
6
41


d 
 
or 
41
41

20. y 
 2x  1
Use (0, 1).
2x  y 
 2 → 2x  y  2 
 0
d


534


or 17

d

d


Ax1  By1  C


A2  
B2
2(0)  (1)(0)  3

(2)2 
 (1
)2

3
5
3
 or 
5
5


Chapter 7




6


Ax1  By1  C


A2  
B2
5(3)  (3)(5)  10


52  (
3)2

13. 2x  y 
 3 → 2x  y  3 
 0
d


Ax1  By1  C


A2  
B2
3
6(0)  (8) 8  1


62  (
8)2
4
8
 or 
5
10

19. 4x  5y 
 12
When x 
 3, y 
 0. Use (3, 0).
4x  5y 
 6 → 4x  5y  6 
 0

534


17

d


3

4

5

Ax1  By1  C


A2  
B2
3(2)  (4)(0)  15


32  (
4)2
21

5

10

34


10


1

10

18. 6x  8y 
 3
When x 
 0, y 
 8. Use 
0, 8.
6x  8y 
 5 → 6x  8y  5 
 0

21

5

12. d 


 18 
 0


d 
  or 
10

Ax1  By1  C


A2  
B2

Pages 475–476

Ax1  By1  C


A2  
B2
2(3)  3(1)(1)  (5)

d


22  (
1)2
0


d

or 0

5
4
y 
 3x  6 → 4x  3y
Ax1  By1  C
d 
 

A2  
B2
4(1)  3(2)  (18)
d 
 

42  32
16
16


d 
 5 or 5
16

5
Ax1  By1  C
d 
 

A2  
B2
3(0)  (1)(0)  1

d


32  (
1)2

d



0

613
x  813
y  513
 
 20x  30y  40
(20  613
)x 
(30  813
)y  40  513
 
 0;

d


2513


d 
  or 13

d2 


6x1  8y1  5


10

Ax1  By1  C


A2  
B2
2(2)  3(3)  (12)

22  32


3
5

5

236

Ax1  By1  C


A2  
B2
2(0)  (1)(1)  (2)


22  (
1)2
3

5


35


or  
5

2

21. y 
 3x  6
Use (0, 6).
3x  y 
 4 → 3x  y  4 
 0
Ax1  By1  C


A2  
B2
3(0)  1(6)(1)  (4)
d 
 

32  12
2
10



d 
 10 or 
5

8
y 
 5x 1
Use

27. y 
 3x  1 → 2x  3y  3 
 0

y 
 3x  2 → 3x  y  2 
 0
3x1  y1  2
2x1  3y1  3
d1 
 
d2 
 

22  (
3)2

32  12

d


22.

2x1  3y1  3




13

(210
  313
)x  (13
  310
)y  310
  213

0
2x1  3y1  3
3x1  y1  2

 


13
10


Ax1  By1  C


A2  
B2
8(0)  (5)(1)  15


82  (
5)2

d


210
x  310
y 310
 
 313
x  13
y  213

(210
  313
)x  (13
  310
)y  310
  213

0

28a. Linda: (19, 112)


2089

20

 or 

d

89
 89


2089


89

d




d


3

23. y 
 2x

Use (0, 0).

d


3

y 
 2x  4 → 3x  2y  8 
 0

d


8
d 
 

13
8
13


13

or

d

d

d


d

d


813

13

24. y 
 x  6
xy1
0

d


Linda
4x  3y  228 
 0
4x  3(140)  228 
 0
4x 
 192
x 
 48
29. Let x 
 1.
y

tan v 
 x
y

tan 40° 
 1

3x1  4y1  10
5x1  12y1  26

d2 
 

32  42

52  (
12)2
3x1  4y1  10
5x1  12y1  26
 
 
5
13

y

0.839  0

m

5x  12y 26

d


1
1


 
13

39x  52y  130 
 25x  60y  130
64x  8y  260 
 0
16x  2y  65 
 0
26.

0.8390996312


m

10

39x  52y  130 
 25x  60y  130
14x  112y 
 0
x  8y 
 0
3x1  4y1  10


5

Ax1  By1  C


A2  
B2
4(45)  (3)(120)  228


42  (
3)2
48
 or 9.6
5

28b.

Use (0, 6).

Ax1  By1  C


A2  
B2
1(0)  1(6)  (1)


12  12
5
5
2
 or 
2

2

25. d1 


Ax1  By1  C


A2  
B2
4(19)  (3)(112)  228


42  (
3)2
32
 or 6.4
5

Father: (45, 120)

Ax1  By1  C


A2  
B2
3(0)  2(0)  8


32  22

d


10


210
x  310
y  310
 
 313
x  13
y  213


(0, 1).

8x  15 
 5y → 8x  5y  15 
 0
d


3x  y  2

1
1
 

d

0.839

Ax1  By1  C


A2  
B2
0.839(16)  1(12)  0

2  12
0.839


y  y1 
 m(x  x1)
y  0.839
y
0.839x  y

0.839(x  1)
0.839x
0

d  1.092068438
1.09 m
30. The radius of the circle is 
[(5) 
(2)]2
 (6 
2)2
or 5. Now find the distance from the center of the
circle to the line.

4x1  y1  6
15x1  8y1  68
d1 
 
d2 
 
2
2
2  82

4  1
(15)

4x1  y1  6
15x1  8y1  68
 
  
17
17


68x  17y  102 
 1512
x  817
y  6817


d


(68  1517
)x  (17  817
)y  102  6817

0
4x1  y1  6
15x1  8y1  68


  
17
17


d


68x  17y  102 
 1512
x  817
y  6817


d


(68  1517
)x  (17  817
)y  102  6817

0

Ax1  By1  C


A2  
B2
5(5)  (12)(6)  32


52  (
12)2
65

13

d
5
Since the distance from the center of the circle to
the line is the same as the radius of the circle, the
line can only intersect the circle in one point. That
is, the line is tangent to the circle.

237

Chapter 7

47

3

33.



31. m1 
 
3  1 or 4

y7


3
(x
4

 1)

3x  4y  25 
 0

Ax1  By1  C
a1 
 

A2  
B2
3(1)  (4)(3)  25
a1 
 

32  (
4)2
34
a1 
 5
3  4
7
 or 
m2 
 
1  (3)
2
7
y  4 
 2(x 

2x  7y 
 5
2x  7y  5 
 0

A2  
B2 
 
22  (
7)2 or 53

2

x
53


34.

5

53
53
2
5




53 x 
53
cos 2A 
 1  2 sin2 A
 2
3

 1  2 6
5

 6
2
60°
 
 2,  
 60°
1
1

a2 


35.

(3))

y

Ax1  By1  C


A2  
B2
7(1)  2(7)  13


72  22

34
 or
a2 
 

53
7  (3)

m3 
 
1  (1)


0


 

7x  2y  13 
 0
a2 


7

53
7


53 y

y  

0

 53
 53

1


3453


53

1

y 
 csc (  60˚)
˚ 300˚ 480˚

O 120

or 5

y  7 
 5(x  1)
5x  y  2 
 0
a3 

a3 


Ax1  By1  C


A2  
B2
5(3)  (1)(4)  2


52  (
1)2

36. 110  3 
 330
180°  (60°  40°) 
 80°
x2 
 3302  3302  2(330)(330) cos 80°
x2 179979.4269
x 424.24 miles


1726

 1726

34 3453
, 
 

5
53 , 26
17




37. T 
 2 g

 or 

a3 
 
26
 26

32.


T 
 2 

9.8
2

T 2.8 s
38. 
2 1
8
k
2
20
1 10 20  k
20  k 
 0
k 
 20
39. 2x  y  z 
 9
2x  y  z 
 9
→
2(x  3y  2z) 
 2(10)
2x  6y  4z 
 20
7y  5z 
 11
x  2y  z 
 7
x  3y  2z 
 10
y  2z 
 3
5(y  z) 
 5(3)
5y  5z 
 15
→
7y  5z 
 11
7y  5z 
 11
2y

 4
y 
 2
yz
3
x  2y  z 
 7
2  z 
 3
x  2(2)  (5) 
 7
5 
 z
x 
 6
(6, 2, 5)

y
10
8
6
4
2

O 2 4 6 8

x

The standard form of the equation of the line
through (0, 0) and (4, 12) is 3x  y 
 0. The
standard form of the equation of the line through
(4, 12) and (10, 0) is 2x  y  20 
 0. The
standard form for the x-axis is y 
 0. To find the
3x  y
bisector of the angle at the origin, set  
 y

10
3
x. To find the
and solve to obtain y 
 
1  10


bisector of the angle of the triangle at (10, 0), set
2x  y  20

5



 y and solve to obtain 2x  (1  5
)y

 20 
 0. The intersection of these two bisectors
is the center of the inscribed circle. To solve the

40. square: A 
 s2

3

x into
system of equations, substitute y 
 
1  10



16 
 s2

the equation of the other bisector and solve for x to

4
s
AE 
 s  h
AE 
 4  3 or 7
EF 
 AE
EF 
 7
The correct choice is C.

20(1  10
)
20(1  10
)
 . Then y 
  
5  3 5
  210

5  3 5
  210

3
60

 
  . This y-coordinate is the
1  
10
5  35
  210


get x 


inradius of the triangle. The approximate value is
3.33.

Chapter 7

238

1

triangle: A 
 2bh
1

6 
 2(4)h
3
h

Chapter 7 Study Guide and Assessment
Page 477

19.

Understanding and Using the
Vocabulary

1. b
5. i
9. e

2. g
6. j
10. c

Pages 478–480

3. d
7. h

4. a
8. f

sin4 x  cos4 x

sin2 x
(sin2 x  cos2 x)(sin2 x  cos2 x)

sin2 x
sin2 x  cos2 x

sin2 x
cos2 x

1
sin2 x
1  cot2 x


3


2

17



2




2

  2

6




4

5 2

11

2

9




tan2 v 
 1
6




1




1

1
    sin x


sin x
sin x


4  23

24. cos x 
 
1  si
n2 x

16. cos2 x  tan2 x cos2 x 
 1

18.

x
1




sin2 x 
 1
1
1




1  cos v

2

tan   tan 
3
4

2

1  tan  tan 
3
4

31

1  (3
)(1)
1  3


1 3


 or 2  3



2


 sin x

1  cos v

1  cos v
sec v  1

tan v
sec v  1

tan v
sec v  1

tan v
sec v  1

tan v





23. tan 1
2 
 tan 
 3  4 

1
  (1  sin2 x)

15. csc x  cos2 x csc x 
 
sin x
sin x




1

  2

6

3





2



 

4

tan v 
 4





3

4

tan2 v  1 
 
4


 (csc v  cot v)2

7


 

2
2  
2
2

tan2 v  1 
 sec2 v

1  cos v

1  cos v
1  cos v

1  cos v
1  cos v

1  cos v
1  cos v

1  cos v
1  cos v

1  cos v



16

cos v 
 5

cos2

7


 
sin 4 cos 6  cos 4 sin 6

 cos2 v 
 1
cos2 v 
 2
5

1

4

5
5

4

7


 sin 
4  6

sin2 v  cos2 v 
 1


35

sin y 
 
1  co
s2 x




1
7
 2
25




 or 

 
625
25
576

2 2

1  
3



5

 or 

 
9
3

24

5

cos (x  y) 
 cos x cos y  sin x sin y

5


 
25
3  
25
3

1
cos v 2
  
sin v
sin v
(1  cos v)2

sin2 v
(1  cos v)2

1  cos2 v
(1  cos v)2

(1  cos v)(1  cos v)





1

17


22. sin 
1
2  
 sin 12

1

17.


2

  2

6

1

5

3
3

5

sin2 x

x
cos2 x
cos2 x 


3




4


14. sec v 
 
cos v

cos2


2


 2  2  2  2

v1

v
42  1 
 sec2 v
17 
 sec2 v
17
 
 sec v
1

13. sin v 
 
csc v




1

21. cos 15° 
 cos (45°  30°)

 cos 45° cos 30°  sin 45° sin 30°
sec2





2

  2

6


2





 1  cot2 x




4

1

1

2

tan2





 1  cot2 x


 
2
2  2  2

Skills and Concepts

1

12.


 1  cot2 x


 1  cot2 x
20. cos 195° 
 cos (150°  45°)

 cos 150° cos 45°  sin 150° sin 45°


11. csc v 
 
sin v





 1  cot2 x

24



2

7


48  75


 7
5

1  cos v

tan v



sec v  1

tan v(sec v  1)




sec2 v  1
tan v(sec v  1)




tan2 v
sec v  1



tan v

239

Chapter 7

1

sin y 
 
1  co
s2 y


25. cos y 
 
sec y





2 2




2

 or 

 
9
3



tan (x  y) 


5


3

2

3


 
 







26. cos 75° 
 cos

5  2
5

4

8  5
5

8
10  45


8  5
5

2  3



2  

3











30.

27. sin

7

8

v  cos2 v 
 1

16

or

4

sin v 
 5
sin 2v 
 2 sin v cos v

 2
5
5
4

(Quadrant I)

3

24


 25




2

31. cos 2v 
 2 cos2 v  1
3 2


 2
5  1



2  
3

2

7


 2
5

7

4

2

2 tan v

sin v


32. tan v 
 
cos v

7
1  cos 4












2




(Quadrant II)


2

1
2

2


 


 
4 2
1  3
24
7


 


 2
25
2
5
24

45°

28. sin 22.5° 
 sin 2
1  cos 45°

or

4
2 3

4

3

33. sin 4v 
 sin 2(2v)

 2 sin 2v cos 2v


 2



 
2

4

5

3

5


tan 2v 
 
1  tan2 v






2  2


7

336



 
625

(Quadrant I)

tan x  1 
 sec x
(tan x  1)2 
 sec2 x
tan2 x  2 tan x  1 
 tan2 x  1
2 tan x 
 0
tan x 
 0
x 
 0°
35.
sin2 x  cos 2x  cos x 
 0
1  cos2 x  2 cos2 x  1  cos x 
 0
cos2 x  cos x 
 0
cos x (cos x  1) 
 0
cos x 
 0
or
cos x  1 
 0
x 
 90° or x 
 270°
cos x 
 1
x 
 0°

34.



2

1
2

2

2


 2

 2

Chapter 7

(2  3
 )2

43

sin2 v 
 25






sin2

180  255

61


3

1  
2


 sin

(2  3
)(2  3
)

(2  3
)(2  3
)

3 2


 





3

1 
2

sin2 v  
5 
 1

1  cos 150°

2





3

1 
2




 2  3


180  825


61
150°

2




(Quadrant I)





 

5 5

1  4 
2






5

or 2

tan x  tan y

1  tan x tan y

5
5
  
4
2


1  cos 
6


1  cos 6




sin y


tan y 
 
cos y



6

2


 tan




5

5


 3

29. tan

1  
3

1

3

2



12

240

36.

cos 2x  sin x 
 1
1  2 sin2 x  sin x 
 1
2 sin2 x  sin x 
 0
sin x (2 sin x  1) 
 0
sin x 
 0
or
x 
 0° or x 
 180°

45.

6x 
 4y  5
6x  4y  5 
 0

A2  
B2 
 
62  (
4)2 or 213

6
4
5

x  
y  

0
2
13
2
13
2
13
13
13
13
3
2
5






13 x  13 y  26 
 0
13
13
2
3



sin f 
 
13 , cos f 
  13 , p

2 sin x  1 
 0
1
sin x 
 2
x 
 30° or
x 
 150°


2

37. sin x tan x  2 tan x 
 0

tan f 


tan x 
sin x  2 
 0
2


tan x 
 0
x 
 k

2


sin x 
 2 
 0

or

3

x 
 4  2k or 4  2k
38.

sin 2x  sin x 
 0
2 sin x cos x  sin x 
 0
sin x (2 cosx  1) 
 0
sin x 
 0
or
x 
 k

9
5
3
x  y   
 0
1

06
1

06
1

06
91

06
51

06
3
106
x  y   
 0
106
106
106
51

06
91

06
31

06


, p 
 ; Quadrant I
sin f 
 106 , cos f 
 
106
106

2 cos x  1 
 0
1

cos x 
 2
2

tan f 


x 
 3  2k
4

or x 
 3  2k
cos2 x 
 2  cos x
x  cos x  2 
 0
(cos x  1)(cos x  2) 
 0
cos x  1 
 0
or
cos x  2 
 0
cos x 
 1
cos x 
 2
x 
 2k
no solution
40.
x cos f  y sin f  p 
 0
39.

47.

cos2






3

tan f 


y  43

0
x  3
41.
x cos f  y sin f  p 
 0
x cos 90°  y sin 90°  5 
 0
0x  1y  5 
 0
y5
0
42. x cos f  y sin f  p 
 0
x cos

43.

2
 y sin 3

1
3
2x  2y

48. d 

d


3
0

f

6
13

49. 2y 
 3x  6 → 3x  2y  6 
 0
Ax  By  C

1
1

d

2
2

B
A


3(3)  2(4)  (6)


32  22
23
2313

d 
 
or 13
13


2313


13

d


xy8
0

tan f 


2(5)  (3)(6)  2


22  (
3)2




A2  
B2 
 
72  32 or 58


358


58

758


58

Ax1  By1  C

2
A
B2


6


2

7
3
8

x  
y  

0
58
58

58


758

358

458

x  y   
 0
58
58
29
358

758



sin f 
 58 , cosf 
 58,

or 7

 or 

d

13
 13

3
0


0
2 x  
2y  42

44.

72


10
—
2


10

2



 2; Quadrant II

98°

f

y  6 
 0
x  3
x cos f  y sin f  p 
 0
x cos 225°  y sin 225°  42

0

2

or 9

f 29°
x  7y 
 5
x  7y  5 
 0

A2  
B2 
 
12  (
7)2 or 52


 2y  23

0

2

3

5

106

91

06

106

1
7
5
x  y   
 0
52

52

52

2

72

2

10x  10y  2 
 0
72

2

sin f 
 10, cos f 
 10, p

x cos 3  y sin 3  23

0
1
x
2

2

or 3

f 146°
46. 9x 
 5y  3
9x  5y  3 
 0

A2  
B2 
 
92  52 or 106


2


sin x 
 2


2
13


13
——
13
3


13

13
5




26 ; Quadrant II

50. 4y 
 3x  1 → 3x  4y  1 
 0
Ax1  By1  C

4
58

p 
 29;

d 
 2 
2
B
A



Quadrant I

d


3(2)  (4)(4)  (1)


32  (
4)2
23

3

23

d 
 5 or 5

or 7

23

5

23°

241

Chapter 7

57. x  3y  2 
 0

1

51. y 
 3x  6 → x  3y  18 
 0
d

d


3

Ax1  By1  C

2
A
B2

1(21)  (3)(20)  18


12  
(3)2

y 
 5x  3 → 3x  5y  15 
 0
d1 



2110
21

10
x
y 
 3  6
Use (0, 6).
x
y 
 3  2 → x  3y  6 

Ax1  By1  C
d 
 2

A 
B2
1(0)  (3)(6)  6
d 
 

12  (
3)2

d 
  or 
10

52.

12
10
24

10
12
10
d 
 
5
3
y 
 4x  3
Use (0, 3).
3
1
y 
 4x  2 → 3x  4y  2
Ax1  By1  C
d 
 2 
A

B2

3(0)  (4)(3)  (2)
d 
 

32  (
4)2
14
14
d 
 5 or 5
14
d 
 5

34
x  334
y  234
 
 310
x  510
y 
1510

(34
  310
)x  (334
  510
)y 
234
  1510


0

0

x1  3y1  2

10



0

Page 481

Ax  By  C

A


B
1(0)  1(1)  (5)


12  12
4
 or 22

2


59. d 

d


d 
 22

2

55. y 
 3x  2
d

d


d


Use (0, 2).
60.

sin2 v
v0 2   
cos2 v
cos2 v


 

1
cos2 v
2g  
cos2 v
v02 sin2 v

 2
g
Ax1  By1  C

2
A

B2
4(1600)  (2)(0)  0


42  (
2)2
6400


20

x


sin 30° 
 
100

13
9

50

50

3


y

cos 15°

y 
 2  2 → x  2y  3 
 0
x1  2y1  3
3x1  y1  2

d2 
 

12  22

32  12
3x1 y1 2
x1 2y1  3

 
 
10
5



y

51.76 yd

Page 481

35
x  5
y  25
 
 10
x  210
y  310

(35
  10
)x  (5
  210
)y  25
  310

0
3x1 y1 2


10


Open-Ended Assessment
30°

1. Sample answer: 15°; 15° 
 2


sin 2 
 
2
1  cos 30°

30°

x 2y  3

1
1


 
5



35
x  5
y  25
 
 10
x  210
y  310

(35
  10
)x  (5
  210
)y  25
  310

0






3

1
2

2


2  

3


 2
30°

cos 2 




1  cos 30°



2



3

1
2

2



2  3



 2
Chapter 7

v 
 15°

cos 15° 
 y

56. y 
 3x  2 → 3x  y  2 
 0
d1 


30°  45°  v 
 90°

100 sin 30° 
 x
50 
 x
x
cos v 
 y


or 
d 
 
13
13
x

sin2 v



cos2 v

 
1
2g  
cos2 v
v0 2

d 
 1431 ft

Ax1  By1  C


2
B2
A

2(0)  (3)(2)  3


22  (
3)2
9


Applications and Problem Solving

v02 tan2 v


2g sec2 v

1
1

d

2
2

d


3x1  5y1  15


34

58. The formulas are equivalent.

54. x  y 
 1
Use (0, 1).
xy
5→xy5
0

d





310
x  510
y  1510
 
 34
x  334
y
 234

(34
  310
)x  (334

510
)y 234
  1510

0

d 
  or 
5

53.

x1  3y1  2
3x1  5y1  15

d2 
 
2  (


(1)2 
 32
3
5)2
3x1  5y1  15
x1  3y1  2
 
  
34

10


242

tan 2 


1  cos 30°


1  cos
30°











30°






3. One way to solve this problem is to label the three
interior angles of the triangle, a, b, and c. Then
write equations using these angles and the
exterior angles.
a  b  c 
 180
x  a 
 180
y  b 
 180
z  c 
 180
Add the last three equations.
x  a  y  b  z  c 
 180  180  180
x  y  z  a  b  c 
 180  180  180
Replace a  b  c with 180.
x  y  z  180 
 180  180  180
x  y  z 
 180  180 or 360
The correct choice is D.
4. Since x  y 
 90°, x 
 90°  y.
Then sin x 
 sin (90°  y).
sin (90°  y) 
 cos y


3

1 
2



3

1 
2

2  3


2

2  3


2


(2  3
)(2  3
)

(2  3
)(2  3
)



4
3
(2  3)2


 2  3


1  cos2 x


2. Sample answer: sin x tan x 
 
cos x
1  cos2 x


sin x tan x 
 
cos x
sin x

sin2 x

sin2 x

cos x




cos x

sin x

cos y



sin x 
cos x 
 cos x
sin2 x

cos y

The correct choice is D.
Another solution is to draw a diagram and notice
b
b
that sin x 
 c and cos y 
 c.

SAT & ACT Preparation
Page 483

sin(90°  y)

 
  
 1


cos y
cos y

sin x

cos y

SAT and ACT Practice

1. The problem states that the measure of ∠A is 80°.
Since the measure of ∠B is half the measure of
∠A, the measure of ∠B must be 40°. Because ∠A,
∠B, and ∠C are interior angles of a triangle, the
sum of their measures must equal 180°.
m∠A  m∠B  m∠C 
 180
80  40  m∠C 
 180
120  m∠C 
 180
m∠C 
 60
The correct choice is B.
2. To find the point of intersection, you need to solve
a system of two linear equations. Substitution or
elimination by addition or subtraction can be used
to solve a system of equations. To solve this
system of equations, use substitution. Substitute
2x  2 for y in the second equation.
7x  3y 
 11
7x  3(2x  2) 
 11
7x  6x  6 
 11
x
5
Then use this value for x to calculate the value
for y.
y 
 2x  2
y 
 2(5)  2 or 8
The point of intersection is (5, 8). The correct
choice is A.




b

c
—
b

c


1

x˚

c

a
b

y˚

5. In order to represent the slopes, you need the
coordinates of point A. Since A lies on the y-axis,
let its coordinates be (0, y). Then calculate the two
y0
y


slopes. The slope of 
AB
 is 
0  (3) 
 3 . The slope
y0
y


of A
D
is 
0  3 
  3 . The sum of the slopes is
y
y
   
 0.
3
3
The correct choice is B.
6. Since PQRS is a rectangle, its angles measure 90°.
The triangles that include the marked angles are
right triangles. Write an equation for the measure
of ∠PSR, using expressions for the unmarked
angles on either side of the angle of x°.
90 
 (90  a)  x  (90  b)
0 
 90  a  b  x
a  b 
 90  x
The correct choice is A.

243

Chapter 7

7. Simplify the fraction. One method is to multiply
both numerator and denominator by
1
y  y

y2

 
 y2
1
2
1  y  
y2







1
y  y



1
2
1  y  
y2
y3  y

y2  2y  1

9. Since the volume V varies directly with the
temperature T, the volume and temperature
satisfy the equation V 
 kT, where k is a constant.
1
When V 
 12, T 
 60. So 12 
 60k, or k 
 5.
1
The relationship is V 
 5T.
To find the volume when the temperature is 70°,
1
substitute 70 for T in the equation V 
 5T.
1
V 
 5(70) or 14. The volume of the balloon is
14 in3.
The correct choice is C.
10. Two sides have the same length. The lengths of all
sides are integers. The third side is 13. From
Triangle Inequality, the sum of the lengths of any
two sides must be greater than the length of the
third side. Let s be the length of the other two
sides. Write and solve an inequality.
2s  13
s  6.5
The length of the sides must be greater than 6.5.
But the length of the sides must be an integer.
The smallest integer greater than 6.5 is 7. The
answer is 7. If you answered 6.5, you did not find
an integer. If you answered 6, you found a number
that is less than 6.5.

y2
.
y2



y(y2  1)




(y  1)(y  1)
y(y  1)(y  1)




(y  1)(y  1)
y2  y




y1

Another method is to write both the numerator
and denominator as fractions, and then simplify.
1
y  y

 

1
2
1  y  
y2

y2  1








y
——
y2  2y  1

y2
y2  1

y2



 y

y2  2y  1 
y(y  1)(y  1)




(y  1)(y  1)
y2  y




y1

The correct choice is A.
8. Since the triangles are similar, use a proportion
with corresponding sides of the two triangles.
BC
BD
 
 
AC
AE
2
4
 
 
23
AE

2AE 
 4(2  3)
AE 
 10
The correct choice is E.

Chapter 7

244

Chapter 8 Vectors and Parametric Equations
11.

Geometric Vectors

8-1


13 x



2z


2.9 cm

Page 490

Check for Understanding

12˚

2.9 cm, 12°
12. h  2.9 cos 55°
h 
 1.66 cm

1. Sample answer:


b


a

2z

 
 13
x

13a.


a 
 b


v  2.9 sin 55°
v 
 2.38 cm
100

5

13b. Use the Pythagorean Theorem.
c2  a2  b2
c2  (100)2  (5)2
c2  10,025
c  10,02
5 or about 100.12 m/s

Draw u
a . Then drawu
b so that its initial point (tip)
is on the terminal point (tail) of u
a . Draw a dashed
line from the initial point ofu
a to the terminal
point ofu
b . The dashed line is the resultant.
2. Sample answer: A vector has magnitude and
direction. A line segment has only length. A vector
can be represented by a directed line segment.
3. Sample answer: the velocities of an airplane and a
wind current
4. No, they are opposites.
5-11. Answers may vary slightly.
5. 1.2 cm, 120° 6. 2.9 cm, 55°
7. 1.4 cm, 20°

Pages 491–492

Exercises

14. 2.6 cm, 128°
15. 1.4 cm, 45°
16. 2.1 cm, 14°
17. 3.0 cm, 340°
18-30. Answers may vary slightly.


r  
s

18.


x 
 y


8.

3 cm
3.5 cm


r


s


y

101˚

3 cm, 101°


x

70˚

19.

3.5 cm, 70°


x

9.
2.6 cm


s

210˚


s  
t

3.4 cm


t
25˚

3.4 cm, 25°
x

 
 
y


y



20.


s

2.6 cm, 210°


z

10.

359˚

4y

  
z

u


3.8 cm


s  u


3.8 cm, 359°

4y


12.9 cm

51˚

12.9 cm, 51°

245

Chapter 8

21. 324˚


t

26.


u

u



r



r

5.5 cm

3.5 cm


r 
 
t 
 u


22˚

3.5 cm, 22°
27. 
r  
s 
 

u


u 
 
r
5.5 cm, 324°
22.


u


5.4 cm
s




r


r 
 
t
3.9 cm

155˚


r
133˚


t
3.9 cm, 155°

5.4 cm, 133°

23.

2s



28.
2r


u



5.2 cm


 12 
r
358˚

5.5 cm
2s

  u
 
 12 
r

5.5 cm, 358°


r

128˚

29. Draw to scale:

2t



24.


r


s



5.2 cm, 128°

301˚

3s




s

3.4 cm

3u



4.2 cm


r 
 2t
 
 
s  3u

3.4 cm, 301°

45˚

30. Draw to scale:
4.2 cm, 45°

3t


25.



2u
11.7 cm

322˚

3u



357˚

11.7 cm, 357°
31. h  2.6 cos 128°
h  1.60 cm
32. h  1.4 cos 45°
h  0.99 cm
33. h  2.1 cos 14°
h  2.04 cm

8.2 cm

2s


3u

 
 2s


8.2 cm, 322°

Chapter 8

246

3t

  2u


v  2.6 sin 128°
v  2.05 cm
v  1.4 sin 45°
v  0.99 cm
v  2.1 sin 14°
v  0.51 cm

34. h  3.0 cos 340°
v  3.0 sin 340°
h  2.82 cm
v  1.03 cm
35. c2  a2  b2
c2  (29.2)2  (35.2)2
c2  2091.68
c  2091.
68
 or about 45.73 m
36. The difference of the vectors; sample answer: The
other diagonal would be the sum of one of the
vectors and the opposite of the other vector, so it
would be the difference.
37. Yes; sample answer:

41. h  47 cos 40°
v  47 sin 40°
h 
 36 mph
v 
 30 mph
42. It is true when k  1 or whenu
a is the zero vector.
2
2
2
43. c  a  b
c2  (50)2  (50)2
c  5000
 or about 71 lb
44.


a

60˚

2.4 cm

u u
b  24
a
equilateral triangleu
a u
b  24 lb
45. The origin is not in the interior of the acute angle.
d1  
d2


r 
 s


r


s 
 
r

60˚

60˚ 60˚


s

r

b



60˚ 60˚


s

x
y2

x
y2



1 
(
1)


2


 or 
d1  
2
2

38.

y
5

d2  2 2 or y 
 5

35 N

0  1


x
y2



2

 
( y 
 5)

(y 
 5)
x 
 y  2  2
x 
 y  2  2
y 
 52

x 
 y  2 
 2
y  52
0
x 
 (1  2
)y  2  52
0

40 N

1

sin v



46. csc v cos v tan v  
sin v  cos v  cos v

60 N

sin v

cos v

 

sin v  cos v

47.

6.1




4

1
 
n where n is an integer

48.
23˚

61 N, 23° north of east
39. Sometimes;



2, 52
 scl

2 by [
3, 3] scl
1

a

 
 
b  5

b



3


a

  2.3 
 2.7

 
 b

x  , 2 for 0  x  2


5

5


49. tan 18°29
  
0.5b


a

5


b
0.5 tan 18°29


b  29.9 cm
5

sin 18°29
  h


a 
 b



a

5

a

  1.5

 
 b


h
sin 18°29


a

  2.3 
 2.7

 
 b

h  15.8 cm

5
b



40a. v  1.5 sin 52°
v 
 1.18 N
40b. h  1.5 cos 78°
h 
 0.31 N

h  1.5 cos 52°
h 
 0.92 N
v  1.5 sin 78°
v 
 1.47 N

247

Chapter 8

u
2. Use XY   
(x2 
 
x1)2 
(y2 

y1)2 and replace
the values for x and y.
x(
5, 
6), y(3, 
4)
u
XY   
[3 
 (

5)]2 
 [
4

 (

6)]2
2
2
 
(8) 
(
2)
 64
4
  or 68

3. Jacqui is correct. The representation is incorrect.
2, 0  0, 
5 is not equal to 51, 0  (
2)0, 1.
u 
 5j
u.
The correct expression is 2i
u
4. MP  
3 
2, 4 
 (
1) or 
5, 5
u
(
5)2 
 (5)2
MP   
 50
 or 52
 units
u
5. MP  0 
 5, 5 
 6 or 
5, 
1
u
MP   
(
5)2 
 (
1
)2
 26
 units
u
6. MP  4 
 (
19), 0 
 4 or 23, 
4
u
MP   
(23)2 
(
4)2
 545
 units
7. u
t u
u u
v
 
1, 4  3, 
2
 
1  3, 4  (
2) or 2, 2
u
u 
u
v
8. t  1u

50. vo  volume of original box
vn  volume of new box
v o  
o  w o  h o
(w  1)  w  2w
(w  1)2w2
 2w3  2w2
vn  
n  w n  h n
 (w  2)  (w  1)  (2w  2)
 (w2  3w  2)(2w  2)
 2w3  8w2  10w  4
2w3  8w2  10w  4  160
w

1
1
2
3

2
2
2
2
2

8
6
10
12
14

10
4
20
34
52


156

160

136

88
0

wo  3

o  2w
 2  3 or 6
ho  w  1
 3  1 or 4
So, the dimensions of the original box are
3 ft  4 ft  6 ft
x2


51. g(x)  
(x 
 1)(x  3)



vertical: As x approaches 1 and 
3, the expression
approaches  or 
. So, x  1 and x  
3 are
vertical asymptotes.




y

2
x

 
x2
x2

x2
2x
3

 

 
x2
x2
x2
1
2
  
x
x2

2
3
1   
 
x
x2

 
8
1, 4
 
8(
1), 
8(4) or 8, 
32
11. 8, 
6  
82  (

6)2
 100
 or 10
u 
 6j
u
8i
12. 
7, 
5  
(
7)2 
 (
5
)2
 74

u 
 5j
u

7i
u
13. Let T represent the force Terrell exerts.
u
Let W represent the force Mr. Walker exerts.
u
u
Tx  400 cos 65°
Wx  600 cos 110°

 169.05

 
205.21
u
u
Ty  400 sin 65°
Wy  600 sin 110°

 362.52

 563.82
u
u
T  169.05, 362.52, W  
205.21, 563.82
u u
T  W  
36.16, 926.34
u u
T  W  
(
36.
16)2 
(926.3
4)2

 927 N

As x increases positively or negatively, the
expression approaches 0. So, y  0 is a horizontal
asymptote.
52. Let x, x  2, and x  4 be 3 consecutive odd
intergers.
3x  2(x  4)  3
x  4  15
3x  2x  8  3
3x 
 2x  11
x  11
The correct answer 15.

8-2

Algebraic Vectors

Pages 496–497

or 
312, 4

u  6u
9. u
t  4u
v
 4 
1, 4  63, 
2
 
4, 16  18, 
12
 
4  18, 16  (
12) or 14, 4
u
u
10. t  
8u

x2


horizontal: y  
x2  2x 
 3

y

2
1

1, 4 
 3, 
2
2

12, 2 
 3, 
2

12 
 3, 2 
 (
2)

Check for Understanding

1. Sample answer:u
a  8, 6,u
b  6, 8; equal
vectors have the same magnitude and direction.

Pages 497–499

Exercises

u
14. YZ  2 
 4, 8 
 2 or 
2, 6
u
YZ  
(
2)2 
 62
 40
 or 210


Chapter 8

248

u
15. YZ  
1 
 (
5), 2 
 7 or 4, 
5
u
YZ  
42  (

5)2
 41

u
16. YZ  1 
 (
2), 3 
 5 or 3, 
2
u
YZ  
32  (

2)2
 13

u
17. YZ  0 
 5, 
3 
 4 or 
5, 
7
u
YZ  
(
5)2 
 (
7
)2
 74

u
18. YZ  0 
 3, 4 
 1 or 
3, 3
u
(
3)2 
 32
YZ  
 18
 or 32

u
19. YZ  1 
 (
4), 19 
 12 or 5, 7
u
YZ  
52  72
 74

u
20. YZ  7 
 5, 6 
 0 or 2, 6
u
YZ  
22 62
 40
 or 210

u
21. YZ  23 
 14, 
14 
 (
23) or 9, 9
u
YZ  
92  92
 162
 or 92

u
22. AB  36 
 31, 
45 
 (
33) or 5, 
12
u
AB  
52  (

12)2
 169
 or 13
23. u
a u
b u
c
 6, 3  
4, 8
 6  (
4), 3  8 or 2, 11
u u
24. u
a  2b
c
 26, 3  
4, 8
 12, 6  
4, 8
 12  (
4), 6  8 or 8, 14
u
u
c
25. a  b  2u
 6, 3  2
4, 8
 6, 3  
8, 16
 6  (
8), 3  16 or 
2, 19
u  3u
26. u
a  2b
c
 26, 3  3
4, 8
 12, 6  
12, 24
 12  (
12), 6  24 or 0, 30
u  4c
u
u
27. a  
b
 
6, 3  4
4, 8
 
6, 
3  
16, 32
 
6  (
16), 
3  32 or 
22, 29
28. u
a u
b 
 2u
c
 6, 3 
 2
4, 8
 6, 3 
 
8, 16
 6 
 (
8), 3 
 (16) or 14, 
13
u
u
29. a  3b
 36, 3
 3  6, 3  3 or 18, 9
u
c
30. a  
1u


2

12
4, 8

12  (
4),



u
32. u
a  0.4b 
 1.2u
c
 0.46, 3 
 1.2
4, 8
 2.4, 1.2 
 
4.8, 9.6
 2.4 
 (
4.8), 1.2 
 9.6 or 7.2, 
8.4
1 u
33. u
a  (2b 
 5u
c)




3
1
(26, 3 
 5
4, 8)
3
1
(12, 6 
 
20, 40)
3
1
(12 
 (
20), 6 
 40)
3
1
32
34
32, 
34 or , 

3
3
3




u u
u
u
34. a  (3b  c )  5b
 36, 3  
4, 8  56, 3
 18, 9  (
4, 8  30, 15
 18  (
4)  30, 9  8  15 or 44, 32
u 
 2.5n
u  3
5, 
6 
 2.56 
 9
35. 3m
 [
15, 
18 
 15, 
22.5
 [
15 
 (15), 
18 
 (
22.5
 
30, 4.5
36. 3, 4  
32  42
 25
 or 5
u  4j
u
3i
37. 2, 
3  
22  (

3)2
 13

u 
 3j
u
2i
38. 
6, 
11  
(
6)2 
 (
1
1)2
 157

u 
 11j
u

6i
39. 3.5, 12  
(3.5)2 
 122
 156.2
5
 or 12.5
u  12j
u
3.5i
40. 
4, 1  
(
4)2 
 12
 17

u u

4i
j
41. 
16, 
34  
(
16)2
 (

34)2
 1412
 or 2353

u 
 34j
u

16i
u
42. ST  
4 
 (
9), 
3 
 2 or 5, 
5
u 
 5j
u
5i
43. Student needs to show that
(u
v 1 u
v 2) u
v 3 u
v 1  (u
v 2 u
v 3)
u
u
u
( v 1  v 2)  v 3  [a, b  c, d]  e, f
 a  c, b  d  e, f
 a  c  e, b  d  f
 a  c  e, b  d  f
 a, b  c  e, d  f
 a, b  [c, d  e, f]
u
v 1  (u
v 2 u
v 3)
44a.
100 N

Fy

20˚

Fx


12

 8 or 2, 
4

u
F y


44b. sin 20°  
100
u
Fy  100 sin 20°

u
u
31. u
a  6b  4c
 66, 3  4
4, 8
 36, 18  
16, 32
 36  (
16), 18  32 or 20, 50


 34 N

249

Chapter 8

45a.

53. Let a  400, b  600, C  46.3
°
c2  4002  6002 
 2(400)(600) cos 46.3
°
c2 
 18.8578.39
c 
 434
Pabc

 400  600  434

 1434 ft
1
s  2(a  b  c)

Surfer

Vk

Vs  15

30˚

Vx

Shore

45b. sin 30° 
u
Vk 

15

u
Vk
15

sin 30°

1

s 
 2(1434) or 717
k 
 s(s


a)(s


b)(s 

c)
k 
 717(7
17
400)(7

 17
600)(7

 17
434)


k 
 7,525
,766,0
79

k 
 86,751 sq ft


 30 mph
u u
u
u
46a. Since QR  ST  0, QR  
ST .
So, they are opposites.
u
u
46b. QR and ST have the same magnitude, but
opposite direction. So, they are parallel.
Quadrilateral QRST is a parallelogram.

54. Sample answer:
f(x)  3x2 
 2x  1
r
3

2
1
3
1
2
3
4

d

47a. t  r


150 m

5 m/s

or 30 s

47b. d  rt
 (1.0 m/s)(30 s) or 30 m
u
u
47c. V B  V C  0.5  1.0
 
12  52
 26
 or about 5.1 m/s
(x2 
 x1)
v cos v
48. cos v   → (x 
 x ) u
sin v 
49.

u
v
(y2 
 y1)

u
v

2

An upper bound is 2.
1
2
9

f(
x)  3x2 
 2x  1
r
3
2
1
1
3
5
6
2
3
8
17

1

A lower bound
is 
1.

55.

→ (y2 
 y1) u
v sin v

u
PQ  
2 
 8, 5 
 (
7)
 
10, 12
u
PQ   
(
10)2
 122
 244

u
RS  7 
 8, 0 
 (
7)
 
1, 7
u
RS   
(
1)2 
 72
 50

none

50. d 
d

[
4, 4] scl
1 by [
4, 4] scl
1
max: (0, 3), min: (0.67, 2.85)
56.

Ax1  By1  C


A2  
B2
3(
1) 
 7(4) 
 1


32  (

7)2

3
322

d   or about 4.2

58

51. sin 255°  sin (225°  30°)
 sin 225° cos 30°  cos 225° sin 30°
2



3

2
 1

2

 
2  2  
2
6
 
 2


 
 
4

y →  as x → , y →  as x → 

57. 7x  1 7x 
 1
1 
1
This statement is true regardless of the value of x,
so it is true for all real values of x.
The correct choice is A.

52. y  A sin (kx  c)
A: A  17
A  17 or 
17
2



k:   
k
4
k8
c
c: 
k  
60°
c

  
60°
8
c  480°
y  17 sin (8x  480°)
Chapter 8

f(x)  x2  3x  1
x
f(x)

10,000
99,970,001

1000
997,001

100
9701

10
71
0
1
10
131
100
10,301
1000
1,003,001
10,000
100,030,001

250

8-3

11. 132, 3454, 0  
1322 
34542
 02
 11,947
,540


 3457 N

Vectors in Three-Dimensional
Space

Pages 502–503

Check for Understanding

1. Sample answer: sketch a coordinate system with
the xy-axes on the horizontal, and the z-axis
u is two units along the
pointing up. Then, vector 2i
u
x-axis, vector 3j is three units along the y-axis,
u is four units along the z-axis. Draw
and vector 4k
broken lines to represent three planes.

Pages 503–504

Exercises

12.

z
y

z
x
B(4,1,
3)
u
42  12
 (

3)2
OB   
 26


y
x
13.
2. Sample answer: To find the components of the
vector, you will need the direction (angle) with the
horizontal axis. Using trigonometry, you can
obtain the components of the vector.
3. Sample answer: Neither is correct. The sign for
theu
j -term must be the same (
), and the
coefficient for theu
k -term is 0, so the correct way
to express the vector as a sum of unit vectors is
u
u.
i 
 4j

z

B(7, 2, 4)
y
x
u
OB   
72  22
 42
 69

14.
z

4.
G(4,
1, 7)

B(10,
3,15)

16
14
12
10
8
6
4
2 
4


12
10
8
6
4
22 2 4 y

2
4

4
6
8

u
OG   
42  (

1)2 
72
 66

u
5. RS  3 
 (
2), 9 
 5, 
3 
 8 or 5, 4, 
11
u
RS   
52  42
 (

11)2
 162
 or 92

u
6. RS  10 
 3, 
4 
 7, 0 
 (
1) or 7, 
11, 1
u
RS   
72  (

11)2
 12
 171
 or 319

u u
7. u
a  3f
g
 31, 
3, 
8  3, 9, 
1
 3, 
9, 
24  3, 9, 
1
 3  3, 
9  9, 
24  (
1) or 6, 0, 
25
u
8. u
a  2u
g 
 5f
 23, 9, 
1 
 51, 
3, 
8
 6, 18, 
2 
 5, 
15, 
40
 6 
 5, 18 
 (
15), 
2 
 (
40) or 1, 33, 38
u
9. EF  6 
 (
5), 
6 
 (
2), 6 
 4
 11, 
4, 2
u 
 4j
u  2k
u
11i
u
10. EF  
12 
 (
12), 17 
 15, 
22 
 (
9)
 0, 2, 
13
u 
 13k
u
2j

x
u
OB   
102 
(
3)2
 152
 334

u
15. TM  3 
 2, 1 
 5, 
4 
 4 or 1, 
4, 
8
u
TM   
12  (

4)2 
(
8)2
 81
 or 9
u
16. TM  
3 
 (
2), 5 
 4, 2 
 7 or 
1, 1, 
5
u
TM   
(
1)2 
 12 
(
5)2
 27
 or 33

u
17. TM  3 
 2, 1 
 5, 0 
 4 or 1, 
4, 
4
u
TM   
12  (

4)2 
(
4)2
 33

u
18. TM  
1 
 3, 1 
 (
5), 2 
 6 or 
4, 6, 
4
u
TM   
(
4)2 
 62 
(
4)2
 68
 or 217

u
19. TM  
2 
 (
5), 
1 
 8, 
6 
 3 or 3, 
9, 
 9
u
TM   
32  (

9)2 
(
9)2
 171
 or 319


251

Chapter 8

u
20. TM  1 
 0, 4 
 6, 
3 
 3 or 1, 
2, 
6
u
TM   
12  (

2)2 
(
6)2
 41

u
21. CJ   3 
 (
1), 
5 
 3, 
4 
 10
or 4, 
8, 
14)
u
42  (

8)2 
(
14
)2
CJ   
 276
 or 269

u  2u
22. u
u  6w
z
 62, 6, 
1  23, 0, 4
 12, 36, 
6  6, 0, 8
 18, 36, 2
1
23. u
u  u
v 
u
w  2u
z

u
35. G1G2  
(x2 
 
x1)2 
(y2 

y1)2 
(z2 

z1)2
u
 
(x1 
 
x2)2 
(y1 

y2)2 
(z1 

z2)2  G1G2
because (x 
 y)2  (y 
 x)2 for all real numbers x
and y.
36. Ifu
m  m1, m2, m3, then
u
u
m   
(m )2 

(m 
)2  (m
)2. If 
m
1

Since m12  (
m1)2, m22  (
m2)2, and m32 
u  m
u.
(
m3)2, 
m
u u
37. 3, 
2, 4  6, 2, 5  F  O
u u
9, 0, 9  F  O
u
F  
9, 0, 9 or

9, 0, 
9
1
38. m  2(x1  x2, y1  y2, z1  z2)

2

 2, 
2, 2 
 2, 6, 
1  6, 0, 8
3 5

 6, 
 72, 112
1

1

24. u
u  4 u
v 
u
w
3

 2(2  4, 3  5, 6  2)

3

 2(6, 8, 8)

1

 44, 
3, 5 
 2, 6, 
1

 (3, 4, 4)
u
39a. OK  1 
 0, 4 
 0, 0 
 0 or 1, 4, 0
u
u
i  4j
u
39b. TK  1 
 2, 4 
 4, 0 
 0 or 
1, 0, 0
u

i
u
40. u
c  b 
u
a
u
c  3, 1, 5 
 1, 3, 1
u
c  2, 
2, 4
z
41a.

 3, 
4, 4 
 2, 6, 
1
9 15

 1, 
84, 44
1

3

2
25. u
u  3u
v 
 3 u
w  2u
z
2

 34, 
3, 5 
 32, 6, 
1  23, 0, 4
 12, 
9, 15 
 3, 4, 
3  6, 0, 8
4

2

 163, 
13, 233
26. u
u  0.75u
v  0.25u
w
2

2

 0.754, 
3, 5  0.252, 6, 
1
 3, 
2.25, 3.75  0.5, 1.5, 
0.25
 3.5, 
0.75, 3.5
u u
27. u
u  
4w
z
 
42, 6, 
1  3, 0, 4
 
8, 
24, 4  (3, 0, 4
 
5, 
24, 8
2
2
28. 3u
f  3u
g 
 5u
h


3


(
m1)2
 (

m2 )2 
(
m3 
)2.

1
u
u  24, 
3, 5 
 2, 6, 
1  23, 0, 4

1

2

u 

m1, 
m2, 
m3, then 
m

2

3,
3

4.5, 
1  3
2, 1, 6

6

O
6

x

y

(15 

0)2 
(15 

0)2 
(15 

0)2
d  
 675
 or about 26 feet



15

41c. sin v  26



u
29. LB  5 
 2, 
6 
 2, 2 
 7 or 3, 
8, 
5
u 
 8j
u 
 5k
u
3i
u
30. LB  
4 
 (
6), 5 
 1, 
1 
 0 or 2, 4, 
1
u  4j
u 
u
2i
k
u
31. LB  7 
 9, 3 
 7, 
2 
 (
11) or 
2, 
4, 9
u 
 4j
u  9k
u

2i
u
32. LB  
8 
 12, 7 
 2, 
5 
 6 or 
20, 5, 
11
u  5j
u 
 11k
u

20i
u
33. LB  
8 
 (
1), 5 
 2, 
10 
 (
4)
or 
7, 3, 
6
u  3j
u 
 6u

7i
k
u
34. LB  6 
 (
9), 5 
 12, 
5 
 (
5)
or 15, 
 7, 0
u 
 7j
u
15i
52 36 278

5, 5, 1
5

Chapter 8

12

12

6




41b. Find distance between (0, 0, 0) and (15, 15, 15).

2

 56, 
3, 
3
12
6
6
, 
, 

5
5
5

 
2, 3, 
3  
6, 3, 18 
 
2

12

v  sin1

15


675

v  35.25°
u
(1 
 2
)2  (
 
 0
3
)2  (0

 0)2
42. AB   
 4
 or 2
u
BC  

(1 
 1)2 

1

3


 3


2



22


3


0

2


36 
 
63


 3 or 
 1.69
u
AC  

(1 
 2)2 

1

3


0

2



22


3


0

2

 2
 or 
 1.41
No, the distances between the points are not
equal. A and B are 2 units apart, B and C are 1.69
units apart, and A and C are 1.41 units apart.
43. 3, 5  
1, 2  3  (
1), 5  2
 2, 7

252

44.

45.

u
AB  
3 
 5, 3 
 2 or 
8, 1
u
CD  d1 
 0, d2 
 0 or d1, d2
u u
AB  CD

8, 1  d1, d2
D  (
8, 1)
sin 2X

1 
 cos 2X
2 sin X cos X

1 
 cos2 X  sin2 X
2 sin X cos X

2 sin2 X
cos X

sin X

u
i
2. u
a u
a  ax
ax
 ay
ay

u
u
j
k
ay az
ay az
ay u
az u
i 
 ax az u
j  ax
k
az
ax az
ax ay
u 
 (a a 
 a a )j
u
 (ayaz 
 ayaz)i
x z
x z
 (a a 
 a a )u
k

 cot X

x y

 cot X
 cot X
 cot X

3.

cot X  cot X
46. cos v 

4.

2

3

sin2 v  1 
 cos2 v
sin2 v  1 


47.

4

9

5.

5
sin2 v  9
5

sin v  3
v
y  6 sin 2

6.

amplitude  6 or 6
2


period  k

7.

2

 or 4

1

2
2
 rad
1 min
   
1 rev
60 sec


rev


48. 16 
min 

8


 15 radians per second

13, 1, 
5  1, 
3, 2
13(1)  1(
3)  (
5)(2)
13 
 3 
 10  0
13, 1, 
5  
2, 1, 
5
13(
2)  1(1)  (
5)(
5)

26  1  25  0
u
i u
j u
k
8. 6, 2, 10  4, 1, 9  6 2 10
4 1
9
2 10 u
6 10 u
6 2u
i 

j 
k

1
9
4
9
4 1
u 
 14j
u 
 2k
u or 8, 
14, 
2, yes
 8i

49. Yes, because substituting 7 for x and 
2 for y
results in the inequality 
2 180 which is true.
y 4x2 
 3x  5

2 4(7)2 
 3(7)  5

2 180
3 31

4

3

2



50. 2

21  3

4

3

So, A, C, and D are not correct.
2 21


3 31

3

2

3

 4

3

4

So, B is not correct.
The correct choice is E.

8-4

8, 
14, 
2  6, 2, 10
8(6)  (
14)(2)  (
2)(10)
48 
 28 
 20  0
8, 
14, 
2  4, 1, 9
8(4)  (
14)(1)  (
2)(9)
32 
 14 
 18  0
9. Sample answer: Let T(0, 1, 2), U(
2, 2, 4), and
V(
1, 
1, 
1)
u
TU  
2, 1, 2
u
UV  1, 
3, 
5
u u
TU  UV 
u
u
i
j u
k
 
2
1
2
1 
3 
5
2u
2u
1u
i 
 
2
j  
2
k
 1

3 
5
1 
5
1 
3
u  5k
u or 1, 
8, 5
u
i 
 8j

Perpendicular Vectors

Pages 508–509

x y

u 
 0j
u  0k
u
 0i
 0, 0, 0
u
0
Sample answer: No, because a vector cannot be
perpendicular to itself.
5, 2  
3, 7  5(
3)  2(7)
 
15  14
 
1, no

8, 2  4.5, 18  
8(4.5)  2(18)
 
36  36
 0, yes

4, 9, 8  3, 2, 
2  
4(3)  9(2)  8(
2)
 
12  18 
 16
 
10, no
u
u
u
i
j
k
1, 
3, 2  
2, 1, 
5  1 
3
2

2
1 
5

3
2u
1
2u
1 
3 u

i 

j 
k
1 
5

2 
5

2
1
u 
u
u or 13, 1, 
5, yes
 13i
j 
 5k

Check for Understanding

1. Sample answer: Vector 
vw
 is the negative of
vector w

v
.
 u u u
i j k
u
v u
w  
1 0 3
1 2 4
0
3u

i 
 
1 3 u
j  
1 0 u
k
2 4
1 4
0 2
u  7j
u 
 3k
u
 
6i
u
u
u
i j k
u
v u
w 
1 2 4

1 0 3
2
4u

i 
 1 4u
j  1 2u
k
0 3

1 3

1 0
u
u
u
 6i 
 7j  3k

253

Chapter 8

u
10. AB  (0.65, 0, 0.3) 
 (0, 0, 0)
 0.65, 0, 0.3
u
F  0, 0, 
32
u u u
T  AB  F 
u
u
u
i
j
k
 0.65 0
0.3
0
0 
32
0
0.3 u
0.65
0.3 u
0.65 0.3 u
i 

j 
k

0 
32
0 
32
0
0
u 
 20.8j
u  0k
u
 0i

u
i u
j u
k
21. 0, 1, 2  1, 1, 4  0 1 2
1 1 4
1 2u
0 2u
0 1u

i 

j 
k
1 4
1 4
1 1
u  2j
u 
u
 2i
k or 2, 2, 
1, yes
2, 2, 
1  0, 1, 2
2(0)  2(1)  (
1)(2)
22
20
2, 2, 
1  1, 1, 4
2(1)  2(1)  (
1)(4)
22
40

u
T   
02  (
20.8

)2  02
 20.8 foot-pounds

Pages 509–511

u
i u
j u
k
22. 5, 2, 3  
2, 5, 0 
5 2 3

2 5 0
2 3u
5 3u
5 2u

i 

j 
k
5 0

2 0

2 5
u 
 6j
u  29k
u or 
15, 
6, 29, yes
 15i

Exercises

11. 4.8  6, 
3  4(6)  8(
3)
 24 
 24
 0, yes
12. 3, 5  4, 
2  3(4)  5(
2)
 12 
 10
 2, no
13. 5, 
1  
3, 6  5(
3)  (
1)(6)
 
15 
 6
 
21, no
14. 7, 2  0, 
2  7(0)  2(
2)
0
4
 
4, no
15. 8, 4  (2, 4  8(2)  4(4)
 16  16
 32, no
16. 4, 9, 
3  
6, 7, 5  4(
6)  9(7)  (
3)(5)
 
24  63 
 15
 24, no
17. 3, 1, 4  2, 8, 
2  3(2)  1(8)  4(
2)
68
8
 6, no
18. 
2, 4, 8  16, 4, 2  
2(16)  4(4)  8(2)

32  16  16
 0, yes
19. 7, 
2, 4  3, 8, 1  7(3)  (
2)(8)  4(1)
 21 
 16  4
 9, no
20. u
a u
b  3, 12  8, 
2
 24 
 24
 0, yes
u
b u
c  8, 
2  3, 
2
 24  4
 28, no
u
a u
c  3, 12  3, 
2
 9 
 24
 
15, no


15, 
6, 29  5, 2, 3
(
15)(5)  (
6)(2)  29(3)

75 
 12  87  0

15, 
6, 29  
2, 5, 0
(
15)(
2)  (
6)(5)  29(0)
30 
 30  0  0
u
i u
j u
k
23. 3, 2, 0  1, 4, 0  3 2 0
1 4 0
2 0u
3 0u
3 2u

i 

j 
k
4 0
1 0
1 4
u 
 0j
u  10k
u or 0, 0, 10, yes
 0i
0, 0, 10  3, 2, 0
0(3)  0(2)  10(0)
0000
0, 0, 10  1, 4, 0
0(1)  0(4)  10(0)
0000

u
u
u
i
j
k
24. 1, 
3, 2  5, 1, 
2  1 
3
2
5
1 
2

3
2u
1
2u
1 
3 u

i 

j 
k
1 
2
5 
2
5
1
u  12j
u  16k
u or 4, 12, 16, yes
 4i
4, 12, 16  1, 
3, 2
4(1)  12(
3)  16(2)
4 
 36  32  0
4, 12, 16  5, 1, 
2
4(5)  12(1)  16(
2)
20  12 
 32  0

u
u
i
j u
k
25. 
3, 
1, 2  4, 
4, 0  
3 
1 2
4 
4 0

1 2 u

3 2 u

3 
1 u

i 

j 
k

4 0
4 0
4 
4
u  8j
u  16k
u or 8, 8, 16, yes
 8i
8, 8, 16  
3, 
1, 2
8(
3)  8(
1)  16(2)

24 
 8  32  0
8, 8, 16  4, 
4, 0
8(4)  8(
4)  16(0)
32 
 32  0  0

Chapter 8

254

u
u
i u
j
k
26. 4, 0, 
2  
7, 1, 0  4 0 
2

7 1
0
0 
2 u
4

2
u
i 

j  4 0u
k

1
0

7
0

7 1
u  14j
u  4k
u or 2, 14, 4, yes
 2i

30. Sample answer:
Let T(
2, 1, 0), U(
3, 0, 0), and V(5, 2, 0).
u
TU  
1, 
1, 0
u
UV  8, 2, 0
u
u
j u
k
i
u u
TU  UV  
1 
1 0
8
2 0

1 0 u
i 
 
1 0 u
j  
1 
1 u
k

2 0
8 0
8
2
u  0j
u  6k
u or 0, 0, 6
 0i

2, 14, 4  4, 0, 
2
2(4)  14(0)  4(
2)
80
80
2, 14, 4  
7, 1, 0
2(
7)  14(1)  4(0)

14  14  0  0
27. Sample answer:
u  
v , 
v , 
v 
Letu
v  v1, v2, v3 and 
v
1
2
3
u
u
u
j
k
i
u
u) 
v  (
v
v
v
v
1

2

31. Sample answer:
Let T(0, 0, 1), U(1, 0, 1), and V(
1, 
1, 
1).
u
TU  1, 0, 0
u
UV  
2, 
1, 
2
u
u
u
i
j
k
u u
TU  UV  1
0
0

2 
1 
2
0u
0u
0u
i 
 1
j  1
k
 0

1 
2

2 
2

2 
1
u  2j
u 
u
 0i
k or 0, 2, 
1

3


v1 
v2 
v3
v
v3 u
v
v2 u
3 u
2

i 
 v1
j  v1
k

v2 
v3

v1 
v3

v1 
v2
u 
 0j
u  0k
u0
 0i
u
u
u
i
j
k
u
u
28. a  (b  c ) 
a1
a2
a3
(b1  c1) (b2  c2) (b3  c3)
u
u
a3
a
a3
a1
2
i 

j

(b2  c2) (b3  c3)
(b1  c1) (b3  c3)
u

2 u
2 (b3 a1 c3) 
 a3 a(b
 [a
ck )]i
(b1  c1) (b2  2c2) 2 u
[a1 (b3  c3) 
 a3 (b1  c1)]j 
u
[a1 (b2  c2) 
 a2 (b1  c1)]k
u

 [(a2b3  a2c3) 
 (a3b2  a3c2)]i
u
[(a1b3  a1c3) 
 (a3b1  a3c1)]j
u
[(a1b2  a1c2) 
 (a2b1  a2c1)]k
u
 [(a2b3 
 a3b2)  (a2c3 
 a3c2)]i 

u
[(a1b3 
 a3b1)  (a1c3 
 a3c1)]j
u
[(a1b2 
 a2b1)  (a1c2 
 a2c1)]k
u
u
 (a2b3 
 a3b2)i  (a2c3 
 a3c2)i 

u 
 (a c 
 a c )u
(a1b3 
 a3b1)j
1 3
3 1 j 
u  (a c 
 a c )k
u
(a1b2 
 a2b1)k
1 2
2 1
u 
 (a b 
 a b )j
u
 [(a2b3 
 a3b2)i
1 3
3 1 
u
u

(a1b2 
 a2b1)k ] [(a2c3 
 a3c2)i
u
u
(a c 
 a c )j  (a c 
 a c )k ]
1 3







 ab

1 2

33a.

elbow

forearm

2 1



a3 u
i 
 a1 a3 u
j  a1 a2 u
k 
b3
b1 b3
b1 b2

a2 a3 u
i 
 a1 a3 u
j  a1
c2 c3
c1 c3
c1
u
(u
a  b ) (u
a u
c)

a2 u
k
c2

0.04 m

u u u
33b. T  AB  F
u
AB  0.04 cos (
30°), 0, 0.04 sin (
30°)
 0.02(3
), 0, 
0.02
u
F  0, 0, 600
u
u
u
i
j
k
u u
AB  F  0.023
 0 
0.02
0
0 600
u 
 123
u
 0i
u
j  0k
u
u u
T  AB  F   123
 or about 21 N-m
u
i u
j u
k
34. u
x u
y  2 3 0

1 1 4
 3 0u
i 
 2 0u
j  2 3u
k
1 4

1 4

1 1
u 
 8j
u  5k
u
 12i



2

600 N

30˚

a2 a3 u
i 
 a1 a3 u
j  a1 a2 u
k 
b2 b3
b1 b3
b1 b2
2



3 1

32. The expression is false.u
m u
n andu
n u
m have
the same magnitude but are opposite in direction.

1 u
A  2x
u
y



1

122 
(
8)2
 (5)2
 2
1

 2233

u
35a. o  120, 310, 60
u
c  29, 18, 21
35b. u
o u
c  120(29)  310(18)  60(21)

29. Sample answer:
Let T(0, 
2, 2), U(1, 2, 
3), and V(4, 0, 
1)
u
TU  1, 4, 
5
u
UV  3, 
2, 2
u
u
u
i
j
k
u u
TU  UV  1
4 
5
3 
2
2
4 
5 u
1 
5 u
1
4u
i 

j 
k


2
2
3
2
3 
2
u 
 17j
u 
 14k
u or 
2, 
17, 
14
 
2i

 $10,320

255

Chapter 8

u
u2 
 2a
u cos v
u2  b
u b
40. BA 2 a

36a.


(a1 
 
b1)2 
(a2 

b2)2 2
2
2
2
 a

a2
 
b12 
b22 2
1 
2
2

 2 a

a22 b

b22 cos v
1 
1 
(a1 
 b1)2  (a2 
 b2)2

F

 a12  a22  b12  b22 
 2 
a12 
a22

45˚


b12 
b22 cos v a12 
 2a1b1  b12

u u
36b. W  F  d
 cos v
W  120  4  cos 45°
W 
 339 ft-lb
u
37a. X  2 
 1, 5 
 0, 0 
 3 or 1, 5, 
3)
u
Y  3 
 2, 1 
 5, 4 
 0 or 1, 
4, 4
u u
u
i
j k
u u
X Y  1
5 
3
1 
4 4
5 
3 u
1 
3 u
1
5u

i 

j 
k

4
4
1
4
1 
4
u 
 7j
u 
 9k
u or 8, 
7, 
9
 8i

 a22 
 2a2b2  b22
 a12  a22  b12  b22

 2 
a12 
a22 
b12 
b22 cos v 

2a1b1 
 2a2b2
 
2 
a12 
a22 
b12 
b22 cos v
a1b1  a2b2
 
a12 
a22 
b12 
b22 cos v
a1b1  a2b2
u cos v u
u b
 a
a u
b
u cos v
u b
 a

u
41. AB  5 
 3, 3 
 3, 2 
 (
1) or 2, 0, 3
42. D(8, 3)
E(0, 
2)
u
DE  0 
 8, 
2 
 3) or 
8, 
5
u
(
8)2 
 (
5
)2
DE   
 89

43. 4x  y 
 6  0

A2  
B2  
42  12 or 17


37b. The cross product of two vectors is always a
vector perpendicular to the two vectors and the
plane in which they lie.
u u
38a. v u
p  (q
r)
u u u
i j
k
u
r  2 1 
4
q u

3 1 
5
1 
4 u
2 
4 u
2 1u

i 

j 
k
1 
5

3 
5

3 1
u 
 22j
u  5k
u or 
1, 
22, 5
 
i

417


17
617

 x  y 
  
17
17
17
617

p  17 
 1.46 units

17
sin f  1
7
1


tan f  4

u
u u
p  (q
r )  0, 0, 
1  
1, 
22, 5
 0(
1)  0(
22)  (
1)(5)
 
5 or 5 units3
0 0 
1
2 1 
4
38b.

3 1 
5
1 
4
2 
4
2 1

0

0
(
1)
1 
5

3 
5

3 1

4
17

cos f  17

f  14°
44. A  36°, b  13, and c  6
a2  b2  c2 
 2 bc cos A
a2  132  62 
 2(13)(6) cos 36°
a 
 8.9
sin 36°
sin B
 
 
8.9
13

 
5 or 5 units3
They are the same.
u u
39. Need (kv
w ) u
u  0.
[k1, 2  
1, 2]  5, 12  0
[k, 2k  
1, 2]  5, 12  0
k 
 1, 2k  2  5, 12  0
(k 
 1)5  (2k  2)12  0
5k 
 5  24k  24  0
29k  19  0
19
k  
29

Chapter 8

0

B 
 sin1

13 sin 36°

8.9

B 
 59.41° or 59°25

C 
 180° 
 36° 
 59°25

C 
 84.59° or 84°35

45.

h

tan 73°  4
4 tan 73°  h

13.1  h; 13.1 m
46. 3  3x

 4 10

3x

4 7

3x 
 4 49
x 17.67
4
47. 81  3
64  26  (22)3 or (23)2
4  22
2  21
9  32
So 64  43  82
The correct choice is B.

256

4

cos 73°  


4




cos 73°


  13.7 m

Page 511

Mid-Chapter Quiz

8-4B Graphing Calculator Exploration:

Finding Cross Products

1.
2.3 cm
46˚

Page 512

Fx 
 2.3 cos 46°

 1.6 cm

Fy 
 2.3 sin 46°

 1.7 cm

1. 49, 32 55
2. 168, 96, 76
3. 0, 0, 0
4. 11, 15, 3
5. 0, 0, 7
6. 0, 40, 0
u
u
7. u  x 
 6, 6, 12
u u
u
v  
 
62  62
 (
12)2

 216

8. u
u u
v 
 1, 13, 20
u u
u
v  
 
12  (
13)2
 (
20)2

 570

9. Sample answer: Insert the following lines after
the last line of the given program.
:Disp “LENGTH IS”
:Disp ((BZ  CY)2  (CX  AZ)2  (AY  BX)2)

2.
115˚
2.7 cm

Fy 
 27 sin 245°
Fx 
 27 cos 245°

 11.4 mm

 24.5 mm
u
3. CD 
 4  (9), 3  2 or 5, 5
u
52  (
5)2
CD  
 

 50 or 52

u
4. CD 
 5  3, 7  7, 2  (1) or 2, 0, 3
u
2  32
CD  
 
22  0

 13

u
u
5. u
r 
 t  2s

 6, 2  2 4, 3

 6, 2  8, 6

 6  8, 2  6 or 14, 8
u u
6. u
r 
 3u
v

 3 1, 3, 8  3, 9, 1

 3, 9, 24  3, 9, 1

 3  3, 9  9, 24  (1) or 6, 0, 25
7. 3, 6  4, 2 
 3(4)  6(2)

 12  12

 0; yes
8. 3, 2, 4  1, 4, 0 
 3(1)  (2)(4)  4(0)

38

 11; no
u
9. 1, 3, 2  2, 1, 1 
 u
u u
j
k
1 3
2
2 1 1
3
2u
1
2u
1
3u


i 
j 
k
1 1
2 1
2 1
u  5j
u  7k
u or 1, 5, 7 , yes

 i

8-5

Applications with Vectors

Pages 516–517

Check for Understanding

1. Sample answer: Pushing an object up the slope
requires less force because the component of the
weight of the object in the direction of motion is
mg sin v. This is less than the weight mg of the
object, which is the force that must be exerted to
lift the object straight up.
2. The tension increases.
3. Sample answer: Forces are in equilibrium if the
u
resultant force is O .
Current

4.
23 knots

17˚

u
u
5. F1 
 300i
u
u
F2 
 (170 cos 55°)u
i  (170 sin 55°)j
u u
F1  F2 
 (300


170 c
os 55°
)2  (1
70 sin
55°)2
421.19 N

1, 5, 7  1, 3, 2
(1)(1)  5(3)  (7)(2)
1  15  14 
 0
1, 5, 7  2, 1, 1
(1)(2)  5(1)  (7)(1)
2  5  7 
 0
10. Let X(2, 0, 4) and Y(7, 4, 6).
XY 
 
(7  2
)2  (4
 0)2
 (6 
 4)2

 45
 or about 6.7 m

tan v 


170 sin 55°

300  170 cos 55°

v 
 tan1 
 
300  170 cos 55° 
170 sin 55°

u
u
6. F1 
 50i
u
u
F2 
 100j
u u
F1  F2 
 
502 
1002
111.8 N
100
tan v 
 5
or
2
0
v 
 tan1 2
63.43°
7. horizontal 
 18 cos 40°
13.79 N
vertical 
 18 sin 40°
11.57 N

257

Chapter 8

u
u  (33 sin 90°)j
u or 33j
u
8. F1 
 (33 cos 90°)i
u
u  (44 sin 60°)j
u
F2 
 (44 cos 60°)i
u
u
or 22i  22 3
j
u u
2
F1  F2 
 
22 
(33 
223

)2
74 N
33  223


u
u
v 1 
 (115 cos 60°)i  (115 sin 60°)j
15. u
u
u
or 57.5i  57.53
j
u  (115 sin 120°)j
u
u
v2 
 (115 cos 120°)i
u
u
or 57.5i  57.53
j
2  (1153
u u
v
v



0

)2
1
2

 1153

199.19 km/h
Since tan v is undefined and the vertical
component is positive, v 
 90°.
16. The force must be at least as great as the
component of the weight of the object in the
direction of the ramp. This is 100 sin 10°, or about
17.36 lb.
u
u
17. F 
 105i

3  23


tan v 
 22 or 2
v 
 tan1 
2
3  23


73°
A force with magnitude 74 N and direction 73° 
180° or 253° will produce equilibrium.
4 mph

9a.

1

u
u  (110 sin 50°)j
u
F2 
 (110 cos 50°)i
u u
F1  F2
 
(105 
110 cos

50°
)2  (1
10 sin
50°)2
194.87 N

12 mph

110 sin 50°

105  110 cos 50°
110 sin 50°
tan1 
105  110 cos 50°

tan v 

9b. If v is the angle between the resultant path of
the ferry and the line between the landings,
1
4
1 1, or about

then sin v 
 1
3
2 or 3 . So v 
 sin
19.5°.

Pages 517–519

sin1 
7
5 
v
52.1

44° v
u
u  (250 sin 25°)j
u
19. F1 
 (250 cos 25°)i
u
u
u
F2 
 (45 cos 250°)i  (45 sin 250°)j
u u
F1  F2


Wind

11.

27˚

42 N

256 mph
53˚

(250
cos

25°
 45 
cos 25
0°)2 
(250 s
in 25°
 45 sin

25
0°)2

342 lb

12.



25.62°
F 
 w sin v
52.1 
 75 sin v
52.1
 
 sin v
75

18.

Exercises

10.




v


220.5 lb
tan v 


94˚

250 sin 25°  45 sin 250°

250 cos 25°  45 cos 250°
250 sin 25°  45 sin 250°

v 
 tan1 
 
250 cos 25°  45 cos 250° 
16.7°
u
u  (70 sin 330°)j
u or 353
u
20. F1 
 (70 cos 330°)i
u
i  35j

454 lb

u
u  (40 sin 45°)j
u or 202
F2
 (40 cos 45°)i
u
i  202
u
j
u
u
u
F 
 (60 cos 135°)i  (60 sin 135°)j or 30 2
  302
u
j
3

u
u
13. F1 
 425i
u
u
F 
 390j

tan v 

v


2

u u
F1  F2 
 
4252 
3902
576.82 N
390



37.5°

78

58.6 lb
u
u
u
21. F1 
 (23 cos 60°)i  (23 sin 60°)j
u
u
or 11.5i  11.53j

u
u  (23 sin 120°)j
u
F2 
 (23 cos 120°)i
u
u
or 11.5i  11.53j

u u
F1  F2 
 
02  (
233

)2

 233

39.8 N
Since tan v is undefined and the vertical
component is positive , v 
 90°. A force with
magnitude 39.8 N and direction 90°  180° or
270° will produce equilibrium.

78

v 
 tan1 85
42.5°
u
u
14. v 1 
 65i

u  (50 sin 300°)j
u or 25i
u 253
u
v 2 
 (50 cos 300°)i
u
j

u u
v
v2 
 
902  (253
 )2
1
99.87 mph
53



tan v 
 9
0 or  18

v 
 tan1 
1
8 
5
3

A positive value for v is about 334.3°.

Chapter 8




u u u
F1  F2  F3 
 
(353

 10
)2  
2
(35 
502

)2



tan v 
 
425 or 85

25
3

35  502


353
  102

35  502

1

tan
353
  102


258

22. a 
 g sin 40°

 32 sin 40°
20.6 ft/s2
u
u  (36 sin 20°)j
u
23. F1 
 (36 cos 20°)i
u
u
F 
 (48 cos 222°u
i  (48 sin 222°)j

F1 


760 lb
cos 174.5°


F2 
 
cos 6.2° F1

761 lb
30. Sample answer: Method b is better. Let F be the
force exerted by the tractor, T be the tension in
the two halves of the rope, and v be the angle
between the original line of the rope and half of
the rope after it is pulled. At equilibrium,
F

2T sin v  F 
 0, or T 
 
v 30°,
2 sin v . So, if 0°
the force applied to the stump using method b is
greater than the force exerted by the tractor.
31. Let T be the tension in each towline and suppose
the axis of the ship is the vertical direction.
2T sin 70°  6000 
 0
6000

T

2 sin 70°

2

u u
F1  F2



cos
20° 
48 co
s 222°
)2  (3
6 sin 2
0°  4
8 sin 2
22°)2
(36
19.9 N

tan v 

v


36 sin 20°  48 sin 222°

36 cos 20°  48 cos 222°
36 sin 20°  48 sin 222°
tan1 
36 cos 20°  48 cos 222°






264.7° or 5.3° west of south
24a. 135 lb
165˚
70 lb

75˚

120˚

3192.5 tons
32. Let T be the tension in each wire. The halves of
the wire make angles of 30° and 150° with the
horizontal.
T sin 30°  T sin 150°  25 
 0

115 lb

u
u
24b. F1 
 70i
u
u  (135 sin 165°)j
u
F2 
 (135 cos 165°)i
u u u
F1  F2  F3 


1
T
2

134.5 lb
135 sin 165°  115 sin 240°

70  135 cos 165°  115 cos 240°
135 sin 165°  115 sin 240°

v 
 tan1 
 
70  135 cos 165°  115 cos 240° 

2v02

2  1002


 3
sin 65° cos 65°
2
239.4 ft
36. Sample answer: A plot of the data suggests a
quadratic function. Performing a quadratic
regression and rounding the coefficients gives
y 
 1.4x2  2x  3.9.
37. b  0.3
b
(0.33, 0.33)
p  0.2
0.5
(0.2, 0.46)
b  p  0.66
bp
0.4
0.3

1


27a. tan v 
 1
8 or 6

0.2

1

v 
 tan1 6
9.5° south of east
27b. s 


(0.3, 0.3)

p
O 0.1 0.2 0.3 0.4
The vertices are at (0.2, 0.3), (0.3, 0.3), (0.33, 0.33)
and (0.2, 0.46).
cost function C(p, b) 
 90p  140b  32(1  p  b)

 32  58p  108b
C(0.2, 0.3) 
 32  58(0.2)  108(0.3) or 76
C(0.3, 0.3) 
 32  58(0.3)  108(0.3) or 81.8
C(0.33, 0.33) 
 32  58(0.33)  108(0.33) or 86.78
C(0.2, 0.46) 
 32  58(0.2)  108(0.46) or 93.28
The minimum cost is $76, using 30% beef and
20% pork.

18.2 mph
28. F cos v 
 100 cos 25°
90.63 N
29. F1 cos 174.5°  F2 cos 6.2° 
 0
F1 sin 174.5°  F2 sin 6.2°  155 
 0
cos 174.5°


The first equation gives F2 
 
cos 6.2° F1.
Substitute into the second equation.
cos 174.5° sin 6.2°

cos 6.2°

(0.2, 0.3)

0.1


182 
32

F1 sin 174.5° 

0.

35. d 
 g sin v cos v

208.7° or 28.7° south of west
u u u
Since F1  F2  F3 0, the vectors are not in
equilibrium.
u
25. W 
 F u
d
u  (1600 sin 50°)j
u]  1500i
u

 [(1600 cos 50°)i

 (1600 cos 50°)(1500)  (1600 sin 50°)(0)
1,542,690 N-m
26a. Sample answer: The horizontal forward force is
u
F cos v. You can increase the horizontal forward
force by decreasing the angle v between the
handle and the lawn.
26b. Sample answer: Pushing the lawnmower at a
lower angle may cause back pain.
3

1

 2T  25 
 0
T 
 25 lb
33. u
u
 u
v 
 9(3)  5(2)  3(5)

 2
The vectors are not perpendicular sinceu
u u
v
u
34. AB 
 0  12, 11  (5), 21  18

 12, 6, 3

(70 
135 c
os 165
°  11
5 cos 2
40°)2 
(135 s
in 165
°  11
5 sin 2
40°)2


tan v 


155

sin 174.5°  cos 174.5° tan 6.2°

F1  155 
 0

F1 (sin 174.5°  cos 174.5° tan 6.2°) 
 155

259

Chapter 8

11b. 50  10t 
 0
50 
 10t
5
t
When t 
 5, the coordinates of the defensive
player are (10  0.9(5), 54  10.72(5)) or (5.5,
0.4), so the defensive player has not yet caught
the receiver.

38. *4  *(3) 
 (43  4)  [(3)3  (3)]

 60  (24)

 84
The correct choice is A.

Vectors and Parametric Equations

8-6

Pages 523–524

524–525

Check for Understanding

1. When t 
 0, x 
 3 and y 
 1. When t 
 1, x 
 7
and y 
 1. The graph is a line through (3, 1) and
(7, 1).
2. Sample answer: For every single unit increment of
t, x increases 1 unit and y increases 2 units. Then,
the parametric equations of the line are x 
 3  t,
y 
 6  2t.
3. When t 
 0, x 
 1 and y 
 0, so the line passes
through (1, 0). When t 
 1, x 
 0 and y 
 1, so
the line passes through (0, 1), its y-intercept. The
10


slope of the line is 
0  1 or 1.

4. x  (4), y  11 
 t 3, 8
x  4, y  11 
 t 3, 8
x  4 
 3t
y  11 
 8t
x 
 3t  4
y 
 8t  11
5. x  1, y  5 
 t 7, 2
x  1 
 7t
y  5 
 2t
x 
 1  7t
y 
 5  2t
6. 3x  2y 
 5
7. 4x  6y 
 12
2y 
 3x  5
6y 
 4x  12
3
5
2
y 
 2x  2
y 
 3x  2
x
t
3
5
y 
 2t  2

x
t
2
y 
 3t  2

x 
 4t  3
x  3 
 4t

8.
1

9. x 
 9t
x
 
 t
9

3

4x  4 
 t
y 
 5t  3

y 
 4t  2
x
y 
 4
9  2

y 
 5
4x  4  3
1

y

10.

5
4x

t
1
0
1
2



3

4

y 
 9x  2

3

4

x
2
2
6
10

y
2
1
0
1

y
11a. receiver:
x  5 
 0t
y  50 
 10t
x
5
y 
 50  10t
O
defensive player:
x  10 
 0.9t
y  54 
 10.72t
x 
 10  0.9t
y 
 54  10.72t

Chapter 8

Exercises

12. x  5, y  7 
 t 2, 0
x  5 
 2t
y  7 
 0t
x 
 5  2t
y
7
13. x  (1), y  4 
 t 6, 10
x  1, y  4 
 t 6, 10
x  1 
 6t
y  4 
 10t
x 
 1  6t
y 
 4  10t
14. x  (6), y  10 
 t 3, 2
x  6, y  10 
 t 3, 2
x  6 
 3t
y  10 
 2t
x 
 6  3t
y 
 10  2t
15. x  1, y  5 
 t 7, 2
x  1 
 7t
y  5 
 2t
x 
 1  7t
y 
 5  2t
16. x  1, y  0 
 t 2, 4
x  1, y 
 t 2, 4
x  1 
 2t
y 
 4t
x 
 1  2t
17. x  3, y  (5) 
 t 2, 5
x  3, y  5 
 t 2, 5
x  3 
 2t
y  5 
 5t
x 
 3  2t
y 
 5  5t
18. x 
 t
y 
 4t  5
19. 3x  4y 
 7
20. 2x  y 
 3
4y 
 3x  7
y 
 2x  3
3
7
y 
 2x  3
y 
 4x  4
x
t
x
t
y 
 2t  3
3
7
y 
 4t  4
21. 9x  y 
 1
y 
 9x  1
x
t
y 
 9t  1

22. 2x  3y 
 11
3y 
 2x  11
2
11
y 
 3x  3
x
t
2
11
y 
 3t  3

23. 4x  y 
 2
y 
 4x  2
x
t
y 
 4t  2

24. 3x  6y 
 8
6y 
 3x  8
1
4
y 
 2x  3
1

The slope is 2.
1

y  5 
 2(x  2)
1

x

y 
 2x  6
x
t
1

y 
 2t  6

260

25. x 
 2t
x
 
 t
2

1

x  7 
 2t
2x  14 
 t
y 
 3t
y 
 3(2x  14)
y 
 6x  42

y
1t
x
y 
 1  2
1

y 
 2x  1
27.

x 
 4t  11
x  11 
 4t
1
11
x    
 t
4
4
y
t3
1
11
y 
 4x  4  3
1

23

y 
 4x  4
29.

33.

1

x 
 7  2t

26.

28.

x 
 4t  8
x  8 
 4t
1
x  2 
 t
4
y
3t
1
y 
 3  4x  2
1

y 
 4x  5

x 
 3  2t
x  3 
 2t
1
3
x   
 t
2
2

[10, 10] Tstep1
[20, 20] Xscl2
[20, 20] Yscl2

y 
 1  5t
1
3
y 
 1  5 
2x  2
5

34.

17

y 
 2x  2
30. Regardless of the value of t, x is always 8, so the
parametric equations represent the vertical line
with equation x 
 8.
31a. x  11, y  (4) 
 t 3, 7
x  11, y  4 
 t 3, 7
31b. x  11 
 3t
y  4 
 7t
x 
 3t  11
y 
 7t  4
31c.
x 
 3t  11
x  11 
 3t
1
x
3

11

 3 
 t

y 
 7t  4
1
11
y 
 7
3x  3  4
7

89

y 
 3x  3

[10, 10] Tstep1
[10, 10] Xscl1
[10, 10] Yscl1
35a. x 
 2  3t and y 
 4  7t
If t  0, then x  2 and y  4, so the part of the
line to the right of point (2, 4) is obtained.
35b.
x 0
2  3t 0
3t 2
2
t 3

32.

36. x  y 
 cos2 t  sin2 t

1
0  cos2 t  1 and 0  sin2 t  1, so the graph is
the segment of the line with equation x  y 
 1
from (1, 0) to (0, 1).

y
[5, 5] Tstep1
[10, 10] Xscl1
[10, 10] Yscl1

1

( 12 , 12 )

x

1

261

Chapter 8

37a. target drone:
x 
 3  (1)t
x
3t
missile:
x
2t
37b. 3  t 
 2  t
1 
 2t
1
 
 t
2

45. The slope is 1.
y  1 
 1[x  (3)]
y1
x3
xy4
0
46. The linear velocity of the belt around the larger

y 
 4  0t
y
4
y 
 2  2t

pulley is (120 rpm)
2  2 in./rev 
 1080
9

in./min. The linear velocity around the smaller
pulley must be the same, so its angular velocity is

1

When t 
 2, the missile has a y-coordinate of 3,
not 4, so it does not intercept the drone.
38a. Ceres: x 
 1  t, y 
 4  t, z 
 1  2t
Pallas: x 
 7  2t, y 
 6  2t, z 
 1  t
38b. Adding the equations for x and y for Ceres gives
x  y 
 3. Subtracting the equations for x and y
for Pallas results in x  y 
 1. The solution of
this system is x 
 1 and y 
 2. Eliminating t
from the equations for y and z results in the
system 2y  z 
 7, y  2z 
 4 which has
solution y 
 2 and z 
 3. Hence, the paths cross
at (1, 2, 3).
38c. 1  t 
 1 v t 
 2
7  2t 
 1 v t 
 4
Ceres is at (1, 2, 3) when t 
 2 but Pallas is at
(1, 2, 3) when t 
 4. The asteroids will not
collide.


(1080 in./min)

2  3 in.  
 180 rpm. The correct
1 rev

choice is D.

8-6B Graphing Calculator Exploration:

Modeling with Parametric
Equations

Page 526
1. 408.7t 
 418.3(t  0.0083)
408.7t 
 418.3t  3.47189
9.6t 
 3.47189
3.47189


t

9.6

0.362 hr or 21.7 min
2. d 
 rt
3.47189


 408.7

9.6 
147.8 mi

39. The line is parallel to the vector 0  
3, 5  1,
1

1
,
3

8  1 or

line is x  


4, 9 . The vector equation of the

, y 
 1, z  1

1
3

1

x  3, y  1, z  1 
 t
1

1

x


1
3

x  3 
 3t

1
,
3


t

1
,
3

500


3. The time for plane 1 to fly 500 miles is 
408.7 . The

4, 9 or

500


time for plane 2 is 
418.3  0.0083. Suppose the
speed of plane 1 is increased by a mph.

4, 9 .

y  1 
 4t


1
t
3

500

408.7  a

y 
 1  4t

500




418.3  0.0083

1

 
500
  0.0083
418.3
500
a 
 
 408.7
500
  0.0083
418.3

408.7  a

500

z  1 
 9t
z 
 1  9t
u
u
u
40. v 1 
 (150 cos 330°)i  (150 sin 330°)j
u  (50 sin 245°)j
u
u
v 
 (50 cos 245°)i
2

6.7 mph

u u
v
v2 

1

(150 c
os 330
°  50
cos 2
45°)2 
 (15
0 sin 
330° 
50 sin
245°)2


162.2 km/h
tan v 


150 sin 330°  50 sin 245°

150 cos 330°  50 cos 245°

8-7

150 sin 330°  50 sin 245°

v 
 tan1 
 
150 cos 330°  50 cos 245° 

Page 531

47°53
4 or 47°53
4 south of east
41. 1, 3  3, 2 
 1(3)  3(2)

 3
Since the inner product is not 0, the vectors are
not perpendicular.
42. Since A 90°, a b, and a b sin A, no solution
exists.
43. A graphing calculator indicates that there is one
real zero and that it is close to 1. f(1) 
 0, so the
zero is exactly 1.
3
3

x  2 
 2y
2
x
3

4

 3 
 y
2

4

y 
 3x  3
Chapter 8

Check for Understanding

1. Sample answer: a rocket launched at 90° to the
horizontal; tip-off in basketball
2. Equal magnitude with opposite direction.
3. The greater the angle of the head of the golf club,
the greater the angle of initial velocity of the ball.
u  
 v
u sin v
u  
 v
u cos v
4. v
5. v
y
x

 50 sin 40°

 20 cos 50°
32.14 ft/s
12.86 m/s
u  
 v
u cos v
u  
 v
u sin v
6. v
v
x
y

 45 cos 32°

 45 sin 32°
38.16 ft/s
23.85 ft/s
u  
 v
u cos v
u  
 v
u sin v
7. v
v
x
y

 7.5 cos 20°

 7.5 sin 20°
7.05 m/s
2.57 m/s

x 
 2y  2

44.

Modeling Motion Using
Parametric Equations

262

16. To find the time the projectile stays in the air, set
y 
 0 and solve for t.
u sin v  1gt2 
 0
tv
2

 
8a. 300 mph 

mile 
 3600 s  
 440 ft/s
u cos v
x 
 tv
5280 ft

h

x 
 t(440) cos 0°
x 
 440t
u sin v  1gt2  h
y 
 tv

u sin v  gt) 
 0
t(v
2
1

u sin v  1gt 
 0
v
2
u sin v 
 1gt
v

2

1

y 
 t(440) sin 0°  2(32)t2  3500

u sin v
2v

g

 3500
y

8b. Sample graph:
y
16t2

4000
3000
Height
(feet) 2000
1000

x

g




3500


t2 
 
16

Pages 531–533




16
3500

14.8 s

1084
u 
 

v
7 sin 78°
u 158.32 ft/s
v
1 u
 cos v  50 yd
17b. x 
 3tv

Exercises

u  
 v
u cos v
9. v
x

 65 cos 60°

 32.5 ft/s
u  
 v
u cos v
10. v
x

 47 cos 10.7°
46.18 m/s
u  
 v
u cos v
11. v
x

 1200 cos 42°
891.77 ft/s
u  
 v
u cos v
12. v
x

 17 cos 28°
15.01 ft/s
u
u cos v
13. vx 
 v

 69 cos 37°
55.11 yd/s
u  
 v
u cos v
14. v
x

 46 cos 19°
43.49 km/h
u
15a. x 
 tv  cos v


 3(7)(158.32) cos 78°  50
127 yd
u cos v
18.
x 
 tv
x
 
 t
u
v  cos v

u sin v  gt2
y 
 tv
2
2
x
u sin v  1g 
x
y 
  v
1

u cos v
v

y 
 x tan v 

2
g

u2 cos2 v
2v
of the x2-term

u cos v
v

The presence
(due to the force of
gravity) means that y is a quadratic function of x.
Therefore, the path of a projectile is a parabolic
arc.
19. To find the time the projectile stays in the air if
the initial velocity isu
v, set y 
 0 and solve for t.
u sin v  1gt2 
 0
tv
2
u sin v  1gt 
 0
t 
v
2
u sin v  1gt 
 0
v
2

2

y 
 175t sin 35° 
y
0
175t sin 35°  16t2 
 0
t(175 sin 35°  16t) 
 0
175 sin 35°  16t 
 0
175 sin 35° 
 16t
175 sin 35°

16

1

u  
 v
u sin v
v
y

 65 sin 60°

 56.29 ft/s
u  
 v
u sin v
v
y

 47 sin 10.7°
8.73 m/s
u  
 v
u sin v
v
y

 1200 sin 42°
802.96 ft/s
u  
 v
u sin v
v
y

 17 sin 28°
7.98 ft/s
u
u sin v
vy 
 v

 69 sin 37°
41.53 yd/s
u  
 v
u sin v
v
y

 46 sin 19°
14.98 km/h
1
u
y 
 tv  sin v  gt2

x 
 175t cos 35°

15b.

u2g sin 2v
v

g

As the angle increases from 0° to 45°, the
horizontal distance increases. As the angle
increases from 45° to 90°, the horizontal distance
decreases.
17a. y 
 300 when t 
 7
u sin 78°  1(32)72 
 300
7v
2
u sin 78°  784 
 300
7v
u sin 78° 
 1084
7v

8c. 16t2  3500 
 0
16t2 
 3500

t
8d. x 
 440t

 440(14.8)

 6512 ft


t

The greater the angle, the greater the time the
projectile stays in the air. To find the horizontal
distance covered, substitute the expression for t in
the equation for x.
u cos v
x 
 tv
u sin v
2v
u cos v

  v

2000 4000 6000
Horizontal Distance
(feet)

t


2

16t2

u sin v 
 gt
v
2
1

u sin v
2v

g
t


t

To find the range, substitute this expression for t
in the equation for x.
u cos v
x 
 tv


t

u

2v  sin v u
 cos v

 g v

x 
 175t cos 35°

u2 sin 2v
v


 175 
1
 cos 35°
6
175 sin 35°


 g

899.32 ft or 299.77 yd

263

Chapter 8

22b.

If the magnitude of the initial velocity is doubled
u)2 sin 2v
(2v
u, the range becomes 
 or
to 2v
g

u2 sin 2v
v

4 g. The projectile will travel four times as
far.
 
20a. 800 km/h 

km 
 3600 s 
u
x 
 tv  cos v
h

1000 m

222.2 m/s

x 
 222.2 t cos 45°

23a. y 
 300 when t 
 4.8
u sin 82°  1(32)(4.8)2 
 300
4.8v
2
u sin 82°  368.64 
 668.64
4.8v

u sin v  gt2
y 
 tv
2
1

1

y 
 222.2t sin 45°  2(9.8)t2

668.64
u 
 

v
4.8 sin 82°
u 140.7 ft/s
v

y 
 222.2t sin 45°  4.9t2
The negative coefficient in the t-term in the
equation for y indicates that the aircraft is
descending. The negative coefficient in the
equation for x is arbitrary.
20b. y 
 222.2t sin 45°  4.9t2

 222.2(2.5) sin 45°  4.9 (2.5)2
423.4
The aircraft has descended about 423.4 m.
20c.

423.4 m

2.5 s

u cos v  100
23b. x 
 3tv
1

131.3 yd
u cos v
24a. x 
 tv

x 
 155t cos 22°
y 
 155t sin 22°  16t2  3
24b.
x 
 420
155t cos 22° 
 420
420

t

155 cos 22°

169 m/s

or
km
3600 s
 
169 m/s 

1000 m 
 h  
 608.4 km/h
21a. 70 mph 


5280 ft

mi







h

3600 s

y 
 155t sin 22°  16t2  3
420


2

t

t

t
3.84 s
u cos v
x 
 tv
323.2 ft
21b.
y
8
308
t
3 sin 35°  16t2  10 
 8
308

n 35° 
 4(16)2



3038 si

2

t 
 
t 3.71 s
u cos v
x 
 tv
312.4 ft
21c. From the calculations in part b, the time is
about 3.71s.
y
22a.

17.4


27. cos A 
 
21.9
17.4


A 
 cos1 
21.9

A

t

O

Chapter 8

155 sin 22°  
(155 s
in 22°
)2  4
(16)
3


3.68 s
u cos v
x 
 tv
528.86 ft
25.
x 
 11  t
x  11 
 t
x  11 
 t
y 
 8  6t
y 
 8  6(x  11)
y 
 6x  58
26a. mg sin v 
 300(9.8) sin 22°
1101.3 N
26b. mg cos v 
 300(9.8) cos 22°
2725.9 N

16t2  t
3 sin 35°  2 
 0
308
3 sin 35° 

420

36.04 ft
Since 36.04  15, the ball will clear the fence.
24c.
y
0
155t sin 22°  16t2  3 
 0

sin 35°  
3 sin 35°  4(16)10
t 
 
308

2




 155 

155 cos 22°  sin 22°  16 
 155 cos 22°   3

308

3

ft/s
y
0
308
t
3 sin 35°  16t2  10 
 0
308

3

u sin v  1gt2  h
y 
 tv
2

x

264

37°

u
5a. BE 
 0  5, 2  5, 4  0 or 5, 3, 4
A(5, 5  (3), 0) 
 A(5, 2, 0)
C(5  (5), 5, 0) 
 C(0, 5, 0)
D(5  (5), 5  (3), 0) 
 D(0, 2, 0)
F(5, 5  (3), 0  4) 
 F(5, 2, 4)
G(5, 5, 0  4) 
 G(5, 5, 4)
H(5  (5), 5, 0  4) 
 H(0, 5, 4)
5 5 0 0 0 5 5 0
The matrix is 2 5 5 2 2 2 5 5 .
0 0 0 0 4 4 4 4

28. 2(2x  y  z) 
 2(2)
x  3y  2z 
 3.25
5x  y

 0.75
1(2x  y  z) 
 1(2)
4x  5y  z 
 2.5
6x  4y

 0.5
4(5x  y) 
 4(0.75)
6x  4y 
 0.5
14x 
 3.5
3.5
x 
 14



x 
 0.25
5x  y 
 0.75
2x  y  z 
 2
5(0.25)  y 
 0.75
2(0.25)  0.5  z 
 2
y 
 0.5
z
1
2
2
29.   5    3 
 25  9

 16
The correct choice is B.

Page 534
1.

7°12


360°

5b.

History of Mathematics

5c.

7.2°




360°
1

5000 stadia


 x

x 
 50(5000)
x 
 250,000 stadia
250,000(500) 
 125,000,000 ft
125,000,000 5280 23,674 mi
The actual circumference of Earth is about
24,901.55 miles.
2. See students’ work. No solution exists.
3. See students’ work.

8-8





1
0 0
0 1 0
0
0 1
5
5

 2 5
0
0



5 5 0 0 0
2 5 5 2 2
0 0 0 0 4
0
0
0
5 2 2
0
0
4

E

F
C

B





G

x

5
2
4
5
2
4



5 0
5 5
4 4
0
5
5 5
4
4



z

O
D

y

A

The image is the reflection over the xz-plane.
5d. The dimensions of the resulting figure are half
the original.
6a. The scale factor of the dilation is 4. The
translation increases x-coordinates by 2. The
matrices are
4 0 0
D 
 0 4 0 and
0 0 4
2 2 2 2 2 2 2 2
T
 0 0 0 0 0 0 0 0 .
0 0 0 0 0 0 0 0

Check for Understanding




1. Matrix T multiplies x-coordinates by 2 and
y- and z-coordinates by 2, so it produces a
reflection over the yz-plane and increases the
dimensions two-fold.
u
2. CC 
 
 8  6, 8  7, 2  3 or 2, 1, 1
2
2
2
2
2
2
The matrix is
1
1
1
1
1
1 .
1 1 1 1 1 1





H

Transformation Matrices in
Three-Dimensional Space

Pages 539–540

54
54
04
04
2  (1) 5  (1) 5  (1) 2  (1)
02
02
02
02
04
54
54
04
2  (1) 2  (1) 5  (1) 5  (1)
42
42
42
42
9 9 4 4 4 9 9 4

 1 4 4 1 1 1 4 4
2 2 2 2 6 6 6 6




 5
0
1

50









1 0
0
0 1
0 
 T, so the transformations
0 0 1
are the same.

3. VU 






4a-c.
Transformation
Reflection
Translation
Dilation

Orientation
yes
no
no

Site
no
no
yes

Shape
no
no
no

265

Chapter 8

z

6b. Sample answer: If the original prism has
vertices A(3, 3, 0) B(3, 3, 3), C(3, 3, 3),
D(3, 3, 0), E(5, 3, 0), F(5, 3, 3), G(5, 3, 3),
and H(5, 3, 0), then the image has vertices
A
(10, 12, 0), B
(10, 12, 12), C
(10, 12, 12),
D
(10, 12, 0), E
(22, 12, 0), F
(22, 12, 12),
G
(22, 12, 12), and H
(22, 12, 0).

F

D

B
D
 C
G D
A
F
 F
E
H

G


H


H
A

B

G

E

z
C


C

O
y

x

B


The result is a translation of 2 units along the yaxis and 4 units along the z-axis.

A


01
01
01
01
11. 0  (2) 3  (2) 3  (2) 0  (2)
1  (2) 2  (2) 5  (2) 4  (2)



21
0  (2)
1  (2)
1 1 1
1

 2 1 1 2
1 0 3
2
z

E

x



Pages 540–542

Exercises

u
7. FB 
 3  3, 1  7, 4  4 or 0, 6, 0
A(2, 3,  (6), 2) 
 A(2, 3, 2
C(4, 7  (6), 1) 
 C(4, 1, 1)
2 3
4
4 2 3
The matrix is 3 1
1
7 3 7 .
2 4 1 1 2 4
u
8. AH 
 4  (3), 1  (2), 2  2 or 7, 3, 4
B(3, 2  3, 2) 
 B(3, 1, 2)
C(3, 2  3, 2  (4)) 
 C(3, 1, 2)
D(3, 2, 2  (4)) 
 D(3, 2, 2)
E(3  7, 2, 2  (4)) 
 E(4, 2, 2)
F(3  7, 2, 2) 
 F(4, 2, 2)
G(3  7, 2  3, 2) 
 G(4, 1, 2)
3 3 3 3
4
4 4
4
The matrix is 2
1
1 2 2 2 1
1
2
2 2 2 2
2 2 2
u
9. CF 
 6  4, 0  (1), 0  2 or 2, 1, 2
D(2  2, 2  1, 3  (2)) 
 D(4, 1, 1)
E(1  2, 0  1, 4  (2)) 
 E(3, 1, 2)
2 1
4
4 3 6
The matrix is 2 0 1 1 1 0 .
3 4
2
1 2 0



F


A






E


Chapter 8



C

G

H


B


y







D


00
00
00
00
10. 0  (2) 3  (2) 3  (2) 0  (2)
14
24
54
44
20
20
20
20
0  (2) 0  (2) 3  (2) 3  (2)
14
44
54
24
0 0 0
0
2
2 2 2

 2 1 1 2 2 2 1 1
5 6 9
8
5
8 9 6







The result is a translation of 1 unit along the xaxis, 2 units along the y-axis, and 2 units
along the z-axis.
01
01
01
01
12.
05
35
35
05
1  (3) 2  (3) 5  (3) 4  (3)
21
21
21
21
05
05
35
35
1  (3) 4  (3) 5  (3) 2  (3)
1
1 1 1
3 3 3
3


5
8 8 5
5 5 8
8
2 1 2 1 2 1 2 1
z





21
21
3  (2) 3  (2)
5  (2) 2  (2)
3 3
1 1
3 0

x





D


21
0  (2)
4  (2)
3
3
2 2
1
2

F

x

A

E




C

G

H


y

B


The results is a translation of 1 unit along the xaxis, 5 units along the y-axis, and 3 units along
the z-axis.

266







1 0 0 0 0 0
13. 0 1 0 0 3 3
0 0 1 1 2 5
0 0 0 0 2

 0 3 3 0 0
1 2 5 4 1



z
F


0
0
4
2
0
4

2
0
1
2
3
5



0.75
0
0
1
0
0
0 0.75
0 

0 1
0
0
0 0.75
0
0 1
0.75
0
0
0 0.75
0 , so the figure is three-fourths
0
0 0.75
the original site and reflected over all three
coordinate planes.
2x
2 0 0 x
19a. 2y 
 0 2 0 y , so the transformation can
5z
0 0 5 z
2 0 0
be represented by the matrix 0 2 0 .
0 0 5
19b. The transformation will magnify the x- and
y-dimensions two-fold, and the z-dimension
5-fold.
23.6 23.6 23.6 23.6 23.6
20a.
72
72
72
72
72
0
0
0
0
0
20  23.6 136  23.6 247  23.6
20b. 58  72
71  72
74  72
27  0
53  0
59  0
302  23.6 351  23.6
83  72
62  72
37  0
52  0
43.6 159.6 270.6 325.6 374.6


14
1
2
11
10
27
53
59
37
52
20c. The result is a translation 23.6 units along the
x-axis and 72 units along the y-axis.
1 0 0
21. The matrix
0 1 0 would reflect the prism
0 0 1
0.5
0
0
over the yz-plane. The matrix
0 0.5
0
0
0 0.5
would reduce its dimensions by half.
0.5
0
0 1 0 0
0.5
0
0
0 0.5
0
0 1 0 

0 0.5
0
0
0 0.5
0 0 1
0
0 0.5
22a. Placing a non zero element in the first row and
third column will skew the cube so that the top
is no longer directly above the bottom.
1 0 1
Sample answer: 0 1 0
0 0 1
22b. Sample graphs:

2 2 2
0 3 3
4 5 2
2
3
2



18.

C


D


G


A


H


E


y

The transformation does not change the figure.
1
0
0 0 0 0 0 2 2 2 2
14. 0 1
0 0 3 3 0 0 0 3 3
0
0 1 1 2 5 4 1 4 5 2
0
0
0
0
2
2
2
2


0 3 3
0
0
0 3 3
1 2 5 4 1 4 5 2
z6
3
3 6 9
96 3O
y

A
9 3
12 6
B

x E
H
D





F





The transformation results in reflections over the
xy  and xz-planes.
1
0
0 0 0 0 0 2 2 2 2
15.
0 1
0 0 3 3 0 0 0 3 3
0
0 1 1 2 5 4 1 4 5 2
0
0
0
0 2 2 2 2


0 3 3
0
0
0 3 3
1 2 5 4 1 4 5 2
z



x

H





G

C


F


 













 



D


The transformation results in reflections over all
three coordinate planes.
16. The matrix results in a dilation of scale factor 2,
so the figure is twice the original size.
3 0
0
1 0
0 3 0 0
17. 0 3
0 
 0 1
0 0 3 0 , so the
0 0 3
0 0 1 0 0 3
figure is three times the original size and reflected
over the xy-plane.









y

A








E


B


 



C








G



 



 

B


x










z
F

G



C
H

x

D

z

B
E

y

F


G


A
C


E


B


y

H


x

267

D


A


Chapter 8

80x3  80x2  80x 
 24.2
80x3  80x2  80x  24.2 
 0
A graphing calculator indicates that there is a
solution between 0 and 1. By Descartes’ Rule of
Signs, it is the only solution. When x 
 0.2, 80x3 
80x2  80x  24.2 
 4.36 and when x 
 0.3,
80x3  80x2  80x  24.2 
 9.16. So the solution
to the nearest tenth is 0.2.
30. Divide each side of the equations by 2, 3, 4, and 6,
respectively, so that the left side is x  2y.
I. x  2y 
 4
II. y 
 4
8
III. x  2y 
 2
IV. x  2y 
 3

23. The first transformation reflects the figure over
all three coordinate planes. The second
transformation stretches the dimensions along the
y- and z-axes and skews it along the xy-plane.
(The first row of T changes the x-coordinate of
(x, y, z) to x  2z.)
24. To multiply the x-coordinate by 3, the first row of
the matrix must be 3 0 0. Since the y-coordinate is
multiplied by 2, the second row is 0 2 0. To convert
a z-coordinate to x  4z, use a third row of 1 0 4.
3 0
0
The matrix is 0 2
0 .
1 0 4
25a. The x-coordinates are unchanged, the
y-coordinates increase, and the z-coordinates
decrease, so the movement is dip-slip.



25b.



123.9
88.0
205.3

41.3
145.8
246.6

29.



201.7
28.3
261.5

73.8
82.6
212.0

129.4
36.4
97.1 123.9
166.4
85.3



123.9 41.3 201.7
73.8 129.4
36.4
 86.4 144.2 29.9 84.2
95.5 125.5
206.5 247.8 262.7 213.2 165.2
84.1
0
0
0
0
0
0


1.6
1.6
1.6
1.6
1.6
1.6
1.2 1.2 1.2 1.2 1.2 1.2




Only I and II are equivalent, so the correct choice
is A.

Chapter 8 Study Guide and Assessment



Page 543



26a. La Shawna
x
0
y 
 16t2  150
16t2  150 
 0
150 
 16t2
150

16

1.
3.
5.
7.
9.

Jaimie
x 
 35t
y 
 16t2  150

3.06

t

x 
 35t


p



 35
16
150

y


4.1 cm

4.1 cm, 23°


q
2p



5.3 cm
25˚


t

5.3 cm, 25°

x1
 
2 
5
2
48
5x  5





2

15.

10

3p

  q


28. sec 
cos1 5 


1

2
cos cos1 
5
1


2

5






q



3p



2.5 cm
98˚

5


 2

Chapter 8

p

  
q


q

14.

y 
 2t  10
y


12. 2.9 cm, 10°

23˚

107 ft
26b. Since the stones have the same parametric
equations for y, they land at the same time. In
part a, it was calculated that the elapsed time is
about 3.06 seconds.
27. x 
 5t  1
x  1 
 5t
x1

5

unit
cross
vector
standard
components

Skills and Concepts

11. 1.3 cm, 50°
13.

150



16 
 t

2.
4.
6.
8.
10.

resultant
magnitude
inner
parallel
direction

544–546


 t2

Understanding and Using the
Vocabulary

2.5 cm, 98°

268

2p

  q


u  5v
u
31. u
u 
 2w
u
u 
 2 4, 1, 5  5 1, 7, 4
u
u 
 8, 2, 10  (5, 35, 20
u
u 
 8  (5), 2  35, 10  (20)
u
u 
 13, 37, 30
u
u
32. u 
 0.25u
v  0.4w
u
u 
 0.25 1, 7, 4  0.4 4, 1, 5
u
u 
 0.25, 1.75, 1  1.6,  0.4, 2
u
u 
 0.25  1.6, 1.75  (0.4), 1  2
u
u 
 1.35, 1.35, 1
33. 5, 1  2, 6 
 5(2)  (1)6

10  6

16; no
34. 2, 6  3, 4 
 2(3)  6(4)

6  24

 18; no
35. 4, 1, 2  3, 4, 4 
 4(3)  1(4)  (2)4

 12  4  8

 0; yes
36. 2, 1, 4  6, 2, 1 
 2(6)  (1)(2)  4(1)

12  2  4

18; no
37. 5, 2, 10  2, 4, 4

 5(2)  2(4)  (10)(4)

 10  8  40

 42; no
u
u
i
j u
k
38. 5, 2, 5  1, 0, 3 
 5 2
5
1
0 3
2 5 u
5
5u
5 2 u
u 
j 
k


0 3
1 3
1
0
u  10j
u  2k
u or 6, 10, 2

 6i

q



16.
4p

  q


3.5 cm

4p



82˚

3.5 cm, 82°
17. h 
 1.3 cos 50°
v 
 1.3 sin 50°
h 
 0.8 cm
v 
 1 cm
18. h 
 2.9 cos 10°
v 
 2.9 sin 10°
h 
 2.9 cm
v 
 0.5 cm
u
19.
CD 
 7  2, 15  3 or 5, 12
u
CD  
 
52  1
22

 169
 or 13
u
20.
CD 
 4  (2), 12  8 or 6, 4
u
CD  
 
62  42

 52
 or 213

u
21.
CD 
 0  2, 9  (3) or 2, 12
u
CD  
 
(2)2 
 122

148
 or 237

u
22.
CD 
 5  (6), 4  4 or 1, 8
u
CD  
 
12  (
8)2

 65

23. u
u 
u
v u
w
u
u 
 2, 5  3, 1
u
u 
 2  3, 5  (1) or 5, 6
24. u
u 
u
v u
w
u
u 
 2, 5  3, 1
u
u 
 2  3, 5  (1) or 1, 4
u
u  2w
u
25. u 
 3v
u
u 
 3 2, 5  2 3, 1
u
u 
 6, 15  6, 2
u
u 
 6  6, 15  (2) or 12, 17
u  2w
u
26. u
u 
 3v
u
u 
 3 2, 5  2 3, 1
u
u 
 6, 15  6, 2
u
u 
 6  6, 15  (2) or 0, 13
u
27. EF 
 6  2, 2  (1), 1  4 or 4, 1, 3
u
EF  
 
42  (
1)2 
(3)2

 26

u
28. EF 
 1  9, 5  8, 11  5 or 10, 3, 6
u
EF  
 
(10)2
 (
3)2  
62

 145

u
29. EF 
 2  (4), 1  (3), 7  0) or 6, 2, 7
u
2  72
EF  
 
62  2

 89

u
30. EF 
 4  3, 0  7, 5  (8) or 7, 7, 13
u
EF  
 
(7)2 
 (7
)2  1
32

 267


6, 10, 2  5, 2, 5
6(5)  10(2)  (2)(5)
30  20  10 
 0
6, 10, 2  1, 0, 3
6(1)  10(0)  (2)(3)
6  0  6 
 0
39. 2, 3, 1  2, 3, 4 


u
i
2
2

u
u
j
k
3
1
3 4
3
1u
2
1u
2 3 u


i 
j 
k
3 4
2 4
2
3
u  6j
u  0k
u or 9,  6, 0

 9i
9, 6, 0  2, 3, 1
9(2)  (6)(3)  0(1)
18  18  0 
 0
9, 6, 0  2, 3, 4
9(2)  (6)(3)  0(4)
18  18  0 
 0

269

Chapter 8

u
u
i u
j
k
1 0
4
5 2 1
0
4 u 1
4 u 1 0 u


i 
j 
k
2 1
5 1
5 2
u
u
u

 8i  19j  2k or 8, 19, 2
8, 19, 2  1, 0, 4
(8)(1)  19(0)(2)(4)
808
0
8, 19, 2  5, 2, 1
(8)(5)  19(2)  (2)(1)
40  38  2 
 0
41. 7, 2, 1  2, 5, 3 
 u
i u
j u
k
7 2 1
2 5 3
2 1u 7 1u
7 2u


i 
j 
k
5 3
2 3
2 5

47. x  4, y  0 
 t 3, 6
x  4, y 
 t 3, 6
x  4 
 3t
y 
 6t
x 
 4  3t
48. x 
 t
49. x 
 t
1
5
y 
 8t  7
y 
 2t  2

 7  8t
u  
 v
u cos v
u  
 v
u sin v
50. v
v
x
y

 15 cos 55°

 15 sin 55°
8.60 ft/s
12.29 ft/s
u  
 v
u cos v
u  
 v
u sin v
51. v
v
x
y

 13.2 cos 66°

 13.2 sin 66°
5.37 ft/s
12.06 ft/s
u  
 v
u cos v
u  
 v
u sin v
v
52. v
x
y

 18 cos 28°

 18 sin 28°
15.89 m/s
8.45 m/s
u
53. CH 
 4  3, 2  4, 2  (1) or 7, 6, 3
A(3, 4  (6), 1  3) 
 A(3, 2, 2)
B(3, 4  (6), 1) 
 B(3, 2, 1)
D(3, 4, 1  3) 
 D(3, 4, 2)
E(3  (7), 4, 1  3) 
 E(4, 4, 2)
F(3  (7), 4, 1) 
 F(4, 4, 1)
G(3  (7), 4  (6), 1) 
 G(4, 2, 1)
The matrix for the figure is
3
3
3 3 4 4 4 4
2 2
4 4
4
4 2 2 .
2 1 1 2
2 1 1
2
The matrix for the translated figure is
5
5 5 5 2 2 2 2
2 2 4 4
4
4 2 2 .
5
2 2 5
5
2
2
5

40. 1, 0, 4  5, 2, 1 


u  31k
u or 1, 19, 31

u
i  19j
1, 19, 31  7, 2, 1
1(7)  (19)2  31(1)
7  (38)  31 
 0
1, 19, 31  2, 5, 3
1(2)  (19)5  31(3)
2  (95)  93 
 0
42. Sample answer:
Let x(1, 2, 3), y(4, 2, 1) and z(5, 3, 0)
u
xy 
 4  1, 2  2, 1  3 or 5, 0, 4
u
yz 
 5 (4), 3  2, 0  (1) or 9, 5, 1
u
u
5, 0, 4  9, 5, 1 
 u
i
j
k
5
0 4
9 5
1
0 4 u
5 4 u
5
0u


i 
j 
k
5
1
9
1
9 5
u  31j
u  25k
u or 20, 31, 25

 20i
u
u
43. F1 
 320i
u
u
F 
 260j









z
E


H

G


A


D


F


C


y

2

u u
F1  F2 
 
3202 
2602
412.31 N
260

B


x

13



tan v 
 
320 or 16

The figure moves 2 units along the x-axis and 3
units along the z-axis.

13

v 
 tan1 1
6
39.09°
u
44. u
v1 
 12j
u  (30 sin 116°)j
u
u
v2 
 (30 cos 116°)i
u u
v
v  
 
(30 co
s 116°
)2  (1
2  3
0 sin 
116°)2
1

2

41 m/s
tan v 


12  30 sin 116°

30 cos 116°
12  30 sin 116°


v 
 tan1 

30 cos 116° 

108.65°
45. x  3, y  (5)
x  3, y  5
x  3 
 4t
x 
 3  4t
46. x  (1), y  9
x  1, y  9
x  1 
 7t
x 
 1  7t
Chapter 8


 t 4, 2

 t 4, 2

y  5 
 2t
y 
 5  2t

 t 7, 5

 t 7, 5
y  9 
 5t
y 
 9  5t

270






1
0 0
54. 0 1 0
0
0 1
3
3
3
2 2
4
2 1 1
3
3
3

 2
2 4
2 1 1
z



3 4 4 4 4
4
4
4 2 2
2
2 1 1
2
3 4 4 4 4
4 4 4
2
2
2
2 1 1
2





tan v 


Page 547

7

18  73


7

Open-Ended Assessment

1a. Sample answer: X(4, 1), Y(1, 1)
u
XY 
 1  4, 1  (1) or 3, 2
u
1b. XY  
 
(3)2 
 22 or 13

u
The magnitude of XY only depends on the
differences of the coordinates of X and Y, not the
actual coordinates.
2a. Sample answer: P(1, 1), Q(3, 3), R(3, 1), S(5, 3)
u
PQ 
 3  1, 3  1 or 2, 2
u
RS 
 5  3, 3  1 or 2, 2
u
u
PQ and RS are parallel because they have the
same direction. In fact, they are the same vector.
2b. Sample answer:u
a 
 8, 4 ,u
b 
 3, 6
u
u
a  b 
 8(3)  (4)6 or 0
u
a andu
b are perpendicular because their inner
product is 0.

G
y

x
B

The figure is reflected over the xz-plane.

Page 547

or

v 
 tan1 
 
18  73


H

F
A O

C

35

90  353


13.1°

E
D

u
58. F1 
 90i
u
u
u
F2 
 (70 cos 30°) i  (70 sin 30°)j or
u
u
353
 i  35j
u u
F1  F2 
 
(90 
353

)2  3
52
154.6 N

Applications and Problem Solving

1
u
3

55. AB 
 1 cos 120°, 0, 1 sin 120° or 2, 0, 2
u
F 
 0, 0, 50
u u u
T 
 AB  F
1 ft
u u u
i j
k
50 lb
1
3

60˚

 2 0 2

0 0 50
3


3


1

1

 u


 u
2 j  2 0u
2 i  2

 0
k
0 50
0 0
0 50
u  25j
u  0k
u or 0, 25, 0

 0i
u
02  (
25)2
 02
T  
 

 25 lb-ft
u sin v  1gt2  h
56. y 
 tv
2

Chapter 8 SAT & ACT Preparation
Page 549

1


 0.5(38) sin 40°  2(32)(0.5)2  2
10.2 ft
57a.

16 km/h

3 km/h
35˚

250 m

u 
 (16 cos 55°)c
u
u  (16 sin 55°)j
b
u
u
c 
 3j
u u
2  (1
b
c  
 
(16 co
s 55°)
6 sin 
55° 
3)2
13.7 km/h
57b.

u

250

16 sin 55°  3




16 cos 55°

1

A 
 2bh

16 sin 55°  3

16 cos 55°

u 
 250 

u

SAT and ACT Practice

1. Recall that the formula for the area of a
parallelogram is base times height. You know the
base is 5, but you don’t know the height. Don’t be
fooled by the segment BD; it is not the height of
the parallelogram. Try another method to find
the area. The parallelogram is made up of two
triangles. Find the area of each triangle. Since
ABCD is a parallelogram, AB 
 DC and AD 
 BC.
The two triangles are both right triangles, and
they share a common side, BD. By SAS, the two
triangles are congruent. So you can find the area
of one triangle and multiply by 2. The hypotenuse
of the triangle is 5 and one side is 3. Use the
Pythagorean Theorem to find the other side.
52 
 32  b2
25 
 9  b2
16 
 b2
4
b
The height is 4.
Use the formula for the area of a triangle.



1

A 
 2(4)(3) or 6

275.3 m

Since the parallelogram consists of two triangles,
the area of the parallelogram is 2  6 or 12. The
correct choice is A.

271

Chapter 8

6. This figure looks more complex than it is. A semicircle is just one half of a circle. Notice that the
answer choices include , so don’t convert to
decimals. Find the radius of each semi-circle.
Calculate the area of each semi-circle.
The area of the shaded region is the area of the
large semi-circle minus the area of the medium
semi-circle plus the area of the small semi-circle.

2. In order to write the equation of a circle, you need
to know the coordinates of the center and the
length of the radius. The general equation for a
circle is (x  h)2  (y  k)2 
 r2, where the center
is (h, k) and the radius is r. From the coordinates
of points A and B, you know the length of the side
is 4. So the center Q, has coordinates (0, 4).
To calculate the length of the radius, draw the
radius OB. This creates a 45°-45°-90° right
triangle. The two legs each have length 2. The
hypotenuse has length 22
.
)2
(x  4)2  (y  0)2 
 (22
(x  4)2  y2 
 4(2)
(x  4)2  y2 
 8
The correct choice is B.
3. Write the equation for the perimeter of a
rectangle. then replace x with its value in terms
of y. Solve the equation for y.
p 
 2x  2y

9

1

Large semi-circle area 
 2 32 
 2
4

1

Medium semi-circle area 
 2 22 
 2
1

1

Small semi-circle area 
 2 12 
 2
9

4

1

6

Shaded area 
 2  2  2 
 2 
 3
The correct choice is A.
7. The only values for which a rational function is
undefined are values which make the
denominator 0. Since f(x) 


x2 –3x  2
 ,
x–1

the denominator is only 0 when x – 1 
 0 or x 
 1.

p 
 2
3 y  2y
2

The correct choice is D.
8. Start by sketching a diagram of the counter

4

p 
 3y  2y
10

p 
 3y
3p

10


y

28  38

30

The correct choice is B.
4. Recall the triangle Inequality Theorem: the sum
of the lengths of any two sides of a triangle is
greater than the length of the third side. Let x
represent the length of the third side.
40  80  x
120  x
40  x  80
x  40
Since x must be greater than 40, x cannot be equal
to 40. The correct choice is A. To check your
answer, notice that the other answer choices are
greater than 40 and less than 120, so they are all
possible values for x.
5. Since the answer choices have fractional
exponents of x, start by rewriting the expression
with fractional exponents. Simplify the fractions
and use the rules for exponents to combine terms.
2
3
3
9



x2  
x3 
 x 3  x 9
2
1



 x3  x3
2
1



x( 3  3 )

x1 or x
The correct choice is E.

40

Use your calculator to find the area of the whole
counter and then subtract the area of the white
tiles in the center. The white tiles cover an area of
(30  2)(40  2) or (28)(38).
(30)(40) 
 1200
(28)(38) 
 1064
Red tiles 
 1200  1064 
 136
The correct choice is B.
9. First, find the slope of the line containing the
points (–2, 6) and (4, –3).
m


–3 – 6

4 – –2
3

–9

m 
 6 or – 2
The point-slope form of the line is
3

y – 6 
 – 2(x – –2).
3

y – 6 
 – 2x – 3
3

y 
 – 2x  3
So the y-intercept of the line is 3.
The correct choice is B.

Chapter 8

272

10. Write an expression for the sum of the areas of the
two triangles. Recall the area of a triangle is one
half the base times the height.
1
(AC)(AB)
2

1
(AC)(AB)
2

1

1

1

 2(CE)(ED) 
 2(AC)2  2(CE)2
1


 2[(AC)2  (CE)2]

1

Using the Pythagorean Theorem for ACE, you
know that (AC)2  (CE)2 
 (AE)2 or 1.

 2(CE)(ED)

From the figure, you know that ABC and CDE
are both isosceles, because of the angles marked x°
and because B
C
D
 is a line segment. These two
triangles have equal corresponding angles.
Since they are isosceles triangles, AC 
 AB and
CE 
 ED. Use these equivalent lengths in the
expressions for the area sum.

1

1

So the sum of the two areas is 2(1) 
 2. You can
grid the answer either as .5 or as 12.

273

Chapter 8

Chapter 9 Polar Coordinates and Complex Numbers
11.

Polar Coordinates

9-1

90˚

120˚

Check for Understanding

180˚

1. There are infinitely many ways to represent the
angle v. Also, r can be positive or negative.
2. Draw the angle v in standard position. Extend the
terminal side of the angle in the opposite
direction. Locate the point that is r units from
the pole along this extension.
3. Sample answer: 
60° and 300°
Plot (4, 120) such that v is in standard position
and r is 4 units from the pole. Extend the
terminal side of the angle in the opposite
direction. Locate the point that is 4 units from the
pole along this extension.
r  
4
v  120 
 180 or v  120  180
 
60
 300
4. The points 3 units from the origin in the opposite
direction are on the circle where r  3.
5. All ordered pairs of the form (r, v) where r  0.
6.

90˚

120˚
150˚

60˚

330˚

210˚
240˚

8.

120˚

270˚
90˚

C

180˚

1 2 3 4

240˚

270˚

19

6



13

7

, 
2,



13



25

→ 2,



6

Chapter 9



5.25  15 cos 
12

5

90˚

120˚

 (1) → 2,
 (3) → 2,

19

6

60˚
30˚

5 10 15 20

0˚

330˚
300˚

270˚

15b. 210 
 (
30)  240
N

2

A
360 (r )
240

2


360 (20 )

 838 ft2

11
6
5
3

Pages 558–560

7

6

, 2, ,

16.

120˚

Exercises
90˚

0˚

330˚
240˚

274

30˚

1 2 3 4

210˚




17.

60˚

E

180˚

(
r, v  (2k  1))
7

6



5

6.25  9  15 cos 
12

180˚

→
2, 6  2(2)) → 
2, 6
→ 2,



2.52 (
3)2 
 2(2.5)(
3) cos 
4 
 6

150˚

→
2, 6  2(1) → 
2, 6



6

300˚

270˚

150˚

(r, v  2k)

0˚



15a.

0

25

6

5
3

3
2

 4.37


3

3
2

11
6

7
6

330˚

14. P1P2 

1 2 3 4

4
3

1 2 3 4

60˚

1 2 3 4

240˚

7
6

0



30˚

240˚


6

D


3

6

4
3

210˚

0˚


90˚

120˚

210˚

5
3

5
6

300˚

10. Sample answer: 
2,


2

270˚

180˚

11
6

3
2

2
3

330˚

210˚

2,

9.
30˚

150˚

B

0˚

300˚

150˚

1 2 3 4

4
3

60˚

13.

0

7
6

300˚

240˚


2

5
6

330˚

210˚


6



1 2 3 4


3

5
6

0˚

1 2 3 4


2

2
3

30˚

A

180˚

7.

2
3

30˚

150˚

Pages 557–558

12.

60˚

270˚

300˚

2
3


2


3

6

5
6

F
0



1 2 3 4
11
6

7
6
4
3

3
2

5
3

18.

90˚

120˚

19.

60˚

180˚

0˚

2 4 6 8


6

H

0



1 2 3 4

G
330˚

210˚
240˚

20.

270˚

2

2
3

7
6

300˚

21.
150˚

90˚

J
4
3

22.

3
2

2

2
3

23.


3

90˚

L
4
3

24.

90˚

120˚

240˚

5
3

3
2

25.

60˚

180˚

330˚

210˚
240˚

26.


2

2
3

27.

0



180˚

1 2 3 4

Q

7
6
4
3

3
2

30˚
1 2 3 4



270˚



3

7

3

 2(1) → 2,


3

6
0



1 2 3 4
11
6

7
6

35.

6


2

5
6

4
3


3

120˚

5
3

3
2
90˚

60˚
30˚

150˚
180˚

0
1 2 3 4

1 2 3 4

4
3

300˚

36.

120˚

90˚

30˚

180˚

1 2 3 4

240˚

275

0˚

330˚

210˚

(
r, v  (2k  1)180°)
→ (
2, 60°  (1)180°) → (
2, 240°)
→ (
2, 60°  (3)180°) → (
2, 600°)

37.

