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Printed in the United States of America.
4 5 6 7 8 9 10 024 08 07 06 05 04
1Chapter 1
Chapter 1 Linear Relations and Functions
xy
42
61
05
84
22
40
xy
13
22
31
40
51
62
73
xy
47
34
21
12
05
18
211
314
417
xy
15
25
35
45
55
65
75
85
O
y
x
4
2
4
2
4
488
O
y
x
4
4
4
8
12
24
O
y
x
O
y
x
Relations and Functions
Pages 8–9 Check for Understanding
1.
2. Sample answer:
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:
Equation: yx 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}
O
y
x
1-1 8.
9.
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)25(3)
108 9 15 or 84
14. g(m1) 2(m1)24(m1) 2
2(m22m1) 4m4 2
2m24m2 4m4 2
2m2
15. x1 0
x1
The domain excludes numbers less than 1.
The domain is {xx1}.
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}
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}
26.
27.
28.
29.
30.
Chapter 1 2
xy
13
26
39
412
515
618
721
824
927
O
y
x
6
246810
12
18
24
xy
611
510
49
38
27
16
O
y
x
O
y
x
xy
44
35
26
17
08
19
210
311
412
16b.
16c. No; a vertical line at x77, x78, x82, or
x83 would pass through two points.
Pages 10–12 Exercises
17. Table Graph:
Equation: y3x
18. Table:
Equation: yx5
19. Table: Graph:
Equation: y8 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}
xy
49
38
27
16
05
14
xy
11
22
33
44
55
66
xy
55
44
33
22
11
00
11
xy
10
23
36
49
512
xy
11 3
11 3
O
y
x
O
y
x
O
y
x
O
y
x
O
y
x
–4
–2
4812
2
4
180
7270
O
7674 78
Height (in.)
8280 8684
200
220
240
260
280
300
Weight
(lb)
31.
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};
Afunction because each x-value is paired with
exactly one y-value.
40. domain: {x8 x8}; range: {y8 y8};
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)23(2) 2
20 6 2 or 12
43. h(0.5)
0
1
.5
2
44. j(2a) 1 4(2a)3
1 4(8a3)
1 32a3
45. f(n1) 2(n1)2(n1) 9
2(n22n1) n1 9
2n24n2 n1 9
2n25n12
46. g(b21)
3
5
(
(
b
b
2
2
1
1
)
)
3
6
b2
b
21
or
2
6
b
b
2
2
47. f(5m) (5m)213
25m213
48. x25 0
x25
x5
; x 5
49. x29 0
x29
3 x3; x3 or x3
50. x27 0
x27
7
x7
; x7
or x7
51a. x1
51b. x5
51c. x2, 2
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.
52c. Yes; no member of the domain is paired with
more than one member of the range.
53. x2m1, so
x
2
1
m.
Substitute
x
2
1
for min f(2m1) to solve for f(x),
24m336m226m
24
x
2
1
336
x
2
1
226
x
2
1
24
36
x22
4
x1
26
x
2
1
3x39x29x3 9x218x9 13x13
3x34x7
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
x33x23x1
8
3Chapter 1
xy
42
42
O
y
x
3
4
5
6
7
12 20 2816 24
Number Applied
(thousands)
Number
Attending
(thousands)
2
O
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
56. P(4)
(1)(2
3
)1
1
P(5)
(2)(3
1
)1
7
P(6)
(3)(1
7
)1
4
7
57. 72(3242) 49 (9 16)
49 25 or 24
The correct choice is B.
Composition of Functions
Page 13 Graphing Calculator Exploration
1.
2.
1-2
3.
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),
(fg)(x) f(x) g(x); (fg)(x) f(x) g(x);
(fg)(x) f(x) g(x); and
g
f
(x)
g
f(
(
x
x
)
)
, g(x) 0
Page 17 Check for Understanding
1. Sample answer: f(x) 2x1 and g(x) x6;
Sample explanation: Factor 2x211x6.
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; [fg](x) is the function f(x) performed on g(x)
and [gf](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) 3x24x5 2x9
3x26x4
f(x) g(x) 3x24x5 (2x9)
3x22x14
f(x) g(x) (3x24x5)(2x9)
6x335x226x45
g
f
(x)
g
f(
(
x
x
)
)
3x2
2
x
4x
9
5
, x
9
2
6. [fg](x) f(g(x))
f(3 x)
2(3 x) 5
2x11
[gf](x) g(f(x))
g(2x5)
3 (2x5)
2x8
Chapter 1 4
7. [fg](x) f(g(x))
f(x22x)
2(x22x) 3
2x24x3
[gf](x) g(f(x))
g(2x3)
(2x3)22(2x3)
(4x212x9) 4x6
4x216x15
8. Domain of f(x): x1
Domain of g(x): all reals
g(x) 1
x3 1
x2
Domain of [fg](x) is x2.
9. x1f(x0) f(2)
2(2) 1 or 5
x2f(x1) f(5)
2(5) 1 or 11
x3f(x2) f(11)
2(11) 1 or 23
5, 11, 23
10a. [KC](F) K(C(F))
K
5
9
(F32)
5
9
(F32) 273.15
10b. K(40)
5
9
(40 32) 273.15
40 273.15 or 233.15
K(12)
5
9
(12 32) 273.15
24.44 273.15 or 248.71
K(0)
5