60˚

150˚



240˚

5
3

3
2

270˚

300˚

0˚

330˚

210˚

11
6

7
6

→ 2, 3  2(0) → 2, 3

→ 2,


2

2
3

0˚

7



300˚

270˚



28. Sample answer: 2, 3, 2, 3, (
2, 240°),
(
2, 600°)
(r, v  2k)


0˚

330˚

2
3

330˚
240˚

5
3

1 2 3 4

5
6

60˚

210˚

11
6

30˚

240˚

34.

33.

60˚

210˚

5
3

150˚ R

90˚

180˚

1 2 3 4

90˚

10

150˚


3

3
2



120˚

11
6

120˚


6

5
6

32.

0

P

4
3


3

300˚

7
6

300˚

270˚

0˚


6



N

4

31. Sample answer: (4, 675°), (4, 1035°), (
4, 135°),
(
4, 495°)
(r, v  360k°)
→ (4, 315  360(1)°) → (4, 675°)
→ (4, 315  360(2)°) → (4, 1035°)
(
r, v  (2k  1)180°)
→ (
4, 315  (
1)180°) → (
4, 135°)
→ (
4, 315  (1)180°) → (
4, 495°)

330˚

5
6

0˚

1 2 3 4


2

2
3

30˚

150˚

270˚



→ 1, 3  (1) → 1, 3

60˚

M

210˚

11
6

7
6

13

→ 1, 3  (3) → 1, 3

1 2 3 4

1 2 3 4



(
r, v  (2k  1))

30˚

180˚

7

→ 
1, 3  2(2) → 
1, 3

300˚

150˚
0



270˚



→ 
1, 3  2(1) → 
1, 3

330˚

120˚


6

5
6

(r, v  2k)

0˚

1 2 3 4

240˚

5
3

10

1, 
3 

60˚

210˚

11
6

7

13
4
 , 1, ,
30. Sample answer: 
1, 3, 
1, 
3  
3

30˚

1 2 3 4

7
6

5
3

K

180˚

0



3
2

120˚


6

5
6

11
6
4
3


3

29. Sample answer: (1.5, 540°), (1.5, 900°), (
1.5, 0°),
(
1.5, 360°)
(r, v  360k°)
→ (1.5, 180°  360(1)°) → (1.5, 540°)
→ (1.5, 180°  360(2)°) → (1.5, 900°)
(
r, v  (2k  1)180°)
→ (
1.5, 180°  (
1)180°) → (
1.5, 0°)
→ (
1.5, 180°  (1)180°) → (
1.5, 360°)


3

5
6

30˚

150˚


2

2
3

2
3

270˚

2

300˚


3

6

5
6

0



1 2 3 4

7
6

11
6
4
3

3
2

5
3

Chapter 9

38.

90˚

120˚

39.

60˚

30˚

150˚
180˚

1 2 3 4

0˚

150˚

240˚

40.

90˚

120˚

30˚
1 2 3 4

0˚

50a.

330˚

210˚
240˚

300˚

270˚

49a. When v  120°, r  17. The maximum speed at
120° is 17 knots.
49b. When v  150°, r  13. The maximum speed at
150° is 13 knots.

60˚

r
0

180˚

330˚

210˚

90˚

120˚

30˚

180˚

1 2 3 4

240˚

270˚

0˚

11
6
4
3

50b.

300˚

1  25 
 10 cos 12



26 
 10 cos 12

2

N

N

120

120

2
2



360 ((300 ) 
 360 ((25) )
120



360  (90,000 
 625)

 93,593 ft2
If each person’s seat requires 6 ft2 of space,



7






 
3  3 or 120°

2
2


A
360 (R ) 
 360 (r )

12  52 
 2(1)(5) cos 4 
 6
3



3

5
3

3
2

Let R  3 100 or 300 and let r  0.25 100
or 25.

41. r  
2
 or r  

2
 for any v.
2 
 2(
42  6
4)(6) c
os (10
5° 
 170°)

42. P1P2  

 
16
36
 48


cos°)
(
65
 
52
48


cos)
(
65
 5.63
43. P1P2 

0
1 2 3 4

7
6

330˚

210˚


6



60˚

150˚


3

5
6

300˚

270˚


2

2
3

93,593

there are 6 or 15,599 seats.

7

51. The distance formula is symmetric with respect to
(r1, v1) and (r2, v2). That is,

 5.35



r22  
r12 
 
2r2r1 
cos (v1

 v2)

44. P1P2 
(
2.5)2  (
1.75)2 
 2(
2.5)(
1.75)cos


21

6.25  30.0625 
 8.75 cos 

40



21

9.3125 
 8.75 cos 

40





2


5 
 8



 

r12  
r22 
 
2r1r2 cos[


(v2 
 
v1)]



 

r12  
r22 
 
2r1r2 
cos(v2

 v1)



52a.



120˚

90˚

60˚
30˚

150˚

 3.16
180˚

45. P1P2  

1.32 
(
3.6
)2 
 2(1.3)(


3.6) co
s (
62
° 
 (
47°))

 
1.69


12.966
 9.3(
62°
cos °)
 47
 
14.65
6
 9.3(
15°
cos )
 4.87

46. r 



(
3)2 
 42

240˚

5

180° 
 53°  127°
Sample answer: (5, 127°)
47. There are 360° in a circle. If the circle is cut into 6
360

equal pieces, each slice measures 6 or 60°.
Beginning at the origin, the equation of the first
line is v  0°. The equation of the next line,
rotating counterclockwise, is v  0  60 or 60°.
The equation of the last line is v  60  60 or
120°. Note that lines extend through the origin, so
3 lines create 6 pieces.




3 mph

54. 3, 
2, 4

1, 
4, 0  (3)(1)  (
2)(
4)  (4)(0)
380
 11
No, the vectors are not perpendicular because
their inner product is not 0.

 r1 
 r2
Chapter 9

300˚

 8 mph

48. P1P2  

r12  
r22 
 
2r1r2 
cos (v 

 v)



r12  
r22 
 
2r1r2
2


(r1
 r
2)

270˚

52b. P1P2 

52  62

 2(
5)(6) c
os (34
5° 
 310°)

 
25
36
 60


cos
(35°)
 
61
60


cos
(35°)
 3.44
No; the planes are 3.44 miles apart.
53. Draw a picture.
sin v  38
Boat
1
sin (sin v)  sin138
v  22.0°

4



0˚

330˚

210˚

sin v  5, v  53°



r12 
 
r22 
 
2r1r2 
cos 0

2 4 6 8

276

55. Rewrite y  9x 
 3 as 9x 
 y 
 3  0.
d





2 4 
1
4 0
1 
1
0

4
 (
1)
1 
1 0  
2 
1

3 5

3 4
5
4

3 4 5
 
2(
5) 
 4(5) 
 1(1)
 
11
64. 11 
 (
3)  14
11 
 (
2)  13
11 
 (
1)  12
11 
 0  11
{(
3, 14), (
2, 13), (
1, 12), (0, 11)}
For each x-value, there ia a unique v-value.
Yes, the relation is a function.
65. Since the two triangles formed are right triangles,
the side opposite the right angles, A
B
, intercept
an arc measuring 180°, or half the circle. 
AB
 is a
diameter.
C  d
50  d
50  d
The correct choice is E.

Ax1  By1  C



A2  
B2
9(
3)  (
1)(2)  (
3)



92  (

1)2

32


82


32

82




82


82


63.


32
82


  
82

56.


16
82

41

1 
 sin2a

sin2a



Distance is always positive.
1
 
 1
sin2a
csc2a 
 1


 cot2a

3


57. Arc cos 
2  30°
1
In a 30°-60°-90° right
triangle, the angle opposite
the smallest leg is 30°.
58. y  5 cos 4v
Amplitude  5; Period  24 or 2

2


3



59. b sin A  18.6 sin 30°
 9.3
Since a  b sin A, there is one solution.
Find B.
Find C.
18.6

sin B

9.3


sin 30°

18.6 sin 30°

9.3

 60°

 sin B

90°  B
Find c.
c

sin 60°

Graphs of Polar Equations

Page 565

Check for Understanding

1. Sample answer: r  sin 2v
The graph of a polar equation whose form is
r  a cos nv or a sin nv, where n is a positive
integer, is a rose.
2. 
1  sin v  1 for any value of v. Therefore, the
maximum value of r  3  5 sin v is r  3  5(1)
or 8. Likewise, the minimum value of r  3  5
sin v is r  3  5(
1) or 
2.
3. The polar coordinates of a point are not unique. A
point of intersection may have one representation
that satisfies one equation in a system, another
representation that satisfies the other equation,
but no representation that satisfies both
simultaneously.
4. Barbara is correct. The interval 0  v   is not
always sufficient. For example, the interval 0  v
  only generates two of the four petals for the
rose r  sin 2v. r  sin 2v is an example where
values of v from 0 to 4 would have to be
considered.

C  180° 
 90° 
 30°

18.6 sin 30°  9.3 sin B

9-2

9.3


sin 30°

c sin 30°  9.3 sin 60°
9.3 sin 60°

c
sin 30°

c  16.1
60. 3 or 1 positive
f(
x)  
x3 
 4x2 
 4x 
 1
0 negative
P
1
Q

Since there are only positive real zeros, the only
rational real zero is 1.
61.
x
 3
x  5x2

2x


3
x2
5
x


3x 
 3

3x 
 15

10
10
 → 0. Therefore, the slant
As x →  , 
x5
asymptote is y  x 
 3.
62. y-axis:
For x:
f(x)  x4 
 3x2  2
For 
x: f(
x)  (
x)4 
 3(
x)2  2
 x4 
 3x2  2
So, in general, point (
x, y) is on the graph if and
only if (x, y) is on the graph.

5.

90˚

120˚

30˚

150˚
180˚

1

2

0˚

330˚

210˚
240˚

270˚

cardioid

277

6.

60˚

300˚

120˚

90˚

60˚
30˚

150˚
180˚

2 4 6 8

0˚

330˚

210˚
240˚

270˚

300˚

limaçon

Chapter 9

7.

2
3


2

8.


3

6

5
6


11
6
4
3

11.

3
2


2

5
3

0

4
3

3
2

If 
 

or 
 

5

6

equation, r  1. If 
 

3

2

2
3

4
3

3
2

330˚

120˚

270˚

300˚

90˚

60˚
30˚

150˚
1

0˚

2

330˚

210˚

300˚

240˚

rose
17.

270˚

300˚

rose
2
3


2

0

4 8 12 16

4
3

3
2

0

19.

2
3



4
3

cardioid

278

3
2

5
3

0˚

330˚
270˚

300˚

2
3


2


3

6

5
6
0

11
6

7
6

4 6 8

limaçon
20.

1 2 3 4

2

240˚


6

5
6

60˚
30˚

180˚

5
3


3

90˚

210˚

Spiral of Archimedes

2

120˚
150˚

11
6

7
6

5
3

18.


3

6



14

10b. Sample answer: 0  v  
3
Begin at the origin and “spiral” twice around it,
or through 4 radians. Move straight up
through 4  2 or 92 radians. Now move to the
left slightly, through approximately 92  6 or
14
 radians.
3

Chapter 9

0˚

is substituted in either

11
6

7
6

270˚

60˚

1 2 3 4

0˚ 180˚

2

330˚

5
6

4 8 12 16

16.

1

90˚

30˚

240˚

30˚

240˚


6



300˚

lemniscate

60˚

210˚


3

5
6

90˚

180˚

1, 6, 1, 56, and 
2, 32.
10a.

270˚

210˚

5
3

150˚

original equation, r  
2. The solutions are

2

3
2

120˚

180˚

0

11
6

120˚

is substituted in either original

330˚

150˚

spiral of Archimedes


  6 or 
  56 or 
  32

0˚

2 4 6 8

210˚

14.


3

5 10 15 20

15.

30˚

240˚


6

4
3

2 sin 
  2 cos 2

sin 
  cos 2

sin 
  1 
 2 sin2 

2 sin2 
  sin 
 
 1  0
(2 sin 
 
 1)(sin 
  1)  0
2 sin 
 
 1  0 or sin 
  1  0
sin 
  12
sin 
  
1


6

2
3


2

7
6

5
3

60˚

cardioid



11
6

90˚

150˚

300˚

5
6

2

7
6

270˚

circle
13.

1

330˚
240˚

120˚

0˚ 180˚

1 2 3 4

210˚


6



12.

60˚

30˚

180˚


3

5
6

90˚

150˚

spiral of Archimedes
2
3

Exercises

120˚

0

11
6
4
3

rose
9.

3 6 9 12

7
6

5
3

3
2


6



0

Pages 565–567


3

5
6

1

7
6


2

2
3



1 2 3 4

0

11
6

7
6
4
3

lemniscate

3
2

5
3

21.

90˚

120˚

22.

60˚

120˚

30˚

150˚
180˚

330˚
240˚

30˚
0˚

1 2 3 4

330˚

210˚

300˚

270˚

28. (1, 0.5), (1, 1.0), (1, 2.1), (1, 2.6), (1, 3.7), (1, 4.2),
(1, 5.2), (1, 5.8)

60˚

150˚

0˚ 180˚

1

210˚

90˚

240˚

270˚

300˚

rose
cardioid
23. Sample answer: r  sin 3v
The graph of a polar equation of the form r  a cos
3v or r  a sin 3v is a rose with 3 petals.
24. Sample answer: r  2v


4

 a2

r  12


1

2

a

r  2




25. 3  2  cos 

1  cos 


0
The solution is (3, 0)

2
3


2


3

6

5
6


1 2 3 4

4
3

26. 1  cos 
  1 
 cos 

2 cos 
  0
cos 
  0

2
3

Substituting each angle
into either of the original
equations gives r  1, so
the solutions of the system
are 1,
27.

2
3


2


3

6



1

2

 and 1, .

0

[
4, 4] scl1 by [
4, 4] scl1

11
6

7
6

3

2


2

5
3

3
2

5
6


  2 or 
  32

[
6, 6] scl:1 by [
6, 6] scl1
30. (3.6, 0.6), (2.0, 4.7)

0

11
6

7
6



2

[
2, 2] scl1 by [
2, 2] scl1
29. (2, 3.5), (2, 5.9)

4
3

3
2

31a. If the lemniscate is 6 units from end to end, then
a  12(6) or 3.

5
3

r2  9 cos 2
 or r2  9 sin 2

31b. If the lemniscate is 8 units from end to end, then
a  12(8) or 4.


3

r2  16 cos 2
 or r2  16 sin 2



6

5
6

32.


1

2

4
3

3
2

90˚

60˚

0
30˚

150˚

11
6

7
6

120˚

180˚

5
3

2 4 6 8

330˚

210˚

2 sin 
  2 sin 2

sin 
  sin 2

sin 
  2 cos 
 sin 

0  2 cos 
 sin 
 
 sin 

0  sin 
 (2 cos 
 
 1)
sin 
  0 or 2 cos 
 
 1  0

0˚

240˚

270˚

300˚

This microphone will pick up more sounds from
behind than the cardioid microphone.
33. 0  v  4: Begin at the origin and curl around
once, or through 2 radians. Curl around a second
time and go through 2  2 or 4 radians.
34. All screens are [
1, 1] scl1 by [
1, 1] scl1

cos 
  12

  0 or  or 
  3 or 
  53
If 
  0 or 
   is substituted in either original
equation, r  0. If 
  3 or 
  53 is substituted in
either original equation, r  
3
 or r  

3
,
respectively. The solutions are (0, 0), (0, ),


3, 3, and 

3, 53.

279

Chapter 9







34a. r  cos 2

r  cos 7







r  cos 4

r  cos 9




When n  11, the innermost loop will be on the
left and there will be an additional outer ring.
35. Sample answer: r  
1 
 sin v
A heart resembles the shape of a cardioid. The
sine function orients the heart so that the axis of
symmetry is along the y-axis. If a  
1, the heart
points in the right direction.
36a. For a limaçon to go back on itself and have an
inner loop, r must change sign. This will happen
if b a.
36b. For the other two cases, a  b.
Experimentation shows that the dimple
disappears when a  2b, so there is a
dimple if b  a  2b.
36c. For this remaining case, there is neither an
inner loop nor a dimple if a  2b.
37a. Subtracting a from v rotates the graph
counterclockwise by an angle of a.
37b. Multiplying v by 
1 reflects the graph about the
polar axis or x-axis.
37c. Multiplying the function by 
1 changes r to its
opposite, so the graph is reflected about the
origin.
37d. Multiplying the function by c results in a
dilation by a factor of c. (Points on the graph
move closer to the origin if c  1, or farther
away from the origin if c 1.)
38. Sample answer: (4, 405°), (4, 765°), (
4, 
135°),
(
4, 225°)
(r, 
  360k°)
→ (4, 45°  360(1)°) → (4, 405°)
→ (4, 45°  360(2)°) → (4, 765°)
(r, 
  (2k  1)180°)
→ (
4, 45°  (
1)180°) → (
4, 
135°)
→ (
4, 45°  (1)180°) → (
4, 225°)

r  cos 6




r  cos 8

When n  10, two more outer rings will appear.



34b. r  cos 3




r  cos 5

Chapter 9

280

i j k
2 3 0

1 2 4
3 0 
2 0 
2 3 

i

j
k
2 4

1 4

1 2



 12i 
 8j  7k
 12, 
8, 7
2, 3, 0
12, 
8, 7  24  (
24)  0 or 0

1, 2, 4
12, 
8, 7  
12  (
16)  28 or 0
40. 3.5 cm, 87°


39. v

x2
r cos 
  2
2

r
cos 

r  2 sec 

4. To convert from polar coordinates to rectangular
coordinates, substitute r and v into the equations
x  r cos v and y  r sin v. To convert from
rectangular coordinates to polar coordinates, use
the equation r  

x2  y2 to find r. If x 0, v 
Arctan yx. If x  0, v  Arctan yx  . If x  0, you
can use 2 or any coterminal angle for v.
3.

 
w

2

sin x
2
41. 
cos4 x  cos2 x sin2 x  tan x
sin2 x

cos2 x (cos2 x  sin2 x)
sin2 x

(cos2 x)(1)
sin2 x

cos2 x

 tan2 x


tan2

y

x

 tan2 x

r

tan2 x  tan2 x


42. Find C.
C  180° 
 21°15
 
 49°40

 109°5

Find b.
b

sin 49°40


O

28.9


sin 109°5


 
4
 or 2

28.9 sin 49°40


sin 109°5


2, 34

b  23.3

6. r  

(
2)2 
 (
5
)2

Find a.
a

sin 21°15


28.9


sin 109°5


a sin 109°5
  28.9 sin 21°15

28.9 sin 21°15



7.

a  sin 109°5

a  11.1
NY LA Miami
$240 $199 $260
$254 $322 $426

43.
Bus
Train

8.

12
1
3
44. 1  6  1  

8

4

8

1


6


8

9.

8

13


4
8
1
3 
16
8
So 
 


3
8
3
3


16

16

10.

2
6

3

The correct choice is A.

9-3
Page 571

11.

Polar and Rectangular
Coordinates

12.

Check for Understanding

x

 34



2


2



5
 
v  Arctan 

2 

  5.39
 4.33
 
29
(5.39, 4.33)
x  
2 cos 43
y  
2 sin 43
1
 
3

(1, 
3
)
x  2.5 cos 250°
y  2.5 sin 250°
 
0.86
 
2.35
(
0.86, 
2.35)
y2
r sin v  2
2

r
sin v
r  2 csc v
x2  y2  16
(r cos v)2  (r sin v)2  16
r2(cos2 v  sin2 v)  16
r2  16
r  4 or r  
4
r6


x2  y2  6
x2  y2  36
r  
sec v
r

r

1. Sample answer: (2
2
, 45°)
22  22
r  


x 
 r cos 

5. r  

(

2

)2  (

)2 v  Arctan 
2

b sin 109°5
  28.9 sin 49°40

b

y 
 r sin 


1



r cos v


1

1
x


  Arctan 22

x  
1

 
8

 45°
 2
2

2. The quadrant that the point lies in determines
y
y
whether v is given by Arctan x or Arctan x  .

281

Chapter 9

13a.

90˚

120˚

22. x  
1 cos 
6

60˚


3


 
1 
2 

30˚

150˚
180˚


3


1 2 3 4



23 , 12

330˚

210˚
240˚

270˚

23. x  
2 cos 270°
0
(0, 2)
24. x  4 cos 210°

300˚

13b. No. The given point is on the negative x-axis,
directly behind the microphone. The polar
pattern indicates that the microphone does not
pick up any sound from this direction.


3


2

 4

25.

Pages 572–573
Exercises

2
2
2

14. r  
2
 (

2)
v  Arctan 2

26.

 
4

 
8
 or 2
2

Add 2 to obtain v  74.

2
2, 74

15. r  

02  12

27.

 
1
 or 1
Since x  0 when y  1, v  2.

1, 2

16. r  

12  (

3
)2

v  Arctan


 
4
 or 2

 3

2, 3
 

1 2

4

17. r 

28.


3


1






3
 2


4

 14
6
 24 or 12

12, 43

18. r  

32  82
 
73
  8.54
(8.54, 1.21)
42  (

7)2
19. r  


v  Arctan


3




4


14

 

29.


3
 Arctan  or 43
1

30.
v  Arctan 83
 1.21

7

v  Arctan 
4 

31.

 
65
  8.06
 
1.05
Add 2 to obtain v  5.23.
(8.06, 5.23)
20. x  3 cos 2
0
(0, 3)

y  3 sin 2
3
32.

21. x  12 cos 34

y  12 sin 34

 12 
2

 12 
2

 
4



2


4



2


2




42 , 
42 

Chapter 9



2

33.

282

 
1
12
 12

 
2

0˚

y  
1 sin 
6



y  
2 sin 270°
2
y  4 sin 210°
 4
12

 
2
3

 
2
(
2
3
, 
2)
x  14 cos 130°
y  14 sin 130°
 
9.00
 10.72
(
9.00, 10.72)
x  
7
r cos v  
7

7
r  
cos v
r  
7 sec v
y5
r sin v  5
5
r  
sin v
r  5 csc v
x2  y2  25
(r cos v)2  (r sin v)2  25
r2(cos2 v  sin2 v)  25
r2  25
r  5 or r  
5
x2  y2  2y
(r cos v)2  (r sin v)2  2r sin v
r2 (cos2 v  sin2 v)  2r sin v
r2  2r sin v
r  2 sin v
x2 
 y2  1
(r cos v)2 
 (r sin v)2  1
r2 (cos2 v 
 sin2 v)  1
r2 (cos 2v)  1
1

r2  
cos 2v
2
r  sec 2v
x2  (y 
 2)2  4
x2  y2 
 4y  4  4
(r cos v)2  (r sin v)2 
 4r sin v  0
r2(cos2 v  sin2 v) 
 4r sin v  0
r2 
 4r sin v  0
r2  4r sin v
r  4 sin v
r2


x2  y2  2
x2  y 2  4
r  
3
2


x  y2  
3
x2  y2  9

v  3

34.

y
43. horizontal distance:
25(4  2 cos 120°)  75 m east
vertical distance:
25(3  2 sin 120°)  118.30 m north

y

y
Arctan x  3
y


3
  

x
1

2
6
3

y  
3
x

1



3

2

1  2y
y2
r  3 cos v
r2  3r cos v
x2  y2  3x
37.
r2sin 2v  8
r22 sin v cos v  8
2r sin v r cos v  8
2yx  8
xy  4
38.
yx
36.

Arctan

6.07

5.47
6.07

5.47

5.47

cos 47.98°

 tan v

 Arctan 1

r  sin v
r2  r sin v
x2  y2  y
40. x  325 cos 70°
 111.16
(111.16, 305.40)

r

8.17  r
8.17∠47.98°
44d. 8.17 sin (3.14t  47.98°)
45.
r  2a sin v  2a cos v
r2  2ar sin v  ar cos v
x2  y2  2ay  2ax
2
2
x 
 2ax  y 
 2ay  0
(x 
 a)2  (y 
 a)2  2a2
The graph of the equation is the circle centered at
(a, a) with radius 
2
a.

1

39.



4

r sin v



r cos v

47.98  v; 47.98°
5.47  r cos 47.98°

v  4

5

4

y  325 sin 70°
 305.40

46.

120˚

90˚

60˚
30˚

150˚

—
41. —
6 
 6 


5

24
4

24






24

180˚

1 2 3 4

0˚

330˚

210˚

 0.52 unit
42. Drop a perpendicular from the point with polar
coordinates (r, v) to the x-axis. r is the length of
the hypotenuse in the resulting right triangle.
x is the length of the side adjacent to angle v, so
cos v  xr. Solving for x gives x  r cos v. y is the
y
length of the side opposite angle v, so sin v  r.
Solving for y gives y  r sin v. (The figure is drawn
for a point in the first quadrant, but the signs
work out correctly regardless of where in the
plane the point is located.)

240˚

270˚

300˚

47. Sample answer: (
2, 405°), (
2, 765°), (2, 225°),
(2, 585°)
(r, v  360k°)
→ (
2, 45°  360(1)°) → (
2, 405°)
→ (
2, 45°  360(2)°) → (
2, 765°)
(
r, v  (2k  1)180°)
→ (2, 45°  (1)180°) → (2, 225°)
→ (2, 45°  (3)180°) → (2, 585°)
48. r2  502  4252  2 50 425 cos 30°
r  382.52 mph

y
(r, )

50

sin v

r

382.52


sin 30°

50 sin 30°  382.52 sin v

r

50 sin 30°

382.52



O

x

44a. x  4 cos 20°
y  4 sin 20°
 3.76
 1.37
3.76, 1.37
x  5 cos 70°
y  5 sin 70°
 1.71
 4.70
1.71, 4.70
44b. 3.76, 1.37  1.71, 4.70
 3.76  1.71, 1.37  4.70
 5.47, 6.07
44c.
5.47  r cos v; 6.07  r sin v



r sin v

y

x
y

x

120˚

O

x

35. r  2 csc v
r

r

2

 sin v



30˚ 425 mph
50 mph

3°45
  v
The direction is 3°45
 west of south.

x

283

Chapter 9

3. The graph of the equation x  k is a vertical line.
Since the line is vertical, the x-axis is the normal
line through the origin. Therefore, f  0° or f 
180°, depending on whether k is positive or
negative, respectively. The origin is k units
from the given vertical line, so p  k. The polar
form of the given line is k  r cos (v 
 0°) if k is
positive or 
k  r cos (v 
 180°) if k is negative.
Both equations simplify to k  r cos v.
4. You can use the extra ordered pairs as a check on
your work. If all the ordered pairs you plot are not
collinear, then you have made a mistake.

sin2 A  cos A 
 1
1 
 cos2 A  cos A 
 1
0  cos2 A  cos A 
 2
0  (cos A  2)(cos A 
 1)
cos A  2  0
or cos A 
 1  0
cos A  
2
cos A  1
A  0°
y
50.

49.

2

y 
 2 cos 

1

O

1

90˚ 180˚ 270˚ 360˚ 


5. 

A2  
B2  

32  (

4)2
 5
Since C is negative, use 5.


2

4

f  Arctan 
43

x
 3,  1
2
2

(

 
53° or 307°
p  r cos (v 
 f)
2  r cos (v 
 307°)

)

6. 

A2  
B2  

(
2)2 
 42
 2
5

Since C is negative, use 2
5
.

52. Enter the x-values in L1 and the f(x)-values in L2
of your graphing calculator. Make a scatter plot.
The data points are in the shape of parabola.
Perform a quadratic regression.
a  
0.07, b  0.73, c  
1.36
Sample answer:
y  
0.07x2  0.73x 
 1.36
53. 2 1 0 0 
3 0 
20
2
4
8
1
0
2
0

1 2 4 5 10  0
x4  2x3  4x2  5x  10
625 
 145

54. m  
25 
 17

 60
55. x y and y
If x

2

4

9

 x   y 
   0


2

2

2

5
5
5


9
5

2
5


5

, sin f  , p  
cos f  

10
5
5

f  Arctan(
2)
 
63°
Since cos f  0, but sin f
the second quadrant.
f  180° 
 63° or 117°
p  r cos (v 
 f)

(y 
 145)  60(x 
 17)


9
5

10

y  60x 
 875
z, so x


 45y 
 2  0

cos f  35, sin f  
5, p  2

30˚


3


cos 210°  
2

3
x
5

y

51. The terminal side is in the
third quadrant and the
reference angle is
210 
 180 or 30°.

0, the normal lies in

 r cos (v 
 117°)

7. 3  r cos (v 
 60°)
0  r cos (v 
 60°) 
 3
0  r (cos v cos 60°  sin v sin 60°) 
 3

z.

z, then 0  xz  1.

The correct choice is C.



3
 r sin v 
 3
0  12r cos v  
2


3
y 
 3
0  12x  
2

9-4

y 
 6 or
0  x  
3
x  
3
y 
 6  0

Polar Form of a Linear Equation

Pages 577–578

8.

Check for Understanding

r cos (v  4)  2

1. The polar equation of a line is p  r cos (v 
 f).
r and v are the variables. p is the length of the
normal segment from the line to the origin and
f is the angle the normal makes with the positive
x-axis.
2. For r to be equal to p, we must have cos (v 
 f) 
1. The first positive value of v for which this is
true is v  f.

Chapter 9

r  2 sec v  4

r cos v cos 4 
 sin v sin 4 
 2  0


2

2



2
 r sin v 
 2  0
r cos v 
 
2


2
x
2



2
y 
 2  0


2

x 
 
2
y 
 4  0

2

284

9.

2
3


2

10.


3

6

5
6

90˚

30˚

180˚

1 2 3 4

2 4 6 8

7
6

240˚

270˚

4

f  Arctan 
3
 
53°
Since cos f 0, but sin f  0, the normal lies in
the fourth quadrant.
f  360° 
 53° or 307°
p  r cos (v 
 f)
2.1  r cos (v 
 307°)

300˚

Since the shortest distance is along the normal,
the answer is (p, f) or 5, 56.
2
3


2

15. 

A2  
B2  

32  22
 
13

Sinc C is negative, use 
13
.


3

6

5
6

3

x


13

0



3
2

5
3

16.

4

x


41

 25
Since C is positive, use 
25.
cos f

24
 25y 
 4  0
7
24
,
 
2
5 , sin f  
25

5

10

y 
 
0



1

41

4



5
41


10
41

4

41
, p  

, sin f  
cos f  
41
41
41

f  Arctan 
4
 
51°
Since cos f 0, but sin f  0, the normal lies in
the fourth quadrant.
f  360° 
 51° or 309°
p  r cos (v 
 f)
5

p4

f  Arctan 
274
 
74°
Since cos f  0, but sin f 0, the normal lies in
the second quadrant.
f  180° 
 74° or 106°
p  r cos (v 
 f)
4  r cos (v 
 106°)
13. 

A2  
B2  

212 
202
  29
Since C is negative, use 29.
21
x
29

5

13
  r cos (v 
 34°)
13
A2  
B2  

42  (

5)2



 
14

Since C is negative, use 
41
.

Pages 578–579
Exercises
2B

2  

12. 
A
72  (

24)2
7

2
5x

5

f  Arctan 23
 34°
p  r cos (v 
 f)

11
6
4
3

2

y 
 
0

13
13





3

2
13
5

13
13

, sin f  , p  
cos f  
13
13
13

2 4 6 8

7
6

4

cos f  5, sin f  
5, p  2.1

11a. p  r cos (v 
 f) → 5  r cos v 
 56

11b.

1

 18
y 
 21
0
0
0
3

5
3

3
2

6
x
10

0˚

330˚

210˚

11
6
4
3

14. 

A2  
B2  

62  (

8)2
 10
Since C is negative, use 10.

60˚

150˚
0



120˚


10
41


41

 r cos (v 
 309°)

A2  
B2  

(–1)2 
32
17. 

 
10

Since C is negative, use 
10
.
1

x


10

cos f 

 2209y 
 8279  0

3

10



10


10

7

10


 y 
   0
3

10
7

10
, p  
sin f  
10
10

f  Arctan (
3)
 
72°
Since cos f < 0, but sin f > 0, the normal lies in
the second quadrant.
f  180° 
 72° or 108°
p  r cos (v 
 f)

cos f  2219, sin f  2209, p  3
f  Arctan 2201
 44°
p  r cos (v 
 f)
3  r cos (v 
 44°)

7

10
 
10

r cos (v 
 108°)

18. 6  r cos (v 
 120°)
0  r (cos v cos 120°  sin v sin 120°) 
 6


3
 r sin v 
 6
0  
 12 r cos v  
2


3
y 
 6
0  
 12x  
2

y 
 12 or
0  
x  
3

x  
3
y 
 12  0

285

Chapter 9

28.

19. 4 
 r cos 
v  4

0 
 r 
cos v cos 4  sin v sin 4  4

x  2
y  8 or
0 
 2
x  2
y  8 
 0
2
20. 2 
 r cos (v  )
0 
 r (cos v cos   sin v sin ) 2
0 
 r cos v  2
0 
 x  2
x 
 2
21. 1 
 r cos (v  330°)
0 
 r (cos v cos 330°  sin v sin 330°)  1
0


r 
cos v cos

r cos 
v 
 sin v sin

7

6



1
y
2

13
5
 

13

30˚

150˚
180˚

3 6 9 12

330˚

210˚
240˚

26.
120˚

270˚
90˚

300˚

0˚

180˚

1 2 3 4
330˚

210˚
240˚

270˚

300˚


3

11
6

7
6

120˚
30˚


2

0

27.

3
2
90˚

5
3
60˚
30˚

150˚
180˚

2 4 6 8

0˚

330˚

210˚
240˚

5

13



0

3
13
5
13
, p 
 
sin f 
 
13
13

r cos (v  56°)

Use a graphing calculator and the INTERJECT
feature to find solutions to the system at (2.25,
0.31) and (5.39, 0.31). Since p, the length of the
normal, must be positive, use f 
 2.25 and p 

0.31.
0.31 
 r cos (v  2.25)
32a. p 
 r cos (v  f) → 6 
 r cos (v  16°)
Since the shortest distance is along the normal,
the closest the fly came was p or 6 cm.
32b. (p, f) or (6, 15°)
33. Since both normal segments have length 2, p must
be 2 in both equations. Since the two lines
intersect at right angles, their normals also
intersect at right angles. This can be achieved by
having the two f-values differ by 90°. To make
sure neither line is vertical, neither f-value
should be a multiple of 90°. Therefore, a sample
answer is 2 
 r cos (v  45°) and 2 
 r cos (v 
135°).
34. m 
 0
(y  4) 
 0(x  5) → y  4 
 0
cos f 
 0, sin f 
 1, p 
 4
Since cos f 
 0 when sin f 
 1, f 
 90°.
p 
 r cos (v  f)
4 
 r cos (v  90°)

1 2 3 4

4
3



→ p 
 2 cos 
76  f


6



3

13


→ p 
 3 cos 
4  f

5
6

60˚

150˚

Chapter 9

0˚

5
3

31. p 
 r cos (v  f)

 11 
 0

2
3

3
2

56°
p 
 r cos (v  f)

  11 
 0

25.



2
13
,
cos f 
 
13
f 
 Arctan 32

 
 11

60˚

4
3


 13

Since C is negative, use 13
.

y  10 
 0
x  3
90˚

11
6


A2  
B2 
 
22  32


3
1
r cos v   r sin v  5 
 0
2
2

3
1
x  y  5 
 0
2
2

120˚

7
6

2

x  y  22 
 0
3
23.
r 
 5 sec (v  60°)
r cos (v  60°) 
 5
r (cos v cos 60°  sin v sin 60°)  5 
 0

24.

0
2 4 6 8

(y  1) 
 3(x  4) → 2x  3y  5 
 0

r 
 11 sec 
v  76

7

6


3

6



5
3

3
2


2

4
2
 or 
30. m 
 
6
3

1

3
 r cos v  r sin v  11 
 0

2
2

3
x

2

11
6

2

13


22.
7

6

0

4
3

2
3
5
6

2 4 6 8

7
6

sin v  1

x  y  2 or
0 
 3
3
x  y  2 
 0

29.


3

6



2
2

x  y  4
0

2
2

0



2

5
6


2
2
 r cos v   r sin v  4
0

2
2

1
3
 r cos v  r
2
2
1
3
x  y  1
2
2

2
3

270˚

300˚

286

35a.

120˚

90˚

40. x  3y 
 6

60˚

x  6

y

3

30˚

150˚

y 
 13x  2

0˚

x 
 t, y 
 13t  2

12
5
25
0
37
505
0

180˚

N
 r2
41. A 
 
360 


330˚

210˚
240˚

270˚

65
 62


360 


300˚

35b. p 
 r cos (v  f)
→ p 
 125 cos (130  f)
→ p 
 300 cos (70  f)
Use a graphing calculator and the INTERSECT
feature to find the solutions to the system at
(45, 124.43) and (135, 124.43). Sinc p, the
length of the normal, must be positive, use f 

135° and p 
 124.43.
124.43 
 r cos (v  135°)
36. k 
 r sin (v  a)
k 
 r [sin v cos a  cos v sin a]
k 
 r sin v cos a  r cos v sin a
k 
 y cos a  x sin a
This is the equation of a line in rectangular
coordinates. Solving the last equation for y yields
k
. The slope of the line shows
y 
 (tan a)x  
cos a
that a is the angle the line makes with the x-axis.
To find the length of the normal segment in the
figure, observe that the complementary angle to a
in the right triangle is 90°  a, so the v-coordinate
of P in polar coordinates is 180°  (90°  a) 

a  90°. Substitute into the original polar
equation to find the r-coordinate of P:
k 
 r sin (a  90°  a)
k 
 r sin 90°
k
r
Therefore, k is the length of the normal segment.

y

20.42 ft2
42. Since 360° lies on the x-axis
of the unit circle at (1, 0),
sin 360° 
 y or 0.

(1, 0)

x

43. 2x3  5x2  12x 
 0
x(2x2  5x  12) 
 0
x(2x  3)(x  4) 
 0
x 
 0 or x 
 32 or x 
 4
44.

c2  d2 
 48
(c  d)(c  d) 
 48
12(c  d) 
 48
cd
4

Page 579
1.

Mid-Chapter Quiz
90˚

120˚

2.

60˚

180˚

2 4 6 8

240˚

120˚

P

270˚
90˚



37. p 
 r cos (v  f)
→ p 
 40 cos (0°  f)
→ p 
 40 cos (72°  f)
Use a graphing calculator and the INTERSECT
feature to find the solutions of the system at
(144, 32.36) and (36, 32.36). Since p, the length
of the normal, must be positive, use f 
 36° and
p 
 32.36.
32.36 
 r cos (v  36°)
38.
r
6

x2  y2 
 6
x2  y2 
 36
39. The graph of a polar equation of the form
r 
 a sin nv is a rose.

0˚

330˚
270˚

300˚

5. r 
 
(2

)2  (
2
)2

 4
 or 2

11
6

7
6

4.

2 4 6 8

240˚

0
1 2 3 4

4
3

60˚

210˚


6



30˚

180˚

x


3

0˚

300˚

150˚


2

5
6

330˚

210˚




2
3

30˚

150˚

3.

O

y

2
3

5
3

3
2

2


3

6

5
6


1

0

11
6

7
6
4
3

5
3

3
2

v 
 Arctan

 4




2





Since (2
, 2
) is in the third quadrant,
v 
   4 or 54.


2, 54

6. r 
 
02  (
4)2

 16
 or 4
Since x 
 0 when y 
 4, v 
 32.


4, 32

287

Chapter 9

x2  y2 
 36

x2  y2 
 36

r 
 6 or r 
 6
8.
r 
 2 csc v
r sin v 
 2
y
2


9.
A2  
B2 
 
52  (
12)2

 13
Since C is positive, use 13.

4. Sample answer: x2  1 
 0
(x  i)(x  i) 
 0, where the solutions are x 

x2  xi  xi  i2 
 0
x2  (1) 
 0
x2  1 
 0
6
4
2
5. i 
 (i )  i2

 12  (1)

 1
6. i10  i2 
 (i4)2  i2  i2

 (1)2i2  i2

 1  (1) or 2
7. (2  3i)  (6  i) 
 (2  (6))  (3i  i)

 4  4i
8. (2.3  4.1i)  (1.2  6.3i)

 (2.3  (1.2))  (4.1i  (6.3i))

 3.5  10.4i
9. (2  4i)  (1  5i) 
 (2  (1))  (4i  5i)

 1  9i
10. (2  i)2 
 (2  i)(2  i)

 4  4i  i2

 3  4i

7.

5

12
3
y   
 0
1
3x  
13
13
5

12
3
, p 
 
cos f 
 1
3 , sin f 
 
13
13
2
f 
 Arctan 
15


67°
Since cos f 0, but sin f  0, the normal lies in
the second quadrant.
f 
 180°  67° or 113°
p 
 r cos ( f)
3

13

10.


 r cos (v  113°)


A2  
B2 
 
(2)2 
 (6
)2

i
i
1  2i
 
   
11. 
1  2i
1  2i 1  2i


 210

Since C is negative, use 210
.
6
2
10

2

i  2i2




1  4i2

2
210



 y   
 0
210
x
cos f 



10
1
,
0

f 
 Arctan 


sin f 


310

10,

p




i2



5

10



10


 25  15i

3

1

72°
Since cos f 0 and sin f
the third quadrant.
f 
 180°  72° or 252°
p 
 r cos (v  f)
10



10

12. (2.5  3.1i)  (6.2  4.3i)

 (2.5  (6.2))  (3.1i  4.3i)

 3.7  7.4i N

0, the normal lies in

Pages 583–585


 r cos (v  252°)

Page 583

Check for Understanding

1. Find the (positive) remainder when the exponent
is divided by 4. If the remainder is 0, the answer
is 1; if the remainder is 1, the answer is i; if the
remainder is 2, the answer is 1; and if the
remainder is 3, the answer is i.
2.

Complex Numbers (a  bi )
Reals
(b 
 0)

Imaginary
(b  0)

19. 
12  i  (2  i) 
 
12  (2)  (i  (i))

Pure
Imaginary
(a 
 0)


 32  2i
20. (3  i)  (4  5i) 
 (3  (4))  (i  (5i))

 7  4i
21. (2  i)(4  3i) 
 8  10i  3i2

 5  10i
22. (1  4i)2 
 (1  4i)(1  4i)

 1  8i  16i2

 15  8i

3. When you multiply the denominators, you will be
multiplying a complex number and its conjugate.
This makes the denominator of the product a real
number, so you can then write the answer in the
form a  bi.
Chapter 9

Exercises

13. i6 
 i4  i2

 1  1

 1
14. i19 
 (i4)4  i3

 14  i

 i
15. i1776 
 (i4)444

 1444

1
16. i9  i5 
 (i4)2  i  (i4)2  i3

 12  i  12  i

 i  (i) or 0
17. (3  2i)  (4  6i) 
 (3  (4))  (2i  6i)

 1  8i
18. (7  4i)  (2  3i) 
 (7  2)  (4i  3i)

 9  7i

Simplifying Complex Numbers

9-5

i.

288

3i
3i
 
 
34. 
(2  i)2
(2  i)(2  i)

23. (1  7
i)(2  5
i)

 2  5
i  27
i  35
i2

 (2  35
)  (27
  5
)i
24. (2  3
)(1  12
) 
 (2  3
i)(1  12
i)

 2  212
i  3
i
 36
i2

 2  43
i  3
i  6

 8  33
i

3i




4  4i  i2

3i



3  4i

3  4i
3i
  


3  4i 3  4i
9  9i  4i2




9  16i2

13  9i



25

1  2i
2i
2i
 
   
25. 
1  2i
1  2i 1  2i
2  3i  2i2



1  4i2


 1235  295i

4  3i



5

35.

(1  i)2

(3  2i)2


 45  35i




3  2i
i
3  2i
 4 


26. 
4  i  
4  i
4  i




5  12i
2i
  


5  12i 5  12i
10i  24i2




25  144i2

11

0

 11
 17i
7

27.










5i

5i



24  10i



169

5i

5i

24
10
  i

 
169
169

25  10i  i2

25  i2

36a. Z 
 R  (XL  XC )j
→ Z 
 10  (1  2)j → Z 
 10  j ohms
→ Z 
 3  (1  1)j → Z 
 3  0j ohms
36b. (10  j)  (3  0j) 
 (10  3)  (1j  0j)

 13  j ohms

24  10i

26
12
5
  i
13
13

28. (x  i)(x  i) 
 0
x2  i2 
 0
x2  1 
 0
29.
(x  (2  i))(x  (2  i)) 
 0
(x  2  i)(x  2  i) 
 0
x2  2x  xi  2x  4  2i  xi  2i  i2 
 0
x2  4x  4  1 
 0
x2  4x  5 
 0
30. (2  i)(3  2i)(1  4i) 
 (6  i  2i2)(1  4i)

 (8  i)(1  4i)

 8  31i  4i2

 12  31i
31. (1  3i)(2  2i)(1  2i) 
 (2  8i  6i2)(1  2i)

 (4  8i)(1  2i)

 4  16i  16i2

 12  16i
32.

1
  3
i
2

1  2
i




1

1

36c. S 
 Z → S 
 
6  3j
1
6  3j
  


6  3j 6  3j
6  3j




36  9j2

6  3j



45
0.13  0.07j siemens

37a. x 



 3  4i
37b. No
37c. The solutions need not be complex conjugates
because the coefficients in the equation are not
all real.
37d. (3  4i)2  8i(3  4i)  25 
 0
7  24i  24i  32  25 
 0
0
0
(3  4i)2  8i(3  4i)  25 
 0
7  24i  24i  32  25 
 0
0
0
38. f(x  yi) 
 (x  yi)2

 x2  2xyi  y2

 (x2  y2)  2xyi
39a. z0 
 2  i
z1 
 i(2  i)  i2 or 1  2i
z2 
 i(2i  1) 
 2i2  i or 2  i
z3 
 i(2  i) 
 2i  i2 or 1  2i
z4 
 i(1  2i) 
 i  2i2 or 2  i
z5 
 i(2  i) 
 2i  i2 or 1  2i

1
  3
i
2
i
1  2
  
i
1  2
1  2
i
1
2
  
i  3i  6i2
2
2
1  2i


12  6  
22  3i


 
3



6
3
2
  
  i

 
16  
3 
3
6
2  2
i

3  
6i

2  2
i
3  6
i

3  6
i
3  6
i


   



6  2
6i  3
2i  
12i2

9  6i2




(6  2
3)  (2
6  3
2)i

15


23

8i 
(8i)2 
4(1)(
25)

2(1)

8i 36




2


 
2

33.

1  2i  i2

9  12i  4i2

2i



5  12i

12  11i  2i2

16  i2

10  11i



17
5i

5i

(1  i)(1  i)

 
(3  2i)(3  2i)


26


2
  

  i

 
25  
15 
15
5

289

Chapter 9

39b. z0 
 1  0i
z1 
 (0.5  0.866i)(1  0i) 
 0.5  0.866i
z2 
 (0.5  0.866i)(0.5  0.866i)

 0.25  0.866i  0.75

 0.500  0.866i
z3 
 (0.5  0.866i)(0.500  0.866i)

 0.250  0.750

 1.000  0.000i
z4 
 (0.5  0.866i)(1.000) 
 0.500  0.866i
z5 
 (0.5  0.866i)(0.500  0.866i)

 0.250  0.866i  0.75

 0.500  0.866i

46. tan a 
 43
tan2 a  1 
 sec2 a
4 2

3


 

1

25

9
9

25
3

5

sin2








2

48. h 
 x3


x3


tan 52° 
 
x  45

x tan 52°  45 tan 52° 
 x3

x tan 52°  x3
 
 45 tan 52°
45 tan 52°
tan 52°  3


x 
 
x 127.40
h 
 x3
 
 127.40(3
) 221 ft

3
10

p

20

8˚

52˚
45 ft

h

60˚
x

49. Enter the x-values in L1 and the f(x)-values in L2
of your graphing calculator. Make a scatter plot.
The data points are in the shape of a parabola, so
a quadratic function would best model the set of
data.
50. Let d 
 depth of the original pool.
The second pool’s width 
 5d  4, the length 

10d  6, and the depth 
 d  2.
(5d  4)(10d  6)(d  2) 
 3420
(50d2  70d  24)(d  2) 
 3420
50d3  100d2  70d2  140d  24d  48 
 3420
50d3  170d2  164d  3372 
 0
25d3  85d2  82d  1686 
 0
Use a graphing calculator to find the solution
d 
 3.
The dimensions of the original pool are 15 ft by
30 ft by 3 ft.
51. 80 
 k(5)(8)
2
k
y 
 2(16)(2)

 64

0

11
6
5
3

44. x  (3), y  6 
 t 1, 4
x  3, y  6 
 t 1, 4

45. u 
 1 8, 6, 4  2 2, 6, 3
4


 2, 32, 1  4, 12, 6

 6, 227, 5

Chapter 9

30˚

120˚


3

3
2






6

4
3

12

y 
 3.5 cos 
6t

r cos (v  162°)

7
6

4

5


period 
 1
2 or 6

18°
Since cos f 0, but sin f  0, the normal lies in
the second quadrant.
f 
 180°  18° or 162°
p 
 r cos (v  f)

7 14 21 28

5

cos B 
 1
3

1

6
210




25


cos2 B 
 
169

47. amplitude 
 2(7) or 3.5

 x  y   
 0

5
6

 cos2 B 
 1

33


62  (
2)2

 210

Since C is positive, use 210
.


2

2


1123


 65


A2  
b2 


2
3


 sin B
B  cos2 B 
 1

16

25
4

5

3

41. c1(cos 2t  i sin 2t)  c2(cos 2t  i sin 2t)

 c1 cos 2t  c1i sin 2t  c2 cos 2t  c2i sin 2t

 (c1  c2)(cos 2t)  (c1  c2)(i sin 2t)

 (c1  c2)(cos 2t) only if c1 
 c2

43.


 sin2 B

 

 
5
1
3   
 5 
 13 

11

3
10
 

20

sin2


 csc2 B

cos (a  B) 
 cos a cos B  sin a sin B

11  2i


11  2i


 

169

144
144

169
12

13


 cos2 a

sin a 


1

(3  4i)(1  2i)

2
3
210

210

310

10



,
cos f 
  10 , sin f 
 
10
1
f 
 Arctan 3

5


 sec2 a

sin2 a 


2

1  
1

 csc2 B
2

a

3 2

2

i

 
12 5  
125

42.

1  cot2 B 
 csc2 B

sin2 a  
5 
 1

1

1

11  2i
1

11  2i
11  2i

125

sec2


 cos a
a  cos2 a 
 1


40. (1  2i)3 
 
(1  2i)3




cot B 
 152

290

52.

4. The conjugate of a  bi is a  bi.
(a
 b
i)(a

bi) 
 
a2  b2, so the friend’s
method gives the same answer.
Sample answer: The absolute value of 2  3i is

22  32 
 13
. Using the friend’s method, the
absolute value is (2
 3
i)(2

3i) 
 4

9

13
.
5. 2x  y  (x  y)i 
 5  4i
2x  y 
 5
xy
4
2x  (x  4) 
 5
y 
 x  4
x
1
y 
 (1)  4 or 3
6.
7.

y 
 7  x2
x 
 7  y2
x  7 
 y2
x  7 
 y2
x
7

y
f1(x) 
 7

x

y

53.

(6, 8)

(6, 1)

O (1, 1)

x

f(x, y) 
 2x  y
f(1, 1) 
 2(1)  1 or 3
f(6, 1) 
 2(6)  1 or 11
f(6, 8) 
 2(6)  8 or 4
The maximum value is 3 and the minimum value
is 11.
54. x  2y  7z 
 14
x  3y  5z 
 21
y  2z 
 7
x  3y  5z 
 21 → 5x  15y  25z 
 105
5x  y  2z 
 7
5x  y  2z 
 7
16y  27z 
 112
y  2z 
 7
→
16y  32z 
 112
16y  27z 
 112
 6y  27z 
 112
1

59z 
 0
z
0
y  2(0) 
 7
→ y
7
x  2(7)  7(0) 
 14 → x 
 0
(0, 7, 0)
55. Since BC 
 BD, m∠BDC 
 m ∠DCB 
 x
m ∠DBC 
 180  120 or 60.
x  x  60 
 180
2x 
 120
x 
 60
x  40 
 60  40 or 100
The correct choice is A.

i

2

2

1

1

O

2 1
(2, 1)

i

1

1

2

2

1



2

z 
 
12  (
2
 )2

 3

2
v 
 Arctan 
2  2

z 
 
(22 
(1)2

 5

2
8. r 
 
2  (
2)2

7

 or 22


 8


 4

v is in the fourth quadrant.
7
7
2  2i 
 22
 
cos 4  i sin 4
5
9. r 
 
42  52
v 
 Arctan 
4

 41


 0.90
4  5i 
 41
(cos 0.90  i sin 0.90)
0

(2)2 
 02
10. r 
 


v 
 Arctan 
2  

 or 2

 4

 is on the x-axis at 2.
2 
 2 (cos   i sin )
11.

The Complex and Polar Form
of Complex Numbers

Pages 589–590

2 1 O

1

2
3


2


3


(4, 3 )

i

5
6

9-6



2

(1, 
2 )



11
6
4
3

1. To find the absolute value of a  bi, square a and
b, add the squares, then take the square root of
the sum.


2. i 
 0  i; cos 2 
 0 and sin 2 
 1


i 
 cos 2  i sin 2
3. Sample answer: z1 
 i, z2 
 i

3
2



5
3


4
cos 3  i sin 3

3
——
2


 4
  i

1

2

i

 2  23



2
3


6

2 4 6 8
7
6

Check for Understanding

12.

0






i


2


3

6

5
6


(2, 3)

0
1 2 3 4

7
6



11
6
4
3

3
2

5
3

2(cos 3  i sin 3)
2(0.99  i(0.14))

 1.98  0.28i

z1  z2 
 z1  z2
i  (i) 
 i  i
0 
 i  i
0 2i

291

Chapter 9

13.

2
3


2

i

19.


3

20.

i

i


6

5
6

(2, 3)

3

( 2 , 2
) 0



1 2 3 4
7
6




O

5
3

3
2

3
(cos 2  i sin
2
3

 2 (1  i(0))
3

 2

14.



O

11
6
4
3

(3, 4)

2)
z 
 
22  32

 13


z 
 
32  (
4)2

 25
 or 5

21.

i

22.

i

1

i

0.63
1

0.90 
0.36 1 1.30

O

0.38i

O




O
(0, 3)

1
(1, 5)

102 
152
15a. magnitude 
 

 325

18.03 N

(1)2 
 (5
)2 z 
 
02  (
3)2
z 
 

 26


 9
 or 3
23.
24.

15

15b. 
 
 Arctan 1
0
56.31°

(1, 
5 )

i

i

2

Pages 590–591

(4, 
2 )

1

Exercises

16. 2x  5yi 
 12  15i
2x 
 12
5y 
 15
x
6
y 
 3
17. 1  (x  y)i 
 y  3xi
1
y
x  y 
 3x
x  (1) 
 3x
1 
 2x
1
 
 x
2
18. 4x  (y  5)i 
 2x  y  (x  7)i
y5
x7
4x 
 2x  y
y 
 x  12
4x 
 2x  (x  12)
3x 
 12
x 
 4
y 
 (4)  12 or 8

2 1 O

1

2





O

1
2

z 
 
(1)2 
 (5

)2

 6

25. r 
 
(4)2 
 62

 52
 or 213


z 
 
42  (
2
 )2

 18
 or 32


v 
 Arctan 
3
3

26. r 
 
32  32




 18
 or 32



 4




3  3i 
 32
 
cos 4  i sin 4
27. r 
 
(1)2 
 (
 )2
3


3


v 
 Arctan 

1   
4

 or 2

 4


 3

v is in the third quadrant.
4

4

i 
 2
cos 3  i sin 3
1  3
28. r 
 
62  (
8)2

 or 10
5.36

 100
v is in the fourth quadrant.
6  8i 
 10(cos 5.36  i sin 5.36)

Chapter 9

292

8

v 
 Arctan 
6  2


v 
 Arctan 

4   
1

29. r 
 
(4)2 
 12


2.90

 17
v is in the second quadrant.
4  i 
 17
(cos 2.90  i sin 2.90)

39.

5


 2

0


 9
 or 3
v is on the x-axis at 3.
3 
 3(cos 0  i sin 0)

41.


0


42

 or 42




 32
v is on the x-axis at 42
.
42
 
 42
 (cos   i sin )
2


34. r 
 0  (2)2

 4
 or 2
3
Since x 
 0 when y 
 2, v 
 2.

i


3

2
3




(3, 4 ) 
6

5
6


1 2 3 4



4
3

5
3

3
2






2

32

3
2


3



7
6



11
6
4
3

3
2

5
3

3(cos   i sin )

 3(1  0)

 3

1

O

0.50 0.39i
0.60 0.39i

0.25 i 1

44.



i i

1

1


 2  2i

32

2
3

i


2

38.


3

2
3


6

5
6

0



1 2 3 4
7
6

0
1 2 3 4

5
3


 2  2 i
37.


6

0.44 0.44i

5
3




2


3

i

1



cos 
6  i sin 
6


 3
2  i 
2


2

6

4
3

3
cos 4  i sin 4

i

(3, )



11
6

(1, 

)
6 11

7
6

2
3

1

0
1 2 3 4

5
3

0.44 0.5i


6



11
6

7
6

i


3

5
6
0

(5, 0) 0

43.

2

3
2

5
6

5(cos 0  i sin 0)

 5(1  0)

5

3

36.

42.


3

2 4 6 8

3
2

11
6

2.5(0.54  i(0.84))


6

4
3

0




 1.35  2.10i

7
6

2i 
 2
cos 2  i sin 2
2
3

i



v 
 Arctan   

3



5
6
0


33. r 
 (4
)2  
2
02


2

2
3


2

6

2.5(cos 1  i sin 1)


2
2

  2
i

 2

v 
 Arctan 
3

32  02
32. r 
 

(2.5, 1) 

4
3

5

i



3

7
6

2
cos 4  i sin 4
2
2


2

1 2 3 4

5
3

3
2

i





11
6

4

 or 25

2.03

 20
v is in the second quadrant.
2  4i 
 25
(cos 2.03  i sin 2.03)

2
3
5
6

0
1 2 3 4

(2, 5
 )
4
7
6
4
3


v 
 Arctan 

2   

31. r 
 
(2)2 
 42

40.


3

6



21

 or 29
5.47

 841
v is in the fourth quadrant.
20  21i 
 29(cos 5.47  i sin 5.47)

35.


2

i

5
6

v 
 Arctan 
2
0   2

202 
(21
)2
30. r 
 

2
3

(2, 4
 )
3
4
3
3
2
4

11
6
5
3
4

2
cos 3  i sin 3

3


 2
2  i 
2
1

i

 1  3



i


2

1


3

1

O


6

5
6

1

0



3 6 9 12

(10, 6)
7
6



45.

4
3

3
2

i
1

11
6

0.5  0.5i

5
3

1

10(cos 6  i sin 6)

O

1



0.5  0.5i
1

10(0.960  i(0.279))

 9.60  2.79i

293

Chapter 9

46a. 40∠30° 
 40(cos 30°  j sin 30°)

3

2


 40


 j


1

2

51. (6  2i)(2  3i) 
 12  22i  6i2

 6  22i
52. x 
 3 cos  135°
y 
 3 sin  135°




 34.64  20j
60∠60° 
 60(cos 60°  j sin 60°)

1
3

 60
2  j
 2


2

49a.
49b.
49c.
49d.
50a.





71.96


v 
 Arctan 
64.64









tan 60°  tan 45°

1  tan 60° tan 45°
3
1
1  (3
)(1)
3
  1 1  
3
  
1  3
 1  3

3
  3  1  3



4  23




2


 2  3

v

55. w 
 t
12(2)


 1 or 24
v 
 rq

 18(24) or 432 cm/s
432 cm/s 
 4.32 m/s
13.57 m/s

14.11



 
 Arctan 
21.69

12

56. sin A 
 1
8
A 
 sin1 
3
2

A 41.8°
47. 2a
  
 3a
1
5

2a  1 
 3a  5
4
a
58. as x → , y → ; as x → , y → 
y 
 2x2  2
x
1000
10
0
10
1000

1.41(cos 0.79  i sin 0.79)

y
2  106
202
2
202
2  106

59. In the fourth month, the person will have received
3 pay raises.
$500(1.10)3 
 $665.50
The correct choice is D.




cos 4  i sin 4

 102


 14.14(cos 0.79  i sin 0.79).
50d. To multiply two complex numbers in polar form,
multiply the moduli and add the amplitudes. (In
the sample answer for 50c, note that 5.18  1.89

 7.07, which is coterminal with 0.79.)

Chapter 9

32


 

z2 
 5(cos 0.93  i sin 0.93)
z1z2 
 52
 (cos 1.71  i sin 1.71)

 7.07(cos 1.71  i sin 1.71)
50c. Sample answer: Let z1 
 2  4i and
z2 
 1  3i. Then
z1 
 25
(cos 5.18  i sin 5.18)
4.47(cos 5.18  i sin 5.18),
z2 
 10
(cos 1.89  i sin 1.89)
3.16(cos 1.89  i sin 1.89), and
z1z2 
 (2  4i)(1  3i)

 10  10i



 2

53. magnitude 
 
(3)2 
 72

 58

u  7j
u
3, 7 
 3i
54. tan 105° 
 tan (60°  45°)

25.88
0.58
21.69  14.11j 
 25.88 (cos 0.58  j sin 0.58) ohms
Translate 2 units to the right and down 3 units.
Rotate 90° counterclockwise about the origin.
Dilate by a factor of 3.
Reflect about the real axis.
Sample answer: let z1 
 1  i and z2 
 3  4i.
z1z2 
 (1  i)(3  4i)

 1  7i

50b. z1 
 2 
cos 4  i sin 4


 2

32 32
 


2 , 2 

96.73
48°
v(t) 
 96.73 sin (250t  48°)
47. The graph of the conjugate of a complex number is
obtained by reflecting the original number about
the real axis. This reflection does not change the
modulus. Since the amplitude is reflected, we can
write the amplitude of the conjugate as the
opposite of the original amplitude. In other words,
the conjugate of r(cos v  i sin v) can be written as
r(cos (v)  i sin (v)), or r(cos v  i sin v).
48a. 10(cos 0.7  j sin 0.7) 7.65  6.44j
16(cos 0.5  j sin 0.5) 14.04  7.67j
48b. (7.65  6.44j)  (14.04  7.67j)

 (7.65  14.04)  (6.44j  7.67j)

 21.69  14.11j ohms

2
48c. r 
 21.69
 14
.112


 3
2

32


 30  51.96j
46b. (34.64  20j)  (30  51.96j)

 (34.64  30)  (20j  51.96j)

 64.64  71.96j
46c. v(t) 
 r sin (250t  v°)
2  71
r 
 
64.64
.962


2


 3
2

294



3

9-6B Graphing Calculator Exploration:

5. r 
 4

Geometry in the Complex Plane

2

v 
 6  3
3




 6 or 2


cos 
2  i sin 
2 
 34(0  (1)i)

3

4

Page 592

3


 4i

1. They are collinear.

9



v 
 4  
2

4

6. r 
 2 or 2

9

2

11


 4  4 or 4
11

11


2


2

2
cos 4  i sin 4 
 2
2  i
2

 2
  2
i
7. r 


1
(6)
2



5

v 
 3  6

or 3

2

5

7


 6  6 or 6

2. Yes. M is the point obtained when T 
 0, and N is
the point obtained when T 
 1.
3. The points are again collinear, but closer together.

7

7


3

3
cos 6  i sin 6 
 3
2  i
2
1

33

3


 2  2i
22  (
23
)2
8. r1 
 

 16
 or 4
r 
 4(23
) or 83


r2 
 
(3)2 
 (3

)2

 12
 or 23


23

3

v1 
 Arctan 
2

  
v2 
 Arctan 
3



5


 6


 3


5

v 
 3  6
2

9-7

7

3

1

11

11






 2
cos 6  j sin 6  3
cos 3  j sin 3

r 
 2(3) or 6

11



11

2

13



v 
 6  3

 6  6




 6 or 6



V 
 6
cos 6  j sin 6 volts

Pages 596–598

Exercises

10. r 
 4(7) or 28



2

v 
 3  3
3


 3 or 

Check for Understanding



7


 12  43
i
9. E 
 IZ

28(cos   i sin ) 
 28(1  i(0))

 28

1. The modulus of the quotient is the quotient of the
moduli of the two complex numbers. The
amplitude of the quotient is the difference of the
amplitudes of the two complex numbers.
2. Square the modulus of the given complex number
and double its amplitude.
3. Addition and subtraction are easier in rectangular
form. Multiplication and division are easier in
polar form. See students’ work for examples.
4. r 
 2  2 or 4

7

83

cos 6  i sin 6 
 83

2  i
2

Products and Quotients of
Complex Numbers in Polar Form

Pages 596

5


 6  6 or 6

4. The points are on the line through M and N.
5. If one of a, b, or c equals 0, then aK  bM  cN is
on KMN. If none of a, b, or c equals 0, then
aK  bM  cN lies on or inside KMN.
6. M is the point obtained when T 
 0 and N is the
point obtained when T 
 1. Thus, a point between
M and N is obtained when 0 T 1.
7. The distance between z and 1  i is 5. This
defines a circle of radius 5 centered at 1  i.
8. The distance between a point z and a point at
2  3i is 2.
z  (2  3i) 
 2

6

3



2



v 
 4  4

11. r 
 2 or 3



 4 or 2



3
cos 2  i sin 2 
 3(0  i(1))

 3i
12. r 


1

2

3



1



v 
 3  6

or 6

2

3






 6  6 or 6

v 
 2  2


cos 6  i sin 6 
 16
23  i
12

1

6

4


 2 or 2

3

4(cos 2  i sin 2) 
 4(1  i(0))

4

1





12  12 i

295

Chapter 9

3

13. r 
 5(2) or 10

24. r1 
 
22  (
2)2 r2 
 
(3)2 
 32

 8
 or 22


 18
 or 32

r 
 22
(32
) or 12

v 
   4
4

3

7


 4  4 or 4
7

7


2

10
cos 4  i sin 4 
 10
2  i
2

2

v1 
 Arctan 
2  2





7

3

7

10



11

3

3


 12i
25. r1 
 
(2
)2
 (2
)2 r2 
 (3


)2  (
2
32

)2

 4
 or 2

 36
 or 6
r 
 2  6 or 12

11

v1 
 Arctan 
  2 v2 
 Arctan 


3
cos 6  i sin 6 
 3
2  i 
2
1

33

7

17. r 


2


2


2




22



2

7



12(cos   i sin ) 
 12(1  i(0))

 12
(3
)2
 (
1)2
r2 
 
22  (23

)2
26. r1 
 

 4
 or 2

 16
 or 4

3

v 
 4  4
4


 4 or 

2

20

4

7





cos


2

 i sin


2






4
(0
3

22

cos

 i sin



3

4

2
5
 4 or 4


2
2
2  i 2



2










3






1

5



v 
 3  3

or 8

4


 3
4

4


3

8
cos 3  i sin 3 
 8
2  i
2
1


 4  43
i

Chapter 9

r 2 
 
62  62

 72
 or 62


 






v2 
 Arctan 
6
6




 4



22



3 (0  i(1))
2
2


 3i

12
cos 6  i sin 6 
 12
2  i
2

 63
  6i

8
62

4

32

42


6

2

2

2
2



or 3
4
2
v1 
 Arctan 
42

3

 4
3

v 
 4  4
2


 4 or 2
2


2
 cos   i sin 
2
2
3







1

r 
 

 i (1))

 
 22





 )2  
2
(42
)2
27. r1 
 (4

 64
 or 8

v 
 4  2
5

4




 4  4i


 6  6 or 6

23. r 


10


3

v 
 3  
6

22. r 
 2(6) or 12

4

1

2

11


cos 6  i sin 6 
 12
23  i
12

7
22
  
6
6
15

6 or 2


 2  2i



5

1

2

4
3i



5

4

11


 6  6 or 6

11

3

) or 22

21. r 
 2(2

5


 3

v 
 6  3

3.10  2.53i



4

3

11

v 
 6  3

20. r 
 15 or 3

v2 
 Arctan 
   2

1
3



 6

v 
 2  3.6 or 5.6

4[cos (5.6)  i sin (5.6)]

23


v1 
 Arctan 
  2

2(cos   i sin ) 
 2(1  i(0))

 2
18. r 
 3(0.5) or 1.5
v 
 4  2.5 or 6.5
1.5(cos 6.5  i sin 6.5) 1.46  0.32i
4

1

r 
 4 or 2

or 2

19. r 
 1 or 4

5

12


 3  33
i
7


 4


 4 or 


3
2

6(cos 300°  i sin 300°) 
 6
  i


5

3
2
32


v 
 4  4

v 
 240°  60°

 300°
1

2

2


2


 4

3


 2  2i
16. r 
 2(3) or 6



12
cos 2  i sin 2 
 12(0  i(1))


 6  6 or 6
11




 4 or 2



14

3

v 
 4  4

v 
 3  2

15. r 
 1 or 3

3


 4


 4


5
v 
 3  6
2
5

 6  6
3


 6 or 2

18
cos 2  i sin 2 
 18(0  i(1))

 18i

3

7


 52
  52
i
14. r 
 6(3) or 18


v2 
 Arctan 

3   

2

296

E

28. I 
 Z

5

34.
13

5




3  2j

r1 
 13
13

13

r cos 
v  6 
 5

r 2 
 
32  (
2)2

 13


r
cos v cos 6  sin v sin 6  5 
 0

v2 
 Arctan 
3

2x  2y  5 
 0

5

5


3

1

2r cos v  2r sin v  5 
 0

r 
  or 13



3

2

v1 
 0

35.

v 
 0  (0.59) or 0.59
I 
 13
(cos 0.59  j sin 0.59)
29. Z 




3  2j amps

130˚

x lb

x lb
23 lb

x lb

23 lb

50˚

r 2 
 
42  (
3)2

 25
 or 5

r1 
 100

x lb

Prop

Since the triangle is isosceles, the base angles are

100

r 
 5 or 20

180  50

congruent. Each measures 2 or 65°.

3

v2 
 Arctan 
4

v1 
 0

v 
 0  (0.64) or 0.64
z 
 20(cos 0.64  j sin 0.64)

 16  12j ohms
30. Start at z1 in the complex
plane. Since the modulus
z1
5
of z2 is 1, z1z2 and z will
6
2
both have the same
modulus
as z1. Then z1z2 
z1
and z can be located by
2

rotating z1 by 6
7
6
counterclockwise and
clockwise, respectively.

23

sin 50°

0.64

23 sin 65°

sin 50°

2
3

i

z1z2


2

z1
z2


3

36.

6
0

1 2 3 4



3
2





5


x

27.21 x; 27.21 lb
cos 2x  sin x 
 1
1  2 sin2 x  sin x 
 1
2 sin2 x  sin x 
 0
sin x (2 sin x  1) 
 0
sin x 
 0 or 2 sin x  1 
 0
x 
 0°

11
6
4
3

x




sin 65°

23 sin 65° 
 x sin 50°

1

sin x 
 2

x 
 30°
y 
 cos x
x 
 cos y
arccos x 
 y
38. BC 
 ED 
 BE 
 AF 
 CD 
 3
AB 
 FE 
 2
AC 
 AB  BC

 2  3 or 5
FD 
 FE  ED

 2  3 or 5
perimeter of rectangle ACDF 
 3  5  3  5
or 16
perimeter of square BCDE 
 4(3) or 12
16  12 
 4
The correct choice is C.

5
3

37.

31a. The point is rotated counterclockwise about the
origin by an angle of v.
31b. The point is rotated 60° counterclockwise about
the origin.
32. Since a 
 1, the equation will be the form z2  bz
 c 
 0. The coefficient c is the product of the
7
7
solutions, which is 6
cos 6  i sin 6, or
33
  3i in rectangular form. The coefficient b
is the opposite of the sum of the solutions, so
convert the solutions to rectangular form to do the
addition.
5

b 
 3
cos 3  i sin 3  2
cos 6  i sin 6
3
3


 
2  2i  (3
  i)
3

2
33


 
2  3
  
2i
3

Therefore, the equation is z2  

2
33

2




1

x  y 10 
 0
3

0.59

E

I
100

4  3j

r 
 5 sec
v  6

3
2

iz  (33  3i) 
 0.

52  (
12)2
33. r 
 

9-8

 3
 

Page 602

12


 
 Arctan 
5  2

Powers and Roots of Complex
Numbers
Graphing Calculator Exploration

1. Rewrite 1 in polar form as 1(cos 0  i sin 0).
Follow the keystrokes to find the roots at 1,
0.5  0.87i, and 0.5  0.87i.

 or 13
5.11

 169
5  12i 
 13(cos 5.11  i sin 5.11)





2. Rewrite i in polar form as 1
cos 2  i sin 2.
Follow the keystrokes to find the roots at
0.92  0.38i, 0.38  0.92i, 0.92  0.38i, and
0.38  0.92i.

297

Chapter 9



3. Rewrite 1  i in polar form as 2

cos 4 
i sin



4

.

v 
 2

1

6


cos 
16
2  i sin 
16
2




 1
cos 1
2  i sin 12 
1

Follow the keystrokes to find the roots at 1.06 
0.17i, 0.17  1.06i, 0.95  0.49i, 0.76 
0.76i, and 0.49  0.95i.
4. equilateral triangle
5. regular pentagon
6. If a  0 and b 
 0, then a  bi 
 a. The principal
roots of a positive real number is a positive real
number which would lie on the real axis in a
complex plane.

Pages 604-605

0.97  0.26i

1

3


5



4

1

4

3

3  4n

3  4n

3

3

x1 
 cos 8  i sin 8
x2 
 cos
x3 
 cos
x4 
 cos

0.38  0.92i

7
7
  i sin 
8
8
11
11
  i sin 
8
8
15
15
  i sin 
8
8



4

i

1

0.92  0.38i
0.38  0.92i
0.92  0.38i

0.38  0.92i

0.92  0.38i
1

1

O

0.92  0.38i
0.38  0.92i
1

10. 2x3  4  2i 
 0 → x3 
 2  i
Find the third roots of 2  i.
(2)2 
 (1
)2 
 5

r 
 

a
a  ai

O

a



1


1

3

)
x2 
 (5



4

. By De Moivre’s Theorem,


the polar form of (a  ai)2 is 2a2
cos 2  i sin 2.
 i sin

)
x3 
 (5





23
cos (3)
6  i sin (3)
6

1
3







 8 cos 
2  i sin 
2

 
 8(0  i (1))

 8i

Chapter 9

v  2
v  2
  i sin 

cos 
3
3
v  4
v  4




cos

i
sin


3
3 

O
1
1.29  0.20i
1

5

32  (
5)2 or 34
 v 
 Arctan 
3
6. r 
 
34
4 (cos (4)(
)  i sin (4)(v))

 644  960i

v

0.47  1.22i

1

v 
 Arctan 
 or 6

(3
)2
 (
1)2 or 2
5. r 
 

1

3

v

i



Since cos 2 
 0, this is a pure imaginary number.



v  2n

x1 
 (5
) 3 
cos 3  i sin 3 
 0.47  1.22i

4. Shembala is correct. The polar form of a  ai is
a2

cos

v  2n


 (5
) 3 
cos 3  i sin 3
1




4

1


v 
 Arctan 

2   

3.605240263 1
1


(2  i) 3 
 [5
(cos (v  2i)  i sin (v  2n))] 3

a  ai

a

1


3


 cos 8  i sin 8

a  ai

a

3

v 
 2

(i) 
 1 (cos 
2  2n  i sin 
2  2n 4


 4  4i
2. Finding a reciprocal is the same as raising a
number to the 1 power, so take the reciprocal of
the modulus and multiply the amplitude by 1.

i


cos 
 (
)  i sin (3)(
)

02  (
1)2 
 1
r 
 


cos (5)
   i sin (5)
 
5
5

 42

cos 4  i sin 4


2
2

 42
 
2  i
2

a  ai

2.677945045

1

3


 0.82  1.02i
9. x4  i 
 0 → x4 
 i
Find the fourth roots of i.

Check for Understanding

(2
 )5

1


v 
 Arctan 

2   

(2)2 
 (1
)2 or 5

8. r 
 

1. Same results, 4  4i; answers may vary.
(1  i)(1  i)(1  i)(1  i)(1  i)

 (1  2i  i2)(1  2i  i2)(1  i)

 (2i)(2i)(1  i)

 4(1  i)

 4  4i
(1  i)5

→ r 
 2
, v 
 4

3.



02  12 or 1
7. r 
 

1.030376827

298

1



0.82  1.02i

1.29  0.20i
0.81  1.02i

11. For w1, the modulus 
 (
0.82 
(0.7
)2)2 or 1.13.
For w2, the modulus 
 1.132 or 1.28.
For w3, the modulus 
 1.282 or 1.64.
This moduli will approach infinity as the number
of iterations increases. Thus, it is an escape set.

21. r 
 
(2)2 
 12 
 5



v 
 Arctan 

2   
1

2.677945045

)
( 5

1

4


cos 
 (v)  i sin 
 (v)
1

4

1

4


 0.96  0.76i

1

v 
 Arctan 
4

42  (
1)2 
 17

22. r 
 

Pages 605–606
12.

33



Exercises


cos (3) 6  i sin


27 cos 2  i sin 2





 






6

(3)




13.






)
(17

(22
)


 













 


 162



 i sin

2

2

 i


2

2

7



1





v 
 Arctan 
1 
 3

4


 16 


i


3

2


 8  83
i


cos 
12
2  i sin 
12
2

2n

2n


 cos 3  i sin 3



x1 
 cos 0  i sin 0 
 1
6

3

v 
 Arctan 


32  (
6)2 
 35

16. r 
 



4

4

1


3

1

2

0.9827937232
(13
)2 (cos (2)(v)  i sin (2)(v))

 0.03  0.07i

2

O

1

v 
 Arctan 
2
4

1

 1  
3 i
2


 1.107148718
4
(25
) (cos (4)(v)  i sin (4)(v))

 112  384i
2


3

 1  
3 i

3

2

1

i

v 
 Arctan 
2

22  42 
 25

18. r 
 

2

x3 
 cos 3  i sin 3 
 2  2i

(35
)4 (cos (4)(v)  i sin (4)(v))

 567  1944i
22  32 
 13

17. r 
 

2

x2 
 cos 3  i sin 3 
 2  2i

1.107148718

1




v 
 2


 0.71  0.71i
26. x3  1 
 0 → x3 
 1
Find the third roots of 1.
r 1
 
12  02 
 1
v
0
1


3
1 
 [1 (cos (0  2n)  i sin (0  2n))] 3


 16 
cos 3  i sin 3
1
2


cos 
14
34  i sin 
14
34


 0.91  0.61i
1

2


3

24 
cos (4)
3  i sin (4)
3

1

4

02  12 
 1
25. r 
 



12  (
3
 )2 
 2
15. r 
 
4




( 2
)

7

4

21

4


 16  16i


1


v 
 Arctan 

1   
3

v 
 Arctan 





cos 
13
4  i sin 
13
4


 4

7


2

2

2


 1.37  0.37i

(22
)3
cos (3)
9  i sin (3)
4

 162
 
cos



v 
 Arctan 
2 
 4

(1)2 
 (1
)2 
  2

24. r 
 


 162
  162
i
2
2

14. r 
 
(2)  2 
 22

21

4

1

3

0.2449786631

1

3

22  22 
 22

23. r 
 



cos (5) 4  i sin (5) 4
5
5
32 cos 4  i sin 4


2
2
32 2  i 2





cos 
 (v)  i sin 
 (v)
1

3


 1.60  0.13i


 27(0  i(1))

 27i
25



1

3

2

1



19. 32 5 
cos 
5
3  i sin 
5
3
1


 2
cos

2

15

1

 i sin

2

15



1.83  0.81i
20. r 
 
(1)2 
 02 
 1
1

4

1

v



cos 
 ()  i sin 
 ()
1

4



1

4




 cos 4  i sin 4
0.71  0.71i

299

Chapter 9

27. x5  1 → x5 
 1
Find the fifth roots of 1.
r 
 
(1)2 
 02 
 1

i



2  

2i

v


1

5



2  

2i

1
1

5

(1) 
 [1 (cos (  2n)  i sin (  2n))]
  2n

  2n


 cos 5  i sin 5
x1 
 cos
x2 
 cos
x3 
 cos
x4 
 cos
x5 
 cos

O

1



  i sin  
 0.81  0.59i
5
5
3
3
  i sin  
 0.31  0.95i
5
5
5
5
  i sin  
 1
5
5
7
7
  i sin  
 0.31  0.95i
5
5
9
9
  i sin  
 0.81  0.59i
5
5



1

1
2i


2  


2i


2  


30. x4  (1  i) 
 0 → x4 
 1  i
Find the fourth roots of 1  i.
1




v 
 Arctan 
1 
 4
1

r 
 
12  12 
 2



1




(1  i) 4 
 2

cos 
4  2n  i sin 
4  2n 4

i

1

0.31  0.95i

1


  8n

  8n


 (2
) 4 
cos 16  i sin 16

0.81  0.59i

i
1

O

1

1.07  0.21i
0.81  0.59i

0.31  0.95i
1

3

1

v
0

2i
2




1


9

9

1

)
x3 
 (2
1

x4 
 (2
)

3

2

3

17



4

1


(1  3
i) 4

4

64 
 [64 (cos (0  2n)  i sin (0  2n))]
n

n

 i sin



2

1


x1 
 2

1


x2 
 2 4

 (cos   i sin ) 
 22

x3 
 22




1


x3 
 2 4

 
 22i

Find the fourth roots of 16.
2  02 
 16

r 
 (16)
v

1

4

1


1

 i sin

4

28

0.59  1.03i

1.03  0.59i

  2n



 i sin

28

1

4

x1 
 2 
cos 4  i sin 4 
 2
  2
i
 i sin

4


cos 12  i sin 12 
 0.59  1.03i
10
10


cos 1
2  i sin 12  
 1.03  0.59i
22
22

cos 12  i sin 12 
 0.59  1.03i
i


 2 
cos 4  i sin 4
3

4
5

4
7

4

4  6n

x4 
 2 4 
cos 12  i sin 12 
 1.03  0.59i

(16) 
 [16 (cos (  2n)  i sin (  2n))]
  2n

4  6n

1

4

 
 22i

3
3
x4 
 22
 cos 2  i sin 2
3x4  48 
 0 → x4 
 16

1



 2 4 
cos 12  i sin 12

x1 
 22
 (cos 0  i sin 0) 
 22



2

4


 2 
cos 
3  2n  i sin 
3  2n 4


 22
 
cos 2  i sin 2

3

4
5

4
7

4

17



cos 1
6  i sin 16  
 1.07  0.21i
25
25

cos 16  i sin 1
6  
 0.21  1.07i


3

1

4



1

4

 
    or 
v 
 Arctan 

3
3
1 

2i
2


1

4

x2 
 22
 
cos

1

4

i 
 0 → x4 
 1  3
i.
31. 2x4  2  23
Find the fourth roots of 1  3
i.
r 
 
(1)2 
 (
 )2 
 2
3

2

Chapter 9



) 4 
cos 16  i sin 16 
 0.21  1.07i
x2 
 (2

3 2 1 O
1

x4 
 2 
cos

1


2

2

2

x3 
 2 
cos

0.21  1.07i

x1 
 (2
) 4 
cos 16  i sin 1
6  
 1.07  0.21i

i

2

2

x2 
 2 
cos

1

O
1
1.07  0.21i

28. 2x4  128 
 0 → x4 
 64
Find the fourth roots of 64.
r 
 
642 
02 
 64

29.

1

0.21  1.07i

1

 
 2  2i
 
 2  2i
 
 2  2i

O

0.59  1.03i
1

300

1
1.03  0.59i

32. Rewrite 10  9i in polar form as
9

10

 cos 
tan1 

181

38a. The point at (2, 2) becomes the point at (0, 2).
From the origin, the point at (2, 2) had a length
of 22
 and the new point at (0, 2) has a length
2

of 2. The dilation factor is 2.

9

10

  i sin 
tan1 
 .

Use a graphing calculator to find the fifth roots at
0.75  1.51i, 1.20  1.18i, 1.49  0.78i,
0.28  1.66i, and 1.66  0.25i.
33. Rewrite 2  4i in polar form as
25
[cos (tan1 (2))  i sin (tan1 (2))].
Use a graphing calculator to find the sixth roots at
1.26  0.24i, 0.43  1.21i, 0.83  0.97i,
1.26  0.24i, 0.43  1.21i, and 0.83  0.97i.
34. Rewrite 36  20i in polar form as

2 y
(0, 2)
1

x
2


2

2




3

4

38b.

For w2, the modulus 
 (0.81)2 or 0.66.
For w3, the modulus 
 (0.66)2 or 0.44.
This moduli will approach 0 as the number of
iterations increase. Thus, it is a prisoner set.
36a. In polar form the 31st roots of 1 are given by
2n

x3 
 cos
x4 
 cos
x5 
 cos
x6 
 cos

 i sin
 i sin
 i sin
 i sin

2

3
3

3
4

3
5

3







1


 2(cos 90°  i sin 90°)



5



10

v 
 6  3
11


3

11


 6  6 or 6
1

45°


3

1
2







3
i
2


 1



2

43. cos 22.5° 
 cos 
2

x2 
 cos 3  i sin 3 
 2  2i
2

3
3

3
4

3
5

3




 33
  3i
41. (2  5i)  (3  6i)  (6  2i)

 (2  (3)  (6))  (5i  6i  2i)

5i
42. x 
 t, y 
 2t  7

n

1



22 (cos 45°  i sin 45°)

11


 cos 3  i sin 3



2

 i sin 
2

6
cos 6  i sin 6 
 6 
2  i 
2

1







40. r 
 2(3) or 6

1 6 
 [1 (cos (0  2n)  i sin (0  2n))] 6
x1 
 cos 0  i sin 0 
 1

(cos 45°  i sin 45°)

 2

2
 
2
2

The square is rotated 90° counterclockwise and
dilated by a factor of 0.5.
39. The roots are the vertices of a regular polygon.
Since one of the roots must be a positive real
number, a vertex of the polygon lies on the
positive real axis and the polygon is symmetric
about the real axis. This means that the non-real
complex roots occur in conjugate pairs. Since the
imaginary part of the sum of two complex
conjugates is 0, the imaginary part of the sum of
all the roots must be 0.


cos 3
1  i sin 31 , n 
 0, 1, . . . , 30. Then
2n


a 
 cos 31 . The maximum value of a cosine
expression is 1, and it is achieved in this
situation when n 
 0.
36b. From the polar form in the solution to part a, we
2n
2n

get b 
 sin 3
1 . b will be maximized when 31 is

as close to 2 as possible. This occurs when n 
 8,
16
so the maximum value of b is sin 3
1 , or about
0.9987.
37. x6  1 
 0 → x6 
 1
Find the sixth roots of 1.
r 
 
12  02 
 1
v
0
n

2


 0.5  0.5i

2



122 

 2 or 0.81.

1


1

2

5

35. For w1, the modulus 


2n

1 O
1

4106
 cos 
tan1 
9  i sin 
tan1 
9.
Use a graphing calculator to find the eighth roots
at 1.59  0.10i, 1.05  1.19i, 0.10  1.59i,
1.19  1.05i, 1.59  0.10i, 1.05  1.19i,
0.10  1.59i, and 1.19  1.05i.
5

(2, 2)
2 
2





1
3
2  2i

1
3
  i
2
2






1  cos 45°



2



2

1
2

2

2  2




4 


2  

2

2

44. Find B.
B 
 180°  90°  81°15


 8°45

Find a.
a

tan 81°15
 
 2
8
28 tan 81°15
 
 a
181.9 
 a

301

Chapter 9

Find c.

15. Sample answer: (4, 585°), (4, 945°), (4, 45°),
(4, 405°)
(r, v  360k°)
→ (4, 225°  360(1)°) → (4, 585°)
→ (4, 225°  360(2)°) → (4, 945°)
(r, v  (2k  1)180°)
→ (4, 225°  (1)180°) → (4, 45°)
→ (4, 225°  (1)180°) → (4, 405°)

90˚
16.
17.
2

60
120
2

28

cos 81°15
 
 c
28


c

cos 81°15


c 
 184.1
45. Let x 
 the number of large bears produced.
Let y 
 the number of small bears produced.
x  300
y
1200
y  400
(300, 900)
1000
x  y  1200
800
f(x, y) 
 9x  5y
(800, 400)
600
f(300, 400) 
 9(300)  5(400) 400

 4700
200 (300, 400)
x
f(300, 900) 
 9(300)  5(900)
O 200 600 1000

 7200
f(800, 400) 
 9(800)  5(400)

 9200
Producing 800 large bears and 400 small bears
yields the maximum profit.
46. 0.20(6) 
 1.2 quarts of alcohol
0.60(4) 
 2.4 quarts of alcohol
1.2  2.4

64

˚

˚

3

180˚

1 2 3 4

240˚

270˚

18.

90˚

120˚

3.6


 1
or 36% alcohol
0

19.

1.
3.
5.
7.
9.

Check for Understanding
2.
4.
6.
8.
10.

absolute value
prisoner
pure imaginary
rectangular
Argand

120˚

Polar
iteration
cardioid
spiral of Archimedes
modulus

11.
120˚

12.

180˚

0˚

1 2 3 4

13.

2
3

270˚

2

240˚

300˚

14.


3

5
6

C


6

1 2 3 4

7
6

11
6
4
3

Chapter 9

3
2

5
3


2

4
3

3
2

270˚


22 


 32

(32
, 32
)
25. x 
 2 cos 330°


 

5
3


2

3


2


 3

(3
, 1)

302

8 16 24 32

0

11
6

7
6

2
3

3
2

5
3


2


3

6

5
6


2 4 6 8

0

11
6

7
6
4
3

300˚


6

11
6

0˚

330˚

limaçon
24. x 
 6 cos 45°

D

7
6



23.

2 4 6 8


3

0


3

6

4
3

30˚

240˚

1 2 3 4

5
3

5
6

0˚

60˚

210˚


6



90˚

300˚

5
6
0



2
3

270˚

3
2

2

Spiral of Archimedes

180˚

330˚

210˚

2
3

300˚

150˚

0˚

1 2 3 4

B

330˚
240˚

30˚

180˚

A
210˚

120˚

11
6

21.

330˚

22.

60˚

150˚

30˚

150˚

7
6

60˚

210˚
270˚

1 2 3 4

4
3

2 4 6 8


3

0



circle
90˚

120˚

60˚

0˚

30˚

180˚

Skills and Concepts

90˚

90˚


2

5
3


6

300˚

150˚

240˚

Pages 608–610

270˚

3
2

5
6

330˚

20.

Pages 607

2
3

60˚

210˚
240˚

11
6
4
3

1 2 3 4

Chapter 9 Study Guide and Assessment

0
1 2 3 4

7
6

30˚

180˚


6



300˚

150˚

The correct choice is A.

0˚

330˚

210˚

3

5
6

30˚

150˚

3
2

rose
y 
 6 sin 45°


22 


6


 32


y 
 2 sin 330°


 1


 2 2

 1

5
3


34
2


 2
2


34
2


 2
2

26. x 
 2 cos


 2

(2
, 2
)

27. x 
 1 cos 






y 
 1 sin 


v 
 Arctan





y  6 or
0 
 x  3
x  3
y  6 
 0

35. 4 
 r cos 
v  2

4

3






3


3



5




 50
 or 52



 4


52, 4

36.

 
v 
 Arctan 

3 
1

30. r 
 
(3)2 
 12
 3.16

 10
(3.16, 2.82)

37.

2.82

38.

v 
 Arctan 
4
2

31. r 
 
42  22

 20
 4.47
(4.47, 0.46)


A2  
B2 


0.46

39.


22  12

40.


 5

Since C is positive, use 5
.
2
5


1
5

25

cos f 
 5,
1
f 
 Arctan 2

3
5


  x   y   
 0

41.

5


35


sin f 
 5, p 
 5

5  2i

20  18i  4i2

16

42.


32  12

5i

1  
2i

1  
2i
1  2
i

5i
1  
2i


   



5  52
i  i  2
i2

1  2i2




(5  2
)  (52
  1)i

3
1  52


5  2



 3  3i

v 
 Arctan 
2
2

22  22
43. r 
 

4

10

x   y   
 0




 8
 or 22


10

22
 
cos 4  i sin 4


2
10

sin f 
 1
, p 
 5
0

18°
Since cos f 0 and sin f
the third quadrant.
f 
 180°  18° or 198°
p 
 r cos (v  f)

18


 29  29i

0, the normal lies in


 10

Since C is positive, use 10
.

210


5

4  2i

  


5  2i 5  2i
16  18i


 r cos (v  207°)


 

4  2i

5  2i


 29

27°
Since cos f 0 and sin f
the third quadrant.
f 
 180°  27° or 207°
p 
 r cos (v  f)

1

10
310

cos f 
 10,
1
f 
 Arctan 3

0 
 0  r sin v  4
0 
 y  4
0 
 y  4 or
y4
0
i10  i25 
 (i4)2  i2  (i4)6  i

 (1)2  (1)  (1)6  i

 1  i
(2  3i)  (4  4i) 
 (2  (4))  (3i  (4i))

 2  7i
(2  7i)  (3  i) 
 (2  (3))  (7i  (i))

 1  6i
3
3
i (4  3i) 
 4i  3i4

 4(i)  3(1)

 3  4i
(i  7)(i  7) 
 i2  14i  49

 1  14i  49

 48  14i



25  4i2


 


A2  
B2 




0 
 r cos v cos 2  r sin v sin 2  4

v 
 Arctan 
5

29. r 
 
52  52

3

10


3


0 
 2x  2 y  3




(3

)2  (
3)2

 or 23


 12

33.

3


1

0 
 2r cos v  2 r sin v  3


1


23, 43

35


5



0 
 r cos v cos 3  r sin v sin 3  3
1



2


0
(0, 1)

32.




 2




2

28. r 


34. 3 
 r cos 
v  3

y 
 2 sin


 4



44. r 
 
12  (
3)2

v 
 Arctan 
1  2
3



 10
10
 (cos 5.03  i sin 5.03)

0, the normal lies in

45. r 
 
(1)2 
 (3

)2

 
v 
 Arctan 

1 
3


2

 or 2

 4
2
cos 3  i sin 3
2


 r cos (v  198°)

5.03


 3

2

46. r 
 
(6)2 
 (4
)2

4


v 
 Arctan 

6   

 or 213


 52
213
 (cos 3.73  i sin 3.73)
47. r 
 
(4)2 
 (1
)2

1


v 
 Arctan 

4   



 17
17
 (cos 3.39  i sin 3.39)

303

3.73

3.39

Chapter 9

48. r 
 
42  02

 16
 or 4
4(cos 0  i sin 0)
(2
 )2  
2
02
49. r 
 

 8
 or 22

22
(cos   i sin )

v
0

02  32
50. r 
 

v


60. r 
 (
3
 )2 
 (1
)2 
 2
27 
cos

v




2
3
5
6


 128 




2


6



2 cos 6  i sin 6
1
3


 2 2  i 2








 

( 3, 5

3)
4
3



2
2
12 cos 3  i sin 3
1
3


 12 2  i 2


 

3




5

11
6


 4
5

15

2


 16  16i
1

4

1



1





0.92  0.38i






2

1

3

3

3

1

1






1.24  0.22i


 162
  162
i
55. r 
 2(5) or 10
10 (cos 2.5  i sin 2.5)
8.01  5.98i

v 
 2  0.5 or 2.5

Page 611

Applications and Problem Solving

65. lemniscate

8

56. r 
 2 or 4

5

7

10





v



3


cos  i sin 
3 1
3


 2 
2  i 
2






2



3

6



r cos 
v  2  5 
 0


r cos 
 cos 2  r sin v sin 2  5 
 0



6


6

or

r sin v  5 
 0
y5
0
y 
 5



3
E

68. I 
 Z
50  180j

32





4  5j


 4  4i
2.2

50  180j

0.5 (cos 0.9  i sin 0.9)
0.31  0.39i
59. r 
 
22  22 
 22










v 
 Arctan 
2 
 4
2

4  5j

 


4  5j  4  5j

v 
 1.5  0.6 or 0.9


58. r 
 
4.4 or 0.5

59.04°


67.

3

2

125


 21,25
0
 145.77
(145.77, 59.04°)



4cos 
2  i sin 
2 
 6  6 or 2

 4(0  i(1))

 4i

v 
 Arctan 
7
5 

752 
1252
66. r 
 

7

v 
 6  3



3



1


 2 3 
cos 
18  i sin 
18


 4  4 or 4

2






1

200  470j  900j2

16  25j2
1100  470j2

41

26.83  11.46j amps



(2
)8 
cos (8)
4  i sin (8)
4

 4096 (cos 2  i sin 2)

 4096

304




v 
 Arctan 
 
3

 6

2 
cos 
3
6  i sin 
3
6

v 
 4  2

2


Chapter 9



1 
cos 
4
2  i sin 
4
2


 32
2  i 
2

3

v 
 2

(3
)2
 12 
 2
64. r 
 

3

or



02  12 
 1
63. r 
 

 cos 8  i sin 8

32
cos 4  i sin 4

3

2


2


 162
 
2  i 
2



54. r 
 8(4) or 32

57. r 


15


 162
 
cos 4  i sin 4


 6  63
i

6

4

5

(22
)3 
cos (3)
4  i sin (3)
4







2


v 
 Arctan 

2   

(2)2 
 (2
)2 
 22

62. r 
 

5
3

3
2




3


 4(cos 3  i sin 3)

 4

5
5
3 cos 3  i sin 3
1
3


 3 2  i 2
3
33


 2  2i


2
v 
 3  3 or 3

53. r 
 4(3) or 12




1

(2
)4 
cos (4)
4  i sin (4)
4

1 2 3 4

7
6


 3
i





v 
 Arctan 

1   
3

0

5
3

3
2


 






 4


3

6



11
6
4
3

i 2

5
6

1 2 3 4 

7
6



2
3

0







 643
  64i

52.

( 2, 
6 )



(1)2 
 12 
 2

61. r 
 


3

i 2





 128 



 9
 or 3


3
cos 2  i sin 2
51.


1
 
 
6
3



(7) 6  i sin (7) 6
7
7
cos 6  i sin 6

1
3
2  i 2

v 
 Arctan

Page 611

180  (c  b) 
 180  x
x
cb
The correct choice is E.
4. Volume 
 
wh
16,500 
 75  w  10
16,500 
 750w
22 
 w

Open-Ended Assessment

1a. Sample answer: 4  6i and 3  2i
(4  6i)  (3  2i) 
 (4  3)  (6i  2i)

 7  4i
1b. No. Sample explanation: 2  3i and 5  i also
have this sum.
(2  3i)  (5  i) 
 (2  5)  (3i  (i))

 7  4i
2a. Sample answer: 4  i
z 
 
42  12

 17

2b. No. Sample explanation: 1  4i also has this
absolute value.
z 
 
12  42

 17


w
5.

8

h

C

The answer choices include sin x. Write an
expression for the height, using the sine of x.
h

1

sin x 
 8

A 
 2bh
1

8 sin x 
 h


 2 (10)(8 sin x)


 40 sin x
The correct choice is B.
3. Since PQRS is a parallelogram, sides PQ and SR
are parallel and m∠Q 
 m∠S 
 b.

M
c˚

P

Q
b˚

T
x˚

R
a˚

S

1

10

The correct choice is A.
6. Consider the three unmarked angles at the
intersection point. One of these angles, say the top
one, is the supplement of the other two unmarked
angles, because of vertical angles. So the sum of
the measures of the unmarked angles is 180°.
The sum of the measures of the marked angles
and the three unmarked angles is 3(180), since
these angles are the interior angles of three
triangles.
m(sum of marked angles) 
m(sum of unmarked angles) 
 3(180)
m(sum of marked angles)  180 
 3(180)
m(sum of marked angles) 
 360
The correct choice is C.
7. Subtract the second equation from the first.
5x2  6x

 70
5x2
 6y 
 10
6x  6y 
 60
x  y 
 10, so 10x  10y 
 100.
The correct choice is E.
8. Since ∠B is a right angle, ∠C is a right angle also,
because they are alternate interior angles.
In the triangle containing ∠C, 90  x  y 
 180 or
x  y 
 90.
The straight angle at D is made up of 3 angles.
120  x  x 
 180
2x 
 60 or x 
 30
x  y 
 90
(30)  y 
 90
y 
 60
The correct choice is B.
9. In the slope-intercept form of a line, y 
 mx  b,
m represents the slope of the line, and b
represents the y-intercept. Since the slope is
3
given as 2, the slope-intercept form of the line is
3
y 
 2x  b.
Since (–3, 0) is on the line, it satisfies the
3
9
equation. 0 
 2(–3)  b. So b 
 2.
The correct choice is D.
10. Note that consecutive interior angles are
supplementary.
110  2x 
 180
y  x 
 180
2x 
 70
y  (35) 
 180
x 
 35
y 
 145
The answer is 145.

B

10

1





1099 
 100100  10100
9

SAT and ACT Practice

x˚

1

100100




10100

1. ∠a and ∠b form a linear pair, so ∠b is
supplementary to ∠a. Since ∠b and ∠d are
vertical angles, they are equal in measure. So ∠d
is also supplementary to ∠a. Since ∠d and ∠f are
alternate interior angles, they are equal. So ∠f is
supplementary to ∠a. And since ∠f and ∠h are
vertical angles, ∠h is supplementary to ∠a. The
angles supplementary to ∠a are angles b, d, f, and
h. The correct choice is A.
2. Draw the given triangle and draw the height h
from point B.

A


 
 75 ft

The correct choice is A.

Chapter 9 SAT & ACT Preparation
Page 613

h 
 10 ft

O
N

In SMO, c  b  a 
 180 or a 
 180  (c  b).
Also, x  a 
 180 or a 
 180  x since consecutive
interior angles are supplementary.

305

Chapter 9

Chapter 10 Conics
10-1

8. AB 
 
(x2  
x1)2 
(y2 
y1)2

Introduction to Analytic Geometry

Pages 619–620


 
(6  3
)2  (2
 4)2

 13

slope of 
AB


Check for Understanding

y2  y1

1. negative distances have no meaning
2. Use the distance formula to show that the
measure of the distance from the midpoint to
either endpoint is the same.
a
5
3a. Yes; the distance from B to A is 2 and the
a
5
distance from B to C is also 2.
3b. Yes; the distance from B to A is 
a2  b2 and
the distance from C to A is also 
a2  b2.
3c. No; the distance from A to B is 
a2  b2, the
distance from A to C is a  b, and the distance
from B to C is b2
.
4. (1) Show that two pairs of opposite sides are
parallel by showing that slopes of the lines
through each pair of opposite sides are equal.
(2) Show that two pairs of opposite sides are
congruent by showing that the distance
between the vertices forming each pair of
opposite sides are equal.
(3) Show that one pair of opposite sides is parallel
and congruent by showing that the slopes of
the lines through that pair of sides are equal
and that the distances between the endpoints
of each pair of segments are equal.
(4) Show that the diagonals bisect each other by
showing that the midpoints of the diagonals
coincide.
5. d 
 
(x2  
x1)2 
(y2 
y1)2
d 
 
(5  5
)2  (1
1  1
)2
2
2
d 
 
0  1
0
d 
 
100 or 10
x1  x2 y1  y2

 
2 ,
2




5  5 1  11
, 
2
2






 (5, 6)


m

x x
2

24

1

2





6  3 or  3

DC 
 
(x2  
x1)2 
(y2 
y1)2

 
(8  5
)2  (7
 9)2

 13

slope of 
DC

y2  y1


m

x x
2

79

1

2





8  5 or  3

Yes; A
 and DC 
 13
,
B
D
C
 since AB 
 13
2
and A
B
D
C
 since the slope of A
B
 is 3 and the
2
slope of 
DC
 is also 3.
(x2  
x1)2 
(y2 
y1)2
9. XY 
 

 
[1 
(3)]2
 (
6  2
)2

 68
 or 217

XZ 
 
(x2  
x1)2 
(y2 
y1)2

 
[5  (
3)]2 
 (0 
2)2

 68
 or 217

Yes; X
 and XZ 
 217
,
Y
X
Z
, since XY 
 217
therefore XYZ is isosceles.
10a.
y
D (c, a)

A(0, a)
E
B (0, 0)



O

C (c, 0)

10b. BD 
 
(c  0
)2  (a
 0)2
2
2

 
c  a

(x2  
x2)2 
(y2 
y1)2
6. d 
 
d 
 
(4 
0)2 
(3 
0)2
2
2
d 
 
(4) 
 (3
)
d 
 25
 or 5

AC 
 
(c  0
)2  (0
 a)2
2
2

 
c  a
Thus, A
C
B
D
.

x1  x2 y1  y2

0  (4) 0  (3)
, 

2, 2 
 

2
2

10c. The midpoint of 
AC
 is 
2, 2. The midpoint of
c a
B
D
 is 
2, 2. Therefore, the diagonals intersect
c a
at their common midpoint, E
2, 2. Thus,
AE

E
C
 and B
E
E
D
.
c


 (2, 1.5)

[0  (
2)]2 
 (4 
2)2
7. d 
 
2
2
d 
 
2  2
d 
 8
 or 22

x1  x2 y1  y2

2  0 2  4
, 

2, 2 
 

2
2

a

10d. The diagonals of a rectangle are congruent and
bisect each other.
11a. Both players are located along a diagonal of the
field with endpoints (0, 0) and (80, 120). The
kicker’s teammate is located at the midpoint of
this diagonal.


 (1, 3)

x1  x2 y1  y2

0  80 0  120
 

2, 2 
 

2 ,
2


 (40, 60)

Chapter 10

x

306

11b. d 
 
(x2  
x1)2 
(y2 
y1)2

19. d 
 
(x2  
x1)2 
(y2 
y1)2

d 
 
(40 
0)2 
(60 
0)2
2
2
d 
 
40 
60
d 
 5200

d 
 2013
 or about 72 yards

Pages 620–622

d 
 
(c  2
 c)2
 (d 
 1 
d)2
2
2
d 
 
2  (
1)
d 
 5

x1  x2 y1  y2

cc2 dd1
, 

2, 2 
 

2
2
2c  2 2d  1






 2 , 2 
1

 
c  1, d  2

Exercises

12. d 
 
(x2  
x1)2 
(y2 
y1)2

20. d 
 
(x2  
x1)2 
(y2 
y1)2

d 
 
[4  (
1)]2 
 (13
 1)2
d 
 
52  1
22
d 
 169
 or 13

d 
 
[w  (
w  2
)]2  (
4w  
w)2
d 
 
22  (
3w)2
d 
 
4  9
w2 or 
9w2 
4

x1  x2 y1  y2

1  4 1  13
, 

2, 2 
 

2
2

x1  x2 y1  y2

w  2  w w  4w
, 

2, 2 
 

2
2
5

 
w  1, 2w


 (1.5, 7)

(x2  
x1)2 
(y2 
y1)2
13. d 
 
d 
 
(1 
1)2 
(3 
3)2
2
2
d 
 
(2) 
 (6
)
d 
 40
 or 210


21.

d 
 
(x2  
x1)2 
(y2 
y1)2

20 
 (2a

 a)2 
 [7 
(9)]2
2
2
20 
 (3a)

 16
20 
 
9a2 
256
400 
 9a2  256
144 
 9a2
a2 
 16
a 
 16
 or 4
22. Let D have coordinates (x2, y2).

x1  x2 y1  y2

1  (1) 3  (3)
, 

2, 2 
 

2
2


 (0, 0)

(x2  
x1)2 
(y2 
y1)2
14. d 
 
d 
 
(0  8
)2  (8
 0)2
d 
 
(8)2 
 82
d 
 128
 or 82


4  x2 1  y2


2, 2 
 
3, 52

x1  x2 y1  y2

80 80
 

2, 2 
 

2 , 2 

4  x2

2


 (4, 4)


 3

1  y2

2

5


 2

4  x2 
 6
1  y2 
 5
x2 
 10
y2 
 6
Then D has coordinates (10, 6).
y
23. Let the vertices
A(2, 3)
of the quadrilateral
D (3, 2)
be A(2, 3), B(2, 3),
C(2, 3), and D(3, 2).
x
O
A quadrilateral is a
parallelogram if one pair
B (3, 2)
of opposite sides are
C (2, 3)
parallel and congruent.
AD

 and B
C
 are one pair of opposite sides.
slope of 
AD
slope of 
BC



(x2  
x1)2 
(y2 
y1)2
15. d 
 
d 
 
[5  (
1)]2 
 [3
 (
6)]2
2
2
d 
 
6  3
d 
 45
 or 35

x1  x2 y1  y2

1  5 6  (3)
 , 

2, 2 
 

2
2


 (2, 4.5)

(x2  
x1)2 
(y2 
y1)2
16. d 
 
d 
 
(72

 3
 )2  
2
[1 
(5)]2
2  42
d 
 
(42
)
d 
 48
 or 43

x1  x2 y1  y2

  72
 5  (1)
32
, 

2, 2 
 
 

2
2

y2  y1


m

x x


 (52
, 3)

2

1

23

(x2  
x1)2 
(y2 
y1)2
17. d 
 




3  (2)

d 
 
(a  a
)2  (
9  7
)2
2
2
d 
 
0  (
16)
d 
 256
 or 16

1


 5

y2  y1


m

x x
2

1

3  (2)




2  (3)
1


 5

Their slopes are equal, therefore A
D
B
C
.
AD 
 
(x2  
x1)2 
(y2 
y1)2

x1  x2 y1  y2

a  a 7  (9)
 

2, 2 
 

2 ,
2


 
[3  (
2)]2 
 (2 
3)2
2
2

 
5  (
1)

 26

BC 
 
(x2  
x1)2 
(y2 
y1)2


 (a, 1)

(x2  
x1)2 
(y2 
y1)2
18. d 
 
d 
 
[r  2
 (6 
 r)]2
 (s 
s)2
2
2
d 
 
(8) 
0
d 
 64
 or 8


 
[2  (
3)]2 
 [3
 (
2)]2
2
2

 
5  (
1)

 26

The measures of 
AD
 and B
C
 are equal.
Therefore A
BC
D
B
C
. Since A
D
B
C
 and A
D

,
quadrilateral ABCD is a parallelogram; yes.

x1  x2 y1  y2

6rr2 ss
, 

2, 2 
 

2
2
2r  4 2s






 2 , 2


 (r  2, s)

307

Chapter 10

24. Let the vertices of the
quadrilateral be
A(4, 11), B(8, 14),
C(4, 19), and D(0, 15).
A quadrilateral is a
parallelogram if both
pairs of opposite sides
are parallel.
A
B
 and D
C
 are one pair
of opposite sides.
slope of 
AB

y2  y1

m

x x
2

20

y C (4, 19)

16
D (0, 15)
12

1

E
FH

G
 since the slope of E
F
 is 2 and the slope
1


of 
HG
 is  2 . Thus the points form a
parallelogram. 
EF
⊥F
G
 since the product of the
1 2
FG
slopes of E
F
 and 
, 2  1, is 1. Therefore, the
points form a rectangle.
28. Let A(0, 0), B(b, c), and C(a, 0) be the vertices of
a triangle. Let D be the midpoint of 
AB
 and E
be the midpoint of B
C
.

B (8, 14)
A (4, 11)

8
4

x
4 O

4

8

12

y

slope of 
DC

y2  y1

m

x x

1

2

14  11

B (b, c)

1

19  15




84




40

3

D

4


 2

DC
Since 
AB
 ⁄ 
, quadrilateral ABCD is not a
parallelogram; no.
25. The slope of the line through (15, 1) and (3, 8)
should be equal to the slope of the line through
(3, 8) and (3, k) since all three points lie on the
same line.
slope through (15, 1)
slope through (3, 8)
and (3, 8)
and (3, k)
y2  y1

8  1



3  15
9
1




18 or 2
k8
1
 
  ⇒ k
6
2

2






DE 


1

k  (8)

3  (3)
k8

6

1


 5

y
D (b  a, c)


 
(4  2
)2  (4
 5)2

 5

HG 
 
(x2  
x1)2 
(y2 
y1)2

O A(0, 0)


 
(2  0
)2  (0
 1)2

 5

slope of 
EF

y2  y1
1

1





4  2 or  2

slope of 
FG

y2  y1


m

x x
1

2





2  4 or 1

slope of 
HG

y2  y1


m

x x
1

1





2  0 or  2

E
FH
 and HG 
 5
.

G
 since EF 
 5
Chapter 10

C (a, c)

B (b, 0)

x

AC 
 
(a  0
)2  (c
 0)2
2
2

 
a  c
BD 
 
(b  a
 b)2
 (c 
0)2

 
a2  c2
AC 
 
a2  c2 
 
a2  c2 
 BD, so the
diagonals of an isosceles trapezoid are congruent.


m

x x

2

a

Since DE 
 2AC, the line segment joining the
midpoints of two sides of a triangle is equal in
length to one-half the third side.
29. In trapezoid ABCD, let A and B have coordinates
(0, 0) and (b, 0), respectively. To make the trapezoid
isosceles, let C have coordinates (b  a, c) and let
D have coordinates (a, c).

Yes, AB 
 4, BC 
 4, and CA 
 4. Thus,
A
B
B
C
  CA. Therefore the points A, B, and C
form an equilateral triangle.
27. EF 
 
(x2  
x1)2 
(y2 
y1)2

01

c

ba
b 2
c
c 2
   

 
2 




2 
2
2
a2

(3 
1)2 
(0  
0)2
CA 
 

 16
 or 4

2

ba

 or 

 
4
2

[1  (
1)]2 
 (0 
23
)2
BC 
 

 16
 or 4

04

ba c0

c

(a  0
)2  (0
 0)2
AC 
 
2

 
a or a

2  (2
AB 
 
[1 
(3)]
3
 
0)2

 16
 or 4

2

b

The coordinates of E are 
2, 2 or 
2, 2.

(x2  
x1)2 
(y2 
y1)2
26. d 
 

45

0b 0c

The coordinates of D are 
2, 2 or 
2, 2.

y2  y1

1

C (a, 0) x

O A (0, 0)


m

x x


m

x x
2

E


 4 or 1

308

30. In ABC, let the vertices be A(0, 0) and B(a, 0).
Since 
AC
 and B
C
 are congruent sides, let the third
a
vertex be C
2, b. Let D be the midpoint of 
AC

and let E be the midpoint of 
BC
.

32. Let the vertices of quadrilateral ABCD be A(a, e),
B(b, f), C(c, g), and D(d, h). The midpoints of 
AB
,
B
CD
C
, 
, and D
A
, respectively,
cd gh

ad eh

N
2, 2, and P
2, 2.

C ( a2, b)

y

bc fg

ab ef

are L
2, 2, M
2, 2,

y
D

B (b, f )

L

A(a, e)
E

O

O A(0, 0)

D (d, h)

x

B (a, 0)

The coordinates of D are:




 or 


The coordinates of E are:




 or 


a

2

0 b0
 , 
2
2
a

2

a b0
 , 
2
2

0 
0 
 

b2


 

 

BD 


b
1 9a

 
 

b2


a  a

0  
4  
2
2 
4 

2

2

b

2

2

.

3a b
, 
4 2

2

.

x

C (c, g)
fg
ef
   
2
2

bc
ab
  
2
2
gh
eh
  
2
2

ad
cd
   
2
2

ge


or 
c  a.

eg


or 
a  c.

These slopes are equal, so 
LM
N
P
.
fg
gh
  
2
2
The slope of M
N
 is 
bc
cd
  
2
2
ef
eh
  
2
2
The slope of P
L
 is 
ad
ab
   
2
2

2

Since AE 
 BD, the medians to the congruent
sides of an isosceles triangle are congruent.
31. Let A and B have coordinates (0, 0) and (b, 0)
respectively. To make a parallelogram, let C have
coordinates (b  a, c) and let D have coordinates
(a, c).
y
D (a, c)

N

The slope of N
P
 is

9a2

4

1

2

M

The slope of L
M
 is

a b
, 
4 2

AE 


3a

4

O

P

f h


or 
bd .

hf


or 
d  b.

These slopes are equal, so 
MN
P
L
. Since
L
M
N
P
, and M
N
P
L
, PLMN is a parallelogram.
33. Let the vertices of the
y
rectangle be A(3, 1),
B
B(1, 3), C(3, 1), and
D(1, 3). Since the area
A
of a rectangle is the length
x
O
times the width, find the

C (a  b, c)

C

measure of two consecutive
sides, 
AD
 and D
C
.

O A(0, 0)

B (b, 0)

(x2  
x1)2 
(y2 
y1)2
AD 
 

x

ab 0c
,  or
2
2
ab0 c0
, 
2
2

The midpoint of 
BD
 is 


D




 
[1  (
3)]2 
 (3
 1)2
2
2

 
4  (
4)

 32
 or 42

DC 
 
(x2  
x1)2 
(y2 
y1)2

ab c
,  .
2
2
ab c



or
2 , 2






The midpoint of 
AC
 is 

 

.
Since the diagonals have the same midpoint, the
diagonals bisect each other.


 
(3  1
)2  [
1  (
3)]2
2
2

 
2  2

 8
 or 22

Area 
 
w

 (42
)(22
)

 16
The area of the rectangle is 16 units2.
34a.
y
20

10
20 10 O

10

20

30

x

10

309

Chapter 10

37a. Find a representation for MA and for MB.
MA 
 
t2  (
3t  1
5)2

 
t2  9
t2  9
0t  2
25

 
10t2 
90t 
225

34b. The two regions are closest between (12, 12)
and (31,0).
d 
 
(x2  
x1)2 
(y2 
y1)2

 
[31 
(12
)]2  (
0  1
2)2
2
2

 
43 
(12
)

 1993
 or about 44,64
The distance between these two points is about
44.64 pixels, which is greater than 40 pixels.
therefore, the regions meet the criteria.
35. Let the vertices of the isosceles trapezoid have
the coordinates A(0, 0), B(2a, 0), C(2a  2c, 2b),
D(2c, 2b). The coordinates of the midpoints are:
P(a, 0), Q(2a  c, b), R(a, 2b), S(c, b).
y
D (2c, 2b)

R

MB 
 
(t  9
)2  (3
t  12
)2
2
2

 
t  1
8t  8
1  9
t  7
2t  144


 
10t2 
90t 
225
By setting these representations equal to each
other, you find a value for t that would make the
two distances equal.
MA 
 MB
2

10t 
90t 
225 
 
10t2 
90t 
225
Since the above equation is a true statement,
t can take on any real values.
37b. A line; this line is the perpendicular bisector of
A
B
.
b
a2  b2
v 
 Arctan a
38. r 
 

C (2a  2c, 2b)

12


 
(5)2 
 122
S

O A (0, 0)

Q

P



 Arctan 
5


 169
 or 13
1.176005207
(5  12i)2 
 132 [cos 2v  i sin 2v]

 119  120i
u
39. If v 
 (115, 2018, 0), then
u 
 
v
1152 
20182
 02

 40855
49
 or about 2021
The magnitude of the force is about 2021 N.

B (2a, 0) x

PQ 
 
(2a 
c  a
)2  (b
 0)2
2
2

 
(a  c
)  b
QR 
 
(2a 
c  a
)2  (b
 2b
)2

 
(a  c
)3  b2
RS 
 
(a  c
)2  (2
b  b
)2
2
2

 
(a  c
)  b
PS 
 
(a  c
)2  (0
 b)2

 
(a  c
)2  b2
So, all of the sides are congruent and
quadrilateral PQRS is a rhombus.
36a. distance from fountain to rosebushes:
d 
 
(x2  
x1)2 
(y2 
y1)2

1

1



40. 2 sec2 x 
 
1  sin x  1  sin x

2 sec2 x 

2 sec2 x 

2 sec2 x 

2 sec2 x 

41.
s 
 rv
11.5 
 12v

(1  sin x)  (1  sin x)

(1  sin x)(1  sin x)
2

1  sin2 x
2



cos2 x
2 sec2 x

11.5

v 
 1
2 radians
11.5

12

180°

  54.9°
42. sin 390° 
 sin (390°  360°)
1

 sin 30° or 2
43.
z2  8z 
 14
2
z  8z  16 
 14  16
(z  4)2 
 2
z  4 
 2

z 
 4 2

44. x2 
 16
x 
 16
 or 4
y2 
 4
y 
 4
 or 2
Evaluating (x  y)2 when x 
 4 and y 
 2
results in the greatest possible value, [4  (2)]2
or 36.

d 
 
[1  (
3)]2 
 (3
 2)2
d 
 41
 or 241
 meters
distance from rosebushes to bench:
d 
 
(x2  
x1)2 
(y2 
y1)2
d 
 
(3  1
)2  [3
 (
3)]2
d 
 210
 or 410
 meters
distance from bench to fountain:
d 
 
(x2  
x1)2 
(y2 
y1)2
d 
 
(3 
3)2 
(2  
3)2
d 
 37
 or 237
 meters
Yes; the distance from the fountain to the
rosebushes is 241
 or about 12.81 meters. The
distance from the rosebushes to the bench is
410
 or about 12.65 meters. The distance from
the bench to the fountain is 237
 or about
12.17 meters.
36b. The fountain is located at (3, 2) and the
rosebushes are located at (1, 3).

10-2

Circles

x1  x2 y1  y2

3  1 2  (3)
 , 

2, 2 
 

2
2
1

 
1, 2

Chapter 10

Page 627

Check for Understanding

1. complete the square on each variable

310

2. Sample answer: (x  4)2  (y  9)2 
 1,
(x  4)2  (y  9)2 
 2, (x  4)2  (y  9)2 
 3,
(x  4)2  (y  9)2 
 4, (x  4)2  (y  9)2 
 5
3. Find the center of the circle, (h, k), by finding
the midpoint of the diameter. Next find the
radius of the circle, r, by finding the distance
from the center to one endpoint. Then write
the equation of the circle in standard form as
(x  h)2  (y  k)2 
 r2.
4. The equation x2  y2  8x  8y  36 
 0 written
in standard form is (x  4)2  (y  4)2 
 4.
Since a circle cannot have a negative radius, the
graph of the equation is the empty set.
5. Ramon; the square root of a sum does not equal
the sum of the square roots.
6. (x  h)2  (y  k)2 
 r2
(x  0)2  (y  0)2 
 92
x2  y2 
 81

9.

y

(5  23, 2)

x2  y2  Dx  Ey  F 
 0
02  02  D(0)  E(0)  F 
 0
42  02  D(4)  E(0)  F 
 0
02  42  D(0)  E(4)  F 
 0
F
0
F
0
4D  F 
 16 ⇒ D 
 4
4E  F 
 16
E 
 4
x2  y2  4x  4y 
 0
x2  4x  4  y2  4y  4 
 0  4  4
(x  2)2  (y  2)2 
 8
center: (h, k) 
 (2, 2)
radius: r2 
 8
r 
 8
 or 22

11.
x2  y2  Dx  Ey  F 
 0
12  32  D(1)  E(3)  F 
 0 ⇒ D  3E  F 
 10
52  52  D(5)  E(5)  F 
 0 ⇒ 5D  5E  F 
 50
52  32  D(5)  E(3)  F 
 0 ⇒ 5D  3E  F 
 34
D  3E  F 
 10
(1)(5D  5E  F) 
 (1)(50)
4D  2E

 40
5D  5E  F 
 50
(1)(5D  3E  F) 
 (1)(34)
2E

 16
E 
 8
4D  2(8) 
 40
4D 
 24
(6)  3(8)  F 
 10
D 
 6
F 
 20
x2  y2  6x  8y  20 
 0
x2  6x  9  y2  8y  16 
 20  9  16
(x  3)2  (y  4)2 
 5
center: (h, k) 
 (3, 4)
radius: r2 
 5
r 
 5

12.
(x  h)2  (y  k)2 
 r2
[x  (2)]2  (y  1)2 
 r2
(x  2)2  (y  1)2 
 r2
(1  2)2  (5  1)2 
 r2
25 
 r2
(x  2)2  (y  1)2 
 25

10.

x

(0, 9)

(x  h)2  (y  k)2 
 r2
[x  (1)]2  (y  4)2 
 [3  (1)]2
(x  1)2  (y  4)2 
 16
(1, 8) y

(1, 4)

(3, 4)

x

O
8.

x2  y2  4x  14y  47 
 0
x2  4x  4  y2  14y  49 
 47  4  49
(x  2)2  (y  7)2 
 100

y

(2, 3)

x

O
(8, 7)

x
(5, 2)

(9, 0)

7.

(5,2  23)

O

y

O

2x2  2y2  20x  8y  34 
 0
2x2  20x  2y2  8y 
 34
2(x2  10x  25)  2(y3  4y  4) 
 34  2(25)
 2(4)
2(x  5)2  2(y  2)2 
 24
(x  5)2  (y  2)2 
 12

(2, 7)

311

Chapter 10

13. midpoint of diameter:
x1  x2 y1  y2

 
2 ,
2




18.

2  10 6  (10)
, 
2
2








 (4, 2)

(x  h)2  (y  k)2 
 r2
9 2
[x  (5)]2  (y  0)2 
 
2
81

(x  5)2  y2 
 4

(x2  
x1)2 
(y2 
y1)2
radius: r 
 

y


 
[4  (
2)]2 
 [6 
(2)]2
2
2

 
6  8

 100
 or 10
(x  h)2  (y  k)2 
 r2
(x  4)2  [y  (2)]2 
 102
(x  4)2  (y  2)2 
 100
14. (x  h)2  (y  k)2 
 r2
(x  0)2  (y  0)2 
 (1740  185)2
x2  y2 
 19252

Pages 627–630

(

5, 9

2

)

(5, 0)

(

19 ,
2

0

x

O

)

19. (x  h)2  (y  k)2 
 r2
(x  6)2  (y  1)2 
 62
(x  6)2  (y  1)2 
 36

y

Exercises

15. (x  h)2  (y  k)2 
 r2
(x  0)2  (y  0)2 
 52
x2  y2 
 25

(6, 7)

y (0, 5)

(6, 1)

(0, 1)

x

O

(5, 0)

x

O

20.

16.

(x  h)2  (y  k)2 
 r2
[x  (4)]2  (y  7)2 
 (3
 )2
(x  4)2  (y  7)2 
 3

y

x
(3, 2)

(4  

3, 7)

(7, 2)

(4, 7 
3)

17.

(3, 2)

O

y
(4, 7)

O

(x  h)2  (y  k)2 
 r2
(x  3)2  [y  (2)]2 
 [2  (2)]2
(x  3)2  (y  2)2 
 16

y

21. 36  x2 
 y2
x2  y2 
 36

x

(0, 6)

(x  h)2  (y  k)2 
 r2
2
 2
[x  (1)]2  [y  (3)]2 
 
 
2 
(x  1)2  (y 

y
O

(6, 0)

1
3)2 
2

x
22.

(1, 6 2
2 )

2  
2
,
2

(

y

3

x2  y2  y 
 4
1

3

1 2

2


1

1

x2  y2  y  4 
 4  4

3 )

x2  
y 

(1, 3)

Chapter 10

x

O

312



(0, 12 )
(0,  12 )

O

x

(1,  12 )

23.

x2  y2  4x  12y  30 
 0
x2  4x  4  y2  12y  36 
 30  4  36
(x  2)2  (y  6)2 
 10

y
O

27.

x2  y2  14x  24y  157 
 0
x2  14x  49  y2  24y  144 
 157  49  144
(x  7)2  (y  12)2 
 36

y
O

x
(2, 6  

10 )

(7, 6)
(1, 12)

(2, 6)
(7, 12)

(2  

10, 6)

x2  y2  Dx  Ey  F 
 0
02  (1)2  D(0)  E(1)  F 
 0 ⇒
E  F 
 1
(3)2  (2)2  D(3)  E(2)  F 
 0 ⇒
3D  2E  F 
 13
(6)2  (1)2  D(6)  E(1)  F 
 0 ⇒
6D  E  F 
 37
 E  F 
 1
(1)(3D  2E  F) 
 (1)(13)
3D  E

 12
3D  2E  F 
 13
(1)(6D  2E  F) 
 (1)(37)
3D  2E

 24
3D  E 
 12
3(6)  E 
 12
3D  E 
 24
E 
 6
6D

 36
D
6
(6)  F 
 1
F 
 7
x2  y2  6x  6y  7 
 0
x2  6x  9  y2  6y  9 
 7  9  9
(x  3)2  (y  3)2 
 25
center: (h, k) 
 (3, 3)
radius: r2 
 25
r 
 25
 or 5
29.
x2  y2  Dx  Ey  F 
 0
72  (1)2  D(7)  E(1)  F 
 0 ⇒
7D  E  F 
 50
112  (5)2  D(11)  E(5)  F 
 0 ⇒
11D  5E  F 
 146
32  (5)2  D(3)  E(5)  F 
 0 ⇒
3D  5E  F 
 34
7D  E  F 
 50
(1)(11D  5E  F) 
 1(146)
4D  4E

 96
11D  5E  F 
 146
(1)(3D  5E  F) 
 1(34)
8D

 12
D 
 14
4(14)  4E 
 96
4E 
 40
7(14)  (10)  F 
 50
E 
 10
F 
 58
x2  y2  14x  10y  58 
 0
x2  14x  49  y2  10y  25 
 58  49  25
(x  7)2  (y  5)2 
 16
center: (h, k) 
 (7, 5)
radius: r2 
 16
r 
 16
 or 4

2x2  2y2  2x  4y 
 1
2x2  2x  2y2  4y 
 1

24.

28.

2
x2  1x  4  2(y2  2y  1) 
 1  2
4  2(1)
1

1

1 2
3
x  2  2(y  1)2 
 2
1 2
3
x  2  (y  1)2 
 4

2









y

(1 2 
3, 1)
( 12 , 1)
3 O
( 12 , 1  

)
2

25.
6(x2

x

6x2  12x  6y2  36y 
 36
 2x  1)  6(y2  6y  9) 
 36  6(1)  6(9)
6(x  1)2  6(y  3)2 
 96
(x  1)2  (y  3)2 
 16

y
O

x
(1, 3)

(3, 3)
(1, 7)

16x2  16y2  8x  32y 
 127
16x2  8x  16y2  32y 
 127

26.

2
2

16
x2  2x  1
6   16(y  2y  1) 
 127  16
 16 
 16(1)
1

1

1

1 2

16
x  4  16(y  1)2 
 144
2


x  14

y

x

 (y  1)2 
 9

( 134, 1)

( 14 , 1)
O

x

( 14 , 2)

313

Chapter 10

x2  y2  Dx  Ey  F 
 0
(2)2  72  D(2)  E(7)  F 
 0 ⇒
2D  7E  F 
 53
(9)2  02  D(9)  E(0)  F 
 0 ⇒
9D  F 
 81
(10)2  (5)2  D(10)  E(5)  F 
 10 ⇒
10D  5E  F 
 125
2D  7E  F 
 53
(1)(9D
 F) 
 (1)(81)
7D  7E

 28
DE
4
10D  5E  F 
 125
(1)(9D
 F) 
 (1)(81)
D  5E

 44
D  5E 
 44
D E
4
4E 
 40
E 
 10
D  (10) 
 4
D 
 6
9(6)  F 
 81
F 
 135
x2  y2  6x  10y  135 
 10
x2  6x  9  y2  10y  25 
 135  9  25
(x  3)2  (y  5)2 
 169
center: (h, k) 
 (3, 5)
radius: r2 
 169
r 
 169
 or 13
31.
x2  y2  Dx  Ey  F 
 0
(2)2  32  D(2)  E(3)  F 
 0 ⇒
2D  3E  F 
 13
62  (5)2  D(6)  E(5)  F 
 0 ⇒
6D  5E  F 
 61
02  72  D(0)  E(7)  F 
 0 ⇒
7E  F 
 49
2D  3E  F 
 13
(1)(6D  5E  F) 
 (1)(61)
8D  8E

 48
D  E 
 6
6D  5E  F 
 61
(1)(7E  F) 
 (1)(49)
6D  12E

 12
D  2E 
 2
D  E 
 6
D  2E 
 2
E 
 4
E 
 4
D  2(4) 
 2
D 
 10
7(4)  F 
 49
F 
 21
x2  y2  10x  4y  21 
 0
x2  10x  25  y2  4y  4 
 21  25  4
(x  5)2  (y  2)2 
 50
center: (h, k) 
 (5, 2)
radius: r2 
 50
r 
 50
 or 52


x2  y2  Dx  Ey  F 
 0
42  52  D(4)  E(5)  F 
 0 ⇒
4D  5E  F 
 41
(2)2  32  D(2)  E(3)  F 
 0 ⇒
2D  3E  F 
 13
(4)2  (3)2  D(4)  E(3)  F 
 0 ⇒
4D  3E  F 
 25
4D  5E  F 
 41
(1)(2D  3E  F) 
 (1)(13)
6D  2E

 28
3D  E 
 14
4D  5E  F 
 41
(1)(4D  3E  F) 
 (1)(25)
8D  8E

 16
D  E 
 2
3D  E 
 14
(1)(D  E) 
 (1)(2)
2D

 12
D 
 6
6  E 
 2
E
4
2(6)  3(4)  F 
 13
F 
 37
x2  y2  6x  4y  37 
 0
x2  6x  9  y2  4y  4 
 37  9  4
(x  3)2  (y  2)2 
 50
center: (h, k) 
 (3, 2)
radius: r2 
 50
r 
 50
 or 52

33.
x2  y2  Dx  Ey  F 
 0
12  42  D(1)  E(4)  F 
 0 ⇒
D  4E  F 
 17
22  (1)2  D(2)  E(1)  F 
 0 ⇒
2D  E  F 
 5
(3)2  02  D(3)  E(0)  F 
 0 ⇒
3D  F 
 9
D  4E  F 
 17
(1)(2D  E  F) 
 (1)(5)
D  5E

 12
D  4E  F 
 17
(1)(3D 
F) 
 (1)(9)
4D  4E

 8
D  E 
 2
D  5E 
 12
D  E 
 2
6E 
 14
7
E 
 3

30.

Chapter 10

32.

D  
3 
 2
7

1

D 
 3
3
3  F 
 9
1

F 
 8
x2  y2  Dx  Ey  F 
 0
1
7
x2  y2  3x  3y  8 
 0
1

7

1

49

1

49

2




x2  3x  3
6  y  3 y  36 
 8  36  36
2


x  16

314

7 2

 
y  6 
 1
8
169

39. (x  h)2  (y  k)2 
 r2
(x  5)2  (y  1)2 
 r2
x  3y 
 2
x  3y  2 
 0 ⇒ A 
 1, B 
 3, and C 
 2
Ax1  By1  C

r


A2  
B2 

center: (h, k) 
 
6, 6
1 7

169

radius: r2 
 1
8

r 
 
18
169

13

132


 or 


6
32


x2  y2  Dx  Ey  F 
 0
02  02  D(0)  E(0)  F 
 0 ⇒

34.

(1)(5)  (3)(1)  2


  


12  32

F
0
(2.8)2  02  D(2.8)  E(0)  F 
 0 ⇒
2.8D  F 
 7.84
(5)2  22  D(5)  E(2)  F 
 0 ⇒
5D  2E  F 
 29
2.8D  0 
 7.84
5(2.8)  2E  (0) 
 29
2.8D 
 7.84
2E 
 15
D 
 2.8
E 
 7.5
x2  y2  2.8x  7.5y  0 
 0
x2  2.8x  1.96  y2  7.5y  14.0625 
 1.96  14.0625
(x  1.4)2  (y  3.75)2 
 16.0225
or about (x  1.4)2  (y  3.75)2 
 16.02
35.
(x  h)2  (y  k)2 
 r2
[x  (4)]2  (y  3)2 
 r2
(x  4)2  (y  3)2 
 r2
(0  4)2  (0  3)2 
 r2
25 
 r2
(x  4)2  (y  3)2 
 25
36. (x  h)2  (y  k)2 
 r2
(x  2)2  (y  3)2 
 r2
(5  2)2  (6  3)2 
 r2
18 
 r2
(x  2)2  (y  3)2 
 18
37. midpoint of diameter:

 or 10

 


10
10

2



(x  5)2  (y  1)2 
 10
40. center: (h, 0), radius: r 
 1
(x  h)2  (y  k)2 
 r2
2


22  h
1

2

 
2  0 
 12
2


1

2  2
h  h2  2 
 1
h 
 1  1
h2  2






 
0
h h  2

h 
 0 or h 
 2
2
2
(x  0)  (y  0) 
 1
x  2





x2  y2 
 1

or

41a. (x  h)2  (y  k)2 
 r2

2



 (y  0)2 
 1
2


x  2

 y2 
 1

12 2

(x  0)2  (y  0)2 
 
2

x2  y2 
 36
41b. x2  y2 
 36
y2 
 36  x2
y 
 
36  
x2
dimensions of rectangle:
36  
x2
2x by 2y ⇒ 2x by 2

x1  x2 y1  y2

2  (6) 3  (5)
, 

2, 2 
 

2
2




41c. A(x) 
 2x 2
36  
x2


 (2, 1)
r 
 
(x2  
x1)2 
(y2 
y1)2


 4x
36  
x2


 
(2 
2)2 
(1 
3)2
2
2

 
(4) 
 (4
)

 32

(x  h)2  y  k)2 
 r2
2
[x  (2)]2  [y  (1)]2 
 
32

2
2
(x  2)  (y  1) 
 32
38. midpoint of diameter:
x1  x2 y1  y2




(x  5)2  (y  1)2 
 10




41d.

3  2 4  1


2, 2 
 
2, 2
1 5

 
2, 2

[0, 10] scl:1 by [0, 100] scl:20
41e. Use 4: maximum on the CALC menu of the
calculator. The x-coordinate of this point is about
4.2. The maximum area of the rectangle is the
corresponding y-value of 72, for an area of
72 units2.

r 
 
(x2  
x1)2 
(y2 
y1)2
2
2
2 
1


12 

52  
5 2
3 2

 

2 
 
2




42a.



 
2
17

(x  h)2  (y  k)2 
 r2
2

2

2

17

x  
12  
y  52 
 

2 
2
2

x  12  
y  52 
 127

[15.16, 15.16] scl:1 by [5, 5] scl:1
42b. a circle centered at (2, 3) with radius 4
42c. (x  2)2  (y  3)2 
 16

315

Chapter 10

42d. center: (h, k) 
 (4, 2)
radius: r2 
 36
r 
 36
 or 6
2nd [DRAW] 9:Circle( - 4 , 2 , 6 )

2(13)  (9)  F 
 5
F 
 30
x2  y2  13x  9y  30 
 0
x2  13x  42.25  y2  9y  20.25 
 30  42.25  20.25
(x  6.5)2  (y  4.5)2 
 32.5
y
45a. (x  h)2  (y  k)2 
 r2
y
x
(x  k)2  (y  k)2 
 22
(x  k)2  (y  k)2 
 4
45b. (x  1)2  (y  1)2 
 4
O
(x  0)2  (y  0)2 
 4
x
(x  1)2  (y  1)2 
 4

[15.16, 15.16] scl:1 by [5, 5] scl:1
43a. (x  h)2  (y  k)2 
 r2

45c. All of the circles in this family have a radius of 2
and centers located on the line y 
 x.
46a. (x  h)2  (y  k)2 
 r2
(x  0)2  (y  0)2 
 142
x2  y2 
 196
y2 
 196  x2
y 
 
196 
x2
2
y 
 
196 
x
46b. No, if x 
 7, then y 
 147
 12.1 ft, so the
truck cannot pass.
47.
x2  y2  8x  6y  25 
 0
x2  8x  16  y2  6y  9 
 25  16  9
(x  4)2  (y  3)2 
 0
radius: r2 
 0
r 
 0
 or 0
center: (h, k) 
 (4, 3)
Graph is a point located at (4, 3).
48a. (x  h)2  (y  k)2 
 r2
(x  0)2  (y  0)2 
 r2
x2  y2 
 r2
2
(475)  (1140)2 
 r2
1,525,225 
 r2
x2  y2 
 1,525,225
48b. r2 
 1,525,225
r 
 1,525
,225
 or 1235
A 
 r2

  (1235)2 or approximately 4,792,000 ft2

24 2

(x  0)2  (y  0)2 
 
2

x2  y2 
 144
43b. x2  y2 
 6.25 ⇒ r12 
 6.25
r1 
 6.25
 or 2.5
If the circles are equally spaced apart then
radius r2 of the middle circle is found by adding
the radius of the smallest circle to the radius of
the largest circle and dividing by two.
12 
 2.5

r2 
 2 or 7.25
area of
area of
area of



region B middle circle smallest circle

 r22  r12

  (r22  r12)

  (7.252  2.52)

  (46.3125) or about 145.50
The area of region B is about 145.50 in2.
44.

y
x  5y  3 
 0
(2, 1)

(12, 3)

x
(5, 1)
2x  3y  7 
 0

O
4x  7y 
 27

x2  y2  Dx  Ey  F 
 0
22  12  D(2)  E(1)  F 
 0 ⇒
2D  E  F 
 5
52  (1)2  D(5)  E(1)  F 
 0 ⇒
5D  E  F 
 26
122  32  D(12)  E(3)  F 
 0 ⇒
12D  3E  F 
 153
2D  E  F 
 5
(1)(5D  E  F) 
 (1)(26)
3D  2E

 21
2D  E  F 
 5
(1)(12D  3E  F) 
 (1)(153)
10D  2E

 148
3D  2E 
 21
10D  2E 
 148
13D

 169
D 
 13
3(13)  2E 
 21
2E 
 18
E 
 9

Chapter 10

48c.

25002  4,792,000

25002


 0.23328

about 23%

y4


49a. 
PA
 has a slope of 
B
 has slope of
x  3 and P
y4
. If P
A

x3
y4 y4
  
x3 x3
y2  16


x2  9
2
y  16

y4

y4

 
⊥P
B
 then 
x  3  x  3 
 1.


 1

 1


 x2  9
x2  y2 
 25
49b. If P
A
⊥P
B
, then A, P, and B are on the circle
x2  y2 
 25.
50. d 
 
(x2  
x1)2 
(y2 
y1)2
d 
 
(2 
4)2 
[6  
(3)]2
d 
 
(6)2 
 92
d 
 117


316

51. (2  i)(3  4i)(1  2i) 
 (6  8i  3i  4i2)(1  2i)

 [6  8i  3i  4(1)](1  2i)

 (10  5i)(1  2i)

 10  20i  5i  10i2

 10  20i  5i  10(1)

 20  15i
u cos v
u sin v  1gt2
y 
 tv
52. x 
 tv

(0, 0), (0, 9), (5, 8), (10, 5)
f(x, y) 
 320x  500y
f(0, 0) 
 320(0)  500(0) 
 0
f(0, 0) 
 320(0)  500(9) 
 4500
f(5, 8) 
 320(5)  500(8) 
 5600
f(10, 5) 
 320(10)  500(5) 
 5700
The maximum profit occurs when 10 cases of
drug A and 5 cases of drug B are produced.
57b. When 10 cases of drug A and 5 cases of drug B
are produced, the profit is $5700.
58.
y

2

1

x 
 t(60) cos 60°

y 
 t(60) sin 60°  2(32)t2

x 
 60t cos 60°
y 
 60t sin 60°  16t2
x 
 60(0.5) cos 60° y 
 60(0.5) sin 60°  16(0.5)2
x 
 15
y 21.98076211
15 ft horizontally, about 22 ft vertically

x

5

53. A 
 2 or 2.5
2

20 
 k




x

x

x



and k 
 1
0

y 
 A cos (kt)

y 
 2.5 cos 
10t
54. s 


x

1
(a  b 
2
1
(15  25
2

y
x  x  x  x  x  y  y 
 5x  2x
The correct choice is A

c)
 35)


 37.5
a)(s

b)(s 
c)
k 
 s(s


 37.5(3
7.5

7.5
15)(3
7.5
25)(3

3.5)

 26,36
7.187
5

162 units2
55. v 
 
v20  6
4h

10-3

Pages2 637–638
1.


152 
64h

95 

952 
 152  64h
952  152
h 
 6
4

O

4
y
0

8

(y  k)2
(x  h)2

 b

1
2
a2
(y  0)2
[x  (7)]2

 3

1
2
62
y2
(x  7)2
   
 1
36
9
a2  b2
foci:
c 
 

x  5y 
 45

(5, 8)
(10, 5)
4
(0,
0)
x
0

3x  5y 
 55
12

16

20

Check For Understanding

y
x2


a2  b2 
 1
y2
x2


82  52 
 1
y2
x2
   
 1
64
25

2. Since the foci lie on the major axis, determine
whether the major axis is horizontal or vertical. If
the a2 is the denominator of the x terms, the major
axis is horizontal. If the a2 is the denominator of
the y terms, the major axis is vertical.
3. When the foci and center of an ellipse coincide,
c 
 0.
c
e 
 a
c2 
 a2  b2
0 
 a2  b2
0
e 
 a
b2 
 a2
e
0
b
a
The figure is a circle.
c
4. In an ellipse, b2 
 a2  c2 and a 
 e.
c
 
 e
b2 
 a2  c2
a
c 
 ae
b2 
 a2  a2e2
2
2
2
c 
a e
b2 
 a2(1  e2)
5. Shanice; an equation with only one squared term
cannot be the equation of an ellipse.
6. center: (h, k) 
 (7, 0)
a 
 0  6 or 6
b 
 7  (4) or 3

h 
 137.5 ft
56. y 
 6x4  3x2  1
b 
 6a4  3a2  1
(x, y)  (a, b)
x-axis: (x, y) 
 (a, b)
b 
 6a4  3a2  1; no
y-axis: (x, y) 
 (a, b)
b 
 6(a)4  3(a)2  1
b 
 6a4  3a2  1; yes
y 
 x: (x, y) 
 (b, a)
a 
 6b4  3b2  1; no
origin: f(x) 
 f(x)
f(x) 
 6(x)4  3(x)2  1 f(x) 
 (6x4  3x2  1)
f(x) 
 6x4  3x2  1
f(x) 
 6x4  3x2  1
no
The graph is symmetric with respect to the y-axis.
57a. Let x 
 number of cases of drug A.
Let y 
 number of cases of drug B.
x 
  10
y
y9
x 
 10
(0, 9)
12
3x  5y  55
y
9
x  5y  45
8

Ellipses

24 x

c 
 
62  32
c 
 33


317

(h, k

c) 
 (7, 0 33
)

 (7, 33
)

Chapter 10

7. center: (h, k) 
 (0, 0)
a2 
 36
b2 
 4
c 
 
a2  b2
a 
 36
 or 6 b 
 4
 or 2 c 
 36
4
  or 42

foci: (h c, k) 
 (0 42
, 0) or ( 42
, 0)
major axis vertices: (h a, k) 
 (0 6, 0) or ( 6, 0)
minor axis vertices: (h, k b) 
 (0, 0 2) or (0, 2)

10.

center: (h, k) 
 (1, 2)
a2 
 9
b2 
 4
a 
 9
 or 3
b 
 4
 or 2
foci: (h, k c) 
 1, 2 5


y
(0, 2)

(6, 0)

O




(6, 0)

(0, 2)



c 
 
a2  b2
c 
 9

 4 or 5


major axis vertices: (h, k a) 
 (1, 2 3) or
(1, 1), (1, 5)
minor axis vertices: (h  b, k) 
 (1 2, 2) or
(3, 2), (1, 2)
y

x

(1, 1)

8. center: (h, k) 
 (0, 4)
a2 
 81
b2 
 49
c 
 
a2  b2
a 
 81
 or 9 b 
 49
 or 7 c 
 81
9
 4 or 42

foci: (h c, k) 
 (0 42
, 4) or ( 42
, 4)
major axis vertices: (h a, k) 
 (0 9, 4) or
( 9, 4)
minor axis vertices: (h, k b) 
 (0, 4 7) or
(0, 11), (0, 3)

y

9x2  4y2  18x  16y 
 11
9(x2  2x  ?)  4(y2  4y  ?) 
 11  ?  ?
9(x2  2x  1)  4(y2  4y  4) 
 11  9(1)  (4)
9(x  1)2  4(y  2)2 
 36
(x  1)2
(y  2)2
   
 36
4
9

(1, 2)

O

x
(3, 2)

(1, 2)

(1, 5)

11. center: (h, k) 
 (2, 3)
8
a 
 2 or 4

(0, 11)

2

b 
 2 or 1

(y  k)2
(x  h)2

 b
2
a2
[y  (3)]2
[x  (2)]2

 1
2
42
(y  3)2
(x  2)2
  
16
1

(0, 4)

(9, 4)

(9, 4)

O

x
(0, 3)

9.


1

1

12. The major axis contains the foci and it is located
on the x-axis.

25x2  9y2  100x  18y 
 116
25(x2  4x  ?)  9(y2  2y  ?) 
 116  ?  ?
25(x2  4x  4)  9(y2  2y  1) 
 116  25(4)  9(1)
25(x  2)2  9(y  1)2 
 225

1  1 0  0

center: (h, k) 
 
2, 2 or (0, 0)
c 
 1, a 
 4
c2 
 a2  b2
12 
 42  b2
b2 
 15

(x  2)2
(y  1)2
   
 1
9
25

center: (h, k) 
 (2, 1)
a2 
 25
b2 
 9
c 
 
a2  b2
a 
 25
 or 5
b 
 9
 or 3 c 
 25
9
  or 4
foci: (h, k c) 
 (2, 1 4) or (2, 5), (2, 3)
major axis vertices: (h, k a) 
 (2, 1 5) or
(2, 6), (2, 4)
minor axis vertices: (h b, k) 
 (2 3, 1) or
(1, 1), (5, 1)
(2, 6)


1

(x  h)2

a2
(x  0)2

42

(y  k)2

 b

1
2
(y  0)2

 15 
 1
x2

16

y2

 15 
 1

13. center: (h, k) 
 (1, 2)

y

y

The points at (1, 4) and
(5, 2) are vertices of the
ellipse.

(1, 4)

(5, 1)

(2, 1)

O

(5, 2)
(1, 1)

(1, 2)

x

a 
 4, b 
 2

(2, 4)

Chapter 10

x

O

(x  h)2

a2
(x  1)2

42
(x  1)2

16

318

(y  k)2

 b

1
2
(y  2)2

 2

1
2
(y  2)2

 4 
 1

14. center: (h, k) 
 (3, 1)
a
6
c
e 
 a
1

3

c


 6


foci: (h, k c) 
 
3, 4 39
19. center: (h, k) 
 (2, 1)
a2 
 4
b2 
 1
c 
 
a2  b2
a 
 4
 or 2
b 
 1
 or 1 c 
 4

 1 or 3

foci: (h, k c) 
 (2, 1 3
)
major axis vertices: (h, k a) 
 (2, 1 2) or
(2, 3), (2, 1)
minor axis vertices: (h b, k) 
 (2 1, 1) or
(1, 1), (3, 1)
y

2
c
c2 
 a2  b2
22 
 62  b2
4 
 36  b2

b2 
 32

(y  k)2
(h  h)2

 b
2
a2
(y  1)2
(x  3)2

 32
62
(y  1)2
(x  3)2
  
36
32


1

1

1

(2, 3)

15. The major axis contains the foci and is located on
the x-axis.
center: (h, k) 
 (0, 0)
c 
 0.141732
1
a 
 2(3.048) or 1.524

(1, 1)

(3, 1)

O

x

(2, 1)

c2 
 a2  b2
(0.141732)2 
 (1.524)2  b2
0.020 
 2.323  b2
b2 
 2.302
b 1.517
(x  h)2
(y  k)2

 b
2
a2
(x  0)2
(y  0)2


1.5242  1.5172
x2
y2


1.5242  1.5172

(2, 1)

20. center: (h, k) 
 (6, 7)
a2 
 121
b2 
 100
a 
 121
 or 11
b 
 100
 or 10
a2  b2
c 
 
c 
 121


100 or 21

foci: (h, k c) 
 (6, 7 21
)
major axis vertices: (h, k a) 
 (6, 7 11) or
(6, 18), (6,  4)
minor axis vertices: (h b, k) 
 (6 10, 7) or
y (4, 7)
(16, 7),


1

1

1

(6, 18)

Pages 638–641

Exercises

16. center: (h, k) 
 (0, 5)
a 
 0  (7) or 7
b 
 5  0 or 5
(x  h)2
(y  k)2

 b
2
a2
(x  0)2
(y  (5)]2

 5
2
72
(y  5)2
x2
  
25
49
a2  b2
c 
 

(4, 7)


1


1

O

x
(6, 4)

21. center: (h, k) 
 (4, 6)
a2 
 16
b2 
 9
c 
 
a2  b2
a 
 16
 or 4 b 
 9
 or 3 c 
 16
9
  or 7

foci: (h c, k) 
 (4 7
, 6)
major axis vertices: (h a, k) 
 (4 4, 6) or
(8, 6), (0, 6)
minor axis vertices: (h, k b) 
 (4, 6 3) or
y 3), (4, 9)
(4,


1

1

1

x

O

c 
 
42  22 or 23

foci: (h c, k) 
 (2 23
, 0)
18. centers: (h, k) 
 (3, 4)
a 
 4  12 or 8
b 
 3  2 or 5
(y  k)2
(x  h)2

 b
2
a2
(y  4)2
[x  (3)]2

 5
2
82
(y  4)2
(x  3)2
  
64
25
a2  b2
c 
 

(16, 7)


1

c 
 
72  52 or 26

foci: (h c, k) 
 (0 26
, 5)

 ( 26
, 5)
17. center: (h, k) 
 (2, 0)
a 
 2  2 or 4
b 
 0  2 or 2
(x  h)2
(y  k)2

 b
2
a2
[x  (2)]2
(y  0)2

 2
2
42
(x  2)2
y2
  
16
4
a2  b2
c 
 

(6, 7)

(4, 3)
(4, 6)
(0, 6)

(8, 6)

(4, 9)


1

1

1

c 
 
82  52 or 39


319

Chapter 10

y

22. (h, k) 
 (0, 0)
a2 
 9
b2 
 4
c 
 
a2  b2
a 
 9
 or 3
b 
 4
 or 2 c 
 9

 4 or 5

foci: (h, k c) 
 0, 0 5
 or 0, 5





 


(3, 8)



(3, 4)

major axis vertices: (h, k a) 
 (0, 0 3) or
(0, 3)
minor axis vertices: (h b, k) 
 (0 2, 0) or
( 2, 0)
y

(8, 4)

(2, 4)

O

x

(3, 0)

(0, 3)

25.
(2, 0)

(2, 0)

x

O

(x  3)2

8

(0, 3)

23.

y2

(x  1)2

1

c 
 
a2  b2
c 
 24
8
  or 4
foci: (h, k c) 
 (3, 1 4) or (3, 5), (3, 3)
major axis vertices: (h, k a) 
 (3, 1 26
)
minor axis vertices: (h b, k) 
 (3 22
, 1)

(y  3)2

 4 
 1

center: (h, k) 
 (1, 3)
b2 
 1
a2 
 4
a 
 4
 or 2
b 
 1
 or 1
foci: (h, k c) 
 1, 3 3







c 
 
a2  b2
c 
 4

 1 or 3


(3, 1  26) y

major axis vertices: (h, k a) 
 (1, 3 2)
or (1, 1), (1, 5)
minor axis vertices: (h b, k) 
 (1 1, 3)
or (2, 3), (0, 3)
y

O
(0, 3)
(1, 3)

24.

(1, 1)

(3, 1) (3  22, 1)
(3  22, 1)

26.

(2, 3)
(1, 5)

16x2  25y2  96x  200y 
 144
16(x2  6x  ?)  25(y2  8y  ?) 
 144  ?  ?
16(x2  6x  9)  25(y2  8y  16) 

144  16(9)  25(16)
16(x  3)2  25(y  4)2 
 400
(y  4)2

 16 
 1

center: (h, k) 
 (3, 4)
b2 
 16
c 
 
a2  b2
a2 
 25
a 
 25
 or 5 b 
 16
 or 4 c 
 25
16
  or 3
foci: (h c, k) 
 (3 3, 4) or (6, 4), (0, 4)
major axis vertices: (h a, k) 
 (3 5, 4) or
(8, 4), (2, 4)
major axis vertices: (h, k b) 
 (3, 4 4) or
(3, 8), (3, 0)

6x2  12x  6y  36y 
 36
6(x2  2x  ?)  6(y2  6y  ?) 
 36
6(x2  2x  1)  6(y2  6y  9) 
 36  6(1)  6(9)
6(x  1)2 2 6(y  3)22 
 96
(x  1)
( y  3)
   
 1
16
16
center: (h, k) 
 (1, 3)
a2 
 16
b2 
 16
c 
 
a2  b2
a 
 16
 or 4 b 
 16
 or 4 c 
 16
16
  or 0
foci: (h c, k) or (h, k c) 
 (1, 3)
Since a 
 b 
 4, the vertices are (h 4, k) and
(h, k 4) or (5, 3), (3, 3), (1, 1), (1, 7)
y
(1, 1)

x

O

(1, 3) (5, 3)
(3, 3)

(1, 7)

Chapter 10

x

O

(3, 1 2
6)

x

(x  3)2

25

(y  1)2

 24 
 1

center: (h, k) 
 (3, 1)
a2 
 24
b2 
 8
a 
 24
 or 26

b 
 8
 or 22


  8x  6y  9 
 0
4(x2  2x  ?)  (y2  6y  ?) 
 9  ?  ?
4(x2  2x  1)  (y2  6y  9) 
 9  4(1)  9
4(x  1)2  (y  3)2 
 4
4x2

3x2  y2  18x  2y  4 
 0
 6x  ?)  (y2  2  ?) 
 4
3(x2  6x  9)  (y2  2y  1) 
 4  3(9)  1
3(x  3)2  (y  1)2 
 24
3(x2

320

27.

18y2  12x2  144y  48x 
 120
18(y2  8y  ?)  12(x2  4x  ?) 
 120  ?  ?
18(y2  8y  16)  12(x2  4x  4) 

120  18(16)  12(4)
18(y  4)2  12(x  2)2 
 216
(y 

12
4)2

center: (h, k) 
 (2, 4)
a2 
 18
a 
 18
 or 32


(x 

29. 49x2  16y2  160y  384 
 0
49x2  16(y2  10y  ?) 
 384  ?
49x2  16(y2  10y  25) 
 384  16(25)
49x2  16(y  5)2 
 784
x2

16

2)2

 18 
 1

( y  5)2

 49 
 1

center: (h, k) 
 (0, 5)
b2 
 16
c 
 
a2  b2
a2 
 49
a 
 49
 or 7 b 
 16
 or 4 c 
 49
16
  or 33

foci: (h, k c) 
 (0, 5 33
)
major axis vertices: (h, k a) 
 (0, 5 7) or
(0, 2), (0, 12)
minor axis vertices: (h b, k) 
 (0 4, 5) or
( 4, 5)

b2 
 12
b 
 12
 or 23


c 
 
a2  b2
c 
 18
12
  or 6

foci: (h c, k) 
 (2 6
, 4)
major axis vertices: (h g, k) 
 (2 32
, 4)
minor axis vertices: (h, k b) 
 (2, 4 23
)
y

y
(0, 2)

(2, 4  23)

x

O
(4, 5)

(2, 4)
(2  32, 4)

(0, 5)

(4, 5)

(2  32, 4)
(0, 12)

O (2, 4  23)

x
9y2  108y  4x2  56x 
 484
 12y  ?)  4(x2  14x  ?) 
 484  ?  ?
9(y2  12y  36)  4(x2  14x  49) 

484  9(36)  4(49)
9(y  6)2  4(x  7)2 
 36

30.
28.

9(y2

4x2  8y  9x2  54x  49 
 0
 2y  ?)  9(x2  6x  ?) 
 49  ?  ?
2
4(y  2y  1)  9(x2  6x  9) 
 49  4(1)  9(9)
4(y  1)2  9(x  3)2 
 36

4(y2

( y  1)2

9



(x  3)2

4

(y  6)2

4


1

center: (h, k) 
 (7, 6)
b2 
 4
c 
 
a2  b2
a2 
 9
a 
 9
 or 3
b 
 4
 or 2
c 
 9

 4 or 5

foci: (h c, k) 
 (7 5
, 6)
major axis vertices: (h a, k) 
 (7 3, 6) or
(10, 6)(4, 6)
minor axis vertices: (h, k b) 
 (7, 6 2) or
(7, 4), (7, 8)

center: (h, k) 
 (3, 1)
b2 
 4
c 
 
a2  b2
a2 
 9
a 
 9
 or 3
b 
 4
 or 2 c 
 9

 4 or 5

foci: ( j, k c) 
 (3, 1 5)
major axis vertices: (h, k a) 
 (3, 1 3) or (3, 4),
(3, 2)
minor axis vertices: (h b, k) 
 (3 2, 1) or
(5, 1), (1, 1)
y

y

(3, 1)

(7, 4)
(5, 1)
(4, 6)

x

O

x

O

(3, 4)

(1, 1)

(x  7)2

 9 
 1

(3, 2)

(10, 6)
(7, 6)
(7, 8)

31. a 
 7, b 
 5

(x  h)2
( y  k)2

 b

1
2
a2
[x  (3)]2
[y  (1)]2

 5

1
2
72
(x  3)2
( y  1)2
   
 1
49
25

321

Chapter 10

32. The major axis contains the foci and it is located
on the x-axis.
2  2 0  0
center: (h, k) 
 
0, 2 or (0, 0)

36. The major axis contains the foci and it is the
vertical axis of the ellipse.
5  (1)

c 
 2 or 3

c 
 2, a 
 7
c2 
 a2  b2
22 
 72  b2
b2 
 45
(x  h)2

a2
(x  0)2

72

33. b 

6


1  1 1  5

center: (h, k) 
 
2, 2 or (1, 2)

(y  k)2
 b
2
(y  0)2
 45
x2
y2
  
49
45

( y  k)2

a2
(2  2)2

a2


1

(x  h)2

a2
(x  0)2

82


1

1

37.

(y  k)2

(y  0)2

 6

1
2
y2

 36 
 1

5
c

1  (1) 1  (5)

center: (h, k) 
 
2, 2 or (1, 2)
c 
 1  k
c 
 1  (2) or 3
c2 
 a2  b2
2
32 
 213
  b2





 52  9
b2 
 43


a 
 213


1
b2 
 a2  c2
b2 
 102  52
b2 
 75

( y  0)2

 75 
 1
x2

100

y2

 75 
 1

(4, 0)


1

4

8

12

x

tangent vertices:
(4, 0), (0, 7)
a 
  7  0 or 7
b 
 4  0 or 4

(4, 7)

12


1

1

39.

(7, 5)

12 8 4 O


1

( y  k)2

O

The horizontal axis of
the ellipse is the major
axis.

12
(2, 9)
8
(11, 5)
4
(2, 1)

9 
 b2

(x  h)2

b2
(x  1)2

9

 b

1
2

4
(0, 7)
8

y

35.


1

y

38.



(y  k)2
(x  h)2

 b
2
a2
[x  (1)]2
[y  (2)]2




43
(21
3)2

(y  2)2
(x  1)2
  
52
43

( y  k)2


a2
( y  2)2
 
18
1
c
 
 
2
a
a
 
 c
2
10
 
 c
2
(x  h)2

a2
(x  0)2

102

34. The major axis contains the foci and it is the
vertical axis of the ellipse.

b2

32

 b2 
 1
9

b2

 b

1
2
x2

64

(y 

c2 
 a2  b2
32 
 a2  9
18 
 a2

1)2

 b

1
2
02

a2

3
a
4
3
a
4

8
a

(x  h)2

 b

1
2

4

( y  k)2
(x  h)2

 b
2
a2
[y  (7)]
(x  4)2

 4
2
72
2
( y  7)
(x  4)2
  
49
16
2
b 
 a2(1  e2)


1

1

1

3 2

b2 
 221  
4

b2

8x




28

16



or 1.75

Case 1: Horizontal axis is major axis.
11  7 5  5
, 
2
2

enter: (h, k) 
 

ha
7
2  a 
 7
a
9
(x  h)2

a2
[x  (2)]2

92
(x  2)2

81

Chapter 10





(y  k)2

b2
(y  5)2

42
(y  5)2

16

(x  h)2
( y  k)

 b
2
a2
(x  0)2
( y  0)2



 1.75
22
y2
x2
   
1.75
4

 or (2, 5)

kb
9
5b
9
b
4


1

1

1

Case 2: Vertical axis is major axis.


1

( y  k)2

a2
( y  0)2

22


1

1

(x  h)2

 b

1
2
(x  0)2



1.75 
 1
y2

4

322

x2



1.75 
 1

44.
4x2  y2  8x  2y 
 1
4(x2  2x  ?)  (y2  2y  ?) 
 1  ?  ?
4(x2  2x  1)  (y2  2y  1) 
 1  4(1)  1
4(x  1)2  (y  1)2 
 4
(y  1)2 
 4  4(x  1)2
y  1 
 
4  
4(x  
1)2
y 
 
4  4
(x  1
)2  1
Vertices: (0, 1),
(2, 1), (1, 1),
(1, 3)

40. The major axis contains the foci and it is the
horizontal axis of the ellipse.
31 55

center: (h, k) 
 
2, 2 or (2, 5)
foci: (3, 5) 
 (h  c, k)
3
hc
3
2c
1
c
c

e 
 a
0.25 


b2 
 a2(1  e2)
b2 
 42(1  0.252)
b2 
 15

1

a

a
4

41.

(x  h)2
( y  k)2

 b
2
a2
(x  2)2
( y  5)2

 15
42
(x  2)2
( y  15)2
  
16
15
20
a 
 2 or 10
b2 
 a2(1  e2)


1

1

2

b2 
 1021  
1
0
7

[4.7, 4.7] scl:1 by [3.1, 3.1] scl:1
45.
4x2  9y2  16x  18y 
 11
2
4(x  4x  ?)  9(y2  2y  ?) 
 11  ?  ?
4(x2  4x  4)  9(y2  2y  1) 
 11  4(4)  9(1)
4(x  2)2  9(y  1)2 
 36
9(y  1)2 
 36  4(x  2)2


1

 or 51

(y  k)2
(x  h)2

 b

1
2
a2
(y  0)2
(x  3)2

 51 
 1
102
y2
(x  3)2
   
 1
100
51

y1

y


c

5


3

b2 
 a2(1  e2)



b2 
 4

[7.28, 7.28] scl:1 by [4.8, 4.8] scl:1
46.
25y2  16x2  150y  32x 
 159
25(y2  6y  ?)  16(x2  2x  ?) 
 159  ?  ?
25(y2  6y  9)  16(x2  2x  1) 

159  25(9)  16(1)
25(y  3)2  16(x  1)2 
 400
25(y  3)2 
 400  16(x  1)2

(y  k)2
(x  h)2

 b

1
2
a2
[y  (1)]2
(x  1)2

 4 
 1
32
2
(y  1)
(x  1)2
   
 1
9
4
x2  4y2  6x

 24y 
 41
 6x  ?) 
 6y  ?) 
 41  ?  ?
(x2  6x  9)  4(y2  6y  9) 
 41  9  4(9)
(x  3)2  4(y  3)2 
 4
4(y  3)2 
 4 (x  3)2
(x2

2


5 2

b2 
 321  
3

5



 a

a
3
major axis: vertical axis

43.

2

Vertices: (1, 1),
(5, 1), (2, 3),
(2, 1)

) 
 (h, k  c)
42. focus: (1, 1  5
1  5

kc
1  5
 
 1  c

c
5
e 
 a

36  4(x  2)





9
36  4(x  2)



1
9

(y  3) 


4(y2

(y  3)2 

y3

y


y


400  16(x  1)




25 
400  16(x  1)




5 3
2

2

Vertices: (4, 3),
(6, 3), (1, 7),
(1, 1)

4  (x  3)2

4

4  (x  3)




4
4  (x  3)




3
4
2

2

vertices: (5, 3),
(1, 3), (3, 2),
(3, 4)

[15.16, 15.16] scl:1 by [10, 10] scl:1

[7.28, 7.28] scl:1 by [4.8, 4.8] scl:1

323

Chapter 10

47. The target ball should be placed opposite the
pocket, 5
 feet from the center along the major
axis of the ellipse. The cue ball can be placed
anywhere on the side opposite the pocket. The
ellipse has a semi-major axis of length 3 feet and a
semi-minor axis of length 2 feet. Using the
equation c2 
 a2  b2, the focus of the ellipse is
found to be 5
 feet from the center of the ellipse.
Thus the hole is located at one focus of the ellipse.
The reflective properties of an ellipse should
insure that a ball placed 5
 feet from the center
of the ellipse and hit so that it rebounds once off
the wall should fall into the pocket at the other
focus of the ellipse.
48. A horizontal line; see students’ work.

8

52a. a 
 2 or 4
b
3
a2  b2
c 
 
c 
 
42  32
c 
 7

foci: (h c, 0) 
 (0 7
, 0) or ( 7
, 0)
, 0) from
The thumbtacks should be placed ( 7
the center of the arch.
52b. With the string anchored by thumbtacks at the
foci of the arch and held taunt by a pencil, the
sum of the distances from each thumbtack to the
pencil will remain constant.
53a. GOES 4; its eccentricity is closest to 0.

96

49a. a 
 2 or 48
b


46

2

c

6955

or 23

(h, k) 
 (0, 0)
(x  h)2

a2
(x  0)2

482

(y  k)2
 b
2
(y  0)2


232
x2
y2
  
2304
529

x


1

a

x2
y2
  
9
4
a2 
 9


1

A 
 r2


1

A
rr
A
ab
A 
 ab

[figure not drawn to scale]
x 
 a  c  Earth’s radius
x 
 6955  361.66  6357
x 
 959.66
x 960 km
54.
x2  y2  Dx  Ey  F 
 0
2
0  (9)2  D(0)  E(9)  F 
 0 ⇒
9E  F 
 81
72  (2)2  D(7)  E(2)  F 
 0 ⇒
7D  2E  F 
 53
(5)2  (10)2  D(5)  E(10)  F 
 0 ⇒
5D  10E  F 
 125
 9E  F 
 81
(1)(7D  2E  F) 
 (1)(53)
 7D  7E

 28
DE
4
7D  2E  F 
 53
(1)(5D  10E  F ) 
 (1)(125)
12D  8E

 72
(8)(D  E) 
 (8)(4)
12D  8E 
 72
4D

 40
D 
 10
DE
4
9E  F 
 81
10  E 
 4
9(6)  F 
 81
E 
 6
F 
 135
x2  y2  Dx  Ey  F 
 0
x2  y2  10x  6y  135 
 0
(x2  10x  25)  (y2  6y  9) 
 135  25  9
(x  5)2  (y  3)2 
 169


1

b2 
 4
a
3
b
2
A 
 ab
A 
 (3)(2)
A 
 6 units2
51. If (x, y) is a point on the ellipse, then show that
(x, y) is also on the ellipse.
x2

a2
(x)2

a2

y2

 b2 
 1
(y)2

 b

1
2
x2

a2

y2

 b2 
 1

Thus, (x, y) is also a point on the ellipse and
the ellipse is symmetric with respect to the origin.

Chapter 10

x


1

49b. c 

c 
 2304


529
c 42.13
He could have stood at a focal point, about 42
feet on either side of the center along the major
axis.
49c. The distance between the focal points is 2c.
2c 
 2(42)

 84
about 84 ft
50a. x2  y2 
 r2
x2
y2

 r2
r2
x2
y2

 b2
a2


 0.052

c 
 361.66

O
c


e


1


a2  b2

50b.

c

a

y

53b.

324

center: (h, k) 
 (5, 3)
radius: r2 
 169
r 
 13
y

60. Let h 
 0.1.
x  h 
 x  0.1

 1  0.1 or 1.1
f(x  0.1) 
 f(1.1)

 (1.1)2  4(1.1)  12

 15.19
x  h 
 x  0.1

 1  0.1 or 0.9
f(x  0.1) 
 f(0.9)

 (0.9)2  4(0.9)  12

 14.79
f(x) 
 16
f(x) f(x  0.1) and f(x) f(x  0.1), so the point
is a location of a minimum.
61. The graph of the parent function g(x) 
 x is
translated 2 units right.

(5, 16)
(5, 3)

(8, 3)

x

O

55. Graph the quadrilateral with vertices A(1, 2),
B(5, 4), C(4, 1), and D(5, 4).
A quadrilateral is a
D (5, 4) y
parallelogram if one
pair of opposite sides
C (4, 1)
are parallel and
congruent.
x
O

g (x )

A(1, 2)
B (5, 4)

slope of 
DA


slope of 
CB


m



m

x x






y2  y1


x2  x1
2  4

1  (5)
6
3
 or 
4
2

y2  y1
2






62. Initial location: (2, 0)
Rot90 
 0 1 2 
 0 or (0, 2)
1
0 0
2
1
0
2
Rot80 


 2 or (2, 0)
0 1 0
0
0
1
2
0
Rot270 



or (0, 2)
1 0 0
2

1

4  1

54
5
1





The slopes are not equal, so D
CB
A
  
. The
quadrilateral is not a parallelogram; no.
56. cos 2v 
 1  2 sin2 v
7 2

cos 2v 
 1  2
8
34

17

2

k

57. A 
 4
c
k
c
2


 180

4

k
2


 20°

h
0


 20°

Page 641

c 
 40°
y 
 A cos(kx  c)  h
y 
 4 cos[2x  (40°)]  0
y 
 4 cos(2x  40°)
58. A 
 180°  (121° 32
  42° 5
) or 16° 23

a

sin A
4.1

sin 16° 23

4.1 sin 42° 5


sin 16° 23


b




sin B
b




sin 42° 5



b

a

sin A
4.1

sin 16° 23

4.1 sin 121° 32


sin 16° 23


   
   
   

63. mQTS  mTSR 
 180
a  b  c  d 
 180
b  b  c  c 
 180
2b  2c 
 180
b  c 
 90
The correct choice is C.


cos 2v 
 6
4 or  32

A


x

O

p  b  c 
 180
p  90 
 180
p 
 90

Graphing Calculator Exploration

1. Sample answer: The graph will shift 4 units to the
right.

c




sin C
c




sin 121° 32



c

9.7 b
12.4 c
59. P(x) 
 x4  4x3  2x2  1
P(5) 
 54  4(5)3  2(5)2  1
P(5) 
 74
P(5) 0; no, the binomial is not a factor of the
polynomial.

[15.16, 15.16] scl:2 by [10, 10] scl:2

325

Chapter 10

2. Sample answer: The graph will shift 4 units to the
left.

2. transverse axis: vertical
2a 
 4
a
2
An equation2in standard
form of the hyperbola
y
y2




must have 22 or 4 as the first term; b.
c

3. e 
 a, so ae 
 c and a2e2 
 c2.
Since c2 
 a2  b2 we have
a2e2 
 a2  b2
2
2
a e  a2 
 b2
a2(e2  1) 
 b2
4. With the equation in standard form, if the first
expression contains “x”, the transverse axis is
horizontal. If the first expression contains “y”, the
transverse axis is vertical.
5. center: (h, k) 
 (0, 0)
a2 
 25
b2 
 4
c 
 
a2  b2
a
5
b
2
c 
 25
4
  or 29

transverse axis: horizontal
foci: (h  c, k) 
 (0 29
, 0) or ( 29
, 0)
vertices: (h a, k) 
 (0 5, 0) or ( 5, 0)

[15.16, 15.16] scl:2 by [10, 10] scl:2
3. Sample answer: The graph will shift 4 units up.

[15.16, 15.16] scl:2 by [10, 10] scl:2
4. Sample answer: The graph will shift 4 units down.

b
(x
a
2
(x
5
2
x
5

asymptotes: y  k 

y0

y

8

 0)

y

(5, 0) 4

(5, 0)

8 4 O
4

[18.19, 18.19] scl:2 by [12, 12] scl:2
5. Sample answer: The graph will rotate 90°.

 h)

4

8

x

8

6. center: (h, k) 
 (2, 3)
a2 
 16
b2 
 4
c 
 
a2  b2
a 
 16
 or 4 b 
 4
 or 2 c 
 16
4
  or 25

transverse axis: vertical
foci: (h, k c) 
 
2, 3 25

vertices: (h, k a) 
 (2, 3 4) or (2, 7), (2, 1)
asymptotes: y  k 


[15.16, 15.16] scl:2 by [10, 10] scl:2
6. For (x  c), the graph will shift c units to the left.
For (x  c), the graph will shift c units to the
right.
7. For (y  c), the graph will shift c units down. For
(y  c), the graph will shift c units up.
8. The graph will rotate 90°.

y3

y3


y
(2, 7)

(2, 3)

10-4

Hyperbolas
x

O

Pages 649–650

(2, 1)

Check For Understanding

1. The equations of both hyperbolas and ellipses
have x2 terms and y2 terms. In an ellipse, the
terms are added and in a hyperbola these terms
are subtracted.

Chapter 10

326

a
(x
b
4
(x
2

 h)
 2)

2(x  2)

7.

11. 2b 
 6
b
3
x1  x2 y1  y2
33 40
center: 
2, 2 
 
2, 2

 (3, 2)
transverse axis: vertical
a 
 4  2 or 2

y2  5x2  20x 
 50
y2  5(x2  4x  ?) 
 50  ?
y2  5(x2  4x  4) 
 50  (5)(4)
y2  5(x  2)2 
 30
y2

30

(x  2)2

 6 
 1

xh
x2
yk
y
h
2
k
0
center: (h, k) 
 (2, 0)
a2 
 30
b2 
 6
c 
 
a2  b2
a 
 30

b 
 6

c 
 30
6
  or 6
transverse axis: vertical
foci: (h, k c) 
 (2, 0 6) or (2, 6)
vertices: (h, k a) 
 
2, 0 30
 or 
2, 30

asymptotes: y  k 

y0

y


( y  k)2

a2
( y  2)2

22
( y  2)2

4

(x  2)
5

x1  x2 y1  y2

0  0 6  (6)

c2

a2 
 2
(2, 
30)
(2, 0)

62

a2 
 2 or 18

( y  k)2

a2
( y  0)2

18

4
8x
30)
(2, 

(x  h)2

 b

1
2
(x  0)2

 18 
 1
y2
x2
   
 1
18
18
x1  x2 y1  y2

 
2 ,
2

8

13. center: 


8. center: (h, k) 
 (0, 5)
transverse axis: horizontal
a 
 5, b 
 3
(x  h)2

a2
(x  0)2

52
x2

25

(x  3)2

 9 
 1


 (0, 0)
transverse axis: vertical
c 
 distance from center to a focus

 0  6 or 6
b2 
 c2  a2
b2 
 a2
2
2
2
a 
c a
b2 
 18
2a2 
 c2

a
(x  h)
b

30
(x  2)

6

8

4 O
4

(x  3)2

 3

1
2

12. center: 
2, 2 
 
2, 2

y

4

(x  h)2

 b

1
2

10  (10) 0  0

 
 
2, 2

( y  5)2


 (0, 0)
transverse axis: horizontal
c 
 distance from center to a focus

 10  0 or 10
c
e 
 a
b2 
 c2  a2

( y  5)2

5

3

( y  k)2

 b

1
2
 3

1
2
 9 
 1

9. c 
 9
quadrants: II and IV
transverse axis: y 
 x
vertices: xy 
 9
3(3) 
 9
(3, 3)

a
6

(x  h)2

a2
(x  0)2

62

xy 
 9
3(3) 
 9
(3, 3)

y

b2 
 102  62
b2 
64
( y  k)2

 b

1
2
( y  0)2

 64 
 1
x2

36

y2

 64 
 1

14a. The origin is located midway between stations A
and B; (h, k) 
 (0, 0). The stations are located at
the foci, so 2c 
 130 or c 
 65.

(3, 3)

y

O

x
(3, 3)

A(65, 0)

10. center: (h, k) 
 (1, 4)
(x  h)2
( y  k)2

 b
2
a2
(x  1)2
[y  (4)]2




52
22
(x  1)2
( y  4)2
  
25
4

10


 a

O

B (65, 0) x

The difference of the distances from the plane to
each station is 50 miles.
50 
 2a (Definition of hyperbola)
25 
 a
b2 
 c2  a2
b2 
 652  252
b2 
 3600


1

1

1

327

Chapter 10

16. center: (h, k) 
 (0, 5)
a2 
 9
b2 
 81
a 
 9
 or 3
b 
 81
 or 9
c 
 
a2  b2
c 
 9
1
 8 or 310

transverse axis: horizontal
foci: (h c, k) 
 (0 310
, 5) or ( 310
, 5)
vertices: (h a, k) 
 (0 3, 5) or ( 3, 5)
b
asymptotes: y  k 
 a(x  h)

transverse axis: horizontal
(x  h)2
(y  k)2

 b
2
a2
(x  0)2
(y  0)2



252
3600
x2
y2
  
625
3600


1

1

1

a, k) 
 (0

14b. Vertices: (h

asymptotes: y  k 

y0

y

Plane
located
on this
branch

60

25, 0) or ( 25, 0)

b
(x  h)
a
3

600
(x 
25
12
x
5

0)

y

9
(x
3

y5


3x

 0)

y

40

(0, 5)
Station B
(65, 0)

20

Station A
(65, 0)

y5


60 40 20 O
20

20

(3, 5)

(3, 5)

60 x

40

O

x

40

17. center: (h, k) 
 (0, 0)
b2 
 49
a2 
 4
a 
 4
 or 2
b 
 49
 or 7
c 
 
a2  b2
c 
 4
9
 4 or 53

transverse axis: horizontal
foci: (h c, k) 
 (0 53
, 0) or ( 53
, 0)
vertices: (h a, k) 
 (0 2, 0) or ( 2, 0)

60

14c. Let y 
 6.
x2

625
x2

625
x2

625

y2



3600 
 1
62



3600 
 1
36



3600 
 1
x2

625
x2

625
x2

36



1
3600

asymptotes: y  k 



 1.01

y0



 625(1.01)
x2 
 631.25
x 
 631.2
5

x
25.1
Since the phase is closer to station A than
station B, use the negative value of x to locate
the ship at (25.1, 6).

y

8

b
(x  h)
a
7
(x  0)
2
7
x
2

y

4
(2, 0)

(2, 0)

4 2 O

2

4x

4

Pages 650–652

8

Exercises

15. center: (h, k) 
 (0, 0)
a2 
 100
b2 
 16
a 
 100
 or 10
b 
 16
 or 4
2
2
c 
 
a  b
c 
 100


16 or 229

transverse axis: horizontal
foci: (h c, k) 
 (0 229
, 0) or ( 229
, 0)
vertices: (h a, k) 
 (0 10, 0) or ( 10, 0)
asymptotes: y  k 

y0

y


b
(x  h)
a
4
(x  0)
10
2
x
5

18. center: (h, k) 
 (1, 7)
a2 
 64
b2 
 4
a 
 64
 or 8
b 
 4
 or 2
c 
 
a2  b2
c 
 64
4
  or 217

transverse axis: vertical
foci: (h, k c) 
 (1, 7 217
)
vertices: (h, k a) 

y
(1, 7 8) or (1, 15), (1, 1)

y

4
(10, 0)

(10, 0)
16 8 O
4

8

y7


4(x  1)

 h)

(1, 15)

 (1)]
(1, 7)

16 x

O
(1, 1)

8

Chapter 10

y7


a
(x
b
8
[x
2

asymptotes: y  k 


8

328

x

19.

x2  4y2  6x  8y 
 11
(x2  6x  ?)  4( y2  2y  ?) 
 11  ?  ?
(x2  6x  9)  4( y2  2y  1) 
 11  9  (4)(1)
(x  3)2  4( y  1)2 
 16
(x  3)2

16

21. 16y2  25x2  96y  100x  356 
 0
16(y2  6y  ?)  25(x2  4x  ?) 
 356
16(y2  6y  9)  25(x2  4x  4) 
 356  16(9)  25(4)
16(y  3)2  25(x  2)2 
 400

( y  1)2

(y  3)2
(x  2)2
   
 1
25
16

 4 
 1

center: (h, k) 
 (3, 1)
b2 
 4
c 
 
a2  b2
a2 
 16
2
a 
 16
 or 4 b 
 4
 or 2 c 
 16
4
  or 25

transverse axis: horizontal
foci: (h c, k) 
 (3 25
, 1)
vertices: (h a, k) 
 (3 4, 1) or (1, 1),
(7, 1)
b
asymptotes: y  k 
 a(x  h)
y  (1) 

y1


center: (h, k) 
 (2, 3)
b2 
 16
c 
 
a2  b2
a2 
 25
a 
 25

b 
 16

c 
 25
16

or 5
or 4
or 41

transverse axis: vertical
foci: (h, k c) 
 (2, 3 41
)
vertices: (h, k a) 
 (2, 3 5) or (2, 8), (2, 2)
asymptotes: y  k 


2
[x  (3)]
4
1
(x  3)
2

y3


 h)
 2)

y

y

(2, 8)

O
(7, 1)

a
(x
b
5
(x
4

(2, 3)

x
(1, 1)

(3, 1)

x

O
(2, 2)

20.

4x  9y2  24x  90y  153 
 0
9( y2  10y  ?)  4(x2  6x  ?) 
 153
9(y2  10y  25)  4(x2  6x  9) 
 153 
9(25)  4(9)
9( y  5)2  4(x  3)2 
 36
( y  5)2

4



(x  3)2

9

22.

36(x2x1)49(y2 6y9)
216936(1)49(9)
36(x1)2 49(y3)2 
1764
(x 1)2
(y 3)2
   
1
49
36


1

center: (h, k) 
 (1, 3)
b2 
 36
c 
 
a2  b2
a2 
 49
a 
 49

b 
 36

c 
 49
36

or 7
or 6
or 85

transverse axis: horizontal
foci: (h c, k) 
 (1 85
, 3)
vertices: (h a, k) 
 (1 7, 3) or (8, 3),
(6, 3)

center: (h, k) 
 (3, 5)
b2 
 9
c 
 
a2  b2
a2 
 4
a 
 4
 or 2
b 
 9
 or 3 c 
 4

 9 or 13

transverse axis: vertical
foci: (h, k c) 
 (3, 5 13
)
vertices: (h, k a) 
 (3, 5 2) or (3, 7), (3, 3)
asymptotes: y  k 

y5

y5


36x2 49y2 72x294y
2169

36(x2 2x?)49(y2 6y?)
2169??

a
(x  h)
b
2
[x  (3)]
3
2
(x  3)
3

asymptotes: y  k 

y  (3) 

y3


y

 h)
 1)
 1)

y

(3, 7)
(3, 5)

O

(3, 3)

b
(x
a
6
(x
7
6
(x
7

x

(6, 3)

O

329

(1, 3)

x
(8, 3)

Chapter 10

23.

27. c 
 49
quadrants: I and III
transverse axis: y 
 x
vertices: xy 
 49
7(7) 
 49
(7, 7)

25y2  9x2  100y  72x  269 
 0
25(y2  4y  ?)  9(x2  8x  ?) 
 269  ?  ?
25(y2  4y  4)  9(x2  8x  16) 
 269  25(4)  9(16)
25(y  2)2  9(x  4)2 
 225
(y  2)2
(x  4)2
   
 1
9
25

center: (h, k) 
 (4, 2)
b2 
 25
c 
 
a2  b2
a2 
 9
a 
 9

b 
 25

c 
 9
5
 2
or 3
or 5
or 34

transverse axis: vertical
foci: (h, k c) 
 
4, 2 34

vertices: (h, k a) 

(4, 2 3) or (4, 5), (4, 1)
asymptotes: y  k 

y2

y2


a
(x
b
3
[x
5
3
(x
5

y
(7, 7)

(7, 7)

 (4)]
28. c 
 36
quadrants: II and IV
transverse axis: y 
 x
vertices:
xy 
 36
6(6) 
 36
(6, 6)

 4)

y

(4, 5)

3
y  2 
 (x  4)
5

O

(4, 1)

y

(6, 6)

24. center: (h, k) 
 (4, 3)
transverse axis: vertical
a 
 4, b 
 3

x

O
(6, 6)

(x  h)2

 b

1
2
(x  4)2

 3

1
2

29. 4xy 
 25
25
xy 
 4

(x  4)2

 9 
 1

25. center: (h, k) 
 (0, 0)
transverse axis: horizontal
a 
 3, b 
 3
(x  h)2

a2
(x  0)2

32

25

c 
 4

(y  k)2

quadrants: II and IV
transverse axis: y 
 x

(y  0)2

vertices:

 b

1
2
 3

1
2
x2

9

(y 
(x 

 b
2
a2
(y  0)2
[x  (4)]2

 1
2
22
y2
(x  4)2
  
4
1
k)2

b)2

25

xy 
 4

25

25

2
2 
 4

xy 
 4


 

5
5
  

2
2
5
5
,  
2
2

y2

 9 
 1




26. center: (h, k) 
 (4, 0)
transverse axis: vertical
a 
 2, b 
 1

5 5

4




52, 52

y


1

( 52, 52 )


1

x


1

O

( 52,  52 )

Chapter 10

xy 
 36
6(6) 
 36
(6, 6)

x

(4, 2  
34)

(y  k)2

a2
(y  3)2

42
(y  3)2

16

x

O

 h)

(4, 2  
34)
3
y  2 
  (x  4)
5
(4, 2)

xy 
 49
7(7) 
 49
(7, 7)

330

25

34. 2b 
 8
b
4

30. 9xy 
 16
xy 

c


16

9
16

9

x1  x2 y1  y2

3  (3) 9  (5)

center: 
2, 2 
 
2, 2

 (3, 2)
transverse axis: vertical
a 
 distance from center to a vertex

 2  9 or 7

quadrants: I and III
transverse axis: y 
 x
16

xy 
 9

16

4 4
16
  
 
3 3
9
4 4
, 
3 3

3
3 
 9

xy 
 9

vertices:





 

4



4

16


43, 43

y
35.

x

O

8  (8) 0  0
,   
 
 

2
2





 (0, 0)
transverse axis: horizontal
c 
 distance from center to a focus

 0  8 or 8
b2 
 c2  a2
b2 
 a2
a2 
 c2  a2
b2 
 32
2a2 
 c2

( 43, 43 )
( 43,  43 )

(y  k)2
(x  h)2

 b

1
2
a2
(y  2)2
[x  (3)]2

 4

1
2
72
(y  2)2
(x  3)2
   
 1
49
16
x1  x2 y1  y2
center: 2, 2

c2

a2 
 2

31. center: (h, k) 
 (4, 2)
(y  k)2

a2
[y  (2)]2

22
(y  2)2

4

82

(x  h)2

a2 
 2 or 32

 b

1
2

(x  h)2

a2
(x  0)2

32

(x  4)2

 3

1
2
(x  4)2

 9 
 1

x1  x2 y1  y2

c

5

4

(x  h)2

a2
[x  (3)]2

42
(x  3)2

16

(x  0)2

 72 
 1
x2

 72 
 1

5  (5) 2  2


 (0, 2)
transverse axis: horizontal
c 
 distance from center to a focus

 0  5 or 5
b2 
 c2  a2
b2 
 52  32 or 16



(y  1)2

 9 
 1

3

(y  k)2



(y  1)2

 9 
 1

y  2 
 4(x  4)
a

b
3

b

 b

1
2
(y  2)2

16
(y  2)2

16

(y  k)2

 b

1
2

37. center: (h, k) 
 (4, 2)
a 
 distance from center to a vertex

 2  5 or 3
transverse axis: vertical
4y  4 
 3x
4y  4  12 
 3x  12
4y  8 
 3x  12
4(y  2) 
 3(x  4)

center: 
2, 2 
 
2, 2

(x  b)2

a2
(x  0)2

32
x2

9

5


 a

a
4
transverse axis: horizontal

(x  h)2

x1  x2 y1  y2

b2 
 c2  a2
b2 
 52  42
b2 
 9

e 
 a

 b

1
2

33. 2a 
 6
a
3

y2

 32 
 1

36. centers: (h, k) 
 (3, 1)
c 
 distance from center to a focus

 3  2 or 5


 (0, 0)
transverse axis: vertical
a 
 distance from center to a vertex

 0  3] or 3
c 
 distance from center to a focus

 0  (9) or 9
b 2 
 c 2  a2
b2 
 92  32 or 72

y2

9

(y  0)2

 32 
 1
x2

32

0  0 3  (3)

32. center: 
2, 2 
 
2, 2

(y  k)2

a2
(y  0)2

32

(y  k)2

 b

1
2


1

3


 4
3


 4

b
4


1

(y  k)2

a2
(y  2)2

32
(y  2)2

9

331

(x  h)2

 b

1
2
(x  4)2

 4

1
2
(x  4)2

 16 
 1

Chapter 10

38. center: (h, k) 
 (3, 1)
a 
 distance from center to a vertex

 3  5 or 2
transverse axis: horizontal
3x  11 
 2y
3x  11  4 
 2y  4
3x  15 
 2y  4
3(x  5) 
 2(y  2)
3
(x
2

transverse axis: horizontal
(x  h)2

a2
(x  0)2
__
81

2

3


 2
3


 2

a

b

b
3




0  0 8  (8)
  , 
2
2





c

4

3




8

a

a
6
transverse axis: vertical
(y 

a2
(y  0)2

62
k)2

40.

(x 



b2 
 c2  a2
b2 
 82  62
b2 
 28

a2 
 4
5

6





1

PV 
 505
(101)V 
 505
V 
 5.0 dm3
43c.
PV 
 505
(50.5)V 
 505
V 
 10.0 dm3
43d. If the pressure is halved, then the volume is
doubled, or V 
 2(original V ).
44. In an equilateral hyperbola, a 
 b and
c2 
 a2  b2.
c 2 
 a2  a2
a
b
c2 
 2a2
c 
 a2

c

9  (9) 0  0
,  
0
2





Since e 
 a, we have



c

e 
 a


 (0, 0)
c 
 distance from center to a focus

 0  9 or 9
b2 
 c 2  a 2
b2 
 a2
a2 
 92  a2
2a2 
 81
81
a2 
 2


1

43b.

a
5
transverse axis: horizontal
(x  h)2
(y  k)2

 b

1
2
a2
(x  4)2
[y  ( 3)]

 11 
 1
52
(x  4)2
(y  3)2
   
 1
25
11
x1  x2 y1  y2



center:
2 ,
2


1

O 2 4 6 8 10 P

10  (2) 3  (3)
, 
 
 

2
2

b2 
 c2  a2
b2 
 62  52
b2 
 11


 a

64

a2 
 5

V

(x  0)2

c

16

250
200
150
100
50

 28 
 1

e 
 a

Chapter 10

a 
 2b
c2 
 a2  b2
42 
 (2b)2  b2
16 
 5b2

43a. quadrants: I and II
transverse axis: y 
 x


 (4, 3)
c 
 distance from center to a focus

 4  10 or 6

41.

a2 
 (2b)2

 b

1
2




1  1 5  (3)
 
 
 

2 ,
2

2

h)2

y2
x2
   
 1
36
28
x1  x2 y1  y2
centers: 2, 2

6

5




16
 
 b2
5
(y  k)2
(x  h)2

 b
2
a2
(x  1)2
(y  1)2
  
64
16


5
5
5(y  1)2
5(x  1)2
  
64
16


 (0, 0)
c 
 distance from center to a focus

 0  8 or 8
e 
 a


1


 (1, 1)
c 
 distance from center to a focus

 1  5 or 4
transverse axis: vertical

3

39.

(y  0)2
__
81

2

42. center: 


 5) 
 y  2

(x  b)2
(y  k)2

 b

1
2
a2
(x  3)2
[y  (1)]2

 3

1
2
22
(x  3)2
(y  1)2
   
 1
4
9
x1  x2 y1  y2



center:
2 ,
2



2x2
2y2
   
 1
81
81
x1  x2 y1  y2

, 
2
2

y  2 
 2(x  5)
b

a
b

2

(y  k)2

 b

1
2

e


a2


a


e 
 2
Thus, the eccentricity of any equilateral hyperbola
is 2
.

81

b2 
 2

332

y

45a.

The lightning is 2200 feet farther from station B
than from station A. The difference of distances
equals 2a.
2200 
 2a (Definition of hyperbola)
1100 
 a
b 2 
 c 2  a2
b 
 
c2  a2
b 
 
10,56
02  1
1002
b 10,503
center: (h, k) 
 (0, 0)
transverse axis: horizontal

150 ft

x

O

2a 
 150
a 
 75
5

3

c

75




(x  h)2

a2
(x  0)2

11002
x2

11002

b2 
 c2  a2
b2 
 1252  752
b2 
 10,000
b 
 100

c

e 
 a

125 
 c
transverse axis: horizontal
center: (h, k) 
 (0, 0)
(x  h)2

a2
(x  0)2

752

(y  k)2

(y  0)2



1002 
 1

45b.

y2



1002 
 1

y
(x, 100)
(0, 100)

48a.

1002



1002 
 1

x2

752

(x, y) 
 (x, 100)

1
1
x2

752
x2

x2

16

(y  k)2

 b

1
2
(y  0)2

 11 
 1
y2

 11 
 1

y2

 9 
 1

asymptotes: y  k 



2

y0



 11,250
x 106.07 ft

x2

y


y2



base: 
752  1002 
 1
x2
(350)2


752  1002
x2

752  12.25
x2

752
x2

y2

9


1

PF1  PF2  2a

center: (h, k) 
 (0, 0)
b2 
 9
a2 
 16
a 
 16
 or 4
b 
 9
 or 3
transverse axis: horizontal
vertices: (h a, k) 
 (0 4, 0) or ( 4, 0)

y2



top: 
752  1002 
 1
x2

752

y2



10,5032 
 1

x2

25

(x, 350)

(0, 350)

x2

(x  h)2

a2
(x  0)2

52

x

O

(y  0)2



10,5032 
 1

47. center: (h, k) 
 (0, 0)
c 
 0  6 or 6
PF1  PF2 
 10
2a 
 10
a
5
b2 
 c2  a2
b2 
 62  52
b2 
 11
transverse axis: horizontal

 b

1
2

x2

752

(y  k)2

 b

1
2

(x, y) 
 (x, 350)



x2

16

b
(x
a
3
(x
4
3
x
4

 h)
 h)


1

center: (h, k) 
 (0, 0)
a2 
 9
b2 
 16
a 
 9
 or 3
b 
 16
 or 4
transverse axis: vertical
vertices: (h, k a) 
 (0, 0 3) or (0, 3)


1

 13.25


 74,531.25
x 273.00 ft
46. The origin is located midway between stations A
and B. The stations are located at the foci, so
2c 
 4 or c 
 2 miles.
c 
 2 mi

asymptotes: y  k 

y0

y


a
(x
b
3
(x
4
3
x
4

 h)
 0)

y

5280 ft


c 
 2 mi  
1 mi

c 
 10,560 ft
d 
 rt
d 
 1100(2) or 2200 ft

y
O
A

(10,560, 0)

x

B

O

(10,560, 0) x

333

Chapter 10

48c.
48d.

(y  2)2

25
(x  3)2

16




(x  3)2

16
(y  2)2

25

y

51.

48b. They are the same lines.

1

A (1, 3)

center: (h, k) 
 (3, 2)
b2 
 25
a2 
 16
a 
 16
 or 4
b 
 25
 or 5
transverse axis: horizontal
vertices: (h a, k) 
 (3 4, 2) or (7, 2), (1, 2)
asymptotes: y  k 

y2

(y  2)2

25



(x  3)2

16

b
(x
a
5
(x
4

C (6, 2)

B (2, 1)

AB 
 
(2  1
)2  (
1  3
)2 
 5
2
BC 
 
(2  6
)  (
1  2
)2 
 5
2
2
(6  3
)  (2
 6) 
 5
CD 
 
AD 
 
(3  1
)2  (6
 3)2 
 5
Thus, ABCD is a rhombus. The slope of A
D



 h)
 3)


1

asymptotes: y  k 

y2


a
(x
b
5
(x
4

63

31

 h)
 3)

8
4
8

x

4
x1  x2 y1  y2

1
x
2

2  2 3  (3)

49. center: 
2, 2 
 
2, 2

50.

4


3

 2 y  3 
 0

y  6 
 0
x  3
x

55.


 (2, 0)
a
4
c 
 distance from center to a focus

 0  3 or 3
b2 
 a2  c2
b2 
 42  32 or 7
major axis: vertical
(y  k)2

a2
(y  0)2

42
y2

16

31

3



or 4 and the slope of 
AB


1  2 or  3 .

Thus, A
D
 is perpendicular to A
B
 and ABCD is a
square.
52. (r, v) 
 (90, 208°)
(r, v  360 k°) 
 (90, 208°  360(1)°)

 (90, 152°)
(r, v  (2k  1)(180°))

 (90, 208  (2(1)  1)(180°))

 (90, 28°)
53. 4(5)  1(2)  8(2) 
 6
No, the inner product of the two vectors is not
zero.
54. x cos f  y sin f  p 
 0
x cos 60  y sin 60  3 
 0

y

4

x

O

center: (h, k) 
 (3, 2)
b2 
 16
a2 
 25
a 
 25
 or 5
b 
 16
 or 4
transverse axis: vertical
vertices: (h, k a) 
 (3, 2 5) or (3, 7), (3, 3)

4 O

D (3, 6)


1

9000 m
30˚
60˚

(x  h)2

 b

1
2

x


tan 30° 
 
9000
9000 tan 30° 
 x
5196 x
d 
 rt

(x  2)2
 7 
 1
(x  2)2
 7 
 1
x2  y2  4x 

14y  28 
 0
(x2  4x  ?)  (y2  14y  ?) 
 28  ?  ?
(x2  4x  4)  (y2  14y  49) 
 28  4  49
(x  2)2  (y  7)2 
 81

d

t
5196

15


r

r

346.4 r
about 346 m/s

y
56.

x

O
(2, 7)
(11, 7)

(2, 16)

Chapter 10

Since 0.2506 is closer to zero than 0.6864, the
zero is about 1.3.

334

7. y2  4x  2y  5 
 0
y2  2y 
 4x  5
y2  2y  ? 
 4x  5  ?
y2  2y  1 
 4x  5  1
(y  1) 
 4(x  1)
vertex: (h, k) 
 (1, 1)
4p 
 4
p
1
focus: (h  p, k) 
 (1  1, 1) or (2, 1)
directrix: x 
 h  p
x
11
x
0
axis of symmetry: y 
 k
y 
 1

Since 0.0784 is closer to zero than 0.2446, the
zero is about 0.6.
57. Case 1: r is positive and s is negative.
Case 2: r is negative and s is positive.
I. r3  s3 is false if r is negative.
II. r3 
 s2 is false for each case.
III. r4 
 s4 is true for each case.
The correct choice is C.

y
x
0

10-5

Parabolas

Pages 658–659

O

x

(2, 1)
(1, 1)

Check for Understanding

1. The equation of a parabola will have only one
squared term, while the equation of a hyperbola
will have two squared terms.
2. vertex: (h, k) 
 (2, 1)
p 
 4
(x  h)2 
 4p(y  k)
(x  2)2 
 4(4)( y  1)
(x  2)2 
 16( y  1)
3. The vertex and focus both lie on the axis of
symmetry. The directrix and axis of symmetry are
perpendicular to each other. The focus and the
point on the directrix collinear with the focus are
equidistant from the vertex.
4. (h, k) 
 (4, 5)
p 
 5
(y  k)2 
 4p(x  h)
(y  5)2 
 4(5)[x  (4)]
(y  5)2 
 20(x  4)
5a. ellipse
5b. parabola
5c. hyperbola
5d. circle
6. vertex: (h, k) 
 (0, 1)
4p 
 12
p
3
focus: (h, k  p) 
 (0, 1  3) or (0, 4)
directrix: y 
 k  p
y
13
y 
 2
axis of symmetry: x 
 h
x
0
y

8. x2  8x  4y  8 
 0
x2  8x 
 4y  8
x2  8x  ? 
 4y  8  ?
x2  8x  16 
 4y  8  16
(x  4)2 
 4(y  2)
vertex: (h, k) 
 (4, 2)
4p 
 4
p 
 1
focus: (h, k  p) 
 (4, 2  (1)) or (4, 1)
directrix: y 
 k  p
y 
 2  (1)
y
3
axis of symmetry: x 
 h
x 
 4

y
y
3
(4, 2)
(4, 1)

O x

9. vertex: (h, k) 
 (0, 0)
opening: downward
p 
 4
(x  h)2 
 4p(y  k)
(x  0)2 
 4(4)(y  0)
x2 
 16y

(0, 4)
(0, 1)

y

O

x

x

O
y 
 2

335

Chapter 10

10.

(y  k)2 
 4p(x  h)
(1  5)2 
 4p[2  (7)]
(6)2 
 36p
1
p
(y  k)2 
 4p(x  h)
(y  5)2 
 4(1)[x  (7)]
(y  5)2 
 4(x  7)

12b. The maximum height is s 
 52 ft.
12c. Let s 
 0.
s 
 56t  16t2  3
0 
 16t2  56t  3

(h, k)  (7, 5);
(x, y)  (2, 1)

y

t


b 
b2  4
ac

2a

t


56 
562 
4(1
6)(3)

2(16)

t 3.6 or 0.05
3.6 s

Pages 659–661
(7, 5)

x
O
11. vertex: (h, k) 
 (4, 3)
opening: upward
(x  h)2 
 4p(y  k)
(5  4)2 
 4p[2  (3)]
12 
 20p
(x 

1

20
h)2

Exercises

13. vertex: (h, k) 
 (0, 0)
4p 
 8
p
2
focus: (h  p, k) 
 (0  2, 0) or (2, 0)
directrix: x 
 h  p
x
02
x 
 2
axis of symmetry: y 
 k
y
0
(h, k)  (4, 3);
(x, y)  (5, 2)

y


p

 4p(y  k)

O

(x  4)2 
 4
2
0 (y  3)
1

(2, 0)

x

x 
 2

1

(x  4)2 
 5( y  3)

y
14. vertex: (h, k) 
 (0, 3)
4p 
 4
p
1
focus: (h, k  p) 
 (0, 3  (1)) or (0, 2)
directrix: y 
 k  p
y
y 
 3  (1)
y
4
y
4
(0, 3)
axis of symmetry: x 
 h
(0, 2)
x
0

x
O
(4, 3)

s 
 v0t  16t2  3
s 
 56t  16t2  3
2
16t  56t 
 s  3

12a.

O

x

16
t2  5t  ? 
 s  3  ?
7

16
t2  2t  16 
 s  3  (16)
16
7

49












opening: downward







50
45
40
35
30
25
20
15
10
5

O
Chapter 10

15. vertex: (h, k) 
 (0, 6)
4p 
 4
p
1
focus: (h  p, k) 
 (0  1, 6) or (1, 6)
directrix: x 
 h  p
x
01
x 
 1
y
axis of symmetry: y 
 k
y
6

7 2
t  4 
 s  52
7 2
1
t  4 
 1
6 (s  52)
7 2
1
t  4 
 4 6
4 (s  52)
7
s
k) 
 4, 52
55

16


vertex: (h,

49

x 
 1

(1, 6)

(0, 6)

1

2

3

4

x

O

336

x

16.

y 2  12x 
 2y  13
y2  2y 
 12x  13
y2  2y  ? 
 12x  13  ?
y2  2y  1 
 12x  13  1
( y  1)2 
 12(x  1)
vertex: (h, k) 
 (1, 1)
4p 
 12
p 
 3
focus: (h  p, k) 
 (1  (3), 1) or (4, 1)
directrix: x 
 h  p
x 
 1  (3)
x
2
axis of symmetry: y 
 k
y
1

18. x2  10x  25 
 8y  24
(x  5)2 
 8(y  3)
vertex: (h, k) 
 (5, 3)
4p 
 8
p 
 2
focus: (h, k  p) 
 (5, 3  (2)) or (5, 1)
directrix: y 
 k  p
y 
 3  (2)
y
5
axis of symmetry: x 
 h
x 
 5
y
y
5

y
(5, 3)

x
2
(1, 1)

O

O

17.

(5, 1)

x

(4, 1)

x
19. y2  2x  14y 
 41
y2  14y 
 2x  41
2
y  14y  ? 
 2x  41  ?
y2  14y  49 
 2x  41  49
(y  7)2 
 (2x  4)
vertex: (h, k) 
 (4, 7)
4p 
 2

y  2 
 x2  4x
x2  4x 
 y  2
x2  4x  ? 
 y  2  ?
x2  4x  4 
 y  2  4
(x  2)2 
 y  2
vertex: (h, k) 
 (2, 2)
4p 
 1

2

focus: (h  p, k) 
 
4  2, 7 or 
2, 7
1

1

focus: (h, k  p) 
 
2, 2 

 or 
2,

1

4

1



x 
 4  2

7
4

9

x 
 2

directrix: y 
 k  p
1

axis of symmetry: y 
 k
y 
 7

y 
 2  4
y


9
4

y

axis of symmetry: x 
 h
x
2
y

O

7

directrix: x 
 h  p

p 
 4

1

1

p 
 4 or 2

1

2

3

x


x

( 72 , 7)

(2,  74 )
(2, 2)

x

O
9
2

y 
  94

(4, 7)

337

Chapter 10

vertex: (h, k) 
 (3, 2)
4p 
 10

20. y2  2y  12x  13 
 0
y2  2y 
 12x  13
y2  2y  ? 
 12x  13  ?
y2  2y  1 
 12x  13  1
( y  1)2 
 12(x  1)
vertex: (h, k) 
 (1, 1)
4p 
 12
p
3
focus: (h  p, k) 
 (1  3, 1) or (4, 1)
directrix: x 
 h  p
x
13
x 
 2
axis of symmetry: y 
 k
y
1

10

5

p 
 4 or 2
focus: (h, k  p) 
 
3, 2  2 or 
3, 2
5

9

directrix: y 
 k  p
5

y 
 2  2
1

y 
 2
axis of symmetry: x 
 h
x
3
y

(3, 92 )

y
x 
 2

O
(1, 1)

(4, 1)

O

x

23. 2y2  16y  16x  64 
 0
2y2  16y 
 16x  64
2(y2  8y  ?) 
 16x  64  ?
2(y2  8y  16) 
 16x  64  2(16)
2(y  4)2 
 16x  32
(y  4)2 
 8x  16
(y  4)2 
 8(x  2)

21. 2x2  12y  16x  20 
 0
2x2  16x 
 12y  20
2(x2  8x  ?) 
 12y  20  ?
2(x2  8x  16) 
 12y  20  2(16)
2(x  4)2 
 12y  12
(x  4)2 
 6(y  1)
vertex: (h, k) 
 (4, 1)
4p 
 6
6

y 
  12 x

(3, 2)

vertex: (h, k) 
 (2, 4)
4p 
 8
p 
 2
focus: (h  p, k) 
 (2  (2), 4) or (4, 4)
directrix: x 
 h  p
y
x 
 2  (2)
x
0
O
x
axis of symmetry: y 
 k
x
0
y 
 4

3

p 
 4 or 2
focus: (h, k  p) 
 
4, 1  2 or 
4, 2
3

1

directrix: y 
 k  p
3

(4, 4)

y 
 1  2
5

y 
 2
axis of symmetry: x 
 h
x
4
y

24. vertex: (h, k) 
 (5, 1)
opening: right
hp
2
5  p 
 2
p
7
(y  k)2 
 4p(x  h)
(y  1)2 
 4(7)[x  (5)]
(y  1)2 
 28(x  5)

(4, 12 )
O

x

(4, 1)

y 
  52

y
8

22. 3x2  30x  18x  87 
 0
3x2  18x 
 30y  87
3(x2  6x  ?) 
 30y  87  ?
3(x2  6x  9) 
 30y  87  3(9)
3(x  3)2 
 30y  60
(x  3)2 
 10y  20
(x  3)2 
 10(y  2)

Chapter 10

4
(5, 1)
12 8 4 O
4
8

338

x

(2, 4)

25. opening: left
focus: (h  p, k) 
 (0, 6)
hp
0
h  (3) 
 0
h
3
(y  k)2 
 4p(x  h)
(y  6)2 
 4(3)(x  3)
(y  6)2 
 12(x  3)

28. vertex: (h, k) 
 (2, 3)
(y  k)2 
 4p(x  h)
[1  (3)]2 
 4p[3  (2)]
42 
 4p
4 
 p
(y  k)2 
 4p(x  h)
(y  3)2 
 4(4)(x  2)
(y  3)2 
 16(x  2)

k
6

y
16
12
8
(3, 6)

O 4

8

y
O

x

(2, 3)

4
8 4

(h, k)  (2, 3);
(x, y)  (3, 1)

x

4

29. opening: upward
p
2
focus: (h, k  p) 
 (1, 7)
h 
 1
kp
7
k2
7
k
5
2
(x  h) 
 4p(y  k)
[x  (1)]2 
 4(2)(y  5)
(x  1)2 
 8(y  5)

26. opening: upward
vertex: (h, k) 
 
4,

51

2

 or (4, 3)

focus: (h, k  p) 
 (4, 1)
kp
1
3  p 
 1
p
2
(x  b)2 
 4p(y  k)
(x  4)2 
 4(2)[y  (3)]
(x  4)2 
 8(y  3)
y

O

(1, 5)

O

x
30. opening: downward
vertex: (h, k) 
 (5, 3)
(x  h)2 
 4p(y  k)
(1  5)2 
 4p[7  (3)]
(4)2 
 16p
1 
 p
(x  h)2 
 4p(y  k)
(x  5)2 
 4(1)(y  3)
(x  5)2 
 4(y  3)

(4, 3)

27. opening: downward
vertex: (h, k) 
 (4, 3)
(x  h)2 
 4p(y  k)
(5  4)2 
 4p(2  3)
12 
 4p

(h, k)  (4, 3);
(x, y)  (5, 2)

1

4 
 p

y
O

(x  h)2 
 4p(y  k)
(x 

4)2


 4


(x 

4)2


 (y  3)

y

O

y

(y  3)

1
4

x

(maximum)
(h, k)  (5, 3);
(x, y)  (1, 7)

x
(5, 3)

(4, 3)

x

31. opening: right
vertex: (h, k) 
 (1, 2)
( y  k)2 
 4p(x  h)
(0  2)2 
 4p[0  (1)]
(2)2 
 4p
1
p
(y  k)2 
 4p(x  h)
(y  2)2 
 4(1)[x  (1)]
(y  2)2 
 4(x  1)

(h, k)  (1, 2);
(x, y)  (0, 0)

y

(1, 2)

O

339

x

Chapter 10

34a. Let y 
 income per flight.
Let x 
 the number of $10 price decreases.
Income 
 number of passengers  cost of a ticket
y 
 (110  20x)  (140  10x)
y

15,400  1100x  2800x  200x2

32. opening: upward
h




x1  x2


2
12
3
 or 
2
2

vertex: (h, k) 
 
2, 0
3

(x  h)2 
 4p(y  k)
2


1  32

(x 

1
 
 4p
4
1
 
 p
16
h)2 
 4p(y

3 2
x  2
3
y  2






y 
 200
x2  2x  15,400
17

(h, k)  
2, 0;
3


 4p(1  0)

y  15,400 
 200
x2  2x
17

(x, y)  (1, 1)

17 2

1


200

 k)

 
 4
 (y  0)
 
 14y
1

16

1

( 32 , 0)

2

x

33a. vertex: (h, k) 
 (0, 0)
depth: x 
 4
p
2
y
(4, y 1)

O

(2, 0)

(4, 0) x

(4, y 2)

( y  k)2 
 4p(x  h)
(y  0)2 
 4(2)(x  0)
y2 
 8x
y 
 8x

y 
 8(4)

y 
 42

y1 
 42
, y2 
 42


(h, k)  (0, 0);
p2

(y  19,012.5) 
 
x 



17 2

4



17

3 2

3 2

y  15,400 100
2 
 100 x2  3x  
2
y  15,625 
 100
x 
1


100

(y  15,625) 
 
x 

3 2

2





3 2

2



The vertex of the parabola is at 
2, 15,625, and
because p is negative, it opens downward. So the
vertex is a maximum and the number of $10
3
price decreases is 2 or 1.5.
number of passengers 
 110  10x

 110  10(1.5)

 125
This is less than 180, so the new ticket price can
be found using 1.5 $10 price decreases.
cost of a ticket 
 140  10x

 140  10(1.5)

 $125

diameter 
 y1  y2

 42
  (42
)

 82
 in.
33b. depth: x 
 1.25(4)
x
5
( y  k)2 
 4p(x  h)
(y  0)2 
 4(2)(x  0)
(h, k)  (0, 0);
y2 
 8x
p2
y 
 8x

y 
 8(5)

y 
 210

y1 
 210
, y2 
 210

diameter 
 y1  y2
  (210
)

 210

 410
 in.

Chapter 10

17

The vertex of the parabola is at 
4, 19,012.5,
and because p is negative it opens downward. So
the vertex is a maximum and the number of $10
17
price decreases is 4 or 4.25.
number of passengers 
 110  20x

 110  20(4.25)

 195
However, the flight can transport only 180
people.
number of passengers 
 110  20x
180 
 110  20x
3.5 
 x
Therefore, there should be 3.5 $10 price
decreases.
cost of ticket 
 140  10x

 140  10(3.5)

 $105
34b. Let y 
 income per flight.
Let x 
 the number of $10 price decreases.
Income 
 number of passengers  cost of a ticket
y 
 (110  10x)  (140  10x)
y 
 15,400  300x  100x2
y 
 100(x2  3x)  15,400
y  15,400 
 100(x2  3x)

1

O

17 2

y  15,400  200
4 
 200 x2  2x  
9

y

340

3

35a. Let (h, k) 
 (0, 0).
x2 
 4py
x2 
 4
8y
1

2x2 
 y
x2 
 4py
x2 
 1(1)y
1
x
4

38b. 4p 
 16
p 
 4
focus of parabola 
 center of circle
vertex: (h, k) 
 (1, 4)
focus: (h, k  p) 
 (1, 4  (4)) or (1, 0)
diameter 
 latus rectum

 16

x2 
 4py

1

p  8

x2 
 4
4y
1

1

p  4

x2 
 y

y

p1

2x 2 
 y

x2 
 y


y

1

radius 
 2(16)

8
(x  h)2  (y  k)2 
 r2
(x  1)2  (y  0)2 
 82
(x  1)2  y2 
 64
39. center: (h, k) 
 (2, 3)
a2 
 25
b2 
 16
a 
 25
 or 5
b 
 16
 or 4
c 
 
a2  b2
c 
 25
16
  or 41

transverse axis: vertical
foci: (h, k c) 
 (2, 3 41
)
vertices: (h, k a) 
 (2, 3 5) or (2, 8), (2, 2)
a
asymptotes: y  k 
 b (x  h)

The opening becomes
narrower.
1 2
x
4

O


y

x

35b. The opening becomes wider.
36a. Sample answer:
opening: upward
vertex: (h, k) 
 (0, 0)

y
(2100, 490)
(0, 0)
2100 10 ft O

500 ft

y3


2100 x

y

roadway

(x  h)2 
 4p(y  k)
(2100  0)2 
 4p(490  0)
(h, k)  (0, 0)
2250 
 p
(x, y)  (2100, 490)
(x  h)2 
 4p(y  k)
(x  0)2 
 4(2250)( y  0)
x2 
 9000y
36b.
x2 
 9000y
(720)2 
 9000y
57.6 
 y
57.6  10 
 67.6 ft
37.
(y  k)2 
 4p(x  h)
y2  2ky  k2 
 4px  4ph
y2  4px  2ky  k2  4ph 
 0
y2  Dx  Ey  F 
 0
(x  h)2 
 4p(y  k)
x2  2hx  h2 
 4py  4pk
x2  4py  2hx  h2  4pk 
 0
x2  Dx  Ey  F 
 0
38a. 4p 
 8
p 
 2 or p 
 2
opening: right
(y  k)2 
 4p(x  h)
(y  2)2 
 4(2)[x  (3)]
(y  2)2 
 8(x  3)
opening: left
(y  k)2 
 4p(x  h)
(y  2)2 
 4(2)[x  (3)]
(y  2)2 
 8(x  3)

5
(x
4

 2)

(2, 8)

(2, 3)

O

x

(2, 2)

40. 4x2  25y2  250y  525 
 0
4x2  25(y2  10y  ?) 
 525  ?
4x2  25(y2  10y  25) 
 525  25(25)
4x2  25(y  5)2 
 100
x2

25

( y  5)2

 4 
 1

center: (h, k) 
 (0, 5)
a2 
 25
b2 
 4
a 
 25 or 5
b 
 4
 or 2
2
2
c 
 
a  b
c 
 25
4
  or 21

foci: (h c, k) 
 (0 21
, 5) or ( 21
, 5)
major axis vertices: (h a, k) 
 (0 5, 5) or
( 5, 5)
minor axis vertices: (h, k b) 
 (0, 5 2) or
(0, 3), (0, 7)
y

O

x

(0, 3)
(5, 5)

(0, 5)
(5, 5)
(0, 7)

341

Chapter 10

41.



0


6

6



3

6


6, 3



2

12


12, 2

2

3

6


6, 23

5

6

6


6, 56

12

(12, )

7

6

6


6, 76

4

3

6


6, 43

3

2

12

3

2


12, 

5

3

6


6, 53

6


6, 116 



11

6


2

2
3

45. 19 t 14  19
19 t 33
t 
 33 or 27
perimeter 
 14  19  t

 14  19  27

 60
The correct choice is C.

(r, 
)
(12, 0)

12 cos 2

12


6, 6

Page 661


 
(6  3
)2  (9
 3)2
2
2

 
3  6

 45

BC 
 
(x2  
x1)2 
(y2 
y1)2

 
(9  6
)2  (3
 9)2
2
2

 
3  (
6)

 45

Since AB 
 BC, triangle ABC is isosceles.
1b. AC 
 
(x2  
x1)2 
(y2 
y1)2

 
(9  3
)2  (3
 3)2
2
2

 
6  0

6
perimeter 
 AB  BC  AC

 45
  45
6
19.42 units
2. Diagonals of a rectangle intersect at their
midpoint.
x1  x2 y1  y2
midpoint of A
C
 
 
2, 2


3

6

5
6


0

3 6 9

11
6

7
6
4
3

3
2

5
3

4  5 9  5


 
2, 2

42. 2n, where n is any integer
43.

3.

The measure of a is
360° 12 or 30°.
a


cos 30° 
 
6.4

a

44.



6.4 cm

6.4 cos 30° 
 a
5.5 a; 5.5 cm


 (0.5, 7)
x2  y2  6y  8x 
 16
(x2  8x  ?)  (y2  6y  ?) 
 16  ?  ?
(x2  8x  16)  (y2  6y  9) 
 16  16  9
(x  4)2  (y  3)2 
 9
center: (4, 3): radius: 9
 or 3
y
(4, 6)

(4, 3)

4


g(x)  
x2  1

x
10,000
1000
100
10
0
10
100
1000
10,000

(7, 3)

y  g(x)
4  108
4  106
4  104
0.04
4
0.04
4  104
4  106
4  108

O

x

(x  h)2  (y  k)2 
 r2
2
[x  (5)]2  (y  2)2 
 
7

(x  5)2  (y  2)2 
 7
y
5a.
Let d1 be the greatest
distance from the
d2
satellite to Earth. Let
d1
O
d2 be the least
x
a
c
distance from the
satellite to Earth.
4.

y → 0 as x → , y → 0 as x →  

Chapter 10

Mid-Chapter Quiz

1a. AB 
 
(x2  
x1)2 
(y2 
y1)2

342

1

7.

a 
 2(10,440)
a 
 5220
c
 
 e
a
c

5220

(y  4)2

2


 0.16


 3960
d1 
 a  c  Earth radius
d1 
 5220  835.20  3960
d1 
 2095.2 miles
d2 
 major axis  d1  Earth diameter
d2 
 10,440  2095.2  7920
d2 
 424.8 miles
5b. (h, k) 
 (0,0)
a 
 5220
b2 
 a2(1  e2)
b2 
 (5220)2 (1  0.162)
b2 
 26,550,840.96

vertices: (h, k

y  (4) 

y4


y

(4, 5)

1)

x

(1, 4  2)
(1, 4)
(1, 4  2)

8. To find the center, find the intersection of the
asymptotes.
y 
 2x  4
2x 
 2x  4
4x 
 4
x
1
y
21
y
2
The center is at (1, 2).
Notice that (4, 2) must be a vertex and a equals
4  1 or 3.
Point A has an x-coordinate of 4.
Since y 
 2x, the y-coordinate is 2  4 or 8.
The value of b is 8  2 or 6.

(y  5)2

 9 
 1

(x  1)2

(y  2)2

The equation is 9  36 
 1.

y

(1, 5)

(1)]

O

center: (h, k) 
 (4, 5)
b2 
 9
c 
 
a2  b2
a2 
 25
a
5
b
3
c 
 25
9
  or 4
major axis vertices: (h a, k) 
 (4 5, 5) or
(9, 5), (1, 5)
minor axis vertices: (h, k b) 
 (4, 5 3) or
(4, 2), (4, 8)
foci: (h c, k) 
 (4 4, 5) or (8, 5), (0, 5)

9. y2  4x  2y  5 
 0
y2  2y 
 4x  5
y2  2y  ? 
 4x  5  ?
y2  2y  1 
 4x  5  1
(y  1)2 
 4(x  1)
vertex: (h, k) 
 (1, 1)
4p 
 4
p
1
focus: (h  p, k) 
 (1  1, 1) or (2,  1)
axis of symmetry: y 
 k
y 
 1
y
directrix: x 
 h  p
x
0
x
11
x
0
(1, 1)
O

x

(4, 2)

2


22


a
(x  h)
b
2

 [x 
6

3

 (x 
3

asymptotes: y  k 


 554 
 0
9(x2  8x  ?)  25(y2  10y  ?) 

544  ?  ?
9(x2  8x  16)  25(y2  10y  25) 

544  9(16)  25(25)
9(x  4)2  25(y  5)2 
 225

O

c 
 
a2  b2
c 
 2

 6 or 22


a) 
 
1, 4

c) 
 
1, 4

foci: (h, k

(x  b)2
(y  k)2

 b

1
2
a2
(x  0)2
(y  0)2
   
 1
5220
26,550,840.96
x2
y2
    
 1
27,248,400
26,550,840.96
9x2  25y2  72x  250y

(x  4)2

25

(x  1)2

 6 
 1

center: (h, k) 
 (1, 4)
b2 
 6
a2 
 2
a 
 2

b 
 6

transverse axis: vertical

c 
 835.20
1
radius of Earth 
 2(7920)

6.

3y2  24y  x2  2x  41 
 0
3(y2  8y  ?)  (x2  2x  ?) 
 41  ?  ?
3(y2  8y  16)  (x2  2x  1) 
 41  3(16)  1
3(y  4)2  (x  1)2 
 6

(9, 5)

(4, 8)

(2, 1)

343

x

Chapter 10

10. vertex: (h, k)  (5, 
1)
(x 
 h)2  4p(y 
 k)
(9 
 5)2  4p[
2 
 (
1)]
42  
4p

4  p
(x 
 h)2  4p(y 
 k)
(x 
 5)2  4(
4)[y 
 (
1)]
(x 
 5)2  
16(y  1)

10-6

3. Sample answer:
rectangular equation: y2  
x
parametric equations: y  t

(h, k)  (5, 1)
(x, y)  (9, 2)

y  t, x  
t2, 
   t  
4. A  1, c  9; since A and C have the same sign
and are not equal, the conic is an ellipse.
x2  9y2  2x 
 18y  1  0
(x2  2x  ?)  9(y2 
 2y  ?)  
1  ?  ?
(x2  2x  1)  9(y2 
 2y  1)  
1  1  9(1)
(x  1)2  9(y 
 1)2  9

Rectangular and Parametric
Forms of Conic Sections

Page 665

y2  
x
t2  
x

t2  x

(x  1)2

9

( y 
 1)2

 1  1

center: (h, k)  (
1, 1)
b2  1
a2  9
a3
b1
vertices: (h  a, k)  (
1  3, 1) or (2, 1), (
4, 1)
(h, k  b)  (
1, 1  1) or (
1, 2), (
1, 0)

Graphing Calculator Exploration

1.

y

(4, 1)

Tmin: [0, 6.28] step: 0.1
[
7.58, 7.58] scl1 by [
5, 5] scl1
1a. (
1, 0)
1b. clockwise
2.

(1, 2)
(2, 1)

(1, 0) O

(1, 1)

x

5. A  1, C  0; since C  0, the conic is a parabola.
y2 
 8x  
8
y
y2  8x 
 8
y2  8(x 
 1)
vertex: (h, k)  (1, 0)
(1, 0)
opening: right
O
x
Tmin: [0, 6.28] step: 0.1
[
7.58, 7.58] scl:1 by [
5, 5] scl1
2a. (0, 1)
2b. clockwise
3.

6. A  1, C  
1; since A and C have different signs,
the conic is a hyperbola.
x2 
 4x 
 y2 
 5 
 4y  0
(x2 
 4x  ?) 
 (y2  4y  ?)  5  ?  ?
(x2 
 4x  4) 
 (y2  4y  4)  5  4 
 4
(x 
 2)2 
 (y  2)2  5
(x 
 2)2

5

center: (h, k)  (2, 
2)
a2  5
a  
5

vertices: (h  a, k)  (2  
5
, 
2)
b
asymptotes: y 
 k   a(x 
 h)

Tmin: [0, 6.28] step: 0.1
[
7.58, 7.58] scl1 by [
5, 5] scl1
an ellipse
4. The value of a determines the length of the radius
of the circle.
5. Each graph is traced out twice.


5


 (x 
 2)
y 
 (
2)   


5

y

Page 667

y  2   (x 
 2)

Check for Understanding

1. For the general equation of a conic, A and C have
the same sign and A C for an ellipse. A and C
have opposite signs for a hyperbola. A  C for a
circle. Either A  0 or C  0 for a parabola.
2. 
   t  

Chapter 10

( y  2)2


 5  1

O

x

(2  

5, 2)

(2  

5, 2)

(2, 2)

344

7. A  1, C  1; since A  C, the conic is a circle.
x2 
 6x  y2 
 12y  41  0
(x2 
 6x  ?)  (y2 
 12y  ?)  
41  ?  ?
(x2 
 6x  9)  (y2 
 12y  36)  
41  9  36
(x 
 3)2  (y 
 6)2  4
center: (h, k)  (3, 6)
radius: r2  4
r2
y

10. Sample answer:
Let x  t.
y  2x2 
 5x
y  2t2 
 5t
x  t, y  2t2 
 5t, 
   t  
11. Sample answer:
x2  y2  36
x2
y2
  
36
36
x 2
y 2

 6
6
cos2 t  sin2 t

 

(3, 8)

x 2

6
x

6

 

(5, 6)

(3, 6)

 



cos2

1
1
1

y 2

6

t

y

6

 cos t

 sin2 t
 sin t

x  6 cos t
y  6 sin t
x  6 cos t, y  6 sin t, 0  t  2
12.

x

O

t2

x  8
0
y2

x  80
80x  y2
y2  80x


t2

8. y 

 6t  2
y  
x2 
 6x  2
t

6

5

4

3

2

1
0

x

6

5

4

3

2

1
0

(x, y)
(
6, 2)
(
5, 7)
(
4, 10)
(
3, 11)
(
2, 10)
(
1, 7)
(0, 2)

y
2
7
10
11
10
7
2

Pages 667–669

Exercises

13. A  1, C  0; since C  0, the conic is a parabola.
x2 
 4y 
 6x  9  0
x2 
 6x  ?  4y 
 9  ?
x2 
 6x  9  4y 
 9  9
(x 
 3)2  4y
vertex: (h, k)  (3, 0)
opening: upward

y

y

O

O
9. x  2 cos t
x

2

cos2

t

x 2

2

x

y

3

sin2

 sin t

y
t


t1



2

y 2

 3  1

x2

4

14. A  1, C  1; since A  C, the conic is a circle.
x2 
 8x  y2  6y  24  0
2
(x 
 8x  ?)  ( y2  6y  ?)  
24  ?  ?
(x2 
 8x  16)  ( y2  6y  9)  
24  16  9
(x 
 4)2  (y  3)2  1
center: (h, k)  (4, 
3)
radians: r2  1
r1
y

y  3 sin t

 cos t

t



y2

 9  1

t
0

O

x

(3, 0)

x
O
(3, 3)

3

t

2

(4, 3)

x

(4, 4)

t
0

x
2

y
0

(x, y)
(2, 0)



2

0

3

(0, 3)




2

0

(
2, 0)

0


3

(0, 
3)

3

2

345

Chapter 10

17. A  1, C  0; since C  0, the conic is a parabola.
x2  y  8x 
 16
x2 
 8x  ?  y 
 16  ?
x2 
 8x  16  y 
 16  16
(x 
 4)2  y
y
vertex: (h, k)  (4, 0)
opening: upward

15. A  
1, C  3; since A and C have different signs,
the conic is a hyperbola.
x2 
 3y2  2x 
 24y 
 41  0

x2  3y2 
 2x  24y  41  0
3(y2  8y  ?) 
 (x2  2x  ?)  
41  ?  ?
3(y2  8y  16) 
 (x2  2x  1)  
41  3(16) 
 1
3(y  4)2 
 (x  1)2  6
(y  4)2

2

(x  1)2


 6  1

center: (h, k)  (
1, 
4)
a2  2
a  
2

vertices: (h, k  a)  (
1, 
4  
2
)

O

a

asymptotes: y 
 k   b(x 
 h)
y 
 (
4)  
y4


2




6

x

18. A  C  D  E  0; the conic is a hyperbola.
2xy  3

[x 
 (
1)]


3

(x
3

(4, 0)

3

xy  2

 1)

quadrants: I and III
transverse axis: y  x

y

vertices: 

 or  62, 62,

 32, 
 32 or 

26 , 

26 

(1, 4  
2)

O

x

3
,
2

3

2

y

(1, 4)

(
26 , 
26 )
(1, 4  
2)

O

(
26 , 
26 )

16. A  9, C  25; since A and C have the same sign
and are not equal, the conic is an ellipse.
9x2  25y2 
 54x 
 50y 
 119  0
9(x2 
 6x  ?)  25(y2 
 2y  ?)  119  ?  ?
9(x2 
 6x  9)  25(y2 
 2y  1) 
119  9(9)  25(1)
9(x 
 3)2  25(y 
 1)2  225
(x 
 3)2

25

19. A  5, C  2; since A and C have the same sign
and are not equal, the conic is an ellipse.
5x2  2y2 
 40x 
 20y  110  0
2
5(x 
 8x  ?)  2(y2 
 10y  ?)  
110  ?  ?
5(x2 
 8x  16)  2(y2 
 10y  25) 

110  5(16)  2(25)
5(x 
 4)2  2(y 
 5)2  20

( y 
 1)2

 9  1

center: (h, k)  (3, 1)
b2  9
a2  25
a5
b3
vertices: (h  a, k)  (3  5, 1) or (8,1), (
2, 1)
(h, k  b)  (3, 1  3) or (3, 4), (3, 
2)

(x 
 4)2
( y 
 5)2
  
4
10
( y 
 5)2
(x 
 4)2
  
10
4

(3, 4)

(2, 1)

O

(4, 5  

10)

y
(2, 5)

(8, 1)

x

(4, 5)

(6, 5)

(3, 2)
(4, 5  

10)

O

Chapter 10

1
1

center: (h, k)  (4, 5)
b2  4
a2  10
a  
10

b2
vertices: (h, k  a)  (4, 5  
10
)
(h  b, k)  (4  2, 5) or (6, 5), (2, 5)

y

(3, 1)

x

346

x

20. A  1, C 1; since A  C, the conic is a circle.
x2 
 8x  11  
y2
(x2 
 8x  ?)  y2  
11  ?
(x2 
 8x  16)  y2  
11  16
(x 
4)2  y2  5
center: (h, k)  (4, 0)
radius: r2  5
r  
5


23.


4y2  10x  16y 
 x2 
 5
x2 
 4y2  10x 
 16y  5  0
A  1, C  
4; since A and C have different signs,
the conic is a hyperbola.
x2 
 4y2  10x 
 16y  5  0
2
(x  10x  ?) 
4(y2  4y  ?)  
5  ?  ?
(x2  10x  25) 
 4(y2  4y  4)  
5  25 
 4(4)
(x  5)2 
 4(y  2)2  4
(x  5)2

4

y
(4, 

5)

center: (h, k)  (
5, 
2)
a2  4
a2
vertices: (h  a, k)  (
5  2, 
2) or (
3, 
2),
(
7, 
2)

(4  
5, 0)

O

x

(4, 0)

( y  2)2


 1  1

b

asymptotes: y 
 k   a(x 
 h)
1

y 
 (
2)   2[x 
 (
5)]

21. A  
9, C  8; since A and C have different signs,
the conic is a hyperbola.
8y2 
 9x2 
 16y  36x 
 100  0
8(y2 
 2y  ?) 
 9(x2 
 4x  ?)  100  ?  ?
8(y2 
 2y  1) 
 9(x2 
 4x  4)  100  8(1) 
 9(4)
8(y 
 1)2 
 9(x 
 2)2  72
( y 
 1)2

9

1

y  2   2(x  5)

y
(5, 2)

(x 
 2)2

x

O


 8  1

center: (h, k)  (2, 1)
a2  9
a3
vertices: (h, k  a)  (2, 1  3) or (2, 4), (2, 
2)

(7, 2)

(3, 2)

a

asymptotes: y 
 k   b(x 
 h)
24.

3

(x 
 2)
y
1
2

2
3

2

y 
 1   4(x 
 2)

y
(2, 4)
(2, 1)

x

O
(2, 2)

Ax2  Bxy  Cy2  Dx  Ey  F  0
2x2  0  2y2  (
8)x  12y  6  0
2x2  2y2 
 8x  12y  
6
A  C; circle
2(x2 
 4x  ?)  2(y2  6y  ?)  6  ?  ?
2(x2 
 4x  4)  2(y2  6y  6y  9) 

6  2(4)  2(9)
2(x 
 2)2  2(y  3)2  20
(x 
 2)2  (y  3)2  10
y
center: (h, k)  (2, 
3)
2
radius: r  10
(2, 3  

10)
r  
10

x
O

22. A  0, C  4; since A  0, the conic is a parabola.
4y2  4y  8x  15
4(y2  y  ?)  
8x  15  ?
4

(2  

10, 3)
(2, 3)

  
8x  15  4 

  
8x  16

  
2x  4

  
2(x 
 2)
1
vertex: (h, k)  2, 
2
1
y2  y  4
1 2
4 y  2
1 2
y  2
1 2
y  2

opening: left

1

4

y

25. y  2t2 
 4t  1
y  2x2 
 4x  1

O

x

y
t

1
0
1
2

( 2 ,  12 )
O

x

347

x

1
0
1
2

y
7
1

1
7

(x, y)
(
1, 7)
(0, 1)
(1, 
1)
(2, 7)
Chapter 10

26. x  cos 2t
y  sin 2t
cos2 2t  sin2 2t  1
x 2  y2  1
t
0


2



3

2

x
1

y
0

(x, y)
(1, 0)

0

1

1
0

(0, 1)
(
1, 0)

0


1

(0, 
1)

29.

x  
 sin 2t

y  2 cos 2t
y

2


x  sin 2t
2t 

cos2

y 2

2

 

sin2



2t  1

(
x)2

y2

4

 cos 2t

y t
0

1



x2

1

x2 

y2

4

1

O

x

y

t
0



4


2

x

O

x
0

y
2

(x, y)
(0, 2)


1

0

(
1, 0)

0


2

(0, 
2)

1

0

t
0

3

4

27.

x  
cos t

x  cos t
cos2 t  sin2 t  1
(
x)2  y2  1
x 2  y2  1
t
0


2



3

2

y  sin t

30.

x  2t 
 1
x  1  2t
x1

2

y
0

(x, y)
(
1, 0)

0
1

1
0

(0, 1)
(1, 0)

0


1

(0, 
1)

t
0
1
2
3
4

t
0

x
31.

28. x  3 sin t

y

2

2

y

x 2

 3  1

y2

4
x2

9




x2

9
y2

4

t
0

O

x 2

5
x

5

 



2



3

2

Chapter 10

x
2

y
0

(x, y)
(0, 2)

3
0

0

2

(3, 0)
(0, 
2)


3

0

(
3, 0)

y

3

 sin 2t

2t  1

y 2
 3  1
x2
y2
    1
9
9
x2  y2  9

 

x2
y2
  
25
25
x 2
y 2

 5
5
cos2 t  sin2 t

x

 

t
0

y  3 sin 2t

32. Sample answer:
x2  y2  25

1
1

(x, y)
(
1, 0)
(1, 1)
(3, 
2
)
(5, 
3
)
(7, 2)

y
0
1

2


3

2

x  
3 cos 2t

 

 cos t

cos2 t  sin2 t  1

x

1
1
3
5
7

x 2

3

x

O

x

3  cos 2t
cos2 2t  sin2

y  2 cos t

 sin t

2y

t
x1

2

y

y

x

3

y

y  
t

x

1

O

(1, 0)



 

cos2

 cos t

t

1
1
1

2

5y

y

5

 sin2 t
 sin t

x  5 cos t
y  5 sin t
x  5 cos t, y  5 sin t, 0  t  2

348

33. Sample answer:
x2  y2 
 16  0
x2  y2  16
x2
y2
  
16
16
y 2
x 2

 4
4
cos2 t  sin2 t

 

x 2

4
x

4

 

 



cos2

39b.

t
0
1
2
3

1
1
1

t

 cos t

y 2

4

y

4

x
0
1
2
3

(x, y)
(0, 0)
(1, 1)
(2, 4)
(3, 9)

y
0
1
4
9

y
 sin2 t
 sin t

x  4 cos t
y  4 sin t
x  4 cos t, y  4 sin t, 0  t  2
34. Sample answer:
x2

4

2

2x

y2

y 2

 5  1

cos2 t  sin2 t  1
2

2x

x

2

x

O

 25  1

 cos2 t
 cos t

y 2

5

y

5

 sin2 t
 sin t

x  2 cos t
y  5 sin t
x  2 cos t, y  5 sin t, 0  t  2
35. Sample answer:
y2

16

 x2  1
y 2

x2  4  1
cos2 t  sin2 t  1
x2  cos2 t
x  cos t

39c.
39d.
y 2

4

y

4

 sin2 t
40a.

 sin t

y  4 sin t
x  cos t, y  4 sin t, 0  t  2
36. Sample answer:
Let x  t.
y  x2 
 4x  7
y  t2 
 4t  7
x  t, y  t2 
 4t  7, 
  t  
37. Sample answer:
Let y  t.
x  y2  2y 
 1
x  t2  2t 
 1
x  t2  2t 
 1, y  t, 
  t  
38. Sample answer:
Let y  t.
(y  3)2  4(x 
 2)
(t  3)2  4(x 
 2)
0.25(t  3)2  x 
 2
0.25(t  3)2  2  x
x  0.25(t  3)2  2, y  t, 
  t  
39a. Answers will vary. Sample answers:
Let x  t.
x  
y
t  
y
t2  y
x  t, y  t2, t  0
Let y  t.
x  
y
x  
t
x  
t, y  t, t  0

40b.

Tmin: [0, 5] step: 0.1
[
7.58, 7.58] scl1 by [
5, 5] scl1
yes
There is usually more than one parametric
representation for the graph of a rectangular
equation.
a circle with center (0, 0) and radius 6 feet
(x 
 h)2  (y 
 k)2  r2
(x 
 0)2  (y 
 0)2  62
x2  y 2  36
Sample answer:
x2  y2  36
x2
y2
  
36
36
x 2
y 2

 6
6
sin2 (qt)  cos2 (qt)

 

 

1
1

1
Since the paddlewheel completes a revolution in
2
2 seconds, the period is q  2, so q  .
sin2 (t)  cos2(t)  1
2

6x

x

6

 sin2(t)
 sin(t)

2

6y

y

6

 cos2 (t)
 cos(t)

x  6 sin(t)
y  6 cos (t)
x  6 sin (t), y  6 cos (t), 0 t  2
40c. C  2r
C  26
C  37.7 ft
The paddlewheel makes 1 revolution, or moves
37.7 ft in 2 seconds.
37.7 ft

2s

60 s  1131 ft

The paddlewheel moves about 1131 ft in
1 minute.

349

Chapter 10

41a. A  2, C  5; since A and C have the same sign
and A C, the graph is an ellipse.
2x2  5y2  0
5y2  
2x

44. After drawing a vertical line through (x, y) and a
horizontal line through the endpoint opposite
(x, y), two right triangles are formed. Both
triangles contain an angle t, since corresponding
angles are congruent when two parallel lines are
cut by a transversal. Using the larger triangle,
x
cos t  a or x  a cos t. Using the smaller triangle,

2

y2  
5x
2

y  
5x

y

sin t  b or y  b sin t.
45. x2 
 12y  10x  
25
x2  10x  ?  12y 
 25  ?
2
x  10x  25  12y 
 25  25
(x  5)2  12y
vertex: (h, k)  (
5, 0)
4p  12
p3
focus: (h, k  p)  (
5, 0  3) or (
5, 3)
axis of symmetry: x  h
x  
5
directrix: y  k 
 p
y0
3
y  
3

This equation is true for (x, y)  (0, 0).
The graph is a point at (0, 0); the equation is
that of a degenerate ellipse.
41b. A  1, C  1; since A  C, the graph is a circle.
x2  y2 
 4x 
 6y  13  0
(x2 
 4x  ?)  (y2 
 6y  ?)  
13
(x2 
 4x  4)  (y2 
 6y  9)  
13  4  9
(x 
 2)2  (y 
 3)2  0
center: (h, k)  (2, 3)
radius: 0
The graph is a point at (2, 3); the equation is
that of a degenerate circle.
41c. A  
9, C  1; since A and C have different
signs, the graph is a hyperbola.
y2 
 9x2  0
y2  9x2
y  3x
The graph is two intersecting lines y  3x; the
equation is that of a degenerate hyperbola.
42. The substitution for x must be a function that
allows x to take on all of the values stipulated by
the domain of the rectangular equation. The
domain of y  x2 
 5 is all real numbers, but
using a substitution of x  t2 would only allow for
values of x such that x  0.
43a. center: (h, k)  (0, 0)
(x 
 h)2  (y 
 k)2  r2
(x 
 0)2  (y 
 0)2  6
x2  y2  36
43b.
x2  y2  36
y2
x2
  
36
36
y 2
x 2

 6
6
sin2 t  cos2 t

 

x 2

6
x

6

 

 

 sin2 t
 sin t

y

O x

(5, 0)

46. c  
25
quadrants: II and IV
transverse axis: y  
x
vertices: xy  
25
5(
5)  
25
(5, 
5)

1

xy  
25

5(5)  
25
(
5, 5)

y

1
1

(5, 5)
y 2

6
y

6

 

 cos2 t

O

 cos t

x
(5, 5)

x  6 sin t
y  6 cos t
x  6 sin t, y  6 cos t
Since the second hand makes 2 revolutions,
0  t  4.

47.

43c.

3x2  3y2 
 18x  12y  9
3(x2 
 6x  ?)  3(y2  4y  ?)  9  ?  ?
3(x2 
 6x  9)  3(y2  4y  4)  9  3(9) 
 3(4)
3(x 
 3)2  3(y  2)2  48
(x 
 3)2  (y  2)2  16
center: (h, k)  (3, 
2)
(3, 2)
y
radians: r2  16
r4

O

(3, 2)

Tmin: [0, 4] step: 0.1
[
9.10, 9.10] scl1 by [
6, 6] scl1

Chapter 10

(7, 2)

350

x

x

48.

y  kxz
y  kxz
16  k(5)(2)
y  1.6(8)(3)
1.6  k
y  38.4
54. 5
9  5(
3) 
 7(9)
7

3
 
78
Yes, an inverse exists since the determinant of the
matrix 0.
53.

60˚

y

30



y

x

cos 60°  30

sin 60°  30

y2 
 y1


55. m  
x 
x

x  30 cos 60°
y  30 sin 60°
x  15 lb
y  15
3
 lb
49.
y  
0.13x  37.8
0.13x  y 
 37.8  0
A  0.13, B  1, C  
37.8
Car 1: (x1, y1)  (135, 19)
d1 
d1 



2




1

7
4

3 
 (
6)
1

3

y 
 y1  m(x 
 x1)
1

1

y 
 4  3(x  6) or y 
 7  3(x 
 3)

Ax1  By1  C



A2  
B2

y  mx  b
4

0.13(135)  1(19)  (
37.8)

2  12
 

(0.13)

Ax2  By2  C



A2  
B2

d2 

0.13(245)  1(16)  (
37.8)

2  12


(0.13)

y  mx  b
1

b

y  3x  6

6b
56. (1 # 4) @ (2 # 3)  1 @ 2
2
The correct choice is B.

d1  
1.24
The point (135, 19) is about 1 unit from the line
y  
0.13x  37.8.
Car 2: (x2, y2)  (245, 16)
d2 

1
(
6)
3

Transformation of Conics

10-7

d2  9.97
The point (245, 16) is about 10 units from the line
y  
0.13x  37.8.
Car 1: the point (135, 19) is about 9 units closer
to the line y  
0.13x  37.8 than the point
(245, 16).

Pages 674–675

Check for Understanding

1. Sample answers:
(h, k)  (0, 0)
x  y2

y

1

50. Let v  Sin1 2.

x 
 y2

1

Sin v  2
v  30

1

 sin (2 30)
 sin 60

3

 
2

(h, k)  (
3, 3)
(x 
 h)  (y 
 k)2
(x  3)  (y 
 3)2
y

1

51. s  2(a  b  c)
1

s  2(48  32  44)
s  62
a)(s


b)(s 

c)
K  
s(s


K  
62(62
)(62

 48

2
32)(64)

 4
K  
46872
0

K  685 units2
52. 
2y


 3 
 
2y

 3  
1

2y

 3  
2y


3
1
2y 
 3  2y  3 
 2
2y
31


7  
2
2y
3

7

2
49

4
37

4
37

8

x

O

sin 2 Sin1 2  sin (2v)

x  3 
 (y  3)2

O

x

2. Replace x with x
 cos 30°  y
 sin 30° or

 
2y

3


3

 x

2

 2y  3

Replace y with 
x
 sin 30°  y
 cos 30° or

 2y


2x
, 

1

1

 2y
.

2

 y
.
2

y

351

Chapter 10

y

3.

x2
100

8. B2 
 4AC  0 
 4(1)(1)
 
4
A  C  1; circle

90° or 270°

y2

 25 
 1


 2
x
 cos 4  y
 sin 4 
 5

90˚

x2
25

y2

 100 
 1

y

7 2

 5  2 4
49
y 
 5  8
9
y  8
9
y  8 
 k
9
y  8 
 5
31
y 
 8
31
y  8
31
y 
 8

 














7
2 x2 
 2x
7
7 2
2 x2 
 2x  4
7 2
2 x 
 4
7 2
2 x 
 4
2
7
2 x 
 4 
 h
2
7
2 x 
 4  4
9 2
2 x  4
18
81
2 x2  4x  1
6
81
2x2  9x  8
2x2  9x 
 y  14












2(x
)2  2(y
)2 
 5
2
x
 
 5
2
y
 
 6  0

 4AC  42 
 4(9)(4)
9.
 
128
A C; ellipse
B2

B


tan 2v  
A
C
4


tan 2v  
9
4

tan 2v  0.8
2v  38.65980825°
v  19°
10. B2 
 4AC  52 
 4(8)(
4)
 153
hyperbola

 




B


tan 2v  
A
C
5


tan 2v  
8 
 (
4)

tan 2v  0.416666667
2v  22.61986495°
v  11°
11.
3(x 
 1)2  4(y  4)2  0
3(x2 
 2x  1)  4(y2  8y  16)  0
3x2 
 6x  3  4y2  32y  64  0
4y2  32y  (3x2 
 6x  67)  0

(h, k)  (4, 5)















0
2x2  9x 
 y  14  0
7. B2 
 4AC  0 
 4(1)(
1)
4
hyperbola

↑
a

1
(x
)2
4


3


12 x
  
23 y


↑
c

b  

b2 
 4ac


2


 
2x
  2y
  9

3


1

3

 2x
y
  4(y
)2 


y


32  

322 

4(4)(3
x2 
 6
x  67)

2(4)

y


32  


48x2
 96
x 
 48

8

y


32  


48(x

 1)2

8

x  1, y  
4; point

3
1

3

(x
)2 
x
y
  (y
)2  9
4
4
2
1
1

2(x
)2  
3
x
y
  2(y
)2  9
x
y
 
 (y
)2  
18
(x
)2 
 2
3

y
x

O

(x
)2 
 2
3
x
y
 
 (y
)2  18  0

Chapter 10

↑
b

y  
 
2a

x 2 
 y2  9
(x
 cos 60°  y
 sub 60°)2 

(
x
 sin 60°  y
 cos 60°)2  9
2



1
1
5
2

5
2

(x
)2  x
y
  (y
)2 
 x
 
  y

2
2
2
2
1
1
 2(x
)2 
 x
y
  2(y
)2  3
5
2

5
2

(x
)2  (y
)2 
 2x
 
 2y
  3
x
 
 5
2
y
  6
2(x
)2  2(y
)2 
 5
2

2

4. Ebony; B2 
 4AC  6
3
 
 4(7)(13)  0
and A C
5. B2 
 4AC  0 
 4(1)(1)
 
4
A  C  1; circle
(x 
 h)2  (y 
 k)2  7
(x 
 3)2  (y 
 2)2  7
(h, k)  (3, 2)
2
x 
 6x  9  y2 
 4y  4  7
x2  y2 
 6x 
 4y  6  0
2
6. B 
 4AC  0 
 4(2)(0)
0
parabola
y  2x2 
 7x  5
y 
 5  2x2 
 7x
y
5

x2 
 5x  y2  3


 y
 sin 4


 2
 
x
sin 4  y
 cos 4  3
2

22 x
  
22 y
 
 5
22 x
  
22 y

2

2


2

 
2x
  2y
  3



x

270˚ O


x
 cos 4

(1, 4)

352

14. B2 
 4AC  0 
 4(4)(5)
 
80
A C; ellipse

1

y  6x2

12a.

1


x
 sin 30°  y
 cos 30°  6(x
 cos 30°  y
 sin 30°)2
2


2x
  2 y
  62x
  2y


3


1

1 
3


1

1


3


1 3
(x
)2
4

 4 x
y

1
 4(y
)2

1


3


1


3



2x
  2y
  6

4x2  5y2  20
4(x 

 5(y 
 k)2  20
4(x 
 5)2  5(y  6)2  20
(h, k)  (5, 6)
4(x2 
 10x  25)  5(y2  12y  36)  20
4x2 
 40x  100  5y2  60y  180  20
4x2  5y2 
 40x  60y  260  0
2
15. B 
 4AC  0 
 4(3)(1)
 
12
A C; ellipse
3x2  y2  9
2
3(x 
 h)  (y 
 k)2  9
3(x  1)2 (y 
 3)2  9 (h, k)  (1, 3)
3(x2  2x  1)  y2 
 6y  9  9
3x2  6x  3  y2 
 6y  9  9
3x2  y2  6x 
 6y  3  0
2
16. B 
 4AC  0 
 4(12)(4)
 
192
A C; ellipse
4y2  12x2  24
2
4(y 
 k)  12(x 
 h)2  24
4(y 
 4)2  12(x  1)2  24
(h, k)  (1, 4)
4(y2 
 8y  16)  12(x2  2x  1)  24
4y2 
 32y  64  12x2  24x  12  24
y2 
 8y  16  3x2  6x  3  6
3x2  y2  6x 
 8y  13  0
2
17. B 
 4AC  0 
 4(9)(
25)
 900
hyperbola
9x2 
 25y2  225
2
9(x 
 h) 
 25(y 
 k)2  225
9(x 
 0)2 
 25(y 
 5)2  225
(h, k)  (0, 5)
9x2 
 25(y2 
 10y  25)  225
9x2 
 25y2  250y 
 850  0
18.
(x 
 3)2  4y
2
x  6x  9 
 4y  0
B2 
 4AC  0 
 4(1)(0)
0
parabola
(x  3)2  4y
(x  3 
 h)2  4(y 
 k)
(x  3  7)2  4(y 
 2)
(h, k)  (7, 2)
(x  10)2  4y 
 8
x2  20x  100  4y 
 8
2
x  20x 
 4y  108  0
19. B2 
 4AC  0 
 4(1)(0)
0
parabola
x2 
 8y  0
2
(x
 cos 90°  y
 sin 90°) 
8(
x
 sin 90°  y
 cos 90°)  0
(y
)2 
 8(
x
)  0
(y
)2  8x  0
h)2

2
3


1

2

2x
  2y
  8(x
)2  12x
y
  2
4 (y
)

↑
a






















12x
  12
3
y
  3(x
)2  2
3
x
y
  (y
)2
0  3(x
)2  2
3
x
y

 (y
)2  12x
 
 12
3
y

3(x
)  2
3
x
y
  (y
)2  12x
 
 12
3
y
  0
12b. 3x2  2
3
xy  y2  12x 
 12
3
y  0
2
1y  (2
3
x 
 12
3
)y  (3x2  12x) 0
↑
b

↑
c

y

b2 
 4
ac

b  


2a

y


(2
3
x 
 12
3
)  

(2
3
x

 12

3
 )2 

 4(1
)(3x2 
12x)

2(1)

y  

3
x  6
3




12x2 

144x 
 432

 12
x2 
 4
8x

2

y  

3
x  6
3




192
x2
 43

2

Pages 675–677

1

and y  6x2

Exercises

13. B2 
 4AC  0 
 4(3)(0)
0
parabola
y  3x2 
 2x  5
y 
 5  3x2 
 2x
y 
 5  3x2 
 3x
2

1 2

1 2

y 
 5  33  3 x2 
 3x  3
2

1 2

y 
 3  3x 
 3
14

2

y 
 3 
 k  3x 
 3 
 h
14

1

2

y 
 3  3  3x 
 3 
 2
14

y

y

y


1

5

3
5

3
5

3

7 2
x 
 3
14
x2 
 3x

 3
 3



(h, k)  (2, 3)

 9
49

49

 3x2 
 14x  3

0  3x2 
 14x 
 y  18
3x2 
 14x 
 y  18  0

353

Chapter 10

20. B2 
 4AC  0 
 4(2)(2)
 
16
A  C; circle

24. B2 
 4AC  0 
 4(16)(
4)
 256
hyperbola
16x2 
 4y2  64
16(x
 cos 60°  y
 sin 60°)2

 4(
x
 sin 60°  y
 cos 60°)  64

2x2  2y2  8

2(x
 cos 30°  y
 sin 30°)2

 2(
x
 sin 30°  y
 cos 30°)2  8









16


 4AC 
1
hyperbola

x
 cos



4

6

y2  8x  0

3

  
8
  
8



23. B2 
 4AC  0 
 4(1)(1)
 
4
A  C; circle

x
 cos


2

3

y

2

1
x

2



1
(x
)2
4

 y
 sin




3


 2

3



 
x
 sin


 5



3

4

tan 2v  1
2v  45°
v  23°
27. B2 
 4AC  (
1)2 
 4(1)(
4)
 17
hyperbola

 
8



3


3


 5x 

 y
 sin
 y
 cos

y2



3
 2

3





tan 2v  
A
C

3


1


tan 2v  
1 
 (
4)
1

tan 2v  
5
2v  
11.30993247°
v  
6°
28. B2 
 4AC  82 
 4(8)(2)
0
parabola
B

tan 2v  
A
C

3


2
1


3
 
2x
  2 y
  3


3

y

2

5
3


5



3

3

1





2(y
)2

(y
)2




5
x

2




5

3
y
2

8


tan 2v  
8
2

3

4

tan 2v  3
2v  53.13010235°
v  27°
29. B2 
 4AC  92 
 4(2)(14)
 
31
A C; ellipse


 5x
 
 5
3
y
 
 6  0

B


tan 2v  
A
C
9


tan 2v  
2 
 14
3

tan 2v  
4
2v  
36.86989765°
v  
18°
Chapter 10


3


2

1


tan 2v  
9
5

 4(x
)2 
 2x
y
  4(y
)2  3
2(x
)2



3

1

 2 x
y
  4(y
)2

B

 2x
y
  4(y
)2 
 2x
 
 2y

(x
)2

3
(x
)2
4

B

x2


 5x
 cos

1
x

2

2

1


tan 2v  
A
C

 16
(x
)2 
 (y
)2 
 16  0



3

1


 2x
y
  4(y
)2  64

26. 32 
 4AC  42 
 4(9)(5)
 
164
A C; ellipse

xy  
8



 y
 sin

x
 sin 4  y
 cos 4

2


2


2


2

x
  y
 
x
  y

2
2
2
2
1
1
1
1

2(x
)2  2x
y
 
 2x
y
  2(y
)2
(x
)2 
 (y
)2




3


3
(x
)2
4

1

3

3
(x
)2 
 x
y
  (y
)2 
4
2
4
18
63
6
5
5
3

(x
)2   x
y
  (y
)2  (x
)2 
 x
y

4
2
4
4
2
15
 4(y
)2 
23

3

21
(x
)2  x
y
  (y
)2 
 30 
4
2
4
23(x
)2  2
3
x
y
  21(y
)2 
 120 

 4(y
)2  4
3
x
  4y
  0



3

5

3(y
)2  16
3
x
  16y
  0

 4(0)(0)



4

2

 2x
y
  4(y
)2


3


2

22.

1

62x
  2y
  5
2x
  2y
  30


x
 sin 6  y
 cos 6  8x
 cos 6  y
 sin 6  0
2

12x
  
23 y
  8
23 x
  12y
  0
12



3


3


x
y
  12(y
)2
4(x
)2  8
3

 3(x
)2  2
3
x
y
 
 (y
)2  64
x
y
  11y
)2 
 64  0
(x
)2  10
3
25.
6x2  5y2  30
2
6(x
 cos 30°  y
 sin 30°)
 5(
x
 sin 30°  y
 cos 30°)2  30


 4AC  0 
 4(1)(0)
0
A C; parabola

B2

1
(x
)2
4


4

B2

1

3

(x
)2 
 
x
y

2
4
x
y
 
(x
)2 
 2
3


3


1

(x
)2  (y
)2 
 4  0

21.

2

162x
  2y
 
 4
2x
  2y
  64

2
2
1
1

3


3

2 2x
  2y
  2 
2x
  2y
  8
3
1

3

2 4(x
)2  2x
y
  4(y
)2
1
3

3

 2 4(x
)2 
2x
y
  4(y
)2  8
3
1
1
(x
)2  
3
x
y
  2(y
)2  2(x
)2
2
3

 
3
x
y
  2(y
)2  8
2(x
)2  2(y
)2  8

354

30

30
0
0

30. B2 
 4AC  42 
 4(2)(5)
 
24
A C; ellipse

36.

B


tan 2v  
A
C
4


tan 2v  
2
5
4

3

tan 2v 
2v  
53.13010235°
v  
27°
2
31. B2 
 4AC  4
3
 
 4(2)(6)
0
parabola

y

0  

0 
 4
(1)(x2)

2(1)




4x2

y  
2
x  0, y  0; point

y

B

A
C

4
3

2
6

(0, 0)

x

O


tan 2v  

3
2v  
60°
v  
30°
32. B2 
 4AC  42 
 4(2)(2)
0
parabola

37.
(1)y2

x2 
 2xy  y2 
 5x 
 5y  0
 (
2x 
 5)y  (x2 
 5x)  0
↑
b

↑
c

↑
a



A  C; v  4 or 45°
33.


b  

b2 
 4
ac

2a









tan 2v 

y









tan 2v 

(x 
 2)2  (y 
 2)2  4(x  y)  8
x2 
 4x  4  y2 
 4y  4  4x  4y  8
(1)y2  0y  x2  0
↑
↑
↑
a
b
c

(x 
 2)2 
 (x  3)2  5(y  2)
2
x 
 4x  4 
 x2 
 6x 
 9  5(y  2)

10x 
 5  5(y  2)

2x 
 1  y  2
y

2x 
 3  y
y  
2x 
 3 line
O
x

y
y
y
y


b  

b2 
 4
ac

2a

(
2x 
 5)  

(
2x 

5)2 

4(1)(x2

 5x)

2(1)
2x  5  

4x2 
20x 
25 

4x2 
20x

2
2x  5  
40x


25

2

y 
 2x  3











34. 2x2  6y2  8x 
 12y  14  0
x2  3y2  4x 
 6y  7  0
2
3y  (
6)y  (x2  4x  7)  0
↑
a

↑
b

↑
c

38.

[
6.61, 14.6] scl1 by [
2, 12] scl1
2x2  9xy  14y2  5
14y2  (9x)y  (2x2 
 5)  0


 b  

b2 
 4
ac

2a

y


(
6)  

(
6)2 

 4(3
)(x2 
4x  
7)

2(3)

y

6  


12x2

 48
x 
 48

6

y


b  

b2 
 4
ac

2a

y

6  


12(x
 2)2

6

y


9x  

(9x)2 

4(14)
(2x2 

5)

2(14)

y


9x  


21x2
 28
0

28









y

↑
a

x  
2, y  1; point
y

↑
b

↑
c

(2, 1)

O

x

35. y2 
 9x2  0
y2  9x2
y  

9x2
y   3x
intersecting lines

y
y 
 3x

O

[
7.58, 7.58] scl1 by [
5, 5] scl1

x

y 
 3x

355

Chapter 10

8x2  5xy 
 4y2  
2
(
4)y2  (5x)y  (8x2  2)  0

42.

↑
a

↑
c

↑
a

↑
b

↑
c

y


b  

b2 
 4
ac

2a

y


b  

b2 
 4
ac

2a

y


5x  

(5x)2 

4(
4
)(8x2 
2)


y


4x  

(4x)2 

4(6)(9
x2 
 2
0)

2(6)

y

153x2
 32

5x  



y


4x  


212
x2  4
80

12

[
7.85, 7.85] scl1 by [
5, 5] scl1
43a.
y

[
7.58, 7.58] scl1 by [
5, 5] scl1
2x2  4
3
xy  6y2  3x  y
2
6y  (4
3
x 
 1)y  (2x2  3x)  0

y

↑
b

(0, 5280)









↑
a









40.

↑
b

9x2  4xy  6y2  20
6y2  (4x)y  (9x2 
 20)  0
















39.

↑
c

(1320, 1320)


b  

b2 
 4
ac

2a

O (5280, 0) x



(4
3
x

 1)2

 4(6
)(2x2 
3x)

y


(4
3
x 
 1) 

2(6)

y


4
3
x  1  

48x2 

8
3
x
 1 

48x2 

 72x

12

y


4
3
x  1  


8
3

x 
 72
x1

12

T(1320, 1320)
43b. circle
center: (h, k)  (1320, 1320)
radius: r  1320
(x 
 h)2  (y 
 k)2  r2
(x 
 1320)2  (y 
 1320)2  13202
(x 
 1320)2  (y 
 1320)2  1,742,400
44a. B2 
 4AC  0 
 4(1)(0)
0
parabola; 360°
44b. B2 
 4AC  0 
 4(8)(6)
 
192
A C; ellipse; 180°
44c. B2 
 4AC  42 
 4(0)(0)
 16
hyperbola; 180°
44d. 32 
 4AC  0 
 4(15)(15)
 
900
A  C; circle; There is no minimum angle of
rotation, since any degree of rotation will result
in a graph that coincides with the original.
45. Let x  x
 cos v  y
 sin v and
y  
x
 sin v  y
 cos v.
x 2  y2  r 2
(x
 cos v  y
 sin v)2  (
x
 sin v  y
 cos v)2  r2
(x
)2 cos2 v  x
y
 cos v sin v  (y
)2 sin2 v
 (x
)2 sin2 v 
 x
y
 cos v sin v  (y
)2 cos v  r2
[(x
)2  (y
)2] cos2 v  [(x
)2  (y
)2] sin2 v  r2
[(x
)2  (y
)2](cos2 v  sin2 v)  r2
[(x
)2  (y
)2](1)  r2
(x
)2  (y
)2  r2
2
2
46a. B 
 4AC  
10
3
 
 4(31)(21)
 
2304
A C; elliptical

[
8.31, 2.31] scl1 by [
2, 5] scl1
x 
 2
2
y  
12
2x2  4xy  2y2  2
2
 (4x 
 2
2
)y  (2x2  2
2
x  12)  0














2y2









41.

↑
b

↑
c

↑
a

y


b  b2 
 4ac

2a

(4x 
 2
2
) 

(4x 
 
2
2
 )2 

 4(2
)(2x2 
2
2
x
 12)

2(2)

y


4x  2
2
  

16x2 

16
2

x8

 16x2

 16

2
x 

96

4

y


4x  2
2
  


32

x 
 8
2
8

4

y

[
10.58, 4.58] scl1 by [
2, 8] scl1

Chapter 10

356

48a. center: (h, k)  (0, 0)
major axis: horizontal
a  
81
 or 9
b  
36
 or 6
c  

a2 
 b2
c  
81
36


c  
45
 or 3
5


31x2 
 10
3
xy  21y2  144
21y2  (
10
3
x)y  (31x2 
 144)  0
↑
a











46b.

↑
b

↑
c

y


b  

b2 
 4
ac

2a

y


(
10
3
x)  

(
10

x)2 

3
4(21)
(31x2 

 144)

2(21)

y

x  

230

4x2 
12,096

10
3

42

y

8
4

y

8


4 O

4

8

x


4

O


8

x
48b. T(3
5, 0)
B


46c. tan 2v  
A
C

tan 2v 


10
3


31 
 21

48c.

x2
y2
    1
81
36
(x 
 h)2
(y 
 k)2
    1
81
36
(x
 
 3
5
 )2
(y
)2
    1
36
81

49. A  
3, C  5; since A and C have different signs,
the conic is a hyperbola.
50. (h, k)  (2, 
3)


tan 2v  

3
2v  
60°
v  
30°
B

c


47a. tan 2v  
A
C

e  a
c
2
6

  
1
5
2
6

  c
5
b2  a2


2

3


tan 2v  
9 
 11

tan 2v  
3

2v  60°
v  30°
The graph of this equation has been rotated 30°.
To transform the graph so the axes are on the
x- and y-axes, rotate the graph 
30°.

1

b2  2
5
major axis: horizontal

2


3


1

2



3

(x 
 h)2
(y 
 k)2

 b
2
a2
(x 
 2)2
[y 
 (
3)]2
  
1
1

25
(x 
 2)2  25(y  3)2

2

3
1

3

(x
)2 
 x
y
  (y
)2 

4
4
2
3
1

3


3

2




2
3
 4 (x
)  4 x
y
 
 4x
y
 
 4(y
)2
1
3


3
 11 4(x
)2  2x
y
  4(y
)2
27
9
6
9
3

(x
)2 
 x
y
  (y
)2 
 (x
)2
4
4
4
2
6
6
3

2

3
2

 4x
y
  4x
y
  4(y
)
11
33
11
3

 4(x
)2  2x
y
  4(y
)2
8(x
)2  12(y
)2

(y 
 k)2

a2
[y 
 (
3)]2

12

(y  3)2 


 24  0
51.

9

(x
)2

3

1
1
1

major axis : vertical

1

12x
  
23 y
  1112x
  
23 y


2
6
 2

b2  12 
 5

47b. B2 
 4AC  
2
3
 
 4(9)(11)
 
384
A C; the graph is an ellipse.
9x2 
 2
3
xy  11y2 
 24  0
9[x
 cos (
30°)  y
 sin (
30°)]2 
 2
3

[(x
 cos (
30°)  y
 sin (
30°)]
[
x
 sin (
30°)  y
 cos (
30°)]  11
[
x
 sin (
30°)  y
 cos (
30°)2] 
 24  0
2x
 
 2y

92x
 
 2y
 
 2
3


 c2

v
r


 24  0

0
1
120˚

(x 
 h)2

 b
1
2

(x 
 2)2
 1
1

25
25(x 
 2)2  1



30°
1.4

60°
3.9

90˚

1 2 3 4

240˚

(y
)2

357

180°

1

0˚

330˚

210˚

 2  1

150°

1

30˚

180˚


 24  0

120°

1.4

60˚

150˚


 24  0

90°

3.9

270˚

300˚

Chapter 10

52.

58.

N
E

W
5

c

S

53. cos 70°  0.34
cos 170°  
0.98
cos 70°
54.

5

16

5

180°



 16

 56.25
15

 566
0

h (x )

 56° 15


55.

2y  5


y2  3y  2
2y  5


y2  3y  2

2y  5



(y  2)(y  1)
A

B




y2  y1

x

2y  5  A(y  1)  B(y  2)
2(
2)  5  A(
2  1)  B(
2  2)
1  
A

1  A
2y  5  A(y  1)  B(y  2)
2(
1)  5  A(
1  1)  B (
1  2)
3B

56.

A
B
  
y2
y1
x1
x2


y2  y1
12

y2




1

5a8b5
180a b

1  b and 2  a, the expression is always larger

3




y2  y1

2213
36

3223
36

72
36

expression is always less than    or 2.
The correct choice is B.

m

3

4



3
n
8



1
p
2

4m  9n 
 2p  
4

44  8n  2p  9n 
 2p  
4
3

1

3

21
n
2

 
7
2
n  
3
6m  12n  5p  
1

8m 
 3n 
 4p  6

8m 
 3
3 
 4p  6 6m  12
3  5p  
1
8m  2 
 4p  6
6m 
 8  5p  
1
8m 
 4p  4
6m  5p  7
2m 
 p  1
2m 
 p  1 →
10m 
 5p  5
6m  5p  7 →
() 6m  5p  7
16m
 12
2

10-8

Systems of Second-Degree
Equations and Inequalities

Page 682

Check for Understanding

1. Possible number of solutions: 0, 1, 2, 3, or 4

3  2n  2p  9n 
 2p  
4

2

3

m  4
8m 
 3n 
 4p  6

 
 3
23 
 4p  6

3

4

6  2 
 4p  6
1

p  2

34, 
23, 12

Chapter 10

1
9

than   . Since b  2 and a  3, the

y2  9.6
57. 8m 
 3n 
 4p  6
8m  6  3n  4p

8

a2b3
36

59. The expression 
6 2 simplifies to . Since

5

4

3

h(x) 
 [[x]]  3

6

5

4

3

2

1
0
1
2
3

x

3  x  
 2

2  x  
1

1  x  0
0x1
1x2
2x3
3x4
4x5
5x6
6x7

c2  a2  b2
c2  82  52
c  
89

c  9.4
about 9.4 m/s

8

358

2. Sample answer: y  x2, x2  (y  3)2  9

7. x2  y2  4
y2  4 
 x2
9x2 
 4y2  36
2
9x 
 4(4 
 x2)  36
13x2 
 16  36
x2  4
x  2
(2, 0)

y
y 
 x2

x

O
x 2  (y  3)2 
 9

x2  y 2  4
22  y2  4
y2  0
y0

y
3. The system contains equation(s) that are
equivalent. The graphs coincide.
4. Graph each second-degree inequality. The region
in which the graphs overlap represents the
solution to the system.
5. x 
 y  0
xy
(x 
 1)2
(y 
 1)2
  
20
5
(x 
 1)2
(x 
 1)2
  
20
5

 1)2  4(x 
 1)2

(2, 0)

(2, 0)

O

8.

1
1

 20
(x
x2 
 2x  1  4(x2 
 2x  1)  20
5x2 
 10x  5 
 20  0
5(x2 
 2x 
 3)  0
5(x 
 3)(x  1)  0
x
30
x10
x3
x  
1
(3, 3), (
1, 
1)

x

xy  1
x(x2)  1
x3  1
x1
(1, 1)

x2  y
12  y
1y

y
(1, 1)

O

x

y
(3, 3)

9.

y

10.

y
4

O
x
(1, 1)

4

8

12

x


4

8

6. x  2y  10
x  10 
 2y
x2  y2  16
(10 
 2y)2  y2  16
100 
 40y  5y2  16
5y2 
 40y  84  0


12

11.

y
8

y


b  

b2 
 4
ac

2a

y

2 
 4(

(
40)  

(
40)
5)(84)

2(5)

y

40  


80

10

4

8 
4 O

4

no solution

4

8

x


8

y

O

O

x

O

x

359

Chapter 10

12a. Let x  side length of flowerbed 1.
Let y  side length of flowerbed 2.
A1  x x or x2
A2  y y or y2
Total Area  x2  y2
680  x2  y2
x2  y2  680
Difference of Areas  x2 
 y2
288  x2 
 y2
x2 
 y2  288
12b.

40

y

y
(2, 1)

O

20

15.

40 x


40

12c. x2  y2  680
y2  680 
 x2
x2 
 y2  288
2
x 
 (680 
 x2)  288
2x2  968
x2  484
x  22
22 ft and 14 ft

x

(2, 1)

Since side length
cannot be negative, an
estimated solution is
(22, 14)
(22, 14).

20

40 
20 O

20

(2, 1), (
2, 
1)

x2  y2  680
222  y2  680
y2  196
y  14


1  2x  y

1 
 2x  y
4x2  y2  25
2
4x  (
1 
 2x)2  25
4x2  1  4x  4x2  25
8x2  4x 
 24  0
4(2x2  x 
 6)  0
4(2x 
 3)(x  2)  0
2x 
 3  0
x20
x  1.5
x  
2

1  2x  y

1  2x  y

1  2(1.5)  y

1  2(
2)  y

4  y
3y
(1.5, 
4), (
2, 3)

y
(2, 3)

Pages 682–684

Exercises

13. x 
 1  0
x1
y2  49 
 x2
y2  49 
 (
1)2
y2  48
y  6.9
(1, 6.9)

O
y

(1, 6.9)
(1.5, 4)

O

x

16. x 
 y  2
x2y
x2  100 
 y2
(2  y)2  100 
 y2
4  4y  y2  100 
 y2
2
2y  4y 
 96  0
2(y2  2y 
 48)  0
2(y 
 6)(y  8)  0
y
60
y80
y6
y  
8
x
y2
x
y2
x
62
x 
 (
8)  2
x8
x  
6
(8, 6), (
6, 
8)

(1, 6.9)

14. xy  2
2

y  x
x2  3  y2

2 2

x2  3  x
4

x2  3  x2
x4  3x2  4
x4 
 3x2 
 4  0
(x2 
 4)(x2  1)  0
x2 
 4  0
x2  1  0
2
x 4
x2  
1
x  2
xy  2
xy  2
2(y)  2
(
2)y  2
y1
y  
1

y
(8, 6)

O

(6, 8)

Chapter 10

x

360

x

19. x  y  
1
y  
1 
 x
(y 
 1)2  4  x
(
1 
 x 
 1)2  4  x
(
2 
 x)2  4  x
4  4x  x2  4  x
x2  3x  0
x(x  3)  0
x0

17. x 
 y  0
xy

(x 
 1)2
 
 y2
9
(y 
 1)2
 
 y2
9

 1)2 
 9y2

1
1

9
(y
y2 
 2y  1 
 9y2  9

8y2 
 2y 
 8  0
4y2  y  4  0
y


b  

b2 
 4
ac

2a

y


1  

12 
 4(4)(4)


2(1)

y


1  

63


2

x30
x  
3
x  y  
1

3  y  
1
y2

x  y  
1
0  y  
1
y  
1
(0, 
1), (
3, 2)

y

no solution
y

(3, 2)

O
O

x

(0, 1)

x

20. xy  6  0
6

18. x2  2y2  10
x2  10 
 2y2
3x2  9 
 y2
3(10 
 2y2)  9 
 y2
30 
 6y2  9 
 y2

5y2  
21
y2  4.2
y  2.0
3x2  9 
 y2
3x2  9 
 (2.0)2
x2  1.6
x  1.3
(1.3, 2.0), (1.3, 
2.0)

y  
x
x2  y2  13
6 2

x2  
x  13
36

x2  x2  13
x4  36  13x2
x4 
 13x2  36  0
(x2 
 9)(x2 
 4)  0
x2 
 9  0
x2  9
x  3
xy  6  0
3y  6  0
y  
2
xy  6  0

3y  6  0
y2
(3, 
2), (
3, 2)

3x2  9 
 y2
3x2  9 
 (
2.0)2
x2  1.6
x  1.3

y
(1.3, 2.0)

(1.3, 2.0)

O

x

x2 
 4  0
x2  4
x  2
xy  6  0
2y  6  0
y  
3
xy  6  0

2y  6  0
y3
(2, 
3), (
2, 3)

y

(1.3, 2.0)

(2, 3)
(3, 2)

(1.3, 2.0)

O

x
(3, 2)
(2, 3)

361

Chapter 10

23. xy  
4

21. x2  y 
 3  0
x2  
y  3
x2  4y2  36

y  3  4y2  36
4y2 
 y 
 33  0
(y 
 3)(4y  11)  0
y
30
y3
x2  y 
 3  0
x2  3 
 3  0
x2  0
x0
(0, 3), (2.4, 
2.8)

4

y  
x
x2  25 
 9y2
144

x2  25 
 x
2

4y  11  0
y  
2.8
x2  y 
 3  0
x2  (
2.8) 
 3  0
x2  5.8
x  2.4

x4  25x2 
 144
x4 
 25x2  144  0
(x2 
 9)(x2 
 16)  0
x2 
 9  0
x2 
 16  0
x2  9
x2  16
x  3
x  4
xy  
4
xy  
4
3y  
4
4y  
4
y  
1.3
y  
1
xy  
4
xy  
4

3y  
4

4y  
4
y  1.3
y1
(3, 
1.3), (
3, 1.3)
(4, 
1), (
4, 1)
24.
25.
y
y

y
(0, 3)

x

O

(2.4, 2.8)

(2.4, 2.8)

4 2

x2  25 
 9
x

8
4

O

22. 2y 
 x  3  0
2y  3  x
x2  16 
 y2
(2y  3)2  16 
 y2
4y2  12y  9  16 
 y2
5y2  12y 
 7  0
y


b  

b2 
 4ac


2a

y


12 

122 

4(5)(

7)

2(5)

y


12  
284


10

y  0.5
or
2y 
 x  3  0
2(0.5) 
 x  3  0
x  4.0
(4.0, 0.5), (
2.8, 
2.9)


8 
4

4

8

x


8

y

26.

x

O

y  
2.9
2y 
 x  3  0
2(
2.9) 
 x  3  0
x  
2.8

y

27.

y
8

(4, 0.5)

O

4

x

8 
4 O

4

(2.8, 2.9)

Chapter 10

O


4


8

362

4

8x

x

28.

37a. 2x  2y  P
xy  A
2x  2y  150
xy  800
37b. A system of a line and a hyperbola may have 0,
1, or 2 solutions.
37c.
y

29.

y

y
8
4

8 
4 O

4

x

O

8x

4

80
40


8


80 
40 O

30.

31.
8

40

80

x


40

80

y

y

4

37d. xy  800
800

8 4 O

4

8

y  x

x

4

2x  2y  150

x

O

2x  2x  150
800

8

1600

2x  x  150

32.

33.
8

2x2  1600  150x
x2 
 75x  800  0

y

y

4


8 
4 O

8x

4


8 
4 O

4


4

4

x


8


8

x

75  

2425

2

x  12.88
or x  62.12
xy  800
xy  800
12.88y  800
62.12y  800
y  62.11
y  12.88
12.9 m by 62.1 m or 62.1 m by 12.9 m
38a. (h, k)  (0, 4)
(x, y)  (6, 0)
(x 
 h)2  4p(y 
 k)
(6 
 0)2  4p (0, 4)
36  
16p

2.25  p
(x 
 h)2  4p(y 
 k)
(x 
 0)2  4(
2.25)(y 
 4)
x2  
9(y 
 4)
2
x  
9(y 
 4), y  0
y
38b.

34. parabola:
vertex: (1, 
3)
(y 
 k)2  4p(x 
 h)
(
5  3)2  4p(
1 
 1)
1


2  p
(y 
 3)2  4
2(x 
 1)
1

(y  3)2  
2(x 
 1)
line:
m  
2, b  
7
y  mx  b
y  
2x 
 7
35. circle:
center: (0, 0), radius: 2
2

x2  y2  r2
x2  y2  8
hyperbola:
(
2)(
2)  4
xy  4
36. large ellipse:
a  5, b  4, center  (0, 0)
y2

a2
y2

25

x


b  

b2 
 4
ac

2a
2 
 4(1)(800

(
75)  

(
75)
)

2(1)

x

4

x
O

x2

 b2  1
x2

 16  1 (interior is shaded)

small ellipse:
a  3, b  2, center  (0, 
1)
x2
(y 
 k)2



a2 
b2
x2
(y  1)2
  
9
4

1
1 (exterior is shaded)

363

Chapter 10

38c. (h, k)  (0, 3)
(x, y)  (6, 0)
(x 
 h)2  4p(y 
 k)
(6 
 0)2  4p(0 
 3)
36  
12p

3  p
(x 
 h)2  4p(y 
 k)
(x 
 0)2  4(
3)(y 
 3)
x2  
12(y 
 3)
2
x  
 12(y 
 3), y  0
39. xy  
12

y

40b.

estimate: (40, 30)

80
40

O

80 
40

40

80x


40

40c. x2  y2  2500
x2  2500 
 y2
x2  (y 
 30)2  1600
2500 
 y2  y2 
 60y  900  1600

60y  1800  0
y  30
(x 
 35)2  (y 
 18)2  169
x2 
 70x  1225  (30 
 18)2  169
x2 
 70x  1200  0
(x 
 30)(x 
 40)  0
x 
 30  0
or
x 
 40  0
x  30
x  40
Check (30, 30) and (40, 30):
x2  y2  2500
x2  y2  2500
2
2
2
30  30  2500
40  302  2500
1800 2500
2500  2500 
(40, 30)
41.
x  3y  k
2y2  3y  k
2y2  3y 
 k  0

12

y  
x
x  
y  1

x  

x  1
12

x2  12  x

 x 
 12  0
(x 
 4)(x  3)  0
x
40
x30
x4
x  
3
xy  
12
xy  
12
4y  
12

3y  
12
y  
3
y4
(4, 
3)
(
3, 4)
Check that (4, 
3) and (
3, 4) are also solutions
of y2  25 
 x2.
y2  25 
 x2
y2  25 
 x2
(
3)2  25 
 (4)2
(4)2  25 
 (
3)2
99 
16  16 
(4, 
3), (
3, 4)
x2

3

1

y2  2y 
 2k  0
3 2

y2  2y  4  0 Complete the square.

y

3

2

y  34
1
3 2

2k  4

(3, 4)

1

9


2k  1
6

x

O

0

9

k  
8

(4, 3)

42a.
8

y

4

40a. first station: (h, k  (0, 0)
x2  y2  r2
x2  y2  502
x2  y2  2500
second station: (h, k)  (0, 30)
(x 
 h)2  (y 
 k)2  r2
x2  (y 
 30)2  402
x2  (y 
 30)2  1600
third station: (h, k)  (35, 18)
(x 
 h)2  (y 
 k)2  r2
(x 
 35)2  (y 
 18)2  132
(x 
 35)2  (y 
 18)2  169


8 
4 O

4

4

8


8

42b. yes; (6, 2) or (6, 
2)
42c. Earth’s surface:
x12  y12  40
x 2
1
40

2




4

0
x1

y12

 4
0 1
2

  1
 

4

0
y1

cos2 t  sin2 t  1
2




4

0
x1

x1

2
10


 cos2 t
 cos t

 cos t
x1  2
10

Chapter 10

x

364

2




4

0
y1

y1

2
10


 sin2 t
 sin t

y1  2
10
 sin t

49. No; the domain value 4 is mapped to two elements
in the range, 0 and 
3.
50. area of rectangle  
q
 8(4) or 32
area of circles  2 (r2)
 2(4p) or 8p
area of shaded region  32 
 8p
The correct choice is E.

asteroid:
Let y2  
t.
x2  0.25y2  5
x2  0.25 t2  5
42d.

10-8B Graphing Calculator Exploration:

Shading Areas on a Graph

Tmin  
8, Tmax  8, Tstep  0.13
[
15.16, 15.16] scl1 by [
10, 10] scl  1

Page 686

x2
   y2  1
9

43.

1.

(x
 cos 30°  y
 sin 30°)2
  (
x
 sin 30°  y
 cos 30°)2  1
9
2


3
1
x
  y

2
2
2
1


3

 
2x
  2y
  1
9





3


3
1
(x
)2  x
y
  (y
)2
4
2
4

9






3


1

3

 4(x
)2 
 2x
y
  4(y
)2  1

3
1
9
27

3

9

3
(x
)2  x
y
  (y
)2  (x
)2 
x
y
   (y
)2  9
4
4
4
4
2
2
x
y
  7(y
)2 
 9  0
3(x
)2 
 4
3

44.

x  4t  1

x
1

4

y  5t 
 7

y7

5

t
x
1

4

[
9.1, 9.1] scl1 by [
6, 6] scl1
2.



t

y7

5

(x 
 1)(5)  (4)(y  7)
5x 
 5  4y  28
5x 
 4y 
 33  0



45. 4 csc v cos v tan v  4
sin v (cos v) cos v 
1

sin v

[
9.1, 9.1] scl1 by [
6, 6] scl1
3.

4
46. r  10 cm or 0.10 m
v  rq
v  (0.10)1
5 2p

v  3.14 m/s
47.

48.

r
0
1
2

1
1
1
1

0
0
1
2


4

4

3
4

0
0
1
4

[
9.1, 9.1] scl1 by [
6, 6] scl1
4.

between 1 and 2
y  (x  2)2 
 3
x  (y  2)2 
 3
x  3  (y  2)2

x

3y2

x
3
2y
y

[
9.1, 9.1] scl1 by [
6, 6] scl1
5a. 3
5b. Find the points of intersection for the boundary
equation by using the TRACE function.

y 
 (x  2)2  3

O
x
y 
  

x32

365

Chapter 10

14. (x 
 h)2  (y 
 k)2  r2
2
(x 
 0)2  (y 
 0)2  3
3

x2  y2  27
y

5c. SHADE(

((36
X2)/9),
((36
X2)/9),
6,
2,3,4);
SHADE(
X22,
((36
X2)/9),
2,2,3,4,);
SHADE(

((36
X2)/9), 
((36
X2)/9),2,6,3,4)

(0, 3
3)
(3
3, 0)

O

[
9.1, 9.1] scl1 by [
6, 6] scl1
6. See students’ work.

x

15. (x 
 h)2  (y 
 k)2  r2
(x 
 2)2  (y 
 1)2  22
(x 
 2)2  (y 
 1)2  4

y

(2, 1)

Chapter 10 Study Guide and Assessment

(4, 1)

O

x
(2, 1)

Page 687

Understanding and Using the
Vocabulary

1. true
3. false; transverse
5. false; hyperbola

16.

2. false; center
4. true
6. false; axis or axis of
symmetry
8. false; parabola
10. false; ellipse

7. true
9. true

x2  y2  6y
x2  (y2 
 6y  ?)  0  ?
x2  (y2 
 6y  9)  0  9
x2  (y 
 3)2  9

y
(0, 6)
(3, 3)

(0, 3)

Pages 688–690

Skills and Concepts

11. d  

(x2 
 
x1)2 
(y2 

y1)2
d  

(
3 

1)2 
[
4 

(
6)]2
d  
20
 or 2
5

x1  x2 y1  y2

 
2 ,
2



1  (
3) 
6  (
4)
, 
2
2



 (
1, 
5)

O
17.



x

x2  14x  y2  6y  23
 14x  ?)  (y2  6y  ?)  23  ?  ?
2
(x  14x  49)  (y2  6y  9)  23  49  9
(x  7)2  (y  3)2  81
(x2

2
x2 
 x
(y2 
 
y1)2
12. d  

1)  

d  

(a  3

 a)2
 (b 
 4 

b)2
d  
25
 or 5

(7, 6)

y

(7, 3)

O

x1  x2 y1  y2

aa3 bb4
, 
2, 2  
2
2

 (a  1.5, b  2)

(16, 3)

(x2 
 
x1)2 
(y2 

y1)2
13. AB  

 

[3 
 (

5)]2 
 [4 

(
2)]2
 
100
 or 10
DC  

(x2 
 
x1)2 
(y2 

y1)2

18.

 

(10 

2)2 
[3 
 
(
3)]2
 
100
 or 10
BC  

(x2 
 
x1)2 
(y2 

y1)2
 

(10 

3)2 
(3 
 
4)2
 
50
 or 5
2

AD  

(x2 
 
x1)2 
(y2 

y1)2

3x2  3y2  6x  12y 
 60  0
x2  y2  2x  4y 
 20  0
(x2  2x  ?)  (y2  4y  ?)  20  ?  ?
(x2  2x  1)  (y2  4y  4)  20  1  4
(x  1)2  (y  2)2  25

y
(1, 3)

 

[2 
 (

5)]2 
 [
3

(
2
)]2
 
50
 or 5
2

Yes; AB  DC  10 and BC  AD  5
2
. Since
opposite sides of quadrilateral ABCD are
congruent, ABCD is a parallelogram.

Chapter 10

x

O
(1, 2)

366

x
(4, 2)

minor axis vertices: (h, k  b)  (3, 
1  2) or
(3, 1), (3, 
3)

x2  y2  Dx  Ey  F  0
12  12  D(1)  E(1)  F  0 ⇒
D  E  F  
2
(
2)2  22  D(
2)  E(2)  F  0 ⇒

2D  2E  F  
8
(
5)2  12  D(
5)  E(1)  F  0 ⇒

5D  E  F  
26
D  E  F  
2
(
1)(
5D  E  F)  (
1)(
26)
6D
 24
D4

2D  2E  F  
8
(
1)(D  E  F)  (
1)(
2)

3D  E
 
6

3(4)  E  
6
E6
D  E  F  
2
4  (6)  F  
2
F  
12
x2  y2  Dx  Ey 
 F  0
x2  y2  4x  6y 
 12  0
2
(x  4x  ?)  (y2  6y  ?)  12  ?  ?
(x2  4x  4)  (y2  6y  9)  12  4  9
(x  2)2  (y  3)2  25
center: (h, k)  (
2 
 3)
r  
25
 or 5
20. center: (h, k)  (5, 2)
a2  36
b2  16
a  
36
 or 6
b  
16
 or 4
a2 
 b2
c  

c  
6


 4 or 
2

foci: (h, k  c)  (5, 2  
2
)
major axis vertices: (h, k  a)  (5, 2  6) or
(5, 8), (5, 
4)
minor axis vertices: (h  b, k)  (5  4, 2) or
(9, 2), (1, 2)
y
(5, 8)

19.

(1, 2)

(5, 2)

O

y
(3, 1)

O

(8, 1) x

(2, 1) (3, 3)

22.

6x2  4y2  24x 
 32y  64  0
6(x2  4x  ?)  4(y2 
 8y  ?)  
64  ?  ?
6(x2  4x  4)  4(y2 
8y  16)  
64  6(4)  4(16)
6(x  2)2  4(y 
 4)2  24
(x  2)2
(y 
 4)2
    24
4
6

center: (h, k)  (
2, 4)
b2  4
a2  6
a  
6

b  
4
 or 2
c  

a2 
 b2
c  
6


 4 or 
2

foci: (h, k  c)  (
2, 4  
2
)
major axis vertices: (h, k  a)  
2, 4  
6

minor axis vertices: (h  b, k)  (
2  2, 4) or
(0, 4), (
4, 4)
y
(2, 4  

6)
(2, 4)
(0, 4)

(4, 4)
(2, 4  

6)

O
23.

x

x2  4y2  124  8x  48y
(x2 
 8x  ?)  4(y2 
 12y  ?)  
124  ?  ?
(x2 
 8x  16)  4(y2 
 12y  36)  
124  16  4(36)
(x 
 4)2  4(y 
 6)2  36
(x 
 4)2
(y 
 6)2
    36
36
9

(9, 2)

center: (h, k)  (4, 6)
a2  36
b2  9
a  
36
 or 6
b  
9
 or 3
c  

a2 
 b2
c  
36
 
 9 or 3
3

foci: (h  c, k)  (4  3
3
, 6)
major axis vertices: (h  a, k)  (4  6, 6) or
(10, 6), (
2, 6)
minor axis vertices: (h, k  b)  (4, 6  3) or
(4, 9), (4, 3)
y

x
(5, 4)

21.

(3, 1)

4x2  25y2 
 24x  50y  39

4(x2 
 6x  ?)  25(y2  2y  ?)  39  ?  ?

4(x2 
 6x  9)  25(y2  2y  1)  39  4(9)  25(1)
4(x 
 3)2  25(y  1)2  100
(x 
 3)2
(y  1)2
    1
25
4

center: (h, k)  (3, 
1)
b2  4
a2  25
a  
25
 or 5
b  
4
 or 2
c  

a2 
 b2
c  
25
4

  or 
21

foci: (h  c, k)  (3  
21
, 
1)
major axis vertices: (h  a, k)  (3  5, 
1) or
(8, 
1), (
2, 
1)

(4, 9)
(2, 6)

(4, 6)
(10, 6)
(4, 3)

O

367

x

Chapter 10

24. (h, k)  (
4, 1)
a9
b6

(y 
 k)2
(x 
 h)2

 b
2
a2
(y 
 1)2
[x 
 (
4)]2

 6
2
92
(y 
 1)2
(x  4)2
  
81
36

center: (h, k)  (0, 
2)
b2  1
a2  4
a  
4
 or 2
b  
1
 or 1
c  

a2  b2
c  
4

 1 or 
5

transverse axis: horizontal
foci: (h  c, k)  (0  
5
, 
2) or (
5
, 
2)
vertices: (h  a, k)  (0  2, 
2) or (
2, 
2),
(2, 
2)

1
1
1

25. center: (h, k)  (0, 0)
b2  16
a2  25
a  
25
 or 5
b  
16
 or 4
c  

a2  b2
c  
25
16
  or 
41

transverse axis: horizontal
foci: (h  c, k)  (0  
41
, 0) or (
41
, 0)
vertices: (h  a, k)  (0  5, 0) or (
5, 0), (5, 0)

b

asymptotes: y 
 k   a(x 
 h)
1

y 
 (
2)   2(x 
 0)
1

y  2   2x

y

b

asymptotes: y 
 k   a (x 
 h)

x

O

4

y 
 0   5(x 
 0)

(2, 2)

(2, 2)
(0, 2)

4

y   5x

y
28.
(5, 0)

(5, 0)

x

O

9x2 
 16y2 
 36x 
 96y  36  0
9(x2 
 4x  ?) 
 16(y2  6y  ?)  
36  ?  ?
9(x2 
 4x  4) 
 16(y2  6y  9)  
36  9(4) 
 16(9)
9(x 
 2)2 
 16(y  3)2  
144
(y  3)2
(x 
 2)2
 
   1
9
16

center: (h, k)  (2, 
3)
b2  16
a2  9
a  
9
 or 3
b  
16
 or 4
c  

a2  b2
c  
9
6
 1 or 5
transverse axis: vertical
foci: (h, k  c)  (2, 
3  5) or (2, 2), (2, 
8)
vertices: (h, k  a)  (2, 
3  3) or (2, 0), (2, 
6)
a
asymptotes: y 
 k   b(x 
 h)

26. center: (h, k)  (1, 
5)
a2  36
b2  9
a  
36
 or 6
b  
9
 or 3
c  

a2  b2
c  
36
9
  or 3
5

transverse axis: vertical
foci: (h, k  c)  1, 
5  3
5

vertices: (h, k  a)  (1, 
5  6) or (1, 1), (1, 
11)
a

y 
 (
3)   4(x 
 2)

3

6

y  3   4(x 
 2)

asymptotes: y 
 k   b(x 
 h)

3

y 
 (
5)   3(x 
 1)
y  5   2(x 
 1)

y

y
(2, 0)

O

(1, 1)

O

x
(1, 5)

(2, 6)

(1, 11)

27.

x2 
 4y2 
 16y  20
x2 
 4(y2  4y  ?)  20  ?
x2 
 4(y2  4y  4)  20 
 4(4)
x2 
 4(y  2)2  4
x2

4

Chapter 10

(2, 3)

(y  2)2


 1  1

368

x

29. c  9
quadrants: I and III
transverse axis: y  x
vertices: xy  9
3(3)  9
(3, 3)

33. vertex: (h, k)  (1, 
2)
4p  
16
p  
4
focus: (h  p, k)  (1  (
4), 
2) or (
3, 
2)
directrix: x  h 
 p
x  1 
 (
4)
x5
axis of symmetry: y  k
y  
2

xy  9
(
3)(
3)  9
(
3, 
3)

y
(3, 3)

y
O

x

(3, 3)

O

(3, 2)

30. 2b  10
b5

x1  x2 y1  y2

1  1 
1  5

center: 2, 2  2, 2

x
5

 (1, 2)
transverse axis: vertical
a  distance from center to a vertex
 2 
 (
1) or 3
(y 
 k)2

a2
(y 
 2)2

32
(y 
 2)2

9

34. y2  6y 
 4x  
25
y2  6y  ?  4x 
 25  ?
y2  6y  9  4x 
 25  9
(y  3)2  4(x 
 4)
vertex: (h, k)  (4, 
3)
4p  4
p1
focus: (h  p, k)  (4  1, 
3) or (5, 
3)
directrix: x  h 
 p
x4
1
x3
axis of symmetry: y  k
y  
3

(x 
 h)2


 b
1
2
(x 
 1)2


 5
1
2
(x 
 1)2


 25  1
x1  x2 y1  y2


2  6 
3  (
3)

31. center: 2, 2  2, 2
 (2, 
3)
a  distance from center to a vertex
 2 
 (
2) or 4
c  distance from center to a focus
 2 
 (
4) or 6
b2  c2 
 a2
b2  62 
 42
b2  20
transverse axis: horizontal

y

(4, 3)

(5, 3)

x
3

35.

32. vertex: (h, k)  (5, 3)
4p  8
p2
focus (h, k  p)  (5, 3  2) or (5, 5)
directrix: y  k 
 p
y3
2
y1
axis of symmetry: x  h
x5
y

x2  4x  y 
 8
x2  4x  4  y 
 8  4
(x  2)2  y 
 4
vertex: (h, k)  (
2, 4)
4p  1
1

p  4
focus: (h, k  p)  
2, 4  4 or (
2, 4.25)
1

directrix: y  k 
 p
1

y  4 
 4
y  3.75
axis of symmetry: x  h
x  
2

(5, 5)

O

x

O

(x 
 h)2
(y 
 k)2


 b
1
2
a2
(x 
 2)2
[y 
 (
3)]2


 20  1
42
(x 
 2)2
(y  3)2
 
   1
16
20

(5, 3)

x

(1, 2)

y
(2, 4.25)

y 
 3.75
(2, 4)

y
1

x
O

369

x

Chapter 10

44. x  2 sin t

36. vertex: (h, k)  (
1, 3)
(y 
 k)2  4p(x 
 h)
(7 
 3)2  4p[
3 
 (
1)]
16  8p
2p
Since parabola opens left, p  
2.
(y 
 k)2  4p(x 
 h)
(y 
 3)2  4(
2)[x 
 (
1)]
(y 
 3)2  
8(x  1)

37. vertex: (h, k)  5,

38.
39.
40.
41.
42.

2
4

2

x

2

t

2

1
0
1
2

x

2

1
0
1
2

y

1
2
3
2

1

y

3

 sin t
t

sin2

x 2

2

 

y 2
 3
x2
y2
  
4
9

 

1
1

x
0

y
3

(x, y)
(0, 3)



2

2

0

(2, 0)



0


3

(0, 
3)

3

2


2

0

(
2, 0)

y

1
1
3.5

(x, y)
(0, 
1)
(2, 1)
(3, 3.5)

y
t
0
t


2
t 
 3

2

x

O
t



45. x  
t
x2  t

(x, y)
(
2, 
1)
(
1, 2)
(0, 3)
(1, 2)
(2, 
1)

t

y  2 
 1
x2

y  2 
 1
t
0
4
9

y

x
0
2
3

y

t
9

x

O

O
43. x  cos 4t
cos2 4t  sin2 4t  1
x2  y 2  1
t
0

y  sin 4t

x
1

y
0

(x, y)
(1, 0)



8


4

0

1

(0, 1)


1

0

(
1, 0)

3

8

0


1

(0, 
1)

x2
y2
  
49
49
x 2
y 2

 7
7
sin2 t  cos2 t

 

x 2

7




t
8



t
4

x

7

t
0

O

 

 sin2 t
 sin t

x  7 sin t

x
t

t
0

x

46. Sample answer:
Let x  t.
y  2x2  4
y  2t2  4, 
  t  
47. Sample answer:
x2  y2  49

y

Chapter 10

 cos t

t1

cos2

t
0

 or (5, 
1)

focus: (h, k  p)  (5, 2)
kp2

1  p  2
p3
(x 
 h)2  4p(y 
 k)
(x 
 5)2  4(3)[y 
 (
1)]
(x 
 5)2  12(y  1)
A  5, c  2; ellipse
A  C  0; equilateral hyperbola
A  C  5; circle
C  0; parabola
y  
t2  3
y  
x2  3

y  3 cos t

3


 8

370

1
1
1

y 2

7

y

7

 cos2 t
 cos t

y  7 cos t, 0  t  2p

54. B2 
 4AC  (
6)2 
 4(1)(9)
0
parabola

48. Sample answer:
x2
y2
  
36
81
x 2
y 2

 9
6
cos2 t  sin2 t

 

x 2

6

x

6

 

1
1

B


tan 2v  
A
C

1

y 2

9

 cos2 t

y

9

 cos t

x  6 cos t
49. Sample answer:
Let y  t.
x  
y2
x  
t2, 
  t  
50. B2 
 4AC  0 
 4(4)(9)
 
144
A C; ellipse


6


tan 2v  
1
9

 sin2 t

3

tan 2v  4

 sin t

4x2  9y2  36

p 2
p
p 2
x
 cos  y
 sin 6  9 
x
 sin 6  y
 cos 6
2
2
1
1

3


3

4 2x
  2y
  9 
2x
  2y

3
1

3

4 4(x
)2  2x
y
  4(y
)2
1
3

3

 9 4(x
)3 
 2x
y
  4(y
)2
3(x
)2  2
3
x
y
  (y
)2

4



p

6






2v  36.86989765°
v  18°
55. (x 
 1)2  4(y 
 1)2  20
(y 
 1)2  4(y 
 1)2  20
5(y 
 1)2  20
(y 
 1)2  4
y 
 1  2
y  3 or 
1
y3
xy
x3
y  
1
xy
x  
1
(3, 3), (
1, 
1)

y  9 sin t, 0  t  2p






9

9
3


27

5
3


31

 36
 36

y

 4(x
)2 
 2 x
y
  4(y
)2  36
21
(x
)2
4

O

2

56. 2x 
 y  0
2x  y
y2  49  x2
(2x)2  49  x2
4x2  49  x2
3x2  49
x  4.04
2x 
 y  0
2(4.04) 
 y  0
y  8.08
(4.0, 8.1), (
4.0, 
8.1)


 42x
 
 2y
  0

2


x

(1, 1)


 2 x
y
  4(y
)2  36
21(x
)2 
 10
3
x
y
  31(y
)2 
 144  0
2
51. B 
 4AC  0 
 4(0)(1)
0
parabola
y2 
 4x  0
2
(
x
 sin 45°  y
 cos 45°) 
 4(x
 cos 45°  y
 sin 45°)  0



22 x
  
22 y


(3, 3)

 36


2


1
1
(x
)2 
 x
y
  (y
)2 
 2
2
x
  2
2
y
  0
2
2
x
  4
2
y
  0
(x
)2 
 2x
y
  (y
)2
 4
2

52. B2 
 4AC  0 
 4(4)(
16)
 256
hyperbola
4x2 
 16(y 
 1)2  64
4(x 
 h)2 
 16(y 
 k 
 1)2  64
4(x 
 1)2 
16(y  2 
 1)2  64
4(x 
 1)2 
 16(y  1)2  64
4(x2 
 2x  1) 
 16(y2  2y  1) 64
4x2 
 8x  4 
 16y2 
 32y 
 16 
 64  0
x2 
 4y2 
 2x 
 8y 
 19  0

2x 
 y  0
2(
4.04) 
 y  0
y  
8.08

y
(4.0, 8.1)

(4.0, 8.1)

O

x

 
 4(6)(8)
53. B2 
 4AC  2
3
2

A

 
180
C; ellipse
B


tan 2v  
A
C
2

3


tan 2v  
6
8

tan 2v  

3

2v  
60°
v  
30°

371

Chapter 10

57. x2 
 4x 
 4y  4
x2 
 4x 
 4  4y
(x 
 2)2  4y  0
2
(x 
 2)  x2 
 4x 
 4  0
2
x 
 4x  4  x2 
 4x 
 4  0
2x2 
 8x  0
2x(x 
 4)  0
2x  0
x
40
x0
x4
(x 
 2)2  4y  0
(x 
 2)2  4y  0
(0 
 2)2  4y  0
(4 
 2)2  4y  0
y  
1
y  
1
(0, 
1), (4, 
1)

61.

Page 691

r  

(12 

0)2 
(16 

0)2
r  20
x2  y2  r2
x2  y2  202
x2  y2  400
63b. area of watered portion  pr2
 p202
 1256.6 ft2
area of backyard  
q
 50(40)
 2000 ft2
area of nonwatered portion  2000 
 1256.6
 743.4 ft2
743.4

percent not watered  
2000

(4, 1)

58. xy  
4
4

y  
x
x2  y2  12
4 2

x2  
x  12
16

x2  x2  12

 0.37

x4  16 
 12x2  0
(x2)2 
12(x2)  16  0

x2 

12  

122 

4(1)(1
6)

2(1)

about 37%

(3.2, 1.2)
(1.2, 3.2)

60.

y
4
4 O

4

x

4

Chapter 10

c2

(y 
 k)2

 b
1
2
(y 
 0)2

34,560,000
y2

34,560,000

1
1

65. a  3.5
b3
a2 
 b2
c  

c  

3.52 

32
c  1.8
about 1.8 feet from the center

x

x

a2

(x 
 h)2

a2
(x 
 0)2

60002 
x2
 
36,000,000

O

O

c


0.2  
6000




b2  60002 
 12002
b2  34,560,000

(1.2, 3.2)

y

e  a
1200  c

b2

y

59.

c

64. 2a  2,000
a  6000

or
x2  1.528
x2  10.472
x   3.236
x  1.236
xy  
4
xy  
4
3.236y  
4
1.236y  
4
y  
1.236
y  
3.236
xy  
4
xy  
4

3.236y  
4

1.236y  
4
y  1.236
y  3.236
(3.2, 
1.2), (
3.2, 1.2), (1.2, 
3.2), (
1.2, 3.2)

(3.2, 1.2)

Applications and Problem Solving

63a. r  

(x2 
 
x1)2 
(y2 

y1)2

x


b  

b2 
 4
ac

2a

O

x

O 2

O

x2 

y

2

y

(0, 1)

62.

y

372

x

Page 691

Open-Ended Assessment

3. The information in the question confirms the
information given in the figure. Recall the formula
for the area of a triangle — one half the base
times the height. The triangle DCB is obtuse, so
the height will lie outside of the triangle.
Let D
C
 be the base. The length of the base is 6.
The height will be equal to 7, since it is a line
segment parallel to A
D
 through point B.

1. Sample answer:
c

e  a
1

9

c

 a

Let a  9.
1

9

c

 9

b2  a2 
 c2
b2  92 
 12
b2  80

c1
x2
y2


a2  b2
x2
y2


92  80
x2
y2
  
81
80

1

1

A  2bh
1

 2(6)(7) or 21

1

The correct choice is A.
4. The problem asks how many more girls there are
than boys. First find how many girls and how
many boys there are in the class.
One method is to find the fraction of girls in the
whole class and the fraction of boys in the whole
class. Since the ratio of girls to boys is 4 to 3, the
4
fraction of girls in the whole class is 7. Find the
number of girls in the class by multiplying this
fraction by 35.
4
(35)  20
There are 20 girls in the class.
7

1

2. Sample answer:
axis of symmetry: x  h
x  2, so h  2
focus: (h, k  p)  (2, 5)
kp5
Let k  2, p  3.
(x 
 h)2  4p(y 
 k)
(x 
 2)2  4(3)(y 
 2)
(x 
 2)2  12(y 
 2)

3

Using the same process, the fraction of boys is 7.
3
(35)
7

SAT & ACT Preparation
Page 693

SAT & ACT Practice

Red
3
60

There are 15 boys in the class.

So there are 5 more girls than boys. The correct
choice is D.
Another method is to use a “Ratio Box.” First
enter the given information, shown in the darker
cells below. Then enter the number for the total of
the first row, 7. To go from the total of 7 to the
total of 35, you must multiply by 5. Write a 5 in
each cell in the second row.

1. Add the two numbers of parts to get the whole, 8.
3
The fraction of red jelly beans to the whole is 8.
The total number of jelly beans is 160. The
3
number of red jelly beans is 8(160) or 60. The
correct choice is C.
Or you can use a ratio box. Multiply by 20.
Green
5

 15

Girls
4
5
20

Whole
8
160

Boys
3
5
15

Total
7
5
35

Then multiply the two numbers in the first
column to get 20 girls, shown with a dark border.
Multiply the second column to get 15 boys.
Subtract to find there are 5 more girls than boys.
5. Set A is the set of all positive integers less than
30. Set B is the set of all positive multiples of 5.
The intersection of Sets A and B is the set of all
elements that are in both Set A and Set B. The
intersection consists of all positive multiples of
5 which are also less than 30. The intersection
of the two sets is {5, 10, 15, 20, 25}.
The correct choice is A.

2. Notice the capitalized word EXCEPT. You might
want to try the plug-in method on this problem.
Choose a value for b that is an odd integer, say 1.
Then substitute that value for b in the equation.
a2b  122
2
a (1)  122
a2  122
a  12
Check the answer choices for divisors of this value
of a. 12 is divisible by 3, 4, 6, and 12, but not by 9.
The correct choice is D.

373

Chapter 10

6. For a quadratic equation in the form
y  a(x – h)2  k, the coordinates of the vertex
of the graph of the function are given by the
ordered pair (h, k). So the vertex of the graph
1
of y  2(x – 3)2 + 4 has coordinates (3, 4).
The correct choice is C.
7. On the SAT, if you forget the relationships for 45°
right triangles, look at the Reference Information
in the gray box at the beginning of each
mathematics section of the exam. The measure of
each leg of a 45–45–90 triangle is equal to the
length of the hypotenuse divided by 
2
. Multiply
both numerator and denominator by 
2
 and
simplify.
BC

9. Let d represent the number of dimes in the jar.
Since there are 4 more nickels than dimes, there
are d  4 nickels in the jar. So, the ratio of dimes
d
to nickles in the jar is . This ratio is less
d4
than 1. The only answer choice that is less than
8
d
8
1 is choice A, . If   , then d  16. So,
10
d4
10
there are 16 dimes and 16  4 or 20 nickels in the
8
16
jar, and   .

total liters

total bottles
8

20

8



2

8

2

 


2
 
2

8
2



 2 or 4
2

1
x
7
1
x
7

x

 1
2

x  0.4 or 5
The correct answer is .4 or 2/5.

100

 x
1

 5(100)
 20

x  140
The correct choice is E.

Chapter 10

x liters



1 bottle

20x  8

The correct choice is D. You could also use the
Pythagorean Theorem and the fact that the two
legs must be equal in length, but that method
might take more time.
8. Form a ratio using the given fractions as
numerator and denominator. Write a proportion,
using x as the unknown. Multiply the crossproducts. Solve for x.
1

7

1

5

10

20

The correct choice is A.
10. Set up a proportion.

374

Chapter 11 Exponential and Logarithmic Function
3


Page 695

3


8. 32 5 
 (25) 5

Real Exponents

11-1

15



 25


 23

8
2
9. (3a )3  3a5 
 33  a6  3a5

 34a1

Graphing Calculator Exploration

1.

81


 81a1 or a
1


1


10. 
m3n2  
m4n5 
 (m3n2) 2  (m4n5) 2
3


2


7


7


4


5



 m2n2  m2n2
2.

11.


 m 2 n 2 or m3n3mn


 
 

8n  27

4 n

1

2

8n  27

4 n



n


7


82  22


 
n


4 2
n


5.

a m

b


 


 bm, when b







2 2  22

2n

3n


3. am  an 
 amn
4. (am)n 
 amn
am

7


(23) 2  2 2

n

(22) 2


2

2

0

7


3n

2

7


2n



 2  22  2 2
5n  7

2


2

Page 700

n1

2

Check for Understanding


 22n3  2

1. The quantities are not the same. When the
negative is enclosed inside of the parentheses and
the base is raised to an even power, the answer is
positive. When the negative is not enclosed inside
of the parentheses and the base is raised to an
even power, the answer is negative.
2. If the base were negative and the denominator
were even, then we would be taking an even root
of a negative number, which is undefined as a real
number.
3. Laura is correct. The negative exponent of 10
represents a fraction with a numerator of 1 and a
denominator of a positive power of 10. The product
of this fraction and a number between 1 and 10 is
between 0 and 1.
1

4. 54 
 54

5.

1




625

2


196 






1

3

6. 216 
 (63)
3

3


6

6

1


1


1


1


4


8



 22  x2  y2
1



 2 2  x2y4 or x2242

1


13. 
169x5 
 (169x5) 2
1


1



 169 2  (x5) 2
5



 13x 2

1


14. 
a2b3c4
d5 
 (a2b3c4d5) 4
4

1


1


1


1



 (a2) 4 (b3) 4 (c4) 4 (d5) 4
1


3


5



 a 2 b 4 c d 4
1


3 1
 

1


1


15. 6 4 b 4 c 4 
 (6b3c) 4

1


5


3


16. 15x 3 y 5 
 15x 15 y 15

1



 
6b3c
4


 15 (x5y3) 15

 15 
x5y3
15

17. 
p4q6r5 
 (p4q6r5)
3

1


1

2

1

2

1

3
1


1



 (p4) 3 (q6) 3 (r5) 3

1

2

7. 27
  3
 
 27  3

4


5



 p 3 q2r 3

1

2


 (33)  3
3

2


 22n3 
2n1

12. (2x4y8) 
 2 2  (x4) 2  (y8) 2


 


 



1

3

1

9 2

16
16 2

9
162

2
9
256

81

1

2


 pq2r
pr2
3

1

2


3 3

18.

4



 32

 32

9

4

5

y 
 34
4 5
 
5 4


y 

5



 34 4

1


y 
 (345) 4
y

375

82.1
Chapter 11

19. A 
 r2
r 
 3.875  107 m
A 
 (3.875  107 m)2

 (3.875)2  (107)2 m2

 (15.015625  1014) m2
4.717  1013 m2

33. 
216
 
 (216 3 )2

Pages 700–703

34. 81 2  81

20.

(6)4






21.

22. (5  3)2 


 225

23.

2


2



 (63) 3

 62

 36


 

1
64 
  64
1


 
1296
24


 24(1)
2 1

 25

3


78

8

35.


 


 

36.

1


7
(128
)4





1

4




(128) 7
1

4

1

1


 y8  y8

1


 
1
1
  
3
9




1


27

3

38.

1

4

9
9

4


(4y4) 2

3

2


1

3



4 (y4) 2
3


6


1


12



 22y 2

1


27. 729 3 
(93) 3

9


 8y6

1

39.


(27p3q6r1) 3

33

2




1

2

3


1

2

1


1


1


1


6


1

1

1


1

1




28x8

1



 28x8 or


 26

1


1

4

30. 64 2 
 (26) 12

31. 16

1



 2 2 or 2


1

3


40. [(2x4]2 
 
[(2x)4]2

1



 2  62




1

1




16 4
1

1




(24) 4
1

2

1


1


1


41. (36x6) 2 
 36 2 (x6) 2




 6x3

1

256x8
b2n

42. 
b2n




1

2



1



 (b2n  b2n) 2
1



 (b4n) 2

 b2n



43.

37(32)4

6

2n

1
4n


2






1
2

2n  n

4
1


(33) 2

2n 2

 
4

315


 3
9

 36

 729

Chapter 11

3



 3pq2r

1

2


 22  22  62




1


1

r3


3

(37)(94)

276


1


1



 33p3q3

29. 2  12 
 2  (2  6)

32.

1


1


33

1


1



 27 3 (p3) 3 (q6) 3 
r 3


 (33) 3 (p3) 3 (q6) 3 
r 3


(33) 3


 32
1

2

3



 (22) 2 (y4) 2

26. 81 2 
 (92) 2

9

3



[(2)7] 7
1



 (2)
4
1

 1
6
2
3
(3n ) 
 33(n2)3

  y8
37. (y2)4  y8 
 
(y2)4

1

27

2




 27n6


25. (31  33)1 
 
31  31

28.




 89

7

8
8 3

7
83

73
512

343

1


1

1

1

1




1



 (92) 2 

 9  9


 
3




1

2

(9 2) 2


 32
24.

1



 216 3

1


Exercises

1

(6)4
1

1296
152

2

3




376

n


2

1


4

1


4

1


4

1


44. 
3m 2  27n 4  
 34
m 2  274
n4 

 34m2(33)4n
45.

1
f 1 6
4

256g4h4







 (f16  2561  g4  h4)
1
4

(2561)

1
4

1
4

(h4)

1

63.

46.


x2
x 
x  
 x2
x


3

4

6

3

4

3

4


 x2  x

1

3

47. 
2x y

1

4

3

3

4

6

4

2

3


3x y  
 6x

4

3

2

2

4


a5b7c
3

1


1


1


1


1


64. 
20x3y6 
 (20) 2 (x3) 2 (y6) 2

1

6



1



 (a5) 3 (b7) 3 (c) 3
3


 2xy35x

1

6


x  1

 x 
x  1
5

4

1

4

x

1



 ab2
a2bc

4 f4g

h

or

1



 defd


4


 f 4 256  g  h1
f 4gh1

1


62. d3e2f 2 
 (d3) 2 (e2) 2 (f 2) 2

4

1

(g4)

 0.33

 0.69

1

4


4

1

5

3

3


 316m2n

 (f16)


)
(245


61. x 


14.2 
 x

65.

3

2

2



(14.2) 3 
 (x
0.17 x

1

6

3

3

66.

3

2

5


2 3

712 
 15a 2

)

712

15
2
712 
 5
15

y


 

1

2


 6x y
48.

m

1 1
 

1



a
 
 
a n  m
n


a

a

a

4.68

1


49. 
m6n 
 (m6) 2 
n 2 

1 1
 
n m


 m3n

5


5 2
 

1


x 3.79
68. d 
 6.794  103 km so r 
 3.397  103 km
4
V 
 3r3

1


1


3


4


 x 2 y 2
1

3

1

3

51. 
8x3y6 
 8 (x3) (y6)


 3(3.397  103 km)3

1

3

1.64  1011 km3
10 1 2 10 5 7
69. y 
 3x; x 
 8, 6, 5, 33, 2, 3, 9, 3, 2


 2xy2

1


1


1


52. 17
x14y7z12
 
 17(x14) 2 (y7) 7 (z12) 7
7




12


17x2yz 7
1


1


1


53. 
a10b2  
c2 
 (a10) 5 (b2) 5 (c2) 4
4

2

5

1

2


 a2b c
1


1


1


8 27 8
54. 60
r80s56
t27 
 60(r80) 8 (s56)
(t )
27

8
10

 60r st
8

5

8

4


2

3

3 2
 

57. p 3 q 2 r 3 
 p 6 q 6 r 6

58.

6

3

21

1

3

2

1


23


 
p4q3r2
1

7

3

8

8

1



 23
3


 2


7

21

59. 13a b 
 13a b
1

60.


 13


(n3m9) 2

3x

y

8

38

6

36

5

35

9

6561
1

729
1

243


a3b7

10


3 33
1

2

3

2

3

3

10

9

3

5

3

3

7

2

3

1.395
1.732
2.080
3.389
14.620
46.765

21

1


69a. If x
0 then y  0. If x 
 0 then y 
 1. Since
x
0, y  0 and y 1. So, 0 y 1.
69b. If x  0 then y  1. If x 1 then y 3. So,
1 y 3.
69c. If x  1 then y  3. So, y  3.

1



 (n3) 2 (m9) 2
3


x

10

33
1

2
2

3
10

9
5

3
7

2

56. (7a) b 
 
75a5b3

5

55. 16 
 16

1 1
 

2


(x 2 ) 5 
 (28) 5

1

mn

50. 
xy3 
 (x) 2 (y3) 2

2


a

x 2 
 28

mn

1

5

5 2
 


 (a 2 ) 5

1


 a


5

5



 a2

x5 
 3.5
67. 8

1

2

1

nm

3

5


724 
 15a 2  12

9



 n2m2


 nm4mn


377

Chapter 11

69d. If the exponent is less than 0, the power is greater
than 0 and less than 1. If the exponent is greater
than 0 and less than 1, the power is greater than 1
and less than the base. If the exponent is greater
than 1, the power is greater than the base.
Any number to the zero power is 1. Thus, if the
exponent is less than zero, the power is less than
1. A power of a positive number is never
negative, so the power is greater than 0.
Any number to the zero power is 1 and to the
first power is itself. Thus, if the exponent is
greater than zero and less than 1, the power is
between 1 and the base.
Any number to the first power is itself. Thus, if
the exponent is greater than 1, the power is
greater than the base.

n factors




















m factors













m factors







m factors

74b. (am)n 
 a  a  ...  a  a  a  ...  a  ...  a  a  ...  a 

m  n factors







a  a  ...  a 
 amn
m factors













74c.

m factors







m factors

(ab)m


 ab  ab  ...  ab 
 a  a  ...  a  b  b  ...  b 
 ambm
m factors







a m

b


 

74e.

am

an




a

am

a

 b  ...  b 
 bm







74d.

a

b

a  a  ...  a

m factors







 mn


n factors a
a  a  ...  a

y

75.

1


70. r 
 (1.2  1015)A 3
If r 
 2.75  1015 then
1

2.75  105 
 (1.2  1015)A 3
2.75  1015

1.2  1015


A

1


2.29 A 3
12.04 A
Since 12.04 12, which is the mass number of
carbon, the atom is carbon.
71.
32(x 4x) 
 16(x 4x3)
(25)(x 4x) 
 (24)(x 4x3)
2(5x 20x) 
 2(4x 16x12)
5x2  20x 
 4x2  16x  12
x2  4x  12 
 0
(x  6)(x  2) 
 0
x6
0
x2
0
x 
 6
x
2
2

76.

2

2

2

2

72a.

2

Wind Speed
5
10
15
20
25
30

G M e t2

42




Wind Chill
0.8
12.7
22.6
29.9
35.3
39.2

O
2

4

1


r




23
  22
2


3


 Arctan 3



4

 6


So, (23
  2i) 
 4
cos 6  i sin 6.
Use De Moivre’s Theorem.

4
cos 6  i sin 6

1

5

1


1 

1 


45 
cos 
5  6sin 
5  6
1

5

G 
 6.67  1011



1

5




 4 cos 30  4 i sin 30

 1.31  0.14i
78.

2
3


2


3

6

5
6


1 2 3 4

7
6

mn factors



















n factors

2


v 
 Arctan 
23


 16


r  42,250,474.31 m
73b. 42,250,474.31 m 
 42,250.47431 km
42.250.47431  6380 
 35870.47431
35,870 km

3
2

Lemniscate

378

0

11
6
4
3

74a. aman 
 a  a  ...  a  a  a  ...  a 
 a  a  ...  a 
 amn

Chapter 11

x

6

77. (23
  2i) 5
Convert to polar form r(cos v  i sin v).

(6.67  1011)(5.98  1024)(86,400)2

42

m factors

y

2
4
6
8

Mt 
 5.98  1024
t 
 1 day 
 86,4000 seconds
r3

y2 
 12x
(y  0)2 
 4(3)(x  0)
Vertex is at (0, 0) and p 
 3. The parabola opens
to the right so the focus is at (0  3, 0) or (3, 0).
Since the directrix is 3 units to the left of the
vertex, the equation of the directrix is x 
 3.
8
6
4
2

72b. A 5-mile per hour increase in the wind speed
when the wind is light has more of an effect on
perceived temperature than a 5-mile per hour
increase in the wind speed when the wind is
heavy.
73a. r3 


x

O

1

3

5
3

u sin v  1qt2  h.
79. Use the equation y 
 tv
2
u
v  
 105
g 
 32
h
3
v 
 42

87. The time it takes to paint a building is inversely
proportional to the number of painters.
k
48 
 8

1

y 
 t(105)(sin 42°)  2(32)t2  3

k 
 384


 16t2  (105 sin 42°)t  3
Find t when y 
 0 (i.e., the ball is on the ground).
t


384

So t 
 1
6
t 
 24
The correct choice is E.

105 sin 42° (105
in
s)
42°16)(
 4(3)


2(16)

t 
 0.04, 4.43
So, the ball hits the ground after about 4.43 s.
u
80. TC 
 (2  3), (6  (4)), (5  6)

 1, 10, 11
u
TC  
 
(2  3
)2  (6
 (
4))2 
(5 
6)2

 222

81. Sample answer:
1
tan S cos S 
 2
sin S

cos S

cos S

11-2
Page 705

Exponential Functions
Graphing Calculator Exploration

1. positive reals
2. (0, 1)
3. For a 
 0.5 and 0.75, y →  as x → , y → 0 as
x → . For a 
 2 and 5, y → 0 as x → , y →  as
x → .
4. horizontal asymptote at y 
 0, no vertical
asymptotes
5. Yes; the range of an exponential function is all
positive reals because the value of any positive
real number raised to any power is positive.
6. Any nonzero number raised to the zero power is 1.
7. The graph of y 
 bx is decreasing when 0 b 1
because multiplying by number between 0 and 1
results in a product less than the original number.
The graph of y 
 bx is increasing when b  1
because multiplying by a number greater than 1
results in a product greater than the original
number.
8. There is a horizontal asymptote at y 
 0 because a
power of a positive real number is never 0 or less.
As a number between 0 and 1 is raised to greater
and greater powers, its value approaches 0. As a
number greater than 1 is raised to powers
approaching negative infinity, its value
approaches 0.

1

 1 
 2
1

sin S 
 2
82. cot v 
 0

v 
 2
Period of cot v is  so
cot v 
 0

v 
 2  n, where n is an integer.
83. r  6h 
 150 m
150 m


r

6h

r 
 25 m/h
84. 90°, 270°
85. x3  25x 
 0
Degree of 3 so there are 3 complex roots.
x3  25x 
 0
x(x  5)(x  5) 
 0
x
0
x5
0
x5
0
x
5
x
5
3; 5, 0, 5
86.

Page 708

Check for Understanding

1. Power; in a power function the variable is the
base, in an exponential function the variable is
the exponent.
2. Both graphs are one-to-one, have the domain of all
reals, a range of positive reals, a horizontal
asymptote of y 
 0, a y-intercept of (0, 1), and no
vertical asymptote. The graph of y 
 bx is
decreasing when 0 b 1 and increasing when
b  1.
3. If the base is greater than 1, the equation
represents exponential growth. If base is between
0 and 1, the equation represents exponential
decay.
4. The graphs of y 
 4x and y 
 4x  3 are the same
except the graph of y 
 4x  3 is shifted down
three units from the graph of y 
 4x.

[5, 5] scl:0.5 by [5, 5] scl:0.5

abs. min; (0.75, 1.88)

379

Chapter 11

5.

x

y

1

1

9

0
1
2

1
3
9

y

11.
y 
 3x

x
2
1
0
1

y

y
9
3
1

12.

y 
 3x

1

9

x

y

2
1

3
34
1
32

0
1
2

4
2
0

Use (0, 0) as a test point.
?

?


20  4

0 14
0  3 
The statement is true so shade
the region containing (0, 0).
14.

y

x

O
y

15.

16.

 0.45%
9b. Use N 
 N0(1  r)t.
N  8,863,052(1  0.0045)20
N 9,695,766

x

y

1

1

2

0
1
2

1
2
4

y 
 2x

y

x y
2 4
1 2
0 1
1
1 2

y 
 2x

x
3
2
1
0

y 
 2x 3

x
y
3 1
2 2
1 4
0 8

O

x

y
O

x

y 
 2x 3

x

y

1

3
14

0
1
2

1
2
14

y

(0, 0)
0

?

?


42

2

0 1  2
0  1; no

O

x
1
0
1

y

y
5
1
y 
 ( 15 ) x

1

5

Exercises
O

y
17.

O

y 
 2x

x

x
2
1
0
1

y
4
2
1
1

2

380

x

4x  2

x

y

(0, 0)


 

? 1 0
0
2
?
0
1 

y  ( 12 ) x

O

Chapter 11

x
y

y
1
2
4
8

y

282,167
 
 40,309.57
7
40,309.57
  0.0045
8,863,052

10.

1
2
4

2x  4

8. Use N 
 N0 (1  r)t where N0 
 3750, r 
 0.25,
and t 
 2.
N 
 3750 (1  0.25)2

 3750 (0.75)2
 2109.38
9a. 9,145,219  8,863,052 
 282,167

Pages 708–711

0
1
2

x

O

O
13.

0

1

y

x

O
7.

y
1
2

x

O
6.

x

x

18.

y

(0, 0)

3

1

2

?

204

4
5
6

1
2
4

?

24

0
0
0

1

16

23b. The graph of y 
 3x is a reflection of the graph
of y 
 3x across the x-axis.

y

x

y  2x  4



x

O
19.

x
1
0
1

y

y
100
1

B

(0, 1)

1
(1, 100
)

x

O
20.

x
1
0
1

y
5
1

y
(1, 5)

y 
 5x

1

5

C
21.

x
1
0
1
A

(0, 1)

y
49
7
1

[10, 10] scl:1 by [10, 10] scl:1
23c. The graph of y 
 7x is a reflection of the graph
of y 
 7x across the y-axis.

y 
 0.01x

1

100

O

(1, 15 )

[10, 10] scl:1 by [1, 9] scl:1

x

1 x

23d. The graph of y 
 
2 is a reflection of the graph
of y 
 2x across the y-axis.

y
(0, 7)

y 
 71  x
(1, 1)

O

x

22.
[10, 10] scl:1 by [1, 9] scl:1
y
24a.
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2

[5, 5] scl:1 by [5, 5] scl:1
22a. The graph of y 
 5x is a reflection of y 
 5x
across the x-axis. The graph of y 
 5x is a
reflection of y 
 5x across the y-axis.
22b. The graph of y 
 5x  2 is shifted up two units,
while the graph of y 
 5x  2 is shifted down
two units.
22c. The graph of y 
 10x increases more quickly
than the graph of y 
 5x. The graphs are not the
same because 52x 10x.
23a. The graph of y 
 6x  4 is shifted up four units
from the graph of y 
 6x.

O 2 4 6 8 1012 141618 2022x
24b. y 
 9.25(1.06)50
y 170.386427 thousand
y $170,400
25a. y 
 (0.85)x

[10, 10] scl:1 by [1, 9] scl:1

381

Chapter 11

25b. y

28d. Sample answer: A borrower might choose the
30-year mortgage in order to have a lower
monthly payment. A borrower might choose the
20-year mortgage in order to have a lower
interest expense.
29a. P 
 4000, n 
 43, i 
 0.0475

1.3
1.2
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1

O

(1  0.0475)43  1

Fn 
 4000 

0.0475
535,215.918; $535,215.92
29b. P 
 4000, n 
 43, i 
 0.0525
(1  0.0525)43  1

Fn 
 4000 

0.0525

1 2 3 4 5 6 7 8x

611,592.1194 or $611,592.12
$611,592.12  535,215.92 
 $76,376.20
30. The function y 
 ax is undefined when a 0 and
the exponent x is a fraction with an even
denominator.
31a. Compounded once:
I 
 1000[(1  0.05)1  1]

 50; $50
Compounded twice:

25c. y 
 (0.85)12
0.14 or 14%
25d. No; the graph has an asymptote at y 
 0, so the
percent of impurities, y, will never reach 0.
26a. N 
 876,156(1  0.0074)15
978,612.2261 or 978,612
26b. N 
 2,465,326(1  0.0053)15
2,668,760.458 or 2,668,760
26c. 152,307 
 139,510(1  r)10
152,307

139,510




0.05 2

I 
 1000[
1  2  1]


 (1  r)10

152,307

139,510



1

10

50.625; $50.63
Compounded four times:


1r

0.05 4

I 
 1000[
1  4  1]

N 
 139,510(1  r)25

50.9453; $50.94
Compounded twelve times:

N 
 173,736.7334 or 173,737
26d. 191,701 
 168,767(1  r)10
191,701

168,767


 (1  r)10



168,767 
191,701

0.05 12

I 
 1000
1  12

1

10

51.1619; $51.16
Compounded 365 times:


1r

0.05 365


I 
 1000
1  
365 

N 
 168,767(1  r)25
N 
 232,075.6889 or 232,076
35


5280 ft

1 hr

  
27b. s 
 
1 hr  1 mi  3600 s


 0.517x
The return is 5.17%
0.05 365

Super Saver: I 
 x
1  
 1
365 


 6.16 ft/s

36.16


O 
 100(3 5 )
 5800.16
5800 units
28a. Pn 121,000, n 
 30  12 or 360,
i 
 0.075 12 or 0.00625


 0.513x
The return is 5.13%
Money Market Savings
31c.

1  (1  0.00625)360

0.00625

121,000 
 P 

P
846.04955; $846.05
28b. Pn 
 121,000, n 
 20  12 or 240,
i 
 0.0725 12 or 0.00604
121,000 
 P



0.0725 240

1  
1  1
2 
———




0.0725

12



365


(1  0.05) 
 
1  
365 
x

1

365

(1.05)
1

365

x



1
365

365[(1.05)  1] 
 x
0.04879 
 x; 4.88%
32. 4x2(4x)2 
 4x2(4)2(x)2
4x2




16x2



1


 4
33. y 
 15
Use y 
 r sin v.
y 
 15
So, 15 
 r sin v.
34. 3, 9  2, 1 
 (3)(2)  (9)(1)

3
3; no because the inner product does not equal 0.

P 956.35494; $956.35
28c. 30 year: I 
 360(846.05)  121,000

 $183.578
20 year: I 
 240(956.35)  121,000

 $108,524
Chapter 11

 1

51.2675; $51.26
31b. Let x represent the investment.
Statement savings: I 
 x[(1  0.051)1  1]

 0.051x
The return is 5.1%
0.0505 12
Money Market Savings: I 
 x
1  1
 1
2 

27a. O 
 100(3 5 )

 100(33)

 2700 units
4.2 mi

 1

382

35.


cos 78  i sin 78  33cos 
4  i sin 
4
1
7

7


 
3  33
cos 
8  4  i sin 
8  4
5
5

 3

cos 8  i sin 8

1

3

y

y 
 1x

y 
 x 1 3

O


 0.66  1.60i
36.
s 
 72t  16t2  4
s  4 
 16t2  72t
s  4  (16)(5.0625) 
 16(t2  4.5  5.0625)
(s  85) 
 16(t  2.25)2
Vertex: (2.25, 85)
Maximum height: 85 feet.
37a.

y

42.

x

The parent graph is translated 3 units left. The
vertical asymptote is now x 
 3. The horizontal
asymptote, y 
 0, is unchanged.
CAB 
 16
43. CAC 
 32
100.53
50.27
100.53  50.27 
 50.26 or about 50.
The correct choice is E.

(12, 8)

(7, 6)

11-3
O

x

Page 714
37b.
38.

39.

(12  (7)) (8  6)
, 
2
2




 
 (2.5, 7)

2



1 rev 
 4800
4800

V 
 9.2 1
138,732.73 or about 139,000 cm/s
40.

21˚
200 ft

x

tan 69° 


Check for Understanding

1. C
2. If k is positive, the equation models growth.
If k is negative, the equation models decay.
3. Amount in an account with a beginning balance of
$3000 and interest compounded continuously at
an annual rate of 5.5%.
4. reals, positive reals
5. Sample answer: Continuously compounded
interest is a continuous function, but interest
compounded monthly is a discrete function.
6a. growth
6b. 33,430
6c. y 
 33,430e0.0397(60)
361,931.0414 or 361,931
7. A 
 12,000e0.064(12)
25,865.412 or $25,865.41

sin4 A  cos2 A 
 cos4 A  sin2 A
(sin2 A)2  cos2 A 
 cos4 A  sin2 A
(1  cos2 A)2  cos2 A 
 cos4 A  sin2 A
2
(1  2cos A  cos4 A)  cos2 A 
 cos4 A  sin2 A
cos4 A  1  cos2 A 
 cos4 A  sin2 A
cos4 A  sin2 A 
 cos4 A  sin2 A
2900 rev

1

The Number e

x

200

200 tan 69° 
 x
521.02 
 x
about 521 feet

Pages 714–717

Exercises

8. p 
 (100  18)e06(2)  18

 42.6 or 43%
9a. y 
 84e0.23(15)  76

 78.66 or 78.7°F
9b. too cold; After 5 minutes, his coffee will be about
90°F.

41.

10a.

[4, 4] scl:1 by [1, 10] scl:1
10b. symmetric about y-axis
Sample answer: y 
 948.4x  4960.6

383

Chapter 11

11a. Annually: I 
 100[(1  0.08)1  1]

 80
$80.00; 8%

14b. 2 decimal places; 4 decimal places; 6 decimal
places
14c. always greater
15a. 5 days: P 
 1  e0.047(5)
 0.20943
20.9%
20 days: P 
 1  e0.047(26)
 0.60937
60.9%
90 days: P 
 1  e0.047(90)
 0.98545
98.5%
20.9%; 60.9% 98.5%

0.08 2

Semi-annually: I 
 1000
1  2  1

 81.6

$81.60; 8.16%

Quarterly: I 
 1000
1 
 82.4316

Daily: I 
 1000
1 



 1
$82.43; 8.243%

Monthly: I 
 1000
1 
 82.9995

0.08 4

4

0.08 12

12



 1
$83.00; 8.3%

0.08 365

365



 1

15b.

 83.2776
$83.28; 8.328%
0.08(1)
Continuously: I 
 1000(e
 1)
 83.2871
$83.98; 8.329%
Interest
Compounded
Annually
Semi-annually
Quarterly
Monthly
Daily
Continuously

Effective
Annual
Yield
8%
8.16%
8.243%
8.3%
8.328%
8.329%

Interest
$80.00
$81.60
$82.43
$83.00
$83.28
$83.29

about 29 days
15c. Sample answer: The probability that a person
who is going to respond has responded
approaches 100% as t approaches infinity. New
ads may be introduced after a high percentage of
those who will respond have responded. The
graph appears to level off after about 50 days.
So, new ads can be introduced after an ad has
run about 50 days.
16a. all reals
16b. 0 f(x) 1
16c. c shifts the graph to the right or left

r n

11b. continuously
11c. E 
 
1  n  1
r
11d. E 
 e  1
12a. y 
 525(1  e0.038(24))
 314.097
314 people
12b.

(1  0.035)8  1


17. 120,000 
 P 
0.035

after about 61h

120,000  P(9.051687)
P  13,257.19725
$13,257.20
8


3


1


18. x 5 y 5 z 5 
 x
x3y3z
5

19. y 
 6x2
v 
 45°
x sin 45°  y cos 45° 
 6(x cos 45°  y sin 45°)2

[0, 100] scl:10 by [0, 550] scl:50
13a. P 
 1  e6(0.5)
 0.95021
95%
13b.
0.02h

1



60 min

1h


 1.2
6x2

 12xy 

6y2

2
2

y
2

1
x2
2

1
y2
2


2


2
y
2


 6
 
2 x


2

2 x



2

y
2


 6



2 x

2

y
2


 3x2  6xy  3y2

2


x 
 0.02; about 0.02 h


2


2 x

 xy 

2
x  2
y 
 6x2  12xy  6y2
 2
x  2
y 
 0
1


(5)2 
 (1
)2
v 
 Arctan 
20. r 
 
5  

 26

 3.34
26
(cos 3.34  i sin 3.34)

[0, 1] scl:0.1 by [0, 1] scl:0.1
2(10)  1 10

2(10)  1

14a. For x 
 10: 




about 1.2 min
21 10

19





2(100)  1 100


For x 
 100: 

2(100)  1 

201 100



 

199 

 2.718304481

2(1000)  1 1000

2(1000)  1

For x 
 1000: 




 2.720551414



2001 1000



 

1999 

 2.718282055
2.720551414; 2.718304481; 2.718282055
Chapter 11

384





y

21.

Page 717

Mid-Chapter Quiz

1. 642 
 8
1

3
2. ()
343 2 
 (343 31 )2


 ((73) 3 )2
2

7
1

 4
9

(0, 150)

(x, y)

3.

d


8x3y6

1

3




27w z 
6 9

8x3z9
1


y
28˚

O




x

x

u
u
d 
 x, y
F 
 0, 150
x


cos 28° 
 10

2xz3

sin 28° 


4.

x

10



x 
 10 sin 28°
x 4.6947

28.

1


3



 a3b 2
1


5. (125a2b3) 3 
 
125a2
b3
3

 
53a2b3
3

 5b
a2
2
6. 1.75  10 
 0.094 
A3

3

2
2
1.75  10 
 0.94 A
3

1.75  102

0.94
2

2

3

3



 A2

3 2
 


 (A 2 ) 3

151.34  A
1.51  102 mm2
7.
1,786,691 
 1,637,859 (1  r)8
1,786,691

1,637,859

3x  2  6
3x  8
8
x  3


 (1  r)8

1

8

1,786,691
8



1,637,859  
 [(1  r) ]
1,786,691



1,637,859   1 
 r

1

8

1

8



4

3

 

 

1


1


r

Store the exact value in
your calculator’s
memory.
N 
 1,637,859 (1  0.011)24
Use the stored
value for r.

 2,216,156.979
2,126,157

3 2 2
3
9 6
6
9
25. 3


2
6 5 1
6 18 15 3
J
(9, 6), K
(6, 18), L
(6, 15), M
(9, 3); The
dilated image has sides that are 3 times the
length of the original figure.
4x  y
6
26.


x
2y  12
4x  y 
 6
4(2y  12)  y 
 6
x 
 2(6)  12
x 
 2y  12
9y 
 54 x 
 0
y
6
(0, 6)
27. {4, 2, 5}; {5, 7}; yes



2


a6b3 
(a4b3) 2

1.75  10



0.94

24. 3x  2  6
3x  2  6
3x  4
4
x  3
x

1



 (a6) 2 (b3) 2


23. 2x
3
4
2x  3 
 16
2x 
 13
13
x 
 2

x

1


(23) 3 (x3) 3 (z9) 3

1
1
1



(33) 3 (w6) 3 ( y6) 3

 or w2xy2z3


3
3w2y2

x 
 10 cos 28°
x 8.8295
d 
 8.8295, 4.6947
u u
W
F d
W 
 0, 150  8.8295, 4.6947

 0  704.205

 704.2 ft-lb
22. y 
 1.5(10)2  13.3(14)  19.4

 2.4

8
3

1


 3

 

27w6y6 





0.011

0.052 (4)(3.5)

8. A 
 3500 
1  4

 3500(1.013)14

 4193.728
$4193.73 48.1

9a. y 
 6.7e 15
 0.271292
271,292 ft3

The correct choice is D.

48.1


9b. y 
 6.7e 50
 2.560257
2,560,257 ft3

200


10. 2 years: n 
 
1  20e0.35(2)

 18.3
200


15 years: n 
 
1  20e0.35(15)

 181

4

200


60 years: n 
 
1  20e0.35(60)

 200
18.3; 181; 200

385

Chapter 11

2

14. log7 n 
 3 log7 8

Logarithmic Functions

11-4

2


log7 n 
 log7 8 3
2


Pages 722–723

n 
 83
n
4
15. log6 (4x  4) 
 log6 64
4x  4 
 64
x 
 15

Check for Understanding

y 
 3x and log3 x are
similar in that they are
y 
 3x
both continuous, one-toone, increasing and
inverses.
x
O
y 
 3x and log3 x are not
y 
 log3 x
similar in that they are
inverses. The domain of
one is the range of
another and the range
of one is the domain of
the other. y 
 3x has a
y-intercept and a
horizontal asymptote
whereas y 
 log3 x has a
x-intercept and a
vertical asymptote.
2. Let bx 
 m, then logb m 
 x.
(bx)p 
 mp
bxp 
 mp
logb bxp 
 logb mp
xp 
 logb mp
p logb m 
 logb mp
1.

y

1

16. 2 log6 4  4 log6 16 
 log6 x
1


log6 42  log6 16 2 
 log6 x
42

1
log6
 
 log6 x
16 4
42
1 
 x

16 4

x
8

17. x
y
1
0
2 1
4 2

y
y 
 log 1 x
2

x

O

18. x
1
6

y
0
1

y

y

3.

y 
 log 1 x
5

O

y  log6 x

x

O
y 
 log5 x

t


19. 16 
 
3.3 log 1024
4

t 
 16(33 log4 1024)
t 
 16(33  5)
t 
 264 h

Log5 x is an increasing function and log1 x is a
5
decreasing function.
4. Sean is correct. The product property states that
logb mn 
 logb m  logb n.

log4 1024 
 x
4x 
 1024
22x 
 210
2x 
 10
x
5

1

5. In half-life applications r 
 2. So, (1  r)
becomes 

N 
 N0(1




 

1
1
1  2 or 2 . Thus, the formula
1 t
 r)t becomes N 
 N0 2 .

3



 


215 

1

2

7.

8. log7 y 
 6

9. log8 4 
 3

1

1

2x 
 1
6
2x 
 24
x 
 4
1


12. log7 
343 
 x

7x




1

343
73



x 
 3

7x

Chapter 11

1

3


5

1

2

21. 16 2 
 4

13. log2 x 
 5
25 
 x
32 
 x

386

5



22. 74 
 
2401

23. 4 2 
 32

24. ex 
 65.98

25. 
6
 
 36

1

11. log10 0.01 
 x
10x 
 0.01
10x 
 102
x 
 2

Exercises
1


20. 27 
 3
1

6. 9 2 
 27

10. log2 16 
 x

Pages 723–725

4

3

26. log81 9 
 2

27. log36 216 
 2

28. log 18 512 
 3

29. log6 3
6 
 2

1

x

30. log16 1 
 0
32. log8 64 
 x

31. logx 14.36 
 1.238

8x 
 64
8x 
 82
x
2
33. log125 5 
 x

x 
 2
3
47. log1010

x
3
10x 
 10

10x 
 10

125x 
 5
(53)x 
 51
3x 
 1

1

3

1

x 
 3

1

1

x 
 3


 32
2x 
 25
x
5
35. log4 128 
 x

34. log2 32 
 x
4x 
 128
22x 
 27
2x 
 7
7
x 
 2 or 3.5

9x 
 96
x
6
37. log49 343 
 x

36. log9 96 
 x
49x 
 343
72x 
 73
2x 
 3

2x

1

1


1


x

2

8 
 84
x

2


4

x
8
40. 104 log102 
 x
41. logx 49 
 2
x2 
 49
x
7

39. log8 4096 
 x
10log1024 
 x
24 
 x
16 
 x

1


log7 (8x) 2 
 log7 16
1


(8x) 2 
 16
8x 
 256
x 
 32
52. 2 log5 (x  2) 
 log5 36
log5 (x  2)2 
 log5 36
(x  2)2 
 36
x2
6
x
8
53. x
y
1
0

42. log3 3x 
 log3 36
3x 
 36
x 
 12
43. log6 x  log6 9 
 log6 54
log6 9x 
 log6 54
9x 
 54
x
6
44. log8 48  log8 w 
 log8 6
48

log8 w 
 log8 6
48

w


6

6w 
 48
w
8
45. log6 216 
 x
6x 
 216
6x 
 63

1


x 
 9 2  27 3
x
33
x
9
49. log5 (x  4)  log5 8 
 log5 64
log5 (x  4)(8) 
 log5 64
(x  4)(8) 
 64
x4
8
x
4
50. log4 (x  3)  log4 (x  3) 
 2
log4 (x  3)(x  3) 
 2
42 
 (x  3)(x  3)
16 
 x2  9
25 
 x2
5
x
1
51. 2(log7 x  log7 8) 
 log7 16
1
(log 8x) 
 log 16
7
7
2

38. log8 16 
 x
x

8  
 4096

4

1


log12 x 
 log12 9 2  27 3

3

x 
 3

1


log12 x 
 log12 9 2  log12 27 3

x 
 2 or 1.5
8x 
 16
23x 
 24
3x 
 4

1

48. log12 x 
 2 log12 9  3 log12 27

2

1

2

4

1

y
y 
 log4 x

O

x

x
3
46. log5 0.04 
 x
5x 
 0.04
5x 
 52

387

Chapter 11

54. x
1
2
4

y
0
3
6

20
15
10
5

O
5
10
15
20
55. x
2
6

y

r 410

62a. 5000  2500
1  4
r 40

27  
1  4

y 
 3 log2 x
5 10 1520 2530 3540 x

r 40

1


2 40 
 
1  4



1

40

r

1.0175  1  4
0.0699  r
6.99%

y

y
0
1

r 40

2 
 
1  4

62b.

r 28

2 
 
1  4

62c.

y 
 log5 (x  1)

1


r

2 28 
 1  4
r

x

1.0251  1  4
0.1004  r
10.04%

56. x
1
2
4

y  log2 x

1

1
—

63a. n 
 log2

y

y
0
1
2

63b. 3 
 log2 p

1

4

1

23 
 p

n 
 log2 4

8 
 p1

2n 
 4
n
2

x

1

8


p
1

less light; 8

57. x
1
2
4

64. Let y 
 loga x, so x 
 ay.
x 
 ay
logb x 
 logb ay
logb x 
 y logb a

y

y
0
2
4

logb x


y

log a
b

loga x 


x
y  2 log2 x

58. x
0
9

logb x

logb a

P

65a.

16
15.5
P
15 log2.72 (14.7) 
 0.02h
14.5
14
13.5
13
12.5

y

y
0
1

y

log10 (x  1)

x

2 O

4

2

4h

P


65b. log2.72 
14.7 
 0.02(1)
P

14.7

59. Use N 
 N0(1  r)t; r 
 1 since the rate of growth
is 100% every t time periods.
64,000 
 1000 (1  1)t
64 
 2t
log2 26 
 log2 2t
6
t
t  15 
 90 min
60. All powers of 1 are 1, so the inverse of y 
 1x is
not a function.
61. Let logb m 
 x and logb n 
 y.
So, bx 
 m and by 
 n.
m
bx
xy
 
 
n
by 
 b
m

n
m

logb n
m
logb n

Chapter 11


 2.720.02

P 
 14.7 (2720.02)
14.4 psi
P


65c. log2.72 
14.7 
 0.02(6.8)

P 
 14.7(2.720.136)
16.84 psi
1 t

6.8 
 38
2

66.

log
log


 bxy

6.8

38
6.8

38
6.8

38

1 t


 
2

1


 t log 2

t



xy

1 t


 log 
2
6.8
log 
38
__
1
log 
2

t 
 2.5
2.5  3.82 
 9.55
about 9 days


 logb m  logb n

388

75. cos (A  B) 
 cos A cos B  sin A sin B
5
35
cos A 
 1
cos B 
 37
3

67. 69.6164
68. 90,000 
 P








0.115 12.30
1  1  
12
______
0.115

12



x2  52 
 132
x2 
 144
x 
 12
12
So, sin A 
 1
3

P  891.262
$891.26
69. ellipse
9x2  18x  4y2  16y  11 
 0
9x2  18x  4y2  16y 
 11
2
9(x  2x  1)  4(y2  4y  4) 
 11  9  16
9(x  1)2  4(y  2)2 
 36
(x 

4
1)2

(y 

x2  352 
 372
x2 
 144
x 
 12
12
So, sin B 
 3
7
5

35

12

12




cos (A  B) 
 1
3  37  13  37
175

144





481  481
31




481

2)2

76. y 
 A sin (kt  c)  h

 9 
 1

90  64

90  64

h 
 2

A 
 2

y


 13

2

k


4



 77

k 
 2


y 
 13 sin 
2t  c  77


64 
 13 sin 
2(1)  c  77

x

O



13 
 13 sin 
2  c


1 
 sin 
2  c


sin1(1)  2 
 c

70. r 
 3, v 
 2
(3 cos 2t, 3 sin 2t)

3.14

1  
(1))2 
 (3 
 3)2
71. AB 
 (
2
2

 
0  (
6)

6
BC 
 (
3  ((
1))2  
0  (
3))2
2
2

 
4  3

5
AC 
 (
3  (
1))2  
(0  3)2
2
2

 
4  3

5
3

3

So, y 
 13 sin 


2

3

2

3

2

2




 5  2 cos 
4  3  i sin 
4  3
17
17

cos 1
2  i sin 12
17
17

cos 1
2  10i sin 12


 10


 10



 2.59  9.66i
17

 3.14  77

77. c2 
 (6.11)2  (5.84)2  2(6.11)(5.84) cos 105.3
c2 
 37.3321  34.1056  71.3648 cos 105.3
c2 90.2689
c 9.5
(6.11)2 (5.84)2  (9.5)2  2(5.84)(9.5) cos A
37.3321 34.1056  90.25  110.96 cos A
cos A 0.7843
A 38.34 or 38° 20

B 180  (105° 18
  38° 20
)
36° 22

c 
 9.5, A 
 38° 20
, B 
 36° 22

78.
M

72. 5
cos 4  i sin 4  2
cos 3  i sin 3




c

k
2

P

c˚
x˚

Q
b˚

17


10
cos 1
2  i sin 12 , 2.59  9.66i

73. (3  4j)(12  7j) 
 36  21j  48j  28j2

 64  27j
64  27j volts
y
74.

S

N
mSMN 
 mQNM alternate interior angles
x° 
 b°  c° Exterior Angle Theorem
The correct choice is E.

C

A

c˚

50˚
50˚

D

O

x

B

Both vectors have the same direction. 50° south of
u
u
east. Therefore, AB and CD are parallel.

389

Chapter 11

Common Logarithms

11-5
Page 730

x1

4



x1

(x  3) log 3  log 2  4 log 4

Check for Understanding

(4x  12)log 3  4 log 2  (x  1) log 4
4x log 3  12 log 3  4 log 2  x log 4  log 4
4x log 3  x log 4  4 log 2  log 4  12 log 3
x(4 log 3  log 4)  4 log 2  log 4  12 log 3
x

4 log 2  log 4  12 log 3

4 log 3  log 4

x  4.84
17.

[10, 10] scl:1 by [20, 100] scl:10
5.5850
200

18a. R 
 log 

1.6   4.2

 6.3
18b. 10 times; According to the definition of
logarithms, R in the equation
R 
 log 
T  B is an exponent of the base of the
logarithm, 10. 105 is ten times greater than 104.
a

y

10.

Pages 730–732

y  log (x  3)

log 18


11. log12 18 
 
log 12

 1.1632
log 15


12. log8 15 
 
log 8

 1.3023
2.2x  5 
 9.32
(x  5) log 2.2 
 log 9.32
(x  5) 


22. log 0.06 
 log 
0.01  2
12

log 9.32

log 2.2


 log 0.01  log

x

7.83
x2
14.
6

 4x
(x  2) log 6 
 x log 4
x log 6  2 log 6 
 x log 4
2 log 6 
 x log 4  x log 6
2 log 6 
 x (log 4  log 6)
2 log 6

log 4  log 6

4.3x
x log 4.3

Chapter 11

1

1


 2  1.0792  2(0.6021)

 1.2218
23. log 36 
 log (4  9)

 log 4  log 9

 0.6021  0.9542

 1.5563
24. log 108,000 
 log (1000  12  9)

 log 1000  log 12  log 9

 3  1.0792  0.9542

 5.0334

8.84 x
76.2
log 76.2
log 76.2

log 4.3

x

2.97

12

1

42


 log 0.01  log 12  2 log 4


x

x

Exercises

19. log 4000,000 
 log (100,000  4)

 log 100,000  log 4

 5  0.6021

 5.6021
20. log 0.00009 
 log (0.00001  9)

 log 0.00001  log 9

 5  0.9542

 4.0458
21. log 1.2 
 log (0.1  12)

 log 0.1  log 12

 1  1.0792

 0.0792

x

15.

4

3x3  2
4

1. log 1 
 0 means log10 1 
 0. So, 10° 
 1.
log 10 
 1 means log10 10 
 1. So, 101 
 10.
2. Write the number in scientific notation. The
exponent of the power of 10 is the characteristic.
3. antilog 2.835 
 102.835 
 683.9116
4. log 15 
 1.1761
log 5 
 0.6990
log 3 
 0.4771
log 5  log 3 
 0.6990  0.4771 
 1.1761
5. log 80,000 
 log (10,000  8)

 log 104  log 8

 4  0.9031

 4.9031
6. log 0.003 
 log (0.001  3)

 log 103  log 3

 3  0.4771

 2.5229
7. log 0.0081 
 log (0.0001  34)

 log 104  4 log 3

 4  4(0.4771)

 2.0915
8. 2.6274
9. 74,816.95

13.

3x3  2 
4x1

16.

390

25. log 0.0048 
 log (0.0001  12  4)

 log 0.0001  log 12  log 4

 4  1.0792  0.6021

  2.3188
26. log 4.096 
 log (0.001  46)

 log 0.001  6 log 4

 3  6(0.6021)

 0.6124
1

27. log 1800 
 log 
100  9  4 2 

46.

log 3  7 log 2


x
log 3  log 2

x 
 9.2571
47. logx 6  1
log 6

log x

1


 log 100  log 9  2 log 4

 3.2553

log 6

29. 2.9515
31. 2.001
33. 2.1745
log 8


undefined. When x 1, 
log x is negative, which is
not greater than 1. So, x must also be greater than
1. Therefore, 1 x 6.
log 625


34. log2 8 
 
log 2


35. log5 625 
 
log 5


3

48.


4
log 24

log 4


36. log6 24 
 
log 6


37. log7 4 
 
log 7

1.7737
38. log6.5 0.0675 


0.7124

1

2

5 log 4  3 log 3


x
2 log 4  log 3

log 15

1
log 
2

40.

2x 
 95
x log 2 
 log 95

49.

log 95


x

log 2

41.

42x5  3x3
(2x  5) log 4  (x  3) log 3
2x log 4  5 log 4  x log 3  3 log 3
2x log 4  x log 3  5 log 4  3 log 3
x(2 log 4  log 3)  5 log 4  3 log 3

log 0.0675

log 0.5

3.8890
39. log 15 


1

log 6  log x
6x
log 6

When x 
 1, log 1 
 0, which means 
log x is

1


 2  0.9542  2(0.6021)
28. 1.9921
30. 0.871
32. 3.2769

3x1  2x7
(x  1) log 3  (x  7) log 2
x log 3  log 3  x log 2  7 log 2
x log 3  x log 2  log 3  7 log 2
x(log 3  log 2)  log 3  7 log 2

x 6.5699
5x 
 4x  3
x log 5 
 (x  3) log 4
x log 5 
 x log 4  3 log 4
x(log 5  log 4) 
 3 log 4

x  2.1719
0.52x4  0.15x
(2x  4) log 0.5  (5  x) log 0.1
2x(log 0.5)  4 log 0.5  5 log 0.1  x log 0.1
2x log 0.5  x log 0.1  5 log 0.1  4 log 0.5
x(2 log 0.5  log 0.1)  5 log 0.1  4 log 0.5
x

Change inequality sign because (2 log 0.5 
log 0.1) is negative.

3 log 4


x

log 5  log 4

x
42.

1

3

50. log2 x 
 3
x 
 23
x 
 0.1250

18.6377

log x 
 log 8
1

3

x 
8
x 512
43.

5 log 0.1  4 log 0.5

2 log 0.5  log 0.1

x  3.8725
51. x

log 52.7

log 3

x

3.6087

52.

0.1643x 
 0.38x
(43x) log 0.16 
 (8  x) log 0.3
4 log 0.16  3x log 0.16 
 8 log 0.3  x log 0.3
3x log 0.16  x log 0.3 
 8 log 0.3  4 log 0.16
x(3 log 0.16  log 0.3) 
 8 log 0.3  4 log 0.16
x


8 log 0.3  4 log 0.16

3 log 0.16  log 0.3

x

0.3434

[10, 10] scl:1 by [3, 3] scl:1

44. 4 log (x  3) 
 9

53.

9

log (x  3) 
 4
9

(x  3) 
 antilog 4
9

x 
 antilog 4 3
x 174.8297
45. 0.25 
 log 16x
0.25 
 x log 16

[1, 10] scl:1 by [1, 3] scl:1

0.25


x

log 16

x

0.2076

391

Chapter 11

54.

t
1 0.8

0.9535 
 
2

60b.

1

log 0.9535 
 0.8t log 2
log 0.9535

1
log 
2



log

x




 t log 0.8

12.0016 t
12 years
61. Sample answer: x is between 2 and 3 because 372
is between 100 and 1000, and log 100 
 2 and log
1000 
 3.

[10, 1] scl:1 by [2, 10] scl:1
55.

log 0.9535

1
log 
2


 0.8t

0.3210

1


62a. L 
 10 log 
1.0  1012


 10(log 1  log (1.0  1012))

 120 dB

62b.

[5, 5] scl:1 by [10, 50] scl:10
56.

x

I


20 
 log 
1.0  1012

2 
 log I  log (1.0  1012)
2 
 log I  12
10 
 log I
1  1010 
 I; 1  1010 W/m2

0.1975

1 t

63. Use N 
 N0
2 .

N 
 630 micrograms 
 63  104 gram
N0 
 1 milligram 
 1.0  103 gram
1 t

6.3  104 
 (1.0  103)
2

[5, 5] scl:1 by [3, 10] scl:1

6.3  104

1



log 
1.0  103 
 t log 2

x
2

57.

0.6666 t
0.6666  5730 3819 yr
64. loga y 
 loga P  loga q  loga r
p

loga y 
 loga q  loga r
pr

loga y 
 loga q
pr

y 
 q

[5, 5] scl:1 by [5, 10] scl:1
100

65. logx 243 
 5
x5 
 243
x
3

10.3


58a. h 
 9 log 
14.7

1.7 mi
100

P


4.3 
 9 log 
14.7

58b.

66.

0.3870 
 log P  log 14.7
0.3870  log 14.7 
 log P
0.7803 log P
6.03 P; 6 psi
59a. M 
 5.3  5  5 log 0.018
1.58
59b.
5.3 
 8.6  5  5 log P
8.3 
 5 log P
1.66 
 log P
0.0219 P

increasing from  to 
1 2
 

9
1 0.8

60a. q 
 
2

1 0.1342


 
2


 0.9112
$91,116

Chapter 11

1


1


1


67. (a4b2) 3 c 3 
 (a4) 3 (b2) 3 (c2) 3
3

 a 
ab2c2
2
2
68. (5)  (0)  D(5)  E(0)  F 
 0
5D  F  25 
 0
2
2
(1)  (2)  D(1)  E(2)  F 
 0
D  2E  F  5 
 0
2
2
(4)  (3)  D(4)  E(3)  F 
 0
4D  3E  F  25 
 0

392

5D  0E  F  25 
 0
() D  2E  F  5 
 0
4D  2E
 20 
 0
4D  3E  F  25 
 0
() D  2E  F  5 
 0
3D  E
 20 
 0
4D  2E  20 
 0
2(3D  E  20 
 0)
10D  60 
 0
D 
 6
4(6)  2E  20 
 0
2E  4 
 0
E
2
5(6)  0(2)  F  25 
 0
F5
0
F
5
x2  y2  6x  2y  5 
 0
(x2  6x  9)  (y2  2y  1) 
 5  9  1
(x  3)2  (y  1)2 
 5
69.
70.

74. f(x) 
 x3  2x2  11x  12
f(1) 
 1  2(1)  11(1)  12
Test f(1).
f(1) 
 0
(x  1) is a factor.
1
1
2
11
12
1
1
12
1
1
12 
0
x2  x  12 
 0
(x  4)(x  3) 
 0
So, the factors are (x  4)(x  3)(x  1).
y
75.

y 
 5x 3  2x  5

Neither; the graph of the function is not
symmetric with respect to either the origin or the
y-axis.
76. 7  5  4  1 
 17
17,000,000
The correct choice is D.

65
  25
 18  4
,  
 (25
, 11)


2
2

r
6
r2 
 36
x2  y2 
 36

71.

90˚

120˚

60˚
30˚

150˚
180˚

11-6
Page 735

330˚

210˚
270˚

Check for Understanding

1. ln e 
 1 is the same as loge e 
 1. And e1 
 e. So,
ln e 
 1.
2. The two logarithms have different bases.
log 17 ⇒ 10x 
 17 or x 
 1.23
ln 17 ⇒ ex 
 17 or x 
 2.83
3. ln 64 
 4.1589
ln 16 
 2.7726
ln 4 
 1.3863
ln 16  ln 4 
 2.7726  1.3863 
 4.1589
4. The two equations represent the same thing,
A 
 Pert is a special case of the equation
N 
 N0ekt and is used primarily for computations
involving money.
5. 4.7217
6. 1.1394
7. 15.606
8. 0.4570

300˚

u
72. AB 
 (6  5), (5  6)

 1, 1
u
(6  5
)2  (
5  6
)2
AB  
 

 2
 1.414
73.



Natural Logarithms

0˚

2 4 6

240˚

x

O

3.65 cm

a

ln 132


9. log5 132 
 
ln 5

b

v 
 36

3.0339
10. log3 64 


10 
 360°

cos 36° 


a

3.65

sin 36° 


a 
 3.65 cos 36°
2.9529
1

Use A 
 2aP, where P
A

b

3.65

11.

b 
 3.65 sin 36°
2.1454
10(2.1454)

3.7856
18 
 e3x
ln 18 
 3x ln e
ln 18

3

21.454.

ln 64

ln 3


x

0.9635 
 x

1
(2.9529)(21.454)
2

31.6758 or 31.68 cm2

393

Chapter 11

12.

10 
 5e5k
2 
 e5k
ln 2 
 5k ln e
ln 2

5

ln 0.512


33. log8 0.512 
 
ln 8

0.3219
34. log6 323 



k

3.2246

0.1386 k
13. 25ex 100
ex 4
x ln e ln 4
x 1.3863
14.
4.5  e0.031t
ln 4.5  0.031t ln e
ln 4.5

0.031

ln 303

1.6
ln 2

88




35. log5 288
ln 5

36.

1.7593
6x 
 72
x ln 6 
 ln 72

37.

2x 
 27
x ln 2 
 ln 27

ln 72

ln 27


x

ln 2


x

ln 6

t
38.

48.5186  t
15.

x

13.57

2.3869
9x4 
 7.13
(x  4) ln 9 
 ln 7.13
x ln 9  4 ln 9 
 ln 7.13
x ln 9 
 ln 7.13  4 ln 9

4.7549

ln 7.13  4 ln 9


x

ln 9

39.

x 4.8940
3x 
 32

x ln 3 
 ln 3  ln 2

ln 3  ln 2



x

ln 3

[20, 20] scl:2 by [4, 20] scl:2
x  26.90

16.

41.

x 1.3155
60.3 e0.1t
ln 60.3 0.1t ln e
ln 60.3

0.1

t

40.9933
42.

450

760
450

ln 
760
450
ln 
760

0.125




1

2

e0.125a



4.1926

a; 4.2 km

Pages 736–737
18.
20.
22.
24.
26.
28.

19.
21.
23.
25.
27.
29.
ln 56


30. log12 56 
 
ln 12

1.6199
32. log4 83 


ln 25

0.075

0.2705
0.9657
2.2322
1.2134
0.9966
0.2417


 2x ln e

x

y

y  42.9183

ln 36


31. log5 36 
 
ln 5

2.2266

ln 83

ln 4

3.1875

Chapter 11


 e2x

0.3466 x
44.
25 e0.075y
ln 25 0.075y ln e

Exercise

5.4931
6.8876
0
10.4395
0.0233
146.4963

 1 
 e2x
1

2
1

ln 2
1
ln 2

2


a


t

0.7613 t
22 
 44 (1  e2x)

43.


  
 0.125a ln e




t
6.2e0.64t 
 3et1
ln 6.2  0.64t ln e 
 ln 3  (t  1) ln e
ln 6.2  ln 3  1 
 0.36t
ln 6.2  ln 3  1

0.36

[15, 30] scl:5 by [50, 150] scl:10
17a. p 
 760e0.125(3.3)

 760e0.4125
503.1 torrs
17b.
450 
 760e0.125a

394

40. 25ex 
 1000
ex 
 40
x ln e 
 ln 40
x 3.6889

45.

5x  76

x ln 5  ln 7  ln 6


x  4.72

52.

ln 7  ln 6



x
ln 5

46.

x  1.7657
12x4  4x
(x  4) ln 12  x ln 4
x ln 12  4 ln 12  x ln 4
x ln 12  x ln 4  4 ln 12
x(ln 12  ln 4)  4 ln 12

[10, 10] scl:1 by [10, 75] scl:5

4 ln 12


x
ln 12  ln 4
2


x  0.37

53.

x  9.0474

47. x 3  27.6 3

x  (27.6) 2
x  144.9985
48.

x

3.76
[5, 5] scl:1 by [2, 5] scl: 0.5
0.6 
 1e

54.

ln 0.6 


t

20,000(4  1011)

ln e

20,000(4  1011) ln 0.6 
 t
4.09  107 t
7
4.09  10 s

[5, 5] scl:1 by [50, 700] scl:50
49.

t

20,000(41011)

x

7.64

1 t

2.8 
 9
2

55.

2.8

1

ln 9 
 t ln 2
2.8
ln 9
__
1
ln 2

1.6845 t
1.6845  8  24 
 323.4236
324 h
56a. ln 180  72 
 k(0)  c
4.6821 c
56b. ln 150  72 
 k(2)  4.6821

[70, 10] scl:1 by [3, 10] scl:1
50.

t


t

133.14

ln 78  4.6821

2


k

0.1627 k
56c. ln 100  72 
 (0.1622)t  4.6821
ln 28  4.6821

0.1627


t

8.3 t
about 6.3 min
57. e2x  4ex  3 
 0
(ex  3)(ex  1) 
 0
ex  3 
 0
ex 
 3
x ln e 
 ln 3
x 1.0986
0 or 1.0986
58a.
2 
 e0.063t
ln 2 
 0.063t ln e

[10, 150] scl:10 by [100, 2000] scl:100
51.
x 2.14

[6, 6] scl:1 by [4, 24] scl:2

ln 2

0.063

8.3  2 
 6.3

ex  1 
 0
ex 
 1
x ln e 
 ln 1
x
0


t

11.0023 t
about 11 years
58b. See students’ work.

395

Chapter 11

1800 
 5000 ln r

59.

1800

5000
1800

5000

68. 2x  5y  3 
 0

A2  
B2 
 
22  (
5)2 
 29



 ln r

2x

60a.
ln

1

2
1

2

1
ln 
2

1622

229



 1ek(1622)

329



 1622 k ln e

k

1.7


t




tan




2



 
29


529


329


3
29

 29y  29 
 0; 29

0.56; 112°

Natural Logarithms and Area

Pages 738–739

y 0.9, 1.1
x2 0.9  4
x
4.9

x 2.2, 2.2

1. 0.69314718
2. 0.6931471806; It is the same value as found in
Exercise 1 expressed to 10 decimal places.
3a. The result is the opposite of the result in
Exercise 1.
3b. Sample answer: a negative value
4a. 0.69314718
4b. 1.0986123
4c. 1.3862944
4d. 0.6931471806, 1.098612289, 1.386294361
4e. The value for each area is the same as the value
of each natural logarithm.
5. 0.5108256238; 0.6931471806; 0.9162907319;
These values are equal to the value of ln 0.6,
ln 0.5, and ln 0.4.
6. If k  1, then the area of the region is equal to
ln k. If 0 k 1, then the opposite of the area is
equal to ln k.
7. The value of a should be equal to or very close to
1, and the value of b should be very close to e. This
prediction is confirmed when you display the
actual regression equation.
8. Sample answer: Define ln k for k  0 to be
1
the area between the graph of y 
 x, the x-axis,
and the vertical lines x 
 1 and x 
 k if k  1 and
to be the opposite of this area if 0 k 1. Define
e to be the value of k for which the area of the
region is equal to 1.

x2
x
x

11  4
2.9

1.7, 1.7

y

x

m3


 146 cm3  
1003 cm3 
 c

0.00765 c; 0.00765 N  m
66. x 
 0.25 cos 
y 
 0.25 sin 

 0.25

0
(0.25, 0)
67. u
a 
 1, 2  3 4, 3

 1, 2  12, 9

 13, 7

Chapter 11

tan

cos

11-6B Graphing Calculator Exploration:

1 1
(4)(4
 4)

8
1 65


8

52.4 N

m2




0.56 units

5


29
529


29
___
229


29
5
2

69. y 
 70 cos 4v
70. d 
 800  (10  55)

 250
The correct answer is 250.

3


65.

329



 112°

63. 16 4 
 8
64. x2 
 y  4
x2  4y2 
 8
(y  4)  4y2 
 8
4y2  y  4 
 0

O

sin

229

29 x

707.9177 t
about 708 yr
61. y is a logarithmic function of x. The pattern in the
table can be determined by 3y 
 x which can be
expressed as log3 x 
 y.
62. 1.2844

y


529


p 
 29


ln 
2.3 
 0.000427t ln e

y


3

29 x  29y  29 
 0

0.000427 k
60b.
1.7 
 23e(0.000427)(t)
1.7
ln 
2.3

0.000427

5y

     
 0


29

29
29

antiln


r
0.6977 r; about 70%

396

Modeling Real-World Data with
11-7 Exponential and Logarithmic
Functions
Page 744

0.415


ln 
1.0091 
 0.0197x
0.415
ln 
1.0091

0.0197

Check for Understanding

1. Replace N by 4N0 in the equation N 
 N0ekt,
where N0 is the amount invested and k is the
interest rate. Then solve for t.
2. The data should be modeled with an exponential
function. The points in the scatter plot approach a
horizontal asymptote. Exponential functions have
horizontal asymptotes, but logarithmic functions
do not.
3. y 
 2e(ln 4)x or y 
 2e1.3863x; ln y 
 ln 2  (ln 4)x or
ln y 
 0.6931  1.3863x
ln 2

2631.74


ln 
2137.52 
 4r

ln 2

5

5
ln 
10.0170

0.0301

0.0301x
x

23.08

x; 23.08 min

Pages 745–748

Exercises

7. t 


ln 2

0.0225

ln 2


8. t 
 
0.05

30.81 yr
9. t 


2631.74

2137.52

4


r
0.0520 r; 5.2%
17. y 
 40  14.4270 ln x
18a. y 
 –826.4217  520.4168 ln x
18b. The year 1960 would correspond to x 
 0 and
ln 0 is undefined.
19. Take the square root of each side.
y 
 cx2
y 
 
cx2
y 
 cx
20a. 1034.34 
 1000(1  r)1
1.03034 
 1  r
0.03034 r; 3.034%
20b. y 
 1000.0006(1.0303)x
20c. y 
 1000.0006(1.0303)x
y 
 1000.0006(eln 1.0303)x
y 
 1000.0006e(ln 1.0303)x
y 
 1000.0006e0.0299x
20d. 1030.34 
 1000er
ln

39.61 yr
8.66 yr
6a. y 
 10.0170(0.9703)x
6b. y 
 10.0170(0.9703)x
y 
 10.0170(eln 0.9703)x
y 
 10.0170e(ln 0.9703)x
y 10.0170e0.0301x
6c.
5 10.0170e0.0301x

ln 
10.0170


x

45.10 x
45.10  10 
 35.10 min
16a. y 
 2137.5192(1.0534)x
16b. y 
 2137.5192(1.0534)x
y 
 2137.5192(eln 1.0534)x
y 
 2137.5192e(ln 1.0534)x
y 
 2137.5192e0.0520x
16c.
2631.74 
 2137.52e4r


5. t 
 
0.08


4. t 
 
0.0175

0.415 
 1.0091e0.0197x

15c.

13.86 yr

ln 2

0.07125

9.73
10. exponential; the graph has a horizontal asymptote
11. logarithmic; the graph has a vertical asymptote
12. logarithmic; the graph has a vertical asymptote
13. exponential; the graph has a horizontal asymptote
14a. y 
 4.7818(1.7687)x
14b. y 
 4.7818(1.7687)x
y 
 4.7818(eln 1.7687)x
y 
 4.7818e(ln 1.7687)x
y 
 4.7818e0.5702x

1030.34


ln 
1000 
 r

0.0299
21a.

x
ln y

r; 2.99%

0
1.81

50
2.07

100
3.24

150
3.75

190
4.25

200
4.38

21b. ln y 
 0.0136x  1.6889
21c. ln y 
 0.0136x  1.6889
y 
 e0.0136x1.6889
21d. y 
 e0.0136(225)1.6889
115.4572
115.5 persons per square mile
22a. The graph appears to have a horizontal
asymptote at y 
 2, so you must subtract 2 from
each y-value before a calculator can perform
exponential regression.
22b. y 
 2  1.0003(2.5710)x
23a. ln y is a linear function of ln x.
y 
 cxa
ln y 
 ln(cxa)
ln y 
 ln c  ln xa
ln y 
 ln c  a ln x

ln 2

14c. Use t 
 k; k 
 0.5702.
ln 2


t

0.5702

1.215 hr
15a. y 
 1.0091(0.9805)x
15b. y 
 1.0091(0.9805)x
y 
 1.0091(eln 0.9805)x
y 
 1.0091e(ln 0.9805)x
y 
 1.0091e0.0197x

397

Chapter 11

32.

23b. The result of part a indicates that we should
take the natural logarithms of both the x- and
y-values.
ln x
ln y

6.21
4.49

6.91
4.84

8.52
5.65

9.21
5.99

28. 5
cos

 i sin



6

 
 5


3


2

5
3



50; 22
33. Circle X contains the regions a, b, d, and e.
Circle Z contains the regions d, e, f, and g. Six
regions are contained in one or both of circles X
and Z.
The correct choice is C.

Chapter 11 Study Guide and Assessment
Page 749
1.
3.
5.
7.
9.

11.



4i11


1 2

4


 

1


12. (64) 2 
 8


 

4



 16

4


5

b




sin 48°

3

4

3



 (44) 4

 43

 64
1


3x2

3

1


3x2


 3
17.

3


12x4

4

1 3


 
2 (x4)3
1


 8x12
18. (w3)4  (4w2)2 
 w12  42  w4

 16w16
1


1


3

3

1


1


2

19. 
(2a) 3 (a2b) 3  
 (2a) 3   (a2b) 3 

 (2a)(a2b)

 2a3b

2i11



 1
or 5
0

1


1


5


9


20. 
3x 2 y 4 (4x2y2) 
 12x 2 y 4

31. 4 units left and 8 units down

398

3

16. 
6a 3  
 63
a 3 

 216a




9x2

42˚

3


14. (256
) 
 (256) 4


15. 3x2(3x)2 
 
(3x)2

x 
 10

Chapter 11




exponential growth
scientific notation
natural logarithm
exponential function
exponential equation

Skills and Concepts

1

1 2

4

13. (27) 3 
 (33) 3

 34

 81

30. 5x2  8x  12 
 0
Discriminant: (8)2  4(5)(12) 
 176
The discriminant is negative, so there are
2 imaginary roots.

8

Understanding the Vocabulary

common logarithm
2.
logarithmic function 4.
mantissa
6.
linearizing data
8.
nonlinear regression 10.

Pages 750–752

18˚

176


(2, 1)

1

b

8


 38
(4, 4): f(x) 
 2(4)  8(4)  10

 50

x

O

60 sin 48° 
 b sin 24°
b 
 109.625
about 109.6 ft

60 ft

(4, 1) (2, 8): f(x) 
 2(2)  8(3)  10

y
1

1
i
2

60

sin 24°


 26

x  2y 
 4


 2  2i
29.

x
4

 22
x
2
(4, 1): f(x) 
 2(4)  8(1)  10
(4, 4)

(2, 3)

9.62
6.19

23c. ln y 
 0.4994 ln x  1.3901
23d. ln y 
 0.4994 ln x  1.3901
eln y 
 e0.4994ln x1.3901
y 
 e0.4994ln x  e1.3901
y 
 (eln x)0.4994  4.0153
y 
 4.0153x0.4994
24.
2 
 ek(85)
ln 2 
 85k
0.0082 k
12 
 e0.0082t
ln 12 
 0.0082t
303 t
303.04 min or about 5 h
25. 0.01
26. log5 (7x) 
 log5 (5x  16)
7x 
 5x  16
2x 
 16
x
8
27a.
y 
 x(400  20(x  3))
y 
 x(460  20x)
y 
 20x2  460x
y  2645 
 20(x2  23x  132.25)
y  2645 
 20(x  11.5)
vertex at (11.5, 2645), maximum at x 
 11.5
$11.50
27b. At maximum, y 
 2645.
$2645


6

(2, 1): f(x) 
 2(2)  8(1)  10

y

21.

22.

1

45. 2 log6 4  3 log6 8 
 log6 x

y

y

1


log6 42  log6 8 3 
 log6 x
log6

42

1



 log6 x

83

y 
 3x

42

1

y 
 ( 12 ) x




x

83

x

O
23.

O

8
x

x

1

46. log2 x 
 3 log6 27

24.

1


y

log2 x 
 log2 27 3

y
y 
 2x 1

1


x 
 27 3
x
3
47. y
y 
 2x  2

O
x

O
25.

O

x

y 
 log10 x

x

26.

y

y
x

O
y

2x

O
27. A 


48. log 300,000 
 log (100,000  3)

 log 100,000  log 3

 5  0.4771

 5.4771
49. log 0.0003 
 log (0.0001  3)

 log 0.0001  log 3

 4  0.4771

 3.5229
50. log 140 
 log (10  14)

 log 10  log 14

 1  1.1461

 2.1461
51. log 0.014 
 log (0.001  14)

 log 0.001  log 14

 3  1.1461

 1.8539
52.
4x 
 6x2
x log 4 
 (x  2) log 6
x log4 
 x log 6  2 log 6
x log 4  x log 6 
 2 log 6
x(log 4  log 6) 
 2 log 6

y  2x  2

1

x

2500e0.065(10)

4788.8520; $4788.85
28. A 
 6000e0.0725(10)
12,388.3866; $12,388.39
29. A 
 12,000e0.059(10)
21,647.8610, $21,647.86
2


1

30. 8 3 
 4

31. 34 
 8
1

32. log2 16 
 4

33. log5 2
5 
 2

34. 2x 
 32
2x 
 25
x
5

35. 10x 
 0.001
10x 
 103
x 
 3

1

36. 4x 
 1
6

1

37. 2x 
 0.5

4x 
 42
x 
 2

2x 
 21
x 
 1
1

38. 6x 
 216

39. 9x 
 9



x
3
40. 4x 
 1024
4x 
 45
x
5



x 
 1
41. 8x 
 512
8x 
 83
x
3

42. x4 
 81

43.

6x

2 log 6


x

log 4  log 6

63

1


x 
 (81) 4
x
3
44. log3 3  log3 x 
 log3 45
log3 3x 
 log3 45
3x 
 45
x 
 15

8.84
53.
120.5x 
 80.1x4
0.5x log 12 
 (0.1x  4) log 8
0.5x log 12 
 0.1x log 8  4 log 8
0.5x log 12  0.1x log 8 
 4 log 8
x(0.5 log 12  0.1 log 8) 
 4 log 8
x

91

9x

4


12


x

16 
 x

399

x


4 log 8

0.5 log 12  0.1 log 8

x

8.04

Chapter 11

3x


14

54.

1

(x  2) log 6

1

x log 6  2 log 6

3x log 4
3x log 4
1

3x log 4  x log 6
x(3 log

1

4

log 100

64.

4x 
 100
ln 100


x

ln 4

x

2 log 6

1
3 log   log 6
4

65.

3.3219
6x2 
 30
(x  2) ln 6 
 ln 30

1

ln 30

Change the inequality because 3 log 4  log 6 is
negative.
x  0.6
55.
0  12x8  7x4
(2x 8) log 0.1  (x  4) log 7
2x log 0.1  8 log 0.1  x log 7  4 log 7
2x log 0.1  x log 7  4 log 7  8 log 0.1
x(2 log 0.1  log 7)  4 log 7  8 log 0.1
x


x2

ln 6
ln 30


x

ln 6  2

66.

4 log 7  7 log 0.1

2 log 0.1  log 7

x 3.8982
3x1 
 42x
(x  1) ln 3 
 2x ln 4
x ln 3  ln 3 
 2x ln 4
x ln 3  2x ln 4 
 ln 3
x(ln 3  2 ln 4) 
 ln 3

ln 3


x

ln 3  2 ln 4

x  4
56.
log (2x  3) 
 log (3  x)
log (2x  3) 
 log (3  x)1
(2x  3) 
 (3  x)1
(2x  3)(3  x) 
 1
2x2  3x  8 
 0
3

x

0.6563
67.

 5x4
4x ln 9 
 (x  4) ln 5
4x ln 9 
 x ln 5  4 ln 5
4x ln 9  x ln 5 
 4 ln 5
x(4 ln 9  ln 5) 
 4 ln 5
94x

73


x 
 4

4 ln 5


x

4 ln 9  ln 5

1.39, 2.89
57. y

x

y 
 3 log (x  2)

68.

24
ln 24

e2x
2x

0.8967
69. 15ex  200
200
ex  15

ln 24

x

O

1.7829

x ln 4 
 ln 100

 2 log 6

x


63. log15 125 
 
log 15

2.0959

 2 log 6

 log 6)

log 125


62. log4 100 
 
log 9

6x2

200

x  2

x  ln 15

x  1.5890

x  2.5903
x 3.333

70.

y

58.

[5, 5] scl:1 by [10, 60] scl:10

y  7x  2

O

71.

x

x

x 
 3.42

59.

[1, 5] scl:1 by [1, 10] scl:1
[5, 5] scl:1 by [5, 5] scl:1
log 15


60. log4 15 
 
log 4

1.9534

Chapter 11

log 24


61. log8 24 
 
log 8

1.5283

400

2.20

ln 2

ln 2


72. t 
 
0.028

24.76
74. 18 

k


Chapter SAT & ACT Preparation


73. t 
 
0.05125

13.52

ln 2

k
ln 2

18

Page 755

0.0385; 3.85%

Page 753
75.

Applications and Problem solving
1 t

0.065 
 
2

1


t

0.6215 t
0.6215  5730
76a.  
 10 log

To find the quadratic equation, multiply these two
factors and let the product equal zero.
(2x  1)(3x  1) 
 0
6x2  5x  1 
 0
The correct choice is E.
3. The result of dividing T by 6 is 14 less than the
correct average.

3561.13 or 3561 yr.

1.15  1010

10 12

20.6

20.6 dB
9

109


76b.  
 10 log 
102

39.5

T

6

39.5 dB
8.95 

103

T

6


76c.  
 10 log 
1012
99.5
99.5 dB
0.014t
77. 200,000 
 142,000e

ln

100

71
100

71

100
ln 
71

0.014

ln

1
ln 
2

0.20

3.47


 correct answer  14

 14 
 correct average

The correct average is the total divided by the
number of scores, 5.


 e0.014t

T

correct average 
 5


 0.014t

T

6

T

 14 
 5

The correct choice is E.
4.
y
B x,y


t

(2 )

24.4 t
1990  24 
 2014
78a. N 
 65  30e0.20(2)
44.89; 45 words per minute
78b. N 
 65  30e0.20(15)

 63.50; 64 words per minute
78c.
50 
 65  30e0.20t
1

2
1

2

1

2. Since one root is 2, x 
 2, 2x 
 1, and 2x 
 0.
1
Similarly for the root that is 3,
1
x 
 3, 3x 
 1, and 3x  1 
 0.

1

log 0.65 
 t log 2
log 0.65

1
log 
2

SAT and ACT Practice

1. To find the greatest possible value, the other 3
values must be as small as possible. Since they
are distinct positive integers, they must be 1, 2,
and 3. The sum of all 4 integers is 4(11) or 44. The
sum of the 3 smallest is 1  2  3 or 6, so the
fourth integer cannot be more than 44  6 or 38.
The correct choice is B.

y
x
2

A

tan A 



 e0.20t

y

x

2

C (x, 0) x
2y


 x

Find the area of ABC.


 0.20t

1

1

xy

A 
 2bh 
 2xy 
 2
Simplify the ratio.


t

area of ABC

tan A

t; 3.5 weeks




xy

2

2y

x


 2y 
2 
 4
x

xy

x2

The correct choice is E.

Page 753

Open-Ended Assessment

5.

1

4

1. Sample answer: (n4) (4m)1
2. Sample answer:

x

y

10


 2y

2yx 
 10y
2x 
 10
x
5
The correct choice is B.

1

log 2  log(x  2) 
 2 log 36

401

Chapter 11

6. C 
 D
2

3

9. A is the arithmetic mean of three consecutive
x  (x  2)  (x  4) 

positive even integers, so A 
 


 D

D


2

3

and, therefore, r 


3
3x 
6 
 x  2, where x is a positive even integer.


3

1
.
3

Then A is also a positive even integer. Since A is
even, when A is divided by 6, the remainder must
also be an even integer. The possible even
remainders are 0, 2, and 4.
The correct choice is C.
10. First notice that b must be a prime integer. Next
notice that 3b is greater than 10. So b could be 5,
since 3(5) 
 15. (b cannot be 3.) Check to be sure
that 5 fits the rest of the inequality.
5
25
1
3(5) 
 15  6(5) 
 6 
 46

Now use this value for the radius to calculate half
of the area.
1
A
2

1 2




 2r2 
 2
3 
 2
9 
 18
1

1

1

1

The correct choice is A.
7. The average of 8 numbers is 20.
20 


sum of eight numbers

8

sum of eight numbers 
 160
The average of 5 of the numbers is 14.
14 


So 5 is one possible answer. You can check to see
that 7 and 11 are also valid answers.
The correct answer is 5, 7, or 11.

sum of five numbers

5

sum of five numbers 

70
The sum of the other three numbers must be
160  70 or 90. Calculate the average of these
three numbers.
average 


sum of three numbers

3

90


 3 
 30

The correct choice is D.
8. The sum of the angles in a triangle is 180°. Since
∠B is a right angle, it is 90°. So the sum of the
other two angles is 90°. Write and solve an
equation using the expressions for the two angles.
2x  3x 
 90
5x 
 90
x 
 18
The question asks for the measure of ∠A.
∠A 
 2x 
 2(18) 
 36
The correct choice is C.

Chapter 11

402

Chapter 12 Sequences and Series
11. an 
 a1  (n  1)d
3 
 a1  (7  1)(2)
3 
 a1  12
15 
 a1
12. an 
 a1  (n  1)d
34 
 100  (12  1)d
66 
 11d
6 
 d
13. an 
 a1  (n  1)d
24 
 9  (4  1)d
15 
 3d
5
d
9  5 
 14, 14  5 
 19
9, 14, 19, 24

Arithmetic Sequences and Series

12-1

Pages 762–763

Check for Understanding

1. a1 
 6  4(1) or 2
a2 
 6  4(2) or 2
a3 
 6  4(3) or 6
a4 
 6  4(4) or 10
a5 
 6  4(5) or 14
2, 2, 6, 10, 14; yes, there is a common
difference of 4.
an
2a.

n

1

14. Sn 
 2[2a1  (n  1)d]

O 1 2 3 4 5 6

35

S35 
 2[2.7  (35  1)  2]

n


 1435

1

n

Sn 
 2[2a1  (n  1)d]

15.

2

n

210 
 2[2  30  (n  1)(4)]

3

420 
 60n  4n2  4n
4n2  64n  420 
 0
4(n  21)(n  5) 
 0
n 
 21 or n 
 5
Since n cannot be negative, n 
 21.
16. n 
 19, a19 
 27, d 
 1
an 
 a1  (n  1)d
27 
 a1  (19  1)1
9 
 a1

2b. linear
2c. The common difference is 1. This is the slope of
the line through the points of the sequence.
3a. The number of houses sold cannot be negative.
n

3b. Sn 
 2[2a,  (n  1)d]
10

S10 
 2[2  3750  (10  1)500]

 $60,000
4. Negative; let n and n  1 be two consecutive
numbers in the sequence.
d 
 (n  1)  n or 1
5. Neither student is correct, since neither sequence
has a common difference. The difference
fluctuates between 1 and 1. The second sequence
has a difference that fluctuates between 2 and 2.
6. d 
 11  6 or 5
16  5 
 21, 21  5 
 26, 26  5 
 31, 31  5 
 36
21, 26, 31, 36
7. d 
 7  (15) or 8
1  8 
 9, 9  8 
 17, 17  8 
 25, 25  8 
 33
9, 17, 25, 33
8. d 
 (a  2)  (a  6)

 a  a  2  6 or 4
a  2  4 
 a  6, a  6  4 
 a  10,
a  10  4 
 a  14, a  14  4 
 4 
 a  18
a  6, a  10, a  14, a  18
9. an 
 a1  (n  1)d
a17 
 10  (17  1)(3)

 38
10. an 
 a1  (n  1)d
37 
 13  (n  1)5
50 
 5(n  1)
10 
 n  1
11 
 n

n

S19 
 2(a1  a19 )
19


 2(9  27)

 342 seats

Pages 763–765

Exercises

17. d 
 1  5 or 6
7  (6) 
 13, 13  (6) 
 19,
19  (6) 
 25, 25  (6) 
 31
13, 19, 25, 31
18. d 
 7  (18) 
 11
4  11 
 15, 15  11 
 26, 26  11 
 37, 37  11 
 48

19.

20.

21.

22.

403

15, 26, 37, 48
d 
 4.5  3 or 1.5
6  1.5 
 7.5, 7.5  1.5 
 9, 9  1.5 
 10.5,
10.5  1.5 
 12
7.5, 9, 10.5, 12
d 
 3.8  5.6 or 1.8
2  (1.8) 
 0.2, 0.2  (1.8) 
 1.6,
1.6  (1.8) 
 3.4, 3.4  (1.8) 
 5.2
0.2, 1.6, 3.4, 5.2
d 
 b  4  b or 4
b  8  4 
 b  12, b  12  4 
 b  16,
b  16  4 
 b  20, b  20  4 
 b  24
b  12, b  16, b  20, b  24
d 
 0  (x) or x
x  x 
 2x, 2x  x 
 3x, 3x  x 
 4x, 4x  x 
 5x
2x, 3x, 4x, 5x
Chapter 12

40. 4 
 5  (4  1)d
9 
 3d
3
d
5  3 
 2, 2  3 
 1
5, 2, 1, 4
41.
12 
 3
  (4  1)d
12  3
 
 3d

23. d 
 n  5n or 6x
7n  (6n) 
 13n  13n  (6n) 
 19n,
19n  (6n) 
 25n, 25n  (6n) 
 31n
13n, 19n, 25n, 31n
24. d 
 5  (5  k) or k
5  k  (k) 
 5  2k, 5  2k  (k) 
 5  3k,
5  3k  (k) 
 5  4k, 5  4k  (k) 
 5  5k
5  2k, 5  3k, 5  4k, 5  5k
25. d 
 (2a  2)  (2a  5) or 7
2a  9  7 
 2a  16, 2a  16  7 
 2a  23,
2a  23  7 
 2a  30, 2a  30  7 
 2a  37
2a  16, 2a  23, 2a  30, 2a  37
26. d 
 5  (3  7
) or 2 7

7  7
  2  7
 
 9  27
, 9  27
  2  7


 11  37
, 11  37
  2 7
 
 13  47

9  27
, 11  37
, 13  47

27. a25 
 8  (25  1)3

 80
28. a18 
 1.4  (18  1)(0.5)

 9.9
29. 41 
 19  (n  1)(5)
60 
 5(n  1)
12 
 n  1
13 
 n
30. 138 
 2  (n  1)7
140 
 7(n  1)
20 
 n  1
21 
 n
31. 38 
 a1  (15  1)(3)
38 
 a1  42
80 
 a1
2


12  3
 
 d
3


12  3
12  23
3
  

 
,
3
3


12  2
3
12  3
24  3
   
 
3
3
3
12  2 
3 24  
3
3
, 
, 
, 12
3
3

42.

3

1

43. d 
 1  2 or 2
a11 
 2  (11  1)
2
3

1


 3.5

S11 
 2
2  
32
11 3

1


 11
44. d 
 4.8  (5) or 0.2
a100 
 5  (100  1)0.2

 14.8
100

S100 
 2(5  14.8)

 490
45. d 
 13  (19) or 6
a26 
 19  (26  1)6

 131

1

32. 103 
 a1  (7  1)3
2

103 
 a1  2
2

83 
 a1

26

S26 
 2(19  131)

33. 58 
 6  (14  1)d
52 
 13d
4
d
34. 26 
 8  (11  1)d
18 
 10d
4
15 
 d


 1456
n

46. 14 
 2[2(7)  (n  1)1.5]
28 
 14n  1.5n2  1.5n
0 
 1.5n2  15.5n  28
n


)  (4  5
) or 3
35. d 
 (1  5
a8 
 4  5
  (8  1)3

 17  5

36. d 
 6  (5  i) 
 1  i
a12 
 5  i  (12  1)(1  i)

 5  i  11  11i

 16  10i
37. d 
 10.5  12.2 or 1.7
a33 
 12.2  (33  1)(1.7)

 42.2
38. d 
 4  (7) or 3
a79 
 7  (79  1)3

 227
39. 21 
 12  (3  1)d
9 
 2d
4.5 
 d
12  4.5 
 16.5
12, 16.5, 21

Chapter 12

5 
 2  (5  1)d
3 
 4d
0.75 
 d
2  0.75 
 2.75, 2.75  0.75 
 3.5,
3.5  0.75 
 4.25
2, 2.75, 3.5, 4.25, 5

15.5 
(15.
5)2  
4(1.5)
(28)

2(1.5)
1

n 
 8 or n 
 23
Since there cannot be a fractional number of
terms, n 
 8.
n

47. 31.5 
 2[2(3)  (n  1)2.5]
63 
 6n  2.5n2  25n
0 
 2.5n2  8.5n  63
n


8.5 
(8.5
)2 4(
2.5)(
63)

2(2.5)

n 
 7 or n 
 3.6
Since n cannot be negative, n 
 7.
48. d 
 7  5 or 2
an 
 5  (n  1)2

 2n  3
49. d 
 2  6 or 8
an 
 6  (n  1)(8)

 8n  14

404

58. Sn 
 a1  a2  (a31  a32)
 (a41  a42)  (a51  a52)  

 a1  a2  a2  a1  a3  a2  a4  a3  

 a1  a2  a2  a1  (a2  a1)  a2  a3  a2
 a3  

0
59. A 
 Pert

 100e0.07(15)

 $285.77
60. 4x2  25y2  250y  525 
 0
4x2  25(y2  10y) 
 525
4x2  25(y  5)2 
 100

50. 9:00, 9:30, 10:00, 10:30, 11:00, 11:30, 12:00
n 
 7, d 
 2, a1 
 3
a7 
 3  (7  1)2

 15 data items per minute
51. Let d be the common difference. Then, y 
 x  d,
z 
 x  2d, and w 
 x  3d. Substitute these
values into the expression x  w  y and simplify.
x  (x  3d)  (x  d) 
 x  2d or z.
52. a1 
 5, d 
 4, n 
 25
a25 
 5  (25  1)4

 101
25

S25 
 2(5  101)

150n 
 180n  360
30n 
 360
n 
 12
54a. S4 
 (4  2)180° or 360°
S5 
 (5  2)180° or 540°
S6 
 (6  2)180° or 720°
S7 
 (7  2)180° or 900°
360°, 540°, 720°, 900°
54b. The common difference between each
consecutive term in the sequence is 180,
therefore the sequence is arithmetic.
54c. a35 
 180  (35  1)180

 5940°
55a. a1 
 1, d 
 2
S5 


y

O

(5, 5)

(5, 5)

(0, 5)
(0, 7)

6

61. r 
 1
2 or 0.5

 (5  1)2]

5





v 
 8  2 or 8




0.5
cos 8  i sin 8

10

55b. S10 
 2[2(1)  (10  1)2]

0.46  0.19i

62. 2, 1, 3  5, 3, 0 
 2(5)  (1)(3)  (3)(0)

7
63. x cos 30°  y sin 30°  5 
 0


 100
55c. Conjecture: The sum of the first n terms of the
sequence of natural numbers is n2.
Proof:
Let an 
 2n  1. The first term of the sequence
of natural numbers is 1, so a1 
 1.
Then, using the formula for the sum of an
arithmetic series,
n
Sn 
 2(a1  an)


3
x
2

1

 2y  5 
 0

x  y  10 
 0
3
64.

y
O


n

2

n

4

Sn 
 2[1  (2n  1)]

 2(2n) or n2

2 x

6

56. a1 
 5, d 
 7, n 
 15
S15 


x
(0, 3)


 25

15
[2(5)
2

 4 
 1


h 
 0, k 
 5, a 
 5, b 
 2, c 
 21
center: (0, 5)
foci: ( 21
, 5)
vertices: major → ( 5, 5)
minor → (0, 3) and (0, 7)

n

53. Sn 
 2(128°  172°); Sn 
 (n  2)180°

5
[2(1)
2

(y  5)2

x2

25


 1325 bricks

 (15  1)7]

65. Find A.
A 
 90°  19° 32


 70° 28

Find a.
a

cos 19° 32
 
 
4.5


 810 feet
57. n 
 10, S10 
 5510, d 
 100
10

5510 
 2[2a1  (10  1)100]
5510 
 10a1  4500
1010 
 10a1
101 
 a1
a10 
 101  (10  1)100

 1001
least: $101, greatest: $1001

4.2 a
Find b.
b

sin 19° 32
 
 
4.5
1.5 b
66. discriminant 
 (3)2  4(4)(2)

 23
Since the discriminant is negative, this indicates
two imaginary roots.

405

Chapter 12

x1
x  3x2
4x

2
(x2  3x)
x  2
(x  3)
1
y
x1
1
1 3 4
0 1
2 0
68.



2 0 1
1 0
1 3 4

64b.

67.



 

 

Value
27,500
25,000
22,500
20,000
17,500
15,000
12,500



10,000
7500
5000
2500
0

A
(1, 2), B
(3, 0), C
(4, 1)
69. a  4b 
 15 → a 
 15  4b
4a  b 
 15
4(15  4b)  b 
 15
60  16b  b 
 15
15b 
 45
b 
 3
4a  (3) 
 15
4a 
 12
a 
3
a  b 
 3  (3) or 6
The correct choice is C.

0

1

2

3
Years

4

5

6

6c. an exponential function
7. r 


4
—
2

3

or 6

24(6) 
 144, 144(6) 
 864, 864(6) 
 5184
144, 864, 5184
3

8. r 
 2

12-2


 

Geometric Sequences and Series
9.

Page 771

Check for Understanding

28.8(4) 
 115.2, 115.2(4) 
 460.8,
460.8(4) 
 1843.2
115.2, 460.8, 1843.2

1. Both arithmetic and geometric sequences are
recursive. Each term of an arithmetic sequence is
the sum of a fixed difference and the previous
term. Each term of a geometric sequence is the
product of a common ratio and the previous term.
2. an 
 (3)11 or 9
a2 
 (3)21 or 27
a3 
 (3)31 or 81
The expression generates the following sequence:
9, 27, 81, . The common ratio is 3, therefore
it is a geometric sequence.
3. If the first term in a geometric sequence were
zero, then finding the common ratio would mean
dividing by zero. Division by zero is undefined.
4. Sample answer: The first term of the series
5  10  20  is 5 and the sum of the first 6
terms of the sequence is 105, but 105 is not
greater than 5.
5a. No; the ratio between the first two terms is 2,
but the ratio between the next two terms is 3.
5b. Yes; the common ratio is 3
.
5c. Yes; the common ratio is x.
6a.
Beginning of
Value of
Year
Computers
1
27,500.00
2
15,125.00
3
8318.75
4
4575.31
5
2516.42
6
1384.03

Chapter 12


  
 881 , 881 
32 
 21463

9 3
27 27 3
  
 ,  
2 2
4
4 2
27 81 243
, , 
4
8
16
7.2


r 
 1.8 or 4

2.1

10. r 
 7 or 0.3
an 
 a1rn1
a7 
 7(0.3)71

 0.005103
11. an 
 a1rn1
24 
 a1(2)51
24 
 16a1
3
 
 a
1
2
2.5

12. a3 
 2 or 1.25
1.25

a2 
 2 or 0.625
0.825

a1 
 2 or 0.3125
0.3125, 0.625, 1.25
13. an 
 a1rn1
27 
 1(r)41
27 
 r3
3
r
1(3) 
 3, 3(3) 
 9
1, 3, 9, 27
1


14. r 
 
0.5 or 2
a1  a1rn


Sn 
 
1r

0.5  0.5(2)9


S9 
 
1  (2)


 85.5

406

15. r 
 1.035
The value of the car after 10, 20, and 40 years will
be the 11th, 21st, and 41st terms of the sequence
respectively.
a11 
 20,000(1.035)111
 $28,211.98
a21 
 20,000(1.035)211
 $39,795.78
a42 
 20,000(1.035)411
 $79,185.19

26. a5 
 8
2

3 51

81


 2
27.

3
8
3
—
r 
 1 or 4

2
1
3 61
a6 
 2 4


 
243



 
2048
0.4


28. r 
 
4.0 or 0.01

Pages 771–773

a7 
 40(0.01)71

 4  1011

Exercises

2

16. r 
 1
0 or 0.2


10

5

29. r 
  or 2


0.4(0.2) 
 0.08, 0.08(0.2) 
 0.016,
0.016(0.2) 
 0.0032
0.08, 0.016, 0.0032
17. r 


20

8


 165


or 2.5

192 
 a1(4)61
192 
 1024a1
0.1875 
 a1
31. 322
 
 a1(2
)51
322
 
 4a1
82
 
 a1
30.

50(2.5) 
 125, 125(2.5) 
 312.5,
312.5(2.5) 
 781.25
125, 312.5, 781.25
18. r 


2

3
2

9

or 3

32.

19. r 


486 
 a1

a2 
 486
3 or 162
1

or

a3 
 162
3 or 54
1

24


  
 61225 , 61225 
25 
 
3125

3 2
6
6 2
  
 ,  
25 5
125 125 5
6
12
24
, , 
125 625 3125
3.5

r

7 or 0.5

486, 162, 54
33. 0.32 
 a1(0.2)51
0.32 
 0.0016a1
200 
 a1
a2 
 200(0.2) or 40
a3 
 40(0.2) or 8
200, 40, 8
34. 81 
 256r51
81
  
 r4
256

1.75(0.5) 
 0.875, 0.875(0.5) 
 0.4375,
0.4375(0.5) 
 0.21875
0.875, 0.4375, 0.21875
6
32



21. r 
 —— or 2
62

2
 
 12, 12
2
 
 122
, 122

2
 
 24

3

4

, 24
12, 122
3
3



3
144
 4 



3


3


3


3

3
3 
 3
, 3

3 
 1, 1
3 
 3

3

3
, 1, 3
1

35.

1

23. r 
 i 
  or i

1

i(i) 
 1, 1(i) 
 i, i(i) 
 1
1, i, 1
t5

24. r 
 t8 or t3

36.

t2(t3) 
 t1, t1(t3) 
 t4, t4(t3) 
 t7
t1, t4, t7

5

2

2

2

2

7

7

3

3

2

5

2

9

1
1 b b2 b3
, 
  
ab a3 , a5 , a7 , a9

407




3

4

144,

108

256, 192, 144, 108, 81
54 
 2r41
27 
 r3
3 
 r
2(3) 
 6, 6(3) 
 18
2, 6, 18, 54
4

7 
 7r31


 

3

192, 192


3

4

49
 
 r2
4
7
 
 r
2
4
7

 

7
2
4
, 2, 7
7


  
 a1b , a1b 
ab 
 a1, a1
ab 
 ab,
b b
b b b
b

 
 , 
 
 
a a
a a a
a
a b


b2 a2


r

256



3

22. r 
 9 or 3

25.

1 51

6 
 8
1 a1

2

5


 

20.

6 
 a1
3
1

2(3) 
 6, 6(3) 
 18, 18(3) 
 54
6, 18, 54
3
——
10
—
3
——
4

91

a9 
 5

2


2

Chapter 12

37. r 


38.

5
—
5

3

5.50

43a. r 
 5 or 1.1

or 3

a10 
 5(1.1)101
 $11.79
a20 
 5(1.1)201
 $30.58
a10 
 5(1.1)401
 $205.72
$11.79, $30.58, $205.72

5
5
  (3)5
3
3
S5 
 
13
605

 3
13
r 
 65 or 0.2
65(0.2)6

S6 
 65  
1  0.2

5  5(1.1)52


43b. S52 
 
1  1.1


 81.2448

 $7052.15
43c. Each payment made is rounded to the nearest
penny, so the sum of the payments will actually
be more than the sum found in b.

3

39. r 


2


1

S10 




3

or 2


 

3 10
1  1 2
——
3
1  2


 

44a.

11,605

512

2
3

40. r 
 2 or 3

S8 






2  2(3
 )8

1  3

160

1  3


160
1  3
  

1  3
3
 1 

44b.

)
160(1  3




2

41a.

41b.
41c.

42a.

42b.

1

4
1

5



 2
0  z

1

1

1


 z
by definition

a2
1

Then an 
 a1rn1
So, an 
 (2)(3)n1
46. a1 
 1, r 
 2.5, n 
 15
a15 
 1(2.5)151

 372,529

47a. 25
1  1
2  
 $25.05
0.024

47b. No; at the end of two years, she will have only
$615.23 in her account.
S24 







0.024 24

25.05  25.05 1  
12

0.024

1 1
12






 $615.23
47c.

3

4 
 3a1

750 











0.024 24

a1  a1 1  
12

0.24

1 1
12

1.5  a1
1  
1  12



0.024 24


 a1

$30.54  a1

a28 
 3
3
281
3

a0
1  1
2   $30.54
0.024


 26,244

Chapter 12


 
2

r 
 a
 3

4 
 a1
3
3

4

1
1
  
z
20

1

8

5
z
45. a2 
 3(a1)

 3(2)

6


 80(1  3
)
The population doubles every half-hour, so r 
 2.
After 1 hour, the number of bacteria is the third
term in the sequence and n  1 
 2. After 2 hours,
it is the fifth term and n  1 
 4. After 3 hours,
it is the seventh term and n  1 
 6. After t
hours it is the 2t  1 term and n  1 
 2t.
bt 
 b0  22t
bt 
 30  22(5)

 30,720
Sample answer: It is assumed that favorable
conditions are maintained for the growth of the
bacteria, such as an adequate food and oxygen
supply, appropriate surrounding temperature,
and adequate room for growth.
a7 
 a4r3
12 
 4r3
3 
 r3
3
3

r
a4 
 a1r41

4

3

1
1
  
1
5
8
 
 x
2
13
1
 
 
80
x
80
 
 x
13

a0  $30.48
The least monthly deposit is $30.48.

408

a2

55. Find the amplitude.

48. r 
 a

86  36

A 
 2 or 25

1

r


1

27
—
1

81

Find h.


3

86  36

h 
 2 or 61
Find k.

an 
 a1 rn1

2

k

1

n1
6561 
 8
1 3



k 
 2

(6561)(81) 
 3n1
(38)(34) 
 3n1
312 
 3n1
12 
 n  1
13 
 n
6561 is the 13th term of the sequence.
49.



y 
 25 sin 
2t  c  61


36 
 25 sin 
2  1  c  61


1 
 sin 
2  c


 
 c

y 
 25 sin 
2t  3.14  61

n

650 
 2[2(20)  (n  1)5]
1300 
 40n  5n2  5n
0 
 5n2  35n  1300
0 
 5(n  13)(n  20)
n 
 13 or n 
 20
Since n cannot be negative, n 
 13 weeks.

56. Since 43° 90°, consider the following.
b sin A 
 20 sin 43°
13.64
Since 11 13.64, no triangle exists.
57. (n2)  49

2
n 7
Solution set: {nn  3 or n  3}

log 26 5


50. log11 265 
 
log
11

2.3269
y

O

52.


A2  
B2 




2 
 2  c

n

Sn 
 2[2a1  (n  1)d]

51.


4

n
5
4
3
3
4
5

x

m
n1
5  1 or 4
4  1 or 3
3  1 or 2
3  1 or 4
4  1 or 5
5  1 or 6

nm
(5)(4) or 20
(4)(3) or 12
(3)(2) or 6
 3(4) or 12
 4(5) or 20
 5(6) or 30

← least
possible
value

The answer is 6.


32  (
52)


 34

Since C is positive, use 34
.
3

5

5
34

5

3

5

 or 


, sin f 
 

p

34 , cos f 
  
34
34


34

tan f 

tan f 


Pages 780–781

5


34


3


34

5
3

1a.

Check for Understanding

an
1.0

f  59°
Since cosine is negative and sine is positive,

 59°  180° or 121°.
p 
 r cos(v  f)
5
34


34

Infinite Sequences and Series

12-3

5

x  
y  

0

34
34
34




0.5


 r cos(v  121°)

O 1 2 3 4 5 6 7 8 9 10n

53. 3x  4y 
 5
3

5

y 
 4x  4
3

1b. The value of an approaches 1 as the value of n
increases.

5

x 
 t, y 
 4t  4
54. csc v 
 3
1

sin v
1

3

1c. lim

n→


3

n1

n


1


 sin v

409

Chapter 12

1d. lim

n→

n1

n

n

n→ n


 lim



10. r 
 
6 or  2


10

1

Sn 



 

 

2b. If r

11. r 



0

1


 18


 no limit

3

3

12. r 
  or 3



 no limit

The sum does not exist since r 
 3
  1.

1, then lim rn 
 0. If r 
 1, then

2

13. a1 
 75, r 
 5

n→

Sn 


n→

3. Sample answer: 2  4  8 
4. Zonta is correct. As n approaches infinity the
expression 2n  3 will continue to grow larger and
larger. Tyree applied the method of dividing by the
highest powered term incorrectly. Both the
numerator and the denominator of the expression
must be divided by the highest-powered term. It is
not appropriate to apply this method here since
the denominator of the expression 2n  3 is 1.
5. 0; as n → , 5n becomes increasingly large and
1
thus the value 5n becomes smaller and smaller,
approaching zero. So the sequence has a limit of
zero.
n→

lim

n→

Pages 781–783
14. lim

n→




5

2

15.

7.

8.

3n  6

7n

0
0

7




n→



6n2  5

3n2



 lim 
3  
3n2 
6

5

n→

5
1
  lim 
2
n→ 3 n→ n



n→

5

17. lim

n→


 2  3  0 or 2

 5n  2

2n3
9n3




 lim 
2  
2n2  n3 
9

2

 lim


 lim 
3  n  n2 
1

1

n→ n


 lim (3)  lim
n→

1

2
n→ n

 lim 4  lim
n→


 3  0  4  0 or 3

126



9. 5.1
26


5
1000  1,000,000  

19. Dividing by the highest powered term, n2, we find

1



a1 
 
1000 , r 
 1000

Sn 
 5 

4

n→

7

126

1

5
1
1
  lim 

2  lim 
3
n→ 2 n→ n
n→ n
5
9
  0  0 or 
2
2
(3n  4)(1  n)
3n2  n  4
 
 lim 
lim 
n2
n2
n→
n→

n→
9

 2 

18.

5

9

n→


 lim


 9
126

n approaches


 lim 2  lim

3
6 1
lim

 lim 7  7  n
n→
n→
3
6
1

 lim 7  lim 7  lim n
n→
n→
n→
3
6
3

 7  7  0 or 7
7
7

0.7
 
 1
0  100  
7
1

a1 
 1
0 , r 
 10
7

10
Sn 
 —
1

1
10

5
2

8  n  
n2
——
3
2
n→


n2  n

lim

126

1000
——
1

1
1000

which as n approaches infinity

800

8



simplifies to 
0  0 
 0 . Since this fraction is
undefined, the limit does not exist.

126



5
999
14


 511

Chapter 12

2

1

n→

1
7
2
  lim   lim 
n→ 5 n→ n
n→ 5
7
2
2

 5  0  5 or 5
n3  2
2
lim n

 lim n  n .
2
n→
n→
2
1
lim n 
 lim 2  n 
 2  0 or 0, but as
n→
n→

16. lim

1




Exercises


 lim 
5  n  5

infinity, n becomes increasingly large, so the
sequence has no limit.

As n approaches infinity, 2n becomes increasingly
large, so the sequence has no limit.
3
,
7

7  2n

5n


 lim

1

5

75

2
1  
5


 125 m


 lim 
2n  2n

n→
5 1

 lim 2  n
n→

5

2n

1

or 3

Sn 
 —
1
1  3


1

lim rn 
 1. If r  1 then lim rn does not exist.

5  n2

2n

1

4
—
3

4


 

3

4


0

n→

6. lim

6
——
1
1  2


 4

The limits are equal.
2a. See students’ work. Student’s should draw the
following conclusions:
1 n
lim 2
n→
1 n
lim 
n→ 4
lim (1)n
n→
lim (2)n
n→
lim (5)n
n→

1

3

1

n→ n

 lim

410

20. lim

n→

4  3n  n2


2n3  3n2  5

4
3n
n2
 

n 3  n3  n 3

 lim 
2n3
3n2
5
n→ 
  
n3  n3  n 3
4
3
1

 
n3  n2  n


 lim
3
5
n→ 2    
n
n3
1
1
1


lim 4  lim 
3  lim 3  lim 
2  lim n
n→
n→ n
n→
n→ n
n→



1
1

lim 2  lim 3  lim n  lim 5  lim 
3
n→
n→
n→
n→
n→ n

259

259

259

1000

Sn 
 6  ——
1

1
1000
259



6
999
7


 62
7
15

15

1

15

100

the value 3n becomes smaller and smaller,
approaching zero. So the sequence has a limit of
zero.
22. Dividing by the highest powered term, n, we find

Sn 
 —
1

1
100
15

23.



2


 lim

n→

5n

n2

 lim

n→

5
  lim
n→ n
n→
(1)n

 lim n
2
n→


 lim

63

Sn 


1

1

5

63

1000

 —
1

1
100

29




110

(1)n


n2

30. The series is geometric, having a common ratio of
0.1. Since this ratio is less than 1, the sum of the
2
series exists and is 9.
12

16
Sn 
 
3
1  4

 64
7.5

32. r 
 5 or 1.5
This series is geometric with a common ratio of
1.5. Since this ratio is greater than 1, the sum of
the series does not exist.
5

4

10

1
Sn 
 —
1  1
0
4


 9

34. The series is arithmetic, having a general term of
7  n. Since lim 7  n does not equal zero, this
n→
series has no sum.

51



25. 0.5
1


100  10,000 …
1



a1 
 
100 , r 
 100
51

100

35. r 


Sn 
 —
1

1
100
370

1

4
—
1

8

or 2

This series is geometric with a common ratio of 2.
Since this ratio is greater than 1, the sum of the
series does not exist.

17


 99 or 3
3
370



26. 0.3
7
0

 
1000  1,000,000 …
370

1


33. r 
 1
0 or 2
10
Sn 
 
1
1  2

 20

1


a1 
 1
0 , r 
 10

51

3


31. r 
 1
6 or 4

4

51

63



 5  
990

(1)n

n2


24. 0.4
 
 1
0  100 …
4

63



a1 
 
1000 , r 
 100

As n increases, the value of the numerator
alternates between 1 and 1. As n approaches
infinity, the value of the denominator becomes
increasingly large, causing the value of the
fraction to become increasingly small. Thus, the
terms of the sequence alternate between smaller
and smaller positive and negative values,
approaching zero. So the sequence has a limit of
zero.
4

63



29. 0.26
3
 
 1
0  1000  100,000 …

1

5n  (1)n
lim n
2
n→

51

5


 99 or 3
3

1

 lim
 
4
n→
  1
n
which as n approaches infinity
1
(2)n

 1, but lim n has no limit since
lim 
4
n→   1
n→
n
o2o  1.

1



a1 
 
100 , r 
 100

21. As n → , 3n becomes increasingly large and thus




15



28. 0.1
5


100  10,000 …

40300

(2)n

n

1



a1 
 
1000 , r 
 1000




2  3  0  5  0 or 0

(2)n

n
lim —
4
n→   1
n

259



27. 6.2
5
9

6
1000  1,000,000 …

1



a1 
 
1000 , r 
 1000
370

1000

Sn 
 ——
1

1
1000
370

10





999 or 27

411

Chapter 12

36. r 


1

9
—
2
3

or




n2

n→ 2n  1

41b. lim

1
6

n2(2n  1)  n2(2n  1)

 


(2n  1)(2n  1)
2n  1  
 lim 
n2

n→

2n3  n2  2n3  n2


 lim 
4n2  1
n→

2

3
——
Sn 

1
1  
6

2n2

2
n→ 4n  1
2n2

n2
—

 lim 4n2
1
n→   
n2
n2


 lim

4


 7
37. r 


4

5
—
6

5

2

or 3


 lim

n→

6

5

Sn 
 —
2
1  3

1

42a. 12 4 
 3
3
r 
 3

a4 
 a1r41 3
3
4 
 a1
3

4 
 a1(3)
4
 
 a
1
3


5

1

 or 
38. r 
 
5
5



5
Sn 
 

5
1  5


 

5

42b. a28 
 a1r281

1  5

5
—

 


5

5
1  5
1  5
5
1
5

 
1  2
5

4


 3
3
27
4


 4(38)

 26,244
n
1
43. No; if n is even, lim cos 2 
 2, but if n is odd,

4

4
3


3

lim cos 2 
 2.

n→

8
Sn 
 

3
1  
2

1

1
D
4


3

2

exists. After 6 hours and before the second
1

1

1

1

1

44b. a1 
 D, r 
 8


3

1  
18n

Sn 
 D ——
1
1  8


3


 32
1  2


 7D1  
8
8


 32  163


1 n



44c. lim Sn 
 S
n→

2

40a. a1 
 35 r 
 5

a1


S

1r

 or 14

2

5

D

 
1
1  8

a3 
 14

a4 
 14
5 or 5.6
2

8


 7 D

a5 
 5.6
35, 14, 14, 5.6, 5.6
14
14

40b. Sn 
 35  
2 
2


1 5
1  5

8

44d. 350  7D
306.25  D
The largest possible dose is 306.25 mg.
20

45a. A side of the original square measures 4 or

70


 35  3  3

5

5 feet. Half of 5 feet is 2 feet.

2


 813 m or about 82 m


522  
522 
 s2

41a. The limit of a difference equals the difference of
the limits only if the two limits exist. Since
n2

2n  1

nor lim

n→

n2

2n  1

50

4
5
2

2

exists, this

property of limits does not apply.

Chapter 12

1

dose, 2  2  2D or 8D exists.

8
1  2

 ——
3
1  4

n→

1

44a. After 2 hours, 2D exists. After 4 hours, 2  2D or


1  
8

   ——


3
1  2

1  23 

neither lim

n→
1

n

39. r 
 8 or 2

20

3


 3(39)


 5(5
  1)

a2 
 35


1

4  n2


 2

3


 35

2



 s2

s
5
2

 feet.
Perimeter 
 4  2 or 102

412

51. vy 
 125 sin 20°

 42.75 miles
vx 
 125 cos 20°
117.46 miles


2

10
2


45b. a1 
 20, r 
 2
0 or 2

20
S
 

2
1  2

225°

20

 
 ——

2

1  2

1  22 



2






46d.
46e.


  
2

2

1

1 2

1


180˚

8

2,

8
1
,
2

1
,
4

1

8

Page 784
1. 1.618181818
2. N 
 24
x2 
 x  1
x2  x  1 
 0
(1) 
(1)2 
 4(1
)(1)

2(1)
1  
5
x 
 
2
1
Since the sum 1  
1

1
1  

x



3

negative, the value of 1 


6

2 4 6 8

1

x 
 1  x

3.

(y  2)2

 1 
 1

7
6

1,

4,

Continued Fractions

50.



p
:
q

12-3B Graphing Calculator Exploration:

16

5
6

2
2,

8


h 
 6, k 
 2, a 
 2, b 
 1, c 
 5
center: (6, 2)
foci: (6 5
, 2)
vertices: (8, 2) and (4, 2)

2

1,
1,

55. If b 
 1, then 4b  26 
 30 which is divisible by 2,
5, and 6.
If b 
 11, then 4b  26 
 70, which is divisible
by 7.
4b  26 is not divisible by 4 since 4 divides 4b
evenly, but does not divide 26 evenly. The correct
choice is B.

48. a16 
 1.5  (16  1)0.5

9
49. x2  4y2  12x  16y 
 16
(x  6)2  4(y  2)2 
 4

2
3

360˚

possible rational zeros,

2, 13, 9, 2
7

(x  6)2

4

180˚

54. possible values of p:
possible values of q:

16

1

O
1

9
3 
 27
8 2

y

53.

47. 3
3 
 2, 2
3 
 13, 13
3 
 9,
2

2  2


4


2  2


17.3032181(0.864605)n
15.0
12.9
11.2
9.7
8.4
7.2
6.2
5.4
4.7
4.0

15.0, 12.9, 11.2, 9.7, 8.4, 7.2, 6.2, 5.4, 4.7, 4.0
The 2000–2001 school year corresponds to the
9th term of the sequence, 4.7. The model is 0.3
below the actual statistic.
The 2006–2007 school year would correspond to
the 15th term of the sequence.
17.3032181(0.864605)15  2.0
Yes; as n → , 17.3032181(0.864605)n → 0.
No, the number of students per computer must
be greater than zero.

46c.

1  2

2



40  202
 ft or about 68 ft

46b.

1  cos 225°





2




20  102


 
2
1  4

n
1
2
3
4
5
6
7
8
9
10

20˚

52. cos 112.5° 
 cos 
2




1  22 

46a.

125 miles

cannot be

1

1

1
1  


1  5

is 
2 .

0

11
6
4
3

3
2

5
3

413

Chapter 12

11. Answers will vary. Sample answers: A 
 1, B 
 14;
A 
 4, B 
 1.
A2  B: If A 
 1 and B 
 14, 
12  1
4 
 15
.

1
1
4. Let x 
 3   , then x 
 3  x.
1


3 3

Solve for x.
x 
 3  1x

42  (1)
 
 15.
If A 
 4 and B 
 1, 

x2 
 3x  1
x2  3x  1 
 0
x




3 
(3)2 
 4(1
)(1)

2(1)
3 
13


2

Pages 790–791

Since the sum of positive numbers must remain
positive, x 


See students’ work.
See students’ work.
See students’ work.
In a given trigonometric series where r  1,
each succeeding term is larger than the one
preceding it. Therefore, the series approaches 
and thus does not converge.
2. As n → , S → 6.

1
1
6. Let x 
 A  
, then x 
 A  x.
1


A A

Solve for x.
1
x 
 A  x




sn

3a.

x2 
 Ax  1
 Ax  1 
 0

x


1.6
1.4
1.2
1
0.8
0.6
0.4
0.2

A 
 4(1
)(1)

2(1)
(A)2

A2  
4
A 

2

Since the sum of positive numbers must remain
positive, x 


A  
A2  
4

2

O 1 2 3 4 5 6 7 8

7. Sample program:
Program: CCFRAC
:Prompt A :Prompt B
:Disp “INPUT TERM”
:Disp “NUMBER N, N  3”
:Prompt N
:1 → K
:B  1/(2A) → C
:Lbl 1
:B  1/(2A  C) → C
:K  1 → K
:If K N  1
:Then: Goto 1
:Else: Disp C  A
8. For large values of N, the program output and the
decimal approximation of 
A2  
B are equal.
B
9.
x  A 
 2A  
B

2A  
2A 
x  A 
 2A 

n

3b. convergent
3c.

n2

3n

3d. We can use the ratio test to determine whether
the series is indeed convergent.
(n  1)2

n2


an 
 3n and an1 
 
3n1

r

(n  1)2


3n1

 lim —
n2
n→

3n

r 
 lim

n→

r 
 lim

n→

(n  1)2


3n1

3n

 n2

n2  2n  1

3n2
2
1
1  n  n

r 
 lim 
3
n→
1

r 
 3

B

xA

Since r

1, the series is convergent.

4. Consider the infinite series an
1. If the lim an 0, the sum of the series does not

(x 
x2  2Ax  A2
x2
x


 2A(x  A)  B

 2Ax  2A2  B

 A2  B

 
A2  
B
B
Since the sum A   cannot be
B

2A  
2A 
A)2

n→

exist, and thus the series is divergent. If the
lim an 
 0, the sum may or may not exit and

n→

therefore it cannot be determined from this test
if the series is convergent or divergent.
2. If the series is arithmetic then it is divergent.
3. If the series is geometric then the series
converges for r 1 and diverges for r  1.
4. Ratio test: the series converges for r 1 and
diverges for r  1. If r 
 1, the test fails. This
test can only be used if all the terms of the

B
negative, the value of A   is
B

2A  
2A 


A2  B
.

10. They will be opposites.
Chapter 12

Check for Understanding

1a.
1b.
1c.
1d.

3  
13

.
2

5. The output and the decimal approximation are
equal.

x2

Convergent and Divergent Series

12-4

414

series are positive and if the series can be
expressed in general form.
5. Comparison test: may only be used if all the
terms in the series are positive.

900

1500  1500
5
S 10 
 ———
3
1  5
3727 m
1500
12b. S 
 
3
1  5
3 10

n1

n


5. an 
 2n , an1 
 
2n1

r 
 lim

n→

an1

an

n1


2n1


 lim n
n→ 2
n
(n  1)2n


 lim n2n
1
n→
(n  1)

 lim 2n
n→
1
1

 lim 2  2n
n→
1

 2





 3750 m
No, the sum of the infinite series modeling this
situation is 3750. Thus, the spill will spread no
more than 3750 meters.


Pages 791–793

4n  1

4(n  1)  1


6. an 
 4n, an1 
 
4(n  1)

r

4(n  1)  1

4(n  1)

 lim ——
4n  1
n→

4n
(4n  3)(4n)


 lim (4n  4)(
4n  1)
n→
16n2  12n



 lim 16n2  12n  4
n→
16n2
12n


n2  n2

r

test provides no information
(n  1)

7. The general term is n.
1

 n for all n, so divergent
1


9. The general term is 
2  n2 .

10. an 

r

10


 5 or 2

for all n, so convergent

1

n  2n ,

2n1

5(n  1)

 lim —
2n
n→

5n
2n  2  5n


 lim 
2n  (5n  5)
n→
10n


 lim 
n→ 5n  5
10n

n


 lim —
5n
5
n→ 


n  n

8. The series is arithmetic, so it is divergent.


2n1

2n

16

1

n

4


3n1
—

 lim 4
n→ 
3n
4  3n

 lim 

n
n→ 4  3  3
1

 3


14. an 
 5n, an1 
 
5(n  1)


 1
6 or 1

1

2  n2

4

convergent


 lim ——
4
16n2
12n
n→ 



n2  n2  n2

n1

n

Exercises

4


13. an 
 3n , an 1 
 
3n1

convergent

r

3



12a. The series is geometric where r 
 
1500 or 5 .

divergent
2n

1


an1 
 
(n  1)2n1

2n1


15. an 
 n2 , an1 
 
(n  1)2

1

(n  1)2n1

 ——
1

n2n
n2n


 lim 
n1
n→ (n  1)2
n


 lim 
n→ 2(n  1)
n

n

r

2n1

(n  1)2

 lim ——
2n
n→

n2
2n  2  n2


n 2
n→ 2 (n  2n  1)
2
2n


 lim 
2
n→ n  2n  1


 lim

2n2

n2


 lim ——
n
1
n→
2
n  n


 lim ——
n2
2n
1
n→ 



n 2  n2  n 2

2
divergent

1




2(1  0)
1


 2
convergent
3

11. The series is geometric where r 
 4.
Since

3

4

1, it is convergent.

415

Chapter 12

2

1


24. The general term is 
2n  1 .

2



16. an 
 
(n  1)(n  2) , an1 
 (n  2)(n  3)

r

1

2n  1

2

(n  2)(n  3)

 lim ——
n→ 2
(n  1)(n  2)


 lim

n→

3

25. The series is geometric where r 
 4.
3

Since 4

2(n2  3n  2)


2(n2  5n  6)
n2

3n

2n  1

2

2n  1

2n  1

17. an 


1

1

5  n2

n→

1

1


n

2n  1

r

1


 lim 
n→ (2n  1)(2n)
1

 lim 
2 
n→ 4n  2n


0
convergent

5n1

1  2  ...  (n  1)

5n

r

2

convergent
31a. No, MagicSoft let a1 
 1,000,000 to arrive at
their figure. The first term of this series is
1,000,000  0.70 or 700,000.
31b. The series is geometric where a1 
 700,000 and
r 
 0.70.
700,000

S

1  0.70

2(n  1)  2(n  2)

2n1

2(n  1)  2(n  2)

2n1
2n  2(n  1)
n→

2n
2(n  1)  2(n  2)  2n

 lim 
2n  2(n  1)  2n  2
n→
n2

 lim 2n
n→
n
2

 lim 2n  2n
n→
1

 2


 $2.3 million

r 
 lim

1

1

1

 n for all n, so convergent
1


21. The general term is 
n3  1 .
1


n3  1

1

 n for all n, so convergent
n


22. The general term is 
n  1.
n

n1

1

 n for all n, so divergent
5


23. The general term is 
n  2.
5

n2

1

 n for all n, so divergent

Chapter 12

1

33a. Culture A: 1400 cells, Culture B; 713 cells
Culture A generates an arithmetic sequence
where a1 
 1000, d 
 200 and n 
 8
a8 
 1000  (8  1)200

 2400
Only considering cell growth, there are
2400  1000 or 1400 new cells.
Culture B generates a geometric sequence where
a1 
 1000 and r 
 1.08
a8 
 1000(1.08)81

1713 A part of a cell cannot be generated.
Only considering cell growth, there are
1713  1000 or 713 new cells.
33b. Culture A: a31 
 1000  (31  1)200

 7000
7000  1000 
 6000 cells
Culture B: a31 
 1000(1.08)311
 10,062
10,062  1000 
 9062
Culture B; at the end of one month, culture A
will have produced 6000 cells while culture B
will have produced 9062 cells.



20. The general term is 
(2n)2 or 4n2 .
1

4n2

1

32. the harmonic series: 1  2  3  4  

convergent
1

1


 4 or 2


0
convergent

2n  2(n  1)

2(n  1)  1


2n2
——

 lim
2n  1
n→


2n1
(2n  1)  2n  2

 lim 
(2n  1)  2n  2  2
n→
2n  1


 lim 
n→ 4n  2
1
2n
  
n
n


 lim —
4n
2
n→ 


n  n

5n1


 lim 1  2  ...n (n  1)
5
n→

1  2  ...  n
5n  5(1  2  ...  n)

 lim 
5n(1  2  ...  (n  1))
n→
5


 lim 
n→ n  1

19. an 
 2n, an1 


2(n  1)  1




30. an 
 
2n1 , an1 

2n  2

(2n  1)(1  2  ...  (2n  1))

(2n  1)(1  2  ...  (2n  1))


18. an 
 
1  2  ...  n , an1 


1

 n for all n, so divergent

29. The series is arithmetic, so it is divergent

2n  1

1  2  ...  (2n  1)

n→


 lim

an1 


1

 n for all n, so convergent

.
28. The general term is 
n


2(n  1)  1

1  2  ...  [2(n  1)  1]

2(n  1)  1

1  2  ...  [2(n  1)  1]

r 
 lim

1

 n for all n, so divergent


27. The general term is 
5  n2 .


1
test provides no information
2n  1
 ,
1  2  ...  (2n  1)

1, it is convergent.


26. The general term is 
2n  1 .

 n2  n2

 lim ——
2
n
5n
6
n→ 



n 2  n2  n 2

n2

1

 n for all n, so convergent

416

34a.

u
41. AB 
 5  8, 1  (3)

 3, 2

2n  1


2n1
1


n1
n→ 2
n
2(2  1)

 lim 2
n
n→
2

 lim 2  2n
n→

34b. S 
 lim Sn 
 lim
n→

2n




42. List all cubes from 1 to 200. There
are five. The correct choice is E.

n
1
2
3
4
5




 2 seconds
35a. When the minute hand is at 4, 20 minutes have

passed. This is 20
6
0  or 3 hour.
1

35b.

35c.

1

1
1
 the distance between 4 and 5 is (5) or
3
3
1
5
5 1
 minutes. This is   or  hour.
36
3
3 60
1
1
65
   (5) 
  minutes
36
3
36
5
65
5
   
  minute
36
36
3
5 1
1
  
  hour
36 60
432
1
1
1
785
     (5) 
  minutes
36
432
3
432
5
785
65
   
  minute
432
432
36
5
1
1
  
  hour
432 60
5184
1
1
, 
432 5184
1 1
1
1
 
The sequence 3, 3
6 , 432 , 5184 is geometric
1
where r 
 1
2.
a1

lim Sn 
 S 
 
1
r
n→
1

3







 




 

Page 793
20

2. S20 
 2[2(14)  (20  1)6]

 860
3. 189 
 56r41



27
 
 r3
8
3
 
 r
2
3
56 2 
 84,


 




6

3  3(2)8


S8 
 
1  (2)


 255
5. lim

n→

hour

36. lim

n→

5

 n2

 lim —
2
3
n
2n
n→   

n2
n2

2

2


137.5

1

1

25

S
 —
1
1  1
0
2


 4
5
1  2   n

1 2 …  (n  1)

, a
n
8. an 
 
n1 

10n
10 1
1  2  …  (n  1)

10n1
r 
 lim
12…n
n→

10n
10n[1  2  …  (n  1)]

 lim 
10n  10(1  2  …  n)
n→
n1

 lim 10
n→
n1
As n → , 10 → . Since

1


 2 r cos v  2 r sin v
1

1


7. a1 
 2
5 , r 
 10

91


 162

38. 19 
 11  (7  1)d
30 
 6d
5
d
11  5 
 6, 6  5 
 1, 1  5 
 4,
4  5 
 9, 9  5 
 14
11, 6, 1, 4, 9, 14, 19
39.
45.9 
 e0.075t
ln 45.9 
 0.075t
51.02 t
40. 6 
 12r cos (v  30°)
1
 
 r(cos v cos 30°  sin v sin 30°)
2

3

5

861 ft

or 2



3

2n

 n2  n2

 lim ——
n2
1
n→


n 2  n2

137.5

a9 
 2

2


1

2

n2

n2



Sn 
 250  
1  0.55  1  0.55

4


 3
37. r 


n2  2n  5


n2  1


1
6. a1 
 250, r 
 0.55
a2 
 250(0.55) or 137.5
a3 
 137.5
a4 
 137.5(0.55) or 75.625
a5 
 75.625 

4

4n2

n2

3

4. r 
 3 or 2

The hands will coincide at 4  11 o’clock,
approximately 21 min 49 s after 4:00.
4n2  5


3n2  2n

84
2 
 126

56, 84, 126, 189


 —
1
1  1
2
4

11

Mid-Chapter Quiz

1. a12 
 11  (19  1)(2)

 25


 

35d.

n3
1
8
27
64
125

r  1, the series is

divergent.

1

0 
 2x  2 y  2

1

9. The series is geometric where r 
 3.

0 
 3
x  y  1

Since r

417

1, it is convergent.

Chapter 12

60

10. 500
1  4 
 515
0.12

13a.

a1 
 515, r 
 1.03 n 
 4
S4 


 389(0.63)n1
n
1
S60 


515  515(1.03)4

1  1.03

389  389(0.63)60

1  0.63

1051 ft


 $2154.57

13b. S 


389

1  0.63

1051 ft

Sigma Notation and the nth term.

12-5

Pages 798–800

Exercises

4

Pages 797–798

14.

Check for Understanding

 (2  4  7)

 (5)  (3)  (1)  1

 8
5
15.  5a 
 5(2)  5(3)  5(4)  5(5)
a
2

 10  15  20  25


70
8
16.  (6  4b) 
 (6  4  3)  (6  4  4)  (6  4  5)
b
3
 (6  4  6)  (6  4  7)  (6  4  8)

 (6)  (10)  (14)  (18)
 (22)  (26)


96
6
17.  (k  k2) 
 (2  22)  (3  32)  (4  42) 
k
2
(5  52)  (6  62)

 6  12  20  30  42

 110

1. The series 4  6  8  10  12 can be
4

5

n
0

n
1

represented by  2n  4 or by  2n  2.
Sample answer: (1)n1
Sample answer: (1)n
9; 2, 3, 4, 5, 6, 7, 8, 9, 10
t
ba1
t
ba1

 3  (2)  1

6

2a.
2b.
3a.
3b.
3c.

3

3d.








k  3 
 2  3  1  3  0  5  1  3  2  3
k
2
1

1

1

1

1

1

1



33
1

1

1

1

1

1


 1  2  3  4  5  6

1

1

2



1

3



1

4



1

5



 (2n  7) 
 (2  1  7)  (2  2  7)  (2  3  7)

n
1

8

18.

1

5







n4 
 54  64  74  84
n
5
n



a
0



7.



p
0

1

1


 123

8

1

1


1


3 p

15
11
6



1

4

3 0



1

8



8.

1

21.

 (8  6n)
n
1

   5


9.

 (3k  1)
k
0

22.

1

1

1

 (0.5)i 
 (0.5)3  (0.5)4  (0.5)5


 8  16  32

 56
23.  k! 
 3!  4!  5!  6!  7!
k
3

 6  24  120  720  5040


5910
10
24.  4(0.75)p 
 4(0.75)0  4(0.75)1  4(0.75)2 
p
0
4(0.75)3    4(0.75)

 4  3  2.25  1.6875    4(0.75)
4

S

1  0.75



7

3



1


 852

5

3
 
4



11.

1

1

3 3

 3
2n
n
2
1


 16

 (3)n

n
1

Chapter 12

1

1



12.

 
2  4r 
 
2  41  
2  42  
2  43
r
1

 42  162  642

135

64


 20

 5n
n
1


 133

i
3

4

10.

1

3

3 2


 

 3m1 
 301  311  321  331
m
0

 3  1  3  9

5
4 
 5
4  5
4  5
4  5
4  
3 10
 5 4
15
45

 5  4  16


 16  32  64  128  256

 496

3

20.

1

16

3 1

 2j 
 24  25  26  27  28

j
4

1

5

Sn 

3
1  4
5

19.


 20  21  22  23  24
1

2

8

1

 (n  3) 
 (1  3)  (2  3)  (3  3) 
n
1
(4  3)  (5  3)  (6  3)

 (2)  (1)  0  1  2  3

3
5
5.  4k 
 4  2  4  3  4  4  4  5
k
2

 8  12  16  20

 56
4.

1

2a

7

7

There are 6 terms.

4

6


 5  3  3  2

6

6.

5

418



25.

2 n

 4
5
n
1

2 1

2 2

2 

2 3

47b. 0  1(1  x)  2(2  x)  3(3  x)  4(4  x)
 5(5  x) 
 25
1  x  4  2x  9  3x  16  4x  25  5x 
 25
55  15x 
 25
15x 
 30
x
2


 4
5  4
5  4
5    4
5



8

5



16

25



32

125

2 

5

   4




8

5

S
 —
2
1  5



5

26.



8

3

or

n

in

n
2

4

27.

7

48a. false;  3k 
 33  34    37

2
23


 (2 
 (3 
 (4 
 (5 

 (1)  (3  i)  (5)  (5  i)

 14
i3)

 (3k  3)
k
1


28.
30.

2k

10

32.



5k  1
k
2
1

34.



35.

 (1)kk2

36.



 (1)n1 2
n
0
32
n



39.



k


2k  3



k
22

k
1

41.



k
11

42.



38.



40.



k
2

2k

7

48c. true; 2
7

5

k!



(k  1)!

44.


 18  32    98 
 270

2n2 
 18  32    98 
 270

9

7

7

n
3

n
3

k
1

 (4  p) 
 4  5    13 
 85

p
0

10

Since 85

105,  (5  n)
k
1

9

 (4  p).
p
0

49a.
49b.
49c.
50a.

(a  1)(a)(a  1)(a  2)!

(a  2)!


 a(a  1)(a  1)



n
3

48d. false;  (5  n) 
 6  7    15 
 105

1
k
3 k!

1

(a  b)!

(a  b  1)!

n
2

n
3

10

k2




a(a  1)




n2

Since 270 
 270, 2 n2 
  2n2.

(a  2)!

(a  1)!

(a  2)!

9

n
3




a(a  1)(a  2)!

43.

8

Since 49 
 49,  (2n  3) 
  (2m  5).


3k!

(a  2)!

a!

(2m  5) 
 1  3    13 
 49

m
3

 k
k
1


b
7

n
2

9



2k  1
k
1



k
3

8

 (13  4k)
k
0

k
0

9

48b. true;  (2n  3) 
 1  3    13 
 49

 (2)3k



k
2

37.

9

 3a.
a
3

 4k
k
0



7

Since there are two 37 terms,  3k   3b

3

 2  5k
k
1

3b 
 37  38  39

b
7

k
0

4

33.



i5)

3

k
4

31.

i4)

4

12

29.

k
3

9

i2)

(a  b)(a  b  1)!

(a  b  1)!


 a

45. 43.64

6!
5! or 120
4! or 24, “LISTEN”
On an 8  8 chessboard, there is 1-8  8 square.
On an 8  8 chessboard, there are 4-7  7
squares, one in each of the four corners.
50b. For the 6  6 squares, begin in one corner.
For different configurations, you can move it
over, up to 2 more spaces, or down, up to 2 more
spaces. Thus, there are 3  3 or 9-6  6 squares.
Continue this procedure for the other sizes of
squares.
9-6  6, 16-5  5, 25-4  4, 36-3  3, 49-2  2,
and 64-1  1
8

50c.


 1  4  9  16  25  36  49  64

 204



46a.

 500,000(0.35)n
n
1

3

175,000


51. The general term is 
n  2.


46b. S 
 
1  0.35

3

n2


 269,239 people
46c.

269,230

500,000

 n2 
 12  22  32  42  52  62  72  82
n
1

1

 n for all n, so divergent

52. 0.42(21) 
 8.82 liters
If with each stroke 20% is removed, then 80%
remains.
a1 
 21(0.80) 
 16.8
a2 
 16.8(0.80) 
 13.44
a3 
 13.44(0.80) 
 10.752
a4 
 10.752(0.80) 
 8.6016
It will take 4 strokes for 42% of the air to remain.

53.8%

46d. The ad agency assumes that the people who buy
the tennis shoes will be satisfied with their
purchase.
47a. (x  3)  (x  6)  (x  9)  (x  12)  (x  15)
 (x  18) 
 3
6x  63 
 3
6x 
 60
x 
 60

419

Chapter 12

51

2c. Even indexed terms are negative and odd
indexed terms are positive.
3. The sum of the exponents of each term is n.
4. The exponents must add to 12, so the exponent of
y is 12  7 or 5.
To find the coefficient of the term use the formula

53. 322
 
 a1
2

322
 
 4a1
82
 
 a1

82

2
 
 16, 16
2
 
 162
,

54.
55.

56.
57.


2
 
 32
162
82
, 16, 162
, 32
log10 0.001 
 3
x2  y2  Dx  Ey  F 
 0
(0, 9) → 81  9E  F 
 0
(7, 2) → 49  4  7D  2E  F 
 0
(0, 5) → 25  5E  F 
 0
9E  F 
 81
9(4)  F 
 81
5E  F 
 25
F 
 45
14E

 56
53  7D  2(4)  45 
 0
E 
 4
7D 
 0
D
0
x2  y2  4y  45 
 0
x2  (y  2)2 
 49
(2
  i)(42
  i) 
 8  52
i  i2

 7  5i2

vy 
 59 sin 63°
52.57 ft/s
59 ft/s
vx 
 59 cos 63°
63˚
26.79 ft/s
2x1  3y1  9

n

(x  y)n 
 

r
0

Evaluate the general term for n 
 12 and r 
 5.
5.

35  6  5  4  3  2
 a
12345
36  6  5  4  3  2  1
 
123456

 a6  18a5  135a4 



1458a  729



3  2  5  (y)2

3  2  1  50 (y)3

321


 125  75y  15y2  y3
8. (3p  2q)4 
 (3p)4 (2q)0  4(3p)3 (2q)1
4  3(3p)2 (2q)2

4  3  2(3p)(2q)3

  

321
21

x1 4y1 4



17

4  3  2  1(3p)0(2q)4

4321
81p4  216p3q  216p2q2




9.

2

 16q4

 96pq3

7!
(a)7r(b)r
r!(7  r)!
7!
75(b)5



5!(7  5)! (a)
7654321

2 5

  
5432121 a b


 21a2b5
10.

The Binomial Theorem

r
9!
(x)9r 3

r!(9  r)!
3
9!
93 3




3!(9  3)! (x)




Check for Understanding










987654321

321654321

x6  33



 2523
 x6

1a. n 
 0: 1
n 
 1: 1  1 or 2
n 
 2: 1  2  1 or 4
n 
 3: 1  3  3  1 or 8
n 
 4: 1  4  6  4  1 or 16
n 
 5: 1  5  10  10  5  1 or 32
1, 2, 4, 8, 16, 32
1b. 2n
2a. The second term of (x  y)3 is 3x2y.
It is negative.
The second term of (x  y)4 is 4x3y.
It is negative.
The second term of (x  y)5 is 5x4y.
It is negative.
2b. The third term of (x  y)3 is 3xy2. It is positive.
The third term of (x  y)4 is 6x2y2. It is positive.
The third term of (y  y)5 is 10x3y2. It is positive.
Chapter 12

540a3  1215a2 


7. (5  y)3 
 53(y)0  3  52(y)  
21


 3

Pages 803–804

32  6  5

 4

12 a

34  6  5  4  3

3(5  m) 
 2(9  m)
15  3m 
 18  2m
m
3
The correct choice is D.

12-6

3

10c2d3  5cd4  d5

6a5

 2

1234 a

  13
x  
413


217
317
y  917
  413

0
5m

9m




a6

33  6  5  4

217
x1  317
y1  917


13
x1  413
y1  413


59.

3)6

 3

123 a

x1  4y1  4




12!
 x7y5 
 792x7y5.
5!7!
c5  5c4d  10c3d2 

6. (a 

, d 
 
58. d1 
 
2

13

17
2x1 3y1 9


1
3

n!
xnryr.
r!(n  r)!

11. (H  T)5 
 H5  5H4T  10H3T2  10H2T3  5HT2  T5
11a. 1
11b. 10
11c. 1  5 or 6
11d. 10  10  5  1 or 26

Pages 804–805

Exercises

12. (a  b)8 
 a8  8a7b  28a6b2  56a5b3  70a4b4
 56a3b5  28a2b6  8ab7  b8
6
13. (n  4) 
 n6  24n5  240n4  1280n3  3840n2
 6144n  4096
14. (3c  d)4 
 81c4  108c3d  54c2d2  12cd3  d4
15. (2  a)9 
 512  2304a  4608a2 5376a3 
4032a4  2016a5  672a6  144a7 
18a8  a9

420

7  6  d5  22

8  7(p2)6(q)2


16. (d  2)7 
 d7  20  7  d6  21  
21
7  6  5  d4  23



321


24. (p2  q)8 
 (p2)8(q)0  8(p2)7(q)1  
21

7  6  5  4  d3  24

4321



7  6  5  4  3  d2  25

54321



7  6  5  4  3  2  d1  26

654321



7  6  5  4  3  2  1  d0  27

7654321

8  7  6(p2)5(q)3

8  7  6  5  4(p2)3(q)5

54321
8  7  6  5  4  3(p2)2(q)6


654321
8  7  6  5  4  3  2(p2)1(q)7


7654321
8  7  6  5  4  3  2  1(p2)0(q)8


87654321




 d7  14d6  84d5  280d4  560d3 
672d2  448d  128
17. (3 

x)5




35(x)0

5



5  4  3  2  1  30(x)5
 
54321
243  405x  270x2





 p16  8p14q  28p12q2  56p10q3
 70p8q4  56p6q5  28p4q6  8p2q7
 q8
3
6
25. (xy  2z ) 
 (xy)6(2x3)0  6(xy)5(2z3)1

5  4  33(x)2


21
5  4  3  2  31(x)4

4321

34(x1)1

5  4  3  32(x)3
 
321

 90x3  15x4  x5
4



3(4a)2(b)2

43
4  3  2  1(4a)0(b)4
  

4321
321

 256a4  256a3b  96a2b2  16ab3



2(4a)1(b)3


 b4


 x6y6  12x5y5z3  60x4y4z6
 160x3y3z9  240x2y2z12
 192xyz15  64z18

19. (2x  3y)3 
 (2x)3(3y)0  3(2x)2(3y)1
3  2(2x)(3y)2






8x3

4

36x2y



3  2  1(2x)0(3y)3

321

54xy2

0



26.

27y3

27.

4  3  2(3m)1(2
 )3

321

m3  108m2 

 81m4  1082
242
m  4
6

0

6  5(c)4(1)2

6  5  4(c)3(1)3

321
6  5  4  3(c)2(1)4

4321
6  5  4  3  2(c)1(1)5

54321
6  5  4  3  2  1(c)0(1)6

654321



21




5




1

3

4

5

23. 
3a  3b 
 (3a)4
3b  4(3a)3
3b
2

0

2


 

1


 

2 2
2 3
4  3(3a)2 3b
4  3  2(3a)1 3b
  
21
321
2 4
0


4  3  2  1(3a) 3 b
 
4321

 81a4  72a3b  24a2b2 




 

32

9

9!
(3c)96(2d)6
6!(9  6)!

30.

1
10!
(x)107(y)7
7!(10  7)! 2

a5  22



 35  16a4  (b3)

 560a4b3


 84  27c3  64d6

 145,152c3d6
1


 120  8x3  (y7)

11!
(2p)115(3q)5
5!(11  5)!

84

4


y


 70x2y2

33. (M  W)8 
 M 8  8M 7W  28M 6W 2  56M 5W 3
 70M 4W 4  56M 3W 5  28M 2W 6
 8MW 7  W 8
70  56  28  8  1 or 163
34. Sample answer: Treat a  b as a single term and
expand [a  b)  c]12 using the Binomial
Theorem. Then evaluate each (a  b)n term in the
expansion using the Binomial Theorem.
35. (T  F)12 
 T12  12T11F  6610F2  220T9F3
 495T8F4  792T7F5  924T6F6
 792T5F7  495T4F8  220T3F9
 66T2F10  12TF11  F12
35a. 495
35b. 924  792  495  220  66  12  1 or 2510
36. Find the term for which both x’s have the same
exponent. This will occur for the middle term of
the expansion, the 4th term when n 
 6. Use the


 

4

29.


x

5
3
2
 4

 3
2 n  8 n  5n  20n  40n  32
2

7!
(2a)73(b)3
3!(7  3)!

8!

4!(8  4)!


 
321
4321
1 0
5


5  4  3  2  1 2 n (2)

54321

1




 462  64p6  (243q5)

 7,185,024p6q5
32. The middle term is the fifth term.


 

4

2




28.

31.

5  4 2n (2)2

21


12n  2 
 
12n (2)0  5
12n (2)1 
1
1
5  4  3
2n (2)
5  4  3  2
2n (2)

x5y4


 15x3y7


 c3  6c2c  15c2  20cc  15c
 6c  1
1 3
22.

3
8!
(a)83 2

3!(8  3)!
87654321

 
32154321

 1122
a5

5

21. 
c  1 
 
c (1)  6
c (1)1



987654321

432154321





 126x5y4

 )4
4  3  2  1(3m)0(2

4321



6



9!
(x)94(y)4
4!(9  4)!

1

20. 
3m  2
 
 (3m)4
2
  4(3m)3
2
 
)2
4  3(3m)2(2

21

6  5  4(xy)3(2z3)3

321
6  5  4  3(xy)2(2z3)4

4321
6  5  4  3  2(xy)1(2z3)5

54321
6  5  4  3  2  1(xy)0(2z3)6

654321

6  5(xy)4(2z3)2



21


18. (4a  b)4 
 (4a)4(b)0  4(4a)3(b)1  
21



21

8  7  6  5(p2)4(q)4

  

4321
321

16

ab3  81b4

421

Chapter 12

Binomial Theorem to find the 4th term for the
1 6
expansion of 3x  4x .






3


41x 

6!
(3x)3
3!3!

12-7

135


 16

Page 809

37a. Sample answer: 1  0.01
37b. Sample answer: 1.04060401
(1  0.01)4 
 14  4  13  0.011  6  12  0.012
 4 11  0.013  0.014

 1.04060401
7

 5  2k 
 (5  2  2)  (5  2  3)  (5  2  4)
x
2

 (5  2  5)  (5  2  6)  (5  2  7)

 1  (1)  (3)  (5)  (7)  (9)

 24

2n

x11


6. 
11!

2n1


39. an 
 n! an1 
 
(n  1)!

r

Pages 811–812

2n1

(n  1)!

 lim ——
2n
n→

n!


 lim

n→

2n  2  n!


2n  (n  1)  n!

2

n→ n  1


0
convergent
1

40. This is a geometric series where r 
 2.



41.

2

3
—
1
1  2
1
13

Pn 
 P

n

1  (1  i)


i

150,000 
 P








0.08 12(30)

1 1
12
———
0.08

12



(0.8)2

(1.36)2

(1.36)3

(1.36)4

1  1.36  0.925  0.419  0.143
3.85
9. sin x

r  0.3183098862 mi
r  0.3183098862(5280)
1681 feet
44. Test all answer choices that are prime integers.
You can eliminate answer choice C.
?
? 5
A: 3(2) 
10 
6 (2)

sin 

x3

x5

x7

x9

x  3!  5!  7!  9!
sin 3.1416
(3.1416)3
(3.1416)5
(3.1416)7
3.1416  3
 5
 7
!
!
!
(3.1416)9

 9
!
3.1416  5.1677  2.5502  0.5993  0.0821

0.0069
3

3


3


cos 4  i sin 4 
 2
 ei 4
10. 2

?


Chapter 12

(0.8)4

 3
 4
8. e1.36 
 1  1.36  2
!
!
!


4

? 5
6 10 
3 ; false
?
? 5
B: 3(3)  10  6 (3)
?
? 5
10 
9
2 ; false
?
? 5
D: 3(11)  10  6 (11)
55
?
? 

10 
33 
6 ; true

(0.8)3



7. e0.8 
 1  0.8  2
!  3!  4!
1  0.8  0.32  0.08  0.017
2.22

150,000 
 P(136.283491)
$1100.65 
 P
u  (8  6)j
u  (2  3)k
u
 
 [4  (2)]i
42. MK
u
u
u

 6i  2j  5k
43.
s 
 rv
0.25 
 r

Check for Understanding

1. The approximation given in Example 1 only used
the first five terms of the exponential series.
Using more terms of the exponential series would
give an approximation closer to that given by the
calculator.
2. Sample answer: 2  x  1.5
3. The problem seems to imply that siblings mate.
Genetically, this can lead to problems. Another
problem is the assumption that each birth
produces only two offspring, one male and one
female. Rabbits are more likely to give birth to
more than two offspring and the ratio of male to
female births is not guaranteed to be 1 to 1.
4. an1 
 an  an1 for n  2
5. ln (7) 
 ln (1)  ln (7)
i  1.9459
6. ln (0.379) 
 ln (1)  ln (0.379)
i  0.9702


 lim

S


Graphing Calculator Exploration

1. Sample answer (without zooming): 2.4  x  2.4
2. Sample answer for greatest difference: about 0.08;
The least difference is 0.
3. Sample answer: 3.4  x  3.4; sample answer:
about 0.15; 0
4. Sample answer: 3.8  x  3.8; sample answer:
about 0.05; 0
5. larger

37c. 1.04060401; the two values are equal.
38.

Special Sequences and Series


3

11. 1  3
i 
 2
2  2i
1

2

2


 2
cos 3  i sin 3

 2e

The correct choice is D.

422

2
i 3

12a.

A 
 Pert
2P 
 Pe(0.06)5
P 
 e0.3



cos 6

(0.5236)2

0.2742



(1.1)3

3!



(0.2)2



(4.2)3



cos 

x6

33. r 


34. r 




sin 4

7

7

7


35. r 
 
32  32 or 32



v 
 Arctan 
3 or 4
3

(2.73)4








3  3i 
 32

cos 4  i sin 4 
 32
 ei 4
36. Sample answer: A transcendental number is one
that cannot be the root of an algebraic equation
with rational coefficients. Examples are  and e.
(3.1416)8

eix  eix

2i






cos x  i sin x  (cos x  i sin x)

2i
2i sin x

2i


 sin x

cos x  i sin x  cos x  i sin x
eix  eix
 
 
2
2
2 cos x

 2

x9

x  3!  5!  7!  9!


 cos x
38. See students’ work.

sin 0.7854
(0.7854)3

7

43
  4i 
 8
cos 6  i sin 6 
 8ei 6

37.

x7


2 
3
(4)2 or 8

4
4

1  4.9348  4.0588  1.3353  0.2353
0.9760
actual value: cos  
 1
x5

3


3

   or 
v 
 Arctan 

6
4
3

cos 3.1416

x3



3

 4
 6
 8
1  2
!
!
!
!

26. sin x

2
2


2

2
 or 2
  


2
  2
i 
 2
cos 4  i sin 4 
 2ei 4

(3.5)4

4!

(3.1416)6







2

x8

(3.1416)4



 or 3
v 
 Arctan 

4
2


1  2!  4!  6!  8!
(3.1416)2






3
  i 
 2
cos 6  i sin 6 
 2ei 6

1  2.73  3.726  3.391  2.314
13.16
x4








(0.2)4

 3
 4
24. e2.73 
 1  2.73  2
!
!
!

x2

5


5

32. r 
 
(3
)2
 12 or 2
1

v 
 Arctan 
or 6
3

(0.55)4

(2.73)3

sin 1.5708



1  3.5  6.125  7.146  6.253
24.02
(2.73)2

x9

1  i 
 2

cos 4  i sin 4 
 2
 ei 4

1  0.55  0.151  0.028  0.004
1.73


0.8660

x7

31. r 
 
12  12 or 2

1

Q 
 Arctan 1 or 4

 3
 4
22. e0.55 
 1  0.55  2
!
!
!

(3.5)3

3!

0.0056

1.5708  0.6460  0.0797  0.0047  0.0002



(4.2)4

(0.55)3

0.0206

30. i 
 cos 2  i sin 2 
 ei 2

1  4.2  8.82  12.348  12.965
39.33
(0.55)2

x5

5

 3
 4
21. e4.2 
 1  4.2  2
!
!
!

25. cos x

(0.5236)8

29. 5
cos 3  i sin 3 
 5ei 3

1  0.2  0.02  0.0013  0.00007
0.82
(4.2)2

0.7516

1.0000

actual value: sin 2 
 1

 3
 4
20. e0.2 
 1  (0.2)  2
!
!
!

23. e3.5 
 1  3.5 

(0.5236)6

x  3!  5!  7!  9!

sin 2

(1.1)4

4!

(0.2)3

x3

28. sin x

1  1.1  0.605  0.222  0.06
2.99

(3.5)2

2!

(0.5236)4

0.8660


3
actual value: cos 6 
 2

ln (4) 
 ln (1)  ln 4 i  1.3863
ln (3.1) 
 ln (1)  ln (3.1) i  1.1314
ln (0.25) 
 ln (1)  ln (0.25) i  1.3863
ln (0.033) 
 ln (1)  ln (0.033) i  3.4112
ln (238) 
 ln (1)  ln (238) i  5.4723
ln (1207) 
 ln (1)  ln (1207) i  7.0959

19. e1.1 
 1  1.1 

x8



1  2  24  
720  40,320

Exercises

(1.1)2

2!

x6

1  2
 4
 6
 8
!
!
!
!

 1.345
approximately 1.345P
12b. No, in five years she will have increased her
savings by about 34.5%, not 100%.
12c. The approximation is accurate to two decimal
places.

13.
14.
15.
16.
17.
18.

x4

cos 0.5236

(0.3)2

2!


 1  0.3 

Pages 812–814

x2

1  2!  4!  6!  8!

27. cos x

(0.7854)5

(0.7854)7

0.7854  3
 5
 7
!
!
!
(0.7854)9

 9
!
0.4845

0.2989

0.1843

0.1137




0.7854  6  
120  5040  362, 880

0.7071


2
actual value: sin 4 
 2

0.7071

423

Chapter 12

39. If you add the numbers on the diagonal lines as
shown, the sums are the terms of the Fibonacci
sequence.

0.01


45a. a1 
 0.005, r 
 
0.005 or 2

a3 
 0.005(2)31

 0.020 cm
a4 
 0.020(2)

 0.040 cm
45b. 0.005(2)n1
45c. a10 
 0.005(2)101

 2.56 cm
a100 
 0.005(2)1001
3.169  1027 cm

1
1

1
1

2
3

1

5

1

2

1

8
13

1

3

1
1
1

4
5

3
6

4

10

6

15

1

10
20

.
.
.

1
5

2


46.

1

15

6

Fn  1

2
1.75
1.5
1.25
1
0.75
0.5
0.25

14

48. r 
 6 or 4




(0.65)2



(0.65)3

(0.65)4

42a.
42b.




75.5

sin 140°



50a.

6

44.







12xy5



y6

v


 t
5(2) radians


 10 radians per second
v

50b. v 
 r t

 
2 ft(10 radians/s)
1

15.7 ft/s
51. Let m 
 multiple choice. Let e 
 essay.
m  e  30
1m  12e  96
e
f(m, e) 
 5m  20e
30
f(24, 6) 
 5(24)  20(6)
m  e 
 30

 240
m  12e 
 96
20
f(0, 8) 
 5(0)  20(8)

 160
10 (0, 8)
(24, 6)
f(30, 0) 
 5(30)  20(0)
m
0

 150
m
10 20
(0, 0)
(30, 0)
e
0
f(0, 0) 
 0
To receive the highest score, answer 24 multiple
choice and 6 essay.

 160x3y3

2k

k
1

Chapter 12

30 sin 140°



75.5




1 second

6  5  4  3  2(2x)1(y)5

54321

60x2y4

30




sin v

v 
 Arcsin

 14°48










30 sin 140°

6  5  4  3(2x)2(y)4

4321

6  5  4  3  2  1(2x)0(y)6

654321
64x6  192x5y  240x4y2

15


sin v 
 
75.5

6  5(2x)4(y)2

6  5  4(2x)3(y)3

15

4 cos 8  i sin 8
49. u
r 2 
 302  502  2(30)(50) cos 140°
u
r 
 75.5 N


43. (2x  y)6 
 (2x)6(y)0  6(2x)5(y)1  
21


321

15


 8 or 8

5000 1  0.65  2
 3
 4
!
!
!
$9572.29
No, the account will be short by more than
$30,000.
about 42 years; 47 years old
40,000 
 Pe0.65
$20,882 
 P
Every third Fibonacci number is an even
number.
Every fourth Fibonacci number is a multiple of 3.

41c.
41d.



v 
 8  4

n

1 2 3 4 5 6 7 8 9 10

40d. yes; 1.618
40e. The two ratios are equivalent to three decimal
places.
40f. See students’ work.
41a. A 
 Pert

 5000e0.05(13)

 5000e0.65

41b.

1



 8 3 or 2

8

40b. neither
40c. Fn

O

1

3

47. y2  Dx  Ey  F 
 0
(0, 0): F 
 0
(2, 1): 1  2D  E  F 
 0
(4, 4): 16  4D  4E  F 
 0
2D  E 
 1
4D  2E 
 2
4D  4E 
 16 → 4D  4E 
 16
2E 
 14
2D  E 
 1
E
7
2
2D  7 
 1
y  3x  7y 
 0
2D 
 6
D
3

1

1 2 3 5 8 13 21 34 55 89
, , , , , , , , , 
1 1 2 3 5 8 13 21 34 55

40a.

83


424

2

52. (DC)2  22 
 
18

(DC)2 
 14
DC 
 14


24

10. pn 
 40 
 0.60
p1 
 0.60  1.75(0.60)(1  0.60)

 1.02
(1.02)(40) 41
p2 
 1.02  1.75(1.02)(1  1.02)

 0.9843
(0.9843)(40) 39
p3 
 0.9843  1.75(0.9843)(1  0.9843)

 1.0113
(1.0113)(40) 40
p4 
 1.0113  1.75(1.0113)(1  1.0113)

 0.9913
(0.9913)(40) 40
p5 
 0.9913  1.75(0.9913)(1  0.9913)

 1.0064
(1.0064)(40) 40
p6 
 1.0064  1.75(1.0064)(1  1.0064)
0.9951
(0.9951)(40) 40
p7 
 0.9951  1.75(0.9951)(1  0.9951)
1.0036
(1.0036)(40) 40
p8 
 1.0036  1.75(1.0036)(1  1.0036)
0.9973
(0.9973)(40) 40
p9 
 0.9973  1.75(0.9973)(1  0.9973)
1.002
(1.002)(40) 40
p10 
 1.002  1.75(1.002)(1  1.002)
0.9985
(0.9985)(40) 40
41, 39, 40, 40, 40, 40, 40, 40, 40, 40

2



14

12-8
Page 819

 52 
 (BC)2
39 
 (BC)2
39
 
 BC
The correct choice is C.

Sequences and Iteration
Check for Understanding

1. Iteration is the repeated composition of a function
upon itself.
2. It is the sequence of iterates produced when a
complex number is iterated for a function f(z).
3. If the prisoner set is connected, then the Julia set
is the boundary between the prisoner set and the
escape set. If the prisoner set is disconnected, then
the Julia set is the prisoner set.
4. f(1) 
 (1)2 
 1
f(1) 
 12 
 1
f(1) 
 12 
 1
f(1) 
 12 
 1
1, 1, 1, 1
5. f(2) 
 2  2  5 
 1
f(1) 
 2  (1)  5 
 7
f(7) 
 2  (7)  5 
 19
f(19) 
 2  (19)  5 
 43
1, 7, 19, 43
6. z0 
 6i
z1 
 0.6(6i)  2i 
 5.6i
z2 
 0.6(5.6i)  2i 
 5.36i
z3 
 0.6(5.36i)  2i 
 5.216i
7. z0 
 25  40i
z1 
 0.6(25  40i)  2i 
 15  26i
z2 
 0.6(15  26i)  2i 
 9  17.6i
z3 
 0.6(9  17.6i)  2i 
 5.4  12.56i
8. z0 
 0, f(z) 
 z2  (1  2i)
z1 
 02  (1  2i) 
 1  2i
z2 
 (1  2i)2  (1  2i)

 1  4i  4i2  1  2i 
 2  6i
z3 
 (2  6i)2  (1  2i)

 4  24i  35i2  1  2i 
 31  22i
9. z0 
 1  2i, f(z) 
 z2  (2  3i)
z1 
 (1  2i)2  (2  3i)

 1  4i  4i2  2  3i 
 1  i
z2 
 (1  i)2  (2  3i)

 1  2i  i2  2  3i 
 2  5i
z3 
 (2  5i)2  (2  3i)

 4  20i  25i2  2  3i 
 19  23i

Pages 820–821

Exercises

11. f(x0) 
 f(4)

 3(4)  7

5
f(x1) 
 f(5)

 3(5)  7

8
f(x2) 
 f(8)

 3(8)  7

 17
f(x3) 
 f(17)

 3(17)  7

 44
12. f(2) 
 (2)2 
 4
f(4) 
 42 
 16
f(16) 
 162 
 256
f(256) 
 2562 
 65.536
13. f(4) 
 (4  5)2 
 1
f(1) 
 (1  5)2 
 16
f(16) 
 (16  5)2 
 121
f(121) 
 (121  5)2 
 13,456
14. f(1) 
 (1)2  1 
 0
f(0) 
 02  1 
 1
f(1) 
 (1)2  1 
 0
f(0) 
 02  1 
 1

425

Chapter 12

19. z0 
 1  2i
z1 
 2(1  2i)  (3  2i)

 2  4i  3  2i

 5  2i
z2 
 2(5  2i)  (3  2i)

 10  4i  3  2i

 13  2i
z3 
 2(13  2i)  (3  2i)

 26  4i  3  2i

 29  2i
20. z0 
 1  2i
z1 
 2(1  2i)  (3  2i)

 2  4i  3  2i

 1  6i
z2 
 2(1  6i)  (3  2i)

 2  12i  3  2i

 5  14i
z3 
 2(5  14i)  (3  2i)

 10  28i  3  2i

 13  30i
21. z0 
 6  2i
z1 
 2(6  2i)  (3  2i)

 12  4i  3  2i

 15  2i
z2 
 2(15  2i)  (3  2i)

 30  4i  3  2i

 33  2i
z3 
 2(33  2i)  (3  2i)

 66  4i  3  2i

 69  2i
22. z0 
 0.3  i
z1 
 2(0.3  i)  (3  2i)

 0.6  2i  3  2i

 3.6  4i
z2 
 2(3.6  4i)  (3  2i)

 7.2  8i  3  2i

 10.2  10i
z3 
 2(10.2  10i)  (3  2i)

 20.4  20i  3  2i

 23.4  22i

15. f(0.1) 
 2(0.1)2  0.1

 0.08
f(0.08) 
 2(0.08)2  (0.08)
0.09
f(0.09) 
 2(0.09)2  0.09
0.07
f(0.07) 
 2(0.07)2  (0.07)
0.08
2

16a. t1 
 1 
 2
2

t2 
 2 
 1
2

t3 
 1 
 2
2

t4 
 2 
 1

2
t10 
 2 
 1
2

1

16b. t1 
 4 
 2
t2 

t3 

t4 


2
 
4
1

2
2
1
 
 
4
2
2
 
4
1

2


t10 


2

1

2


4

2

16c. t1 
 7
t2 

t3 

t4 


2

2

7
2

7
2

2

7


t10 


2

2

7


7


7


7
2

16d. The values of the iterates alternate between x
0
and x0.

1

z1 
 3
3  3i  2i
1

17. z0 
 5i
z1 
 2(5i)  (3  2i)

 3  8i
z2 
 2(3  8i)  (3  2i)

 6  16i  3  2i

 9  14i
z3 
 2(9  14i)  (3  2i)

 18  28i  3  2i

 21  26i
18. z0 
 4
z1 
 2(4)  (3  2i)

 11  2i
z2 
 2(11  2i)  (3  2i)

 22  4i  3  2i

 25  6i
z3 
 2(25  6i)  (3  2i)

 50  12i  3  2i

 53  14i

Chapter 12

2

23. z0 
 3  3i
2


 1  2i  2i

1
z2 
 3(1)  2i

 3  2i
z3 
 3(3  2i)  2i

 9  6i  2i

 9  8i
24. z0 
 0  i, f(z) 
 z2  1
z1 
 (i)2  1

 2
z2 
 (2)2  1

3
z3 
 32  1

8

426

25. z0 
 i, f(z) 
 z2  1  3i
z1 
 i2  1  3i

 3i
z2 
 (3i)2  1  3i

 8  3i
z3 
 (8  3i)2  1  3i

 64  48i  9  1  3i

 56  45i
26. z0 
 1, f(z) 
 z2  3  2i
z1 
 12  3  2i

 4  2i
z2 
 (4  2i)2  3  2i

 16  16i  4  3  2i

 15  18i
z3 
 (15  18i)2  3  2i

 225  540i  324  3  2i

 96  542i
27. z0 
 1  i, f(z) 
 z2  4i
z1 
 (1  i)2  4i

 1  2i  1  4i

 2i
z2 
 (2i)2  4i

 4  4i
z3 
 (4  4i)2  4i

 16  32i  16  4i

 28i

2

31.


2


2

2

1

x 
 percent
0.10
0.325
0.8734
1.1498
0.7192
1.2241
0.5383
1.1596
0.6969
1.225
0.5359
1.1577
0.7013
1.225
0.5359
1.1577
0.7013
1.225

x  2.5x(1  x)
0.325
0.8734
1.1498
0.7192
1.2241
0.5383
1.1596
0.6969
1.225
0.5359
1.1577
0.7013
1.225
0.5359
1.1577
0.7013
1.225
0.5359

2002  1984 
 18; After 18 years, about 54% of
the maximum sustainable population is present.
32.
f(z) 
 z2  c
1  15i 
 (2  3i)2  c
1  15i 
 4  12i  9i2  c
4  3i 
 c

33. 2
2
x1 
 

 2

2

28. z0 
 2  2i, f(z) 
 z2


 2

Iteration
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

1

z1 
 2  2i 
 2  i  2i2 
 1
z2 
 (i)2 
 1
z3 
 12 
 1
29. z0 
 1  i, f(z) 
 z2  2  3i
z1 
 (1  i)2  2  3i

 1  2i  1  2  3i

2i
z2 
 (2  i)2  2  3i

 4  4i  1  2  3i

 5  7i
z3 
 (5  7i)2  2  3i

 25  70i  49  2  3i

 22  73i
30. p1 
 p0  rp0
p1 
 2000  (0.052)(2000)

 $2104
p2 
 2104  (0.052)(2104)

 $2213.41
p3 
 2213.41  (0.052)(2213.14)

 $2328.51
p4 
 2328.51  (0.052)(2328.51)

 $2449.59
p5 
 2449.59  (0.052)(2449.59)

 $2576.97

x1 



2


2


 2



34. See students’ work. Sample topics for discussion
are judging soil quality and detection of heat
stress in cows.
35a. 1.414213562, 1.189207115, 1.090507733,
1.044273782
35b. f(z) 
 z, z0 
 2
35c. 1
35d. 1







36. 2
cos 3  i sin 3 
 2ei 3
37.

8!
(2a)84(3b)4
4!(8  4)!


 70  16a4  81b4

 90,720a4b4

38. Convergent; the series is geometric with
1

r 
 4

1.

39. The distance between the vertices is 130 ft.
2a 
 130, so a 
 75.
c

7

91

e 
 a 
 5 
 65
b2 
 c2  a2
b2 
 912  652

 4056
x2

a2

427

y2

x2

y2



 b2 
 1 ⇒ 
4225  4056 
 1

Chapter 12

40.

n1 sin I 
 n2 sin r
1.00 sin 42° 
 2.42 sin r
r 
 Arcsin
r

12-9





Page 826

16°

40 ft

56˚
42˚

41b. Let h 
 height of the building.
Let x 
 distance from the point of elevation to
the center of the base of the building.
h  40

h

tan 42° 
 x

tan 56° 
 x
h  40

h


x

tan 42°

tan 56° 
 h
(h  40) tan 42°

tan 56° 
 h
1.6466

40

1  h
40

0.6466 
 h
h

62 feet

No, the height of the building is about 62 feet for
a total of about 102 feet with the tower.
42. x 2 for all x, so infinite discontinuity
43. y
x  y 
 14
(5, 9)

y
9
(8, 6)

y
5
(3, 5)

(8, 5)

x
3

x
8
x

O

f(x, y) 
 2x  8y  10
f(3, 9) 
 2(3)  8(9)  10

 88
f(5, 9) 
 2(5)  8(9)  10

 92
f(8, 6) 
 2(8)  8(6)  10

 74
f(8, 5) 
 2(8)  8(5)  10

 66
f(3, 5) 
 2(3)  8(5)  10

 56
max: 92, min: 56
44.

HL

H  L 
 2
2H  2L 
 H  L
H 
 3L
H
 
 3
The correct choice is D.
L

Chapter 12

Check for Understanding

1. The n  1 case shows that the premise is true for
an infinite number of cases.
2. Provide a counterexample.
3a. n(n  2)
3b. Since 3 is the first term in the sequence of
partial sums and 1(1  2) 
 3, the formula is
valid for n 
 1.
Since 8 is the second term in the sequence of
partial sums and 2(2  2) 
 8, the formula is
valid for n 
 2.
Since 15 is the third term in the sequence of
partial sums and 3(3  2) 
 15, the formula is
valid for n 
 3.
3c. Sk ⇒ k(k  2); Sk1 ⇒ (k  1)(k  3)
4. 8n  1 
 7r for some integer r.
5. Sample answer: If we wish to prove that we can
climb a ladder with an indefinite number of steps,
we must prove the following. First, we must show
that we can climb off the ground to rung 1. Next,
we must show that if we can climb to rung k, then
we can climb to rung k  1.
6. Step 1: Verify that the formula is valid for n 
 1.
Since 3 is the first term in the sequence and
1(1  2) 
 3, the formula is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
Sk ⇒ 3  5  7 … (2k  1) 
 k(k  2)
Sk1 ⇒ 3  5  7 …
 (2k  1)  (2k  3) 
 k(k  2)  (2k  3)

 k2  4k  3

 (k  1)(k  3)
Apply the original formula for n 
 k  1.
(k  1)[(k  1)  2] 
 (k  1)(k  3)
The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
 3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
7. Step 1: Verify that the formula is valid for n 
 1.
Since 2 is the first term in the sequence and
2(21  1) 
 2, the formula is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
Sk ⇒ 2  22  23 … 2k 
 2(2k  1)
Sk1 ⇒ 2  22  23 …  2k  2k1

 2(2k  1)  2k1

 2  2k1  2

 2(2k1  1)
When the original formula is applied for n 
 k  1,
the same result is obtained. Thus if the formula is
valid for n 
 k, it is also valid for n 
 k  1. Since
the formula is valid for n 
 1, it is also valid for
n 
 2, n 
 3, and so on indefinitely. Thus, the
formula is valid for all positive integral values
of n.

41a.

(3, 9)

Mathematical Induction

1.00 sin 42°

2.42

428

8. Step 1: Verify that the formula is valid for n 
 1.
1
Since 2 is the first term in the sequence and
1

The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
 3, and so on indefinitely. Thus, the formula
is valid for all positive integral values of n.

1

1  21 
 2, the formula is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
1
1
1
1
1
Sk ⇒ 2  22  23 … 2k 
 1  2k
1

1

1

1

1

1

1




Sk1 ⇒ 2  22  23 … 2k  
2k1 
 1  2k  2k1
2

Pages 826–828

1




1
2  2k  2k1
2

1




1
2k1  2k1
1



1
2k1

When the original formula is applied for n 
 k  1,
the same result is obtained. Thus if the formula is
valid for n 
 k, it is also valid for n 
 k  1. Since
the formula is valid for n 
 1, it is also valid for
n 
 2, n 
 3, and so on indefinitely. Thus, the
formula is valid for all positive integral values
of n.
9. Sn: 3n  1 
 2r for some integer r
Step 1: Verify that Sn is valid for n 
 1.
S1 ⇒ 31  1 or 2. Since 2 
 2  1, Sn is valid for
n 
 1.
Step 2: Assume that Sn is valid for n 
 k and show
that it is also valid for n 
 k  1.
Sk ⇒ 3k  1 
 2r for some integer r
Sk1 ⇒ 3k1  1 
 2t for some integer t
3k  1 
 2r
3(3k  1) 
 3  2r
3k1  3 
 6r
3k1  1 
 6r  2
3k1  1 
 2(3r  1)
Thus, 3k1  1 
 2t, where t 
 3r  1 is an
integer, and we have shown that if Sn is valid,
then Sk1 is also valid. Since Sn is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on
indefinitely. Hence, 3n  1 is divisible by 2 for all
integral values of n.
n(n  1)
10a. 6  4 
 10
10b. an 
 2
10  5 
 15
15  6 
 21
21  7 
 28
28  8 
 36
10, 15, 21, 28, 36
10c. Step 1: Verify that the formula is valid for n 
 1.
Since 1 is the first term in the sequence and
1(1  1)(1  2)

6

(1)[3(1)  1]

2


 1, the formula is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
k(3k  1)

Sk ⇒ 1  4  7 … (3k  2) 
 2
Sk1 ⇒ 1  4  7 … (3k  2)  (3k  1)
k(3k  1)


 2  (3k  1)
k(3k  1)

3k2  5k  2


 2
(k  1)(3k  2)


 2
Apply the original formula for n 
 k  1.
(k  1)[3(k  1)  1]

2

k(k  1)

(k  1)(k  2)

(k  1)(3k  2)


 2

The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
 3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.

Step 2: Assume that the formula is valid for
n 
 k and derive a formula for n 
 k  1.
(k  1)(k  2)

2(3k  1)


 2  2


 1, the formula is valid for n 
 1.

k(k  1)

Exercises

11. Step 1: Verify that the formula is valid for n 
 1.
Since 1 is the first term in the sequence and
(1)[2(1)  1] 
 1, the formula is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
Sk ⇒ 1  5  9 … (4k  3) 
 k(2k  1)
Sk1 ⇒ 1  5  9 … (4k  3)  (4k  1)

 k(2k  1)  (4k  1)

 2k2  3k  7

 (k  1)(2k  1)
Apply the original formula for n 
 k  1.
(k  1)[2(k  1)  1] 
 (k  1)(2k  1)
The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2.
Since it is valid for n 
 2, it is also valid for n 
 3,
and so on indefinitely. Thus, the formula is valid
for all positive integral values of n.
12. Step 1: Verify that the formula is valid for n 
 1.
Since 1 is the first term in the sequence and

Sk ⇒ 1  3  6 … 2 
 6
Sk ⇒ 1  3  6 … 2  2
k(k  1)(k  2)

(k  1)(k  2)


 6  2





k(k  1)(k  2)  3(k  1)(k  2)

6
(k  1)(k  2)(k  3)

6

Apply the original formula for n 
 k  1.
(k  1)[(k  1)  1][(k  1)  2]

6




(k  1)(k  2)(k  3)

6

429

Chapter 12

13. Step 1: Verify that the formula is valid for n 
 1.
1
Since 2 is the first term in the sequence and
1
1


21  1 
  2 , the formula is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
1
1
1
1
1
Sk ⇒ 2  4  8 … 2k 
 2k  1
1

1
1
1
   … 
2k
4
8
1
1


2k  1  2k1
2
1


2  2k  1  2k1
2
1
k1
  1  k1

2
2
1
k1
1
2

Sk1 ⇒ 2 











15. Step 1: Verify that the formula is valid for n 
 1.
Since 1 is the first term in the sequence and
1[2(1)  1][2(1)  1]

3

1


2k1

Sk1 ⇒ 12  32  52 … (2k  1)2  (2k  1)2









When the original formula is applied for n 
 k  1,
the same result is obtained. Thus if the formula is
valid for n 
 k, it is also valid for n 
 k  1. Since
the formula is valid for n 
 1, it is also valid for
n 
 2, n 
 3, and so on indefinitely. Thus, the
formula is valid for all positive integral values of n.
14. Step 1: Verify that the formula is valid for n 
 1.
Since 1 is the first term in the sequence and




(k  1)[2(k  1)  1][2(k  1)  1]

3

k2(k  1)2

Sk ⇒ 1  8  27 … k3 
 4
Sk1 ⇒ 1  8  27 … k3  (k  1)3
k2(k  1)2


 4  (k  1)3






k2(k  1)2  4(k  1)3

4
(k  1)2[k2  4(k  1)]

4
(k  1)2(k2  4k  4)

4
(k  1)2(k  2)2

4

Apply the original formula for n 
 k  1.
(k  1)2[(k  1)  1]2

4

(k  1)2(k  2)2


 4

The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
 3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.

Chapter 12




(k  1)(2k  1)(2k  3)

3

The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
 3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
16. Step 1: Verify that the formula is valid for n 
 1.
Since S1 ⇒ 1 and 21  1 
 1, the formula is valid
for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
Sk ⇒ 1  2  4 … 2k1 
 2k  1
Sk1 ⇒ 1  2  4 … 2k1  2k 
 2k  1  2k

 2(2k)  1

 2k1  1
When the original formula is applied for n 
 k  1,
the same result is obtained. Thus if the formula is
valid for n 
 k, it is also valid for n 
 k  1. Since
the formula is valid for n 
 1, it is also valid for
n 
 2, n 
 3, and so indefinitely. Thus, the
formula is valid for all positive integral values of n.
17. Sn ⇒ 7n  5 
 6r for some integer r
Step 1: Verify that Sn is valid for n 
 1.
S1 ⇒ 71  5 or 12. Since 12 
 6  2, Sn is valid for
n 
 1.
Step 2: Assume that Sn is valid for n 
 k and show
that it is also valid for n 
 k  1.
Sk ⇒ 7k  5 
 6r for some integer r
Sk1 ⇒ 7k1  5 
 6t for some integer t
7k  5 
 6r
7(7k  5) 
 7  6r
7k1  35 
 42r
7k1  5 
 42r  30
7k1  5 
 6(7r  5)
k1
Thus, 7
 5 
 6t, where t 
 7r  5 is an
integer, and we have shown that if Sn is valid,
then Sk1 is also valid. Since Sn is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on
indefinitely. Hence, 7n  5 is divisible by 6 for all
integral values of n.


 1, the formula is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.




k(2k  1)(2k  1)
  (2k  1)2
3
k(2k  1)(2k  1)  3(2k  1)2

3
[k(2k  1)  3(2k  1)](2k  1)

3
(2k2  5k  3)(2k  1)

3
(2k  3)(k  1)(2k  1)

3

Apply the original formula for n 
 k  1.

12(1  1)2

4





 1, the formula is valid for n 
 1.

Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
k(2k  1)(2k  1)
Sk ⇒ 12  32  52 … (2k  1)2 
 
3

430

18. Sn ⇒ 8n  1 
 7r for some integer r
Step 1: Verify that Sn is valid for n 
 1.
S1 ⇒ 81  1 or 7. Since 7 
 7  1, Sn is valid for
n 
 1.
Step 2: Assume that Sn is valid for n 
 k and show
that it is also valid for n 
 k  1.
Sk ⇒ 8k  1 
 7r for some integer r
Sk1 ⇒ 8k1  1 
 7t for some integer t
8k  1 
 7r
8(8k  1) 
 8  7r
8k1  8 
 56r
8k1  1 
 56r  7
8k1  1 
 7(8r  1)
Thus, 8k1  1 
 7t, where t 
 8r  1 is an
integer, and we have shown that if Sn is valid,
then Sk1 is also valid. Since Sn is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on
indefinitely. hence, 8n  1 is divisible by 7 for all
integral values of n.
19. Sn ⇒ 5n  2n 
 3r for some integer r
Step 1: Verify that Sn is valid for n 
 1.
S1 ⇒ 51  21 or 3. Since 3 
 3  1, Sn is valid for
n 
 1.
Step 2: Assume that Sn is valid for n 
 k and show
that it is also valid for n 
 k  1.
Sk ⇒ 5k  2k 
 3r for some integer r
Sk1 ⇒ 5k1  2k1 
 3t for some integer t
5k  2k 
 3r
5k 
 2k  3r
5k  5 
 (2k  3r)(2  3)
5k1 
 2k1  3(2k)  6r  9r
k1
5
 2k1 
 2k1  3(2k)  6r  9r
 2k1

 3(2k)  15r

 3(2k  5r)
Thus, 5k1  2k1 
 3t, where t 
 2k  5r is an
integer, and we have shown that if Sn is valid,
then Sk1 is also valid. Since Sn is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on
indefinitely. Hence, 5n  2n is divisible by 3 for all
integral values of n.
20. Step 1: Verify that the formula is valid for n 
 1.
Since a is the first term in the sequence and
1
[2a  (1  1)d] 
 a, the formula is valid for
2
n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
Sk ⇒ a  (a  d)  (a  2d) … [a  (k  1)d]

Apply the original formula for n 
 k  1.
(k  1)
{2a
2

(k  1)

 [(k  1)  1]d} 
 2(2a  kd)

The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
3, and so on indefinitely. Thus , the formula is
valid for all positive integral values of n.
21. Step 1: Verify that the formula is valid for n 
 1.
1
Since 2 is the first term in the sequence and
1

11

1


 2, the formula is valid for n 
 1.

Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
1

1

1

k





Sk ⇒  
2  3  3  4 … k(k  1) 
 k  1
1

1

1

1

1






Sk1 ⇒ 
1  2  2  3  3  4 … k(k  1)  (k  1)(k  2)
k

1





k  1  (k  1)(k  2)
k(k  2)  1




(k  1)(k  2)
k2  2k  1




(k  1)(k  2)
(k  1)2




(k  1)(k  2)
k1




k2

Apply the original formula for n 
 k  1.
(k  1)

(k  1)  1

k1




k2

The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
 3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
22. Sn ⇒ 22n1  32n1 
 5r for some integer r
Step 1: Verify that Sn is valid for n 
 1.
S1 ⇒ 22(1)1  32(1)1 or 35. Since 35 
 5  7, Sn is
valid for n 
 1.
Step 2: Assume that Sn is valid for n 
 k and show
that it is also valid for n 
 k  1.
Sk ⇒ 22k1  3k1 
 5r for some integer r
Sk1 ⇒ 2k3  32k3 
 5t for some integer t
22k1  32k1 
 5r
22k1 
 5r  32k1
22k1  22 
 (5r  32k1)(32  5)
22k3 
 45r  25r  32k3  5(32k1)
2k3
2
 32k3 
 45r  25r  32k3  5(32k1)
 32k3

 20r  5(32k1)

 5(4  32k1)
Thus, 22k3  32k3 
 5t, where t 
 4  32k1 is
an integer; and we have shown that if Sn is valid,
then Sk1 is also valid. Since Sn is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on
indefinitely. Hence, 22n1  32k1 is divisible by 5
for all integral values of n.

k


 2[2a  (k  1)d]
Sk1 ⇒ a  (a  d)  (a  2d) … [a  (k  1)d]
k

 (a  kd) 
 2[2a  (k  1)d]  (a  kd)
k[2a  (k  1)d]  2(a  kd)


 
2
2ak  k(k  1)d  2a  2kd


 
2
(k  1)2a  [k(k  1)  2k]d


 
2
(k  1)2a  (k2  k)d


 
2
(k  1)2a  k(k  1)d


 
2
(k  1)


 2(2a  kd)

431

Chapter 12

23. Step 1: Verify that the formula is valid for n 
 1.
Since S1 ⇒ [r(cos v  i sin v]1 or r(cos v  i sin v)
and r1[cos (1)v  i sin (1)v] 
 r(cos v  i sin v), the
formula is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
That is, assume that [r(cos v  i sin v)]k 

rk(cos kv  i sin kv).
Multiply each side of the equation by
r(cos v  i sin v).
[r(cos v  i sin v)]k1

 [rk(cos kv  i sin kv]  [r(cos v  i sin v)]

 rk1[cos kv cos v  (cos kv(i sin v)
 i sin kv cos v  i2 sin kv sin v]

 rk1[(cos kv cos v  sin kv sin v)
 i(sin kv cos v  cos kv sin v)]

 rk1[cos (k  1)v  i sin (k  1)v]
When the original formula is applied for
n 
 k  1, the same result is obtained. Thus if
the formula is valid for n 
 k, it is also valid for
n 
 k  1. Since the formula is valid for n 
 1, it
is also valid for n 
 2, n 
 3, and so on
indefinitely. Thus, the formula is valid for all
positive integral values of n.
24a.

(k  1)2  5(k  1) 
 k2  2k  1  5k  5

 (k2  5k)  (2k  6)

 2r  2(k  3)

 2(r  k  3)
Thus, k2  5k 
 2t, where t 
 r  k  3 is an
integer, and we have shown that if Sn is valid,
then Sk1 is also valid. Since Sn is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on
indefinitely. Hence, n2  5n is divisible by 2 for all
positive integral values of n.
26a. Number of people
Number of Interactions
2
1 2
1
3

1 2

3
3

4

1 2 3

4
6





n(n  1)

2

n

26b. Step 1: Verify that Sn ⇒ 0  1  2  3 …
n(n  1)

 (n  1) 
 2 is valid for n 
 1.
Since 0 is the first term in the sequence and
1(1  1)
 
 0, the formula is valid for n 
 1.
2
Step 2: Assume that the formula is valid for
n 
 k and derive a formula for n 
 k  1.
k(k  1)

Sk ⇒ 0  1  2 … (k  1) 
 2

24b. 4  1 
 3
94
5
16  9 
 7
1, 3, 5, 7, …
24c. 2n  1
24d. n2
24e. Step 1: Verify that the formula is valid for n 
 1.
Since 1 is the first term in the sequence and
12 
 1, the formula is valid for n 
 1.
Step 2: Assume that the formula is valid for
n 
 k and derive a formula for n 
 k  1.
Sk ⇒ 1  3  5  7 … (2k  1) 
 k2
Sk1 ⇒ 1  3  5  7 … (2k  1)  (2k  1)

 k2  (2k  1)

 k2  2k  1

 (k  1)2
When the original formula is applied for
n 
 k  1, the same result is obtained. Thus if
the formula is valid for n 
 k, it is also valid for
n 
 k  1. Since the formula is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on
indefinitely. Thus, the formula is valid for all
positive integral values of n.
25. S1 ⇒ n2  5n 
 2r for some positive integer r
Step 1: Verify that S1 is valid for n 
 1.
S1 ⇒ 12  5  1 or 6. Since 6 
 2  3, S1 is valid for
n 
 1.
Step 2: Assume that Sn is valid for n 
 k and show
that it is valid for n 
 k  1.
Sk ⇒ k2  5k 
 2r for some positive integer r
Sk1 ⇒ (k  1)2  5(k  1) 
 2t for some positive
integer t

Chapter 12

k(k  1)

Sk1 ⇒ 0  1  2 … (k  1)  k 
 2  k
k(k  1)  2k


 2
k2  k  2k


 2
k(k  1)


 2
Apply the original formula for n 
 k  1.
(k  1)[(k  1)  1]

2

k(k  1)


 2

The formula gives the same result as adding the
(k  1) term directly. Thus if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for
n 
 2, n 
 3, and so on indefinitely. Thus, the
formula is valid for all positive integral values
of n.
15(14)

26c. Yes; 15 people will require 2 or 105
interactions and last approximately 105(0.5) or
52.5 minutes.

432

27. Step 1: Verify that Sn ⇒ (x  y)n 
 xn  nxn1y 
n(n  1)
xn2y2
2!



n(n  1)(n  2)

3!

When the original formula is applied for
n 
 k  1, the same result is obtained. Thus if
the formula is valid for n 
 k, it is also valid for
n 
 k  1. Since the formula is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on in
indefinitely. Thus, the formula is valid for all
positive integral values of n.

xn3y3 … yn is

valid for n 
 1. Since S1 ⇒ (x  y)1 
 x1  y1 or
x  y, Sn is valid for n 
 1.
Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
k(k  1)

Sk ⇒ (x  y)k 
 xk  kxk1y  2
xk2y2 
!
k(k  1)(k  2)

3!

28e. lim

n→

k(k  1)(k  2)

k(k  1)

xk2y2 
(x  y)k1 
 x(xk  kxk1y  2
!

k(k1)

2!

… yk)  y(xk  kxk1y 

k(k  1)(k  2)

xk2y2  3
xk3y3 … yk)
!
k(k  1)

xk1y2 …

 xk1  kxky  2
!
k(k  1) k2 3
k
k
k1
2
xy  x y  kx
y  2
x
y
!
k1
… y
k(k  1)

 xk1  (k  1)xky  kxk1y2  2
!
xk1y2 … yk1

31. 25x2  4y2  100x  40y  100 
 0
25(x2  4x)  4(y2  10y) 
 100
25(x  2)2  4(y  5)2 
  100  100  100
25(x  2)2 2 4(y  5)22 
 100
(x  2)
(y  5)
   
 1
4
25

k(k  1)


 xk1  (k  1)xky  2
xk1y2
!
k1
… y
When the original formula is applied for n 
 k  1,
the same result is obtained. Thus if the formula is
valid for n 
 k, it is also valid for n 
 k  1. Since
the formula is valid for n 
 1, it is also valid for
n 
 2, n 
 3, and so on indefinitely. Thus, the
formula is valid for all positive integral values of n.
28a. 0.9  0.09  0.009 …
28b.

9

10n

28c. S 


ellipse
32. 
 
 250, k 
 4
(250)2 


A

4(4)

12,271.85 m2
33.
34.

a1  a1 rn


1r
9
9 1 n
   
10
10 10


 


1
10




n




 
9





2




 2; y 



10
1

1  101
10n  1

10n


  
 30° Since 23 is

H

9



Sk1 ⇒ 0.9  0.09  0.009 …  
10k  10k1
9





10k  10k1

60˚
1

C

F
D

B

G

perimeter is 2(2
  42) 
 2 (52) 


10(10k  1)  9




10k1

102
 units.

10k1  10  9




10k1





3

Since A
BF
E
 and B
E
 measure 1 unit, A
H
 and 

each measure 4 units. 
AB
 is the hypotenuse of
an isosceles right triangle with legs that measure
1 unit. So A
2 units. 
CA
B
 measures 
 is the
hypotenuse of an isosceles right triangle with
legs that measure 4 units. So 
CA
 measures
4
2 units. Since ABCD is a rectangle, the

Step 2: Assume that the formula is valid for
n 
 k and derive a formula for n 
 k  1.
10k  1
9


Sk ⇒ 0.9  0.09  0.009 … 
10k 
 10k
9

30˚

2

A

Since 0.9 is the first term in the sequence and
101  1

101 
 0.9, the formula is valid for n 
 1.

1

sin 2x

3

cos1 2

28d. Step 1: Verify that Sn: 0.9  0.9  0.009 …
10n  1
9


10n 
 10n is valid for n 
 1.

10k

3

4

negative, x is
in the second
or third quadrants.
x 
 180°  30° or 150°
x 
 180°  30° or 210°
35. Since the area
E
of the square
is 25, each side
measures 5.


 
1
9
1
 1  
10
10

n→

Thus, 0.999… 
 1.
29. z1 
 2(4  i)  i

8i
z2 
 2(8  i)  i

 16  i
z3 
 2(16  i)  i

 32  i
30. 64 
 6  (29  1)d
70 
 28d
5
 
 d
2

 3
xk3y3 … yk)
!

k(k  1)(k  2)
xk3y3
3!

1



 lim 1  
10n


 1  0 or 1

xk3y3 … yk

Sk1 ⇒ (x  y)k(x  y) 
 (x  y)(xk  kxk1y 
k(k1)
xk2y2
2!

10k  1

10n

The correct choice is B.

10k1  1
k
10 1

433

Chapter 12

4
2

21. r 
 4 or 2


Chapter 12 Study Guide and Assessment

)8
4  4(2

S8 
 

Page 829

1  2


Understanding the Vocabulary

1. d
5. k
9. b

2. i
6. f
10. h

Pages 830–831

3. m
7. c

60

2
1  

)
60(1  2




12


 60
1

3n

n

22. lim

 lim 4n 1
n→
n→   
n
n
3



4
6n  3
6n
3


23. lim n 
 lim n  lim n
n→
n→
n→

Skills and Concepts


60

6

24. Does not exist; lim

n→

25. lim

n→

4n3  3n


n4  4n3

123

1

Sn 
 5 

5
41

504


27. r 
 
1260 or 0.4
1260


Sn 
 
1  0.4

1


 2100


16. r 
 4
9 or 7








17. a15 
 2.2(2)151

 36,044.8
18.
8 
 a1(0.2)71
8 
 0.000064a1
125,000 
 a1
19. 125 
 0.2r51
625 
 r4
5
r
0.2( 5) 
 1, 1( 5) 

0.2, 1, 5, 25, 125
20.

1

7




n2

1

343

r

(n  1)2


5n1

 lim 
n2
n→

5n


 lim

n→

5n(n2  2n  1)


5n  5  n2

n2
2n
  lim 
2
2
n→ 5n
n→ n
1

 5  0  0
1

 5


 lim

5,

5( 5) 


(n  1)2


28. an 
 5n , an1 
 
5n1

25

1

2
n→ n

 lim

convergent
n5

2.4

r

1.2 or 2
1.2  1.2(2)9

S9 
 
1  (2)
1.2  614.4

 3

5

29. The general term is n or 1  n.
5

1

1  n  n for all n, so divergent
2

30. The general term is n.
2

n


 205.2

Chapter 12

123

1000

1

1
1000
123

999



 5
333

n 
 18 or n 
 19.86
Since n is a positive whole number, n 
 18.
1
1
, 
49 49

123



a1 
 
1000 , r 
 1000

1.3 
(1.3)3 
 4(0
.7)(2
50.2)

2(0.7)
1.3 26.5

1.4

1

7

since lim

4n3
3n
 
n4  n 4



 lim n4 4n3
n→   
n4
n4
0

 1
123

250.2 
 2n  0.7n(n  1)
0 
 0.7n2  1.3n  250.2

1
1 1
1  7 
 7, 7
1 1
1
, , 
7 49 343

n→

2n
;
3



26. 5.1
2
3

5
1000  1,000,000 …

n

7


 lim


0

250.2 
 2[2(2)  (n  1)(1.4)]




2nn3

3n3

n→

2n

3

becomes increasingly large as n approaches
infinity, the sequence has no limit.


 217
n
Sn 
 2[2a1  (n  1)d]

n


2


3n

4n  1

11. d 
 4.3  3 or 1.3
5.6  1.3 
 6.9, 6.9  1.3 
 8.2,
8.2  1.3 
 9.5, 9.5  1.3 
 10.8
6.9, 8.2, 9.5, 10.8
12. a20 
 5  (20  1)(3)

 52
13. 4 
 6  (5  1)d
10 
 4d
2.5 
 d
6  (2.5) 
 3.5, 3.5  (2.5) 
 1,
1  (2.5) 
 1.5
6, 3.5, 1, 1.5, 4
14. d 
 23  (30) or 7
a14 
 30  (14  1) 7

 61
14
S14 
 2(30  61)
15.

1  
2



   

1  2

4. j
8. e

434

1

 n for all n, so divergent

9

31.

46. f(3) 
 (3)2  4 
 13
f(13) 
 132  4 
 173
f(173) 
 1732  4 
 29,933
f(29,933) 
 29,9332  4 
 895,984,493
13; 173; 29.933; 895, 984, 493
47. z0 
 4i
z1 
 0.5(4i)  (4  2i) 
 2i  4  2i 
 4
z2 
 0.5(4)  (4  2i) 
 2  4  2i 
 6  2i
z3 
 0.5(6  2i)  (4  2i)

 3  i  4  2i 
 7  3i
48. z0 
 8
z1 
 0.5(8)  (4  2i) 
 4  4  2i 
 2i
z2 
 0.5(2i)  (4  2i) 
 i  4  2i 
 4  3i
z3 
 0.5(4  3i)  (4  2i)

 2  1.5i  4  2i 
 6  3.5i
49. z0 
 4  6i
z1 
 0.5(4  6i)  (4  2i)

 2  3i  4  2i 
 2  i
z2 
 0.5(2  i)  (4  2i)

 1  0.5i  4  2i 
 5  1.5i
z3 
 0.5(5  1.5i)  (4  2i)

 2.5  0.5i  4  2i 
 6.5  2.75i
50. z0 
 12  8i
z 
 0.5(12  8i)  (4  2i)

 6  4i  4  2i 
 10  6i
z2 
 0.5(10  6i)  (4  2i)

 5  3i  4  2i 
 9  5i
z3 
 0.5(9  5i)  (4  2i)

 4.5  2.5i  4  2i 
 8.5  4.5i
51. Step 1: Verify that the formula is valid for n 
 1.
Since the first term in the sequence is 1 and

 (3a  3) 
 (3 5  3)  (3  6  3)  (3  7  3)
a
 5
 (3.8  3)  (3.9  3)


 12  15  18  21  24

 90



32.

 (0.4)k 
 (0.4)1  (0.4)2  (0.4)3 … (0.4)
k
1
0.4


S

1  0.4
2


 3



33.

9

 (2n  1)
a
0

34.

 
1(n2  1)
a
1

6!

6!

5
1
4


35. (a  4)6 
 a6  
1!(6  1)!  a  (4)  2!(6  2)!  a
6!

6!

3
3 

 (4)2  
3!(6  3)!  a  (4)  4!(6  4)!
6!

1
5

 a2  (4)4  
5!(6  5)!  a  (4) 
6!

6!(6  6)!

 a0  (4)6


 a6  24a5  240a4  1280a3  3840a2
 6144a  4096
4!

3
1

36. (2r  3s)4 
 (2r)4  
1!(4  1)! (2r) (3s)
4!

2
2


2!(4  2)! (2r) (3s)
4!

3


3!(4  3)! (2r)(3s)
4!

0
4


4!0! (2r) (3s)


 16r4  96r3s  216r2s2  216rs3
 81s4
37.

10!

4!(10  4)!

38.

8!

2!(8  2)!

39.

10!

7!(10  7)!

 x107  (3y)7 
 120  x3  2187y7

 262,440x3y7

40.

12!

5!(12  5)!

 (2c)125  (d)4 
 792  128c7  (d)5

 101,376c7d5

 x104  (2)4 
 210  x6  16

 3360x6

41. 2
cos

3

4

 4m82  12 
 28  4096m6  1

 114,688m6

3

4

 
 2e


42. 4i 
 4
cos 2  i sin 2
 i sin

1(1  1)

2

Step 2: Assume that the formula is valid for n 
 k
and derive a formula for n 
 k  1.
k(k  1)

Sk ⇒ 1  2  3 … k 
 2
Sk1 ⇒ 1  2  3 … k  (k  1)

3
i4

k(k  1)
k2

k2  k  2k  2
k2  3k  2


 2
(k  1)(k  2)

7


 2
7

2  2i 
 22

cos 4  i sin 4

 22
e

Apply the original formula for n 
 k  1.
(k  1)[(k  1)  1]

2

7
i 4



3



(k  1)(k  2)


 2
The formula gives the same result as adding the
(k  1) term directly. Thus, if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
 3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.

44. r 
 
(33
)2
 32 or 6
 or 
v 
 Arctan 

6
3
3

2k  2


 2

2  (2)
43. r 
 2
2 or 22


7

k


 2  2




2

2(k  1)


 2  2


 4ei 2

v 
 Arctan 
2 or 4


 1, the formula is valid for n 
 1.



33
  3i 
 6
cos 6  i sin 6
i6


 6e
45. f(2) 
 6  3  2 
 0
f(0) 
 6  3  0 
 6
f(6) 
 6  3  6 
 12
f(12) 
 6  3  (12) 
 42
0, 6, 12, 42

435

Chapter 12

52. Step 1: Verify that the formula is valid for n 
 1.
Since the first term in the sequence is 3 and
1(1  1)(2  1  7)

6

12

54b. S12 
 2(16  368)

 2304 ft
55. If the budget is cut 3% each year, 97% remains
after each year.
a1 
 160,000,000, r 
 0.97
a11 
 160,000,000(0.97)111

 $117,987,860.30
6
56a. One side of the original triangle measures 3 or
2 units. Half of 2 units is 1 unit. Each side of the
new triangle measures 1 unit, so its perimeter is
1  1  1 or 3 units.


 3, the formula is valid for n 
 1.

Step 2: Assume that the formula is valid for n 
 k
and derive formula for n 
 k  1.
k(k  1)(2k  7)

Sk ⇒ 3  8  15 … k(k  2) 
 6
Sk1 ⇒ 3  8  15 … k(k  2)  (k  1)(k  3)
k(k  1)(2k  7)


 6  (k  1)(k  3)
k(k  1)(2k  7)

6(k  1)(k  3)


 6  6

3




k(k  1)(2k  7)  6(k  1)(k  3)

6




(k  1)[k(2k  7)  6(k  3)]

6




(k  1)(2k2  7k  6k  18)

6




(k  1)(2k2  13k  18)

6

S


Page 833




Apply the original formula for n 
 k  1.



(k  1)(k  2)(2k  9)

6

Open-Ended Assessment

2. Sample answer: 3n; lim
lim

n→

2

n


 0, but since

n→
5n
lim 3
n→

6  5n2

3n




2

n→ n


 lim

 3
5n

becomes increasingly

large as n approaches infinity, the sequence has
no limit.

Chapter 12 SAT & ACT Preparation
Page 835

SAT and ACT Practice

1. If 40% of the tapes are jazz then 60% of the tapes
must be blues. There are 80 tapes. Find 60% of 80.
0.60(80) 
 48
The correct choice is D.
2. Because of alternate interior angles,
150 
 130  unmarked angle of right 
20 
 unmarked angle of right 
In the right triangle, x  20 
 90 or x 
 70.
The correct choice is C.
d gallons

d



3. fraction pumped 
 
k gallons 
 k

Change this fraction into a percent by multiplying
by 100.
d
100d
percent pumped 
 k  100 or k%
The correct choice is A.
4. First calculate the number of caps Andrei has now.
He starts with 48 and gives away 13, so he has
48  13 
 35 left. Then he buys 17, so he has
35  17 
 52. Then he trades 6 caps for 8 caps.
this leaves him with 52  6  8 or 54. His total is
now 54.

Applications and Problem Solving

54a. a1 
 16, d 
 48  16 or 32
A12 
 16  (12  1)32

 368 ft

Chapter 12


1

6  5n2

The formula gives the same result as adding the
(k  1) term directly. Thus, if the formula is valid
for n 
 k, it is also valid for n 
 k  1. Since the
formula is valid for n 
 1, it is also valid for n 
 2,
n 
 3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
53. Sn ⇒ 9n  4n 
 5r for some integer r
Step 1: Verify that Sn is valid for n 
 1.
S1 ⇒ 91  41 or 5. Since 5 
 5  1, Sn is valid for
n 
 1.
Step 2: Assume that Sn is valid for n 
 k and show
that it is also valid for n 
 k  1.
Sk ⇒ 9k  4k 
 5r for some integer r
Sk1 ⇒ 9k1  4k1 
 5t for some integer t
9k  4k 
 5r
9k 
 4k  5r
9(9k) 
 (4k  5r)(4  5)
9k1 
 4k1  5(4k)  20r  25r
k1
9
 4k1 
 4k1  5(4k)  20r  25r 
4k1

 5(4k)  45r

 5(4k  9r)
k1
k1
Thus, 9
4

 5t, where t 
 4k  9r is an
integer, and we have shown that if Sn is valid,
then Sk1 is also valid. Since Sn is valid for n 
 1,
it is also valid for n 
 2, n 
 3, and so on
indefinitely. Hence, 9n  4n is divisible by 5 for all
integral values of n.

Page 833

6

1
1  2

1a. Arithmetic; arithmetic sequences have common
differences, while geometric sequences have
common ratios.
1b. Sample answer: 1, 4, 7, 10, …; an 
 1  3(n  1)

(k  1)(k  2)(2k  9)

6
(k  1)[(k  1)  1][(2(k  1)  7]

6

1

56b. a1 
 6, r 
 6 or 2

436

54  48

9. The increase from 99 to 100 is 1. So the percent
1
increase from 99 to 100 is .

Percent increase 
 4
 100
8

 12.5%
The correct choice is B.
5. Since the figure is not drawn to scale, do not
assume that the two lines are parallel, even
though they may appear parallel. Since AB 
 AC,
ABC is isosceles. So m∠B 
 m∠ACB.
m∠B  m∠ACB  80 
 180
2  m∠ACB 
 100
m∠ACB 
 50
Since AD is a line segment, x  70  50 
 180.
So, x 
 60.
Consider right triangle CDE.
x  y 
 90
60  y 
 90
y 
 30
x  y 
 60  30 or 30
The correct choice is C.
6. The population of Rockville is now 20,000 and will
double every 8 years. So in 8 years the population
will be 40,000 and in 16 years will be 80,000. So
f(8) 
 40,000 and f(16) 
 80,000. Choice A is
incorrect since f(8) 
 20,000. Choice B is incorrect
since f(16) 
 2(20,000)2 or 800,000,000. Choice C

99
1
1
 is greater than , or 1%.
99
100

The correct choice is A.
10. Choose a number for the total number of cars in
the parking lot. Since the fractions have
denominators of 2, 4, and 5, choose a number that
is divisible by 2, 4, and 5. Let the number of cars
in the parking lot equal 40.
1

5
1

2
1

4

 8 blue cars 
 4 blue convertibles
 the number of convertibles 
 4
the number of convertibles 
 16

Not Blue and Not Convertible

8 Blue

4 Convertible
16

The number of cars that are neither blue nor
convertible is the total number minus the blue
cars minus the convertibles plus the number that
are both blue and convertible.
Neither blue nor convertible

 40  8  16  4

 20
percent that are neither blue nor convertible

20,000

is incorrect since f(8) 
 
64 . Choice E is
incorrect since f(8) 
 20,000. For Choice D,
f(8) = 40,000 and f(16) = 80,000.
The correct choice is D.
7. Notice that the figure is not drawn to scale. ∠A
could be a right angle. To be sure it is, find the
slope of AB and compare it to the slope of AC.
10  4

 40 
 8 blue cars

20


 40  100

6



Slope of AB 
 
65 
 1


 50%
The answer is 50.

1

Slope of AC 
 6
Since the slopes are negative reciprocals, the line
segments are perpendicular, so m∠A 
 90.
Therefore, ABC is a 30°-60°-90° right triangle.
The hypotenuse, AC, is twice the length of the leg
opposite the 30° angle, AB.
2  (1
AB 
 
(6  5)
0  4)2 
 1
6
 3 or 37

AC 
 237

The correct choice is D.
8. Method 1: Substitute each answer choice for x to
test both inequalities.
A: (6)  6  0 and 1  2(6)  1.
0  0 and
13  1; false
Method 2: Solve each inequality for x.
x60
and
1  2x  1
x  6
2x  2
x 1
The solution is 6 x 1. All of the answer
choices except A are in this range.
The correct choice is A.

437

Chapter 12

Chapter 13 Combinatorics and Probability
4!

13-1

Pages 842–843

4321


 4  10 or 40
13. Using the Basic Counting Principle,
10  9  8  7  6  5  4  3  2  1 
 3,628,800.

Check for Understanding

15!


14. C(15, 9) 
 
(15  9)! 9!




Pages 543–545
7!


17. P(7, 7) 
 
(7  7)!




18a. Using the Basic Counting Principle,
9  10  10  10  10  10  10 
 9,000,000.
18b. Using the Basic Counting Principle,
5  5  5  5  5  5  5 
 78,125.
18c. Using the Basic Counting Principle,
10  10  10  10  10  10  1 
 1,000,000.
18d. Using the Basic Counting Principle,
1  1  1  10  10  10  10 
 10,000.
19. dependent
20. independent
21. dependent

654321

1
5!


8. P(5, 3) 
 
(5  3)!

54321




21

8!


22. P(8, 8) 
 
(8  8)!


 60









12!

(12  8)!
——
6!

(6  4)!
12! 2!

6! 4!
12  11  10  9  8  7  6  5  4  3  2  1  2  1

6  5  4  3  2  1  4  3  2 1




6!


23. P(6, 4) 
 
(6  4)!




5!


24. P(5, 3) 
 
(5  3)!

7!

(7  4)! 4!
7654321

3214321

54321




21


 60

7!


25. P(7, 4) 
 
(7  4)!

11. C(20, 15)




20!




(20  15!) 15!

7654321

321


 840

20  19  18  17  16  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1

5  4  3  2  1  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1


 15,504

Chapter 13

654321

21


 360


 35




87654321

1


 40,320


 55,440
10. C(7, 4) 


7654321

1


 5040


 720




Exercises

16. Using the Basic Counting Principle, 2  6  4 
 48.

6!

P(12, 8)

P(6, 4)

15  14  13  12  11  10  9  8  7  6  5  4  3  2  1

654321987654321


 5005
15a. Using the Basic Counting Principle,
10  10  10  10  10 
 100,000.
15b. Using the Basic Counting Principle,
1  9  9  9  10 
 7290
15c. Using the Basic Counting Principle,
(10  10  10  10  10  10  10  10  10) 
(10  10  10  10  10) 
 999,900,000


7. P(6, 6) 
 
(6  6)!

9.

54321

 


1321  32121

1. Sample answer: Both are used to determine the
number of arrangements of a group of objects.
However, order of the objects is important in
permutations. When order of the objects is not
important, combinations are computed.
2. Select 2 jacks out of 4—C(4, 2)
Select 3 queens out of 4—C(4, 3)
number of hands—C(4, 2)  C(4, 3)
3. Sam is correct. The room assignments are an
ordered selection of 5 rooms from the 7 rooms. A
permutation should be used.
4.
blue ————— S-blue
S
green ———— S-green
gray ————— S-gray
blue ———— M-blue
M
green———— M-green
gray ———— M-gray
blue ————— L-blue
L
green ———— L-green
gray ————— L-gray
blue ———— XL-blue
XL
green ——— XL-green
gray ———— XL-gray
5. Using the Basic Counting Principle,
4  3  5  5 
 300.
6. independent



5!

 
12. C(4, 3)  C(5, 2) 
 
(4  3)!3!  (5  2)! 2!

Permutations and Combinations

438

38. C(7, 3)  C(8, 5)

9!


26. P(9, 5) 
 
(9  5)!




7!

7654321


 
4321321 


 15,120
10  9  8  7  6  5  4  3  2  1

321


 604,800
28.

P(6, 3)

P(4, 2)








54321

P(6, 4)

P(5, 3)








14!


40. C(14, 4) 
 
(14  4)! 4!




14  13  12  11  10  9  8  7  6  5  4  3  2  1

10  9  8  7  6  5  4  3  2  1  4  3  2  1


 1001
14!


41. C(14, 5) 
 
(14  5)! 5!

6!

(6  4)!
—
5!

(5  3)!
6! 2!

5! 2!
654321

54321

P(6, 3)  P(7, 5)

P(9, 6)

14  13  12  11  10  9  8  7  6  5  4  3  2  1


 
98765432154321

 2002
42. C(18, 12)
18!




(18  12)! 12!










18  17  16  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1

6  5  4  3  2  1  12  11  10  9  8  7  6  5  4  3  2  1


 18,564
43. C(3, 2)  C(5, 1)  C(8, 2)
3!
5!
8!
  


(3  2)! 2!  (5  1)! 1!  (8  2)! 2!

6!
7!
   
(6  3)! (7  5)!
———
9!

(9  6)!
6! 7! 3!

9! 3! 2!
6543217654321

98765432121




321

54321

 


121  43211 

11!


44. P(11, 11) 
 
(11  1)!

5!

11  10  9  8  7  6  5  4  3  2  1


31. C(5, 3) 
 
(5  3)! 3!


 
1

54321


 39,916,800




21321


 10

13!

10!

13  12  11


 286  78 or 22,308
45b. C(4, 1)  C(4, 2)  C(4, 2)
4!

4!

4!

  


(4  1)! 1!  (4  2)! 2!  (4  2)! 2!

4!


33. C(4, 2) 
 
(4  2)! 2!

4321

4321

4 321

4321

  


3211  2121  2121




2121


 4  6  6 or 144


6

12!

12!

(12  4)! 4!
12  11  10  9  8  7  6  5  4  3  2  1

876543214321


45c. C(12, 5) 
 
(12  5)! 5!




12  11  10  9  8  7  6  5  4  3  2  1

765432154321


 792
46a. Using the Basic Counting Principle,
10  10  10  10  10 
 100,000
46b. Using the Basic Counting Principle,
10  9  8  7  6 
 30,240
46c. Using the Basic Counting Principle,
5  5  4  4  4 
 1600.


 495
35. C(9, 9) 


13  12

 


321  21

10  9  8  7  6  5  4  3  2  1

5432154321


 252




13!

 
45a. C(13, 3)  C(13, 2) 
 
(13  3)! 3!  (13  2)! 2!


32. C(10, 5) 
 
(10  5)! 5!

34. C(12, 4) 


87654321

65432121


 3  5  28 or 420


5




87654321

65432121


 5  6  28 or 840


6
30.

4321

 


43211  2121 

6!

(6  3)!
—
4!

(4  2)!
6! 2!

4! 3!
65432121

4321321


 10
29.

87654321

32154321


 35  56 or 1960
39. C(5, 1)  C(4, 2)  C(8, 2)
5!
4!
8!
  


(5  1)! 1!  (4  2)! 2!  (8  2)! 2!

10!


27. P(10, 7) 
 
(10  7)!




8!

 


(7  3)! 3!  (8  5)! 5!

987654321

4321

9!

(9  9)! 9!


1

14!


36. C(14, 7) 
 
(14  7)! 7!
14  13  12  11  10  9  8  7  6  5  4  3  2  1


 
76543217654321

5!


 3432

4!

 
P(5, 2)  P(4, 3) 
 
(5  2)!  (4  3)!
3!

54321

8!

 
37. C(3, 2)  C(8, 3) 
 
(3  2)! 2!  (8  3)! 3!
321




121 

4321

  


321
1

87654321

54321321


 20  24 or 480


 3  56 or 168

439

Chapter 13

47. C(3, 1)  C(4, 1)  C(6, 1)  C(14, 6)

53b. Yes; let h, t, and u be the digits.
100h  10t  u
100h  10u  t
100t  10h  u
100t  10u  h
100u  10t  h
 100u  10h  t
200(h  t  u)  20(h  t  u)  2(h  t  u)

 222(h  t  u)

3!
4!
6!
14!
   


(3  1)! 1!  (4  1)! 1!  (6  1)! 1!  (14  6)! 6!
654321
321 4321
  


2  1  1  3  2  1  1  5  4  3  2  1 1
14  13  12  11  10  9  8  7  6  5  4  3  2  1
 
87654321654321


 3  4  6  3003 or 216,216
42!


48a. P(42, 42) 
 
(42  42)!

222(h  t  u)

6


 42!
1.4  1051

54. 2140  (1.058) 
 $2264.12
2264.12(1.058) 
 $2395.44
2395.44(1.058) 
 $2534.38

42!


48b. C(42, 30) 
 
(42  30)! 30!
42!




12! 30!

10

55.

1.1 
48c. C(5, 3)  C(12, 6)  C(10, 6)  C(15, 5)
1010

5!

12!

10!

56.
15!

   


(5  3)! 3!  (12  6)! 6!  (10  6)! 6!  (15  5)! 5!

n!
n!
  
 
[n  (n  1)]!
(n  n)!
n!
n!
 
 
0!
1!
0!

 n3 
 13  2 3  .
n
1
7.1x 
 83.1
x ln 7.1 
 ln 83.1
ln 83.1

x
57. x 


e0.346

2.26
Use a calculator.

1.4
58.

1

n! 
 n!

y 
 4x2
x
 sin 45°  y
 cos 45° 
 4(x
 cos 45°
 y
 sin 45°)2
2


2!

2!

21

59.

90˚

120˚

2!

30˚

180˚

1 2 3 4

330˚

210˚
240˚

270˚

11!


51a. C(11, 4) 
 
(11  4)! 4!

6!

5!

 
51b. C(6, 2)  C(5, 2) 
 
(6  2)! 2!  (5  2)! 2!
654321

432121

54321



32121


 28
2


 15  10 or 150

2


10!


52. C(10, 2) 
 
(10  2)! 2!


 28
2

61.

or 37(2  5  9) 
 592

360°

period 
 1 or 360°
phase shift 
 30°

Chapter 13

2


19.80 ft/s
19.80 ft/s
sin 2x  2 sin x 
 0
2 sin x cos x  2 sin x 
 0
2 sin x (cos x  1) 
 0
2 sin x 
 0 or cos x  1 
 0
sin x 
 0
cos x 
 1
x 
 0°, x 
 180°, x 
 360°, or x 
 180°
So, x 
 0°, 180°, 360°
62. y 
 8 cos (v  30°)
amplitude 
 8

10  9  8  7  6  5  4  3  2  1

8765432121


 45
3552

6

300˚

r 
 2, r 
 2 cos 2v
2 
 2 cos 2v
1 
 cos 2v
2v 
 0° or 2v 
 360°
v 
 0°
v 
 180°
(2, 180°), (2, 0°)
60. vx 
 28 cos 45°, vy 
 28 sin 45°

11  10  9  8  7  6  5  4  3  2  1

76543214321


 330

53a.

0˚

21


 6  2  2  2 or 48




60˚

150˚


 1  1  1  1




2


1

0 
 4(x
)2  8x
y
  4(y
)2
 2
x
  2
y


     


(3  3)!  (2  2)!  (2  2)!  (2  2)!
21

2

1


 720
50b. There are 3 ways to arrange the 3 couples, and 2
ways to arrange each of the two members within
a couple.
P(3, 3)  P(2, 2)  P(2, 2)  P(2, 2)
3!

2


2
x
  2
y
 
 8
2x
2  x
y
  2y
2

654321

1




2

2x
  2y
 
 4
2x
  2y


6!


50a. P(6, 6) 
 
(6  6)!

321

. .  103 
 3025


x

ln 7.1


 10  924  210  3003
5.8  109
49. P(n, n  1) 
 P(n, n)





 37(h  t  u)

440

26!

63. Find B.
B 
 180°  90°  27° or 63°
Find a.


22. linear; 
6! 3! 7! 10!

a

15.2

tan 27° 

15.2 tan 27° 
 a
7.7 a
Find c.
15.2

cos 27° 
 c
15.2


c

cos 27°

c 17.1
360
64. Each hour, an hour hand moves through 1
2 or
1
30°. Since 12 minutes is 5 of an hour, the hour
1
hand moves through an additional 5(30) or 6°.
2(30°)  6° 
 66°
The correct choice is A.

5.1  1012

23.
24.
25.
26.
27.
28.

circular; (9  1)! 40,320
circular; (5  1)! 
 24
circular; (8  1)! 
 5040
linear; 6! 
 720
linear; 10! 
 3,628,800
circular; (9  1)! 
 40,320

29.
30.
31.
32.

circular;  
 3,113,510,400
2
circular; (20  1)! 1.22  1017
circular; (32  1)! 8.22  1033
linear; 25! 1.55  1025

33.

8!

2!2!2!2!

(14  1)!


 2520

34a. (7  1)! 
 720
34b. 7! 
 5040
11!
  
 46,200
3!4 !3!
11!

36a. 
2! 2 ! 2! 
 4,989,600

35.

Permutations with Repetitions
and Circular Permutations

13-2

36b. integral calculus
37a.

Pages 848–849

Check for Understanding

4.
5.
6.

n(n  1)(n  2)(n  3)!

(n  3)!
n3  3n2  2n  210

6!


39. C(6, 3) 
 
(6  3)! 3!




5!
2


 756

Pages 849–851

Exercises

12.

8!

2! 2!


 10,080

13.

10!

2! 2!


 907,200

14.

8!

2!

15.

10!

3! 2!

16.

12!

2! 2! 2!


 59,875,200

17.

9!

4! 2! 2!


 3780

18.

7!

2! 2! 2!


 630

19.

9!

5! 4!

20.

654321

321321


 20
40. (5x  1)3 
 C(3, 0)  (5x)3  (1)0
 C(3, 1)  (5x)2  (1)1
 C(3, 2)  (5x)1  (1)2
 C(3, 3)  (5x)0  (1)3

 125x3  75x2  15x  1
41. x log2 413

10. linear;  or 60
9!

5! 2! 2!


 210


0
Use a graphing calculator to find the solution at
n 
 7.

2

7. circular; (11  1)! or 3,628,800
8. circular; (8  1)! or 5040
9. circular; (12  1)! or 39,916,800

11.

7.85  1017

37b. (43  1)! 1.41  1051
37c. 43! 6.04  1052
38. Let n 
 total number of symbols.
Let n  3 
 number of dashes.
n!
 
 35
3! (n  3)!

1. The circular permutation has no beginning or end.
Therefore, the number of different arrangements
1
is always n of a revolution.
2. Sample answer: house or phone numbers where
some of the digits repeat
3. Sample answer: The number of permutations of n
n!
charms on a bracelet with a clasp is .
8!
 
 10,080
2! 2!
9!
 
 22,680
2! 2! 2! 2!
14!
 
 7,567,560
2! 4! 5! 2!

43!

20! 14! 9!

x

log10 413


log10 2

x 8.69
42. h 
 6, k 
 1
3
hp
3
6p
3 
 p
(y  k)2 
 4p(x  h)
(y  1)2 
 12(x  6)
43. 2(4  3i)(7  2i) 
 2(28  29i  6i2)

 56  58i  12(1)

 44  58i


 20,160

 302,400


 126


50!
50!

4! 3! 4! 2! 8! 8! 3! 2! 2! 3!
2! 4!

2.35  1048

21. circular; (12  1)! 
 39,916,800

441

Chapter 13

u
u
j k
0 3
5 0
3u 2 3u
2 0u
i 
j 
k
0
2 0
2 5
u
u  6j
u  10k

 15i

u
i
44. u
v u
w 
 2
2
0


5

C(4, 1)  C(3, 2)


 15, 6, 10
since 2, 0, 3  15, 6, 10 
 30  0  30 or
0 and 2, 5, 0  15, 6, 10 
 30  30  0 or 0,
then the resulting vector is perpendicular to u
v
and u
w.
45. x cos 45°  y sin 45°  8 
 0
2

x
2

11.

2


 2y  8 
 0

x  2
y  16 
 0
2
y 45˚

12.

8
4
4 O

8

4

P(f) 
 1  P(s)


10. P(s) 
 
C(7, 3)

8 x

4

43

 3
5
12

 35
12

35
12
—
odds 
 23 or 23

35
C(3, 1)  C(4, 2)

P(s) 
 
C(7, 3)
36

 3
5
18



 35
18

35
18
odds 
 —
or 1
7
17

35
4
80


P(rain) 
 
100 or 5
4
P(not rain) 
 1  5
1

 5
1

5
1

odds 
 —
4 or 4

5

12


 1  35
23


 35

P(f) 
 1  P(s)
18


 1  35
17


 35

8

Pages 856–858

The sparks will be highest at the y-intercept, 82

inches above the center of the wheel. This is
82
  8 or about 3.31 inches above the wheel.
46. If x2 
 36, then x  6 or x 
 6.
2x1 
 261 or 32

Exercises

13. P(face card) 




444

52
12
3
 or 
52
13

6666

14. P(a card of 6 or less) 
 5
2
24

1


2x1 
 261 or 
128

6


 52 or 1
3
10  10

15. P(a black, non-face card) 
 5
2

The correct choice is E.

20

5



 5
2 or 13

13-3
855–856

17.

Check for Understanding

18.

1. The probability of the event happening is 5050.
2. Answers will vary; see students’ work.
3. Sample answer: The probability of the successful
outcome of an event is the ratio of the number of
successful outcomes to the total number of
outcomes possible. The odds of the successful
outcome of an event is the ratio of the probability
of its success to the probability of its failure.
3

19.
20.
21.

22.

3



4. Geraldo is correct. P(win) 
 
2  3 
 5 or 60%.
7

7

1




5. P(softball) 
 
3  7  11 
 21 or 3
37

23.
10



6. P(not a baseball) 
 
3  7  11 or 21
0


7. P(golf ball) 
 
3  7  11 or 0
7

7



8. P(woman) 
 
7  4 or 11
C(4, 3)


9. P(s) 
 
C(7, 3)
4

odds 

Chapter 13

4

35
—
31

35

24.
P(f) 
 1  P(s)
4

31



 1  3
5 or 35


 3
5

10  10  10  10

52
40
10





 52 or 13
5
5
1



P(red) 
 5  2  3 
 1
0 or 2
1
2
2



P(white) 
 
5  2  3 
 10 or 5
52
7


P(not pink) 
 
5  2  3 or 10
53
8
4



P(red or pink) 
 
5  2  3 
 10 or 5
C(4, 2)

P(2 pop) 
 
C(40, 2)
6
1




780 or 130
C(8, 2)

P(2 country) 
 
C(40, 2)
28
7




780 or 195
C(10, 1)  C(18, 1)
P(1 rap and 1 rock) 
 
C(40, 2)
10  18



735
3
180
 or 


13
780
C(22, 2)

P(not rock) 
 
C(40, 2)
231
77




780 or 260

16. P(not a face card) 


Probability and Odds

25. Using the Basic Counting Principle, there are 1  1
or 1 way to roll both fives. Using the Basic
1
Counting Principle, P(both fives) 
 3
6.

4

or 3
1

442

C(3, 2)


26. P(s) 
 
C(6, 2)

27.

28.

29.

C(3, 1)  C(24, 2)

P(f) 
 1  P(s)


33. P(s) 
 
C(27, 3)

3
1

 1  5

 1
5
1
4

 5

 5
1

5
1

odds 
 —
4 or 4

5
C(4, 2)

P(s) 
 C(6
P(f) 
 1  P(s)
, 2)
6
2



 1  5

 15
2
3

 5

 5
2

5
2
—
odds 
 3 or 3

5
C(1, 1)  C(3, 1)

P(s) 
 
P(f) 
 1  P(s)
C(6, 2)
13
1

 1  5

 15
3
1
4


 1

 5
5 or 5
1

5
1
—
odds 
 4 or 4

5
C(1, 1)  C(2, 1)
C(1, 1)  C(3, 1)
C(2, 1)  C(3, 1)
    
P(s) 
 
C(6, 2)
C(6, 2)
C(6, 2)
2
3
6
11
   or 

 1

5
15
15
15

34.
35.

30.

31.

32.

233




325

P(f) 
 1  P(s)
4

36.

1

4

5
4
—
odds 
 1 or 1

5
C(13, 3)  C(13, 2)
P(s) 
 
C(52, 5)
286  78
  12


2,598,960
267,696


 2,598
,960
429



4165

 P(4, 2)

P(f) 
 1  P(s)
3736

429




1
4165 or 4165

37.

4


 1  15 or 1
5
11

15
11
—
odds 
 4 or 4

15
C(11, 3)

P(s) 
 
C(27 , 3)
165



2925
11



19 5
11

19 5
11

odds 
 —
184 or 18 4

195
C(13, 2)  C(11, 1)

P(s) 

C(27, 3)
78  11



2925
22
858




2925 or 75
22

75
22
—
odds 
 53 or 53

75
C(14, 3)

P(s) 
 
C(27, 3)
364



2925
28



225
28

225
28
—

odds 
 197 or 
197

225

92



1
325


 1  5 or 5

P(f) 
 1  P(s)
11

P(f) 
 1  P(s)

3  276



2925
828
92




2925 or 325
92

325
92

odds 
 —
233 or 233

325
1
1


P(s) 
 1  
249 or 250
4
P(s) 
 5

429

4165
429

odds 
 —
3736 or 3736

4165
1
1


P(s) 
 1 
4 or 5

38. P(s) 
 0.325

P(f) 
 1  P(s)

 1  0.325 or 0.675

0.325

13



odds 
 
0.675 or 27

P(f) 
 1  P(s)

1

1

1

1

1

1

 

39a. P(s) 
 1
0  9  8 or 720

11



1
19 5

1

1

 

39b. P(f) 
 1
0  10  10 or 1000

184

P(s) 
 1  P(f)
1
999



1
1000 or 1000




195

odds 

P(f) 
 1  P(s)

999

1000
—
1

1000

999

or 1
1

1

1

40a. P(both males) 
 2  2 or 4

22


 1  75

1

40b. P(s) 
 1  2

53


 75




P(f) 
 1  P(s)
41a.

28



1
225
197




225

41b.

443

1

2

1

1
2
_
odds 
 1 or 1

2
C(15, 10)  C(5, 0)
P(s) 
 
C(20, 10)
3003  1
21
 or 


184,756
1292
C(15, 8)  C(5, 2)

P(s) 
 
C(20, 10)
6435  10



184,756
64,350



184,756
225



646
225

646
225

odds 
 —
421 or 421

646

P(f) 
 1  P(s)
1

1


 1  2 or 2

P(f) 
 1  P(s)
225



1
646
421




646

Chapter 13

42a. P 




51. Drawing the altitude from one vertex to the
opposite side forms a 30°-60°-90° right triangle
with hypotenuse 2s. The short side of this right
triangle measures s. So the altitude drawn has
length s
3. This is the height of the equilateral
triangle. The base measures 2s. So the area of the
1
equilateral triangle is   2s  s
3 
 
3s2.
2
The correct choice is B.

179,820  151,322

84,475  3273  179,820  151,322
331,142

418,890

0.791
151,322

P(s) 
 1  P(f)


42b. P(f) 
 
418,890

151,322



1
418,890
267,568




418,890

odds 




267,568

418,890
——
151,322

418,890
267,568
133,784
 or 
151,322
75,661

43. P A

2s
3
s 

Q

x

2s

s

s

8x

u
Given a pipe PQ and a random cut point, A,
APAQ 
 18. If AP is x inches long, then AQ is 8x
u
inches long. Now, the cut must be made along AP
so that the longer piece will be 8 or more times as
long as the shorter piece. Thus, the probability
1
x
u


that the cut is on AP is 
x  8x 
 9 . Since the cut

Page 858

15!




20!


2. C(20, 9) 
 
(20  9)! 9!


 167,960
3. Using the Basic Counting Principle,
26  26  26  10  10  10  10 
 175,760,000.

44. This is a circular permutation.
(6  1)! 
 5! or 120

12!


4. P(12, 5) 
 
(12  5)!

10!


45. C(10, 4) 
 
(10  4)! 4!




10  9  8  7  6  5  4  3  2  1

6543214321

n

46. Sn 
 2[2a1  (n  1)d]
14

S14 
 2[2(3.2)  (14  1)1.5]

6.


 7(25.9)

 181.3
47. Let y 
 7log7 2x
log7 y 
 log7 2x
y 
 2x
So, 7log7 2x 
 2x.
48. Center: (7, 2)
r2 
 (10  7)2  (8  2)2

 32  (10)2 or 109
(x  7)2  (y  2)2 
 109
49. r 
 3  2 or 6


5

7.




9!
 
 181,440
2!
10!
 
 4200
3! 4! 3!

8. This is a circular permutation.
(8  1)! 
 5040
C(13, 2)


9. P(both hearts) 
 
C(52, 2)
78

1





1326 or 17
C(3, 1)  C(3, 1)


10. P(s) 
 
C(12, 2)
33

 6
6
9
3



 66 or 2
2
3

22
3

odds 
 —
19 or 19

22

5

5

12  11  10  9  8  7  6  5  4  3  2  1

7654321


 95,040
5. Using the Basic Counting Principle,
18  18  3  6 
 5832.


 210

v 
 (  4) or 4

15  14  13  12  11  10  9  8  7  6  5  4  3  2  1

10  9  8  7  6  5  4  3  2  1


 360,360

can be made on either end of the pipe, the actual
2
probability is 9.




Mid-Chapter Quiz


1. P(15, 5) 
 
(15  5)!



6(cos 4  i sin 4) 
 6 2  
2i
2


2



 32
  32
i
u
50. u 
 3, 5  4, 2

 3  (4), 5  2

 1, 3

13-4

P(f) 
 1  P(s)
3


 1  22
19


 22

Probabilities of Compound
Events

Pages 863–864

Check for Understanding

1. The occurrence of one event does not affect
another for independent events. The occurrence of
the first event affects the occurrence of a second
for dependent events.

Chapter 13

444

2a.

diamonds

spades,
clubs,
hearts

aces

4

4

7

8

26

25

24

 

20. dependent, 1
5  14  13 
 195
23

22

21

20

19

18

21. dependent, 52  51  50  49  48  47  46  45  44 
17

43

16

15

14

19

 

 4
2  41  40 
 1,160,054
12

8

8

32



22. dependent, 28  2
7  26 
 819
5

2b. No, one of the aces can be an ace of diamonds.
2c. P(ace or diamond) 
 P(ace)  P(diamond) 
P(ace and diamond)
3. Answers will vary; see students’ work.
4. independent,
5.
6.
7.
8.
9.

6

36



3

36

4

1

1

1

11

4

26

2

7

26

12

6

8




26. inclusive, 5
2  52  52 
 13

27. inclusive, 52  52  5
2 
 13

36

35

34

28. exclusive
P(at least 3 males)

 P(3 males)  P(4 males)  P(5 males)

33

C(5, 3)  C(4, 2)

C(5, 4)  C(4, 1)

60

20

81

9

1






126  126  126




126 or 14

0.518
12. P(5 odd numbers or 5 multiples of 4)

 P(5 odd numbers)  P(5 multiples of 4)

29. exclusive
P(sum of 6 or sum of 9) 
 P(sum of 6)  P(sum of 9)

17

16

15

14

13

35

34

33

29

28

27

26

C(7, 3)  C(7, 3)

25

13

12

94

93

92

435,643

1 6

34

39

or

1 6

15

6

22

11

1


 64 or 32

C(8, 2)  C(5, 0)

32. inclusive


 78  78



1 6



 64  6
4  64

  


C(13, 2)
C(13, 2)

68

78

302

7


 C(6, 4)
2  C(6, 5)
2  C(6, 6)
2

15. P(at least 1 right handed pitcher 

P(1 right-handed pitcher)  P(2 right-handed
pitchers)
28

2114

147

31. exclusive
P(at least 4 tails)

 P(4 tails)  P(5 tails)  P(6 tails)

91




560,175

40

735





300 3 or 429

     


100  99  98  97  96  95

C(8, 1)  C(5, 1)

1225







3003  3003  3003  3003

14. P(none if 6 clocks are damaged)
95

C(7, 5)  C(7, 1)



C(14, 6)

0.025  0.007  0.001
0.032
96

C(7, 4)  C(7, 2)

C(7, 6)  C(7, 0)

10

11

1

    


C(14, 6)
C(14, 6)
C(14, 6)

    
 
7
5  74  73  72  71 
14

9

30. exclusive
P(at least three women)

 P(3 women)  P(4 women)  P(5 women) 
P(6 women)

 

 
75  74  73  72  71  
75  74  7
3  72  71 
36

4



 3
6 or 4

0.029  0.0004
0.029
13. P(5 even numbers or 5 numbers less than 30) 

P(5 even numbers)  P(5 numbers less than 30) 
P(5 even numbers and 5 numbers less than 30)
37

5



 3
6  36

   
    

 
7
5  74  73  72  71   
 75  74  73  72  71 
34

C(5, 5)  C(5, 0)

    


C(9, 5)
C(9, 5)
C(9, 5)

66 65 64 63 62
P(selecting 5 two digit numbers) 
 75  74  73  72  71

35

4


25. inclusive, 6  6  3
6 
 36

0.025

36

4

256

37

37

4




2401

4
3
2


dependent, 1
0  9 
 15
4 3
2
dependent, 6  5 
 5
4
6
10


exclusive, 1
3  13 
 13
15
11
6
20

inclusive, 27  27  2
7 
 27
4
4
8
2



exclusive, 5
2  52 
 52 or 13

38

35

P(winning next four games) 
 7  7  7  7

10. P(selecting 5 even numbers) 
 75  74  73  72  71
11.

7

24. independent
4
P(winning) 
 7

1

72




4

 

23. independent, 1
6  16  16 
 1024


542  531   
2562  2551  
522  511
12

650

660

55

2






2652  2652  2652




2652 or 221

Pages 864–867
16. dependent,

Exercises
5

9



4

8

33. exclusive
P(at least 2 rock) 
 P(2 rock)  P(3 rock)

5


 1
8

C(6, 2)  C(5, 1)

5

25

  


C(11, 3)
C(11, 3)

1

1

1





165  165

18. independent, 6  6 
 3
6
4

C(6, 3)  C(5, 0)

5

17. independent, 9  9 
 8
1
3

75

20

95

19





165 or 33

2

19. dependent, 7  6 
 7

445

Chapter 13

26

25

24

23

46. P(word processing or playing games)

 P(word processing)  P(playing games)  P(both)

22

34. P(all red cards) 
 52  51  50  49  48
253




9996

2

35. P(both kings or both aces)

 P(both kings)  P(both aces)
3

4

12

12

24

2652

2

221

29

47. P(rain or lightning)

 P(rain)  P(lightning)  P(both)

3





2652  2652




or

3
4



48. P(even sum)

 P(3 even cards)  P(2 odd cards and 1 even card)
C(5, 3)  C(4, 0)

650

12

660

2652

55

221

3

2

1

or

2

38. P(2 pennies) 




10

30

40

10


 84  84






2652  2652  2652





 84 or 2
1

5

21
1

21



49. P(at least 3 women)

 P(3 women)  P(4 women)  P(5 women)

4

20

C(6, 3)  C(7, 2)

240

16

15

7

40. P(at least 1 nickel) 
 1  P(no nickels)

1











59

93



3

958

97

7

93

97

9021





10,000  10,000

17

30

9979




10,000

51. P(supplies or money)

 P(supplies)  P(money)  P(both)

C(5, 1)  C(7, 1)





C(21, 2) 
C(21, 2)
36

210
71

210

531

6

 


 

 

100  100  100  100   
 100  100 

13

20

41. P(2 dimes or 1 penny and 1 nickel)

 P(2 dimes)  P(1 penny and 1 nickel)
C(9, 2)

105

50. P(at least 1 doctor)

 P(1 doctor)  P(both doctors)

4

14

21

420





1287 or 143

6





420 or 7




C(6, 5)  C(7, 0)






1287  1287  1287


 
 

 
2
1  20   
 21  20   
 21  20 
6

C(6, 4)  C(7, 1)

    


C(13, 5)
C(13, 5)
C(13, 5)

39. P(2 nickels or 2 silver-colored coins)
7

C(4, 2)  C(5, 1)

  


C(9, 3)
C(9, 3)


 
 

 
5
2  51   
 52  51   
 52  51 
4

1


 5

13 10

52

37. P(both red or both queens)

 P(both red)  P(both queens)  P(both red queens)
25

2


 5  5  5

36. P(all diamonds) 
 


26

1


 60


 

 
5
2  51   
 52  51 
4

1


 5  3  4

812

625

1062

531

375






2500  2500  2500

35

210





2500 or 1250
5

4

3

52a.

2

  
42. P(all female) 
 1
0  9  8  7
1

P (A )


 4
2

P (B )

43. P(all female or all male)

 P(all female)  P(all male)
4

5

3

2

5

4

3

2

  
   

 1
0  9  8  7  10  9  8  7
1

1

2

42

1

21

P (C )



 4
2  42




or

52b. P(A or B or C) 
 P(A)  P(B)  P(C)  P(A and
B)  P(A and C)  P(B and C)  P(A and B and
C). You must add the intersection of all three
sets which have not been accounted for.
53. P(action video or pop/rock CD or romance DVD)

44. P(at least 3 females)

 P(3 females)  P(4 female)
C(5, 3)  C(5, 1)

C(5, 4)  C(5, 0)

  


C(10, 4)
C(10, 4)
50

5

55

11





210  210




210 or 42

4

7

45. P(at least 2 females and at least 1 male)

 P(2 females and 2 males) 
P(3 females and 1 male)
C(5, 2)  C(5, 2)

C(5, 3)  C(5, 1)

  


C(10, 4)
C(10, 4)
100

50

150

5





210  210




210 or 7

Chapter 13

446

1

5

2

1

1

1

 2  11  9  7  4  4
4

0.93

54a. First consider the probability that no 2 students
have the same birthday. The first person in the
class can have any birthday; there are 366
choices out of 366 days. The second person has
only 365 choices out of 366 days, and so on.
So, P(2 students with the same birthday)

 1  P(no 2 students have the same birthday).

60.

x


365


1

366!

348!

36618


1

366!

(366  348)!

36618

364

349

P(366, 18)



1
36618

0.346
P(366, n)

4 log 3  2 log 12

log 12  log 3

x 6.7549
61. Let y 
 income and x 
 number of $1.00 increases.
income 
 (number of customers)  (cost
of a ticket)
y 
 (400  20x)(3  x)
y 
 1200  340x  20x2
y  1200 
 20(x2  17x)
y  1200  1445 
 20(x2  17x  72.25)
(y  2645) 
 20(x  8.5)2
The vertex of the parabola is (8.5, 2645). An
increase of $8.50 will give a maximum profit of
$2645. The price of each ticket should be 3  8.5
or $11.50.

  


 1  
366  366  366  . . .  366 
366

12x  2 
 3x  4
(x  2) log 12 
 (x  4) log 3
x log 12  2 log 12 
 x log 3  4 log 3
x log 12  x log 3 
 4 log 3  2 log 12
x (log 12  log 3) 
 4 log 3  2 log 12

1

  
54b. 1  
366n
2

54c. In part a, there is only a 0.346 probability that 2
students have the same birthday. This is too
small. Substitute numbers greater than 18 for n
in the inequality of part b. When n is 23, P is
about 0.51. So, 23  18 or 5 more students are
needed in the class.
55a. inclusive


2

A 
 4
k

62.


2 


8270 
 
4(7)
231,560


2 
 

 271.5 yards
x 
 x1  ta1
y 
 y1  ta2
x 
 1  t(2)
y 
 5  t(4)
x  1 
 2t
y  5 
 4t
x  1, y  5 
 t 2, 4
64. 2 tan x  4 
 0
tan x 
 2
tan1(tan x) 
 tan1 2
x 
 63° 26

65. Since ADB  CBD and they are alternate
interior angles, 
AD
BC
L
. Simply because a 
 45
does not mean b 
 45, so you cannot conclude that

3 bisects ABC.
The correct choice is B.

55b.

63.

Dead
Battery

Flooded
Engine

55c. P(flooded engine or dead battery)

 P(flooded engine)  P(dead battery)  P(both)
1

2

1


 2  5  10
4


 5
1

56. P(two threes given a sum of six) 
 5

13-5

57. (7  1)! 
 720
58. Let b 
 basket. Let m 
 miss.
20

Expand (b  m)20 
 

r
0

20!

r!(20  r)!

b20  r mr

Pages 870–871

Check for Understanding

1. Sample answer: If A and B are independent
events, then P(AB) 
 P(A). Thus, the formula
for conditional probability becomes P(A) 


Find the coefficient of the b15m5 term where r 
 5.
20!

5!(20  5)!

Conditional Probability

b205m5 
 15,504b15m5

P(A and B)

P(B)

15,504
59. No, the spill will spread no more than 2000 meters
away.

or P(A)  P(B) 
 P(A and B). This is the

formula for the probability of independent events.
2. S 
 {J spades, Q spades, K spades, J clubs,
Q clubs, K clubs}
3. Answers will vary; see students’ work.

480


a1 
 1200; r 
 
1200 or 0.4
a1


s

1r
1200




1  0.04

4. P(cubes match  sum greater than 5) 



 2000




447

4

36
—
26

36
2

13

Chapter 13

5. P(queen  face card) 



6. P(all heads 

7. P(all heads 

8. P(all heads 

4

52
—
12

52
1

3

13c. P(not rejected  counterfeit) 




1

8
—
first coin is a head) 
 4

8
1

4
1

8
at least 1 head) 
 —
7

8
1



7
1

8
at least 2 heads) 
 —
4

8
1



4

Pages 872–874






Exercises

14. P(1 head  at least 1 tail) 



15. P(Democrat  man) 




2

4
—
3

4
2

3

4

12
—
8

12
1

2

16. P(first bag  first chip is blue) 


9. P(numbers match  sum greater than or equal to 9)
2

36
—
10

36
1

5

1

100
—
25

100
1

25




4

16
—
10

16
2

5

17. P(girls are separated  girl at an end) 


10. P(sum is even  sum greater than or equal to 9)








4

36
—
10

36
2

5

18. P(number end in 52  number is even)



11. P(numbers match or sum is even  sum greater
than or equal to 9) 







4

36
—
10

36
2

5

19. P(2 odd numbers  sum is even) 




68  62


12a. P(disease prevented) 
 
100  100
13


 20

20.

12b. P(disease prevented  vaccine) 




68

200
—
100

200
17

25

21.

12c. P(disease prevented  conventional treatment)




13a.

13b.

62

200
—
100

200
31

50

2

52
—
P(ace  black) 
 26

52
1


 1
3
2

52
P(4  black) 
 —
26

52
1

 1
3

22. P(face card  black) 




69

100
—
P(legal  accepted) 
 70

100
69

 70
6

100
—
P(rejected  legal) 
 75

100
2


 2
5

Chapter 13

321

5!
43214321

5!
1

8

6

52
—
26

52
3

13

23. P(queen of hearts  black) 


0

52
—
26

52


0
24. P(6 of clubs  black) 




448

1

52
—
26

52
1

26

20

72
—
32

72
5

8

12

24
—
20

24
3

5

25. P(jack or ten  black) 




36. A 
 the sum of the cards is 7 or less
B 
 at least one card is an ace
B
 
 both cards not an ace

4

52
—
26

52
2

13

C(48, 2)

188

26. P(second marble is green  first marble was
green) 







3 2
  
8 7
—
3

8
2

7

P(A  B) 




3 5
  
8 7
—
3

8
5

7




37. P(sum greater than 18  queen of hearts)
1

19

  
52 51

 —
51
1
  
52 51

5 4
  
8 7
—
5

8
4

7




C(4, 2)  C(4, 1)  C(20, 1)

C(52, 2)
43



 663
P(A and B)

P(B)
43

663
—
33

221
43

99

P(A and B) 


28. P(second marble is yellow  first marble is yellow)



19


 51
38a.
C(1, 1)  C(1, 1)  C(4, 1)

C(6, 3)

S U

C

29. P(salmon  bass) 
 



15

C(1, 1)  C(5, 2)

C(6, 3)
2
4
 or 
5
10

25

5
155

30. P(not walleye  trout and perch)
C(1, 1)  C(1, 1)  C(3, 1)

C(6, 3)






38b. P(cancer  smokes) 


C(1, 1)  C(1, 1)  (4, 1)

C(6, 3)
3

4




C(1, 1)  C(1, 1)  C(3, 1)

C(6, 3)

P(A  B) 


C(5, 3)

C(6, 3)
3


 1
0
C(1, 1) C(1, 1)  (2, 1)

C(6, 3)

2


 4 or
33.

C(4, 3)

C(6, 3)
1

2

Brown Hair

4

or 5

41. P(passes  studied) 


Brown Eyes

50% 10%

120

500
—
150

500

Four out of five people who ask questions will
make a purchase. Therefore, they are more likely
to buy something if they ask questions.
40. Sample answers: The rolls are independent. The
number cubes do not have a memory, whether
they are fair or biased. Probability does not
guarantee an outcome.

32. P(perch and trout  neither bass nor walleye)



4

5

20%




P(passes and studied)

P(studied)
2

3

P(studied)
2

66

33

42b. P(chip from 3-D Images  defective)

0.10


P(brown eyes  brown hair) 
 
0.60

35. P(no brown

5



42a. P(defective) 
 
1000 or 500

Exercises 33–35

34. P(no brown

5

P(studied) 
 3  4 or 6

20%




25

200
—
30

200
5

6

39. A 
 person buys something
B 
 person asks questions

31. P(bass and perch  not catfish)



33



P(B) 
 1  P(B
) 
 1  
221 
 221

27. P(second marble is yellow  first marble was
green) 


188



P(B
) 
 
C(52, 2) 
 221

1

6




0.20

hair  brown eyes) 
 
0.30
2

 3
0.20

eyes  no brown hair) 
 
0.40
1

 2




21

1000
—
66

1000
7

22
934

467



42c. P(functioning) 
 
1000 or 500

449

Chapter 13

42d. Sample answer: A chip from CyberChip Corp.
has the least probability of being defective.

53.

6 ft2


 0.05
21

P(defective from 3-D Images) 
 
300

x
x

4 ft
If x 
 3 ft then the photo would have a negative


 0.07
20

P(defective from MegaView Designs) 
 
200

1

length and width. So, x 
 2 ft or 6 in.


 0.10

P(A and B

5

 by definition.
43. P(A  B) 
 
P(B)


54. f(x) 
 
x2  4

So, if P(A) 
 P(A  B) then by substitution P(A) 


5




(x  2)(x  2)

or P(A and B) 
 P(A)  P(B). Therefore,

f(x) is discontinuous when x 
 2;
f(x) is undefined when x 
 2.

the events are independent.
44. P(at least 3 women)

 P(3 women)  P(4 women)  P(5 women)
C(6, 3)  C(7, 2)

C(6, 4)  C(7, 1)

55. Drop an altitude
from B to 
AE
, from
u
C to ED , and from
E to B
C
. Label
the diagram as
shown.

C(6, 5)  C(7, 0)

    


C(13, 5)
C(13, 5)
C(13, 5)
420

105

531

59

6






1287  1287  1287




1287 or 143


 3(0.5)b 
 3(0.5)1  3(0.5)2  3(0.5)3  .

S




a1

r
1.5

0.5

h

h

E
1

D

1
(BC)h
2
1

area of shaded region 
 2(AE)h  2(ED)h
1

a1 
 1.5, r 
 0.5


 2h(AE  ED)
BC 
 AE  ED since opposite sides of a
parallelogram are equal. So, the ratio of the areas
1
1
is 2h(BC)2h(BC) or 11.

or 3

47. They are reflections of each other over the x-axis.
y

48.

C

h

area of unshaded region 


. .


 1.5  0.75  0.375  . . .

b
1

B

A

45. C(9, 4) 
 126
46.

x

x

25


P(defective from CyberChip) 
 
500

P(A and B)

P(B)

(4  2x)(3  2x) 
 6
12  14x  4x2 
 6
x
4x2  14x  6 
 0
2(2x  1)(x  3) 
 0
x
1
x 
 2 or x 
 3
x

x

The correct choice is B.
6
4
2
64 O

13-6

x

2 4 6

4
6

Page 877

The Binomial Theorem
and Probability
Graphing Calculator Exploration

1. S 
 {0, 1, 2}
2

r 
cos v 

49.





2

2. P (Bobby wins) 
 3

5
0

3. Answers will vary. In 40 repetitions, it may be
around 0.22. This means that there were exactly
5 wins for 8 or 9 of the 40 repetitions.



r
cos v cos 2  sin v sin 2 
 5
0  r sin v 
 5
y 
 5
y 
 3  2t

50. x 
 4t
x

4
x

4

y3

2


t


t

y3


 2
x

y 
 2  3

2 5

y

0.26
5. There is not a large enough sample of trials.
6. Increase the number of repetitions.

(4, 5)

(0, 3)
(4, 1)

O

Page 878

x

1



 2(8)2
1  
180 
98

71

54.7 ft2
x

52. tan 27° 
 25
12.7

Chapter 13

Check for Understanding

1a. Yes, it meets all three conditions.
1b. No, there are more than 2 possible outcomes.
1c. No, the events are not independent.
2. Sample answer: the probabilities derived from a
simulation rather than an actual event

51. A 
 2r2 v
1

1 1

4. P(winning 5 games) 
 C(6, 5)  
3  
3

x; 12.7 m

450

3. First, determine P(right) and P(wrong). Second,
set up the binomial expansion (pr  pw)5. Third,
determine the term of the expansion. Fourth,
substitute the probability values for pr and pw.
Last, compute the probability of getting exactly 2
correct answers.

15. P(no more than 3 times correct)

 1  P(correct 4 times)
2 4 1 0

3


 1  C(4, 4)
3

11

1

1 1 5 4

6

4. P(only one 4) 
 C(5, 1) 
6




 









6

1 1 5 4


6
6

17. P(7 correct) 


1250

7776






6. P(at least three 4s)

 P(three 4s)  P(four 4s)  P(5 fours)







2 2 1 2

3

16. P(correct exactly 2 times) 
 C(4, 2)
3


  
   C(5, 1)
  
 
3

  
56

1 3 5
C(5, 3) 6 6
1 5
 C(5, 5) 6
250
25
   
7776
7776
23

648

2

1

5

193


11
1

20. P(at least half correct)

7203

1


170 
2835
243



 1  
100,000  100,000 

5

 C(5, 5)  
1
0
3

0


170  

48,461

21. P(4
4 4 1 2
C(6, 4) 5 5
3840
768
 or 
15,625
3125


 
 

5 10 3 3

8

12. P(10 stocks make money) 
 C(13, 10) 
8


 

22. P(3

0.1372

Exercises

13. P(never the correct color) 






8


 4  2
7 


 

to 18


 
 


 
 

23. P(at least 2 heads)

2 3 1 1


 C(4,
3
3
1
16
  1    1
3
81


 
 








2 4 1 0


3
3

4)


1 2 2 2

 answer
3
8
1
1 4

6  9  9  8
1  81
24
9
  
81
81
11

27


 C(4, 2)
3

2 0 1 4
C(4, 0) 3 3
1
1  1  8
1
1

81

14. P(correct at least 3 times)


 

1 5 1 5

 answer
2
1
1
386







 252  32  32  1024
252
386




1024  1024
319



512
1 4 2 0
heads) 
 C(4, 4) 3 3
1

 1  8
1 1
1

 8
1
1 3 2 1
heads) 
 C(4, 3) 3 3
1
2


 4  2
7  3
8

 8
1


 C(10, 5)
2

10. P(having rain no more than three days)

 1  [P(rain on 4 days)  P(rain on 5 days)


 C(4, 3)



 

1

1024




1024

4


170 





100,000 or 20,000

Pages 878–880

1 0 1 10

2

19. P(all incorrect) 
 C(10, 0)
2

9. P(having rain on exactly one day)




1

1

1




512

0


130 

16,807

11. P(4 do not collapse) 


1

1

386




100,000

96,922

1




1024
7





100,000 or 50,000

1

1

8. P(not having rain on any day) 
 C(5, 5)
1
0

4

3

 

 10  
512  2  1  1024  1


 

1

3

7


 
  

 210  6
4  16  120  128  8  45  256  4
1 5 5 0

6


 1  C(5, 4)  
1
0

4


   C(10, 7)
12 
12
1 2
1 9 1 1
 C(10, 
  
2  C(10, 9)
2 
2
1 10 1 0
 C(10, 10)
2 
2




7776

36,015


 
 

1 6 1

2
1 8


8) 2

1

7776

1




 

1

9

8

 27
1 7 1 3
C(10, 7) 2 2
1
1
 
120  
128  8
15

128


 C(10, 6)
2

1 4 5 1


6
6

7. P(exactly five 4s) 
 C(5, 5)
6

3

4

9

18. P(at least 6 correct)


  
   C(5, 4)
  
 
0

  
56


 C(5, 1)
1
0

1

16

81

65

5. P(no more than two 4s)

 P(no 4s)  P(one 4)  P(two 4s)
5


 


 81

3125

7776

1 0 5
C(5, 0) 6 6
1 2
 C(5, 2) 6
3125
3125
   
7776
7776
625

648

16

81


 


 

to 22  answer to 21

1 6 3 4

4
1
81
  
4096 256

24. P(6 correct answers) 
 C(10, 6)
4

 210 

16


 27


 

0.016

451

Chapter 13

1 5 3 5

4
1
243
  
1024 1024

25. P(half answers correct) 
 C(10, 5)
4

 252 

35. Enter 16 nCr X on the Y-menu of your calculator.
16 nCr X represents the coefficients of the
binomial expansion where X is the number of
games won.
P(winning at least 12 games)


 

0.058
26. P(from 3 to 5 correct answers)





1 3 3 7
C(10, 3) 4 4  C(10, 4)
1 5 3 5
 C(10, 5) 4 4
1
2187
1


120  6
4  16,384  210  256 
1
243
 
 252  
1024  1024


 
 

 
 

729

4096

1

 
 

36b. P(between 2 and 6 missiles hit the target)

 1  [P(0 missiles hit the garget)  P(1 missile
hits the target)]

1

1.049  104

 120 


 


 1  
1  1 


 252 

2 5 3 5

5
32
243
  
3125 3125


 

37. P(all men or all women)
4

7

3


 1  1020
38b. P(exactly half have the disease)
1

1

10

0

75

0.807
40. P(less than or equal to 3 policies)

 1  P(4 policies)


 

1 4 1 0

2


 1   C(4, 4)
2

 1  
1 

32. P(at least 2 heads)
1 2 1 1
1 3 1 0

 C(3, 2) 2 2  C(3, 3) 2 2
1 1
1

 3  4  2  1  8  1
4

 8
1

 2
1 2 1 1
P(exactly 2 tails) 
 C(3, 2) 2 2
1 1

 3  4  2
3

 8


 
 




15

16

1

16

 1


 

or about 0.94

41a. If Trina walks 100 meters, then she has flipped
the coin 10 times. To end up where she began,
she walked north and south 5 times each.
1 5 1 5

2
1
1
  
32 32

P(back at her starting point) 
 C(10, 5)
2


 
 


 252 


 

0.246
41b. The closest Trina can come to her starting point
is if she flips 6 heads and 4 tails or 4 heads and
6 tails. However, this places her 20 meters from
her starting point. The answer for part b is the
same as that for part a, 0.246.

34a. The values of the function for 0  x  6 are the
coefficients of the binomial expansion.
34b. Change 6 nCr X to 8 nCr X on the Y-menu.

Chapter 13

0


190 


19060 
4 1 96 74


 C(75, 1)

100  
 100  
4

1

33.

20


190 



 1  C(75, 0)

100 


 4


 
 

0


140

6.4  106
39. P(at least 2 people do not show up)

 1  [P(0 people do not show up)
 P(1 person does not show up)]

1 3 1 0

2

3)
2

10


 C(20, 10) 
1
0

0.166
31. P(3 heads or 3 tails)

 1  8 

6

38a. P(all carry the disease) 
 C(20, 20)
1
0

64
81
128
27
 
  

 210  
15,625  625  120  78,125  125
3
256
9
512
 
 
 45  
390,625  25  10  1,953,125  5
1024

1
9,765,625  1


 

10

 C(10, 10)
1
0
1

4

1 3 1 0

 C(3,
2
1
1  1  8  1

0


160 

0.0062


   C(10, 7)
25 
35
3 2
2 9 3 1
 C(10, 
  
5  C(10, 9)
5 
5
2 10 3 0
 C(10, 10)
5 
5


 C(3, 3)
2

10


 C(10, 10)
1
0

30. P(at least 6 point up)
2 6 3

5
2 8
8) 5


 
 


1077

0.201

 C(10, 6)
5


 




3125

0.215
29. P(exactly 5 point up) 
 C(10, 5)
5

1 0 4 6
1 1 4 5

 C(6, 1) 5 5
5
4096
1 1024
  6    
15,625
5 3125


 1  C(6, 0)
5

2 3 3 7

5
8
2187
   
125 78,125

28. P(exactly 3 point up) 
 C(10, 3)
5

3

on each trial is 20% or 5.

2 10 3 0


5
5


1

13

0.45 or 45%
36a. A success means that a missile hits its target.
There are 6 trials and the probability of success

0.25  0.15  0.06
0.46
1024

9,765,625

4


130   560  
170  
130 
7 14 3 2

 120  
1
0  
 10 
7 15 3 1
7 16 3 1



 16  
1
0  
 10   1  
 10  
 10 
7


 
 

27. P(all point up) 
 C(10, 10)


12


 1820  
1
0

1 4 3 6


4
4

452

50. 4

12
21
62
72
4
32
44
72
1
8
11
18
0
Since the remainder is 0, x  4 is a factor of the
polynomial.

41c. P(exactly 20 meters from the starting point)
1 6 1 4
1 4 1 6

 C(10, 4) 2 2
2
1
1
1
1
    210    
64 16
16 64


 C(10, 6)
2

 210 


 


 
 

0.41
42. P(sum less than 9  both cubes are the same)





51.

4

36
—
6

36
2

3

1

Area of page for text

Area of entire page








(9  1  1)(12  1.5  1.5)

9  12
63

108
7

12

The answer is 7/12.

43. P(letter is contained in house or phone)
5
5
3



 2
6  26  26
7


 2
6

Chapter 13 Study Guide and Assessment

44a. 80, 75, 70, . . .
44b. T 
 80  5n
44c. T 
 125
40,000

n

1000 or 40

Page 881

125 
 g 5(40)
75 
 g; 75° F
45.
3x  1 
 6x
(x  1)log 3 
 x log 6
x log 3  log 3 
 x log 6
x log 3  x log 6 
 log 3
x(log 3  log 6) 
 log 3
log 3

x

log 3  log 6

7. mutually exclusive
9. conditional

Pages 882–884


cos 
47. 2

 i sin 



2

6!




 
 2  0  2  i

654321

3  2 1


 120
8!


15. P(8, 6) 
 
(8  6)!




87654321

21


 20,160


 
(2)2 
 82

5!


16. C(5, 3) 
 
(5  3)! 3!


 68
 or 217


54321




21321

55 cm


 10

106˚

71 cm

Skills and Concepts


14. P(6, 3) 
 
(6  3)!

i

 0  2
u
48. WX 
 6  8, 5  (3) or 2, 8
u
(6  8
)2  (5
 (
3))2
WX  
 

49.

2. failure
4. probability
6. permutation with
repetitions
8. sample space
10. combinatorics

11. Using the Basic Counting Principle, 3  2  1 or 6.
12. Using the Basic Counting Principle, 5  4  3  2  1
or 120.
13. Using the Basic Counting Principle,
6  5  4  3  2 1 or 720.

x 0.38
46. h 
 0, k 
 3, a 
 7, b 
 5, c 
 26

Center: (0, 3)
Foci: ( 26
, 3)
Vertices: major axis → (7, 3) and (7, 3)
minor axis → (0, 2) and (0, 8)


2

Understanding the Vocabulary

1. independent
3. 1
5. permutation

11!


17. C(11, 8) 
 
(11  8)! 8!

d




11  10  9  8  7  6  5  4  3  2  1

32187654321


 165
d2 
 712  552  2(71)(55) cos 106°
d 101.1 cm

18.






106˚

71 cm

P(6, 3)

P(5, 3)

6!

(6  3)!

5!

(5  3)!
65432121

54321321


2

d

5!

3!

 
19. C(5, 5)  C(3, 2) 
 
(5  5)! 5!  (3  2)! 2!

74˚

1

321



 1  
121

55 cm


3

d2 
 712  552  2(71)(55) cos 74°
d 76.9 cm

453

Chapter 13

1

20. There are P(5, 5) ways to arrange the other books
if the dictionary is on the left end. The same is
true if the dictionary is on the right end.
5!

2  P(5, 5) 
 2  
(5  5)!

odds 


54321


 2  1





 240
3!

321

7654321

5432121

5




1296


 3  21 or 63

36. dependent, P(two yellow markets)
4
3


 1
0  9

54321

22.

5!

2! 2!

23.

10!

2! 3! 3!

1

14
—
13

14
1

13

35. independent, P(sum of 2)  P(sum of 6)
1
5


 3
6  36

7!

 
21. C(3, 2)  C(7, 2) 
 
(3  2)! 2!  (7  2)! 2!



121 

13


34. P(s) 
 1
4 ; P(f) 
 14




2121


 30



2


 1
5

10  9  8  7  6  5  4  3  2  1

21321321

37. P(selecting a prime number or a multiple of 4)

 P(prime number)  P(a multiple of 4)


 50,400

24.

8!

2!

25.

6!

3! 2!

26.

9!

3! 2!




87654321

21

6
9


 20,160




 1
4

654321

32121

38. P(selecting a multiple of 2 or a multiple of 3)

 P(multiple of 2)  P(multiple of 3)
 P(multiple of 2 and 3)


 60



987654321

32121

7

30.

31.

32.

33.

Chapter 13

2

9


 1
4

C(7, 3)  C(4, 0)  C(5, 0)

C(16, 3)
35  1  1



560
1

 1
6
C(7, 2)  C(4, 1)  C(5, 0)
P(2 pennies and 1 nickel) 
 
C(16, 3)
21  4  1



 560
3

 2
0
C(7, 0)  C(4, 3)  C(5, 0)
P(3 nickels) 
 
C(16, 3)
141



 560
1



140
C(7, 0)  C(4, 1)  C(5, 2)
P(1 nickel and 2 dimes) 
 
C(16, 3)
1  4  10



560
1



 14
1
15




P(s) 
 16 ; P(f) 
 16
1

16
—
odds 
 15

16
1

 1
5
3
17
 P( f) 
 
P(s) 
 
20;
20
3

20
odds 
 —
17

20
3


 1
7
1
139



P(s) 
 140 ; P(f) 
 
140
1

140
—
odds 
 139

140
1



139

27. P(3 pennies) 


29.

4




 1
4  14  14


 30,240

28.

3



 1
4  14

39. P(selecting a 3 or a 4) 
 P(3) or P(4)
1

1



 1
4  14
1


 7
40. P(selecting an 8 or a number less than 8)

 P(8)  P(less than 8)
1

7



 1
4  14
4


 7
41. P(sum less than 5  exactly one cube shows 1)





4

36
—
10

36
2

5

42. P(different numbers  sum is 8) 




4

36
—
5

36
4

5

43. P(numbers match  sum is greater than or equal
to 5)





4

36
—
30

36
2

15

1 1 1 3

2

44. P(exactly 1 head) 
 C(4, 1)
2

4



1

4

1

2

1 0 1 4

2

45. P(no heads) 
 C(4, 0)
2

11
1


 1
6

454



1

8

1

16


 


 

1 2 1 2

2

46. P(2 heads and 2 tails) 
 C(4, 2)
2
1

2. 4x
–
3  8 
 32
4x

–
3 
 24
x

–
3
6
(x–3 )2 
 62
x – 3 
 36
x – 3 
 39
The correct choice is E.
3. Find the probability of selecting a green marble
from the jar now.


 

1


 6  4  4
3


 8
47. P(at least 3 tails)

 P(3 tails)  P(4 tails)

1 3 1 1

 C(4,
2
1 1
1
4  8  2  1  1
6 1
1
1
5
   or 
4
16
16


 C(4, 3)
2






Page 885


 

1 4 1 0

2

4)
2


 

number of green marbles

total marbles

3

1



 1
5 or 5

Let x represent the number of green marbles
1
2
added so the probability equals 2  5 or 5.

Applications and Problem Solving

3x

48. C(1, 1)  C(6, 4) 
 1  15 or 15

number of green marbles

total marbles

7!
49.  
 2520

5(3  x) 
 2(15  x)
15  5x 
 30  2x
3x 
 15
x
5
The correct choice is C.

2

50. P(at least 1 good chip)

 1  P(both defective chips)
C(3, 2)



1
C(15, 2)
1

sum of terms


Average 
 
number of terms

4.


 1  3
5
34

20 



 35
51b. P(Reba’s name, then a male name) 




Page 885



7

14

Open-Ended Assessment
1

1

sum of five terms

5

sum of five terms 
 100
Since one of the numbers is 18, the sum of the
other four is 100  18 or 82. The correct choice
is C.
5. Select specific numbers for the problem. Let
x  y 
 8. Let x  z 
 12. Let z 
 7.
x  7 
 12, so x 
 5, an odd number.
5  y 
 8, so y 
 3, an odd number.
Statement I is false, and statement III is true.
Since y  z or 3  7 
 10, an even number,
statement II is true also. The correct choice is E.
6. Since the probability of selecting a blue marble is

7

51a. P(female name excluding Reba) 
 1
5
1

15
1

30

2





15  x 
 5

1

1. Yes; sample answer: x  2 
 12, so x 
 6; Two
marbles are chosen from a box containing 6 red, 4
blue, and 2 green marbles. What is the probability
of choosing a red and a green marble?
2. Sample answer: In a permutation, the order of
objects is very important. In a combination, the
order of objects is not important.

1

5

and the total number of blue and white marbles

is 200, the number of blue marbles must be 40. So
the number of white marbles must be 160. After
100 white marbles are added, the total number of
white marbles is 260 and the total of all marbles
is 300. The probability of selecting a white marble

Chapter 13 SAT & ACT Preparation

260

13



is 
300 or 15 . The correct choice is E.

Page 887

7. ∠A and ∠C must be equal because they are
corresponding angles. The correct choice is A.
8. The sample space, or total possible outcomes, is
the 52 cards in a deck. The outcome “drawing a
diamond” consists of the 13 cards that are
diamonds.
13
1
P(diamond) 
 52 or 4

SAT and ACT Practice

1. You might want to draw a diagram of the 20 coins.
The first and last coins are heads. The total
number of heads is 10. There could be 9
consecutive heads followed by 10 tails and then
the final head. The correct choice is D.
H

H

H

H

H

H

H

H

H

The correct choice is C.
9. 7 different entrees are offered. 3 are chosen.
The number of combinations that can be chosen
7!
3!(7  3)!

7!
3!4!

is C(7, 3) 
  
  
 35.
The correct choice is B.
H

455

Chapter 13

10.

1
2

The probability that a dart thrown randomly at
the target will land in the shaded region is equal
to the ratio of the area of the shaded region to the
area of the entire target.
area of shaded region

P 
 
area of target
 (1)2

P

 (2)2


P

4 or

1

4
1

The answer is 
4.

Chapter 13

456

Chapter 14 Statistics and Data Analysis
14-1

6f. Sample answer:

The Frequency Distribution

Pages 892–893

Ages of Presidents
12

Check for Understanding

Frequency

1. A line plot, a bar graph, a histogram, and a
frequency polygon all show data visually. A line
plot shows the frequency of specific quantities by
using symbols and a bar graph shows the
frequency of specific quantities by using bars. A
histogram is a special bar graph in which the
width of each bar represents a class interval. A
frequency polygon shows the frequency of a class
interval using a broken line graph.
2. Choose an appropriate class interval. Use tally
marks to determine the number of elements in
each class interval.
3a. No; there would be too many classes.
3b. Yes; there would be 9 classes.
3c. Yes; there would be 5 classes.
3d. No; there would only be 3 classes.
3e. No; there would only be 2 classes.
4. See students’ work.
Age
5a.

4
0

60-69

4

40

50

Pages 893–896
7a.



43026




43212




















43214

43221
45414
43220
43229
ZIP Codes

7b. 43220
7c. Sample answer: to determine where most of
their customers live so they can target their
advertising accordingly

2000

8a.

Men

Age
65+

30-39

50-64

20-29

35-49

10-19

20-34

0-9

16-19
4

70

Exercises

40-49

0
0
Percent

60

6g. Sample answer: 50–60

50-59

16 12 8

0

Age

70+

1900

8

Women

80 60 40 20 0
0 20 40 60 80
Minutes Behind the Wheel

8 12 16

8b. Sample answer: Men spend more hours driving
than women.

5b. In 1999, there are larger percents of older
citizens than in 1990.
6a. range 
 69  42 or 27
6b. Sample answer: 5
6c. Sample answer: 40, 45, 50, 55, 60, 65, 70
6d. Sample answer: 42.5, 47.5, 52.5, 57.5, 62.5, 67.5
6e. Sample answer:

9a. Rental Revenue Year Sales Revenue
2003
2000
1997
1990

Ages
40–45
45–50
50–55
55–60
60–65
65–70

Frequency
2
6
12
12
7
3

1985
10 8 6 4 2 0
0 2 4 6 8 10 12 14
Dollars (in billions)

9b. Sales; the sales revenue is increasing, and the
rental revenue has started to decrease.
10a. range 
 53  4 or 49
10b. Sample answer: 10
10c. Sample answer: 0, 10, 20, 30, 40, 50, 60
10d. Sample answer: 5, 15, 25, 35, 45, 55

457

Chapter 14

11g. Sample answer:

10e. Sample answer:
Grams of Fat
0–10
10–20
20–30
30–40
40–50
50–60

Olympic Winter
Games

Frequency
7
11
10
7
2
1

8
6
Frequency 4
2
0

10f. Sample answer:

0 10 20 30 40 50 60 70 80
Number of Nations

12a. range 
 1023  404 or 619
12b. Sample answer: 100
12c. Sample answer:

Grams of Fat in
Fast-Food Sandwiches
10

Height (feet)
400–500
500–600
600–700
700–800
800–900
900–1000
1000–1100

8
Frequency 6
4
2
0

10g.
11a.
11b.
11c.
11d.
11e.

0 10 20 30 40 50 60
Grams of Fat

Sample answer: 10–20
range 
 72  16 or 56
Sample answer: 10
Sample answer: 10, 20, 30, 40, 50, 60, 70, 80
Sample answer: 15, 25, 35, 45, 55, 65, 76
Sample answer:
Number of Nations
10–20
20–30
30–40
40–50
50–60
60–70
70–80

Frequency
5
4
2
3
1
2
3

12d. Sample answer:
5
4
Frequency 3
2

Frequency
2
3
8
1
1
2
1

1
0

0 400 600 800 1000
Height (feet)

12e. Sample answer: 400–500
13a. American League

Year
2003
2002
2001

11f. Sample answer:

2000

Olympic Winter
Games

1999

8

1998

6

1997

Frequency 4

1996

2

1995

0

National League

1994

0 10 20 30 40 50 60 70 80
Number of Nations

80 60 40 20 0

0 20 40 60 80

Greatest Number of Stolen Bases
for a Single Player

Chapter 14

458

13b. Sample answer:

16b.
$1,200,000

Stolen Bases
30–40
40–50
50–60
60–70
70–80

Frequency
1
6
6
4
3

Sales $800,000
$400,000
$0

1999 2000
Year

16c. See students’ work.
17. See students’ work.
18. This is a biannual experiment, where 8 mums are
involved and there are only two possible outcomes,
survival S and not survival N.
P(exactly 6 surviving) 
 C(8, 6)(S)6(N)2

 28(0.8)6(0.2)2
0.29 or 29%

13c. Sample answer:

6
Frequency

1998

4

7

 7r(2d)r
19. (c  2d)7 
  
r!(7  r)! (c)

2

7!

r
0

0

0

To find the second term, evaluate the general term
for r 
 1.

30 40 50 60 70 80
Stolen Bases

7!
(c)7r(2d)r
r!(7  r)!

13d. 3 players
13e. 7 players
14. Sample answer: 0.1, 0.2, 0.3, 0.4, 0.6, 0.7, 0.8, 0.9,
1.1, 1.2, 1.3, 1.4, 1.6, 1.7, 1.8, 1.9, 2.1, 2.2, 2.3, 2.4


 7c6  2d

 14c6d
20.

15.
Millions
of
Tons

300
250

7!

 71(2d)1


1!(7  1)! (c)

3.6x 
 58.9
x ln 3.6 
 ln 58.9
x 3.18

21.

y
9xy 
 36

x
O

200
150
100
50
0

Wheat

Rice

Corn

22. (x  y)2 
 x2  2xy  y2

 (x2  y2)  2(xy)

 16  2(8)

 32

China
India
United States

16a. The first interval on the vertical axis represents
$800,000, but the other intervals represent only
$200,000. Therefore, the sales for 1999 appear to
be twice the sales of 1998, but in reality they are
not.

14-2
Page 903

Measures of Central Tendency
Check for Understanding

1. Mean, median, mode; to find the mean, add the
values in a set of data and divide the sum by the
number of values in the set. To find the median,
arrange the values in a set of data from least to
greatest. If there is an odd number of values in
the set, the median is the middle value. If there is
an even number of values in the set, the median is
the mean of the two middle values. To find the
mode, find the item of data that appears more
frequently than any other in the set.
2. Sample answer: {1, 2, 2, 3, 4, 4, 5}
The modes are 2 and 4.

459

Chapter 14

1

3. Write the stems 9, 10, 11, 12, 13, and 14 on the
left. Write the tens digits as leaves to the right of
the appropriate stems. Be sure to order the leaves.
4. Tia; the median 2.5 and the mode 2 do not
represent the greater numbers. The mean 8.5 is
more representative of all 8 items in the data.

9b. X
 
 4
0 (6  3(7)  9  2(13)  2(14)  15  16 
17  18  2(19)  3(20)  5(21)  2(23) 
28  30  3(31)  32  2(34)  36  38 
3(41)  42  47)

 23.55
9c. Md 
 21
9d. Mode 
 21
9e. Since the mean 23.55, the median 21, and the
mode 21 are all representative values, any of
them could be used as an average.

1

5. X
 
 4(10  10  45  58) or 30.75
10  45

Md 
 2 or 27.5
Mode 
 10
1

6. X
 
 1
0 (21  22  23  24  28  29  31  31
 34  37)

 28

Pages 904–907

28  29

Md 
 2 or 28.5

10.

Mode 
 31
1

7. X
 
 3(91  94  95  98  99  105  105  107
 107  107  111  111  112)  100
103.23  100
10,323
Md 
 10,500
Mode 
 10,700
8a. 2  8  15  6  38  31  13  7 
 120
120 members

1

Md 
 3
Mode 
 3
1

12. X
 
 4(17  19  19  21) or 19
Md 
 19
Mode 
 19

 fi 
 120

1

13. X
 
 8(3  5  5  8  14  15  18  18)

 10.75

 (fi  Xi) 
 2(3)  8(7)  15(11)  6(15) 

8  14

Md 
 2 or 11

38(19)  31(23)  13(27)  7(31)

 2320
2320

X


120 or about 19.3

i
1

Visits
1–5
5–9
9–13
13–27
17–21
21–25
25–29
29–33

Members
2
8
15
6
38
31
13
7

Mode 
 5 and 18
1

14. X
 
 1
2 (54  58  62  63  64  70  76  76
 87  87  98)

 73.5

Cumulative Members
2
10
25
31
69
100
113
120

70  76

Md 
 2 or 73
Mode 
 87
1

15. X
 
 1
2 (5  6  6  6  7  8  9  10  11  11
 11  12)

 8.5
89

Md 
 2 or 8.5
Mode 
 6 and 11
1

16a. X
 
 9(117  124  139  142  145  151

Half of the data has been gathered in the 17–21
class. This is the median class.
8d. 69  31 
 38
21  17 
 4
60  31 
 29
Md  17 
 x
38

4

 155  160  172)

 145 lb
16b. Md 
 145 lb

29


 x

1

16c. 
X 
 9(122  129  144  147  150  156

x 3.052631579
Md  17 
 x
Md  17 3.1
Md 20.1
9a. stem

160  165  177)

 150 lb
Md 
 150 lb
Each will increase by 5 lb.

leaf

0
6
1
3
2
0
3
0
4
1
1|3 
 13
Chapter 14

155

11. 
X 
 5(3  3  3  6  12) or 5.4

i
1
8

8c.

 160  170) or 155

Mode: none

8

8b.

Exercises

1
X
 
 4(140  150
150  160
Md 
 2 or

1

7
3
0
1
1

7
4
0
1
1

7
4
1
1
2

17. X
 
 2(35  2(38)  39  44  3(45)  48  2(57)

9
5 6 7 8 9 9
1 1 1 1 3 3 8
2 4 4 6 8
7

 59)
45.8
Md 
 45
Mode 
 45

460

1

24. Order the values from least to greatest. The
median lies between the fourth and fifth terms.
2, 3, 3.2, 8, x, 11, 13, 14

18. X
 
 1
4 (5.2  5.4  5.6  6.0  6.1  6.7  6.8
 6.9  7.1  7.6  8.0  8.2  8.6  9.1)

 6.95

88

x 
 8, since Md 
 2 or 8.

6.8  6.9

Md 
 2 or 6.85

1


25a. 
X

179 [9(245)  14(275)  23(325)  30(375)

Mode: none

 33(425)  28(475)  18(525)  12(575)
 7(625)  3(675)  1(725)  1(775)]
425.6

1

19. X
 
 1
5 (90  91  97  98  99  105  106
 109  113  3(118)  120  2(125))  10

 1088
Md 
 1090
Mode 
 1180
20. stem leaf
1 0 5 5 5 5 7 7
2 0 0 0 5 5 5 5 7 8
3 0 0 5 5 5
4 6
5 5
1|0  10
21a. 135(11) 
 $1485; 145(24) 
 $3480;
155(30) 
 $4650; 165(10) 
 $1650;
175(13) 
 $2275; 185(8) 
 $1480;
195(4) 
 $780
21b. 1485  3480  4650  1650  2275  1480
 780 
 $15,800
21c. 11  24  30  10  13  8  4 
 100
100 employees

25b.

Weekly
Wages
$130–$140
$140–$150
$150–$160
$160–$170
$170–$180
$180–$190
$190–$200

Frequency
11
24
30
10
13
8
4

Cumulative
Frequency
11
35
65
75
88
96
100

9
14
23
30
33
28
18
12
7
3
1
1

x 
 20.4
5

Md  400 
 x
Md  400 20.5
Md 420.5
1

26a. X
 
 5(3.6  3.6  3.7  3.9  4.8)

 3.92
Md 
 3.7
Mode 
 3.6
26b. Only the mean would change. It would increase
to 4.6.
1
X
 
 5(3.6  3.6  3.7  3.9  8.2)

 4.6
26c. The mean increases slightly; the median
increases slightly; the mode stays the same.
1
X
 
 4(3.6  3.6  3.7  3.9)

x
5
Md  150 x
Md  150 5
Md 155
21g. Both values represent central values of the data.
22. 7.5 


200–250
250–300
300–350
350–400
400–450
450–500
500–550
550–600
600–650
650–700
700–750
750–800

Cumulative
Number of
Students
9
23
46
76
109
137
155
167
174
177
178
179


 89.5; Half of the data has been gathered in
the 400–450 class. This is the median class.
25c. 109  76 
 33
450–400 
 50
89.5  76 
 13.5
Md  400 
 x
33
13.5
 
 
50
x

Half of the data has been gathered in the
$150–$160 class. This is the median class.
21f. 65  35 
 30
160  150 
 10
50  35 
 15
Md  150 
 x
30
15
 
 
10
x

1
(2
5

Number
of Students

179

2

15,800


21d. X


100 or about $158

21e.

Scores


 3.7
3.6  3.7
Md 
 2 or 3.65

 4  5  8  x)

37.5 
 19  x
18.5 
 x
1
23. 6 
 4(x  2x  1  2x  3x  1)

27a.
27b.
27c.
27d.

24 
 8x
3
x

461

Mode 
 3.6
Sample answer: {1, 2, 2, 2, 3}
Sample answer: {4, 5, 9}
Sample answer: {2, 10, 10, 12}
Sample answer: {3, 4, 5, 6, 9, 9}

Chapter 14

28a.
stem
0

leaf
1 1
3 3
8 8
0 1
5 9
2

1
2
3
4
5 3
5|3  53

30a. Let 
X 
 50
9

1
4
8
3

1
4
9
3

1
4
9
3

1
5
9
5

 (X  fi) 
 0 
 (50  5)  (50  20)  (50
i
1

1 2 2 2 2 2 3 3 3
5 5 5 6 6 7 7 7 8
9
8 9 9

 37)  (50  44)  (50  52)
 (50  68)  (50  71)
 (50  85)  (50  x)
0 
 68  x
x 
 68
The weight should be hung 68 cm from the end.
 
 50
30b. Let X
9

 (X  fi) 
 0 
 (50  5)  (50  20)  (50
i
1

 37)  (50  44)  (50  52)
 (50  68)  (50  71)
 (50  85)  (50  x) 
(50  x)

1

28b. 
X 
 5
0 [7(1)  5(2)  5(3)  3(4)  4(5)  2(6)
 3(7)  4(8)  4(9)  10  11  3(13)
 15  18  2(19) 25  29  32  53]

 8.7
28c. Md 
 6
28d. Mode 
 1
28e. The mean 8.7 and the median 6 are
representative of the data, but the mode 1 is not
representative of the data.
1
[1(170)
27

29a. X


0 
 118  2x
118 
 2x
59 
 x
The weight should be hung 59 cm from the end.
1

31a. X
 
 1
0 (54  55  59  59  61  62  65  75
 162  226)  1000

 $87,800

 6(190)  10(210)  6(230)

 3(250)  1(270)]
215.2
29b.
Goals
160–180
180–200
200–220
220–240
240–260
260–280

Number of
Teams
1
6
10
6
3
1

61,000  62,000

31b. Md 
 2 or $61,500
31c. Mode 
 $59,000
31d. The mean, since it is the greatest measure of
central tendency.
31e. The mode, since it is the least measure of central
tendency.
31f. Median; the mean is affected by the extreme
values of $162,000 and $226,000, and only two
people make less than the mode.
31g. Sample answer: I have been with the company
for many years, and I am still making less than
the mean salary.

Cumulative
Number of Teams
1
7
17
23
26
27

27

2


 13.5; Half of the data has been gathered in
the 200–220 class. This is the median class.
29c. 17  7 
 10
220  200 
 20
13.5  7 
 6.5
Md  200 
 x
10
6.5
 
 
20
x

32a. X


 15(2.50)  31(3.00)

 37(3.50)  5(4.00)
30.4
32b.

x 
 13
Md  200 
 x
Md  200 
 13
Md 213
1

29d. 
X 
 2
7 (268  248  245  242  239  239
 237  236  231  230  217  215
 214  211  210  210  207  205
 202  200  196  194  192  190
 189  184  179)
215.9
Md 
 211
29e. The mean calculated using the frequency
distribution is very close to the one calculated
with the actual data. The median calculated
with the actual data is less than the one
calculated with the frequency distribution.

Chapter 14

1
[12(2.00)
100

Grade Point
Averages
1.75–2.25
2.25–2.75
2.75–3.25
3.25–3.75
3.75–4.25

Frequency
12
15
31
37
5

Cumulative
Frequency
12
27
58
95
100

The median class is 2.75  3.25.
58  27 
 31
3.25  2.75 
 0.50
50  27 
 13
Md  2.75 
 x
13
31
 
 
x
0.50
x 0.2096774194
Md  2.75 
 x
Md  2.75 0.21
Md 2.96

462

33. He is shorter than the mean (5
 11.6) and the
median (5
 11.5).

Measures of Variability

14-3

1

X 
 1

0 (67  68  69  69  71  72  73  74
 75  78)

 71.6 or 5
 11.6

Page 914

71  72

Md 
 2

 71.5 or 5
 11.5
34.
22
20
18

Frequency

16
14
12
10
8
6
4
2
0

0 60 65 70 75
Speed Limit
3

2

Check for Understanding

1. The median of the data is 70, Q1 is 60, and Q3 is
100. The interquartile range is 40 and the semiinterquartile range is 20. The outliers are 170 and
180. The data in the first two quartiles are close
together in range. The last two quartiles are more
diverse.
2. square the standard deviation
3. Both the mean deviation and the standard
deviation are measures of the average amount by
which individual items of data deviate from the
mean of all the data. The mean deviation uses the
absolute values of the deviations. Standard
deviation uses the squares of the deviations.
4. See students’ work.
5. interquartile range 
 Q3  Q1

 41  25

 16
16
Semi-interquartile range 
 2

8

3


35. dependent; 11  1
0 
 55
n1

n


36. an 
 3n ; an1 
 
3n1

r

n1


3n1
—

 lim n
n→ 3
n

15
6. X



37. Fn 




45

 5.50  5.50  6.30  7.80  11.00

$4.11
1


7. X


200 [15(5000)  30(15,000)  50(25,000)

 60(35,000)  30(45,000)  15(55,000)]

 30,250





864 2
4
6
8

40

1

1, the series is convergent.

8
6
4
2

35

2  (3.24)2 … 8.462

8 (4.29










 $40,305.56
38.

30

$3.54

(1  i)n  1
P i
(1  0.03)20  1

1500 
0.03



25

 12.20  17.20)

 8.74
1
MD 8(4.29  3.24 … 8.46)

3n(n  1)


n1
n→ 3 (n)
n
3 (n  1)

 lim 
n 
n→ 3  3n
n1

 lim 3n
n→
n
1

 lim 3n  3n
n→
1
1

 3  0 or 3


 lim

Since r

20
1
(4.45
8

(25,250)
 15  (15,250)  30 … 24,750  15


200
2

2

2

13,226.39
1

y

8a. X
 
 1
2 (65.7  65.9 … 65.9)

 70.375

69.0  70.3

Md 
 2 or 69.65

O
2 4 6 8x




(4.875)2  (4.475)2 … 6.2252

12

4.25

8b. X



39. To find the area of the triangle use Hero’s formula:

1
(57.3
12

 63.3 … 57.5)

80.48

s(s  
a)(s 
b)(s 
c), where
Area 
 

77.5  82.1

Md 
 2 or 79.8

1
s 
  (a  b  c) and a 
 10, b 
 7, and c 
 5.
2
1
1
So, s 
  (10  7  5) 
 (22) 
 11.
2
2

(23.18)  (22.98) … 25.32


12
2

2

2

17.06

 10)
(11 
7)(11
 5)
Area 
 11(11


 
11  1
 4  6 or 264

The correct choice is A.



16.25.

463

Chapter 14

8c. Los Angeles

1

14. X
 
 1
0 (5.7  5.7 … 3.8)

 4.89
1

MD 
 1
0 (0.81  0.81 … 1.09)

55 60 65 70 75 80 85 90 95 100 105


 0.672

Las Vegas




0.81  0.81 … (1.09)


10
2

2

2

0.73
1

15. X
 
 1
2 (369  376 … 454)

55 60 65 70 75 80 85 90 95 100 105


 403.5

8d. Los Angeles
8e. Los Angeles is near an ocean; Las Vegas is in a
desert.

1

MD 
 1
2 (34.5  27.5 … 50.5)

 20.25



Pages 915–917

Exercises

(34.5)  (27.5) … 50.5


12
2

2

2

25.31

9. interquartile range 
 Q3  Q1

 24  17

7
7
semi-interquartile range 
 2 or 3.5

1

16. X
 
 5(13  22  34  55  91)

 43
Variation 


(30)2  (21)2  (9)2  122  482

5


 774
1


17. X


120 [2(3)  8(7) … 7(31)]

19.33

14 16 18 20 22 24 26 28 30

(16.33)  2  (12.33)  8 … 11.67  7


120
2

10. interquartile range 
 Q3  Q1

 21.5  12

 9.5
9.5
semi-interquartile range 
 2 or 4.75

2

6.48
1

18. X
 
 9
0 [3(57)  7(65) … 12(97)]

 81.8



0

2

(24.8)  3  (16.8)  7 … 15.2  12


90
2

2

2

9.69

10 20 30 40 50 60 70 80

1

19. X
 
 8
5 [2(80)  11(100) … 7(180)]

11. interquartile range 
 Q3  Q1

 10.5  7.6

 2.9
2.9
semi-interquartile range 
 2 or 1.45

129.65


(49.65)  2  (29.65)  11 … 50.35  7


85
2

2

2

23.29
20a. Md 
 259 mi
20b. Q1 
 129 mi; Q3 
 360 mi
20c. interquartile range 
 Q3  Q1

 360  129

 231 mi

5 6 7 8 9 10 11 12 13 14 15 16
12.

231

20d. semi-interquartile range 
 2

 115.5 mi
20e. An outlier would lie 231  115.5 or 346.5 mi
outside of Q1 or Q3. There are no such points.
20f.
1

13. X
 
 6(152  158 … 721)

 381

0 100 200 300 400 500 600 700

1

MD 
 6(229  223 … 340)

20g. The data in the upper quartile is more diverse
than the other quartiles.
21. Sample answer: {15, 15, 15, 16, 17, 20, 24, 26, 30,
35, 45}
22a. Md 
 282
22b. Q1 
 42; Q3 
 770


 211



(229)  (223) … 340


6
2

2

2

223.14

Chapter 14

464

22c. interquartile range 
 Q3  Q1

 770  42

 728
semi-interquartile range 


1

25a. 
X 
 5
0 [26(9)  12(11) … 2(21)]

 11

728

2

25b.


 364
22d. An outlier would lie 728  364 or 1092 points
outside of Q1 or Q3. There are no such points.




(9  11)  26  (11  11)  12 … (21  11)  2


50
2

2

2

2.94
26. yes; when the standard deviation is less than 1;
when both equal 0 or 1
27. See students’ work.

22e.

1

28a. X
 
 3
5 [2(4.4)  4.9  5.4  5.5  2(6.2)  6.4
0
22f. X



200
1
(22
19

400

600

800

1000

 23 … 966)

404.42
1

22g. MD  1
9 (382.4  381.4 … 561.58)
316.97

28b.
28c.
29a.
29b.
29c.
29d.


…

19
382.42

22h. Variance 

381.42

561.582

118,712.56
22i.
22j.
23a.
23b.

23c.

2.56

 118,71
344.55
There is a great variability among the number of
teams in women’s sports.
Q1 
 $3616, Md 
 $4125, Q3 
 $5664
interquartile range 
 Q3  Q1

 5664  3616

 2048
An outlier would lie 2048  1024 or $3072
outside of Q1 or Q3. There are two such values,
$26,954 and $27,394.

 6.5  6.9  7.1  7.4  7.5  7.6  7.7
 7.8  7.9  8.0  8.2  8.4  8.5  8.6
 8.7  8.8  3(8.9)  9.0  9.2  2(9.3)
 9.5  9.6  9.8  9.9]
7.75
Md 
 8.0
Mode 
 8.9
range 
 68  23 or 45
Sample answer: 10
Sample answer: 20, 30, 40, 50, 60, 70
Sample answer:
Programs Sold
20–30
30–40
40–50
50–60
60–70

Frequency
2
1
2
5
2

29e. Sample answer:

23d.
4

0

5000 10,000 15,000 20,000 25,000 30,000

Frequency

1

23e. X
 
 1
9 (2684  2929 … 27,394)

2

6775.95
MD

0

1
(4091.95  3846.95 … 20,618.05)
19

4463.39
(4091.95)  (3846.95) … 20,618.05


19
2

23f.

2

2

1

30. 9(9!) 
 40,302 ways
31. x1 
 0.5(8) 1 or 3
x2 
 0.5(3)  1 or 0.5
x3 
 0.5(0.5)  1 or 0.75
32. 7 ft 
 7(12) or 84 in.
84  9 
 93 in.
93
 
 31 in.
3

7103.45
23g. The data in the upper quartile is diverse.
24a.
0

100 200 300 400 500 600
1

24b. X
 
 4
2 (0  0 … 635)

31

31 in. 
 1
2 or 2 ft 7 in.

60.40
MD

0 10 20 30 40 50 60 70
Programs
Sold

The correct choice is C.

1
(60.40
42

 60.40 … 574.60)

67.87
24c. Variance 

(60.40)2  (60.40)2 … (514.60)2

42

14,065.48
5.48

 14,06
118.60
24e. The data in the upper quartile is diverse.
24d.

465

Chapter 14

Page 917

Mid-Chapter Quiz

3.

1. Sample answer: 10
2. Sample answer:
Exam Scores
50–60
60–70
70–80
80–90
90–100

Frequency
2
4
6
10
8

45 55 65 75 85 95 105

45 55 65 75 85 95 105

The second curve is less variable.
4. Sample answer:

3. Sample answer:
Physics Exam
10
8
Frequency 6
4
2
0

5. 50th percentile; it contains half of the data.
6a.
0 50 60 70 80 90 100
Exam Score

4. stem
leaf
5
4 5
6
2 2
7
1 5
8
0 2
9
0 2
54 
 54

445 480 515 550 585 620 655

4
6
4
3

5
7 8 9
5 6 7 8 9 9 9
3 5 6 8 9

6b. Since 515 and 585 are within are standard
deviation of the mean, it contains 68.3% of the
data.
6c. 99.7% of the data lie within 3 standard
deviations of the mean.
550 3(35) 
 445  655
6d. 550  480 
 70, 620  550 
 70
tj 
 70
t(35) 
 70
t 
 2 → 95.5%
0.995(200) 
 191 values
7a. Since 22 and 26 are within one standard
deviation of the mean, it contains 68.3% of the
data.
7b. 24  20.5 
 3.5, 27.5  24 
 3.5
tj 
 3.5
t(2) 
 3.5
t 
 1.75 → 92.9%
7c. 24  0.7(2) 
 22.6 and 24  0.7(2) 
 25.4
22.6  25.4
7d. 24  1.96(2) 
 20.08 and 24  1.96(2) 
 27.92
20.08  27.92

1

5. X
 
 3
0 (54  55 … 99)

 81.1

84  85

6. Md 
 2 or 84.5
7. Mode 
 89
8.
50

60

9. MD 


70

80

1
(27.1
30

90

100

 26.1 … 17.9)


 10.42
10. Sample answer: The data that are less than the
median are more spread out than the data greater
than the median.

14-4

The Normal Distribution

Pages 922–923

8a.

Check for Understanding

1. The median, mean, and mode are the same.
2. X
 1.5

64 67 70 73 76 79 82

8b.

65 70 75 80 85 90 95

Chapter 14

466

8c. Chemistry; the chemistry grade is 3 standard
deviations above the class mean, while the
speech grade is only 2 standard deviations above
the class mean.

13b.

X  tj 
 180

140  t(20) 
 180
20t 
 40
t 
2
95.5%

2

Pages 923–925

X
  tj 
 150
140  t(20) 
 150
20t 
 10
t 
 0.5
38.3%

2


 47.75%


 19.15%

47.75  19.15 
 28.6%

Exercises

13c.

9a.

50%
25%

25%

7.5 9 10.5 12 13.5 15 16.5

9b. 12 1(1.5) 
 10.513.5
9c. 12  7.5 
 4.5, 16.5  12 
 4.5
tj 
 4.5
t(1.5) 
 4.5
t 
 3 → 88.7%
9d. 12  9 
 3, 15  12 
 3
tj 
 3
t(1.5) 
 3
t 
 2 → 95.5%
10a. 0.683(200) 
 136.6; about 137
10b. 0.955(200) 
 191
10c.

0.683

2

80

100

120

140

160

180 200

t 
 0.7 corresponds with 50% of the data
centered about the mean.
The upper limit results in 75% of the data.
140  0.7(20) 
 154
14a.
X
X
  tj 
 7
  tj 
 6.5
6  t(35) 
 7
6  t(3.5) 
 6.5
3.5t 
 1
3.5t 
 0.5
t 
 0.29
t 
 0.14
23.6%
8%
 
 11.8%
 
 4%
2
2

(200) 
 68.3; about 68

11a. 45% corresponds to t 
 0.6.
82 0.6(4) 
 79.6  84.4
11b. 80% corresponds to t 
 1.3
82 1.3(4) 
 76.8  87.2
11c. 82  76 
 6, 88  82 
 6
tj 
 6
t(4) 
 6
t 
 1.5 → 86.6%
11d. 82  80.5 
 1.5, 83.5  82 
 1.5
tj 
 1.5
t(4) 
 1.5
t 
 0.375 → 31.1%
12a. 25% corresponds to t 
 0.3.
402 0.3(36) 
 391.2412.8
12b. 402  387 
 15, 417  402 
 15
tj 
 15
t(36) 
 15
t 
 0.416
 → 31.1%
12c. 402  362 
 40, 442  402 
 40
tj 
 40
t(36) 
 40
t 
 1.1
 → 72.9%
12d. 45% corresponds to t 
 0.6.
402 0.6(36) 
 380.4  423.6
13a.
X
X
  tj 
 150
  tj 
 100
140  t(20) 
 150
140  t(20) 
 100
20t 
 10
20t 
 40
t 
 0.5
t 
2
38.3%
95.5%
 
 19.15%
 
 47.75%
2
2

14b.

11.8  4 
 7.8%
X
  tj 
 6.2
6  t(0.35) 
 6.2
0.35t 
 0.2
t 
 0.57
45.1%
 
 22.55%
2

X
  tj 
 5.5
6  t(0.35) 
 5.5
0.35t 
 0.5
t 
 1.43
83.8%
 
 41.9%
2

22.55  41.9 
 64.45%
14c.

80%

10%

10%

4.95 5.30 5.65 6.00 6.35 6.70 7.05

t 
 1.3 corresponds with 80% of the data
centered about the mean. The lower limit results
in the value above which 90% of the data lies.
6  1.3(0.35) 
 5.545
1 6

15a. P(no tails) 
 
2 or 64
1

1 5

P(one tail) 
 6
2
2 or 3
2
1

3

1 2 1 4
15
 or 
2
64
1 3 1 3
5




P(three tails) 
 20 2 2 or 1
6
1 4 1 2
5
P(four tails) 
 15 2 2 or 64
1 5 1
3
P(five tails) 
 6 2 2 or 3
2
1 6
1




P(six tails) 
 2 or 64

P(two tails) 
 15
2


 

 
 

 
 

 
 

 

19.15  47.75 
 66.9%

467

Chapter 14

15b.

18.

96%

20
15
Frequency

10
5
0

0

1 2 3 4 5
Number of Tails

6

2%
46

1

15c. 
X 
 6
4 [0(1)  1(6)  2(15)  3(20)  4(15)

(0  3)  (1  3) … (6  3)


64
2

2

2

56

61

66

71

76

20%

20%

1.2
15e. They are similar.
16a.

51

96% corresponds to t 
 2.1.
61  2.1(5) 
 50.5 months
19a.
30%

 5(6)  6(1)]

3
15d. j 


2%

15%

15%

84%

44
8%

3

2

8%

1

0

1

2

19b.
19c.

3 t

The 92nd percentile is the upper limit to 84% of
the data that is centered about the mean. 84%
corresponds to t 
 1.4. The 92nd percentile is
1.4 standard deviations above the mean.
16b.

20a.

57.6%

20b.

21.2%

21.2%

20c.

20d.
3

2

1

0

1

2

21a.

3 t

t 
 0.8 corresponds to 57.6% of the data centered
100  57.6

about the mean. 2 
 21.2
21.2  57.6 
 78.8 percentile
17a.
X
  tj 
 22.3
20.4  t(0.8) 
 22.3
0.8t 
 1.9
t 
 2.38 → 98.4%
100  98.4
 
 0.8%
2

21b.

17b. 100  0.8 
 99.2%

51

58

65

72

79

86

70% corresponds to t 
 1.0.
65  1.0(7) 
 72
65  1.0(7) 
 58
30% corresponds to t 
 0.4
65  0.4(7) 
 67.8 68
The lowest score for an A is 72, so the highest
score for a B is 71. The interval for B’s is 6871.
a normal distribution with a small standard
deviation
a normal distribution with a large standard
deviation
a distribution where values greater than the
mean are more spread out than values less than
the mean
a distribution where all values occur with the
same frequency
95% corresponds to t 
 1.96
X
X
  tj 
 260
  tj 
 250
255  1.96j 
 260
255  1.96j 
 250
1.96j 
 5
1.96j 
 5
j 
 2.55
j 2.55
about 2.55 mL
X
X
  tj 
 357
  tj 
 353
355  t(2.55) 
 357
355  t(2.55)
 353
2.55t 
 2
2.55t 
 2
t 
 0.78
t 
 0.78
57.6%
147  150

22a. Md 
 2 or 148.5
22b. Q1 
 110, Q3 
 200
22c. interquartile range 
 Q3  Q1

 200  110

 90
90

22d. semi-interquartile range 
 2 or 45
22e.
100

Chapter 14

468

200

300

400

500

1

3. About 68.3%; the answer for Exercise 2 can be
rounded to 0.683, which equals 68.3%.

23. X
 
 9(19  33  42  42  45  48  55  71  79)
48.2
Md 
 45
Mode 
 42
24. y 
 sec(k
  c)  h
2

k: k 
 2
k
4
c
c: k 
 

4.

0.95449974

c

4


 
c 
 4
h
3
y 
 sec(4
  4)  3
25a. Sample answer: Use a graphing calculator to
enter the year data as L1 and the Enrollment
data as L2. Then make a scatter plot. The
scatter plot indicates that a cubic function would
best fit the data. Perform a cubic regression to
find the equation
y 
 0.05x3  2.22x2  29.72x  366.92.
25b. Sample answer:
2015  1965 
 50
f(50) 
 0.05  503  2.22  502  29.72  50  366.92

 2553 students
26.

60˚

x˚

3

5.

0.9973002
6. The answer for Exercise 3 can be rounded to
0.954, which is about 95.5%. The answer for
Exercise 4 can be rounded to 0.997, which equals
99.7%.
7. 0.9999; t 
 4 corresponds to P 
 0.999.
8.

4
60˚
1 2

5

30˚

1; no; since the curve is approaching the x-axis
asymptotically, the area is probably not exactly
equal to 1.

Since vertical angles are equal, m2 
 60.
m1  60  30 
 180
m1 
 90
Since an exterior angle of a triangle is equal to the
measure of the sum of the two remote interior
angles, x  60 
 m1. So, x  60 
 90 and x 
 30.
The correct choice is E.

14-5

Sample Sets of Data

Page 930

14-4B Graphing Calculator Exploration:
The Standard Normal Curve

Check for Understanding

1. A sample is a subset of a population. However, a
sample must be similar in every way to the
population.
2. Divide the standard deviation of the sample by the
square root of the number of values in the sample.
3. Use a larger sample.
4. Tyler; every twentieth student to enter the school
should produce a representative sample. The
senior English class will not represent
underclassmen. The track team will not represent
students who prefer other types of activities.

Page 926
1. The domain of the function is the set of all real
numbers. The range is the set of all real numbers
1
. The graph is symmetric
y such that 0 y  

2
with respect to the y-axis and has the x-axis as a
horizontal asymptote. the value of f(x) approaches
0 as x approaches .
2.

73

 or 7.3
5. jX 
 
00

1
3.4


6. jX 
 
50

2

0.22

0.68268949

469

Chapter 14

7. A 1% confidence level is given when P 
 99% and
t 
 2.58.
5
 or 0.83
jX 
 5 

36

internal: X


11.12


20. jX 
 
1

000

0.3516452758
interval: X
 tjX 
 110 2.58jX

 109.09110.91
21. P 
 90% corresponds to t 
 1.65.
4
 or 0.4
jX 
 
1

00

tjX 
 45 2.58(0.83
)
42.8547.15

8. A 5% confidence interval is given when P 
 95%
and t 
 1.96.

interval: X


5.6


jX 
 

300

2.4

 or 0.24
22. jX 
 

100

0.323316156
internal: X


interval: X


tjX 
 55 1.96jX


54.3755.63

17.1

0.9140334473
interval: X
 tjX 
 4526 1.96jX

 4524.214527.79

0.29
9b. P 
 50% corresponds to t 
 0.7.
interval: X
 tjX 
 27.5 0.7jX
27.3027.70 min
9c. A 1% confidence level is given when P 
 99%
and t 
 2.58.
interval: X
 tjX 
 27.5 2.58jX
27.7628.24 min

Pages 930–932

28


24. jX 
 
3

70

1.455650686
interval: X
 tjX 
 678 1.96jX
675.15680.85
0.67


25. jX 
 
8

0

0.0749082772
interval: X
 tjX 
 5.38 1.96jX
5.235.53
1
26a. 
X 
 6
[1(4)

3(6)
… 2(20)]
4

 12.375

Exercises

1.8

 or 0.2
10. jX 
 
8

1
5.8


11. jX 
 

250

0.37

26b. j 


7.8


12. jX 
 

140

2

2

3.37

0.42
26d. P 
 0.95 corresponds to t 
 1.96.
interval: X
 tjX 
 12.375 1.96(0.42)
11.5513.20 min
26e.
tjX 
 1
t(0.42) 
 1
t 2.38 → about 98.4%

0.53
2.7


14. jX 
 
1

30

0.24
13.5


15. jX 
 

375

0.70

12

5.6


27. jX 
 

45


16. 0.056 
 
N


1.788854382
tjX 
 3
t 
 1.68 → 91.1%
100  91.1 8.9%

N
 
 100
N 
 100,000
5.3


17. jX 
 
5

0

0.7495331881
interval: X
 tjX 
 335 2.58jX
333.07336.93
40
 or 5
18. jX 
 
6

4

1.4


28a. jX 
 

50

0.1979898987 or about 0.20
28b. A 5% confidence interval is given when P 
 95%
and t 
 1.96.
interval: X
 tjX 
 16.2 1.96j X

15.8116.59 mm
28c. P 
 99% corresponds to t 
 2.58.
interval: X
 tjX 
 16.2 2.58jX
15.6916.71 mm
28d. P 
 0.80 corresponds to t 
 1.3
interval: X
 tjX 
 16.2 1.3jX
15.9416.46 mm

tjX 
 200 2.58(5)

 187.1212.9


19. jX 
 

200

0.8485281374
interval: X
 tjX 
 80 2.58jX
77.8182.91

45

 or 4.5
29a. jX 
 

100

Chapter 14

2


26c. jX  

64

14

7

00

12

(4  12.375)  (6  12.375) … (20  12.375)


64

3.37

0.66

interval: X


tjX 
 24 1.96(0.24)
23.5324.47


23. jX 
 
3

50

3.5


9a. jX 
 

150

13. jX 


tjX 
 68 1.65(0.4)

 67.3468.66 in.

470

29b. A 1% confidence level is given when P 
 99%
and t 
 2.58.
interval: X
 tjX 
 350 2.58(4.5)

 338.39361.61 hours
29c. Sample answer: 338 hours, there is only 0.5%
chance the mean is less than this number.
30. 10.2064  9.7936 
 0.4128

34c. 50% of 10,000 
 0.50(10,000)

 5000 tires
34d.
X
  tj 
 50,000
40,000  t(5000) 
 50,000
5000t 
 10,000
t 
 2 → 95.5%
100%  95.5%
 
 2.25%
2

0.4128

2


 0.2064
tjX 
 0.2064
2.58tjX 
 0.2064
jX 
 0.08

34e.

j


jX 
 
N

0.8


0.08 
 
N


0.0015(10,000) 
 15 tires

N
 
 10
N 
 100 packages

1

35. X
 
 8(44  49  55  58  61  68  71  72)

 59.75
1
MD 
 8(15.75  10.75 … 12.25)

1.8


31a. jX 
 

10

0.57
31b. interval: X



 8.25

tjX 
 4.1 1.96(0.57)

 2.985.22 hours
With a 5% level of confidence, the average family
in the town will have their televisions on from
2.98 to 5.22 hours.
31c. Sample answer: None; the sample is too small to
generalize to the population of the city.

j


2

2

2

a1  a1rn


36. Sn 
 
1r
1

n 
 10, a1 
 1
6, r 
 4
1

1

  (4)10
16
16

S10 

14

3.2

0.45
32b. P 
 50% corresponds to t 
 0.7.
interval: X
 tjX 
 42.7 0.7jX
42.3843.02 crackers
32c. Sample answer: No; there is a 50% chance that
the true mean is in the interval. However, since
43 is near one end of the interval, they may
want to take another sample in the near future.
33a. A 5% confidence interval gives a P 
 95% and
t 
 1.96.

349,525


 16 or 21,845.3125
37.

x
y
r cos v 
 r sin v
sin v


1

cos v

1 
 tan v
tan1(1) 
 v
45° 
 v
38.
tan x  cot x 
 2
1

tan x  
tan x 
 2


 3.136

1.960jX 
 3.136
jX 
 1.6
X
 tjX 
 753.136
X
  1.96(1.6) 
 753.136
X
 
 750 h

(15.75)  (10.75) … (12.25)


8

9.59


32a. jX 
 

50

753.136  746.864


0.0225(10,000) 
 225 tires
X
  tj 
 25,000
40,000  t(5000) 
 25,000
5000t 
 15,000
t 
 3 → 99.7%
100%  99.7%
 
 0.15%
2

tan2 x  1

tan x
tan2 x  1

tjX


2


 2 tan x
tan2 x  2tan x  1 
 0
(tan x  1)2 
 0
tan x  1 
 0
tan x 
 1
x 
 45°
39. *2 
 22  2(2) or 0
*1 
 12  2(1) or 1
*2  *1 
 0  (1) or 1
The correct choice is C.

j


33b. jX 
 
N

j


1.6 
 
1

600

64 
 j; 64 h
34a. 40,000  35,000 
 5000, 45,000  40,000 
 5000
tj 
 5000
t(500) 
 5000
t 
 1 → 68.3%
0.683(10,000) 
 6830 tires
34b.
X
  tj 
 30,000
40,000  t(5000) 
 30,000
5000t 
 10,000
t 
 2 → 95.5%
95.5%
 
 47.75%
2

Chapter 14 Study Guide and Assessment
Page 933

Understanding the Vocabulary

1. box-and-whisker plot
2. median
3. standard error of the mean

0.4775(10,000) 
 4775 tires

471

Chapter 14

4.
5.
6.
7.
8.
9.
10.

2
 (
2.4)2 
… 2
.62
22. j 
 2.4)
1.74
23. 88  78 
 10 98  88 
 10
tj 
 10
t(5) 
 10
t 
 2 → 95.5%
24. 88  86 
 2
90  88 
 2
tj 
 2
t(5) 
 2
t 
 0.4 → 0.311
25. 90% corresponds to t 
 1.65.
interval: X
 tj 
 88 1.65(5)

 79.7596.25
26. 0.683(150) 
 102.45
27. 0.955(150) 
 143.25

range
measure of central tendency
population
bimodal
inferential statistics
histogram
standard deviation

Pages 934–936

Skills and Concepts

11. range 
 14.0  9.0 or 5
12. 9.5, 10.5, 11.5, 12.5, 13.5
13.

Women's Tennis Shoes
18
16
14
12
Frequency 10
8
6
4
2
0

28.
29.

30. jX 


4.9


120

0.45
25

0

 or 1.25
31. jX 
 

400

9 10 11 12 13 14
Weight (ounces)

18

 or 3.6
32. jX 
 

25

1

15


33. jX 
 

50


5
Md 
 5
Mode 
 4

2.121320344
interval: X
 tjX 
 100 2.58jX
94.53105.47

1

15. X
 
 5(160  200  200  240  250)

30


34. jX 
 
1

5


 210
Md 
 200
Mode 
 200

7.745966692
interval: X
 tjX 
 90 2.58jX
70.02109.98

1

16. X
 
 5(11  13  15  16  19)

24


35. jX 
 

200


 14.8
Md 
 15
Mode: none

1.697056275
interval: X
 tjX 
 40 2.58jX
35.6244.38

1

17. X
 
 8(5.9  6.3  6.3  6.4  6.6  6.6  6.7  6.8)

0.5


36. jX 
 
2

00


 6.45

0.035
37. P 
 0.90 corresponds to t 
 1.65.
range: X
 tjX 
 1.8 1.65(0.035)
1.741.86 h
38. A 5% confidence level is given when P 
 95% and
t 
 1.96.
range: X
 tjX 
 1.8 1.96(0.035)
1.731.87 h
39. A 1% confidence level is given when P 
 99% and
t 
 2.58.
range: X
 tjX 
 1.8 2.58(0.035)
1.711.89 h
40. P 
 0.90 corresponds to t 
 1.65.

6.4  6.6

Md 
 2 or 6.5
Mode 
 6.3 and 6.6
1

18. X
 
 8(122  128  130  131  133  135  141
 146)

 133.25
131  133

Md 
 2 or 132
Mode: none
19. interquartile range 
 Q3  Q1

52

3
3

20. semi-interquartile range 
 2 or 1.5

1.4

 or 0.14
jX 
 
1

00

1

21. X
 
 1
0 (1  1 … 6)

range: X



 3.4
1
(2.4
10

 2.4 … 2.6)


 1.6

Chapter 14


 51.225

0.16

14. 
X 
 9(2  4  4  4  5  5  6  7  8)

MD 


0.683
(150)
2
1.5

jX 
 

90

472

tjX 
 4.6 1.65(0.14)
4.374.83 h

Page 937

m 
 x%(10n)

Applications and Problem Solving

x

41a. stem leaf
1 0 3 5 6 7 9
2 1 3 4 5
3 9 9
10 
 10


2000 
 
100 (10  20)

1000 
 x
The correct choice is D.
4. The numbers in S are positive numbers that are
less than 100 and the square root of each number
is an integer. So the set S contains perfect squares
between 0 and 100.
Make a list of the
n
n

numbers, n, in set S.
1
1
From your list, you can
4
2
see that the median, or
9
3
middle value for n, is 25.
16
4
The correct choice is C.
25
5
36
6
49
7
64
8
81
9

1

41b. 
X 
 2(10  13  15  16  17  19  21  23
 24  25  39  39)

 21.75
19  21

41c. Md 
 2 or 20
41d. Mode 
 39
42. X
  tj 
 80
75  t(2) 
 80
2t 
 5
t 
 2.5 → 98.8%
100%  98.8%

2

Page 937


 0.6%

Open-Ended Assessment

5. m∠DBA 
 90  30 
 60
m∠EBC 
 90  40 
 50
m∠ABC 
 180  m∠DBA  m∠EBC

 180  (60)  (50) or 70
The correct choice is E.

1a. Sample answer: {2, 3, 10, 20, 40}
1b. Sample answer: 15
2. See students’ work.

1

6. X
 
 6(10  20  30  35  35  50)

 30
There are 3 numbers larger than 30: 35, 35, and
50. The correct choice is D.

Chapter 14 SAT & ACT Preparation
Page 939

SAT and ACT Practice

7.

y

1. The percent increase is the ratio of the number
increase to the original amount.
Amy
Brad
Cara
Dan
Elsa

10

80
30

70
20

80
30

60
20

90

1


 8
3


 7
1

O


 4

The line of best fit has a rise of 2 and a run of 5. So
2
the slope of the line of best fit is 5. The closest
2
1
answer choice to 5 is 2.
The correct choice is D.
8. Each year the number of employees increases by
300. The last year of data is 2005. The expected
employment in 2007, two years later, will be
2  300 more employees than in 2005.
3100  600 
 3700
The correct choice is D.

1


 2
2


 9
1

The largest fraction is 2. Dan has the greatest
percent increase. The correct choice is D.
2. a 
 b  bc
b

a

b




(b  bc)
b




b(1  c)
1




1c

The correct choice is B.
3. Method 1:
0.1%(m) 
 10%(n)
0.001m 
 0.1n

x

m 
 x%(10n)
x

m

100  10n

m 
 100n
m 
 0.1xn
100n 
 0.1xn
100 
 x
Method 2
Let m represent a large number such as 2000.
0.1% of m 
 0.001(2000) or 2
10% of n 
 2, so 0.1n 
 2 or n 
 20.

473

Chapter 14

9.To find the median of Set A, first rewrite the
elements of Set A in order: 4, 1, 2, 3, 7, 11.
Since the number of elements is even, the median
is the average of the middle two elements: 2 and 3.
So the median of Set A is 2.5. To find the mean of
Set B, add all of the elements together and divide
by the number of elements in the set. The sum of
the elements is 15, and there are 6 terms. So the

1

10. X
 
 1
0 [820  (65)  (32)  0  1  2  3
 32  64  820]
1


 1
0 (1  2  3)
6


 1
0
The answer is 0.6, 6/10, or 3/5.

15
6

mean is  or 2.5.
The difference between the median of Set A and
the mean of Set B is 2.5  2.5 or 0.
The correct choice is C.

Chapter 14

474

Chapter 15 Introduction to Calculus
6. lim (1  x  2x  cos x) 
 1  0  20  cos 0
x→0

111

1

Limits

15-1
Page 945

x2

x2

 
 lim 
7. lim 
(x  2)(x  2)
x2  4

Graphing Calculator Exploration

x→2

x→2

1



 lim 
x2

1.

lim

x→0

ex  1

x

x→2
1
1




2  2 or  4
2
x(x  3)
x  3x
 
 lim 

8. lim 
2
3
x→0 x  4x
x→0 x(x  4)
x3


 lim 
2
x→0 x  4
03
3
 or 


02  4
4
x2  3x  10
(x  5)(x  2)



9. lim x2  5x  6 
 lim (x  3)(
x  2)
x→3
x→3
(3  5)(3  2)



 (3  3)(3  2)
8(1)
4




6(5) or 15
2
2x  5x  2
(2x  1)(x  2)
 
 lim 
10. lim 
2
x→2 x  x  2
x→2 (x  1)(x  2)
2x  1


 lim 
x→2 x  1


1

2.

2(2)  1




2  1 or 1

11a. v (r )
x2  4

lim 
2
x→2 x  3x  2

0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02


4

3. y is undefined when x 
 1.
4.

x2  4

lim 
2  3x  2
x
x→2


 lim

x→2

(x  2)(x  2)

(x  2)(x  1)
x2



 lim 
x1

x→2
22



21

v (r ) 
 0.65(0.52  r 2)

O

or 4

Yes, the limit is the same.
5. No; if the exact answer is a complicated fraction or
an irrational number, you may not be able to tell
what it is from the decimals displayed by a
calculator.

0.2

0.4 0.6

11b. As the molecules get farther from the center and
closer to the pipe, r is increasing. As r increases,
v(r) gets closer and closer to 0 in./s.

Pages 946–948
Page 946

0.8 r

Exercises

12. The closer x is to 2, the closer y is to 1. So,
lim f(x) 
 1. Also f(2) 
 1.

Check for Understanding

x→2

1. Sample answer: The limit of f(x) as x approaches a
is the number that the values of f(x) get closer to
as the values of x gets closer and closer to a.
2. Sample answer: lim f(x) is the number that the

13. The closer x is to 0, the closer y is to 0. So,
lim f(x) 
 0. However, there is point discontinuity
x→0

when x 
 0. So f(0) is undefined.
14. The closer x is to 3, the closer y is to 4. So,
lim f(x) 
 4. However, there is a point at (3, 2).

x→1

values of f(x) approach as x approaches 1. f(1) is
the number that you get if you actually plug 1 into
the function. They are the same if f(x) is
continuous at x 
 1.
3. Sample answer: If f(x) is continuous at x 
 a you
can plug a into the function. If the function is not
continuous, you may be able to simplify it and
then plug in a. If neither of these methods work,
you can use a calculator. Examples will vary.
4. The closer x is to 0, the closer y is to 3. So, lim f(x)

x→3

So f(3) 
 2.
15. lim (4x2  3x  6) 
 4(2)3  3(2)  6
x→2

 16  6  6

 16
16. lim (x3  3x2  4) 
 (1)3  3(1)2  4
x→1

134

0
17. lim

x→0

x→


 3. However, there is a point at (0, 1), so f(0) 
 1.
5. lim (4x2  2x  5) 
 4(2)2  2(2)  5
x→2

 16  4  5

 17

sin x

x

sin 


 
0


  or 0
18. lim (x  cos x) 
 0  cos 0
x→0

 0  1 or 1

475

Chapter 15

x2  25

(x  5)(x  5)

x→5

1x

1

x



19. lim 
x  5 
 lim
x5


1
x

31. lim 


lim
x

1
x

1
x→1
x→1

x→5


 lim (x  5)
x→5

1x


 5  5 or 10

x→1

20. lim n 
 lim 2n
x→0


 2(0) or 0
21.

x(x  3)



 lim 
(x  5)(x  3)

x→3

32.

x



 lim 
x5

x→3
3

3





3  5 or 8

22. lim

x→1

x3  3x2  4x  8

x6




23. lim

h→2

1348

 4h  4

h2


 lim

h→2

x→1
1

 1 or 1
x  2
x4
x4



lim x  2 
 lim x  2  x  2
x→4
x→4
(x  4)(x  2)

 lim 
x4
x→4


 lim 
x  2

13  3(1)2  4(1)  8

16

x→4


 4
  2 or 4

8


 7 or 7
h2

33. lim

h→0

(h  2)(h  2)

h2

2h3  h2  5h

h

h→0

h→2

 3x

x→3

x

x→0

26.

27.






cos () or 

1





27  18  9 or 2

25.

0



34. lim 
cos (x  ) 
 cos (0  )

2(3)2  3(3)

33  2(3)2  3  6
18  9

h→0

35.

x3  x2  2x

lim 
3  4x2  2x
x
x→0

x(x2  x  2)


 lim 
2
x
x→0 (x  4x  2)
x2  x  2


 lim 
2
x→0 x  4x  2
2
0 02
 or 1


02  4(0)  2
x cos x
x cos x

lim 
2  
 lim 
x→0 x  x
x→0 x(x  1)
cos x


 lim 
x→0 x  1
cos 0



0  1 or 1
2
(x  2)  4
x2  4x  4  4
lim 4 
 lim x
x→0
x→0
x(x  4)

 lim x
x→0

tan 2x

lim x 
 2

x→0

36.


 lim (x  4)
x→0


 0  4 or 4
28. lim

x→2

(x  1)2  1

x2

x2  2x  1  1

x2
x→2
x(x  2)


 lim 
x→2 x  2


 lim

ln x
lim 
x→1 ln (2x  1)


 lim x

x→2

x3  8


x2  4


 lim

(x  2)(x2  2x  4)

(x  2)(x  2)


 lim

x2  2x  4

x2

x→2


 0.5

37.

x→2


2
29. lim

x→2
(2)2  2(2)  4

 
2  2
444

 or 3


4

30.

2x  8

lim 
3
x→4 x  64

Chapter 15

h(2h2  h  5)

h


 2(0)2  0  5 or 5


 2  2 or 0


24. lim 
x3  2x2  x  6


lim


 lim (2h2  h  5)


 lim h  2
2x2

1


 lim  x

x→0

x2  3x

lim 
2  2x  15
x
x→3

1



 lim x  
(1  x)

2n2

1  8

lim 
x→1 x  1

2(x  4)

 lim 
(x  4)(x2  4x  16)
x→4
2


 lim 
2
x→4 x  4x  16
2



42  4(4)  16
2
1


 4
8 or 24

476


 0.5

45. When P 
 0.99, t 
 2.58.
Find X.

38.

X


1.4

50


 

0.20
Find the range.
X
 tj X 
 16.2 2.58(0.20)


 15.68416.716 mm
9

46. P(not getting a 7) 
 1
0

3x  sin 3x

 
 4.5
lim 
x2 sin x

0.1

0.06908

0.1

0.06956

1

0.07177

9

59,049

t→0

2 10  1

t

47. (x  3y)5 
 

r
0

5!

r!(5  r)!


 10(x3)(9y2)

 90x3y2
48. (16y8) 4 
 
16
  y8
 4 
3


5  5 5  1

49. center 
 
2, 2

x

1

0.5

0.1

0.01

0.001

0.540302

0.877583

0.995004

0.999950

1.000000

x2
1  2

0.5

0.875

0.995

0.99995

1.000000


 (5, 2)
The foci are on the x-axis.
91

a 
 2 or 4
1  (5)

b 
 2 or 3
(x  5)2

42

50.

120˚

(x  5)2

90˚

60˚

1 2 3 4

240˚

16(t  2)(t  2)



 lim 
t2

0˚

330˚

210˚

16t2  64



 lim 
t2

(y  2)2

30˚

180˚

agree with those of cos x.

t→2

(y  2)2

 3

 1 → 16  9 
 1
2

150˚

42b. yes; in the last three columns, all the decimal

t→2

3


3

4


 23  y6

 8y6

cos x

43. lim

5!

52(3y)2

(x)5r(3y)r 
 
2!(5  2)! (x)
5!

1

d(t)  d(2)

t2

(x)5r (3y)r

2
 3


2! 3! (x )(9y )

41. No; the graph of f(x) 
 sin
x oscillates infinitely
many times between 1 and 1 as x approaches 0,
so the values of the function do not approach a
unique number.
42a.

places of 1 

5!

r!(5  r)!

To find the third term, evaluate the general term
for r 
 2.


 0.07 or 7%

x2

2




100,000

5

x


lim

5

P(never getting a 7 in five spins) 
 
1
0

x→0

39. lim a
a2  c2 
 a
a2  02
c→0
2

 a
Letting c approach 0 moves the foci together, so
the ellipse becomes a circle. a2 is the area of a
circle of radius a.
x
40.

2 10  1
f(x) 
 
x
x
y
1
0.06697

270˚

300˚

u
51. WX 
 3  4, 6  0

 7, 6
u
WX  
 
(3 
4)2 
(6 
0)2

 49
36


 85


t→2


 lim 16(t  2)
t→2


 16(2  2) or 64 ft/s
44a. As x approaches 0, the decimals for the values of
1


f(x) 
 (1  x) x approach 2.71828 ... , which is the
decimal expansion of e.
44b. He ignored the exponent. As x approaches 0 from

25 in.

1 yd

1 mi

 
52. C 
 1  
36 in.  1760 yd

0.0012395804 mi

65
v 
 C 52437.09741
v
v

w 
 t 
 
3600 s

1

the positive side, x approaches infinity. A
number close to 1 raised to a large power need
not be close to 1. If x approaches 0 from the
1
negative side, x approaches negative infinity. A
number close to 1 raised to a large negative
power need not be close to 1, either.

14.6 rps

477

Chapter 15

y

1


53. csc 270° 
 
sin 270°

3. 0.5

1


 y

270˚

1




1 or 1

x

54. Use a graphing calculator to
find the rational roots at
1
4
 and .
4
3
55.

(0, 1)

4. 1

y 
 4x5  2x2  4
x
y
100
10
1

4  100
4  105
6

10

399804

100

4  1010

5. 1

y →  as x → , y →   as x → 
1 2 
 (1)(6)  (2)(3)
3 6 
 6  (6)

 12
57. yes; opposite sides have the same slope
1
y
m 
 
56.

AB

mDC 


8
1

8

(8, 4)

1

mBC 
 4

A
(0, 3)

1

mAD 
 4
58. If

 8, then n 
 3.
3n2 
 332

 35 or 243

6. 3
B

x

2n

D (2, 5)

C
(10, 4)

7. The graph of a linear function is a line and the
methods in this function will result in the
calculation of the slope of that line.
8. If you zoom in on a maximum or minimum point,
the graph appears flat. The slope is 0.

15-2A Graphing Calculator Exploration:

dy

9. At (0, 1), dx 1. For other points on the curve, the
dy
values for y and dx are approximately the same.

The Slope of a Curve

Page 950
1-6. Exact answers are given. Accept all reasonable
approximations.
1. 4

15-2

2. 1

Derivatives and Antiderivatives

Pages 957–958

Check for Understanding

1. 4x3 is the derivative of x4. x4 is an antiderivative
of 4x3.
1

 n1
2. Letting n 
 1 in the expression 
n1 x
x0

results in 0, which is undefined.
Chapter 15

478

3. f(x  h) means substitute the quantity x  h into
the function. On the other hand, f(x)  h means
substitute x into the function, then add h to the
result. Using f(x)  h instead of f(x  h) in the
definition of the derivative results in:
lim

h→0

f(x)  h  f(x)

h

13.

h

h→0 h


 lim

C(x) 
 1000  10x  0.001x2
C(x) 
 0  10  1x11  0.001(2)x21
C(x) 
 10  0.002x
C(1000) 
 10  0.002(1000)

8
The marginal cost is $8.


 lim 1
h→0

Pages 958–960


1
You would always get 1.
4. f (x) 
 lim

h→0

f(x  h)  f(x)

h


 lim

3(x  h)  2  (3x  2)

h


 lim

3x  3h  2  3x  2

h


 lim

3h

h

h→0
h→0
h→0


2
15. f(x) 
 lim

h→0


3
5. f(x) 
 lim

h→0

f(x  h)  f(x)

h

f(x  h)  f(x)

h


 lim

7(x  h)  4  (7x  4)

h


 lim

7x  7h  4  7x  4

h

h→0


 lim

(x  h)2  x  h  (x2  x)

h


 lim

x2  2xh  h2  x  h  x2  x

h


 lim

2xh  h2  h

h

16. f
(x) 
 lim


 lim

h(2x  h  1)

h


 lim

 3(x  h)  (3x)

h


 lim

3x  3h  3x

h


 lim

3h

h

h→0
h→0
h→0
h→0

h→0

7h

h→0 h


 lim

7

h→0
h→0


 lim 2x  h  1
h→0

h→0


 2x  0  1 or 2x  1
6. f(x) 
 2x2  3x  5
f 
(x) 
 2  2x21  3  1x11  0

 4x  3
7. f(x) 
 x3  2x2  3x  6
f(x) 
 1  3x31  2  2x21  3  1x11  0

 3x2  4x  3
8. f(x) 
 3x4  2x3  3x  2
f (x) 
 3  4x41  2  3x31  3  1x11  0

 12x3  6x2  3
9.
y 
 x2  2x  3
dy

dx
dy

dx

Exercises

f(x  h)  f(x)

h
h→0
2(x  h)  2x



 lim
h
h→0
2x  2h  3x

 lim h
h→0
2h

 lim h
h→0

14. f (x) 
 lim

h→0


 3
17. f
(x) 
 lim

h→0

4h

h

h→0

2(x  h)2  5(x  h)  (2x2  5x)

h


 lim

2x2  4xh  2h2  5x  5h  2x2  5x

h


 lim

4xh  2h2  5h

h


 lim

h(4x  2h  5)

h

h→0
h→0

1

h→0


 4x  2(0)  5 or 4x  5

11. f(x) 
 x3  4x2  x  3

19. f
(x) 
 lim

 31  4
x21  x11
F(x) 
 
31 x
21
11
1

1

1

h→0

 3x  C
12. f(x) 
 5x5  2x3  x2  4

(x  h)3  5(x  h)2  6  (x3  5x2  6)

h


 lim

x3  3x2h  3xh2  h3  5x2  10xh  5h2  6  x3  5x2  6
–––––––––––––––
h


 lim

h(3x2  3xh  h2  10x  5h)

h

h→0

 51  2
x31  x21
F(x) 
 5

5  1 x
31
21
1

1

h→0

 4x  C

f(x  h)  f(x)

h


 lim

h→0

1


 4x4  3x3  2x2  3x  C
1

f(x  h)  f(x)

h


 lim

h→0


 3x3  C

1


 lim
18. f
(x) 
 lim

1

5

4x  4h  9  4x  9

h


 4

 21  C
F(x) 
 
2  1x

4


 lim

h→0

f (1) 
 2(1)  2

4
10. f(x) 
 x2

1

4(x  h)  9  (4x  9)

h

h→0


 2x  2

f(x  h)  f(x)

h


 lim

h→0


 2  1x21  2  lx11  0

f(x  h)  f(x)

h


 3x2  3x(0)  02  10x  5(0)

 3x2  10x

1


 6x6  2x4  3x3  4x  C

479

Chapter 15

20. f(x) 
 8x
f
(x) 
 8  1x11

8
21. f(x) 
 2x  6
f
(x) 
 2  1x11  0

2
1

y 
 (5x2  7)2

 25x4  70x2  49
dy
 
 100x3  140x
dx
f 
(1) 
 100(1)3  140(1)

 240
35. f(x) 
 x6
1
 61  C
F(x) 
 
61 x

34.

4

22. f(x) 
 3x  5
1

f 
(x) 
 3  1x11  0

1


 7x7  C

1


 3

36. f(x) 
 3x  4
1

3x2

23. f(x) 

 2x  9
f
(x) 
 3  2x21  2  1x11  0

 6x  2

 11  4x  C
F(x) 
 3  
1  1x
3


 2x2  4x  C
37. f(x) 
 4x2  6x  7

1

24. f(x) 
 2x2  x  2

1

4


 3x3  3x2  7x  C


x1
25. f(x) 
 x3  2x2  5x  6
f
(x) 
 3x31  2  2x21  5  1x11  0

 3x2  4x  5
26. f(x) 
 3x4  7x3  2x2  7x  12
f
(x) 

3  4x41  7  3x31  2  2x21  7  1x11  0

12x3  21x2  4x  7
27. f(x) 
 (x2  3)(2x  7)

 2x3  7x2  6x  21
f 
(x) 
 2  3x31  7  2x21  6  1x11  0

 6x2  14x  6
28. f(x) 
 (2x  4)2

 4x2  16x  16
f 
(x) 
 4  2x21  16  1x11  0

 8x  16
29. f(x) 
 (3x  4)3

 27x3  108x2  144x  64
f
(x) 
 27  3x31  108  2x21  144  1x11  0

 81x2  216x  144
30. f(x) 

f
(x) 



31.

y

dy

dx

2
1
x3  x2  x  9
3
3
2
1
  3x31    2x21
3
3
2
2x2  3x  1
x3

38. f(x) 
 12x2  6x  1
1


 4x3  3x2  x  C
39. f(x) 
 8x3  5x2  9x  3
F(x) 

8

1

1

 21  9  x11  3x  C
5
2  1x
11
5

9

1

2

40. f(x) 
 4x4  3x2  4
1

2

1

1

 41    x21  4x  C
F(x) 
 4  
3 21
4  1x
2

1

5
 3

 2
0 x  9 x  4x  C

41. f(x) 
 (2x  3)(3x  7)

 6x2  5x  21
1

1

 21  5  x11  21x  C
F(x) 
 6  
2  1x
11
5


 2x3  2x2  21x  C
42. f(x) 
 x4(x  2)2

 x6  4x5  4x4
1

1

1

 61  4  x51  4  x41  C
F(x) 
 
6  1x
51
41

 1  1x11  0

1

2

4


 7x7  3x6  5x5  C
x3  4x2  x

43. f(x) 
 x


 3x2


 x2  4x  1
1
1
 21  4  x11  x  C
F(x) 
 
2  1x
11
1


 3x3  2x2  x  C
2x2  5x  3


44. f(x) 
 
x3


 3x2  14x  4


 2x  1
1
 11  x  C
F(x) 
 2  
1  1x

 x2  x  C
1

1

1

45. Any function of the form F(x) 
 6x6  4x4  3x3 
x  C, where C is a constant.


 2x  1

1

f
(1) 
 2(1)  1

1

Chapter 15

1
x31
31


 2x4  3x3  2x2  3x  C

f
(1) 
 3(1)2  14(1)  4

 7
33.
y 
 (x  1)(x  2)

 x2  x  2
dy

dx

1

 21  6   x11  x  C

 12  
21x
11

f 
(1) 
 3(1)2

3
32.
y 
 x3  7x2  4x  9
dy

dx

1

 21  6  x11  7x  C
F(x) 
 4  
2  1x
11

1

f
(x) 
 2  2x21  1  1x11  0

46a. v(12) 
 15  4(12)  8(12)2

 81 ft/s

480

49a. h(t) 
 3  80t  16t2
h
(t) 
 v(t) 
 0  80  1t11  16  2t21

 80  32t
49b. v(1) 
 80  32(1)

 48 ft/s
49c. At the ball’s maximum height, the velocity is 0.
0 
 80  32t
80 
 32t
2.5 
 t; 2.5 s
49d. h(2.5) 
 3  80(2.5)  16(2.5)2

 103 ft

1

46b. v(t) 
 15  4t  8t2
1

v
(t) 
 0  4  1t11  8  2t21
1


 4  4t
1

v
(12) 
 4  4(12)

 7 ft/s2
46c. When t 
 12 the car’s velocity is increasing at a
rate of 7 ft/s per second.
1

46d. v(t) 
 15  4t  8t2
1

1

1

 11    t21  C
s(t) 
 15t  4  
8 21
1  1t


 15t 

2t2



1
t3
24

50. f
(x) 
 lim

h→0

C

When t 
 0, s(t) should equal 0, so C 
 0.


 lim

exh  ex

h


 lim

ex  e h  e x

h


 lim

ex  (eh  1)

h

h→0

1

3
s(t) 
 15t  2t2  2
4t

h→0

1

3
46e. s(12) 
 15(12)  2(12)2  2
4 (12)

h→0


 540 ft
47. f 
(x) 
 lim

h→0


 ex  lim

f(x  h)  f(x)

h
1

xh

h→0

h→0

f
(x) 


h

x(x  h)


 lim 
h
h→0
1

h→0 x(x  h)
1



x(x  0)


 lim

1

 
x2
48a. When y 
 2010, I(2010) 2.75.
The total amount spent in 2010 on health care
will be about $2.75 trillion.
48b. T(y) is approximately linear near (2010, 2.75).

i. e.

x2  3x  3

ex


 1, so

is its own derivative.

(x  3)(x  1)

x→3

x→3


 lim x  1
x→3


 3  1 or 4
53a.

$3.5

170 180 190 200 210 220 230 240 250 260
53b. See students’ work.
54. List all pairs of matching numbers and their
sums.
1  1 
 2; 2  2 
 4; 3  3 
 6; 4  4 
 8;
5  5 
 10; 6  6 
 12
There are 3 sums out of 6 that are greater than
seven.
3
P(sum  7 given that the numbers match) 
 6

Projections

2.5
2.0
1.5
1.0
0.5

1

2001

2004

2007

2010


 2

2013

55. an 
 a1rn1

Source: Centers for Medicare & Medicaid Services

1 61

a6 
 9
3

Find the slope of the tangent line at (2010, 2.75).
T 
(2010)

1


ex.

 
 lim 
52. lim 
x3
x3

Health Care Spending
(Trillions of Dollars)

0

ex

eh  1

h

51a. total revenue 
 cost per cup  number of cups
r(p) 
 p(100  2p)
51b. When r(p) is at a maximum, the derivative
equals zero.
r(p) 
 p(100  2p)

 100p  2p2
r
(p) 
 100  1p11  2  2p21

 100  4p
0 
 100  4p
100 
 4p
25 
 p; 25 cents

xh

x(x  h)



 lim 
h
h→0

3.0

eh  1

h

A calculator indicates that lim

1

x



 lim 
h
h→0
x

x(x  h)

f(x  h)  f(x)

h

2.75 – 2.56

2010 – 2009

9

1




 
243 or  27

56. y 
 136e0.06(30)  74
96°

0.19
In 2010 the amount spent on health care will be
increasing at a rate of about $190 billion per
year.

481

Chapter 15

57. x2  y2  Dx  Ey  F 
 0
22  (1)2  2D  E  F 
 0
→ 2D  E  F  5 
 0
(3)2  02  3D  F 
 0 → 3D  F  9 
 0
12  42  D  4E  F 
 0 → D4E F  17 
 0
2D  E  F  5 
 0 3D
F 9 
0
(3D  E  F 9) 
 0 ( D  4E  F  17) 
 0
5D  E
 4 
 0 4D  4E
8 
0

x→3


 18  12  6

 36
x2  9x  14

x→2

x→2

x7



 lim 
2x  3

x→2
27



2(2)  3

28

7

3


1

3

E 
 3

1

3

F9
0

1

7

The solution of the system is D 
 3, E 
 3, and
F 
 8
7

x2  y2  3x  3y  8 
 0
2


x  16  
y  76 
 11689
5
5
1

3
58. 5
cos 6  i sin 6 
 5
2  i 
2
53


1
0.1
0.01
0.01
0.1
1

0.9093
1.9867
1.9999
1.9999
1.9867
0.9093

h→0

5


 2  2i
59. x 
 x1  ta1 → x 
 8t  3
y 
 y1  ta2 → y 
 3t  2


 lim

x2  2xh  h2  3  x2  3

h


 lim

2xh  h2

h


 lim

h(2x  h)

h

h→0



O

61.

90˚

180˚

d

H1

270˚

360˚

h→0

450˚


 2x  0 or 2x
5. f(x) 
 
f
(x) 
 0  is a constant.
6. f(x) 
 3x2  5x  2
f
(x) 
 3  2x21  5  1x11  0

 6x  5

H2
319 m

253 m

7. R(M) 
 M 2
2  3
C

42˚12

C

p
:
q

C

2532

1
,
2

1

2

3

2

1

2

2

3192

1,

3
,
2

3
2
5
6

1

1

 21  7   x11  6x  C
F(x) 
 1  
21x
11

3
8
5
3
3

1


 CM  M 2
8. f(x) 
 x2  7x  6
1

7


 3x3  2x2  6x  C

3
3
0

9. f(x) 
 2x3  x2  8
1

1

 31   x21  8x  C
F(x) 
 2  
31x
21
1

1


 2 x4  3 x3  8x  C

1
0
1
2
0
1
The rational roots are 3, 2, and 1.

10. f(x) 
 2x4  6x3  2x  5
F(x) 

1

1

1

 41  6  x31  2  x11  5x  C
2  
4  1x
31
11

63. 90  x  y  z 
 360
x  y  z 
 270
The correct choice is D.

Chapter 15

1

R
(M) 
 2  2M 21  3  3M 32




 2(253)(319) cos 42° 12

d 
 165,7
77
161,4
 14

cos2

42° 1
d 214.9 m

62.

M


 2M 2  3M 3

A
d2

f(x  h)  f(x)

h
(x  h)2  3  (x2  3)

h

h→0

y 
 3 sin(  45˚)

x→0


 lim

h→0

60. y

sin 2x

sin x

x

4. f
(x) 
 lim

or 5
lim x 
 2

x

3.

F 
 8

2

(x  7)(x  2)

 
 lim 
2. lim 
2x2  7x  6
(2x  3)(x  2)

4E 
 3

8 
0

1

 4x  6) 
 2(3)2  4(3)  6

1

(4D  4E  8) 
 0
D


1. lim

Mid-Chapter Quiz

(2x2

4
3  4E  8 
 0

20D  4E  16 
 0
24D

Page 960

2

3


 5x5  2x4  x2  5x  C

482



3

8.

Area Under a Curve

15-3

0

n

3i 3 3

 
 
n
n→ i 
 1 n

x3dx 
 lim

n (n  1)



4
2

81

4
n→ n
81

 lim 4
n→
81

 lim 4
n→
81

 4


 lim

Page 966

Check for Understanding

1. Sample answer: y 

2. Sample answer: Subdivide the interval from a to b
into n equal subintervals, draw a rectangle on
each subinterval that touches the graph at its
upper right corner, add up the areas of the
rectangles, and then find the limit of the total
area of the rectangles as n approaches infinity.
3. Lorena is correct. If the function is decreasing,
then the graph will always be above the tops of
the rectangles, so the total area of the rectangles
will be less than the area under the graph.
x4



2

4.

0

n

6

0








1

2

3

1

0





9b.

0

 lim

n→
9

 lim 2
n→

1

n→


 1600 ft
Yes; integration shows that the ball would fall
1600 ft in 10 seconds of free-fall. Since this
exceeds the height of the building, the ball must
hit the ground in less than 10 seconds.

1



1

6.

0



6

0



2

10.

0

n

 
n  1 
n
n→ i 
 1

2


  
 



 lim

1

4

n  2n  1



n


 lim

1

4


1 

2

2

n→
1

 4 unit2

 n2 
1

0

n

2







n







3i 2 3

n(n  1)(2n  1)



6

27

3
n→ n
9

 lim 2
n→

2n  3n  n



n



9

n→ 2




 lim

9


 lim 36 
2  n  n2 
3

...

 
 
n
n→ i 
 1 n

x2dx 
 lim

 lim

216 n(n  1)(2n  1)


3
6
n→ n
2n3  3n2  n

 lim 36 n
3
n→


 lim



3

11.

6i
6
 
 
n
n→ i 
 1 n

x2dx 
 lim

 





i 3 1


n
n→ i 
 1 n
n2(n  1)2
1

 lim n4 4
n→

2

n

2







26

n

x3dx 
 lim

2i

2
2(1)
2(2)
   1    1
n
n
n→ n
2n
 n  1
2
2

 lim n n  n(1  2  ...  n)
n→
2
2 n(n  1)

 lim n n  n  2
n→
2

 lim n(2n  1)
n→
4n
2

 lim n  n
n→

 4 units2


 lim

3
1
1
lim 6 
2  n  n
  n→

1

n2

Exercises

(x  1)dx 
 lim


 9  3 or 3 unit2

n→



Pages 966-968






7.



2n3  3n2  n

n3


2 

n(n  1)



2

n→



3

n

3200

n2

10


 lim 1600 
1  n






10i

n2  n

 
 
 




n

 32 
n
n
n→ i 
 1

32tdt 
 lim


 lim 1600
n
2 

n(n  1)(2n  1)
1


3
6
n→ n
9 2n3  3n2  n

 lim 2 n
3
n→
1

6


 576 ft

n→

n




6

6i

1152 n(n  1)


2
n2


 lim


 
 

 lim

1

2

2




 n2 

2

i 2 1
3i 2 3

  lim


n
n
n→ i 
 1 n
n→ i 
 1 n
27 n(n  1)(2n  1)

 lim n3 6
n→


 lim


1 




n n

 lim 576 
n
n→
1

 lim 576
1  n
n→
10

0

n

2

2

n

n

n→




2

2

 32 
n
n
n→ i 
 1


 lim

 x dx 
  x dx   x dx
3

n  2n  1



n

32tdt 
 lim

2i 2 2

 
 
n
n→ i 
 1 n

x2dx 
 lim

8
n(n  1)(2n  1)

 lim n3 6
n→
4 2n3  3n2  n

 lim 3 n
3
n→
3
1
4

 lim 3 2  n  n2
n→
8

 3 units2

5.



9a.

2

1

3

2

3


2  n3  n1
2

units2

n→


 72

483

Chapter 15



2

12.

1





0

x2dx 


 x dx
x dx   x dx

1
1

0

2

x2dx 

0
2

2



4

2

17.

0

2

n

n

n

i 2 1

1
n(n  1)(2n  1)


3
6
n→ n
8
n(n  1)(2n  1)




 lim n3
6
n→
3
2
1 2n  3n  n
4 2n3  3n2 n

 lim 6 n
 lim 3 n
3
3
n→
n→















 lim

n→
1
8

 3  3



3

13.

1

6

1

xdx 




3

0




2 

3

n










lim 
2 
  n→
4

3

1

n2

3

n



n





1

n2

14.

n

3

i

1








0



















 lim

125

6

5

1

0

1

3

n

1

2x3
n











16.











6 


 lim


 625

units2

n→

Chapter 15

0



10

n2










  n



 



 



n

 8 
n
n
n→ i 
 1
2i

8xdx 
 lim

2

32 n(n  1)


2
2
n→ n
n2  n

 lim 16 n
2
n→
1

 lim 16 1  n
n→


 16



15

n






 lim








2

20.

 
 
 

625

6

3

3i

 

 

 


n






48
1  n

 n  1 
n




19. lim








 
sin in  n
n→ i 
 1

i 3 1


n
n

250 n2(n  1)2
n2(n  1)2
2

  lim 

4
4
4
n
4
n
n→
n→
625 n2  2n  1
1 n2  2n  1

 lim 2 n
 lim 2 n
2
2
n→
n→
1
1
625
1
2
2

 lim 2 1  n  n2  lim 2 1  n  n2
n→
n→
625
1

 2  2 or 312 units2
n
5
5i 4 5


x4dx 
 lim
n
n→ i 
 1 n
0
6n5  15n4  10n3  n
3125


 lim n
5
30
n→


 lim

2

2

3
9
3 n(n  1)
n(n  1)(2n  1)
 
   
2
n
6
2
n→ n n
27 2n2  3n2  n
n2  n
9









 lim n3
 n2
3
6
2
n→
9
9
3
1
1

 lim 2 2  n  n2  2 1  n  3
n→
9

 9  2  3
9
15

 12  2 or 2 units2


 lim

lim  2 
  
 
 2
  
   n→
n→ i 
 1
i
1


 lim

2

3


2  n  n2 

5i 3 5


n
n

2

 n (1  2  . . .  n)  n



0

(1  2  . . .  n)

4

n



2

5

3i 2














3




31 2
31
3


 n  1
n
n→ n
32 2
32
 n  n  1  . . .
3n 2
3n
 n  n  1
3
9

 lim n n2 (12  22  . . .  n2)
n→

n→
125

 3 units2

 2x dx

  2x  

n

 
n
n→ i 
 1









(x2  x  1)dx


 lim




15.



3

18.

 
 
 


 lim






3
1
32
(2    
)
n
n2
n→ 3
64
208

 3  48 or 3 units2


 lim

xdx

9
1
n(n  1)
n(n  1)

 lim n2 2  lim n2 2
n→
n→
9 n2  n
1 n2  n

 lim 2 n
 lim 2 n
2
2
n→
n→
9
1
1
1

 lim 2 1  n  lim 2 1  n
n→
n→
9
1

 2  2 or 4 units2
n
5
5i 2 5


x2dx 
 lim
n
n→ i 
 1 n
0
125 n(n  1)(2n  1)


 lim n
3
6
n→
125 2n3  3n2  n

 lim 6 n
3
n→









3

0






3

1

3i

4

4i

16 n(n  1)(2n  1)
 


6
n
n→
24 n(n  1)




 n 
 2 
64 2n  3n  n
96 n  n

 lim n 
6  n 
2
n→

lim  
n
n
 

n  n→
n→ i 
 1 n
i
1


 lim



24

n


 lim

or 3 units2

xdx 












2i 2 2

lim  
n 
n
 
 
n  n→
n→ i 
 1 n
i
1


 lim

 6 
n
n

4
41 2
41


 6 n
n
n→ n
42 2
42
 n  6 n  . . .
2
4n
4n
 n  6 n
4 16

 lim n n2 (12  22  . . .  n2)
n→


 lim

By symmetry

0


 lim

4i 2

 
n
n→ i 
 1

(x2  6x)dx 
 lim



1

n4

484

















1

 (x  2)dx 
 lim
4

21.

n

 
1  n  2
n
n→ i 
 1

1

3

3i

n






3
3i

3  n
n→ n i 
 1
3
31

 lim n 3  n 
n→


 lim



3n

 
3  n

 lim

3

n

3n  n3(1  2  . . .  n)


 lim

3

n

n(n  1)

3n  n3 

2

n→
n→

n(n  1)(2n  1)
64 n(n  1)
   
  8
6n4

6
2
n
n(n  1)(2n  1)
n(n  1)
1
4
 lim n 
6  n 
2  2
n→
32 2n  3n  n
n n

 lim 3 
n  32 
n  8
n→
1 2n  3n  n
n n
 lim 6 
n  2 
n  2
n→
32
3
1
1

 lim 3 
2  n  n  32 
1  n  8
n→
1
3
1
1
 lim 6
2  n  n  2
1  n  2
n→

22.



n



5

23.

3

8x3dx 




5

0

64

1


 3  32  8  3  2  2





 24 

 (x
2

25.



0

63

3

or 45
n

5




 lim


2  n3  n1


 lim




n


 lim

n→

8x3dx
n

5i 3 5

5000 n2(n  1)2


4
n4


 lim 1250 
1  n  n2 
1

24.

2

n

1

n2

 (x  4x  2)dx

  (x  4x  2)dx   (x



0

2

n

0

4i 2

 
n
n→ i 
 1


 lim

n

2

 4x  2)dx

4i

26.


n4 

4n 2

0

n→

2



n1

x3dx 
 lim

4

8

3

or

2 2

40

3

5i 3 5


n
n

 
 
 

n→  i 
 1
625 n2(n  1)2


 lim n
4
4
n→ 




 lim

625

4

 2n  1



n


 lim

625

4


1  n2  n1

n→ 
625

 4

 4
n  2  

n
n 2



n

n→ 

4n

1




32

3




 4 
n  2  . . .  

n  4
n
1

n



5

2

 2  lim

6

2

2

 4
n  2

42

 5n4  n2

2

i
i
1
 
n  4 
n  2
n
n→ i 
 1
41 2
41
42 2
4

 lim n 

n  4 
n  2  

n
n→

lim



6n5

3

2

1



2n6

3

1

4



16 
 

n
n→ 3
4 2n  3n  n
 3 
n
16
6
1
4
5

 lim 3
2  n  n  n  3
n→

2  n3  n1


 1250  162 or 1088

4




 lim

n→



 ...
 




n2  2n  1

 lim 162 
n

2
2



22 2

n
2n 2

n

 


n2  2n  1

n→

n→

5

21 2

 
n



2

4


 lim 1250 
n

2

 lim 162 
1 

5

2i 2

 
n

 

5
 



 lim

648 n (n  1)
lim n 
4
  n→

n→

2i 2 2

2 32
5
5
5
 
5 (1  2  . . .  n )
n→ n n
4
 n2 (12  22  . . .  n2)
2n6  6n5  5n4  n2
64

 lim n6 
12
n→
n(n  1)(2n  1)
8
 n3 6

3i 3 3

2

2i 5

21




n 








0




n

 
n
i
1

2

n→ n
22
 n
2n
 n

lim  8 
n 
n
 8 
n 
n  n→
n→ i 
 1
i
1


 lim

2

n

n→

2



2i 5 2

 
 
n  
n 
n
n→ i 
 1 n

 x2)dx 
 lim

2n3  3n2  n

n3

8x3dx 

2

2

 22  32  . . .  n2)

3

2

2

27

2

n(n  1)(2n  1)

6

64

 lim n3
n→
32

 lim 3
n→
32

 lim 3
n→
64

 3

2

2

3

1

 
 
 

x2dx 
 lim

2

3

4i 2 4


n
n

n→ i 
 1
64

 lim n3 (12
n→

0

or

2

2

3

n→


9

2

3

n→

4

3

3


 lim 9  2
1  n
9

2

(1  2  . . .  n)  2n

n→

n2  n

9

4

n


 lim


 lim 9  n2 
2
9





32

 
3  
n ...






4 16 2
2
2
 
2 (1  2  . . .  n )
n→ n n
16
 n (1  2  . . .  n)  2n
1
1
 lim n n2 (12  22  . . .  n2)
n→


 lim

n2

2

2

 4 
n  2  . . .  

n  4 
n  2
2

n

485

Chapter 15



3

27.

0

n

3i 3 3

  
 
n
n→  i 
 1 2 n

1
x3dx
2

1


 lim

31a. r (t )
160

81
n2(n  1)2



 lim 
4
4
n→  2n
81 n2  n  1

 lim 8 n
2
n→ 
1
1
81

 lim 8 1  n  n2
n→ 









120






80



40

81

28a. f(20)
80  2(20)

 $40



40

20

0

n

20i
20
 
80  2 
20  n 
n
n→  i 
 1

n→ 

n

40i

41
42



40  
n   
40  n   . . .
4n




 600  2592  1728

 $1464
31c.

(6  0.06x2)dx
n


 lim

$1464

12


 $122

32a. v (t )

2

10i
10
 
6  0.06 
n 
n

n→  i 
 1
10

 lim n
n→ 

2

3

2

1

n


 800  400 or $400
0

3

2

2

10

3

2

2



2

12

n

2

2

n→ 

12

2

40n  4n0 (1  2  . . .  n)
800 n(n  1)

 lim 800  n 
2
n→ 
n n

 lim 800 400 
n
n→ 
20

n→  n


 lim


 lim 800  400
1 

12i 2

12i

12  1
12  1
50  36
n  3
n 


n→ 
12  2
12  2 2
 
50  36
n  3
n   . . .
12  n
12  n 2
 
50  36
n  3
n 
12
432

 lim n 50n  n(1  2  . . .  n)
n→ 
432
 n(12  22  . . .  n2)
5184 n(n  1)
5184 n(n  1)(2n  1)

 lim 600  n
2  n
6
n→ 
n n
2n  3n  n

 lim 600  2592
n  864
n
n→ 
1
3
1

 lim 600  2592
1  n  864
2  n  n
n→ 

 
40  n

29.

n

 50  36
n  3
n 
n
n→  i
1


 lim

 
40  n
i
1

20

n→  n


 lim

(50  36t  3t2)dt


 lim


 lim

20

n



12

31b.

(80  2x)dx


 lim

t

O


 8 or 10.125 ft2

28b.

r (t ) 
 50  36t  3t 2

15

2

10  1


6  0.06 

n  
2
10  2
 
6  0.06 
n   . . .
10  n 2
 
6  0.06 
n 
6
10

 lim n6n  n (12  22  . . .  n2)
n→ 
60 n(n  1)(2n  1)

 lim 60  n 
6
n→ 
2n  3n  n

 lim 60  10 
n
n→ 
3
1

 lim 60  10 
2  n  n
n→ 

v (t ) 
 3.5t  0.25t 2
10

2

5

3

3

v (t ) 
 1.2t  0.03t 2

2

3

2

O


 60  20 or 40



10

10

6  0.06x2dx 
 2(40) or 80

32b.

By symmetry

0

To make a tunnel 100 ft long, multiply 80 by 100.
80(100) 
 8000 ft3
30. Setting the two functions equal 2 y
to each other and solving for x,
y 
 x2
we find that the curves cross
when x 
 0 and x 
 1. x  x2
y
x
1
for 0  x  1, so we can find
the desired area by subtracting
the area between the graph of
x
y 
 x2 and the x-axis from the
O
1
area between the graph of
1
y 
 x and the x-axis. These areas are 3 unit2 and
1

2
1

6

1

1 2 3 4 5 6 7 8 9 10t

(3.5t  0.25t2)dt
n

10i 2

 3.5
n  0.25
n 
n
n→  i
1
10i


 lim

10

2

10  1
10  1



3.5

n   0.25
 n  
n→ 
10  2
10  2 2
 
3.5
n  0.25
n   . . .
10  n
10  n 2
 
3.5
n  0.25
n 
10 35
25

 lim nn(1  2  . . .  n)  n(12  22
n→ 
 . . .  n2)
350 n(n  1)
250 n(n  1)(2n  1)

 lim n
2  n
6
n→ 
n n
125 2n  3n  n

 lim 175
n  3
n
n→ 
125
1
3
1

 lim 175
1  n  3
2  n  n
n→ 


 lim

10

n

2

2

3

2

3

2

1

unit2, respectively, so the answer is 2  3 


2

3

2

unit2.

Chapter 15



10


 175 

486

250

3

or

275

3

m



10

0

40. Use a graphing calculator to find the maximum
width of x 
 5.4 cm.

(1.2t  0.03t2)dt
n

2

10i
10i
10
 1.2
n  0.03
n 
n
10  1
10  1 2



 
1.2

n   0.03
 n  
10  2
10  2 2
 
1.2
n  0.03
n   . . .
10  n
10  n 2
 
1.2
n  0.03
n 
10 12

 lim nn(1  2  . . .  n)
n→ 
3
 n(12  22  . . .  n2)
120 n(n  1)
30 n(n  1)(2n  1)

 lim n
2  n
6
n→ 
n n
2n  3n  n

 lim 60
n  5
n
n→ 
1
3
1

 lim 60
1  n  5
2  n  n
n→ 


 lim

n→  i
1
10

 lim n
n→ 

[5, 10] scl1 by [30, 90] scl1

2

2

41. 62  82 
 102
36  64 
 100
100 
 100
ABC is a right triangle with a base of 6 inches
and a height of 8 inches. The area of

3

2

3

2

2

3

2


 60  10 or 70 m

1

ABC 
 2(6)(8) or 24 square inches. If the

The first one results in a greater distance
covered.
33. The equation y 
 
r2  x2 can be rearranged to
obtain x2  y2 
 r2, which is the circle centered at
the origin of radius r. In the equation y 
 
r2  x2,
y must be nonnegative, so the graph is only the
top half of the circle. Therefore, the value of the
1
integral is 2 the area of a circle of radius r,
1
or 2r2.

rectangle has a width of 3 inches, then it has a
24

length of 3 or 8 inches, since A
w. The perimeter
of the rectangle 
 2(3)  2(8) or 22 inches. The
correct choice is C.

Page 969

34. f(x) 
 3x2  x2  7x
f
(x) 
 3  3x31  1  2x21  7  1x11

 9x2  2x  7
x2

x→2 x  2

35. lim

22




22
0


 4 or 0
36. log1x 
 3
3

1 3

x 
 
3

15-4

x 
 32 or 27
u
37. u 
 2, 5, 3  3, 4, 7

 2  (3),  5  4, 3  (7)

 1, 1,  10
u u
u

 i
j  10k
38. cos 2v 
 1  2 sin2 v

The Fundamental Theorem of
Calculus

Pages 972–973
1.

Check for Understanding

 f(x)dx represents all of bthe function that have
f(x) as their derivative. a f(x)dx is a number; it

gives the area under the graph of y 
 f(x) from
x 
 a to x 
 b.
2. Sample answer: Let a 
 0, b 
 1, f(x) 
 x, and

3 2


 1  2
5

g(x) 
 x. Then  f(x)g(x)dx 
  x2dx 
 3, but
a
0
b

18


 1  25

b

1

1

1

a f(x)dx  a g(x)dx 
 0

7


 2
5

b

1

xdx  0 xdx 
 2  2 
 4.
1

1

1

1

3. If the “C ” is included in the antiderivative, it
will appear as a term in both F(b) and F(a) and
will be eliminated when they are subtracted.
4. Rose is correct; the order does matter.
Interchanging the order multiplies the result by
1. In symbols, F(a)  F(b) 
 (F(b)  F(a)). So
unless the answer is 0, interchanging the order
will give the opposite of the right answer.

39. y 
 2 sin 10v
1

Amplitude 
 2
2

History of Mathematics

1. See students’ work. The difference in area should
decrease as the number of sides of the polygon
increases.
2. The roots of the resulting equation are the zeros of
the derivative of the original function.
3. See students’ work.



Period 
 10 or 5

5.

(2x2  4x  3)dx 
 2  13x3  4  12x2  3x  C
2


 3x3  2x2  3x  C
6.

(x3  3x  1)dx 
 14x4  3  12x2  x  C
1

3


 4x4  2x2  x  C

487

Chapter 15



0

7.

1

2

(4  x2)dx 
 4x  3x3

0

Pages 973–976

2




C
1 8
7
16. 6x dx 
 6  8 x  C
15.


 
40 3 03  
4(2) 3 (2)3
1

1


 0  
8  3
8



2

8.

0



x4dx 







1

3

17.

(x2  2x  4)dx 
 13x3  2  12x2  4x  C
1

 


1
  25
5
32
  0
5


 3x3  x2  4x  C



1
 5  05
32
or 5 units2

(x2  4x  4)dx
1

18.

1

19.

1

1

(x4  2x2  3)dx 
 15x5  2  13x3  3x  C
1

20.

1

(4x5  6x3  7x2  8)dx
1


 
3  13  2  12  4  1

2

 
3  (1)3  2  (1)2  4  (1)


 3  
3 or 3 units2

 2x dx 
 2 
3

3

1

1







4

11.

1




26

21.

22.



0

1




1

2




 3x3  3x2  3x  C




 
  33  
3  03



1

2



42



3

1

23.

 6  4

2

1

(x2  2)dx 
 3x3  2x


 
1

3

 
3  13  2  12  6  1
1

1

112

35

224

35

12.



0

(2x2  3x  2)dx 
 2 

24.

2

0

3

1
x2
2

3


 3x3  2x2  2x

 2x
2

2






4

13.



(x3

2

1
x4
4

 x  6)dx 




0

 0 or

1
x2
2

3

2

1

0



4



0.1

0

25.

 0  2  0

0

4



1

2

26.

1

2
1
x4
4
0

0

1

4
0


 
4  44  
4  04
1


 64  0, or 64 units2
1

3x6dx 
 3  7x7



1

1
1

1
3
x7
7
1


 
7  17  
7  (1)7
3

3


 7  
7 or 7 unit2
3



0

0.1
1
x2
2
0

27.

2

3

1

1
x3
3




0







488

6

1

(x2  2x)dx 
 3x3  2  2x2

0.1


 (250  (0.1)2)  (250  02)

 2.5  0 or 2.5 J

Chapter 15

1

x3dx 
 4x4

1m


 250x2



1



100 cm 
 0.1 m

500xdx 
 500 

2x2


 4  0 or 4 units2


 48  (6) or 54
10 cm

1

2

1

 
4  24  2  22  6  2

14.

1

(4x  x3)dx 
 4  2x2  1  4x4


 
2  22  4  24  
2  02  4  04

1

1

13





 
4  44  2  42  6  4
1


3  
3 or 3 units2

14

3

 6x

1

0

3

2
3

2

 2  3  
3  23  2  2

33

2


 
3  23  2  22  2  2
14

3



2

63


 6  6 or 2
1
x3
3

3

4


 3  6
2

2


 18  0 or 18 units2

4

1



 

0

2

3

(x2  x  6)dx 
 3x3  2x2  6x
43

0

3

2


 3x3

or 40

1

3

3

1

2x2dx 
 2  3x3

14

1

7

(x2  6x  3)dx 
 13x3  6  12x2  3x  C
3

3

1
  34
2
81
1
  
2
2

3

1

3
1
x4
4
1


 2x4

1


3x6  2x4  3x3  8x  C

1

10.

1


 4  6x6  6  4x4  7  3x3  8x  C

1

7

2


 5x5  3x3  3x  C

1

1


 3x3  2x2  4x

(3x2  x  6)dx 
 3  13x3  1  12x2  6x  C

 x3  2x2  6x  C

1


 3x3  4  2x2  4x

19

Exercises

1
x6
6


 4x8  C

units2

2
1
x5
5
0





1

9.

16

3

x5dx



x2

0
2

0
2

 




1
1
  03  02    (2)3  (2)2
3
3
20
20




0   3 or 3 units2






 (x
3

28.

2

1

1

1

 2x  3)dx 
 1  3x3  2  2x2  3x
1


 3x3  x2  3x

3

3



0

37.

1

1

1

1

1

1

1

29.



0

(x3  x)dx 




16



0

30.

1



1

1

3

1

39.

1




1
x4
4



3

 10x

4x2



3

2

1

1

1

1

0






4

32.

2

2x3

7

3x4dx 
 3 



3

1

1



1

35.



1



1
x2
2

 4x

0

2

1


 












135

12

1

or

1

1
x3
3

2


 5x5  3x3  x




28

15

2

 0 or

1
x4
4



x3



3
x2
2

x

3

2

1

3

3
2

2

 32  3
3

15



1

42.

0

15


 4  0 or 4

1

1

1

 
4  24  23  2  22  2

1

5

1

x2  x  2
dx
x2







1

0
1

(x  2)(x  1)
dx
x2

(x  1)dx

1


 2x2  x

1
0


 
2  12  1  
2  02  0
1

34

3

1

3

43.

1



0


 2
x(4x2  1)dx 




2

0

(4x3  x)dx
1

1



2
1
x2
2
0

1


 4  4x4  2x2

0




0


 
5  15  3  13  1  
5  05  3  03  0
1

1


 
  34  33 

5

2

x

(x3  3x2  3x  1)dx

1

4

1



3

2

0

1

1

12






20

3
1
x3
3
1



1

1

(x4  2x2  1)dx
2

1


 4x4  3  3x3  3  2x2  x


 
4  34  3  33  
4  14  3  13




1

3

32

(x  1)3dx 


1
  (1)2  4  1
2


 (53  52  5)  (13  12  1)

 105  1 or 104
1
x4
4



3

41.

1

1
x5
5

3


 9  
3 or 3
5

3

1
  32  4  3
2
33
7
   or
2
2

45



1

1


 4  
1
2
36.

1

1


 
3  33  32  3  3  
3  13  12  3  1

1

1

(x2  2x  3)dx

3

2

(3x2  2x  1)dx 
 3  3x3  2  2x2  x




1

 


x2)dx

(x  3)(x  1)dx


 3x3  x2  3x





(x3

413

4


 x3  x2  x

3

148


 3x3  2  2x2  3x

3
3
  45    25
5
5
3072
96
2976
   or 
5
5
5

(x  4)dx 


74


 6  6 or 6




4
1
x5
5
2



34.



3

40.





5

0


 (2  73)  (2  03)

 686  0 or 686

3



265

7

0


 5x5

33.

265

23

1

3


 6  
3


 4  
4 or 92 units2
6x2dx 
 6  3x3

2

3

 
3  (2)3  2  (2)2  8  2

1

7

2

5

3


 
3  53  2  52  8  5

 
4  (1)4  4  (1)2  10  1



5

1

1

1

1

(x2  3x  8)dx


 3x3  2x2  8x


 
  34  4  32  10  3

31.

1


 3x3  3  2x2  8x

1

4

345

0

1


 8  0 or 8

5

(x3  8x  10)dx

 4x4  8  2x2  10x

2

1


 
4  24  2  22  2  
4  04  2  02  0

1


 4  0 or 4 unit2

3

(x3  x  1)dx
1

1

1

9


 4x4  2x2  x


 
4  14  2  12  
4  04  2  02
3

1



 0  
2
0  or 20

2

38.

1
1
x2
2
0

1

1

9


 9  3 or 3 units2
1
x4
4

1


 
5  05  4  04  
5  (1)5  4  (1)4

 
3  13  12  3  1

1

0

1


 5x5  4x4


 
3  33  32  3  3
11

(x4  x3)dx

1

2

x4

2
0


 
24  2  22  
04  2  02
1

28

15

1


18  0 or 18

489

Chapter 15



1

44.

1



1







x3





110
100
90
80
70
60
50
40
30
20
10

(3x2  5x  2)dx

1


3

1
x3
3



5

5
x2
2



13

5

2

1
x2
2

 2x






(1)3

 2x

1
1

1
1

12

 2  1



5

2

2

 (1)  2  1



20.5

0

1

1

x3dx 
 4x4

0

1

4

 i3 






100.5

45b.

0




1



i2 


i
1






2

46.

1


 6(558  0)

 $93
1

12  6

49c.

2
1
x2
2
0







2
4
6

1


 $105



R

50.

1

R

(R2  x2)dx 
 R2x    3x3

 
R2  R  3  R3
1

 
R2  R  3  (R)3
1


 3R3  
3R3
2

2

4


 3R3
1000 
 k(6.4  106)2

51a.

1


1000 
 k  
40.96  1012

 2  
 
1

3

value is

03

4.1  1016

 5  0

1
(3)(9)
2






3.8  108

51b.

6.4  106

k; 4.1  1016Nm2

4.1  1016x2dx


 4.1  1016  (1)x1
4.1 

4.1  1016


3.8  108





1.1 

27
.
2

490

6.4  106

6.4  106

4.1  10

  

6.4  10 

108


 6.3  109 J

1 2 3 4 5x
y 
 3x  6

3.8  108

3.8  108

1016


 x

O

Chapter 15

1


 6(1188  558)

0

47c. Since the function is negative, the integral in
part b gives the opposite of the area of the
22
region. The area is 3.
y
48.
The integral represents the
10
area of a right triangle. The
8
6
4
2

6

6

1

(x2  5)dx



12

1

12

 
75  6  4  62  6  63

2

negative. Therefore, the sum is negative. since
b
a f(x)dx is a limit of negative sums, it is also
negative.
2
1
x3  5x
3
0
1
  23  5
3
22
3

1

1

i
1

0

1


 6
75  12  4  122  6  122

positive, so each term in the sum  f(xi)x is

2

1

1

n



6


75  8x  12x2dx


 6
75x  4x 2  6x3


 (245,000  22)  (245,000  02)

 980,000  0 or 980,000 J
47a. Since the graph is below the x-axis, f(x) is
negative. Each f(xi) is negative and x is

47b.

12

1

or 338,350


 245,000 x2




 6
75x  8  2x2  2  3x3

100(100  1)(2  100  1)

6

490,000 xdx 
 490,000 

0

0

1

1

  
13  03

100(101)(201)

6

0

6

 
75  0  4  02  6  03

338,358.38  0 or 338,358.38
100

1

6


 6
75  6  4  62  6  63

100.5
1
x3
3
0
1
  100.53
3





0

1

202(20  1)2

4
400(441)
 or 44,100
40

x2dx




 6
75x  4x2  6x3

44,152.52  0 or 44,152.52
i
1

x
6


75  8x  12x2dx
1
1
1 1

 6
75x  8  2x2  2  3x3


 
  20.54  
  04
20

f (x ) 
 75  8x  12 x 2

1

60

49b.

20.5

1

4

f (x )

O


 2  
2 or 6
11

45a.

49a.

(x  1)(3x  2)dx

16
6

 64.1 

108

R
R



2

52.

0

1
x2dx
2

n

1 2i 2 2

 
 
n
n→  i
1 2 n


 lim

Pages 978–980

4  2 n(n  1)(2n  1)


3
6
n→  2n
3
4 2n  3n2  n

 lim 6 n
3
n→ 
2
3
1

 lim 3 2  n  n2
n→ 
4

 3
2x6  3x2  2





 lim











x→2

12. lim (x3  x2  5x  6) 
 (2)3  (2)2  5(2)  6
x→2

 8  4  10  6

4
13. lim (2x  cos x) 
 2(0)  cos 0
x→0

01

 1

53. f(x) 

f
(x) 
 2  6x61  3  2x21  0

 12x5  6x
54. In a normal distribution, 68.3% of the data lie
within 1 standard deviation of the mean. So,
100  68.3 or 31.7% of the data lie outside
1 standard deviation of the mean. Thus, 31.7% of
the test-takers scored more than 100 points above
or below the mean of 500.
55. To begin with 25 out of the 50 numbers are odd.
With each consecutive draw, there is 1 fewer odd
numbers and 1 fewer tickets.
P(four odd numbers) 




25 24
  
50 49
253

4606



23

48



14. lim

x→1

15. lim

x→0

5x2

2x

45 ft/s

Vy

5




x→0
5
(0)
2


0

16.

22

47

x2  2x

lim 
2  3x  10
x
x→4

x→4

x

x→4 x  5
4



4  5 or 4

17. lim (x  sin x) 
 0  sin 0
x→0

0
18. lim

x→0

x2  x cos x

2x


 lim

x→0

x(x  cos x)

2x

x  cos x

2
x→0
0  cos 0

 2
1

 2
x3  2x2  4x  8
(x  2)(x2  4)

lim 

 lim 
x2  4
x2  4
x→2
x→2


 lim

1

2

 i



3

2

 
 1  3i


 lim x  2
x→2


 2  2 or 4

sin 52° 
 4
5
45 sin 52° 
 vy
35.46 vy; 35.46 ft/s

60. mAE
+ 
 m∠AXE, so ∠AXE is the smallest angle.
If the circle was divided into 5 equal parts, each
angle would measure 72°. Since the circle is not
divided evenly, the smallest angle AXE is less
than 72°. The correct choice is A.

21.

22.

Chapter 15 Study Guide and Assessment
Understanding the Vocabulary
2.
4.
6.
8.
10.

x2  6x  9  9

2x
x→0
x→0
x(x  6)

 lim 2
x
x→0
x6

 lim 2
x→0
06

 2 or 3
x2  9x  20
(x  5)(x  4)
 
 lim 
lim 
x2  5x
x(x  5)
x→5
x→5
x4

 lim x
x→5
54
1

 5 or 5
f(x  h)  f(x)
f 
(x) 
 lim h
h→0
2(x  h)  1  (2x  1)

 lim 
h
h→0
2x  2h  1  2x  1

 lim 
h
h→0
2h

 lim h
h→0

20. lim

Vx

false; sometimes
false; indefinite
false; secant
false; derivative
false; rate of change

x(x  2)



 lim4 
(x  5)(x  2)


 lim

52˚

Page 977

(x  6)(x  6)

x1


 lim 2x

vy

59.

1.
3.
5.
7.
9.

x→1

x→1

19.

; 2



 lim


 1  6 or 5

58. r 
  or 2




x2  36

x6


 lim x  6

56. A 
 Pert

 600e0.06(15)

 $1475.76
57. h 
 6; k 
 1
hp
3
6  p 
 3 or p 
 3
(y  k)2 
 4p(x  h)
(y  1)2 
 12(x  6)
22


2
2


v 
 3  3 or 3


2 cos 3  i sin 3

Skills and Concepts

11. There is a point at (2, 1) so f(2) 
 1. However,
the closer x is to 2, the closer y is to 3.
So, lim f(x) 
 3.

true
true
false; tangent
true
false; derivative

(x  3)2  9

2x


 lim


2

491

Chapter 15

23. f 
(x) 
 lim

h→0


 lim

h→0








f(x  h)  f(x)

h

1

35. f(x) 
 2x3  2x2  3x  2
1

4(x  h)2  3(x  h)  5  (4x2  3x  5)

h

1

1

 6x  C
1
5

 5x5  4x4  x2  6x  C


 8x  3


 lim

x3  3x2h  3xh2  h3  3x  3h  x3  3x

h


 lim

3x2h  3xh2  h3  3h

h

h→0

h→0


lim

h→0

37. f(x) 
 (x  4)(x  2)

 x2  2x  8
1
1
 21  2  x11  8x  C
F(x) 
 
2  1x
11

f(x  h)  f(x)

h
(x  h)3  3(x  h)  (x3  3x)

h

h→0

1


 3x3  x2  8x  C
x2  x

38. f(x) 
 x

x1
1
 11  x  C
F(x) 
 
1  1x

 3xh 
 3)

h
h(3x2

h2

1


 lim 3x2  3xh  h2  3

h→0

 3x2 



2

39.

3

0

25. f(x) 
 2x6
f 
(x) 
 2  6x61

 12x5
26. f(x) 
 3x  7
f 
(x) 
 3  1x11  0

 3
27. f(x) 
 3x2  5x
f
(x) 
 3  2x21  5  1x11

 6x  5



1

40.

0

1

f
(x) 
 4  2x21  1  1x11  0
1


 2x  1
1

29. f(x) 
 2x4  2x3  3x  4
f
(x) 




4

1

41.

 4x41  2  3x31  3  1x11  0

3

1

2i

n(n  1)
8

2 2
n→ n
n2  n

 lim 4 n
2
n→
1

 lim 4 1  n
n→

 4 units2








2






n

i 3 1

 
 
n
n→ i 
 1 n

x3dx 
 lim

n2(n  1)2
1

4 4
n→ n
1 n2  2n  1

 lim 4 n
2
n→
1
2
1

 lim 4 1  n  n2
n→
1

 4 unit2








x2dx 




4

0






3

x2dx 
n


 2x3  6x2  3



x2dx

0
n

4i 2 4

3i 2 3

lim  
n 
n
 
 
n  n→
n→ i 
 1 n
i
1


 lim

30. f(x) 
 (x  3)(x  4)

 x2  7x  12
f
(x) 
 2x21  7  1x11  0

 2x  7
31. f(x) 
 5x3(x4  3x2)

 5x7  15x5
f 
(x) 
 5  7x71  15  5x51

 35x6  75x4
32. f(x) 
 (x  2)3

 x3  6x2  12x  8
f
(x) 
 3x31  6  2x21  12  1x11  0

 3x2  12x  12
33. f(x) 
 8x

64 n(n  1)(2n  1)


3
6
n→ n
27 n(n  1)(2n  1)
 lim n3 6
n→
64 2n3  3n2  n

 lim 6 n
3
n→





 lim









 lim






1


 4x2  C
34. f(x) 
 3x2  2
1

 21  2x  C
F(x) 
 3  
2  1x


 x3  2x  C

492

64

3
37

3







2n3  3n2  n

n3




27

6

n→
64

 lim 6(2
n→

 11  C
F(x) 
 8  
1  1x

Chapter 15

n

 2 
n
n
n→ i 
 1

2x dx 
 lim


 lim

1

1


 2x2  x  C


 lim

28. f(x) 
 4x2  x  4

1

2

3

36. f(x) 
 x4  5x3  2x  6
1
1
1
 41  5  x31  2  x11
F(x) 
 
41 x
31
11

h→0


 lim

2


 8x4  3x3  2x2  2x  C


 lim 8x  4h  3

h→0

1

 11  2x  C
3
1  1x

4x2  8xh  4h2  3x  3h  5  4x2  3x  5
lim 
h
h→0
8xh  4h2  3h
lim h
h→0
h(8x  4h  3)
lim h
h→0

24. f
(x) 
 lim

1

 31  2  x21
F(x) 
 2  
3  1x
21



27

3

units2



3

n



1

n2 )

27

n→ 6

 lim


2  n3  n1
2



2

42.

1

6x2dx 

n


 lim



2



1

6x2dx 

0

n

2i 2 2


n
n

6




Page 981

6.x2dx

 
   lim  6  
  
 










 lim 8(2 
n→

3

n

1

n2 )





4

2

n→

2

3x2dx 
 3  3x3

 x3

1h
60 
mi
5280 ft
r
     

3600 s
1
hr
1
mi
88 ft/s
2
a 
 5 
s 
 17.6 ft/s

53a.

53b. v(t) 


2

1

3



2


 (3 
 (3 

 48  12 or 36





1

n2

4

42)

44.

3

n

2

22)

45.



2

2

3

53c. d(t) 





1
x3
3

1

1
x2
2

0

0

t

0

 3x

2
2
1 2
3


x  2 x  3x
2
1
(23  2  22  3  2) 
1
( 2)3  2  (2)2  3  (2)

(x  2)(2x  3)dx 


Page 981

4

 7x  6)dx

2

0


2

1
x3
3

2

1

 7  2x2  6x
7


 3x3  2x2  6x

4



4

1

7

0

Open-Ended Assessment

0

0

16x3dx 
 4x4

1
0


 4  14  4  04

4

0


 
3  43  2  42  6  4 
2

t

1. Sample answer: f(x) 
 x2  2x  2;
lim(x2  2x  2) 
 12  2(1)  2
x→1

5
2. Sample answer: g(x) 
 16x3;



 (2x

t
0


 8.8t2  8.8(0)2

 8.8t2

2


 12  (16) or 28

4

t

 17.6xdx


 8.8x2






0

1




46.

t

 11

 17.6  
1  1x

3

 x  3)dx 
 3

88 ft/s


 17.6t  17.6(0)

 17.6t


 8  (27) or 35
(3x2

2

 17.6dx


 17.6x


 23  (3)2
2

2

2

0.0000125 m
c(x) 
 9x5  135x3  10,000
c
(x) 
 9  5x51  135  3x31  0

 45x4  405x2
c
(2.6) 
 45(2.6)4  405(2.6)2 or $681.41

4

1

6xdx 
 6  2x2

 3x2



52.



 lim 1 
2 


 16  2 or 14 units2
43.




2

51. lim

i 2 1


n
n

n→i 
 1
n→ i 
 1
6
48 n(n  1)(2n  1)
n(n  1)(2n  1)

 lim n3 6  lim n3 6
n→
n→
48 2n3  3n2  n
6 2n3  3n2  n

 lim 6 n
 lim 6 n
3
3
n→
n→




Applications and Problem Solving

50
50
1



1m  

1  t   
 2 m  
 1  100 
t→100 2

0


23  03  72  02  6  0

368

368

Chapter 15 SAT & ACT Preparation


 3  0 or 3
47.



6x4dx


6



1
x5
5

6
x5
5

C

Page 983

C

48.

(3x2  2x)dx 
 3 

49.



1
x3
3

2

1
x2
2

C

1.


 x3  x2  C
(x2

1

5


 3x3  2x2  2x  C
50.

4

SAT and ACT Practice
1

 
1
1
  
2
3

1

 
1

(3x5  4x4  7x)dx 
 3  16x6  4  15x5  7  12x2  C
1



1

3

1

 
3
2
  
6
6

1

 5x  2)dx 
 3x3  5  2x2  2x  C
1

1

2


6

7


 2x6  5x5  2x2  C


6
The correct choice is A.

493

Chapter 15

2. Let x represent the length of the second side of the
triangle. Then the first side has length 2x.

x

2x

Since 16 is composite,
5

?
Clearly the perimeter must be greater than 3x, so
eliminate answer choices A and B. Use the
Triangle Inequality Theorem.
x  2x  ?
P x  2x  3x
3x  ?
P 6x
?  x  2x
P  x  2x  x
?x
P  4x
Since the perimeter cannot equal 6x or 4x,
eliminate answer choices C and E as well.
The only possible answer choice is D.
3. Draw a figure. Begin with the 3 parallel lines.
Draw the 3 nonparallel lines in positions that are
as general as possible. For example, do not draw
perpendicular lines or concurrent lines. Draw the
first nonparallel line, and mark the intersections.
Then draw the second line, making sure it
intersects each of the other lines. Then draw the
third line, making sure it intersects each of the
others.

5.

23


 69

46


 23

69


 34.5

31


 63

The correct choice is B.
7. The nth term of an arithmetic sequence with first
term a1 and common difference d is given by
an 
 a1  (n – 1)d. a1 
 4, and d 
 3, so
a37 
 4  (37 – 1)(3)

 4  (36)(3) or 112
The correct choice is C.
8. Draw a figure.

Of the 16 cubes on the front, only the 4 in the
center have just one blue face. It is the same on
each of the faces. There are 6 faces, so there are
6  4 or 24 cubes with just one blue face. The
correct choice is A.
9. To find the value of {{x}}, first find the value of {x}.
{x} 
 x2  1.
So {{x}} 
 {x2  1}.
And {x2  1} 
 (x2  1)2  1

 (x4  2x2  1)  1

 x4  2x2
The correct choice is D.
10. There are two distinct prime factors of 20. They
are 2 and 5 and their product is 10. There is only
one distinct prime factor of 16. It is 2.

2


triangles is 2  9 or 18. The correct choice is B.

16



x2 x  6
x6 0
x2  x  6 0
(x  3)(x  2) 0
At x 
 3 and x 
 2,
x
y
the inequality equals
3
6
0. Test values that are
2
0
greater than and less
0
6
than 3 and 2 to
3
0
determine for which
values the inequality
4
6
is less than 0.
For values of x that are between 2 and 3, the
inequality is true.
The correct choice is C.

The answer is 5.

x2

Chapter 15

1


 2(16) or 8.


 15  8 or 23.


 10.5

20

6

2

16

16

21

triangle is 2

 or 9. The area of both
6
2





 3(5) or 15.

Now, you need to determine which of the choices is
equal to 23. Calculate each one.

There are 12 intersections.
The correct choice is D.
4. The triangles are right triangles. The vertical
angles formed by the two triangles each measure
45°, since 180  135 
 45. The hypotenuse of the
triangles is the radius of the circle, which is 6,
since the diameter is 12. Using the relationships
of 454590 right triangles, the length of each
6
height and base must be . The area of one
1

5

6. Since 5 is prime,

494

Extra Practice
4. [f  g](x) 
 f(g(x))

 f(2x3  x2  x  1)

 2(2x3  x2  x  1)

 4x3  2x2  2x  2
[g  f](x) 
 g(f(x))

 g(2x)

 2(2x)3  (2x)2  (2x)  1

 16x3  4x2  2x  1

Lesson 1-1
Page A26
D 
 {2, 1, 2}; R 
 {4, 2, 4}; no
D 
 {3, 0.5, 0.5, 3}; R 
 {0.5, 3}; yes
D 
 {1, 0, 2, 5, 7}; R 
 {1, 2, 3, 5, 7}; yes
D 
 {2, 2.3, 3.2}; R 
 {4, 1, 3, 4}; no
f(4) 
 4(4)  2

16  2 or 14
6. g(3) 
 2(3)2  (3)  5

 2(9)  3  5

 18  3  5 or 26
1.
2.
3.
4.
5.

Lesson 1-3
Page A26

3


7. h(1.5) 
 
2(1.5)

1. x  2 
 0
x
2

3


 3 or 1
8. k(5m) 
 3(5m)2  3

 3(25m2)  3

 75m2  3

f (x )

x

O

Lesson 1-2

f (x ) 
 x  2

Page A26

2. 3x  4 
 0
3x 
 4

1. f(x)  g(x) 
 2x  1  x2  3x  1

 x2  5x  2
f(x)  g(x) 
 2x  1  x2 3x  1

x2  x
f(x)  g(x) 
 (2x  1)(x2  3x  1)

 2x3  5x2  5x  1

4

x 
 3

f (x )


g(x) 
 g(x)
f

f(x)

2x  1




x2  3x  1

2. [f  g](x) 
 f(g(x))

 f(4x2)

 3  4x2
[g  f ](x) 
 f(f(x))

 g(3  x)

 4(3  x)2

 36  24x  4x2
3. [f  g](x) 
 f(g(x))

f(x  9)

O

f (x ) 
 3x  4

x

3. 1 
 0, false
none

4. 4x 
 0
x
0

f (x )

f (x )

f (x ) 
 4x

1


 3(x  9)  1

O

1


 3x  2

x

O

x

f (x ) 
 1

[g  f ](x) 
 g (f(x))


 g 
3x  1
1

1


 3x  1  9
1


 3x  8

495

Extra Practice

5. 2x  1 
 0
2x 
 1

1

6. x  5 
 0
x 
 5
x 
 5

1

x 
 2



5. m 
 

3  or  3
1

y  6 
 3(x  (2))
3y  18 
 x  2
x  3y  16 
 0
6. x 
 10 is a horizontal line; perpendicular slope is
undefined.
y 
 15 or y  15 
 0

f (x )

f (x )
f (x ) 
 2x  1

O
x

O

1

x

f (x ) 
 x  5

2



7. m 
 

5  or  5
2

2

y  (7) 
 5(x  3)
5y  35 
 2x  6
2x  5y  29 
 0

Lesson 1-4
Lesson 1-6
Page A26
1. y 
 mx  b → y 
 2x  1
2. y  2 
 1(x  1)
y  2 
 x  1
y 
 x  3

Page A27
1.

y
48

1

3. y 
 mx  b → y 
 4x  3

47

4. y  (4) 
 0(x  (2))
y4
0
y 
 4

46
45

31
2

Enrollment
(millions)


5. m 
 
2  2
1





4 or  2
1

y  1 
 2(x  2)

42

1

41

y 
 2x  2
60

40


6. m 
 
0  (1)




0

or 6

y 
 mx  b → y 
 6x  6
7. y 
 0
8. y  0 
 1.5(x  10)
y 
 1.5x  15

x
0 1990 19911992 199319941995199619971998 1999 2000 2001
Year

2. Sample answer: y 
 0.6091x  1171.6
47.2 – 40.5


m

2001 – 1990

0.6091
y  40.5 
 0.6091(x  1990)
y 
 0.6091x  1171.6
3. Sample answer: y 
 0.6125x  1178; r 0.99
Enter the School Year data as List 1.
Enter the Enrollment data as List 2.
Perform a linear regression on the graphing
calculator.
4. Sample answer: 53.7 million; yes; the correlation
coefficient shows a strong correlation.
f(2011) 
 0.6125(2011)  1178
53.7 million

Lesson 1-5
Page A27
1. None of these; the slopes are neither the same nor
opposite reciprocals.
2. y  (2) 
 1(x  0)
y2
x
xy2
0
3.
y  3 
 2(x  1)
y  3 
 2x  2
2x  y  1 
 0
4. y 
 1 is a vertical line; parallel slope is undefined.
y 
 12 or y  12 
 0

Extra Practice

43

1

y  1 
 2x  1

6

1

44

496

Lesson 1-7

y

5.

y

6.

1  x  y  4

Page A27
f (x )

1.

x

g (x )

2.

O

g (x ) 
 |x  2|

x

O

O
h (x )

3.

x

y  |x | O

4.

Lesson 2-1

x

f (x )

Page A28
y

1.

2x  y  1 
 0

x

O

x

O
O

5.

6.

g (x )

x

3y 
 6x  3

k (x )

y 
 2x  1; consistent and dependent

y

2.

g (x ) 
 |5  2x |

O

x

(1, 5)

xy
4

y  3x 
 8

x

O

x
O
(1, 5); consistent and independent

Lesson 1-8

3.

3x  y 
 1

Page A27
y

1.

2y  6x 
 4

4.

y

y

x1

x
no solution; inconsistent
4. 5x  2y 
 1
x  2y 
 5
x  2y 
 5
1  2y 
 5
4x

 4
2y 
 6
x 
 1
y
3
(1, 3)
5. 2x  4y 
 8
2x  3y 
 8
2x  3y 
 8
2x  3(0) 
 8
y
0
2x 
 8
x
4
(4, 0)

xy3

x

O

x

2

O
O

O

y

2.
y

3.

y

y

x
O

x

4x  2y  6

497

Extra Practice

6. 8x  2y 
 2
3x  4y 
 23

→

8x  2y 
 2
8(1)  2y 
 2
2y 
 10
y
5
(1, 5)

16x  4y 

4
3x  4y 
 23
19x

 19
x

 1

Lesson 2-3
Page A28
1. 2x  y 
 1
y 
 1  2x

y 
 1  2(1)

1
(1, 1)
2. x  2y 
 5
2x  2y 
 2
3x

 3
x
1
(1, 2)
3.
3x 
 y  7
3x  7 
 y

Lesson 2-2
Page A28
1. x  y 
 6
x  z 
 2
yz
 8
xy
6
x5
6
x
1
(1, 5, 3)
2. 2x  2y  z 
 6
x  y  2z 
 6

yz
 8
yz
 2
2y

 10
y
5
x  z 
 2
1  z 
 2
z 
 3
→

x  2y  z 
 7
2x  2y  z 

6
x  4y

 13
3x  5y 
 18
→
x  4y 
 13
x  4y 
 13
x  4(3) 
 13
x
1
(1, 3, 2)
3. 2x  3y  z 
 1
x  y  z 
 4
3x  2y

 3
x y z 
 4
3x  2y  2z 
 3
3x  2y 
 3
3(1)  2y 
 3
2y 
 0
y
0
(1, 0, 3)

x  2y 
 1
x  2(1  2x) 
 1
3x  2 
 1
3x 
 3
x 
 1

x  2y 
 5
1  2y 
 5
2y 
 4
y
2
4y 
 5x
4(3x  7) 
 5x
12x  28 
 5x
7x 
 28
x 
 4

y 
 3(4)  7

 5
(4, 5)
4  7 1  (5)
4. A  C 
 1  0
5 1
28
6 4

4x  4y  2z 
 12
x  y  2z 
 6
3x  5y

 18






11 6
1
6
10 10
5. D  E 
 4  1 1  (5)
2  (3) 3  2
5 6


5 1
4(2)
4(0)
4(3)
6. 4B 

4(4) 4(3) 4(2)
0 12

 8
16 12
8



3x  5y 
 18
3x  12y 
 39
7y 
 21
y 
 3
x  y  2z 
 6
1  (3)  2z 
 6
2z 
 4
z
2



















7. impossible





 

2(7)
8. 2C  3A 
 2(0)
2(8)

→ 2x  2y  2z 
 8
3x  2y  2z 
 3
5x

 5
x
1
x  y  z 
 4
1  0  z 
 4
z 
 3
z
3



2(5)
3(4) 3(1)
2(1)  3(1)
3(5)
2(4)
3(2)
3(6)

14  12 10  (3)
03
2  15
16  6
8  18
26 13

 3
17
22
26
9. impossible







5
4 1

2
2 3
1(4)

(5)(2)
1(1)  (5)(3)


3(4)  2(2)
3(1)  2(3)

10. ED 


31










 14 14
16
3



Extra Practice

498





 

7
0 3
 0
3
2
8
2(7)

0(0)

(3)(8)


4(7)  (3)(0)  2(8)
10
22


44 15
BC  D 
 10  (4)
44  2
6
23


42 18

11. BC 


24



5
1
4
2(5)  0(1)  (3)(4)
4(5)  (3)(1)  2(4)





0
4. 1
 3 6 5 2
0
0 1
3
2 1
3
6
5
2


3 2 1 0
W
(3, 3), X
(6, 2), Y
(5, 1), Z
(2, 0)



 















y
8

W (3, 3)
Z (2, 0)
4
X (6, 2)
Y (5, 1)
Z (2, 0)
8 4 O
4
8x
Y (5, 1)
X
(6,
2)
4

22  1
15  3





W (3, 3)

8

Lesson 2-4

5. Rot90  Ry 
 x 
 0 1  0 1
1 0
0

 1
1 0
0 1





Page A28
5 2 2


4 6

3

1. 0.5




1



1



1
22
1
12



1 1
2



A
 22, 12 , B
(1, 2), C
(1, 3)





10 01  42

3

 

2 6


3 7

  42

6
2
3 7



F
(4, 2), G
(2, 3), H
(6, 7)

y

y

B (2, 4)

8

B (1, 2)

F (4, 2) 4

x

O
C (1, 3)

2 12,

A (

F (4, 2)

O
8 4
G (2, 3)4

112

)

A (5, 3)

H (6, 7) 8

4

8x

G (2, 3)
H (6, 7)

C (2, 6)

2.



Lesson 2-5

6 3 4 1
2 2 2 2

2 5
7 5
3 3 3 3

 



Page A29



4 5 6 1
5 8
4
2
J
(4, 5), K
(5, 8), L
(6, 4), M
(1, 2)





L (4, 7) y
8
L(6, 4)
4

3 7

 3(2)  (11)(7)
11 2

 83
3
5
2.

 3(2)  (7)(5)
7 2

 29
5
0
1

 5(6)  2 (0)
3.
1
2 6
1.

M (1, 5)
M (1, 2)

8 4 O
4
K (3, 5)
8
K (5, 8)

4


 

8x

J (6, 2)
J (4, 5)


 30
0
2
1
4. 3
1 2
5 1 3
1 2
1

 1
0 3 2 2 3
1 3
5 3
5 1

 1(5)
 0(19)  2(2)

1
3
2
1
5.
1
4 2
3 3 4
134
1  2 4 2

 1 2
3 4
3 3
3 4

0
5 3
0
3. 1 0  5 3


2
0 1
2 8 4
8 4



 

 



N
(5, 2), P
(3, 8), Q
(0, 4)
P (3, 8)
N (5, 2)

y
8
4

P (3, 8)
N (5, 2)

O

4
8x
4 Q (0, 4)
Q (0, 4)

8 4


 1(11)  3(19)  2(6)

 34

499

Extra Practice

6.

7.

4
0 1
5
3
6
2 5
2
3 6
5 6
5
3

4
0
 (1)
5 2
2 2
2 5

 4(36)  0(22) 1(19)

 163



1
5





9.

54

1
10
5

0
4





2

5





(0, 1)

1

2.

3

2

5

2

y
(4, 7)

 
3

(0, 3)

61 53 
 61 53
1

33

5
1




31



1

33
2

11

5

33
1

11

(0, 2)



2
2



1
8 2 2
1
3



31





1
2
x


 8 2 2
2
y
1
3
x
4


y
5

 



22
 6


(1, 1)

 

O

4 2
x


 6
3
y
1
7
1
1 2
1
2

 1
0 3 4
3 4



12.

1

10



   





2
1

(2, 1)

24

13.
1
2
4

1
3

1
3 1
1
0
4
2



 24



1

Extra Practice

vertices: (1, 1), (1, 6), (1, 2), (1, 1)
f(x, y) 
 x  y
f(1, 1) 
 (1)  1 or 0 → minimum
f(1, 6) 
 (1)  (6) or 7 → maximum
f(1, 2) 
 1  (2) or 1
f(1, 1) 
 1  (1) or 0 → minimum

1

10

xy 
 2
1

x
4
1



y
13
3
3 1 
 1 3 1
10 4
4
2
2

   







 xy 
  10 3
4
1

1
3

xy 




x

(1, 6)

 xy 
 31 24 6

7

31 24 43

5
,
2

(1, 1)
(1, 2)

(4, 5)

1
4 2
3
1

x

vertices: (0, 3), (4, 7), (4, 2), (0, 2)
f(x, y) 
 2x  y
f(0, 3) 
 2(0)  3 or 3 → minimum
f(4, 7) 
 2(4)  7 or 1
f(4, 2) 
 2(4)  2 or 6 → maximum
f(0, 2) 
 2(0)  2 or 2
y
3.

   



1

(4, 2)

O

2
x


 22
2
y
6
2 2 
 1 2 2
8 1
1
3
3

11.

x

vertices: (3, 1), (0, 1), (3, 7)
f(x, y) 
 4x  3y
f(3, 1) 
 4(3)  3(1) or 15
f(0, 1) 
 4(0)  3(1) or 3 → minimum
f(3, 7) 
 4(3)  3(7) or 33 → maximum



2

(3, 1)

O



43 65 
 2 43 65

1

3
1

1

5



1
5
6
3
4

3
6

y
(3, 7)

1
0
4
0

 4
0 5 10
10
1
0


 101
1

8
4



10.

1.



3

5
1
5




8.

Page A29

3 1 
 1 3 1
5 1
1
2
2

1
2
1

Lesson 2-6



1
2

 134

 
5

2

1

500

4. Let x 
 the number of video store hours.
Let y 
 the number of landscaping company
hours.
y
y0
16
x
4
x4
x  y  15
(4, 11)
12

Lesson 2-7
Page A29
1. Let x 
 the number of cars.
Let y 
 the number of buses.
y
6x  30y  600
x  y  60
60 x  y 
 60
x0
6x  30y 
 600
40
y0
20
x
0
(0, 0)

O

(0, 20)

8
4

O

(50, 10)

1. f(x) 
 4x
f(x) 
 4(x)
f(x) 
 4x
yes
2. f(x) 
 x2  3
f(x) 
 (x)2  3
f(x) 
 x2  3
no

x  y 
 45
20

30

40x

C(x, y) 
 2.95x  2.95y
C(10, 20) 
 2.95(10)  2.95(20) or 88.50
C(15, 30) 
 2.95(15)  2.95(30) or 132.75
C(25, 20) 
 2.95(25)  2.95(20) or 132.75
alternate optimal solutions
3. Let a 
 the number of company A’s cards.
Let b 
 the number of company B’s cards.
a  b  90
b
80
a 
 40
a  40
b  25
60

40

60

f(x) 
 (x2  3)
f(x) 
 x2  3


f(x) 
 

3x3 

1

1


f(x) 
 
3(x)3
1

1


f(x) 
 
3x3

yes
4. xy 
 2
x-axis

(40, 50)
a  b 
 90
(65, 25)
(40, 25)
20 b 
 25
20

f(x) 
 (4x)
f(x) 
 4x

1


3. f(x) 
 
3x3


f(x) 
 
3x3

→

y-axis

40

O

16x

Page A30

(10, 20)

10

8
12
y
0

Lesson 3-1

(25, 20)

10

4

C(x, y) 
 5x  7y
C(4, 0) 
 5(4)  7(0) or 20
C(4, 11) 
 5(4)  7(11) or 97
C(15, 0) 
 5(15)  7(0) or 75
The maximum earnings is $97.

C(x, y) 
 3x  8y
C(0, 0) 
 3(0)  8(0) or 0
C(0, 20) 
 3(0)  8(20) or 160
C(50, 10) 
 3(50)  8(10) or 230
C(60, 0) 
 3(60)  8(0) or 180
The maximum income is with 50 cars and 5 buses.
2. Let x 
 number of gallons of black walnut.
Let y 
 number of gallons of chocolate mint.
y  2x
y
40
y 
 2x
y  20 thousand
x  y  45 thousand
(15,
30)
30
20

(15, 0)

(4, 0)

40 60 x
(60, 0)

20
y
0

x  y 
 15

y
x

80a

y 
 x

C(a, b) 
 0.30a  0.32b
C(40, 25) 
 0.30(40)  0.32(25) or 20
C(40, 50) 
 0.30(40)  0.32(50) or 28
C(65, 25) 
 0.30(65)  0.32(25) or 27.5
The maximum profit is with 40 cards from
company A and 50 cards from company B.

y 
 x, y 
 x
5. y  x2 
 3
→
x-axis
y-axis
y
x
y 
 x

ab 
 2
a(b) 
 2
ab 
 2
ab 
 2; no
(a)b 
 2
ab 
 2
ab 
 2; no
(b)(a) 
 2
ab 
 2; yes
(b)(a) 
 2
ab 
 2; yes
b  a2 
 3
(b)  a2 
 3
b  a2 
 3; no
b  (a)2 
 3
b  a2 
 3; yes
(a)  (b)2 
 3
a  b2 
 3; no
(a)  (b)2 
 3
a  b2 
 3; no

y-axis

501

Extra Practice

2x2

6. y2 
 7  1

2a2

→

2a2

(b)2 
 7  1

x-axis

b2 

b2 


y-axis

b2 

y
x

(a)2 

a2 


y 
 x

(a)2 

a2 


x-axis, y-axis
7. x 
 4y
x-axis

→

y-axis
y
x
y 
 x
y-axis
8. y 
 3x
x-axis

→

y-axis
y
x
y 
 x

none of these
9. y 
 
x2  1
x-axis

y-axis
y
x
y 
 x

4. g(x) is a translation of f(x) right 1 unit and
compressed vertically by a factor of 2.

b2 
 7  1

→

5. y 
 2x  5
1

2a2
  1; yes
7
2(a)2
  1
7
2a2
  1; yes
7
2(b)2
  1
7
2b2
  1; no
7
2(b)2
  1
7
2b2
  1; no
7

Lesson 3-3
Page A30
y

2.
y

a 
 4b
a 
 4(b)
a 
 4b; no
(a) 
 4b
a 
 4b; yes
(b) 
 4(a)
b 
 4a; no
(b) 
 4(a)
b 
 4a; no

|x  4|

y  x2  1

O
3.

x

y

O

x

y

4.

x

O

y  2|x  1|

b 
 3a
(b) 
 3a
b 
 3a; no
b 
 3(a)
b 
 3a; no
(a) 
 3(b)
a 
 3b; no
(a) 
 3(b)
a 
 3b
a 
 3b; no

y  

x2

5.

y

x

O
y

6.
y

(x  2)2  1

y  |x |  2

O

b 
 
a2  1
(b) 
 
a2  1
b 
 
a2  1
b 
 
a2  1; yes
b 
 
(a)2
1

7. Case 1
x  2  3
(x  2)  3
x  2  3
x  1
{x1  x  5}
8. Case 1
4x  2  18
(4x  2)  18
4x  2  18
4x  16
x  4
{xx  4 or x  5}
9. Case 1
5  2x 9
(5  2x) 9
5  2x  9
2x  14
x 7
{x2 x 7}


a2  1; yes
2 
(b)

1
2
a 
 (b)


1; no
2


1
(a) 
 (b)
2
a 
 b

1
2  1; no
a 
 b

b

(a) 


x-axis, y-axis

Lesson 3-2
Page A30
1. g(x) is a translation of f(x) up 2 units.
2. g(x) is the graph of f(x) expanded vertically by a
factor of 3.
3. g(x) is a translation of f(x) left 4 units and down
3 units.
Extra Practice

y

1.

502

O

x
Case 2
x  2  3
x23
x5

Case 2
4x  2  18
4x  2  18
4x  20
x5

Case 2
5  2x 9
5  2x 9
2x 4
x  2

x

10. Case 1
x  1  3  1
(x  1)  3  1
(x  1)  4
x  1 4
x 5
{xx  5 or x  3}
11. Case 1
2x  3 27
(2x  3) 27
2x  3  27
2x  30
x  15
{x15 x 12}
12. Case 1
3x  4  3x  0
(3x  4)  3x  0
6x  4  0
6x  4
2
x  3
all real numbers

f(x) 
 x2  6
y 
 x2  6
x 
 y2  6
x  6 
 y2
y 
 x

6
f 1(x) 
 x  6; No, it is not a function.
7.
f(x) 
 (x  2)2
y 
 (x  2)2
x 
 (y  2)2
x

y2
y 
 2 x
f 1(x) 
 2 x; No, it is not a function.
x
8.
f(x) 
 2

Case 2
x  1  3  1
(x  1)  3  1
x3

Case 2
2x  3
2x  3
2x
x

6.

27
27
24
12

x

y 
 2

Case 2
3x  4  3x  0
(3x  4)  3x  0
4  0; true

y

x 
 2
2x 
 y
y 
 2x
f 1(x) 
 2x; Yes, it is a function.
9.

1


f(x) 
 
x4
1


y

x4
1


x

y4

Lesson 3-4

1

y  4 
 x
1

y 
 x  4

Page A30
1.

1

f 1(x) 
 x  4; Yes, it is a function.

f (x )

2.
f 1(x )

f 1(x )

x
O

3.

10.
f (x )

f (x )

O

f(x) 
 x2  8x  2
y 
 x2  8x  2
x 
 y2  8y  2
x  2 
 y2  8y
x  2  16 
 (y  4)2
x

 18 
 y  4
4 x
 18 
 y
f 1(x) 
 4 x;
 18 No, it is not a function.
11.
f(x) 
 x3  4
y 
 x3  4
x 
 y3  4
x  4 
 y3
3

x4
y
3
f 1(x) 
 ;
x  4 Yes, it is a function.

f (x )

x

f(x )
f 1(x )

O

x

12.

f (x )

3


f(x) 
 
(x  1)2
3


y 
 
(x  1)2
3

4.


x 
 
(y  1)2

f(x) 
 4x  5
y 
 4x  5
x 
 4y  5
x  5 
 4y

3

(y  1)2 
 x


x
3
y 
 1 
x
3
f 1(x) 
 1 
x; No, it is not a function.
y1


x5

y 
 4
x5

f 1(x) 
 4; Yes, it is a function.
5.

3

f(x) 
 2x  2
y 
 2x  2
x 
 2y  2
x  2 
 2y
x2


y

2
x2


f 1(x) 
 
2 ; Yes, it is a function.

503

Extra Practice

4. yes y 
 x  5

Lesson 3-5

x5
x  
3x2 
2x  1

x2  3x

Page A31
1. Yes; the function approaches 1 as x approaches 2
from both sides.
2. No; the function is undefined when x 
 3.
3. No; the function is undefined when x 
 1.
4. Yes; the function approaches 1 as x approaches 3
from both sides.
5. jump discontinuity
6. y →  as x → , y →  as x → 
7. y →  as x → , y →  as x → 
8. y → 0 as x → , y → 0 as x → 

5x  1
5x  15
16

Page A31
1. y 
 kx
8 
 k(2)
4
k
2.

g 
 kw
10 
 k(3)
10

 3 
 k

Page A31
3.

1. abs. max.: (1, 2)
2. rel. min.: (3, 0), rel max.:
(1, 3), abs. min.: (2, 1)

5.

6.

x  x

3

2
1  x

 84

rt 
 84

84 
 k

r(7) 
 84
r 
 12
y 
 3xz
y 
 3(5)(10)
y 
 150

k

y 
 x2
243 
 k

x
2
y
 —



6

 5 
 w

k

k

3x

x

10

4 
 3 w
t 
 r


27 
 
(3)2

3x


y

x2

10

g 
 3 w

k

4. y 
 kxz
60 
 k(5)(4)
3
k

Page A31

y 
 4x
y 
 4(9)
y 
 36

t 
 r
6 
 1
4

Lesson 3-7

y


16


As x → , 
x  3 → 0.
So, the graph of f(x)
will approach that of
y 
 x  5.

Lesson 3-8

Lesson 3-6

1. x 
 2

16


→ y
x5
x3

a 
 kbc3
36 
 k(3)(2)3
1.5 
 k

243

y 
 x
2
yx2 
 243
y(5)2 
 243
y 
 9.72
a 
 1.5bc3
a 
 1.5(5)(3)3
a 
 202.5

as x → , y → 3; y 
 3
2x2

2. x 
 3


y

x3

y


2x2

x2
——
x
3


x2  x2

y


2

3
1
  
x2
x

Lesson 4-1
Page A32
1. yes; f(x) 
 x3  7x2  2x  40
f(2) 
 (2)3  7(2)2  2(2)  40

 8  28  4  40

0
2. no; f(x) 
 x3  7x2  2x  40
f(1) 
 (1)3  7(1)2  2(1)  40

 1  7  2  40

 36
3. no; f(x) 
 x3  7x2  2x  40
f(2) 
 (2)3  7(2)2  2(2)  40

 8  28  4  40

 24
4. yes; f(x) 
 x3  7x2  2x  40
f(5) 
 (5)3  7(5)2  2(5)  40

 125  175  10  40

0

no horizontal
asymptotes since as
x → , y is undefined
x5


3. h(x) 
 
x2  6x  5




x5

(x  5)(x  1)

x 
 5, x 
 1

x5


y

x2  6x  5

y


y


x
5


x2  x2
——
x2
6x
5

 
 
x2
x2
x2
1
5
  
x
x2
——
6
5
1    
x
x2

as x → , y → 0; y 
 0

Extra Practice

504

5. (x  3)(x  4) 
 0
x2  7x  12 
 0; even; 2
6. (x  (2))(x  (1))(x  2) 
 0
(x  2)(x  1)(x  2) 
 0
(x  2)(x2  x  2) 
 0
x3  x2  4x  4 
 0; odd; 3
7. (x  (1.5))(x  (1))(x  1) 
 0
(x  1.5)(x  1)(x  1) 
 0
(x  1.5)(x2  1) 
 0
x3  1.5x2  x  1.5 
 0; odd; 3
8. (x  (2))(x  (i))(x  i) 
 0
(x  2)(x  i)(x  i) 
 0
(x  2)(x2  1) 
 0
3
x  2x2  x  2 
 0; odd; 1
9. (x  (3i))(x  3i)(x  (i))(x  i) 
 0
(x  3i)(x  3i)(x  i)(x  i) 
 0
(x2  9)(x2  1) 
 0
x4  10x2  9 
 0; even; 0
10. (x  (1))(x  1)(x  2)(x  3) 
 0
(x  1)(x  1)(x  2)(x  3) 
 0
(x2  1)(x2  5x  6) 
 0
4
x  5x3  5x2  5x  6 
 0; even; 4

h2  12h 
 4
h2  12h  36 
 4  36
(h  6)2 
 40
h  6 
 210

h 
 6 210

6. x2  9x  1 
 0
x2  9x 
 1
5.

81

2


x  92

7. b2  4ac 


8.

(3) 
121

x

2(4)
3 11
x 
 8
7
x 
 4 or x 
 1
b2  4ac 
 (2)2 

w


or 121; 2 real

4(1)(10) or 44; 2 real

2 
44

2(1)

w 
 1 11

9. b2  4ac 
 (5)2  4(12)(6) or 263; 2 imaginary
t


 (5) 263


2(12)
5

i 263


t 
 24
10. b2  4ac 
 (6)2  4(1)(13) or 88; 2 real

1. x2  4x  5 
 0
x2  4x 
 5
x2  4x  4 
 5  4
(x  2)2 
 9
x2
 3
x2
3
x
5
2. x2  6x  8 
 0
x2  6x 
 8
x2  6x  9 
 8  9
(x  3)2 
 1
x3
 1
x3
1
x 
 2
3. m2  3m  2 
 0
m2  3m 
 2
9
 3m  4
3 2
m  2
3
m  2




2

6

x 
  3 22

11. b2  4ac 
 (4)2  4(4)(1) or 0; 1 real
 (4)

0



n

2(4)

x  2 
 3
x 
 1

1

n 
 2
12. b2  4ac 
 (6)2  4(4)(15) or 276; 2 real
6


276

6

2 69


3

69



x

2(4)

x 
 8
x 
 4

x  3 
 1
x 
 4

Lesson 4-3

9

4

17





88


x

2(1)


 4

m

4.


77

2
9
77

x 
 2 2
(3)2  4(4)(7)

9

Page A32




77


 4

x  2 


Lesson 4-2

m2

81

x2  9x  4 
 1  4

Page A32
1. 2 1 10
8
2 16
1
8  8
x  8, R8
2. 1 1 3
4 1
1 2
2
1 2
2 1
x2  2x  2, R1
3. 1 1
0 3 5
1
1
2
1 1 2 3
x2  x  2, R3

17


2
3
17



2
2

 8a  6 
 0
a2  4a  3 
 0
a2  4a 
 3
2
a  4a  4 
 3  4
(a  2)2 
 7
a  2 
 7

a 
 2 7


2a2

505

Extra Practice

4. 4 1 2 7 3 4
4
8
4
4
1
2
1
1 0
x3  2x2  x  1
5. f(x) 
 x2  2x  8
f(4) 
 (4)2  2(4)  8

 16  8  8

 0; yes
6. f(x) 
 x3  12
f(1) 
 (1)3  12

 1  12 or 13; no
7. f(x) 
 4x3  2x2  6x  1
f(1) 
 4(1)3  2(1)2  6(1)  1

 4  2  6  1

 7; no
8. f(x) 
 x4  4x2  16
f(4) 
 (4)4  4(4)2  16

 256  64  16

208; no

3. p:
q:

1,
1

2

p
:
q

1,

2

r
1
1
2
2

p
:
q

1,

2,

4,

1
,
2

8,

1
,
3

2
,
3

4
,
3

8
,
3

6
6
6


1
7
5


22
29
27


4
33
23


8
25
15


3

6

3

24

12

0

1

2

6

4

24

16

0

r
1
1


Lesson 4-4

2 1

rational roots: 3, 2

Page A32
1. p:
q:

1,
1

2,

3,

6

p
:
q

1,

2,

3,

6

r
1
1
2
2
3
3

1
1
1
1
1
1
1

2
3
1
4
0
5
1

5. 2 or 0 positive
f(x) 
 x3  4x2  x  4
1 negative
5
2
6
3
5
10
2

r
1

6
8
0
0
16
24
0

r
1

10

1

2

r
1
1

2
2
2

1
1
3

2
3
5

3
0
8

1
1
9

1

2

2

0

2

2

0

2

2

3

 2

9

13

4

1

2

1
1

4
5

1
4

1
1

1

1
2

9

3
5

93

5
0

935

rational zeros: none
7. 2 or 0 positive
f(x) 
 4x3  7x  3
1 negative

1

rational root: 2

r

4

0

7

3

3
2

4

6

2

0

4x2  6x  2 
 0
(4x  2)(x  1) 
 0
1

x 
 2, x 
 1
3 1

rational zeros: 2, 2, 1

Extra Practice

4
0

x2  5x  4 
 0
(x  4)(x  1) 
 0
x 
 4, x 
 1
rational zeros: 1, 1, 4
6. 2 or 0 positive
f(x) 
 x4  x3  3x2  5x  10
2 or 0 negative

rational roots: 3, 1, 2
2. p: 1
q: 1, 2
1,

2
0
2
10
6

0
2
0
6
2

rational root: 1
4. p: 1, 2, 4, 8
q: 1, 2, 3, 6

2

p
:
q

1
2
0
3
1

1
1
1
1
1

506

10
10

9360

1

6

Lesson 4-6

8. 3 or 1 positive
f(x) 
 x4  x3  4x  4
1 negative
r
1

4

1
1

1

1
0

5

4
4

76

0
0

20

Page A33

4
0

300

6

x

1.

rational zero: 1

7

y1

2.

Page A33
r
2
1
0
1
2
3

2
2
2
2
2
2
2

4
8
6
4
2
0
2

5
11
1
5
7
5
1

r
2
1
0
1
2

1
1
1
1
1
1

0
2
1
0
1
2

0
4
1
0
1
4

y


 y 
 
y1

5

r1

3.



5
13
6
5
4
3

4

1

r1

r 1


  
 
2

1

4

4

1



2

2t  t2

4.



2

t2  t2

2(t  2) 
 4  1
2t  4 
 3
2t 
 7
t 
 3.5
1
4
1
   
 
5. 
3w
5w
15
5  12 
 w
17 
 w; w



Test w 
 1:

r
2
1
0
1
2

0

1

3(1)

1 and 2
3.

y1
0
y 
 1

5(r  1)  4(r  1) 
 1
5r  5  4r  4 
 1
r 
 10

1 and 0, 2 and 3
2.

4

x3
0
x
3

7y  4(y  1) 
 y2
3y  4 
 y2
y2  3y  4 
 0
(y  4)(y  1) 
 0
y4
0
or
y
4

Lesson 4-5

1.

x
5

6  x2 
 5x
x2  5x  6 
 0
(x  2)(x  3) 
 0
x2
0
or
x
2

4

1

4

 3  5
1
1
1
1
1
1

0
2
1
0
1
2

1
3
0
1
0
3

4
2
4
4
4
10

2
6
2
2
6
22

17

 1
5




Test w 
 1:

w

507

true

1

3(1)

Test w 
 18:

2 and 1, 1 and 0
4-6. Use the TABLE feature of a graphing calculator.
4. 0.3, 1.3
5. 2.2, 0.3, 1.2
6. 1.3, 1.3

? 1
15
? 1
 15
? 1
 1
5;



5(1) 

1 ? 1



5(1)  15
1
1 ? 1
   

3
5
15
8 ? 1
  ; false
15
15
1
1
? 1
   
3(18)
5(18)
15
1
1 ? 1
   

54
90
15
4 ? 1
 
; true
135
15

0 or w  17

Extra Practice

x2

x

6.

x4

5. 2x
32

2x  3  4
2x  1




x6

(x  6)(x  2) 
 x(x  4)
x2  8x  12 
 x2  4x
12 
 4x
3 
 x; x 0 or 6

1

x  2
2x  3  0
2x  3
3
x  2

(1)  2 ? (1)  4

(1)  6
? 5
3 7 ; false
12 ? 14


1
16
3
?
1  5; true
42
44
 ? 
4
46
1
 ? 0; false
2
72 ? 74


7
76
5
 ? 3; true
7


Test x 
 1: 
1

Test x 
 1:
Test x 
 4:
Test x 
 7:
0

x

?

Test x 
 2: 2(2)

32
?
1
  2; meaningless
?
Test x 
 0: 2(0)


32
?
3
  2; true
?
Test x 
 1: 2(1)


32
?

  2; false
5
3

4

6. 2
6a    4
6a  2  256

3 or x  6

6a  258
a  43
6a  2  0
6a  2
1
a  3

Lesson 4-7
Page A33
1. 2
t
 3 
 4
2  3t 
 16
3t 
 14
14
t 
 3

1

Solution: 2  x  2

?
Test a 
 0: 
6(0) 
2
4
4
?
2
  4; meaningless
4

Check:



14
2  3

3


4


4
16
4
4

?
6(1) 
2
4
Test a 
 1: 
4
?
4
  4; false
4

?


2
4
Test a 
 44: 6(44)
4
?
262
  4; true
4

2. 4  x
2
1
Check: 4  11

2
1
x
 2 
 3

43
1
x2
9
1
1
x 
 11
3
3
3. 
y  7  10 
 2 Check: 505

 7  10 
 2
3

y  7 
 8
8  10 
 2
y  7 
 512
2
2
y 
 505
4.
a
 1  5 
 a


6
a  1  10a

 1  25 
 a  6
10a
 1 
 30

a
1
3

a1
9
a 
 10
Check: 10
   5 
 10
1
6

35
2
2 2
no real solution

Solution: a  43

Lesson 4-8
Page A33
1. f(x) 
 0.75x  2
2. f(x) 
 x3  x2  x  2
3. Sample answer: f(x) 
 0.51x2  0.02x  0.79
4a. Sample answer: y 
 1.632x  99.275
4b. Sample answer: about 122.123 thousand
f(x) 
 1.632x  99.275
f(15) 
 1.632(14)  99.275 
 122.123

Lesson 5-1
Page A34
1. 13.75° 
 13°  (0.75  60)


 13°  45


 13° 45


Extra Practice

508

2. 75.72° 
 75°  (0.72  60)


 75°  43.2


75°  43
  (0.2  60)

 75°  43
  12

 75° 43
 12
3. 29.44° 
 29°  (0.44  60)


 29°  26.4


 29°  26
  (0.4  60)

 29°  26
  24

 29° 26
 24

1


2. sin v 
 
csc v
1


sin v 
 
2.5 or 0.4

3. (AC )2  (BC )2 
 (AB)2
142  162 
 (AB)2
452 
 (AB)2
452
 
 AB; 452
 or 2113

side opposite


sin A 
 
hypotenuse




6. 38° 15
 10 
  38°  15



  10


1°

3600


  38.253°




8. 51° 14
 32 
 51°  14



 51.242°
9. 850°  360° 
 490°
490°  360° 
 130°
130°; II
10. 65°  360° 
 295°
295°; IV
11. 1012°  360° 
 652°
652°  360° 
 292°
292°; IV
12. 578°  360° 
 218°
218°; III
13. 180°  126° 
 54°
14. 480°  360° 
 120°
120°  360° 
 240°
240°  180° 
 60°
60°
15. 642°  360° 
 282°
360° 282° 
 78°
78°
16. 1154°  360° 
 794°
794°  360° 
 434°
434°  360° 
 74°
74°

  32


1°

60


1°

3600

21

13

hypotenuse

21

13

1

13

sec A 
 14 or 7

1

13

side adjacent


cot A 
 
side opposite
14

7


cot A 
 1
6 or 8

4. (AC )2  (BC )2 
 (AB)2
252  (BC )2 
 282
BC 2 
 159
BC 
 159






side opposite


cos A 
 
hypotenuse


159

cos A 
 28

side opposite


csc A 
 
side opposite


159

 or 
csc A 
 
159

159

hypotenuse


cot A 
 
side opposite


sin A 
 
hypotenuse



side adjacent
25

sin A 
 2
8

tan A 
 
side adjacent

hypotenuse

281

59

28

tan A 
 2
5

sec A 
 
side adjacent
28

side adjacent

251

59

25

sec A 
 25

 or 
cot A 
 

159
159

5. (AC)2  (BC )2 
 (AB)2
92  62 
 (AB)2
117 
 (AB)2
117
 
 AB; 117
 or 313

side opposite


sin A 
 
hypotenuse
213


6

 or 
sin A 
 
313

13

side adjacent


cos A 
 
hypotenuse
313


9

 or 
cos A 
 
313

13

side opposite


csc A 
 
side opposite

6

csc A 
 6 or 2


tan A 
 
side adjacent

hypotenuse


sec A 
 
side adjacent
313


hypotenuse

313


2

tan A 
 9 or 3

13

sec A 
 9 or 3

13


side adjacent


cot A 
 
side opposite
9

3

cot A 
 6 or 2

Lesson 5-3
Page A34

y

1. tan v 
 x
Since tan v 
 0, y 
 0.
x

x

cot v 
 y 
 0
cot v is undefined.
2. Sample answers: 90°, 270°

Page A34
1


1. cot v 
 
tan v
1
—
5

6

csc A 
 16 or 8

8

hypotenuse

Lesson 5-2

cot v 


16


sec A 
 
side adjacent

1°


 107.213°


csc A 
 
side opposite


tan A 
 1
4 or 7



7. 107° 12
 45 
  107°  12


60
   45
 3600 
1°

71

13

14

 or 
cos A 
 
113
21

13

side opposite


tan A 
 
side adjacent

4. 87.81° 
 87°  (0.81  60)


 87°  48.6


 87°  48
  (0.6  60)

 87°  48
  36

 87° 48
 36
1°
1°


5. 144° 12
 30 
 144°  12


60
   30
 3600 

 144.208°
1°

60


81

13

16

 or 
sin A 
 
113
21

13

side adjacent


cos A 
 
hypotenuse

6

or 5

x

cos v 
 r
Since cos v 
 0, x 
 0.
On the unit circle, x 
 0 when v 
 90° or v 
 270°.

509

Extra Practice

3. r 
 
x2  y2

 
(1)2 
 (2
)2

 5

y

2.

a

cos 87° 
 19
19 cos 87° 
 a
1.0  a

x

sin v 
 r

5
2

2

sin v 
  or 
5

y
tan v 
 x
2

tan v 
 
1
r
sec v 
 x

5

sec v 
 
1

5

cos v 
 r
1

csc v 

cot v 


or  5


5


cos v 
  or  
csc v 


or 2

cot v 


5

r

y
5

 2
x

y
1
1
 or 
2
2

3.

5

y

16.5

16.5


c

cos 65.4°

c  39.6
4.

12

12


a

tan 42.5°

a  13.1

x

2


2

22

2
y
tan v 
 x
2

tan v 
 
2 or 1
r
sec v 
 x
22


sec v 
 

2 or 2

2


2

cos v 
  or  
csc v 

csc v 

cot v 

cot v 


y

b


sin 75° 
 
5.8

2
2
2
r

y
22

 or 2

2
x

y
2
 or 1
2

5.8 sin 75° 
 b
5.6  b
20  b

20  b

b

tan 42° 
 x
tan 42° 


or

5

cos v 
 
29

r
csc v 
 y
29

csc v 
 2
x
cot v 
 y
5
cot v 
 2

or

b

48˚


x

tan 48°

42˚

b
––
20  b

tan 48°

x

20 tan 42°  b tan 42° 
 b tan 48°
20 tan 42° 
 b(tan 48°  tan 42°)

x

229


29

20 feet

6. tan 48° 
 x

cos v 
 r

2

sin v 
 
29

y
tan v 
 x
2
tan v 
 5
r
sec v 
 x
29

sec v 
 5

b

sin B 
 c

5.

x2  y2
5. r 
 

 
(5)2 
(2)2

 29

sin v 
 r

b

tan B 
 a
tan 42.5° 
 a

cos v 
 r

sin v 
  or 

a

cos B 
 c
cos 65.4° 
 c

4. r 
 
x2  y2

 
(2)2 
 (2)2

 22

sin v 
 r

a

cos B 
 c

529


29

20 tan 42°

(tan 48°  tan 42°)

85.7


b
b; 85.7 ft

Lesson 5-5


x2  y2

6. r 


 
(4)2 
 (3)2

5
y
sin v 
 r
3
sin v 
 5
y
tan v 
 x

tan v 
 
sec v 


Page A35
3

cos v 


cos v 
 




csc v 

cot v 


5



3

1

1


sec A 
 
cos A

5

3
x

y

sec A 


1

1

2

or 2




4

sec v 
  4

1

2. Let A 
 cos1 2. Then cos A 
 2.

r

r

x

3

sin arcsin 4 
 4

4

5

csc v 
 y
3

4

3

1. Let A 
 arcsin 4. Then sin A 
 4.

x

r

cot v 
  3

sec cos
tan1

1. Then tan A 
 1.
3. Let A 

tan(tan1 1) 
 1
a

b

5. sin B 
 c

4. tan A 
 b

Lesson 5-4

38

17

tan A 
 2
5
38

Page A34
1.

a

tan A 
 b
a

15

tan 38° 

15 tan 38° 
 a
11.7  a
Extra Practice

sin B 
 1
9

510

17

A 
 tan1 25

B 
 sin1 1
9

A

B

56.7°

63.5°

a

a

cos B 


7. A 
 180°  (60°  75°) or 45°

7. cos B 
 c

6. cos B 
 c
24

30

cos B 


1

sin B sin C


K 
 2a2 
sin A

9.2

12.6

1

sin 60° sin 75°

B 
 cos1 30


B 
 cos1 
12.6


K 
 2(8)2 
sin 45°

B

B

K

9.2

24

36.9°

8. tan A 

tan A 


43.1°

1
1

K 
 2(16)(12) sin 43°
K

28.4


A 
 tan1 
36.5

A

37.9 units2

8. K 
 2bc sin A

a

b
28.4

36.5

65.5 units2

37.9°

Lesson 5-7
Lesson 5-6
Page A35
1. Since 145°  90°, consider Case II.
5  10; no solution
2. Since 25° 90°, consider Case I.
b sin A 
 10 sin 25°
4.226182617
9  4.226182617; 2 solutions
a
b
 
 
sin A
sin B

Page A35
1. C 
 180°  (75°  50°) or 55°
a

sin A
7

sin 75°

b




sin B


 b sin 50°

a

sin A
7

sin 75°

7 sin 50°


c

sin 75°

b 5.551472956
C 
 55°, b 
 5.6, c 
 5.9
2. B 
 180°  (97°  42°) or 41°
b




sin B
b




sin 41°
12 sin 41°


b

sin 42°






b


b

sin B
b

sin 49°
10 sin 49°

sin 99°

a




sin A
a




sin 22°
25 sin 22°


a

sin 117°

6

sin 25°

5.936340197

B 
 sin1

a

c

c


6
10 sin 25°

B 44.77816685
B 180°  44.8° or 135.2°
Solution 1
C  180°  (25°  44.8°) or 110.2°

17.80004338

a

sin A
6

sin 25°

c

c




sin C
c

sin 110.2°
6 sin 110.2°

sin 2 5°

c 13.32398206
B 
 44.8°, C 
 110.2°, c 
 13.3
Solution 2
C 180°  (25°  135.2°) or 19.8°

5.365247745

a

sin A
6

sin 25°

b
c
 
 
sin B
sin C
25
c
  
 
sin 117°
sin 41°
25 sin 41°

c

sin 117°

a 10.51077021
B 
 117°, a 
 10.5, c 
 18.4

10




sin B

6 sin B 
 10 sin 25°

a
c
 
 
sin A
sin C
10
c
 
 
sin 99°
sin 32°
10 sin 32°

c

sin 99°

b 7.641171301
A 
 99°, b 
 7.6, c 
 5.4
4. B 
 180°  (22°  41°) or 117°
b

sin B
25

sin 117°

c

c
a
 
 
sin C
sin A
12
a
 
 
sin 42°
sin 97°
12 sin 97°

a

sin 42°

b 11.76557801
B 
 41°, a 
 17.8, b 
 11.8
3. A 
 180°  (49°  32°) or 99°
a

sin A
10

sin 99°

c




sin 55°
7 sin 55°


b

sin 75°

c

sin C
12

sin 42°

c




sin C

c

c




sin C
c

sin 19.8°
6 sin 19.8°

sin 25°

c 4.809133219
B 
 135.2°, C 
 19.8°, c 
 4.8
3. Since 56° 90°, consider Case I.
C sin B 
 50 sin 56°
41.45187863
34 41.5; no solution

18.40780654

1

5. K 
 2bc sin A
1

K 
 2(12)(6) sin 34°
K 20.1 units2
6. C 
 180°  (87°  56.8°) or 36.2°
1

sin A sin B


K 
 2c2 
sin C
1

sin 87° sin 56.8°


K 
 2(6.8)2 
sin 36.2°

K

32.7 units2

511

Extra Practice

4. C 
 180°  (45°  85°) or 50°
a

sin A
a

sin 45°

a


c

sin C
15

sin 50°
15 sin 45°

sin 50°

a

13.8459352

b

sin B
b

sin 85°











b


3. b2 
 a2  c2  2ac cos B
b2 
 142  182  2(14)(18) cos 48°
b2 182.7581744
b 13.51880817
a
b
 
 
sin A
sin B
14

sin A

c

sin C
15

sin 50°
15 sin 85°

sin 50°

14 sin 48°

sin A 
 b
A 
 sin1

Lesson 5-8

14.2

sin A

Page A35
1. a2 
 b2  c 2  2bc cos A
a2 
 62  82  2(6)(8) cos 62°
a2 54.93072997
a 7.411526831
a
b
 
 
sin A
sin B

B


A 
 sin1

A

1
1




s 
 2(4  7  10)



6 sin 62°

a

s 
 10.5
a)(s

b)(s 
c)
K 
 s(s

K 
 10.5(1
0.5

.5
4)(10)(10.5
 7)
 10
K 
 119.43
75

K 10.9 units2
1

6. s 
 2(a  b  c)

112

1




16 8 

s 
 2(4  6  5)
s 
 7.5
a)(s

b)(s 
c)
K 
 s(s

K 
 7.5(7.
5
)(7.5
 4

6)(7.5
 5)
K 
 98.43
75

K 9.9 unit2

48.1896851





sin B 


b

sin B
7

sin B
7 sin A

9

B 
 sin1

14.2 sin 85.3°

5. s 
 2(a  b  c)

B 45.62599479
C 180°  (62°  45.6°) or 72.4°
a 
 7.4, B 
 45.6°, C 
 72.4°
2.
a2 
 b2  c2  2bc cos A
92 
 72  122  2(7)(12) cos A
112 
 168 cos A

a

sin A
9

sin A


c

A 31.23444201
B 180°  (31.2°  85.3°) or 63.5°
c 
 27.3, A 
 31.2°, B 
 63.5°

6 sin 62°

a

A 
 cos1

c




sin 85.3°
14.2 sin 85.3°

6

sin1

14 sin 48°

sin A 
 c




sin B

sin B 



b

A 50.31729382
C 180°  (50.3°  48°) or 81.7°
b 
 13.5, A 
 50.3°, C 
 81.7°
4. c2 
 a2  b2  2ab cos C
c2 
 (14.2)2  (24.5)2  2(14.2)(24.5) cos 85.3°
c2 744.8771857
c 27.29243825
a
c
 
 
sin A
sin C

b 19.50659731
C 
 50°, a 
 13.8, b 
 19.5

a

sin 62°

b




sin 48°

1


9
7 sin A

7. s 
 2(a  b  c)
1

s 
 2(12.4  8.6  14.2)

B 35.43094469
C 180°  (48.2°  35.4°) or 96.4°
A 
 48.2°, B 
 35.4°, C 
 96.4°

s 
 17.6
K 
 s(s
a)(s


b)(s 
c)
K 
 17.6(1
7.6

(17.6
12.4)
)(17.6
8.6.2)
 14
K 
 2800.
512

K 52.9 units2
1

8. s 
 2(a  b  c)
1

s 
 2(150  124  190)
s 
 232
a)(s

b)(s 
c)
K 
 s(s

K 
 232(2
32
150)(2
 32
124)(2
 32
190)

K 
 86,29
2,864

K 9289.4 units2

Extra Practice

512

Lesson 6-2

9a.
28 in.

Page A36

110˚

1.
2.
3.
4.

35 in.

d2 
 282  352  2(28)(35) cos 110°
d2 2679.359481
d 51.8 in.
9b. Area 




1
2 2(28)(35)
920.9 in2

sin 110°

5  2 
 10 or about 31.4 radians
3.8  2 
 7.6 or about 23.9 radians
14.2  2 
 28.4 or about 89.2 radians
2.1  2 
 4.2
v

q 
 t



4.2

q 
 5
q 2.6 radians/s
5. 1.5  2 
 3
v

q 
 t

Lesson 6-1

3

q 
 2

Page A36

q 4.7 radians/min
6. 15.8  2 
 31.6




1. 120° 
 120°  
180°




v

q 
 t

2

3

31.6

q 
 1
8



180°

2. 280° 
 280° 

q 5.5 radians/s
7. 140  2 
 280

14


 9

v

q 
 t




3. 440° 
 440°  
180°

280

q 
 2
0

22


 9


q


4. 150° 
 150°  
180°

v

5


 6
5.
6.

8

3
5

12

8

2

q 
 30

180°


 3  

q


 480°
5

44.0 radians/min

8. q 
 t
about 0.2 radian/s

180°


 12  

 75°

Lesson 6-3

180°

7. 2 
 2  

 114.6°

Page A36

180°

8. 10.5 
 10.5  

 601.6°

5



5

1

1. 1
2. 0
3. 1
4.

9. reference angle:   6 
 6; Quadrant II
sin 6 
 2
10. reference angle:

4

3



4


;
3

y
y 
 sin x

Quadrant III

1

3


sin 3 
 2
11.

9

4



4

is coterminal with 4; Quadrant I

3

2

2


9

cos 4 
 2
12.

3

2

is coterminal with




3

1



y

5.

13. If the diameter 
 10 in., the radius 
 5 in.

y 
 cos x



180

4


 9
s 
 rv
s
5
s

x



2

cos 2 
 0
80° 
 80° 

O

4

3

1

O

x

1

4


9

7.0 in.

513

Extra Practice

Lesson 6-4

Lesson 6-5

Page A36

Page A37

1. 2 
 2;

2

1


 

c


 2

y

y

2



1. k 
  2 or 2
y 
 sin(2  
)

1

y 
 2 cos 

1

O



1

2

3

O



2

2

c


2. 3 
 3; 
0.5 
 4

2. k 
 1 or 2

y

y
3
2
1

y 
 3 sin 0.5 

O



1
2
3

2

3



O

3.

4

c
k

O

2

4

6



8

O

(2  
2 )



2

3



1
2

k

4. A 
 0.5
A


0.5

y


0.5 sin 3


 6

k


2

6

or

4. A 
 2

1

3

v

2






3



c


 2

1 
 

A


2

y


2 sin (v  )  1

5. A 
 0.5

2
 or 6
k
 

3

2 sin 6v
2

k

3

5
3

5
3
v
 cos 
5
2

h 
 1

2 or 1 c 
 

k
 


2

4

or

1

2

2

k



c


 4

k


8 
 0

2

4

A


0.5

y


0.5 sin 8v  3

6. A 
 20


 4

k


2

k

c
0

or 8



c


 2

h
3

4 
 2

A


20

y


20 cos (4v  8)  4

h
4

2 or 4 c 
 8

k
 

2

2

k

7. A 
 0.25
A


0.25

y


0.25 cos 4 v




8

k


2

8

or

3

7. A 
 4



4

A

y


Extra Practice

2

k

2

2

k

5. A 
 2

y


or 

y 
 sin

1

1

A


3 

2

y
y 
 12 cos 4

6. A 




2

1

2


 




1

y




2

y

A


y 
 2 cos(  2
)

2

1 2

 2; 
1 
 8


1

2

3 

2

1

2

3.



514

2

k

c
 
0




 10

3
2

k 
 10
4
3

1
 cos  v  
4
5
2



or 5

5

c
0

1

h 
 2

3

Lesson 6-6

4. If y 
 tan 4, then y 
 1.
3

Cos1
tan 4 
 Cos1 y


 Cos1 (1)



Page A37
1a. 12.1  2.7 
 9.4 h
1b. 12.1  2.7 
 14.8 h
1c. m 
 10 represents the middle of October.
d 
 2.7 sin (0.5(10)  1.4)  12.1
d 10.9 h
2. A 


6

2

y
3

2
or 3; 14

cos 7t




1

5. Let a 
 Cos1 2 and b 
 Sin1 0.
1

Cos a 
 2

Sin b 
 0



3

b
0

a




7

sin 


 Sin1 0 
 sin (a  b)

1
Cos1 2


 




 sin 
3  0



 sin 3
3



 2

Lesson 6-7

3


6. Let v 
sin1 2.

3


sin v 
 2

Page A37



2

v 
 3

y


y 
 cot   4

8

(




 cos 
2  3


3

1.

cos 
2 Sin1 2  
 cos (2v)

 cos 3
1


 2

)

4


O

2

3



Lesson 7-1

4
8

2.

Page A38
1


1. csc v 
 
sin v

y

1


 
2

1  co
s v

4



O

3.

1


 
1 2
1  4

y 
 sec   2

2






2

3






1

15

16



1


15


y


4

4
2


O

4


15


15


15


   

y 
 csc(2  2
)
2
3 




2

4
15

15

1

2. tan v 
 
cot v

4

1

 
6




Lesson 6-8



v 
 2
3. Let v 
 Tan1 1.
Tan v 
 1


6

6


 

6



 
2
13
13


3. cos

 cos 6  2
6


 cos 6




Page A37
1. Let v 
 Cos1 0.
Cos v 
 0


3
3

 
6


2. Let v 
 Arcsin 0.
Sin v 
 0
v
0

sin (315°)


4. tan (315°) 
 
cos (315°)





cos(Tan1 1) 
 cos 4





2




2

sin (45°  (360°))

cos (45°  (360°))
sin 
45°

cos 45°


 tan 45°

v 
 4

515

Extra Practice

5. Sample answer: cos x 
 1

1


5. csc (930°) 
 
sin (930°)










cot x

csc x
cos x

sin x

1

sin x

1

sin (360°(2)  210°)
1
sin (210°)
1
sin (30°)
1
sin 30°


1

cos x 
 1
6. Sample answer: cot x 
 2
2 tan x sin x  2 cos x 
 csc x


 csc 30°
6.


1

sin v

tan
v

c
os
v



sin v
sin v
1


cos v

2





cos x  sin x  cos x 
 sin x
1
sin x  cos x
 
 
2 

 sin x
cos x
1
1


2 

cos x  
 sin x
sin x

1

2


 sec v
cos v

sin v

1

2

cos x

    sin v  
7. cot v tan v  sin v sec v 
 
sin v cos v
cos v


2

sin x

sin v

2 
 cot x



1
cos v


 1  tan v
8. (1  sin x)(1  sin x) 
 1  sin2 x

 cos2 x
9.

cot x 
sin x

csc x cos x






Lesson 7-3

cos
x

 sin x
sin x

1
  cos x
sin x
cos x

cos
x

sin x

Page A38
1. cos 75° 
 cos (45°  30°)

 cos 45° cos 30°  sin 45° sin 30°
2


3


2


1


 2  2  2  2


 sin x

6  2



 4
2. sin 105° 
 sin (60°  45°)

 sin 60° sin 45°  cos 60° cos 45°

Lesson 7-2


3

2


1

2



 2  2  2  2
6  2



 4

Page A38
csc2 v 
 cot2 v  sin v 

1

sin v




csc2 v 
 cot2 v  1
csc2 v 
 csc2 v
2.

sec v  csc v

csc v sec v
sec v
csc v
  
csc v sec v
csc v sec v
1
1
  
csc v
sec v





 sin v  cos v


p
tan 4  tan 6



1  tan  tan 
4
6
1  3


3

1  1  3

3


 sin v  cos v




1  3


3


3



1  3

4. tan

7

12



 2  3



 tan A sin A

 tan A sin A

 tan A sin A

 tan A sin A

1  3


3

1  3



 tan 
3  4




 tan A sin A



tan   tan 
3
4



1  tan  tan 
3
4

3

1




1





1  1  1
3
1
1  3

  


3



 2  3

tan A sin A 
 tan A sin A

Extra Practice




3


 sin v  cos v

sin v  cos v 
 sin v  cos v
3. sin2 x  cos2 x 
 sec2 x  tan2 x
1 
 tan2 x  1  tan2 x
1
1
4. sec A  cos A 
 tan A sin A
1
  cos A
cos A
cos2 A
1
  
cos A
cos A
1  cos2 A

cos A
sin2 A

cos A
sin A
 sin A
cos A





3. tan 12 
 tan 
4  6

1. csc2 v 
 cot2 v  sin v csc v

516

29

5

4

5. sec 12  sec 1
2
1
 


cos 4  6

sec2 y  1  tan2 y
2

54

1




 




cos 4 cos 6 
 sin 4 sin 6

9

16
3

4

1
 


2 
3


2
 1
  
  
2
2
2
2

 1  tan2 y
 tan2 y
 tan y
tan x  tan y


tan (x  y)  
1 
 tan x tan y
3
3
  
4
4

1
 

6
 
 
2


4
4

6
  

2
 



6
 
 
2

  
2


6
 
6
  
2


 ——
3 3
1 
 4 4


6. cot 375°  cot 15°

24

1

6

85

8



82 
 52  
39


8

5
5

8

cos y  8

csc y  5

1  tan 45° tan 30°

1

sin y

1



sin y 


3

1 
 3


3

1  1
3



1


3

1  
3
——

3

1 
 
3




3

1  
3
——

3

1  
3

2 2

1 
 5



cos y 

21

25



7

16



21

5



7

4


7


4

2

 5

3

4



122 

112  
23



1

19

12

sin y 
11

12

 
 280
85

48
3315

179

30°





23

12

cos (x 
 y)  cos x cos y  sin x sin y
5



6
3315
 
 35
85


6
3315
 
 35
85


1. sin 15°  sin 2

8. 

122 

52  
119


 1
2

680

6
3315
  35
85


Page A39

6  7
3


 20
sin x 

1

6
85
 
39

7
85
 5
    
85
8
85
8

Lesson 7-4

sin (x  y)  sin x cos y  cos x sin y

21


1

cos x cos y  sin x sin y


3 2

1 
 4



 5


39



sec (x 
 y)  
cos (x 
 y)

 2  
3

7. sin x 

7

85

10. 

72  62  
85
; cos x  85; sin x  85

1

tan 45° 
 tan 30°



6

4
—
7

16

 7



tan (45° 
 30°)



3

9. If cot x  3, then tan x  4.


1

19

 12





23

12

1 
 cos 30°

2

3

1 
 
2

2




2 
 
3


55  
2737


  2



144




2 
 
3


Since 15° is in Quadrant I, sin 15°  2.
150°

2. cos 75°  cos 2


1  cos 150°

2




3

1 
 
2

2




2 
 
3


  2




2 
 
3


Since 75° is in Quadrant I, cos 75°  2.

517

Extra Practice

3. tan



12

 tan



6
—
2

3
5


sin 2v  2 sin v cos v


1 
 cos ——
6


1  cos —6—



 277
3
5


12
5


cos 2v  cos2 v 
 sin2 v
2 2

41




(2 
 
3
 )2

1

 
 4
9
2 tan v


tan 2v  
1 
 tan2 v







45°

1  cos 45°

2


5


 2 3 3

5


2

4
5


 9
cos 2v  cos2 v 
 sin2 v




2  
2



5
 2

2 2

 3 
 3



2  


2

Since 22.5° is in Quadrant I, cos 22.5°  .

1

 9

2

5

6
—
2

2 tan v


tan 2v  
1 
 tan2 v

2 5
2
5


5

1 
 cos 6

2



1 
 5
2
5


2

 4
5



3


1  2

9.



2



(3)2 
(
1)2

3

10

 2 10
3
10


5

5p




2  
3




10


 
10 

3


.
Since 1
2 is in Quadrant I, sin 12 
2

 
 5
cos 2v  cos2 v 
 sin2 v

6. tan 112.5°  tan 225°


1 
 cos 225°

1  cos 225°





2
1  
2


2

1 
 2



10 2

4

 
 5
2 tan v

2 (
 3)



1 
 (
 3)2

2  
2
 2  
2

 
 2  
2

2 
 
2



(2  
2
 )2

2

3
10
 2


 10
 
 1
0 


tan 2v  
1 
 tan2 v



3

 4

or 1  
2


Since 112.5° is in Quadrant II,
tan 112.5°  
1 
 
2
.

518



10

 
10
; sin v  10; cos v  
1
0

sin 2v  2 sin v cos v




2  
3


  2

Extra Practice

2
5


sin 2v  2 sin v cos v

1  2

2



2

8. 

32 
 22  
5
; cos v  3, tan v  5

  2



3
5


 
 4
1


2


5

1 
 2
12

5

4. cos 22.5°  cos 2

5. sin 12  sin

2 2
3
5


or 2 
 
3


Since 12 is in Quadrant I, tan 12  2 
 
3
.



3

5 2

 7 
 7

2 
 
3
 2 
 
3

 
2  
3
 2 
 
3




2

 49



3
1 
 
2



3
1  
2



3
5


7. 

72 
 22  3
5
; sin v  7; tan v  2

3



32 
 (

 2)2  
5


3

2
2

3


5

cos v  3
2

5
tan v  
 5

10. csc v  
 2
1

sin v




sin v  


Lesson 7-6
Page A39
1. x cos 30°  y sin 30° 
 12  0

sin 2v  2 sin v cos v
 2 
 3


3


2x


35 

2

4

5

 
 9

x  y 
 24  0

3





2. x cos 3  y sin 3 
 2  0

cos 2v  cos2 v 
 sin2 v

5
 2

1

2

2 2

 3 
 
 3



3

x  2 y 
 2  0
y 
 4  0
x  
3

1

 9

1

3. x cos 150°  y sin 150° 
 2  0

2 tan v


tan 2v  
1 
 tan2 v



1

 2 y 
 12  0

2 



3


1

1


 2 x  2 y 
 2  0


2
5
 2
1 
 
 5
2
5


5

x  y 
 1  0

 
3
4. 




A2  
B2

4x


 2

29

 
 4
5


 
 

42  1
02  
2
29


10y

10

    0


 2
29


 2

29

2
29


5
29


5
29



 29x 
 29y 
 29  0
5
29


p  29

Lesson 7-5

5

29

2
29


sin f  
 29, cos f  
 29
5

tan f  2

Page A39

f  Arctan 2  180°
f  248°
5

1. 4 cos2 x 
 2  0
1

cos2 x  2
cos x  

5. 

A2  
B2  

12  (

1)2  
2



2


2

x


2




2
 x
2

x  45°, 135°, 225°, 315°
2. sin2 x csc x 
 1  0
sin2 x

1

sin

x  90°



3

cos x

sin x


2


cos x 

 2 cos x

cos x  0
x  90°, 270°


 2  0
or


2


tan f  
1
f  Arctan (
1)  360°
f  315°
6. 

A2  
B2  

22  32  
13



 2 cos x  0


3

sin x

4.

cos x

sin x


2



 2 y 
 
2
0

sin f  
 2, cos f  2

 cos x  2 cos x

3


3

2


p  
2

x
10
sin x  1

3.

y

 
   0



2



2

2



13

3


sin x

12

2
13

3
13

12
13

 x   y 
  
13
13
13
12
13

p  13
3
13

2
13

sin f  13, cos f  13
3
tan f  2
3
f  Arctan 2


20


3

sin x

3


2

3

 y 
   0
x

13



13

2
 sin x

x  60°, 120°
3 cos2 x  6 cos x 
 3
3 cos2 x 
 6 cos x  3  0
3(cos2 x 
 2 cos x  1)  0
3(cos x 
 1)(cos x 
 1)  0
cos x 
 1  0
cos x  1
x  0°

0

f  56°

519

Extra Practice

Lesson 7-7

10.

1.7 cos 70°  h
0.58  h

Page A39
1.

3(2) 
 2(1) 
 2

2  (

2)2

3



11.

2

 

13



2.

2(3)  4(0) 
 2

2  42

2



h


cos 76°  
2.1

2.1 cos 76°  h
0.51  h

2

13

13
4

12.

 

20


h


cos 30°  
3.6

3.6 cos 30°  h
3.12  h

4

 
2
5


3.

h


cos 70°  
1.7

2
5

 5

3(
1)  (
4) 
 1

2


 



(
3)2 
 12
10

10


x


sin 70°  
1.7

1.7 sin 70°  v
1.60  v
v


sin 76°  
2.1

2.1 sin 76°  v
2.04  v
v


sin 30°  
3.6

3.6 sin 30°  v
1.8  v

Lesson 8-2


10


 
 5; d  5
2

4. y  3 x 
 2 → 2x 
 3y 
 6  0
2(4) 
 3(
2) 
 6

2  (

2


3)2




Page A40

u
1. AB  4 
 3, 1 
 6
 1, 
5
u
AB   

12  (

5)2
 
26

u
2. AB  
2 
 (
1), 2 
 3
 
1, 
1
u
AB   

(
1)2 
 (
1
)2
 
2

u
3. AB  
1 
 0, 
8 
 (
4)
 
1, 
4
u
AB   

(
1)2 
 (
4
)2
 
17

u
4. AB  3 
 1, 
9 
 10
 2, 
19
u
AB   

22  (

19)2
 
365

u
5. AB  
3 
 (
6), 
6 
 0
 3, 
6
u
32  (

6)2
AB   

 
45

 3
5

u
6. AB  0 
 4, 7 
 (
5)
 
4, 12
u
AB   

(
4)2 
 122
 
160

 4
10

7.  5, 6   

52  62
 
61

u  6j
u
5i
8.  
2, 4   

(
2)2 
 42
 
20

 2
5

u  4j
u

2i
9.  
10, 
5   

(
10)2
 (

5)2
 
125

 5
5

u 
 5j
u

10i
10.  2.5, 6   

(2.5)2 
 62
 
42.25

 6.5
u  6j
u
2.5i

8

13


8

13

13

5. (0, 
1)
0  2(
1) 
 4

d  2 
2
 2

1



6

 



5

6

5

 5;

6
5


d  5

6. (0, 3)
3(3) 
 2(0) 
 7


d
2
2


3

 (


2)
2

 

13


2

13

 13
7. (2, 
1)
2(2)  5(
1) 
 4

d  2 
2
2  5




5

 


29

 
0.9; d  0.9 unit

Lesson 8-1
Page A40
1.

2.
1.7 cm
2.1 cm

70˚

104˚

3. 330˚

3.6 cm

4. 3.6 cm; 89°
6. 3.7 cm; 357°
8. 7.2 cm; 330°

Extra Practice

5. 2.6 cm; 23°
7. 1.2 cm; 342°
9. 8.8 cm; 340°

520

11.  2, 
6   

22  (

6)2
 
40

 2
10

u 
 6j
u
2i
2  (

12.  
15, 
12   

(
15)
12)2
 
369

 3
41

u 
 12j
u

15i

u

1, 5, 2  u
i
j u
k
3 0 4

1 5 2
u 
 10j
u  15k
u
 
20i

 
20, 
10, 15
3, 0, 4  
20 3  (
10) 0
 15 4
 0; yes

1, 5, 2  
20 (
1)  (
10)
5  15 2
 0; yes
u
2, 1, 3  u
i
j u
k


20, 
10, 15

20, 
10, 15

Lesson 8-3
1. u
p  2 1, 2, 
1  3(
4, 
3, 0
 2, 4, 
2  
12, 
9, 0
 
10, 
5, 
2
1
u
2. p  1, 2, 
1 
 2 
2, 2, 4  
4, 
3, 0
 1, 2, 
1  1, 
1, 
2  
4, 
3, 0
 
2, 
2, 
3
3. u
p  
2 
2, 2, 4  
4, 
3, 0
 4, 
4, 
8  
4, 
3, 0
 0, 
7, 
8
u  3 
4, 
3, 0  2 1, 2, 
1
4. p
4



9

 
3, 
 4, 0  2, 4, 
2

8, 10, 19

3

 
1, 14, 
2

8, 10, 19
0, 0, 0
0, 0, 0
0 
 7, 0 
 0, 0 
 4

7, 0, 
4

4. 8, 6
5. 3, 4, 0
6. 4, 5, 1
7. 1, 0, 3

 8, 10, 19

1, 
3, 2  8 (
1)  10 (
3)  19 2
 0; yes
6, 
1, 
2  8 6  10 (
1)  19 (
2)
 0; yes

Page A41
1. magnitude  

2002 
2202
 297.32 N

Page A41

3. 
5, 3



Lesson 8-5

Lesson 8-4

2. 
3, 2




1 1 0
2 1 3
u  3j
u 
 3k
u
 3i
 3, 3, 
3
3, 3, 
3

1, 1, 0  3 (
1)  3 1  1 (
3) 0
 0; yes
3, 3, 
3 (2, 1, 3  3 2  3 1  (
3) 3
 0; yes
u
u
10. 
1, 
3, 2
6, 
1, 
2  u
i
j
k
2

1 
3
6 
1 
2
u  10j
u  19k
u
 8i

Page A40

1. 3, 4



9. 
1, 1, 0

5. 2, 
4, 1  5, 4, 3 u
v 
7, 0, 4 u
v 
u
v 
u
v 





8. 3, 0, 4

200


direction: tan v  
220

2, 5  3 2  4 5
 26; no
4, 6  
3 4  2 6
 0; yes
2, 
3  
5 2  3 (
3)
 
19; no

2, 
3  8 (
2)  6 (
3)
 
34; no
4, 
3, 6  3 4  4 (
3)  0 6
 0; yes

1, 
2, 3  4 (
1)  5 (
2)  1 3
 
11; no
1, 1, 2  u
i u
j u
k



v  42.3°
u2  502  502 
 2(50)(50) cos 120°
2. r
u2  7500
r
u  86.60 mph; 30°
r
u
3. r 2  3502  2802 
 2(350)(280) cos 135°
u  339,492.9291
r
u  582.66 N
r
582.66

sin 135°

280 sin 135°


sin v  
582.66

sin v  0.34
v  19.9°
u
902 
1102
4. r   

u  142.13 N
r



1 0 3
1 1 2
u u
 
3i
j u
k


3, 1, 1

3, 1, 1

280



sin v

 
3, 1, 1
(1, 0, 3  
3 1  1 0  1 3
 0; yes
1, 1, 2  
3 1  1 1  1 2
 0; yes

90 N

110 N

521

Extra Practice

2c. y  5  75t sin 25° 
 16t2

Lesson 8-6

2.13

 20.6 ft

Page A41
1. x 
 2, y 
 3  t 1, 0
x
2t
y
30
x2t
y3
2. x 
 (
1), y 
 (
4)  t 5, 2
x  1, y  4  t 5, 2
x  1  5t
y  4  2t
x  
1  5t
y  
4  2t
3. x 
 (
3), y 
 6  t 
2, 4
x  3, y 
 6  t 
2, 4
x  3  
2t
y 
 6  4t
x  
3 
 2t
y  6  4t
4. x 
 3, y 
 0  t 0, 
1
x
30
y 
 0  
t
x3
y  
t
5. x  3t
y2t
y
2
y

x

3
1
x
3





1. The figure is 4 times the original size and
reflected over the xy-plane.
2. The figure is half the original size.
3. The figure is 1.5 times the original size and
reflected over the yz-plane.

Lesson 9-1

x

→

t  3

→

ty
2
1.

→

t

→

t

x1

2
y

4

→

yt
1
→
1
10
y  1  3x  3

180˚



t

2
3

K

1 2 3 4

0˚

330˚

x  10

3

240˚

300˚

270˚

3.

90˚

120˚

7

3

N

150˚

330˚
240˚

5.

2
3

300˚

270˚

Page A41
1. vy  70 sin 34°
 39.14 yd/s
70 yd/s
vx  70 cos 34°
34˚
 58.03 yd/s
u
2a. x  t v  cos v
 75t cos 25°
1
u
y  t v  sin v  2gt2  h


2

0

 5  75t sin 25° 
 16t2
2b. y  0 when t  ?

7.

2 
 4(

75 sin 25°  

(75 si
n 25°)

16)(5
)

2(
16)

2
3

3
2

2

t

0

11
6
4
3

522

3
2

P

5
3

11
6

120˚

3
2
90˚

5
3

60˚
30˚
0˚

1 2 3 4

330˚

210˚
240˚

2
3

270˚

2

300˚


3

6

5
6

1 2 3 4

7
6

0
1 2 3 4

7
6

8.


3

6




0.1468595989 or t  2.127882701
t  2.13 s
x  75t cos 25°
 75(2.13) cos 25°
 145 ft


3

6



5
3

5
6


2

5
6

180˚

11
6
4
3

5
3

3
2

150˚

1 2 3 4

7
6

1

2
3

6.

6



11
6

4
3


3

5
6

 75t sin 25°  2(
32)t2  5

0˚

0

1 2 3 4

7
6

30˚

210˚

M



4.

1 2 3 4


3

6

4
3

60˚


2

5
6

ty1

Lesson 8-7

Extra Practice

2.
30˚

180˚

t

60˚

210˚

7. x  3t 
 10

1
x
3

90˚

120˚
150˚

1

2

y  2x  2

y

Page A41

2

y  4t
x

2

Lesson 8-8

Page A42

6. x  
1  2t
y

4

2.13 2

 5  752 sin 25° 
 162

0



1 2 3 4

7
6

11
6
4
3

3
2

5
3

9. r  
5
 or r  

5


6. x  5 cos 240°



Lesson 9-2

90˚

120˚

2.

60˚

30˚

150˚
180˚

1 2 3 4

0˚

330˚

210˚
240˚

270˚

300˚

90˚

0



5 10 15 20
11
6

7
6
3
2

5
3

30˚
1 2 3 4

240˚

270˚

Lesson 9-4

300˚

cardioid

Page A42
1. 


A2  
B2  


62  (

5)2
 

61


Lesson 9-3

6

61


1. r  

12  (

1)2

 
2


1

v  Arctan 1  2


7

4


2, 

v  Arctan 3
0

22  (

2
 )2
3. r  

 
6
 or about 2.45
(2.45, 0.62)

v  Arctan 2
 0.62

 2 2



5. x  4 cos 2
1

 3
10




3

 4

1



3

1

0  2 x  2 y 
 8



x  2y 
 16
0  
3
4. 1  r cos (v  )
0  r (cos v cos  
 sin v sin ) 
 1
0  
r cos v 
 0 
 1
0  
x 
 1
x1

1

0

3

10

0  2 r cos v  2 r sin v 
 8

y  4 sin 2
 4 (1)

90



10

 
2


 4 (0)

9

 y 
   0

3
10

3
10


cos f  1
, sin f  10, p  3
10

0
f  Arctan 3
 72°
3
10
  r cos (v 
 72°)
3. 8  r cos (v 
 30°)
0  r(cos v cos 30°  sin v sin 30°) 
 8


2


1

6
61

61

6
61

  r cos (v 
 140°)
61


A2  
B2  

32  92
3
 x
3
10




 2 2

 
2


2, 
2

2.


2


y  2 sin 4

5

61


sin f  , p  
5

0

4. x  2 cos 4

2


6


61

f  Arctan 
6  180°
 140°

7

4

2. r  

32  02
 
9
 or 3
(3, 0)



5

61

6

,

61



x  y 
   0
cos f 

Page A42

1

6


r
sin v

0˚

330˚

210˚

8. y  6
r sin v  6

r  
2 sec v
r  6 csc v
2
2
9.
x  y  36
(r cos v)2  (r sin v)2  36
r2 (cos2 v  sin2 v)  36
r2  36
r  6 or r  
6
10.
x2  y2  3y
2
(r cos v)  (r sin v)2  3r sin v
r2 (cos2 v  sin2 v)  3r sin v
r2  3r sin v
r  3 sin v
11. r  4
12.
r  4 cos v
r2  16
r2  4r cos v
x2  y 2  16
x2  y2  4x


6

5
6

4. Sample answer:
r  sin 5v

60˚

150˚
180˚

3
5


 
2 or about 
4.33


r  

cos v


3

spiral of Archimedes
120˚

 5 
2

2


2

2
3

4
3

circle
3.



3

 

(
2.5, 
4.33)
7. x  
2
r cos v  
2

Page A42
1.

y  5 sin 240°

1
5 
2
5

2 or 
2.5

1

0, 14

523

Extra Practice

5.

90˚

120˚

Lesson 9-6

60˚
30˚

150˚
180˚

1 2 3 4

Page A43

0˚

1. 4x 
 6yi  14  12i
4x  14

330˚

210˚
240˚

270˚

x

300˚

3.

3

3. i1000  (i4)250
 1250
1

(0, 5)

42  12
z  

 
17


4. i12  i4  (i4)3  (i4)
1
 13  11
2

4.

11.

(i 
 2)2

4  2i



 i sin 4

1

42  (


2
)2
v  Arctan 
4  2
7. r  

 
18
 or 3
2

 5.94
4 
 
2
i  3
2
 (cos 5.94  i sin 5.94)
8a. 5(cos 0.9  i sin 0.9)  3.11  3.92j
8(cos 0.4  j sin 0.4)  7.37  3.12j
8b. (3.11  3.92j)  (7.37  3.12j)
 (3.11  7.37)  (3.92j  3.12j)
 10.48  7.04j ohms

2



2 
 i


2 
 i

12 
 10i 
 2i2

4 
 i2

10 
 10i

5


8c. r  
(10.48

)2  (7
.04)2
v  Arctan 
10.48 
 12.63
 0.59
10.48  7.04j  12.63 (cos 0.59  j sin 0.59) ohms

i2 
 4i  4

7.04



4  2i
3 
 4i

4 
 2i

4 
 2i
12 
 22i  8i2

16 
 4i2
4 
 22i

20
1
11
 
 
5
10 i



4  2i






4



4


6. r  

(
2)2 
 12
v  Arctan 

2   
 
5

 2.68

2  i  
5
(cos 2.68  i sin 2.68)

6  2i



4

4  4i  4
2
cos




2  i



v  Arctan 4

 
32
 or 4
2



 
2 
 2i
12.



5. r  

42  42

1i

1i
4  5i  i2

1 
 i2
3  5i

2
3
5
  i
2
2



z  

22  (


3
)2
 
7


i

(2, 
3)

4i



z  

02  (

5)2
 
25
 or 5

O



1
i

6  2i


2  i



4

2. i17  (i4) i
 14 i
i

5. (4 
 i)  (
3  5i)  (4  (
3))  (
i  5i)
 1  4i
6. (6  6i) 
 (2  4i)  (6  (
2))  (6i  (
4i))
 4  2i
7. (3  i)(5 
 3i)  15 
 4i 
 3i2
 18 
 4i
8. (2  5i)2  (2  5i)(2  5i)
 4  20i  25i2
 
21  20i
9. (1 
 
2
i)(
3 
 
8
i)  
3 
 
8
i  3
2
i  
16
i2
 
3 
 2
2
i  3
2
i  4i2
 
7  
2
i





O

O



i

(4, 1)

1. i10  (i4)
i2
3
1
(
1)
 
1

4i

1
i

y  
2

i

Lesson 9-5

10.

12

y  
6

x  3.5
2.

Page A43


6y  12

14

4

Lesson 9-7
Page A43



1. r  6 4 or 24



v  2  4
3

3

3



 4



24cos 4  i sin 4  24 
2  i2

2



2


 
12
2
  12
2
i
Extra Practice

524

3
2. r  
1 or 6

2

6(cos

7

4

3. d  

(x2 
 
x1)2 
(y2 

y1)2

3

v  4  

d  

(r 
 r
)2  (

2 
 6
)2

7

 4
 i sin

7
)
4



6

 i


2


2


2


2

d  

02  (

8)2
d  
64
 or 8



x1  x2 y1  y2

r  r 6  (
2)
 
2, 2  
2 ,
2

 3
2
 
 3
2
i
3. r  5 2 or 10
v  135°  45°
 180°
10(cos 180°  i sin 180°)  10(
1  i(0))
 
10

4.





1. 44 cos (4)2  i sin (4)2
 256(cos 2  i sin 2)
 256(1  i(0))
 256

2. r 

v  Arctan 

12


5



122 
(
5)2



v  Arctan 

 
2


2


1

3





4

6a.



1

1



1






(5, 2)

x

0  50 0  40
, 
2
2



  (25, 20)

v
1



(2, 5)

d  

252 
202
d  
1025

d  5
41
 or about 32 ft

Lesson 10-2



1 5 cos 5()  i sin 5()
1

(5, 10)
(2, 8)

d  

(25 

0)2 
(20 

0)2

 
2
 3 cos 12  i sin 12
 1.08  0.29i
(
1)2 
 02
4. r  

1

y

6b. d  

(x2 
 
x1)2 
(y2 

y1)2

cos 134  i sin 1312 
1


2  y2

2

O

 13
 1.965587446
133(cos (3)(v)  i sin (3)(v))  2035 
 828i
12  12
3. r  


  (
5, 8)

5. A quadrilateral is a
parallelogram if both
pairs of opposite sides
are parallel. Since only
one pair of opposite sides
are parallel, the
quadrilateral is not a
parallelogram.







8
2  y2  16
6  x2  
10
x2  
16
y2  14
Then A has coordinates (
16, 14).

Lesson 9-8
Page A43

 (r, 2)

6  x2 2  y2


2 ,
2
6  x2
  
5
2



 1cos 5  i sin 5

Page A44
1.

 0.81  0.59i

Lesson 10-1

(x 
 h)2  (y 
 k)2  r2
2
[x 
 (
2)]2  (y 
 2)2  
2

(x  2)2  (y 
 2)2  2
y
(x  2)2  (y  2)2 
 2

Page A44
1. d  

(x2 
 
x1)2 
(y2 

y1)2
d  

(4 
 (

2))2 
 (5 

2)2

O

x



62  32

d
d  
45
 or 3
5


2.

x1  x2 y1  y2


2  4 2  5
, 
2, 2  
2
2

 (1, 3.5)
2. d 
d



(x2 
 
x1)2 
(y2 

y1)2


(8 
 (

3))2 
 (
1

 6)2

d  

112 
(
7)2
d  
170


(x 
 h)2  (y 
 k)2  r2
(x 
 0)2  (y 
 (
4))2  42
x2  (y  4)2  16
y
O
x
x 2  (y  4)2 
 16

x1  x2 y1  y2


3  8 6  (
1)
 , 
2, 2  
2
2

 (2.5, 2.5)

525

Extra Practice

3. x2  49 
 y2 → x2  y2  49
y2
2
8

7. r  

(4 
 2
)2  (0

 (

5))2
r  

22  52

x  y 
 49

r  
29
; r2  29
(x 
 4)2  (y 
 0)2  29
(x 
 4)2  y2  29

4

8 
4 O

4

4

8x

Lesson 10-3


8

4.

  6x 
 8y  18  0
(x2  6x  9)  (y2 
 8y  16)  
18  9  16
(x  3)2  (y 
 4)2  7
y
2
2
x2

y2

Page A44
1. center: (h, k)  (0, 0)
10

(x  3)  (y  4) 
 7

a  2 or 5
6

b  2 or 3
(x 
 h)2

a2
(x 
 0)2

52

O

(y 
 0)2

 3
1
2
x2

25

x

y2

 9  1

c2  a2 
 b2
c2  25 
 9
c2  16
c4
foci: (4, 0)
2. center: (h, k)  (
2, 1)

5. x2  y2  Dx  Ey  F  0
22  (
2)2  2D 
 2E  F  0

→ 2D 
 2E  F  
8
02  (
4)2  0(D) 
 4E  F  0
→ 
4E  F  
16
(
2)2  (
2)2 
 2D 
 2E  F  0
→ 
2D 
 2E  F  
8
2D 
 2E  F  
8
2(2 0 
 2E  F )  2(
8)

2D 
 2E  F  
8

 4E  F )  
16
4D
 
0
F0
D0

4E  (0)  
16

4E  
16
E4
x2  y2  4y  0
center: (0, 
2)
x2  (y2  4y  4)  0  4
radius: 2
x2  (y  2)2  4
2
6. x  y2  Dx  Ey  F  0
(
1)2  32 
 D  3E  F  0
→ 
D  3E  F  
10
(
4)2  62 
 4D  6E  F  0
→ 
4D  6E  F  
52
(
7)2  32 
 7D  3E  F  0
→ 
7D  3E  F  
58
2(
D  3E  F )  2(
10) 
4D  6E  F )  2(
52

4D  6E  F )  2(
52 2(
7D  3E  F )  2(
58)
2D
 F )  2(
32
10D

 F )  2(
64
2D  F  32

(8)  3E  F  
10
10D 
 F  64

4(8)  6E  F  
52
12D
 96
24) 
 3E
 
42
D8

3E  18
E  
6

(8) 3(
6)  F  
10

26  F  
10
F  16
x2  y2  8x 
 6y  16  0
(x2  8x  16)  (y2 
 6y  9)  
16  16  9
(x  4)2  (y 
 3)2  9
center: (
4, 3)
radius: 3
Extra Practice

(y 
 k)2

 b
1
2

8

a  2 or 4
4

b  2 or 2
(y 
 k)2
(x 
 h)2

 b
2
a2
(y 
 1)2
[x 
 (
2)]2




42
22
(y 
 1)2
(x  2)2
  
16
4
2
c  a2 
 b2

1
1
1

c2  16 
 4
c2  12
c  
12
 or 2
3

foci: (
2, 1  2
3
)
3. The major axis contains the foci and it is located
on the x-axis.
y
center: (h, k)  (4, 1)
8 (x  4)2 (y  1)2


1
c2  a2  b2
64
16
c2  64  16
4
c2  80
c  
80
 or 4
5

4
8 12x

4 O

4

foci: (h  c, k)  (4  4
5
, 1)
major axis vertices: (h  a, k)  (4  8, 1)
 (12, 1) and (
4, 1)
minor axis vertices: (h, k  b)  (4, 1  4)
 (4, 5) and (4, 
3)

526

Lesson 10-4

Lesson 10-5

Page A45

Page A45
1. vertex  (h, k)  (0, 0)
4p  4
p1
focus: (h  p, k)  (0  1, 0) or (1, 0)
directrix: x  h 
 p
x0
1
x  
1
axis of symmetry: y  k
y0

y

1.

(1, 2  

58)

8
y  2 
  73 (x  1)
4

y  2 
 73 (x  1)
(1, 2)

4
8 x
(1,
2



58)

4


4 O

y

y

2.

8

y 2 
 4x

4

8 
4 O

4

x

O

8x

4

xy 
 16


8

3. center: (h, k)  (
4, 3)
(x 
 h)2

a2
[x 
 (
4)]2

32
(x  4)2

9








(y 
 k)2

b2
(y 
 3)2

22
(y 
 3)2

4

2. x2 
 4x  4  12y 
 12
(x 
 2)2  12(y 
 1)
vertex  (h, k)  (2, 1)
4p  12
p3
focus: (h, k  p)  (2, 1  3) or (2, 4)
directrix: y  k 
 p
y1
3
y  
2
axis of symmetry: x  h
x2
y

1
1
1

4. transverse axis: x  h  2
foci: (h, k  c)  (2, 7)
k  c  
7
(h, k 
 c)  (2, 
3)
k 
 c  
3
2k
 
4
k  2; c  5
vertices: (h, k  a)  (2, 5)
2a5
(h, k 
 a)  (2, 
1)
a3
a2  b2  c2
32  b2  52
b2  16
b4
(y 
 k)2

a2
(y 
 2)2

32
(y 
 2)2

9

x 2  4x  4 
 12y  12

(x 
 h)2


 b
1
2





(x 
 2)2

42
(x 
 2)2

16

O

x

1
3. vertex: (h, k)  (
2, 3)
focus: (h  p, k)  (0, 3)
h  p  0, k  3

2  p  0
p2
(y 
 k)2  4p(x 
 h)
(y 
 3)2  4(2)[x 
 (
2)]
(y 
 3)2  8(x  2)
4. directrix: y  k 
 p  
3
focus: (h, k  p)  (0, 
2)
k 
 p  
3

2.5  p  
2
k  p  
2
p  0.5
2k
 
5
k  
2.5
(x 
 h)2  4p(y 
 k)
(x 
 0)2  4(0.5)[y 
 (
2.5)]
x2  2(y  2.5)

1

527

Extra Practice

4. B2 
 4AC  02 
4(4)(
25)
 0  400 or 400
Since 400 0, the graph is a hyperbola.
4x2 
 25y2  64




2
4(x cos 90°  y sin 90°) 
 25(
x
 sin 90°
 y
 cos 90°)2  64
4(0  y
)2 
 25(
x
  0)2  64
4(y
)2 
 25(x
)2 
 64  0
2
2
 
 4(1)(2)
5. B 
 4AC  
2
2
 8 
 8 or 0
parabola

Lesson 10-6
Page A45
1. A  1, C  1; since A  C, the conic is a circle.
x2  y2 
 8x  2y  13  0
2
(x 
 8x  16)  (y2  2y  1)  
13  16  1
(x 
 4)2  (y  1)2  4
2. A  1, C  
 4; since A and C have different
signs, the conic is a hyperbola.
x2 
 4y2  10x 
 16y  
5
2
(x  10x  25) 
 4(y2  4y  4)  
5  25 
 16
(x  5)2 
 4(y  2)2  4
(x  5)2

4

B


tan 2v  
A
C

2
2


(y  2)2


tan 2v  
1
2


 1  1
3. A  0, C  1; since A  0, the conic is a parabola.
y2 
 5x 
 6y  9  0
(y2 
 6y  9)  5x 
 9  9
(y 
 3)2  5x
4. A  1, C  2; since A and C have the same sign
x2  2y2  2x  8y  15
(x2  2x  1)  2(y2  4y  4)  15  1  8
(x  1)2  2(y  2)2  24
(x  1)2

24

tan 2v  2
2

2v  70.52877937°
v  35, 
35°
6. B2 
 4AC  52 
 4(15)(5)
 25 
 300 or 
275
Since 
275  0 and A C, the graph is an ellipse.
B


tan 2v  
A
C
5


tan 2v  
15 
 5

(y  2)2

 12  1

1

tan 2v  2
2v  26.56505118°
v  13°

Lesson 10-7
Page A45
1. B2 
 4AC  02 
 4(1)(1)
 0 
 4 or 
4
Since 
4  0 and A  C, the graph is a circle.
x 2  y2  9
2
(x 
 1)  [y 
 (
1)]2  9
(x 
 1)2  (y  1)2  9
x2 
 2x  1  y2  2y  1  9
x2  y2 
 2x  2y 
 7  0
2
2. B 
 4AC  02 
 4(4)(1)
 0 
 16 or 
16
Since 
16  0 and A C, the graph is an ellipse.
4x2  y2  16
4[x 
 (
3)]2  [y 
 (
2)]2  16
4(x  3)2  (y  2)2  16
2
4(x  6x  9)  (y2  4y  4)  16
4x2  y2  24x  4y  24  0
2
3. B 
 4AC  02 
 4(49)(
16)
 0  3136 or 3136
Since 3136 0, the graph is a hyperbola.
49x2 
 16y2  784
 2



Lesson 10-8
Page A45
1. xy  3
y

x 2 
 y2  8

3

x

x2

3 2


 x  8
9

x2 
 x2  8
x4 
 9  8x2

 8x2 
 9  0
2
(x 
 9)(x2  1)  0
x2 
 9  0 or x2  1  0
x2  9
x2  
1
x  3
x  

1
 or i
x4

3

If x  3, then y  (3) or 1.
3


If x  
3, then y  
(
3) or 
1.

 is an imaginary number, disregard
Since x  

1
this solution.
y
(3, 1), (
3, 
1)



49 x
 cos 4  y
 sin 4 
 16 
x
 sin 4

 2

2

2


2

49 2x
  2y
 
 16
1
1
49 2(x
)2  x
y
  2(y
)2



49
(x
)2
2



 49x
y
 

 y
 cos 4  784

2

2


2


2x
  2y
 
1

 16 2(x
)2 
 x
 y

1
 2(y
)2 
49
(y
)2 
 8(x
)2  16x
y

2

 8(y
)2 





(3, 1)

O

784
(3, 1)

784

784
49(x
)2  98x
y
  49(y
)2 
 16(x
)2  32x
y


16(y
)2  1568
33(x
)2  130x
y
  33(y
)2 
 1568  0
Extra Practice

528

x

2. x 
 y  4
xy4

3

11. ((3f)2)  (3f)6

x2  10y2  10
(y  4)2  10y2  10
y2  8y  16  10y2  10
9y2 
 8y 
 6  0

1



(3f)6
1



36 f 6
1


(
8)  

(
8)2 

 4(9
)(
6)


y



729f 6

8  2
70


12.

y  1
8


3a

2

cc 
4a

4  
70


y  9
4  
70


c 6 a

 c8
a
 c
14a
1



c14a

4 
 
70


y  9 or y  9

1


1


6



1

h


216h 
6

14.

y

3

3


1


 
216 
h9

3


9

8
4

6


13. (2n 3 3n 2 )6  26n 3 36n 2
 64n2 799n3
 46,656n5

y  1.4
y  
0.5
If y  1.4, then x  (1.4)  4 or 5.4.
If y  
0.5, then x  (
0.5)  4 or 3.5.
(5.4, 1.4), (3.5, 
 0.5)

h 3
 1


216 3

(5.4, 1.4)

1


216 3
 
h3


8 
4 O
4
8x

4 (3.5, 0.5)

3



216

h
3


8

6

 h3

Lesson 11-1

15.

3

1


z4(z4) 2  

z4 z2
3

 

z6
3

Page A46
1.

6


(
12)2




1

(
12)2
1

144


122

2.

3. (4 6)3  43 63
64 216
 13,824

4

23

4.

 z3
 z2

1

 

122
1



144
24

 34
16

 81

3


3

 4m3n2
12 2
 

19. 15

r12t2  15r 3 t 3
3

1

2

1

2

1

2

1

2

(32)

(22 5)
1

2

1


(22) 2

3

2


15r4t 3

1

2

5

1


2


1


1


16


20. 

256x2
y16  256 8 x 3 y 8
8

1

2

2

6


1


20  (32 3)

 3 2 15
 6
15


 (28) 8 x 4 y2
1


2


7. 
625
  625 4
4

9


 (43) 3 m3n2

16



1


18. 

64m9n6  64 3 m 3 n 3

 4 or 4
6. 27

1 4 8
  

5


17. 

a3b5  a 2 b 2

16
16
5. 1  1


16 2
(42) 2

1

2

1


16. (4r2t5)(16r4t8) 4  4r2t5(16 4 r 4 t 4 )
 (4r2t5)(2rt2)
 8r3t7

 2x 4 y2

1


 625 2
 
625
 or 25
1
1
8. 
 6
3



(15)6
15 3

Lesson 11-2

1



152


9.

(2a4)2

10. (x4)3

Page A46

1

225

1.

y


 4a8
x5  x12 x5
 x17
22

y

2.

(a4)2

y 
 3x

O

529

y 
 3x

x

O

x

Extra Practice

y

3.

9. log36 6  x
36x  6
(62)x  6
62x  61
2x  1
1
x  2
10. log3 y  4
34  y
81  y
11. log5 r  log5 8
r8
12. log5 35 
 log5 d  log5 5

x

O

y 
 3x  1

Lesson 11-3

35

log5 d  log5 5

Page A46

35

d

5
7d

1. p  (100 
 a)e
bt  a
p  (100 
 18)e
0.6(2)  18
 42.7%
2. y  ae
kt  c
y  140e
0.01(10)  70
 197° F
3a. y  6.7e

13. log4 
4
x
1


log4 4 2  x
1
 log
4
2


48.1

t

48.1


y  6.7e 15
y  0.271292 millions of cubic feet
y  271,292 ft3
3b. y  6.7e


48.1

t

1


48.1


15. 4 log8 2  3 log8 27  log8 a



1


log8 24  log8 27 3  log8 a
log8 16  log8 3  log8 a
log8 48  log8 a
48  a

r nt



A  Pert

A  P 1  n

A  5000e0.058(20)

A  5000 1  2



0.058 2(20)



A  $15,949.67
A  $15,688.63
The account that compounds continuously would
earn $261.04 more than the account compounded
semiannually.

Lesson 11-5
Page A47
1. log 5000  log (5 1000)
 log 5  log 103
 log 5  3 log 10
 0.6990  3
 3.6990
2. log 0.0008  log (0.0001 8)
 log 104  log 8
 
4 log 10  log 8
 
4  0.9031
 
 3.0969
3. log 0.14  log (0.01 14)
 log 102  log 14
 
2 log 10  log 14
 
2  1.1461
 
0.8539

Lesson 11-4
Page A47
1


1. 16 4  2
1 3

2

8
1

4

41


3.
4. log8 x  
2
5. logx 32  5
6. log1 16  
2
4

1

7. log5 5  x

log 81


4. log3 81  
log 3

1

5x  5
5x  51
x  
1
8. log3 27  x
3x  27
3x  33
x3
Extra Practice

x

14. log4 (2x  3)  log4 15
2x  3  15
2x  12
x6

y  6.7e 50
y  2.560257 millions of cubic feet
y  2,256,275 ft3
4. Continuously
Semiannually

2.

4x
1

2

4
log 12


5. log6 12  
log 6

 1.3869
log 29


6. log5 29  
log 5

 2.0922

530

7.

3x  45
x log 3  log 45

12.

log 45


x
log 3

8.

x  3.4650
6x  2x 
 1
x log 6  (x 
 1) log 2
x log 6  x log 2 
 log 2
x log 6 
 x log 2  
log 2
x (log 6 
 log 2)  
log 2

16  10(1  ex)
1.6  1  ex
0.6  ex
ln 0.6  x ln e

0.5108  x

Lesson 11-7


log 2


x
log 6 
 log 2

Page A47

x  
0.6309
9. 5 log y  log 32
log y5  log 32
y5  32
y5  25
y2

ln 2


1. t  
0.045

 15.40 yr
ln 2


2. t  
0.06

 11.55 yr
ln 2


3. t  
0.08125

Lesson 11-6

 8.53 yr
4a. y  5.2449(1.5524)x
4b. y  5.2449(eln 1.5524)x
y  5.2449e0.4398x

Page A47

4c. Use t  k; k  0.4398

ln 2

1. 3.5553
2. 
0.5763
3. 3.4398

ln 2


t
0.4398

 1.58 hr

4. log15 10 

ln 10

ln 15

 0.8503

Lesson 12-1

ln 14


5. log3 14  
ln 3

 2.4022
ln 350

Page A48


6. log8 350  
ln 8

1. d  3 
 7 or 
4

1  (
4)  
5, 
5  (
4)  
9,

9  (
4)  
13, 
13  (
4)  
17

5, 
9, 
13, 
17
2. d  
1 
 0.5 or 
1.5

2.5  (
1.5)  
4, 
4  (
1.5)  
5.5,

5.5  (
1.5)  
7, 
7  (
1.5)  
8.5

4, 
5.5, 
7, 
8.5
3. d  
8 
 (
14) or 6

2  6  4, 4  6  10,
10  6  16, 16  6  22
4, 10, 16, 22
4. d  2.8 
 3 or 
0.2
2.6  (
0.2)  2.4, 2.4  (
0.2)  2.2,
2.2  (
0.2)  2, 2  (
0.2)  1.8
2.4, 2.2, 2, 1.8
5. d  
x 
 4x or 
5x

6x  (
5x)  
11x, 
11x  (
5x)  
16x,

16x  (
5x)  
21x, 
21x  (
5x)  
26x

11x, 
16x, 
21x, 
26x
6. d  (2y 
 2) 
 (2y 
 4)
 2y 
 2y  (
2) 
 (
4)
2
2y  2, 2y  2  2  2y  4,
2y  4  2  2y  6, 2y  6  2  2y  8
2y  2, 2y  4, 2y  6, 2y  8

 2.8171
7.
5x  90
x ln 5  ln 90
ln 90


x
ln5

8.

x  2.7959
7x  2  5.25
(x  2) ln 7  ln 5.25
x ln 7  2 ln 7  ln 5.25
x ln 7  ln 5.25 
 2 ln 7
ln 5.25 
 2 ln 7


x
ln 7

9.

x  
1.1478
4x  4
3

x ln 4  ln 4
3

ln 4
3



x
ln 4

x  1.3962
6ex  48
ex  8
x ln e  ln 8
x  2.0794
11.
50.2  e0.2x
ln 50.2  0.2x ln e

10.

ln 50.2

0.2

x

x
19.5801

531

Extra Practice

a8

7. an  a1  (n 
 1)d
a16  2  (16 
 1)5
 77
8. 
20  6  (n 
 1)(
2)

26  
2n  2

28  
2n
14  n
9. 42  a1  (12 
 1)4
42  a1  44

2  a1
10. 30  7  (13 
 1)d
23  12d
11
1 12  d

1



6. r  
a10 or a 2

a6 a2   a4, a4a2   a2, a2 a2   1
1

1

1

a4, a2, 1
7. an  a1rn
1
a6  9(2)6
1
 288
8. 100  a1(4)8
1
100  16,384 a1
25

4096

 a1
1 5
1

9. 10  a1
2
1

10  1
6 a1

11. d  10 
 10.5 or 
0.5
a24  10.5  (24 
 1)(
0.5)
a24  
1

160  a1

a2  160 
2 or 
80
1

a3  
80 
2 or 40
1

n

12. Sn  2[2a1  (n 
 1)d]; d  2.8 
 2 or 0.8

160, 
80, 40
10. 256  4r4
1
64  r3
4r
4(4)  16, 16(4)  64
4, 16, 64, 256

12

S12  2[2 2  (12 
 1)(0.8)]
 6(4  8.8)
 76.8
n

13. Sn  2[2a1  (n 
 1)d]
n

80  2[2 (
4)  (n 
 1)4]

a1 
 a1rn

9



11. Sn  
1
 r ; r  3 or 3

160  n(
8  4n 
 4)
160  
12n  4n2
0  4(n2 
 3n 
 40)
0  4(n 
 8)(n  5)
n  8 or n  
5
Since n cannot be negative, n  8.

3 
 3(3)6


S6  
1
3
3 
 2187




2

 1092
12a. There are four 15-minute periods in an hour, so
r  24 or 16.
bt  b0 16t
12b. b4  12 164
 786,432

Lesson 12-2
Page A48
7

1. r  1
4 or 0.5

Lesson 12-3

3.5(0.5)  1.75, 1.75(0.5)  0.875,
0.875(0.5)  0.4375
1.75, 0.875, 0.4375

Page A49

4


2. r  

2 or 
2

1. lim

n→


8(
2)  16, 16(
2)  
32, 
32(
2)  64
16, 
32, 64
3. r 

3

8

1

2

3

or 4

2n

n→ 3n

 lim

n4 
 3n

n3
n→

 lim n 
 n2 
3

n→

 lim n. But as n approaches infinity, n becomes

   392 , 392 34  12278 , 12278 34  58112

n→

increasingly large, so there is no limit.

9
27
81
, , 
32 128 512

5
r  10 or 
0.5

3. lim

n→

2.5(
0.5)  
1.25, 
1.25(
0.5)  0.625,
0.625(
0.5)  
0.3125

1.25, 0.625, 
0.3125

4.

8
2


5. r  8 or 
2

16
2
  16
2
, 16
2

2
  32, 32
2
  32
2


8n2  6n 
 2

4n2

4n2 
 2n  1

lim 
n2  2
n→

8n3

2
n→ 4n

 lim

4n2

n2

2n

1


2

2
n→ 4n

 lim


 n2  n2
––––
 lim
n2
2
n→

 n2
n2
4
00

532

6n

2
n→ 4n

 lim

200
2



10
4

16
2
, 32, 32
2


Extra Practice

4

n→ 3n
2
 0  3
2
 3

 lim

2. Limit does not exist. lim

3 3
 
8 4

4.

4  2n

3n

5.

n 3 
 n2  4

lim 
3
n→ 5  2n

n3

n3

n2

2. an  3, an1  3
r

1
00



02
1

 2
6. Limit does not exist.

lim (2n

1


3. an  
3n(n  1)3 , an1 

n→

becomes increasingly large, so there is no limit.
9

9

—
——
7. 0.0
9
—
100  10,000  . . .
9

r

1

—
——
a1  —
100 , r  100
9


100
Sn  —
1
—
1
—
100
1

3

3



8. 0.13
  1
0  100  1000  . . .
3

1



a1  
100 , r  10

1

convergent

1
—
Sn  
 1
1
0 
1 
 1
0
2

407

r

407



9. 7.4
0
7
7
1000  1,000,000  . . .
1



a1  
1000 , r  1000
407


1000
Sn  7  ——
1

1

1000

7


5. The general term is 
6n  1 .

11

10. a1 

r

7

6n  1

or

or

for all n, so divergent
1



6. The general term is 
(2n)2 or 4n2 .
1

4n2

1

 n for all n, so convergent.
1

1



7. an  
2n  1 , an1  2n1  1

Sn  —
1
1 
 2


1

n

1

1

2

1

20

2

20

4


2n1
 lim —
4
n→ 
2n
2n
 lim 
n
n→ 2 2
1
 2

convergent

 727
1
,
20

4

4


4. an  2n , an1  
2n1

 1
5

1

40
—
1

20

1


3n1(n  2)3
––––
 lim
1
n→ 

3n(n  1)3
n
3 (n3  3n2  3n  1)
 lim 
3n 3(n3  6n2  12n  8)
n→

 3

3

100

407

1

3n1(n  1  1)3

n3
3n2
3n
1

  
n 3  n 3  n3  n 3
———
 lim 3n3 6n2 12n 8
   
n→ 
n 3  n3  n 3  n 3
1000


3000

 —11—
1

n1

3
—
 lim n

n→
3
n1
 lim n
n→
1
n
 lim n  lim n
n→
n→

10
1
test provides no information

1
n
n  
—
 lim 2n —

2
2  n   lim 2
n→
n→
——  1
n
1) or lim 2n As n approaches infinity, 2n

2n n
lim ——
n→  2  n

n→

n1

n

4


 n3  n3
 lim 
5
2n3
n→
  
n3
n3

1

10

r

11. Sn does not exist. This series is geometric with a
common ratio of 2. Since this ratio is greater than
1, the sum of the series does not exist.

Lesson 12-4
Page A49

1


2n1  1
——
 lim
n→ 1
2n  1
2n  1

 lim 2n 2
1
n→
2n
1



2n
2n
——
 lim 2n 2 1
n→   
2n
2n
10


20
1
 2

convergent

1. an  (2n)2, an1  (2n1)2
 22n
 22n2
22n2
2
n
n→ 2
2
n
2
22

 lim 
22n
n→

r  lim

4
divergent

533

Extra Practice

2n1

2n

Lesson 12-6



8. an  
n  2 , an1  (n  1)  2

r

2n1

n3
 lim —
n
n→ 2
n2
2n 2(n  2)

 lim 
n
n→ 2 (n  3)
2n  2

 lim 
n→ n  3
2n
2
  
n
n
 lim —
n→ n  3
n
n
20


10

Page A50
1. (2  x)4  16  32x  24x2  8x3  x4
2. (n  m)5  n5  5n4m  10n3m2  10n2m4 
5nm4  m5
3. (4a 
 b)3  64a3 
 48a2b  12ab2 
 b3
6 5 m4(
3)2


4. (m 
 3)6  m6(
3)0  6 m5 (
3)1  
2 1

2
divergent



6 5 4 m3 (
3)3

3 2 1



6 5 4 3 2 m1(
3)5

5 4 3 2 1



6 5 4 3 2 1 m0(
3)6

6 5 4 3 2 1



6 5 4 3 m2 (
3)4

4 3 2 1

 m6 
 18m5  135m4 
 540m3
 1215m2 
 1458m  729
4 3(2r)2(s)2


5. (2r  s)4  (2r)4(s)0  4(2r)3(s)1  
2 1

Lesson 12-5

4 3 2 (2r)1(s)3

4 3 2 1(2r)0(s)4

4 3 2 1
24r2s2  8rs3  s4



3 2 1

 16r4  32r3s 

Page A49

6. (5x 
 4y)3  (5x)3(
4y)0  3(5x)2 (
4y)1

5

1.



n1

3 2(5x)1(
4y)2

(3n 
 1)  (3 1 
 1)  (3 2 
 1) 
(3 3 
 1)  (3 4 
 1) 
(3 5 
 1)
 2  5  8  11  14
 40

3 2 1(5x)0(
4y)3

3 2 1
 240xy2 
 64y3



2 1

 125x3 
 300x2y

8 7 6 5 4 3 2 1

5 4 3 2 1 3 2 1
 56x3y5

7.

8!

5!(8 
 5)!

x8
5y5 

34 44 54 6
 12  16  20  24
 72

8.

7!

4!(7 
 4)!

3 9
b7
4 

34  
4 3 2 1 3 2 1 b
3
 315b

 (k2 
 2)  (32 
2 2)  (42 
2 2)  (52 
 2)
k3

9.

10!

2!(10 
 2)!

6

 4a  4
k3

2.

7

3.

 (6 
 2)  (7 
 2)
 7  14  23  34  47
 125

8

4.

j

5

8


8 3
4
5
6
 7  8  9
2683

 3
9240

4

6

7 6 5 4 3 2 1

(4z)10
2 (
w)2
10 9 8 7 6 5 4 3 2 1

8
2
 
2 1 8 7 6 5 4 3 2 1 (4z) (
w)
8
2
 45 65,536z w
 2,949,120z8 w2







j3  43  53  63  73
j4
4

7

10.

12!

7!(12 
 7)!

(2h)12
7(
k)7




7

10



x3y5

8

11

12 11 10 9 8 7 6 5 4 3 2 1

7 6 5 3 2 1 5 4 3 2 1
792 32h5 (
k)7

(2h)5(
k)7


 
25,344h5k7

 3p  30  31  32  33  34
p0

5.

 1  3  9  27  81
 121



6.



2

n1

3 n

4

 




S



n1

 
 

 

Lesson 12-7
2

3 3

4

 

. .2

. . .

Page A50
1. ln (
3)  ln (
1)  ln (3)
 i  1.0986
2. ln (
4.6)  ln (
1)  ln (4.6)
 i  1.5261
3. ln (
0,75)  ln (
1)  ln (0.75)
 i 
 0.2877

3 

4

 

1
12
—
3
1 
 4

6

4

7.

3 1
3 2
2 4  2 4
3 
 2 4
1
1
7
12  18  23
2 .

(3n  2)

9

 4m
1
m0

9.

Extra Practice

(1.2)2

10

8.



r2

(1.2)3

(1.2)4



4. e1.2  1  1.2  2
!  3!  4!
 1  1.2  0.72  0.288  0.0864
 3.29


4r



 w!
w1

10.

(
0.7)2

(
0.7)3

(
0.7)4

 3
 4
5. e
0.7  1  (
0.7)  2
!
!
!
 1 
 0.7  0.245 
 0.057  0.010
 0.50

534

(3.65)2

(3.65)3

4. z0  4  4i
z1  0.5(4  4i)  i
 2  2i  i
2  3i
z2  0.5(2  3i)  i
 1  1.5i  i
 1  2.5i
z3  0.5(1  2.5i)  i
 0.5  1.25i  i
 0.5  2.25i
2  3i, 1  2.5i, 0.5  2.25i
5. p1  p0  rp0
p1  4000  (0.054)4000
 4216
p2  4216  (0.054)4216
 4443.66
p3  4443.66  (0.054)4443.66
 4683.62
p4  4683.62  (0.054)4683.62
 4936.54
p5  4936.54  (0.054)4936.54
 5203.11
$4216, $4443.66, $4683.62, $4936.54, $5203.11

(3.65)4

6. e3.65  1  3.65  2
 3
 4
!
!
!
 1  3.65  6.661  8.105  7.395
 26.81
x2

x4

x6

x8

7. cos x  1 
 2!  4! 
 6!  8!


cos 4  cos 0.7854
(0.7854)2

(0.7854)4

(0.7854)6

(0.7854)8

 4

 6
 8
 1 
 2
!
!
!
!
0.6169

0.3805

0.2347

0.1448



 1 
 2  24 
 
720  40,320

 0.7071





2

x5

x7

actual value: cos 4  2  0.7071
x3

x9

8. sin x  x 
 3!  5! 
 7!  9!


sin 6  sin 0.5236
(0.5236)3

(0.5236)5

(0.5236)7

 5

 7
 0.5236 
 3
!
!
!
(0.5236)9

 9
!
0.1435

0.0394

0.0108

0.0030




 0.5236 
 6  
120 
 5040  362,880

 0.5000



actual value: sin 6  0.5


x2

x4

x6

x8

9. cos 3  1 
 2!  4! 
 6!  8!


cos 3  cos 1.0472
(1.0472)2

(1.0472)4

(1.0472)6

Lesson 12-9

(1.0472)8

 4

 6
 8
 1 
 2
!
!
!
!
1.0966

1.2026

1.3188

1.4462



 1 
 2  24 
 
720  40,320

 0.5000
actual value: cos



3

Page A50
1. Step 1: Verify that the formula is valid for n  1.
Since S1  2 and 1(1  1)  2, the formula is valid
for n  1.
Step 2: Assume the formula is valid for n  k and
show that it is also valid for n  k  1.
2  4  6  . . .  2k  k(k 
 1)
Derive the formula for n  k  1 by adding
2(k  1) to each side.
2  4  6  . . .  2k  2(k  1)
 k(k  1)  2(k  1)
2  4  6  . . .  2k  2(k  1)
 (k  1)(k  2)
Apply the original formula for n  k  1.
Sk1  (k  1)((k  1)  1) or (k  1)(k  2)
The formula gives the same result as adding the
(k  1) term directly. Thus, if the formula is valid
for n  k, it is also valid for n  k  1. Since Sn is
valid for n  1, it is also valid for n  2, n  3,
and so on indefinitely. Hence the formula is valid
for all positive integral values of n.

 0.5

Lesson 12-8
Page A50
1. f(
2)  2 (
2) or 
4
f(
4)  2 (
4) or 
8
f(
8)  2 (
8) or 
16
f(
16)  2 (
16) or 
32

4, 
8, 
16, 
32
2.
f(4)  42 or 16
f(16)  162 or 256
f(256)  2562 or 65,536
f(65,536)  65,5362 or 4,294,967,296
16, 256, 65,536, 4,294,967,296
3. z0  2i
z1  0.5(2i)  i
 2i
z2  0.5(2i)  i
 2i
z3  0.5(2i)  i
 2i
2i, 2i, 2i

535

Extra Practice

2. Step 1: Verify that the formula is valid for n  1.

4. dependent
5. dependent

1(1  1)(1  2)

6

 1, the formula is
Since S1  1 and
valid for n  1.
Step 2: Assume the formula is valid for n  k and
show that it is also valid for n  k  1.
k(k  1)

5!


6. P(5, 5)  
(5 
 5)!
5 4 3 2 1

 1
 120

k(k  1)(k  2)

1  3  6  . . .  2  6

8!


7. P(8, 3)  
(8 
 3)!

Derive the formula for n  k  1 by adding
(k  1)(k  2)

2



to each side.
k(k  1)

 336

(k  1)(k  2)

1  3  6  . . .  2  2


4!


8. P(4, 1)  
(4 
 1)!

k(k  1)(k  2)

6

136. . .

(k  1)(k  2)
 2
k(k  1)
(k  1)(k  2)
 2  2

4 3 2 1



3 2 1
4

k(k  1)(k  2)  3(k  1)(k  2)

6
k(k  1)
(k  1)(k  2)
 6  . . .  2  2


13



10!


9. P(10, 9)  
(10 
 9)!



(k  1)(k  2)(k  3)

6

(k  1)((k  1)  1)((k  1)  2)

6

or

(k  1)(k  2)(k  3)

6

10 9 8 7 6 5 4 3 2 1

1

 3,628,800
9!


10. P(9, 6)  
(9 
 6)!

Apply the original formula for n  k  1.
Sk1 

8 7 6 5 4 3 2 1

5 4 3 2 1



9 8 7 6 5 4 3 2 1

3 2 1

 60,480
7!


11. P(7, 3)  
(7 
 3)!

The formula gives the same result as adding the
k  1 term directly. Thus, if the formula is valid
for n  k, it is also valid for n  k  1. Since the
formula is valid for n  1, it is also valid for n  2,
n  3, and so on indefinitely. Hence, the formula
is valid for all positive integral values of n.
3. Sn: 5n 
 1  2r for some integer r.
Step 1: Verify that Sn is valid for n  1.
S1  51 
 1 or 4. Since 4  2 2, Sn is valid for
n  1.
Step 2: Assume that Sn is valid for n  k and show
that it is also valid for n  k  1.
Sk → 5k 
 1  2r for some integer t
Sk1 → 5k1 
 1  2t for some integer t
5k 
 1  2r
5(5k 
 1)  5(2r)
5k1 
 5  10r
5k1 
 1  10r  4
5k1 
 1  2(5r  2)
Thus, 5k1 
 1  2t, where t  (5r  2) is an
integer. Thus if Sk is valid, then Sk1 is also
valid. Since Sn is valid for n  1, it is also valid for
n  2, n  3, and so on indefinitely. Hence, 5n 
 1
is even for all positive integral values of n.



7 6 5 4 3 2 1

4 3 2 1

 210
12.

P(5, 2)

P(2, 1)





5!

(5 
 2)!
—
2!

(2 
 1)!
5! 1!

3! 2!
5 4 3 2 1 1

3 2 1 2 1

 10
13.

P(8, 6)

P(7, 4)





8!

(8 
 6)!
—
7!

(7 
 4)!
8! 3!

7! 2!
8 7 6 5 4 3 2 1 3 2 1

7 6 5 4 3 2 1 2 1

 24
14.



P(5, 2) P(8, 4)

P(10, 1)

5!
8!
 
(5 
 2)! (8 
 4)!
———
10!

(10 
 1)!

9! 8! 5!

 
10! 4! 3!



Lesson 13-1

9 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 5 4 3 2 1

10 9 8 7 6 5 4 3 2 1 4 3 2 1 3 2 1

 3360
4!


15. C(4, 2)  
(4 
 2)! 2!

Page A51

4 3 2 1



2 1 2 1
6

1. Using the Basic Counting Principle,
6 6 6  216.

10!


16. C(10, 7)  
(10 
 7)! 7!

8!


2. P(8, 8)  
(8 
 8)!





8 7 6 5 4 3 2 1

1

 120

 40,320
3. independent
Extra Practice

10 9 8 7 6 5 4 3 2 1

3 2 1 7 6 5 4 3 2 1

536

6!

4


17. C(6, 5)  
(6 
 5)! 5!



42

4!

4 3 2 1

1 3 2 1

4

9. P(s)  1
6

7 6 5 4 3 2 1

4 3 2 1 3 2 1

1
 2
0
1

20
odds  —
19

20

 140
3!

8!

(8 
 7)! 7!


19. C(3, 1) C(8, 7)  
(3 
 1)! 1!
3 2 1

2 1 1

8 7 6 5 4 3 2 1

1 7 6 5 4 3 2 1

6

10. P(s)  1
6

 24
9!

4!

(4 
 3)! 3!


20. C(9, 5) C(4, 3)  
(9 
 5)! 5!

9 8 7 6 5 4 3 2 1

4 3 2 1 5 4 3 2 1


4 3 2 1

1 2 1

3

15

1

1

8
—
7

8

2 4


120
1
 —1—
5
1

15
odds  —
14

15

Lesson 13-2
Page A51

2.

8!

2! 2!

3.

7!

3!

4.

10!

2!

5.

4!

2!

7 6 5 4 3 2 1

1

 5040

1

or 1
9
5

15

P(f )  1 
 P(s)
1

 10,080

1

or 7

1

 1 
 1
5
14

 1
5
1

or 1
4

C(2, 1) C(4, 1)

C(4, 1) C(10, 1)

C(2, 1) C(10, 1)

2 4

4 10

2 10





120  120  120

 840


P(f )  1 
 P(s)

    

C(16, 2)
C(16, 2)
C(16, 2)

7 6 5 4 3 2 1

3 2 1



7

 1 
 8 or 8

12. P(s)  P(1 white, 1 yellow)  P(1 yellow, 1 red) 
P(1 white, 1 red)

8 7 6 5 4 3 2 1

2 1 2 1



19


 1 
 2
0 or 20

C(2, 1) C(4, 1)



6

P(f )  1 
 P(s)


11. P(s)  
C(16, 2)

1.

12

1

8

odds 

 504

7!

0!

3




8. P(red or green)  
8  4 2  14 or 7

7!

(7 
 3)! 3!


18. C(4, 3) C(7, 3)  
(4 
 3)! 3!



6

84

6



2




7. P(not red)  
8  4  2  14 or 7

6 5 4 3 2 1

1 5 4 3 2 1



4




6. P(green)  
8  4  2  14 or 7

1

1

1



 1
5  3  6

10 9 8 7 6 5 4 3 2 1

2 1

17

 30

 1,814,400

P(f)  1 
 P(s)

4 3 2 1



2 1

17

13

 1 
 30 or 30

 12

odds 

6. circular; (4 
 1)! or 6
7. circular; since the bracelet can be turned over
(9 
 1)!
there are 2 or 20,160 permutations
8. linear; 5! or 120

17

30
—
13

30

17

or 1
3

Lesson 13-4
Lesson 13-3

Page A52
6

2
3
  
7
14
10 10
25


independent; 1
8 18  81
1
6
3

independent; 2 1
0  10
1
1
1
11

inclusive; 6  6 
 3
6  36
4
4
8
2



exclusive; 5
2  52  52 or 13

1. dependent; 8

Page A51
4

2.

1


1. P(ace)  5
2 or 13

3.

44444

52
20
5



 52 or 1
3
6
3
 or 
52
26

2. P(a card of 5 or less) 
3. P(a red face card) 

4.
5.

4. P(not a queen)  1 
 P(a queen)
4

 1 
 5
2
48

12

 52 or 1
3
2

2

1




5. P(blue)  
8  4  2  14 or 7

537

Extra Practice

Lesson 13-5

200

1

5

5

Page A52



256



3


5. P(exactly 3 hits)  C(5, 3)
1000 
200

8
4
  
12 11
—
8

12
4

11

1

125

 10

2

800


1000 

16

25

32



625 or 0.0512

6. P(at least 4 hits)
 P(4 hits)  P(5 hits)

2. P(second clip is bluefirst clip was blue)


4

800


1000 

256

625



625 or 0.4096

1. P(second clip is bluefirst clip was red)


1


4. P(exactly 1 hit)  C(5, 1)
1000 

4


 C(5, 4)
1000 

4
3
  
12 11
—
4

12
3

11

200

5


1

625

21

3125

4

5

1

800


1000 

1

1

3125

200

1

or 0.00672

3. P(numbers on dice matchsum is greater than 7)



3

36
—
15

36
1
3
 or 
5
15

Lesson 14-1
Page A53
1. range  70 
 22 or 48

4. P(sum is greater than 7numbers match)



3

36
—
6

36
3
1
 or 
6
2

2. Sample answer: 10
3. Sample answer: 20, 30, 40, 50, 60, 70, 80
4. Sample answer: 25, 35, 45, 55, 65, 75
5.

5. P(ball is from second boxball is white)


P(2nd box and white)

P(white)
1 3
 
2 8

 ——


1 4
1 3
    
2 7
2 8
21

53

6.

Lesson 13-6

Grams of Fat Consumed
by Adults

6
1 3 1 0


2
2

1. P(all heads)  C(3, 3)
1


1

8

  

Frequency 4

1

2

1

8

1 2

2. P(exactly 2 tails)  C(3, 2)2
3

1

4

0

1

12

1

2

3

 8

1 2 1 1

2
1
1
  1 
2
8

 C(3, 2)2
1

4

3
3

8
4

8

 

1 3 1 0

2

 C(3, 3)2

 

1

1
 8
1
or 2

Extra Practice

0 10 20 30 40 50 60 70 80
Grams of Fat

7. Sample answer: 40–50

3. P(at least 2 heads)
 P(2 heads)  P(3 heads)



Frequency
6
7
8
3
5
1

8

Page A52



Grams of Fat
20-30
30-40
40-50
50-60
60-70
70-80

538

5


 C(5, 5)
1000 

0

800


1000 

Lesson 14-2

11. Order the values from least to greatest. The
median lies between the fourth and fifth terms.
3, 4, 4, 7, x, 12, 16, 19
7x

Page A53
1. X


1
(130
4

7.5  2

 150  180  190)

15  7  x
8x

 162.5

150  180

Md  2 or 165
Mode: none

Lesson 14-3

1

2. 
X  6(15  16  17  18  18  19)
 17.2

17  18

Md  2 or 17.5

Page A54
1. interquartile range  Q3 
 Q1
 58 
 39
 19

Mode  18
1

3. X
  6(25  28  30  36  38  42)
 33.2

30  36

2

Md 

19

semi-interquartile range  2 or 9.5
or 33

Mode: none
1

4. 
X  1
0 (1  2  3  4  5  5  6  9  9  10)
 5.4
Md  5
Mode  5 and 9

40

50

60

70

80

2. interquartile range  Q3 
 Q1
 7.65 
 2.75
 4.9

1

5. X
  1
2 (2.3  2.5  4  2(5.6)  6  6.4 
6.5  2(7)  8  10)

4.9

semi-interquartile range  2 or 2.45

 5.9

6  6.4

Md  2 or 6.2

1

Mode  5.6 and 7

3. X


2

4

5

6

7

8

9 10

 175  180  180  195  200  212
 220  250)
 195.7

1


MD  9 (
45.8  
20.8  
15.8
 . . .  54.2)
 21.98

1

6. X
  1
2 (14  2(15)  16  20  21  24  27
 28  36  2(39))
 24.5

21  24

Md  2 or 22.5
Mode  15 and 39
1

7. X
  1
8 (3.0  3.4  3.6  5.2  2(5.4)  2(5.6)
5.7  6.2  6.3  6.8  7.0  7.1  7.6 
7.7  8.2)
 5.9
Md  5.7
Mode  5.4 and 5.6

(
45.8)2  (
20.8)2  . . .  54.22

9

j

 27.56
1

4. X
  11(1.4  2  2.4  2.9  3  3.5  3.7  4.2
 4.6  5.3  5.5)
 3.5
1
MD  11(
2.1  
1.5  
1.1  . . .  2)
 1.05

1

8. X
  1
4 (800  820  830  890  960  970  980
 1040  2(1050)  1080  1110  1170 
1180)

j

 995

(
2.1)2  (
1.5)2  (
1.1)2  . . .  22

11

 1.26
5a. Md  18

980  1040

Md  2 or 1010
Mode  1050
9.
stem 
1 
2 
3 
4 
5 
12  12

3

1
(150
9

16  18

5b. Q1  2 or 17
leaf
2 7
3 4
4 4
0 2
5

Q3  20
5c. interquartile range  Q3 
 Q1
 20 
 17
3

9
4 6 8 9 9
5 6 7 9
4

3

5d. semi-interquartile range  2 or 1.5
5e. Any points less than 17 
 1.5(3) or 12.5 and any
points greater than 20  1.5(3) or 24.5 are
considered outliers. There are no such points.

1

10. 8  6(4  5  6  9  10  x)
48  34  x
14  x

539

Extra Practice

Lesson 15-1

5f.
15 16 17 18 19 20 21 22 23 24

Page A55
1. lim (x2  2x 
 2)  42  2(4) 
 2
x→4
 22

Lesson 14-4

2. lim (
x4  x3 
 2x  1)  
(
1)4  (
1)3 
 2(
1)  1
x→
1

Page A54
1a. 25% corresponds to t  0.3.
10  0.3(2)  9.4 
 10.6
1b. 10 
 8  2, 14 
 10  4
tj  2
tj  4
t(2)  2
t(2)  4
t1
t2
68.3%

2

 34.15%

95.5%

2

4. lim

x→
4

x→
4

(x  4)(x 
 4)

x4

x→
4

x2  5x  6


2
x→
2 x  x 
 2

 47.75%

(x  3)(x  2)

x→
2 (x 
 1)(x  2)
x3

 lim 
x→
2 x 
 1

2  3



2 
 1
1
 
3
3x  9
3(x  3)
  lim 
lim 
2
x→2 x 
 5x 
 24
x→2 (x 
 8)(x  3)
3

 lim 
x→2 x 
 8
3


2
8
3
1
 
6 or 
2

5. lim

6.

 43.3%

2a. 0.683(400)  273.2
2b. 0.955(400)  382
0.683
(400)
2

 lim

 
4 
 4
 
8

1d. 80% corresponds to t  1.3.
10  1.3(2)  7.4 
 12.6

2c.

x2 
 16

x4

 lim (x 
 4)

34.15%  47.75%  81.9%
1c. 10 
 7  3, 10 
 10  0
tj  3
tj  0
t(2)  3
t(2)  0
t  1.5
t0
86.6%

2

1

3. lim (x  sin x)  0  sin 0
x→0
0

 lim

 136.6

Lesson 15-2
Lesson 14-5
Page A55

f(x  h) 
 f(x)

h
h→0
5(x  h) 
 5x


 lim
h
h→0
5x  5h 
 5x
 lim h
h→0
5h
 lim h
h→0

1. f
(x)  lim

Page A54
1.2

1. jX   or about 0.13
 
90
3.4
2. jX   or 0.34
00
 
1
12.4
3. jX   or about 0.80
40
 
2
4. A 1% confidence level is given when P  99% and
t  2.58.

5
2. f
(x)  lim

h→0

4.2

jX   
40
 

 0.6640783086
interval: X
  tjX  150  2.58jX


 148.29 
 151.71

 lim

9(x  h) 
 2 
 (9x 
 2)

h

 lim

9x  9h 
 2 
 9x  2

h

 lim

9h

h

h→0
h→0

5. A 1% confidence level is given when P  99% and
t  2.58.

f(x  h) 
 f(x)

h

h→0

9

10

jX  
78
 

 1.132277034
interval: X
  tjX  320  2.58jX


 317.08 
 322.92

1

2

3. f(x)  2x  3
1

f
(x)  2 1x1
1  0
1

 2
4. f(x)  x2  4x  8
f
(x)  2x2
1  4 1x1
1  0
 2x  4

Extra Practice

540

5. f(x)  x5
F(x) 

6. f(x) 

1

21

F(x)  2

7. f(x) 
F(x) 

8. f(x) 


1
x3
3

n

1

11

 lim

1

11

n→
2 2

x11  2x  C

0

x11 
 x  C

 lim

x11  x  C

n→
8
 3 
56
 3





5

2.

1

n




lim

n→
5
 lim n
n→


 lim

n→
5
 lim n
n→



1

n

1

n

0



5 7
 n
n 2

 lim



35

2



25

2n

n→
35
3
 2  0 
 2
32
 2 or 16



n

1

3n 
 n(1  2  . . .  n)
n(n  1)

 1500 
 250 or $1250



1

(x  1)dx 


0

n

(x  1)dx

Lesson 15-4

5 n

Page A55

x6dx  17x7  C
1
2. 5x4dx  5 5x5  C

n

1.

 x5  C

 2  . . .  n)

n(n  1)

2


 lim

1

n



1

2

n→
1 3

 lim n 2n 
n→
3
1

 lim 2  2n
n→

 2
5

. .  3 
 n

1

2

5

3 
 n1  3 
 n2  .

n→

 1  n  1  . . .  n  1

n  n

i

 lim 1500 
 2501  n

5 2

n

n

5n00

 3 
 n
i1

500

 1  n  1  . . .  n  1

1
(1
n

500i

n→

5

 lim

n→

5

500

n

6 
 0.0021500  n

 
 lim 1500 
 n
2 
2

n  n (1  2  . . .  n)
1

n

n

500

n→ n
500
 lim n
n→

5
i
1
lim  n  1 n
    1 n 
 n→
i1

5i

n
n→ i1
5 1
5
 lim n n
n→

 lim



2n   8

2

 lim

0

(x  1)dx 

  4n

n2

(6 
 0.002x)dx

n→





88

 lim

   



8 n(n  1)
 
n
2

2  n3  n1  81  n1  8


n→ i1

3i
3
5 n n
n→ i1
45 n(n  1)
 lim n2 2
n→
45
2

 lim 
2 (n  n)
n→ 2n
2
45n
45n


 lim 
2  lim 2n2
n→ 2n
n→





4

3

 lim

45

2
45

2

 4n





2 3

. .  n2)

2 4
n(n  1)(2n  1)
 
 
2
6
n→ n n
8
2n3  3n2  n
16
 lim n3 6  n2
n→

n



2 3 2

 lim

1500



2 1

2 4
2
2
 
2 (1  2  .
n→ n n
8
 n(1  2  . . .  n)

4.

5x dx  lim

2 2 2

 4n  4  n 

 lim


 x2  x  C

Page A55



2

2 1


n 

2

n

2ni   4 n2

4

4n  4  . . .  n  4n  4

Lesson 15-3

1.

2

2i
 n
n→ i1

2000

3

 4x  4)dx

 lim

2
x3 
 4x2  2x  C
3
1
3
x3 
 x 
 1
5
4
1
3
1
1
  x31 
  
5 31
4 11
3
1
 x4 
  x2 
 x  C
8
20
x3 
 2x2  x

x
2
x 
 2x  1
1

2

0

x21 
 8

 21 
 2
F(x)  
21 x



 (x
2

3.

1
x51  C
51
1
x6  C
6
2x2 
 8x  2



(x2 
 x  5)dx  13x3 
 12x2  5x  C
1
1
4. (
4x4  x2 
 6)dx  
4 5x5  3x3 
 6x  C

3.
1

n  n

n(n  1)

2

4





2

5.



2

14x6dx  14
 2x7


0

1

 
5x5  3x3 
 6x  C
2
1
x7
7
2
2
2

 (2 27) 
 (2 (
2)7)
 256 
 (
256)
 512

541

Extra Practice



6

6.

0

1

(x  2)dx  2x2  2x

6
0

7.



2

4

 2 62  2 6 
 2 02  2 0
1

4

1

1



Extra Practice

5

4

(x2 
 2x 
 8)dx

1

1
x2
2

1


 8x

 3x3 
 x2 
 8x

4



1

3

2

1
 43 
 4
3
16
16
 
 

3
3
32

3





 3x3 
 2




(x 
 4)(x  2)dx


 30 
 0
 30
(x2 
 4)dx  3x3 
 4x



5

8.

4 
 

1

3

23


 4 2

53




52

70

10

 3

542

80

4

4


 8 5 
 3 43 
 42 
 8 4

 
3 
 
3



5

5

1

Chapter Tests
Chapter 1 Test

12.
16
14
12
10
y  4x  12 8
6
4
2

Page A56
1. D  [
1, 0, 2, 3}; R  {2, 4, 5}; yes
2. D  {
5, 4, 6, 7}; R  {
3, 
2, 0, 2, 7}; no
3. f(4)  4 
 3 42
 4 
 48 or 
44
4. f(
7)  
7 
 3(
7)2
 
7 
 147 or 
154
5. f(a  2)  a  2 
 3(a  2)2
 a  2 
 3(a2  4a  4)
 a  2 
 3a2 
 12a 
 12
 
3a2 
 11a 
 10

O


2 
 4

5


15. m  
8
0

6

5

y 
 3  3(x  1)
5

5

y 
 4  
4(x 
 0)

3

5

14

y 
 4  
4x

3

y  3x  3

1200

3

 117.89 Im/m2
6b. d cannot be negative or zero.
7. f(x)  g(x)  x2 
 7  x  3
 x2  x 
 4
f(x) 
 g(x)  x2 
 7 
 (x  3)
 x2 
 x 
 10
f(x) g(x)  (x2 
 7)(x  3)
 x3  3x2 
 7x 
 21

y  
4x  4
16. parallel: y 
 2  4(x 
 0)
y 
 2  4x
4x 
 y  2  0
1

perpendicular: y 
 2  
4(x 
 0)
1

y 
 2  
4x
4y 
 8  
x
x  4y 
 8  0

x2 
 7



x3

8. [f 
 g](x)  f(g(x))
 f(4x 
 5)
 4x 
 5  1
 4x 
 4
[g 
 f ](x)  g(f(x))
 g(x  1)
 4(x  1) 
 5
 4x  4 
 5
 4x 
 1
9. [f 
 g](x)  f(g(x))
 f(2x2  6)
 5(2x2  6)
 10x2  30
[g 
 f ](x)  g(f(x))
 g(5x)
 2(5x)2  6
 50x2  6
y
10.
11.

3

 8 or 
4

y 
 3  3x  3



4(0.9)2

x

O

1 2 3 4 5x

14. y 
 3  3(x 
 (
1))

P

O

y
|x |  2

y


5
4
3
2
1

2

4


6a. E  
4d2

f(x)

g(x)

13.

y

1

17. x 
 5y  3 ⇒ m  5
1

parallel: y 
 2  5(x 
 (
1))
1

y 
 2  5(x  1)
5y 
 10  x  1
x 
 5y  11  0
5

perpendicular: y 
 2  
1(x 
 (
1))
y 
 2  
5(x  1)
y 
 2  
5x 
 5
5x  y  3  0
18.

19.

g (x )

f (x )

g (x ) 
 x  1

O

x

O

x

y

x
O

x
2x  y 
 1

y 
 3x  6

543

Chapter Tests

20a.
100
80
Economics 60
Grade
40
20
0

20

→

x 
 5y  z  
15

7x  y  2z  1

→

7x 
 9y  
25
9x 
 11y  
31

→

7x 
 9y  
25
7x 
 9(2)  
25
7x  
7
x  
1
( 
1, 2, 
4)
5. y  12 
 2x

40 60 80 100
Statistics Grade

20b. Sample answer: Using (47, 67) and (95, 88),
y  0.44x  46.44
88 
 67


m
95 
 47

 0.4375
y 
 67  0.4375(x 
 47)
y 
 67  0.4375x 
 20.5625
y  0.4375x  46.4375
y  0.44x  46.44

4x  6y 
 3z  20
3x 
 15y  3z  
45
7x 
 9y
 
25
2x 
 10y  2z  
30

7x  y  2z 
1
9x 
 11y
 
31
63x 
 81y  
225
63x 
 77y  
217

4y  
 8
y2
x 
 5y  z  
15

1 
 5(2)  z  
15

11  z  
15
z  
4
3x 
 4  2y
3x 
 4  2(12 
 2x)
3x 
 4  24 
 4x
7x  28
x4

y  12 
 2(4)
4
(4, 4)
6. A  B  
5  (
2) 1  3
4  (
3) 2  0

7 4

1 2
7. 
3C  
3(
1) 
3(0) 
3 (3)

3 (3) 
3(2) 
3(
4)
3
0 
9


9 
6 12

Chapter 2 Test
Page A57
1.

4. 4x  6y 
 3z  20
x 
 5y  z  
15

y
(1, 3)

2(
2) 2(3)
2(
3) 2(0)

4
6


6 0

2x  y 
 1

8. 2B 

x

O

xy
4

2B 
 A  
4 
 (
5) 6 
 1

6 
 4 0 
 2
2. 3x  y  7
5x  2y  12

→

3x  y  7
3(2)  y  7
y1
(2, 1)
3. 4x 
 5y  
2
3x  2y  
13
4x 
 5y  
2
4x 
 5(
2)  
2
4x  
12
x  
3
(
3, 
2)

→

6x  2y  14
5x  2y  12
x
 2


9. BC 

1
5

10 
2


2 3

3 0


1 0
3
3 2 
4

 
2(
1)  3(3) 
2(0)  3(2) 
2(3)  3(
4)

3(
1)  0(3) 
3(0)  0(2) 
3(3)  0(
4)
11
6 
18

3 0

9

12x 
 15y  
 6
12x  8y  
52

23y  46
y  
2

1 
4 6 
1.5 
6 9

3
3 2

4.5 4.5 3
A
(1.5, 
4.5), B
(
6, 4.5), and C
(9, 3)
y

10. 1.5

B
B

8
6
4
2

C

O C


6
4
2 A 2 4 6 8 10 x

4

6 A

Chapter Tests

544

11.

19.

3
1 
6 
3
3 1 
6 
3

2 
5
1
5

2 5 
1 
5

1
0
0 
1

y 
 12 x  3

F
(3, 
2), G
(1, 5), H
(
6, 
1), J
(
3, 
5)

y

J

y
(2, 4)
(2, 2)
( 43, 0)

(6, 0) y 
 0 O
y 
 3x  4

G

x
x
2

F
H

f(x, y)  2x 
 y
f(
6, 0)  2(
6) 
 0 or 
12 c minimum
f(2, 4)  2(2) 
 4 or 0
f(2, 2)  2(2) 
 2 or 2

x

O
H

F

f 3, 0  23 
 0 or 3 c maximum
4

G

J

2 4
 2(9) 
 (
10)(4)

10 9
 58

y
4

15.

2 1
1 3

16.

1
3 
4
4 
2

x
3

3

5
1


 5

f(x, y)  12,000x  16,000y
f(3, 5)  12,000(3)  16,000(5)
 116,000
f(3, 4)  12,000(3)  16,000(4)
 100,000
f(5, 4)  12,000(5)  16,000(4)
 124,000
The company reaches the most people with 5 ads
and 4 commercial minutes.

1


5
2

5

1 
2 4

2 4
 1
0 
4 3

4 3
1




5
2


5

2

5
3

10

Chapter 3 Test


5
4
17. The inverse does not exist since
 0.

15 12
2 
1
x
7

18.
3
1
y
3
1
1
1 1
1
1
 5
2 
1

3 2

3 2
3

1

5

Page A58
1. y  2x  1
x-axis

1

1 1 2 
1

3 2 3
1

x

O

3 
1  1
3 
1
5 
1

1
2
2


(5, 4)

(3, 4)

1 2 
1
0
1
3
1
3 0

2
 (
1)
14. 3 0
1 1
1 
2
4 
2
4 1
4 1 
2
 1(
1) 
 2(
10) 
 1(3)
 16
1

8

20. Let x  number of ads.
Let y  number of commercial minutes.
x3
y
100x  200y 
 1300
y4
(3, 5)
100x  200y  1300

12. 5 7  5(6) 
 3(7)
3 6
9
13.

4

y-axis

1
x
1 1 7
 5
y

3 2 3

yx

x
2

y

3

y  
x

→

b  2a  1

b  2a  1
b  
2a 
 1 no
b  2(
a)  1
b  
2a  1 no
a  2b  1
a
1
b  2 no

a  2(
b)  1
a  2b 
 1
a1
b  2 no

none of these

545

Chapter Tests

2

2

→

2. y  
x2


b 

x-axis

b
b

y-axis

b
yx

a
b

y  
x


a 

a 
b
→

3. x  y2  3
x-axis

2

a2
2

no
a2
2



 (a)2
2

a2 yes
2

b2
2

a no
2



(
b)2
2

b2
2
 no; y-axis
a
b2  3

yx
y  
x

→

y-axis
yx
y  
x

f 1(x) 

10.


5
4
3
2 O

Chapter Tests

f (x ) 
 5x  4

3


f(x)  
x
2

3


x
y
2
3

y 
 2  x
3

y  x  2
3

f 1(x)  x  2; Yes, it is a function.
22

4



11. f(2)  
2
2  0

No; the function is undefined when x  2.
12. Yes; the function is defined when x  0, the
function approaches 1 as x approaches 0 from both
sides, and f(0)  1.
13. y →  as x → , y →  as x → 

14.

15.

x
3
4
5

y

11

12

11

x
0
1
2

y

1

2

3

16. x  1

At x  4, y is at a minimum
value, since f(4)  f(3) and
f(4)  f(5).

At x  1, y is at a point of
inflection, since
f(0) f(1) f(2).

4x


y
x
1

y

as x → , y → 4; y  4

x

50
45
40
35
30
25
20
15
10
5

x

O

3

17.

8.

f 1(x ) 
 15 x  45


y
x
2

|x  4|

O

x4

5
1
4
x  
5
5

f (x )

Yes, it is a function.

y  x, y  
x
5. The graph of g(x) is the graph of f(x) translated
left 3 units and reflected over the x-axis.
6. The graph of g(x) is the graph of f(x) stretched
vertically by a factor of 4 and then translated
down 2 units.
y
7.

y

f(x)  5x 
 4
y  5x 
 4
x  5y 
 4
x  4  5y
y

a
a  (
b)2  3
a  b2  3 yes

a  b2  3
a  
b2 
 3 no
b  a2  3
a  
b

 3 no


b  (
a)2  3

b  a2  3
a  

b


3
no; x-axis
ab  5
a(
b)  5

ab  5 no
(
a)b  5

ab  5 no
ba  5
ab  5 yes
(
b)(
a)  5
ab  5 yes

y-axis

4. xy  5
x-axis

9.

b  
a2

x2 
 4  0
(x  2)(x 
 2)  0
x  2

y
as x → , y → 0; y  0

y  2x 2  3
1 2 3 4 5x

546

4x

x

x
1
 
 
x
x

4
y 
1
1 
 x
x


y
x2 
 4

y

x

x2

4
x2


 
x2
x2
1

x

y 
4
1 
 x2

0.25  0.004x

6. 
1


5 
13
53
60

1
6
7 
60
1 
6

7
60 
0
x3 
 6x2 
 7x  60
7. f(x)  x3  8x2  2x 
 11
f(
2)  (
2)3  8(
2)2  2(
2) 
 11
 
8  32 
 4 
 11
 9; no
8. f(x)  4x4 
 2x2  x 
 3
f(1)  4(1)4 
 2(1)2  1 
 3
4
2
2
 0; yes
9. 1 positive
f(
x)  
6x3  11x2  3x 
 2
2 or 0 negative


18a. y  
0.25  0.001x
0.25
0.004x
   
x
x

0.25
0.001x
   
x
x

y

y
as x

0.25
  0.004
x

0.25
  0.001
x
0.004

→ , y → 
0.001

or 4; y  4

18b. As the amount of 4 molar solution added
increases, the molarity of the mixture
approaches 4.
19.
y  kx
y  0.25x
0.5  k(2)
y  0.25(10)
0.25  k
y  2.5
20.

72

k

y  x2
8

y  x2

k

(3)2

18 

72  k

1

r

72

x2

1

2
1

3

4
x  2

x2


2

6

11


3


2

6

14

4

0

6

9


6

0

6


1


1

0

1 1

rational zeros: 
2, 
3, 2

Chapter 4 Test

10. 1 positive
f(
x)  x4 
 x3 
 9x2  17x 
 8
3 or 1 negative

Page A59
1. (x 
 4)(x 
 i)(x 
 (
i)  (x 
 4)(x 
 i)(x  i)
 (x 
 4)(x2 
 i2)
 (x 
 4)(x2  1)
 x3 
 4x2  x 
 4
2
2
2. b 
 4ac  (
5) 
 4(1)(4)
9
Since b2 
 4ac 0, there are 2 real roots.
n
n
n
n

x


17
487

393


8
3888
3136

1

1

1
1

2
0


7

9


24

8


32
0



r
1
2









1
1
1


5

1
5

3
4
5


9

10
1

upper bound: 2
f(
x)  
x3  3x2  5x 
 9
r
1
2
3
4
5

4(2)(4)

 
7
Since b2 
 4ac  0, there are 2 imaginary roots.

x


9
63
47

rational zero: 
1

z

x

1
9

7

11. Use the TABLE function of a graphing calculator;

0.8, 3.8.
12. Use the TABLE function of a graphing calculator:

1.3.
13. Sample answer: 2; 
5

n4
n1
3. b2 
 4ac  (
7)2 
 4(1)(
3)
 61
Since b2 
 4ac 0, there are 2 real roots.

4.

1
1
1



b2 
 4
ac

b  


2a

(
5)  

9

2(1)
53

2
53
5
3
 or n  
2
2

b2 
 4
ac

b  


2a

(
7)  
61


z
2(1)
7  
61

z  2
b2 
 4ac  (
5)2 


r
8

8


b  

b2 
 4
ac

2a

(
5)  

7


2(2)

7
i
5  4


1

1

1

1

1

1

3
2
1
0

1

2

5
7
7
5
1

5


9

2
5
6

5

34

lower bound: 
5


3 3 
4
4 2
10
2
1 5 6
2x2  x  5, R6

5. 2

2

547

Chapter Tests

14. Sample answer: 1; 
2
r
1

2
2


1
4

3
5

1
5

19. 
2x

 2  
3x


5
2x  2  3x 
 5
7x
Check: 
2(7)


2  
3(7)



5

16
  
16

20. 
11

 1 9
0m
11 
 10m 81

10m 70
m  
7
11 
 10m  0

10m  
11
11
m  10

1
6

upper bound: 1
f(
x)  2x4 
 3x3 
 x2 
 x  1
r
1
2


3

1
1

2
2
2


1

2
1


1

3
1

1

2
3

lower bound: 
2
15.

1

80

1

1

Test m  
8: 
11
0(
8)

 1 ?

91
 ?
9.54 ?
Test m  0: 
11

 1 ? 9
0(0)

11
 ? 9
3.32 ? 9
Solution: m  
7

 a  1
0
1

a



a
16.

7

80
80

7
4

x
2
4

x
2

3

1

 
 

x2 
 4
4
3

1




(x  2)(x 
 2) 
 4
(x  2)(x 
 2)

21.

4(x  2)  3 
 4
16(x  2)  12 
 (x  2)(x 
 2)
16x  32  12 
 x2  4
x2  16x  16  0
x
x

5

x2

5

Test x 

Test x 

3

7


 3  B2  2
3

7


1  B
Let z  
2.
5(
2) 
 11  A(2 (
2) 
 3)  B(
2  2)

21  
7A
3A

5z 
 11


2z2  z 
 6

22.

3

5
 ?

3  2

1

 
 

z2
2z 
 3

7x2  18x 
 1


(x2 
 1)(x  2)



7x2  18x 
 1


(x2 
 1)(x  2)
2
7x  18x 
 1


0.3148 true

5
2
   

3:

3
3(
3)
5
2

5 ? 
3 
 9
8

5 ? 
19 false
5
5
2
 ?    

1: 

1  2

1
3(
1)
2
5 ? 
5 
 3
2
5 ? 
53 true
5
5
2
 ?   
1: 
12
1
3(1)
5
2
 ? 5  
3
3
2 ?
2
13
53 false

Chapter Tests

3

2


2  2B

5
5
2
 ?   

18  2

18
3(
18)
5 ?
5
1


1

1
6
8 
 27
?

x  
17, 
2  x  0
18. 
y

2
30

y

23
y
29
y  11

B

52 
 11  A2

15x  15(x  2)  2(x  2)
15x  15x  30  2x  4

2x  34
x  
17; x 
2 or 0

Test x 

A

  

z2
2z 
 3

3

2


0.3125

5z 
 11



(z  2)(2z 
 3)

Let z  2.

 x  3x

Test x  
18:

false

5z 
 11  A(2z 
 3)  B(z  2)


16  

162 

4(1)(16)


2(1)

16  8
3


2

x  
8  4
3

17.

5z 
 11


2z2  z 
 6
5z 
 11


2z2  z 
 6

9
9
9 true

7x2  18x 
 1

(x  1)(x 
 1)(x  2)
A

B

C

    

x1
x2
x
1

 A(x 
 1)(x  2)  B(x  1)(x 
 1)
 C(x  1)(x  2)

Let x  1.
7(1)2  18(1) 
 1  A(1 
 1)(1  2)  B(1  1)(1 
 1)
 C(1  1)(1  2)
24  6C
4C
Let x  
2.
7(
2)2  18(
2) 
 1  A(
2 
 1)(
2  2)
 B(
2  1)(
2 
 1)
 C(
2  1)(
2  2)

9  3B

3  B
Let x  
1.
7(
1)2  18(
1) 
 1  A(
1 
 1)(
1  2)
 B(
1  1)(
1 
 1)
 C(
1  1)(
1  2)

12  
2A
6A

Check: 
11


2
30

9

30
3
30
00

7x2  18x 
 1


(x2 
 1)(x  2)

548

6

3

4

 
   

x1
x2
x
1

5. (RS)2  (ST )2  (RT )2
(RS )2  122  152
RS2  225 
 144
RS  
81
 or 9 in.

23. quadratic
24. Let x  the height.
Then 6x  the length and 6x 
 7  the width.
V  x(6x)(6x 
 7)
120  36x3 
 42x2
0  6x3 
 7x2 
 20
Use a graphing calculator to graph the related
function.
V(x)  6x3 
 7x2 
 20
It appears as if the zero is at x  2.
Use the Factor Theorem to check
V(2)  6(2)3 
 7(2)2 
 20  0.
The height is 2 cm, the length is 6(2) or 12 cm and
the width is 12 
 7 or 5 cm.
12 cm by 5 cm by 2 cm
25. Let s  the speed of the freight train.
Then 30  s  the speed of the car.
500

Car: 500  (30  s)t ⇒ t  
30  s
Train: 350  st ⇒ t 
500

30  s

side opposite


cos R  
hypotenuse

12


cos R  1
5 or 5

4

9


sin R  1
5 or 5


csc R  
side opposite

12


csc R  1
2 or 4

4

15

15

5

9

y


3


tan 60°  1 or 
3

r

7. sec 270°  x
1

sec 270°  0; undefined

350

s

8. sin (
405°)  sin (
45°)
sin (
45°)  
y

350

 s


2


sin (
45°)  
2
9. r  

x2  y2
r  

32  52
r  
34


cos v  

34


5
34


cos v  34

r

sec v  x


34


sec v  3

csc v  y
csc v  5

3

5

tan v  3

3
34


x

r

cot v  y


34


cot v  5

3

10. r  

x2  y2
r  

(
4)2 
 22
r  
20
 or 2
5


 
0.65

y

x

sin v  r
2


4


cos v  
2
5

2
5


2


tan v  

4

cos v  
5

r

sec v  x

cot v  y

csc v  2

2
5


sec v 

cot v  2

csc v  
5


sec v  
2

csc v  y

 
3.46

tan v  x



5

sin v  5

a  360° (
4)  
1245°
a 
 1440°  
1245°
a  195°; III

y

cos v  r


sin v  
2
5


 1.14


1245°

360°

tan v  x

5

sin v  34

 2.76

y

cos v  r


sin v  


34

a  360°(1)  410°
a  360°  410°
a  50°; I
4.

x

sin v  r

a  360°(
1)  
234°
a 
 360°  
234°
a  126°; II
3.

3


cos R  1
2 or 4

6. tan 60°  x

a  360°(2)  995°
a  720°  995°
x  275°; IV

410°

360°


cot R  
side opposite

sec R  9 or 3

Page A60


234°

360°

5

side adjacent

hypotenuse


sec R  
side adjacent

Chapter 5 Test

2.

hypotenuse

tan R  9 or 3

y

995°

360°

3

side opposite


tan R  
side adjacent

500 s  10,500  350s
150s  10,500
s  70 km/h

1.

side adjacent


sin R  
hypotenuse

1

tan v  
2
x

r

2
5



4

5



4

cot v  
2

11. r  

x2  y2
r  

02  (

3)2
r  
9
 or 3
y

sin v  r

3

tan v  x

y

0


3

sin v  3

cos v  3

tan v  0

sin v  
1

cos v  0

undefined

r

y
3


3

r

x
3

0

csc v 
csc v 

csc v 
1

549

x

cos v  r

sec v 
sec v 

undefined

x

cot v  y
0


cot v  

3

cot v  0

Chapter Tests

21. Since 98°  90°, consider Case II.
c a; one solution

b

12. cos A  c
42

cos °  c

sin 98°

90

42

cos 77°

c

64 sin 98°  90 sin A

c  186.7
13.

64 sin 98°

A  sin1 90

b

a
b

13

tan B 
tan 27° 

A  44.8°
B  180° 
 98° 
 44.8° or 37.2°
sin 37.2

b

b  13 tan 27°
b  6.6
14.
sin

90 sin 37.2°


b
sin 98°

b  54.9
A  44.8°, B  37.2°, b  54.9
22. Since 31°  90°, consider Case I.
a  b sin A
9  20 sin 31°
9  10.3; none
23.
a2  b2  c2 
 2bc cos A
132  72  152 
 2(7)(15) cos A
169  274 
 210 cos A

105  
210 cos A

a  14 sin 32° 17°
a  7.5
a

15. cos B  c
23

cos B  37
23

B  cos1 37
B  51.6°
a

16. tan A  b


A  cos1 
210 
105

3

tan A  11

A  60°

3

A  tan1 11

sin 60°

13

A  15.3°
17. Let h  the height.

7 sin 60°

B  sin1 13
65 m

B  27.8°
C  180° 
 60° 
 27.8° or 92.2°
A  60°, B  27.8°, C  92.2°
24. b2  a2  c2 
 2ac cos B
b  

202 
242 

2(20)(
24) co
s 47°
b  17.92432912

h

70˚

18. B  180° 
 36° 
 87° or 57°

sin 47°

b

sin B sin C


K  2a2 
sin A
sin 57° sin 87°

K  410.4

24 sin 47°

C  sin1 b

units2

C  78.3°
A  180° 
 47° 
78.3° or 54.7°
b  17.9, C  78.3°, A  54.7°
25. Let x  the distance between the transmitters.

1

19. K  2bc sin A
1

K  2(56.4)(92.5) sin 58.4°
K  2221.7 units2

20.

180° 
 36°

2
sin 36°

22

 72°


s

22 sin 72°

sin 36°

x  35.6 cm
Perimeter  22  35.6  35.6
 93.2 cm

Chapter Tests

70 miles

sin 72°

x

x sin 36°  22 sin 72°

sin C

 24

24 sin 47°  b sin C


K  2(24)2
sin 36°
1

sin B

 7

7 sin 60°  13 sin B

h

65

h  65 sin 70°
h  61.1 m

1

sin 98°

 90

90 sin 37.2°  b sin 98°

a
sin A  c
a
32° 17
  14

sin 70° 

sin A

 64

130 miles

130˚

36˚

x

x

x

 70° 

 2(70)(130) cos 130°
x  
21,80
0
8,200

 130°
cos 1
x  183.0 miles

x2

72˚
22 cm

550

1302

Chapter 6 Test
Page A61

14.

y  Arccsc x
x  Arccsc y
Csc x  y or y  Csc x

y



180°

1. 225°  225°



5

 4


180°

2. 480°  480°

y 
 Arccsc x



8

 3
5

5

5



4. reference angle: 4 
   4; Quadrant 3

5.

15.

1

7.

1


2

x

y


2

y 
 cos x


 
  O
2

1


2

O 

 

2
2


2

x



8. 3  3;

9. 
2  2;

or

2

3

2



2

16. sin Arccos 2  sin 60°

4

1

or 3
2

k

10. A  4



3

2

k

A  0.5



2



2

c


3

 or 



3
3

18.

v

v  r t
2

2300  240,000 t
240,000(2)


t
2300

t  656 hours or 27.3 days

y

5 y 
 tan (2  
 )
4
4
3
2
1 O


2


3


1



c  


or 4

y 
 3 cos 2


7

 tan 6


4  4

y  0.5 cos (4v 
 )  1
12.
13.
y


2 
 O

1

 2

k



17. tan   sin1 2  tan   6

k  4 or 0.5

y  4 sin 0.5v 
 2

3

 2

 4
2

A  4
11. A  0.5

x

y 
 tan x



2

4

x



y 
 sin x


1



y 
 arctan x

y

v  85.2 cm/s
y
6.
1

O


2

y  tan x
x  tan y
arctan x  y or y  arctan x

71.1

v  12


 
 
2


2



1

sin 6  2
5
tan 4 
v
v  r t

O



3. reference angle:  
 6  6; Quadrant 2

y 
 Csc x


2


2


4 
1

2

3

4

5


4


2

73 
 21

73  21

19. A  2

2

k

h  2

A  26

2

h  47

k  1
2


A  26



 12

k  6


y  26 sin 6t  c  47


21  26 sin 6 1  c  47



26  26 sin 6  c



1  sin 6  c


sin1 (
1)  6  c





2  6  c
4


6  c
2


3  c


2

Sample answer: y  26 sin 6t 
 3  47

551

Chapter Tests

20.

65 miles

1 hour

tan v 
v

5280 feet
1 hour
 
1 mile
3600 seconds
2
v

rg
(95.3
 )2

tan1 
1200(32)

 95.3
 ft/s

8.

sec x

sin x

sin x




cos x  cot x

sec x cos x 
 sin2 x

sin x cos x
1 
 sin2 x

sin x cos x
cos x

sin x

v  0.23 radians

 cot x
 cot x
 cot x

cot x  cot x
9.

Chapter 7 Test
Page A62
1. sin2 v  cos2 v  1
1 2

3

 cos2 v  1
8

cos x
cos x
  
1  sin x
1 
 sin x
cos x 
 cos x sin x  cos x  sin x cos x

1 
 sin2 x
2 cos x

1 
 sin2 x
2 cos x

cos2 x
2

cos x

cos2 v  9
cos v 

2

10. csc (A 
 B) 
csc (A 
 B) 
csc (A 
 B) 

sec v 

1

sin2

cos v

cos2 v 
 sin2 v

or

cos 2v


cot 2v  
sin 2v

cot 2v  cot 2v
12. sin 255°  sin (225°  30°)
 sin 225° cos 30°  cos 225° sin 30°

3

or 
5

 
2 2  
2


6 
 
2
 4

2


v

v1



cos2

v1

cos2

v  25

 

5



sin (
420°)



sin (
360° 
 60°)

cos (
360° 
 60°)

sin 60°



cos 60°



 
tan 60°
6. tan v(cot v  tan v)  sec2 v
tan v cot v  tan2 v  sec2 v
1  tan2 v  sec2 v
sec2 v  sec2 v
2
7.
sin A cos2 A  (1 
 sin A)(1  sin A)
cos2 A
2
sin A
cos2 A



13. tan 12  tan 6  4


5. tan (
420°)  
cos (
420°)

sin2 A


2



2
  
6


4




3


 
4

16

cos v  5





 (1 
 sin A)(1  sin A)

 (1 
 sin A)(1  sin A)
1 
 sin2 A  (1 
 sin A)(1  sin A)
(1 
 sin A)(1  sin A)  (1 
 sin A)(1  sin A)

Chapter Tests

sin v


cot 2v  
2 sin v cos v

5

3

cos2

3 2

5

1



cos 2v  
2 sin v 
 2 cos v

1

1

5

3

sec B

sin A 
 cos A tan B
1

cos B

sin B

sin A 
 cos A 
cos B
1

sin A cos B 
 cos A sin B

11. cot 2v  2 cot v 
 2 tan v

3

5


4. sin v  
csc v

sin v 

 2 sec x

csc (A 
 B)  csc (A 
 B)

9

sec v 

 2 sec x

1

cos2 v  2
5
cos v 

 2 sec x


csc (A 
 B)  
sin (A 
 B)

 cos2 v  1

1

cos v
1

3

5

 2 sec x

2 sec x  2 sec x

2
2


3

2. tan2 v  1  sec2 v
tan2 v  1  (
2)2
tan2 v  3
tan v  

3

3. sin2 v  cos2 v  1


45

 2 sec x





tan 6  tan 4



1 
 tan 6 tan 4

1
  1


3

1
 1
1 


3

1
1



1  
3
1

3



1
1


1

1


3


3

1
2




1  
3  3

1
1 
 3
3 4
2

3
   
2 3
3



 2  
3


552



1

2

sin v

sin2

3 2

4

v



1
7

sin2 v  1
6
sin v 

20.


tan v  
cos v

14. sin2 v  cos2 v  1


7


4

sin 2v  2 sin v cos v




7


4

3

4



A2  
B2  

(
1)2 
 12 or 
2

1





2




7

tan f 

1


 1
6 
 16

 1
6 or 8










7

1 
 

3

2
7



3

2

9
45°

1  cos 45°

2




2

1  2

2



2  
2


4



2  
2


2

0

5


p  
5; Quadrant IV

 

15. cos (22.5°)  cos 2


0


5

2
5

sin f  
5, cos f  5,

5


5
1

tan f 
or 
2
2
5


5
1
f  tan1 
2  2

 
3
7


 333°
22. 2x  y  6 → 2x  y 
 6  0
Ax1  By1  C

d  2 
2

A

B


d
d
8
5


5

2(
5)  1(8) 
 6

2  1
2

2

8
8
5

 or 




5

5
Ax1  By1  C

23. d  2
2



A

B

x  
3
 tan x
x 

3
 tan x  0
tan x(tan x 
 
3
)  0
tan x  0 or tan x 
 
3
0
x  0°
tan x  
3

x  60°
17.
cos 2x 
 cos x  0
2 cos2 x 
 cos x 
 1  0
(2 cos x  1)(cos x 
 1)  0
2 cos x  1  0 or cos x 
 1  0
1
cos x  
2
cos x  1
x  120°
x  0°
18. sin x 
 cos x  0
sin x  cos x
tan2

16.


2


2
 or 
1

2



2
tan1 (
1) 

10
5
5
x 
 y 
 
5
5

5

5
5

5
2
5


5



5
x 
   y 
  
5
5
5

 
2



3
2




A2  
B2  

102 
(
5)2 or 5
5


2 tan v


tan 2v  
1 
 tan2 v

7

2 

3


2



f
 135°
21. 5  5y  10x
0  10x 
 5y 
 5


7
 2

 4 
 
4
2

3
2




2

cos 2v  cos2 v 
 sin2 v
7


2


sin f  2, cos f  
2, p  2; Quadrant II

3

3 2

3


2x  2y 
 2  0

 
3

3

7

8

9

1

x  y 
   0



2


2



2

 2
44

7


yx3

x  y 
 3  0

tan2

d
d

3(
6)  4(8)  2

2  42




3
16



5

16

5

24. 5x  2y  7 → 5x  2y 
 7  0
3

y  
4x  1 → 
3x 
 4y  4  0
5x1  2y1 
 7


3x1 
 4y1  4

d1  2 
2

d2  2 
2


5

 2

5x1  2y1 
 7




29




3x1 
 4y1  4


5



3  4

25x1  10y1 
 35  
3
29
x1 
 4
29
y1  4
29

(25  3
29
)x (10  4
29
)y

 35 
 4
29
0

sin x

cos x

1
tan x  1
x  45° or x  225°
19.
2 cos2 x  3 sin x  3
2(1 
 sin2 x)  3 sin x 
 3  0
2 sin2 x 
 3 sin x  1  0
(2 sin x 
 1)(sin x 
 1)  0
2 sin x 
 1  0 or sin x 
 1  0

v0 2

25. R  g sin 2v
v0 2

R  g 2 sin v cos v
R  32 255
882

3

4

R  232.32 ft

1

sin x  2
sin x  1
x  30°, 150°
x  90°

553

Chapter Tests

16. u
r u
s  
1(4)  3(3)  4(
6)
 
19
4u
4u
17. u
r u
s  3
i 
 
1
j  
1 3 u
k
3 
6
4 
6
4 3
 
30u
i  10u
j 
15u
k

Chapter 8 Test
Page A63
1. 2.5 cm, 60°

2. 1.6 cm, 25°

 
30, 10, 
15
18. No; sinceu
r u
s  
19 and not 0.
19.

a  b

3.
3.9 cm

125˚

a
46˚
b

125 lb

165˚

3.9 cm, 46°




60 lb

2b


4.

40˚

x2  1252  602 
 2(125)(60) cos 55°
x  
19,22
5
5,000

 15°
cos 5
x  103.06 lb

334˚

sin 55°

103.06

1.8 cm

sin v

 60

103.06 sin v  60 sin 55°

2b
  a

60 sin 55°


v  sin1 
103.06

v  28.48°
v  40°  28.48°  40° or 68.48°
20. x  x1  ta1
y  y1  ta2
x  3  t(2)
y  11  t(
5)
x  2t  3
y  
5t  11
21.
x  2t  3
yt1

a


1.8 cm, 334°
u
5. AB  
1 
 3, 9 
 6
 
4, 3
u  

AB
(
4)2 
 (3)2
 
25
 or 5
6. u
AB  3 
 (
2), 10 
 7
 5, 3
u  

AB
52  32
 
34

7. u
AB  9 
 2, 
3 
 (
4), 7 
 5
 7, 1, 2
u
72  12
 22
AB   

 
54
 or 3
6

u
8. AB  
8 
 (
4), 
10 
 (
8), 2 
 (
2)
 
4, 
2, 4
u
AB   

(
4)2 
 (
2
)2  42
 
36
 or 6
9. u
r 
u
s  
1 
 4, 3 
 3, 4 
 (
6)
 
5, 0, 10
10. 3u
s  3 4, 3 3, 3 
6
 12, 9, 
18
2u
r  2 (
1), 2 3, 2 4
 
2, 6, 8
3u
s 
 2u
r  12 
 (
2), 9 
 6, 
18 
 8
 14, 3, 
26
11. u
r  3u
s  
1  12, 3  9, 4  (
18)
 11, 12, 
14
12. u
r   

(
1)2 
 32 
42
 
26

13. u
s   

42  32
 (

6)2
 
61

14. u
r  
u
i  3u
j  4u
k
u
u
15. u
s  4 i  3 j 
 6u
k
Chapter Tests

55˚

x

x
3

2

t

y
1
y

y
1t
x
3

2
1
1
x 
 
2
2

22. vx  100 cos 2°
vy  100 sin 2°
vx  99.94 mph
vy  3.49 mph
23. The figure is four times the original size and
reflected over the yz-plane.
u
24. AB  1.5 cos 60°, 0, 1.5 sin 60°
 0.75, 0, 0.75
3

u
110 lb
F  0, 0, 
110

1.5 ft
60˚

u u
T  AB
u
i
 0.75
0

u
F
u
u
j
k
0 0.75
3

0

110
u
u
 0 0.75
3
i 
 0.75 0.75
3
j  0.75 0 u
k
0
0
0 0

110

110
u  82.5j
u  0k
u
 0i

u
T   

02  8
2.52 
02
 82.5 lb-ft

554

u sin v 
 1gt2
25. y  tv
2

u cos v
x  tv

10. r  

(
2)2 
 (
3
)2

y  28t sin 35° 
 2(32)t2

x  28t cos 35°

0  4t(7 sin 35° 
 4t)
4t  0 or 7 sin 35° 
 4t  0

x  23.02 feet


 
13
 3.61
(3.61, 4.12)
11. x  r cos v

1


7 sin 35°

t0


t

4

 4.12

5

4

 3 cos 

t  1.003758764


3


v  Arctan 

2   

y  r sin v



 3
2

 32


2



2


3
2


3
2


 
2

 2

3
2
 3
2

, 


2
2

Chapter 9 Test

12. x  r cos v
90˚

120˚
150˚

2.

60˚

1 2 3 4

0˚

330˚

210˚
240˚

2
3

270˚

2

300˚

0

5.

2
3

7.

120˚

90˚

x  
2
y 
 10
0  
2

0˚

5
34


300˚

1 2 3 4

240˚

270˚

8. r  

22  22
 
8
 or 2
2

2
2, 4

3
34



3


f  Arctan 

5   180°

 211°
p  r cos (v 
 f)
3
34


34

0˚

 r cos (v 
 211°)

A2  
B2  

22  (

4)2
19. 

 
20
 or 2
5

2
4
1
x 
 y 
   0
2
5

2

5
2

5

330˚

210˚

3
34


cos f  
34, sin f  
34, p  34

30˚

180˚


2


A2  
B2  


52  32
18. 


 

34

5
3
3
x 
 y 
   0



34


34



34

60˚

150˚

15. x2  y2  3x
r2  3r cos v
r  3 cos v


2




2

330˚
270˚

y  r sin v
 
4 sin 1.4
 
3.94

0  2x  2y 
 5

2 4 6 8

240˚

5
3

3
2

 
1

y  
3
r sin v  
3

3

r
sin v



2

60˚

210˚

11
6

1

0  2r cos v  2r sin v 
 5
30˚

180˚

0

4
3

90˚

 2
2

r  
3 csc v
r7
r2  49
x2  y2  49
17. 5  r cos (v 
 45°)
0  r cos (v 
 45°) 
 5
0  r (cos v cos 45°  sin v sin 45°) 
 5

5
3

150˚

1 2 3 4

7
6

120˚

3
2

 2
2

16.

11
6

6.

6




3

0

4
3


3

14.

1 2 3 4

5
3

5
6


2

7
6

6


2

5
3


6



C 11
3
2

11
6

5
6

1 2 3 4

4
3

7
6

2
3

 2 sin 6

 

3

(

3
, 
1)
13. x  r cos v
 
4 cos 1.4
 
0.68
(
0.68, 
3.94)


6

1 2 3 4

3
2

7

 2 cos 6

3



3

0



4.

6

7
6


2

B

4
3


3

5
6


2
3
5
6

30˚

A

180˚

3.

y  r sin v

7

Page A64
1.

5

 3 sin 
4

300˚


5


v  Arctan

 4

2
5



5


cos f  5, sin f  
5, p  10

2

2


2

f  Arctan 1  360°
 297°
p  r cos (v 
 f)

9. r  

(
6)2 
 02
 
36
 or 6
Since x  0 and y  0, v  .
(6, )

20.



5
  r cos (v 

10
i93  (i4)23 i

297°)

 123 i
i

555

Chapter Tests

32. x3 
 i  0 → x3  i
Find the cube roots of i.

21. (2 
 5i)  (
2  4i)  (2  (
2))  (
5i  4i)
 0  (
i)
 
i
22. 
6i 
 (
3  2i)  
6i  3 
 2i
 8i  3
23. (3  5i)(3 
 2i)  9  9i 
 10i2
 19  9i
24. (1 
 3i)(2 
 i)(1  2i)  (2 
 7i  3i2)(1  2i)
 (
1 
 7i)(1  2i)
 
1 
 9i 
 14i2
 13 
 9i
25.

6 
 2i

2i



r  

02  12
 
1
 or 1
1


v  2





1




2n



  4n

  4n






3


1

x1  cos 6  i sin 6  2  2i

2
i

2
i
12 
 10i  2i2

4 
 i2
10 
 10i

5

5

5

9

9


3


1

x2  cos 6  i sin 6  
2  2i



x3  cos 6  i sin 6  
i
1

v  Arctan 

(
4)2 
 42
26. r  

 or 4
2

 
32
4
2
cos

3

4



 i sin

3

4



3

4

3


2






1


v  Arctan 

5   

i

3
2

1
i
2

O

 12 i

1

0

27. r  

(
5)2 
 02
 or 5
 
25
5(cos   i sin )


1 i


3

28. r  4 3

33. E  I Z
 8 (cos 307°  j sin 307°)
20 (cos 115°  j sin 115°)
 8 20 [cos (307°  115°)  j sin (307°  115°)]
 160 (cos 422°  j sin 422°)
 160 (cos 62°  j sin 62°)



v  2  4
6  

 12

7

 4 or 4
7

7

12cos 4  i sin 4  122  i
2


2


2


 6
2
 
 6
2
i
2

2
3



3



v  3  6

29. r  

3

2

 6 or




2

Chapter 10 Test



2cos 2  i sin 2  2(0  i)
 2i
12  (

1)2
30. r  


Page A65


1

v  Arctan 1  2

1. d  

(x2 
 
y1)2 
(y2 

y1)2

7


 
2

 4
7

d  

3 
 (

1))2 
(1 
 
2)2

7

(1 
 i)8  (
2
)8 cos (8)4  i sin (8)4

d  

42  (

1)2
d  
17


 16 (cos 14  i sin 14)
 16 (1  0)
 16

x1  x2 y1  y2


1  3 2  1
, 
2, 2  
2
2
3
 1, 2



31. r  

02  (

27)2
v  
2
 
729
 or 27
1

3


27i
  (0 
 27i) 3



1

3



2

2. d  

(x2 
 
y1)2 
(y2 

y1)2


2

    i sin  
   i sin 
6
1

3

  i 

2
2

1
27 cos 3

3 cos 
6

3

3
3


1

3

d  

(2k 

3k)2 
(k 
 
1 
 (k
 1))2



d  

(
k)2 
 (
2
)2
2
d  

k 4
x1  x2 y1  y2

3k  2k k  1  k 
 1
, 
2, 2  
2
2
5


  2 k, k

3

 2 
 2i

3. r  

(
8 

(
6)2
 (3 

 (
4
))2
r  

(
2)2 
 (7)2
r  
53

(x 
 h)2  (y 
 k)2  r2
2
(x 
 (
8)2  (y 
 3)2  
53

(x  8)2  (y 
 3)2  53

Chapter Tests

2n

 cos 6  i sin 6

6 
 2i

4


4

1

3

 1 5 cos 6  3  i sin 6  3



2i


 2 
 2i





(0  i) 3  1 cos 2  2n  i sin 2  2n

556

4. center: (h, k)  (0, 0)
a2  10
b2  6
c  

a2 
 b2
a  
10

b  
6

c  
10
6

  or 2
foci: (h, k  c)  (0, 2)
major axis vertices: (h, k  a)  0, 
10

minor axis vertices: (h  b, k)  (
6
, 0)
5. e 

c

a



(x 
 h)2

a2

( y 
 k)2
 b

2
( y 
 0)2
(x 
 0)2
   
3
1

4
4y2
x2  3 

1

2

1

If c  2, then a  1.



a2 
 b2

c
 

12 
 b2

1

2
1

4

10. A  1, C  0; since C  0, the conic is a parabola.
x2  6x 
 8y 
 7  0
x2  6x  9  8y  7  9
(x  3)2  8y  16
(x  3)2  8(y  2)

y

1
1

O

1

x

 1 
 b2
3

b2  4
6. center: (h, k)  (
4, 2)
a2  16
b2  7
b2  c2 
 a2
a4
b  
7

7  c2 
 16 → c  
23

foci: (h, k  c)  
4, 2  
23

vertices: (h, k  a)  (
4, 2  4)
 (
4, 6) and (
4, 
2)
a
asymptotes: y 
 k  b(x 
 h)
y
y
7. foci: (h, k  c)

11. A  4, C  
1; since A and C have opposite signs,
the conic is a hyperbola.
4x2 
 y2  1
x2

1

4

y

4
(x 
 (
4))

2

7

4

7

 2   7(x  4)
3
c
⇒ h  
5
e  2  a
b2  c2 
 a2

kc 4
k 
 c  
2
2k
 2
k  1; c  3

(y 
 k)2
(x 
 h)2


 b
2
a2
(y 
 1)2
(x 
 (
5))2





22
5
(y 
 1)2
(x  5)2
 
 
4
5

y2


 1  1

O

x

b2  32 
 22
b2  5

1

12. A  1, C  1; since A  C, the conic is a circle.
x2  y2  4x 
 12y  36  0
x2  4x  4  y2 
 12y  36  4
(x  2)2  (y 
 6)2  4

1
1

8. vertex: (h, k)  (0, 
3)
4p  8 ⇒ p  2
focus: (h  p, k)  (0  2, 
3)  (2, 
3)
directrix: x  h 
 p
x0
2
x  
2
axis of symmetry: y  k
y  
3
9. 2k  p  
5
foci: (h, k  p)  (3, 
5)
2k 
 p  
2
directrix: y  k 
 p  
2
2k  p  
7
7

y

O

x

3

k  
2, p  
2
(x 
 h)2  4p(y 
 k)


32y 
 
72
7
(x 
 3)2  
6y  2

(x 
 3)2  4

557

Chapter Tests

13. A  3, C  
16; since A and C have opposite
signs, the conic is a hyperbola.
3x2 
 16y2 
 18x  128y 
 37  0
3(x2 
 6x  9) 
 16(y2 
 8y  16)  37  27 
 256
3(x 
 3)2 
 16( y 
 4)2  
192
( y 
 4)2

12

14
12
10
8
6
4
2

O

6
4
2

2

4

6

16. A  0, C  1; since A  0, the conic is a parabola.
y2 
 2x  10y  27  0
y2  10y  25  2x 
 27  25
(y  5)2  2x 
 2
(y  5)2  2(x 
 1)

(x 
 3)2

O y


 64  1

x

y

x
2 4 6 8 1012 14

17. A  1, C  2; since A and C have the same sign
and A C, the conic is an ellipse.
x2  2y2  2x 
 12y  11  0
2
(x  2x  1)  2(y2 
 6y  9)  
11  1  18
(x  1)2  2(y 
 3)2  8

14. A  9, C  
1; since A and C have opposite signs,
the conic is a hyperbola.
9x2 
 y2 
 90x  8y  200  0
9(x2 
 10x  25) 
 (y2 
 8y  16)  
200  225 
 16
9(x 
 5)2 
 ( y 
 4)2  9
(x 
 5)2

1

(x  1)2

8

(y 
 3)2

 4  1

y

( y 
 4)2


 9  1

y

x

O

18. y  2x2  x
19. x  2 cos t

x

O

15. A  2, C  
13; since A and C have opposite
signs, the conic is a hyperbola.
2x2 
 13y2  5  0
2x2 
 13y2  
5
13y2

5
y2

5

13

x

2

x2

5

2

Chapter Tests

sin2

 sin t

t1

y 2

 2  1

x2

4
x2

y2

 4  1

 y2  4
20. B2 
 4AC  02 
 4(4)(1)
 
16
A C; ellipse
4(x  1)2  (y 
 3)2  36
4(x  1 
 3)2  (y 
 3 
 (
5))2  36
4(x 
 2)2  (y  2)2  36
4x2 
 16x  16  y2  4y  4  36
4x2  y2 
 16x  4y 
 16  0

1

y

O

t

x 2

2

2x2




 cos t

cos2


 5  1

y  2 sin t
y

2

x

558

21. B2 
 4AC  02 
 4(2)(
1)
8
hyperbola

25.

1



3

Replace x with x
 cos 60°  y
 sin 60° or 2x
  2y
.
Replace y with 
x
 sin 60°  y
 cos 60° or

3


2x








2 2



8

8

39

y  2(
3) 
 3 or 
9

Chapter 11 Test

8

Page A66
2


2

1


3

1. 343 3  
343

2
7
 49
3

2. 64

1
3

64

1

4

 


3

2

5


4

12


20


4. x y2z 4   x 2 y8z 4
 x6y8z5

3. ((2a)3)2  (2a)6
1



(2a)6
1



26a6

If x  3, y   
3  23   0.9.
2

1



64a6


(2)2 
 2(2)  0.
If x  2, y  

(2, 0), (0.7, 0.9)
23.

27

If a person is walking northeast, the person will
first hit the motion detector at (
3, 
9).

0
(x
)2 
 6
3
x
y
 
 5( y
)2  32  0
x2  4y2  4
22.
(x 
 1)2  y2  1
x2 
 2x  1  y2  1
x2  4(
x2  2x)  4
x2  y2 
 2x  0
x2 
 4x2  8x  4
2
2
y  
x  2x
3x2 
 8x  4  0
(3x 
 2)(x 
 2)  0
2
x  3 or x  2
2

y  25 
 3 or 5

27

2
2
1

3



3
 2y
 
 
2x
  2y
 
2
1
3

3

2 4(x
)2  2x
y
  4(y
)2
3
1

3


 4(x
)2 
 2x
y
  4(y
)2 
1
3
3
1

3

(x
)2  
3
x
y
  2(y
)2 
 4(x
)2  2x
y
 
 4(y
)2 
2
x
y
  5( y
)2 
 32 

(x
)2  6
3



y  2x 
 3

x 5 or x  
3

1
y
.
2

1
x

2

x2  y2  90
x2  (2x 
 3)2  9
x2  4x2 
 12x  9  90
5x2 
 12x 
 81  0
(5x 
 27)(x  3)  0

1


6


12


12  (33) 3 a 3 b 3
5. 

27a6b
 3a2b4
3

y

1


2


3


4


6. m 2 n 3  m 6 n 6
6
 

m3n4
y
7.

O

x

24. x2  y2  Dx  Ey  F  0
02  12  D(0)  E(1)  F  0

⇒ E  F  
1
(
2)2  32  D(
2)  E(3)  F  0
⇒ 
2D  3E  F  
13
42  52  D(4)  E(5)  F  0
⇒ 4D  5E  F  
41

4D  6E  2E  
26
3E  3F  6
3
4D  5E  F  
41
11E  3F  
67
11E  3F  
67

8E
 64
E  
8
E  F  
1

2D  3E  F  
13

8  F  
1

2D  3(
8)  7  13
F7

2D  4
D  
2
2
2
x  y 
 2x 
 8 y  7  0
x2 
 2x  1  y2 
 8y  16  
7  1  16
(x 
 1)2  (y 
 4)2  10
center: (h, k)  (1, 4)
radius  
10


8.

O
y

x

O

x

1

2

9. 4  2

10.

11. log5 625  4
13. logx 32  
5
x5  32
1

x5
1

32
1 5

2
1

2

 
559

3

16

 216

12. log8 m  5

 32
 x5
 x5
x
Chapter Tests

14. log5 (2x)  log5 (3x 
 4)
2x  3x 
 4
4x
15.
3.6x  72.4
x log 3.6  log 72.4
x
16.

Chapter 12 Test
Page A67
1. d  4.5 
 2 or 2.5
7  2.5  9.5, 9.5  2.5  12, 12  2.5  14.5,
14.5  2.5  17
9.5, 12, 14.5, 17
2. d  
1 
 (
6) or 5
a24  
6  (24 
 1)5
a24  109
3. 8  
4  (5 
 1)d
12  4d
3d

4  3  
1, 
1  3  2, 2  3  5

4, 
1, 2, 5, 8

log 72.4

log 3.6

x  3.3430
6x
1  82
x
(x 
 1) log 6  (2 
 x) log 8
x log 6 
 log 6  2 log 8 
 x log 8
x log 6  x log 8  2 log 8  log 6
x(log 6  log 8)  2 log 8  log 6
2 log 8  log 6


x
log 6  log 8
log 15


17. log4 15  
log 4

 1.9534

n

4. 345  2(2(12)  (n 
 1)5)

log 0.9375


18. log3 0.9375  
log 3

690  24n  5n2 
 5n
0  5n2  19n 
 690
0  (5n  69)(n 
 10)

 
0.0587
log 3


19. log81 3  
log 81

69

1

n  
5 or n  10

 4
20. log 542  2.7340
21. ln 0.248  
1.3943
22. antiln (
1.9101)  0.1481

Since there cannot be a fractional number of
terms, n  10.

ln 2

23. t  k

5.

ln 2


t
0.054

 

t  12.84 yr
24. Let x  the original number of bacteria.
3x  xek(6)
3  e6k
ln 3  6k
k

6.

1

16
1
2

ln 3

6

k  0.1831020481
8x  xe0.1831t
8  e0.1831t
ln 8  0.1831t
t

1

106
1

5

 (5
e

ln 0.2 

r
1

1

16, 8, 4, 2, 1
7. r 

5

5

2

S10 

t


BC



t



106)e 3(2 106)



t



6 106

or 2
5
5
 
 (2)10
2
2

1
2
5
5120
 
 
2
2


1
5115

2

8. does not exist;

t


6 106

n3  3

lim 
2
3
n→ n  1

t  ln 0.2(
6 106)
t  9.66 106 s

3
n  
n2
 lim 
, lim
n→ 3  1 n→
n2

1

2  0,
n→ n

3  3, and lim

so the denominator approaches 3. As n approaches
infinity, the n term in the numerator makes the
whole numerator approach infinity, so the entire
fraction has no limit.
9.

Chapter Tests

 r4
1

ln 8

0.1831

Q(t)  Qe

8

   6425, 6425 25  
3125

162  8, 82  4, 42   2

t  11.3568626 hours
0.3568626(60)  21 minutes
11 hours, 21 minutes
25.

1

10
2
r 
or 5
1

4
1 2
2
2 2
   ,  
25 5
125 125 5
2
4
8
, , 
125 625 3125
1  16r5
1

560

n3  4

lim 
3
n→ 2n  3n

n3
4

 
n3
n3

 lim 2n3 3n
n→   
n3
n3
10


20
1
 2

Evaluate the original formula for n  k  1.

1

.
10. The general term is 
3n2
1

3n2



1

n

(k  1)[(k  1)  1][4(k  1)  5]

3

11. The series is arithmetic, so it is divergent.
19

12. Sample answer:   1 5k
k


3 k
1

13. Sample answer:  62
k1

14. (2a 
 3b)5
5 4(2a)3(
3h)2

 (2a)5(
3b)0  5(2a)4(
3b)1  
2 1


15.
16.

5 4 3 2 (2a)1(
3b)4
5 4 3(2a)2(
3b)3
   
3 2 1
4 3 2 1
5 4 3 2 1(2a)0(
3b)5

5 4 3 2 1
32a5 
 240a4b  720a3b2 
 1080a2b3 

810ab4 
 243b5

10 !
 a10
5
5!(10 
 5)!

10 9 8 7 6 5 4 3 2 1
25  
5 4 3 2 1 5 4 3 2 1
 8064a5

8!
(3x)8
4(
y)4
4!(8 
 4)!
8 7 6 5 4 3 2 1
 
4 3 2 1 4 3 2 1
 70(81x4)(y4)

(k  1)(k  2)(4k  9)

 
3
The formula gives the same result as adding the
(k  1) term directly. Thus, if the formula is valid
for n  k, it is also valid for n  k  1. Since the
formula is valid for n  1, it is also valid for n  2.
Since it is valid for n  2, it is also valid for n  3,
and so on, indefinitely. Thus, the formula is valid
for all positive integral values of n.
20. There are 4 groups of three months in a year.
There are 4 10 or 40 groups of three months in
ten years. So, n  40. Since the interest is
0.08
compounded quarterly, the rate per period is 4
or 0.02. The common ratio r is 1.02.
a1  200(1.02)  204

for all n, so convergent

a1 
 a1rn


Sn  
1
r

a5 32

S40 

204 
 204(1.02)40

1 
 1.02

S40  $12,322.00

(3x)4 (
y)4

Chapter 13 Test

 5670x4y4

17. r  

(
2)2 
 22 or 2
2


3



v  Arctan 

2    or 4
2


2  2i  2
2
cos

3

4

Page A68

 i sin

3

4

  2
2 e

6!


1. P(6, 2)  
(6 
 2)!

3
i4



18. z0  2i
z1  3(2i)  (2 
 i)
 6i  2 
 i
 2  5i
z2  3(2  5i)  (2 
 i)
 6  15i  2 
 i
 8  14i
z3  3(8  14i)  (2 
 i)
 24  42i  2 
 i
 26  41i
19. Step 1: Verify that the formula is valid for n  1.

 30
7!


2. P(7, 5)  
(7 
 5)!



7 6 5 4 3 2 1

2 1

 2520
8!


3. C(8, 3)  
(8 
 3)! 3!



8 7 6 5 4 3 2 1

5 4 3 2 1 3 2 1

 56
5!


4. C(5, 4)  
(5 
 4)! 4!

1(1  1)(4 1  5)

Since 2 1(2 1  1)  6 and 
 6,
3
the formula is valid for n  1.
Step 2: Assume that the formula is valid for n  k
and prove that it is valid for n  k  1.
2 3  4 5  6 7  . . .  2k(2k  1)

5 4 3 2 1



1 4 3 2 1
5
5. Using the Basic Counting Principle,
5 4 3 2 1  120.

k(k  1)(4k  5)

5!

 3
2 3  4 5  6 7  . . .  2k(2k  1)
 2(k  1)(2k  3)


6. P(5, 3)  
(5 
 3)!
5 4 3 2 1



2 1
 60
7. (7 
 1)!  6!  720

k(k  1)(4k  5)

 3  2(k  1)(2k  3)
k(k  1)(4k  5)

6 5 4 3 2 1

4 3 2 1

6(k  1)(2k  3)

 3  3

8. C(3, 1) C(12, 8)
3!



(3 
 1)! 1!

k(k  1)(4k  5)  6(k  1)(2k  3)

 
3

3 2 1



2 1 1
 1485

(k  1)(k(4k  5)  6(2k  3)
 
3

12!

(12 
 8)! 8!
12 11 10 9 8 7 6 5 4 3 2 1

4 3 2 1 8 7 6 5 4 3 2 1
4!

6!

(6 
 3)! 3!
6 5 4 3 2 1
4 3 2 1
 
3 2 1 3 2 1
2 1 2 1


9. C(4, 2) C(6, 3)  
(4 
 2)! 2!

(k  1)(4k2  17k  18)

 
3


 120

(k  1)(k  2)(4k  9)

 
3

561

Chapter Tests

4. Sample answer: 1.5, 4.5, 7.5, 10.5, 13.5, 16.5, 19.5
5. Sample answer:

5

9

6

10. P(2 white)  1
0
1

 3
1

11. P(s)  4

1

odds 

1

4

3

4

Number
of
Absences
0–3
3–6
6–9
9–12
12–15
15–18
18–21

P(f )  1 
 P(s)
3

 1 
 4 or 4
1

or 3

12. P(5 clubs or 5 hearts or 5 spades or 5 diamonds)
 P(5 clubs)  P(5 hearts)  P(5 spades) 
P(5 diamonds)
 452
13



12

51

10

49

11

50



9

48

33

16,660
5

6.

3

36

13. P(sum of 8) P(sum of 4)  3
6
15

5




1296 or 432
1

1

2

14. P(3 odd digits)  2

1

2

1

 8
15. P(all red or all blue)  P(all red)  P(all blue)
3

2

11

 1
2



1

220
1

20



1

10

5

 1
2

4

11

3

10

1

22

26

28

7

2

15  1

5

18. P(both eveneven product)
P(both even and even product)

P(even product)
3


 2
6 or 13

0.006

semi-interquartile range  2

19. P(no more than two heads)
 P(0 heads)  P(1 head)  P(2 heads)

 0.003

2 0 1 5
2 1 1 4
 C(5, 0) 3 3  C(5, 1) 3 3
2 2 1 3
 C(5, 2) 3 3
1
10
40




243  243  243
51
17



243 or 81

  
  

 

  

12.
2.338 2.340 2.342 2.344 2.346 2.348 2.350
1

13. X
  1
5 (2.334  2.338  . . .  2.350)
 2.34393333
1

MD  1
5 ( 2.34393333 
 2.334  . . . 
2.34393333 
 2.350 )
 0.0025

1792



15,625 or 0.114688

14. j 

Chapter 14 Test

(2.34393333 
 2.334)2  . . .  (2.34393333 
 2.350)2

15

 0.0031
15. 68.3% of the data lie within 1 standard deviation
of the mean.
24  2.8  21.2 
 26.8
16. 90% corresponds to t  1.65.
24  1.65(2.8)  19.38
28.62

Page A69
1. range  20 
 1 or 19
2. Sample answer: 3
3. Sample answer: 0, 3, 6, 9, 12, 15, 18, 21
Chapter Tests

6 9 12 15 18 21
Number of Absences

there are 15 terms, the median is the 2 or
8th term.
Md  2.344
10. Q1  2.341; Q3  2.347
11. interquartile range  Q3 
 Q1
 2.347 
 2.341
 0.006

 1
8

4 4 1 3

5

3

9. Order the values from least to greatest. Since

P(3’s together and odd number)

P(odd number)

20. C(7, 4)5

0

10  10

17. P(3’stogetheroddnumber)

6

3
6
16
25
17
8
5

Md  2 or 10

 52 or 1
3



|||
|||||
|||||||||||||
||||||||||||||||||||
||||||||||||||
|||||||
||||

1



 5
2  52 
 52



Frequency

7. X
  8
0 (6  16  12  7  . . .  8  9  7  9)
 10.44
8. Order the data from least to greatest. Since there
are 80 terms, the median is the average of the
40th term and the 41st term.

16. P(ace or black card)
 P(ace)  P(black card) 
 P(black ace)
4

27
24
21
18
Frequency 15
12
9
6
3
0

Tallies

562

17. 24 
 18.4  5.6
tj  5.6
t(2.8)  5.6
t2
95.5%

2

 47.75%
47.75%  49.85%  97.6%
18. 29.6 
 24  5.6
19.
tj  5.6
t(2.8)  5.6
t2
95.5%

2

10. f(x)  6x4 
 2x2 
 30
f
(x)  6 4x4
1 
 2 2x2
1 
 0
 24x3 
 4x

32.4 
 24  8.4
tj  8.4
t(2.8)  8.4
t3

 47.75%

99.7%

2

2

11. f(x)  2x5 
 4x3  5x2 
 6

 49.85%

2

f
(x)  2 5x5
1 
 4 3x3
1  5 2x2
1 
 0
4

 10x4 
 12x2  5x

j


jX  

 
N
3.6


jX  
4
00

jX  0.18


12. f(x)  2x4(x3  3x2)
 2x7  6x6
f
(x)  2 7x7
1  6 6x6
1
 14x6  36x5
13. f(x)  (x  3)2
 x2  6x  9
f
(x)  2x2
1  6 1x1
1  0
 2x  6
14. f(x)  
2x  6

20. A 5% level of confidence is given when P  95%.
95% corresponds to t  1.96.
X
  tjX  57  1.96(0.18)

 56.65
57.35


2

 11  6x  C
F(x)  
1  1x

 
x2  6x  C
15. f(x)  
x3  4x2 
 x  4

Chapter 15 Test


1

Page A70

1

1



x→2

4. lim

x→1



 lim (x  3)
 lim

x→1

(x 
 2)(x 
 1)

x
1

1

1

x→1

6.

7.

 lim

x→3

4

 21 
 x11  x  C
F(x)  
2  1x
11

 3x3 
 2x2  x  C

 
1
x2 
 9

3
x→3 x 
 27

 5x  C

17. f(x)  x
 x2 
 4x  1

 lim (x 
 2)
5. lim

2
1
1
x31 
  x11
7 11
31
1
2
x4 
 x2  5x  C
8
14
1
1
x4 
 x2  5x  C
8
7
x3 
 4x2  x

x→0

x2 
 3x  2

x
1

2

F(x)  2

So f(2)  
1.
3

1

1

2. The closer x is to 2, the closer y is to 4.
So, lim f(x)  4. However, there is a point at (2, 
1).
x2  3x

x

4

16. f(x)  2 x3 
 7x  5

x→
1

x→0

1

 
4x4  3x3 
 2x2  4x  C

1. The closer x is to 
1, the closer y is to 1.
So, lim f(x)  1. Also, f(
1)  1.

3. lim

4

 31  x21 
 x11  4x  C
F(x)  
3  1x
21
11

 x dx  lim  
2

(x  3)(x 
 3)

(x 
 3)(x2  3x  9)

18.

0

x3

 lim 
2
x→3 x  3x  9
33


32  3(3)  9
2
6

 2
7 or 9
x2 
 2x  3
12 
 2(1)  3

lim 2   
3(1)2 
 5
x→1 3x 
 5
2



2 or 
1
f(x  h) 
 f(x)
f 
(x)  lim h
h→0
(x  h)2 
 2(x  h) 
 (x2 
 2x)
 lim 
h
h→0
x2  2xh  h2 
 2x 
 2h 
 x2  2x
 lim 
h
h→0
2xh  h2 
 2h
 lim h
h→0

n

3

n→ i  1

2i 3 2


n
n

 

16 n2 (n  1)2


4
4
n→ n
n2  2n  1
 lim 4 n
2
n→



 lim







 lim 41  n  n2 
2

1

n→



3

19.

1

 4 units2
3x2dx  3
 x3



1

20.

0

3
1
x3
3
1

3

1

 33 
 13
 26 units2
(2x  3)dx  2

1
x2
2

 3x

 x2  3x

 lim (2x  h 
 2)
h→0

1

1
0

0

 [12  3(1)] 
 [02  3(0)]
4

 2x 
 2
1

8. f(x)  2x2 
 7x  1
1

f 
(x)  2 2x21 
 7 1x11  0
x
7
9. f(x)  4x3 
 4
f 
(x)  4 3x3
1 
 0
 12x2

563

Chapter Tests



3

21.

1

1

1

(
x2  x  6) dx  
3x3  2x2  6x
1

1

1

1

3

24.

1

 
3(3)3  2(3)2  6(3) 


3(1)3  2(1)2  6(1)
27

37

25. V  2

 2 
 6
22

 3
22.

x

23.



(3x2

 4x  7) 


Chapter Tests



r

0


brx2  hxdx
h

1
x3
3

 2 
r

(1 
 2x)dx  x 
 22x2  C
x2

h(t)  3  95t 
 16t2
h
(t)  v(t)  0  95 1t1
1 
 16 2t2
1
 95 
 32t
h
(2)  v(2)  95 
 32(2)
 31 ft/s

h


3hr r3  h2
hr
hr
 2 
3  2
 2

C

2

3
4
x3  x2  7x  C
3
2
x3  2x2  7x  C

r
1
x2
2
0

r2 
 
3r 03  2 02
h

2

V(h, r)  V(2, 3)  2 
3  2
2 32

 6 units3

564

2 32

h



---

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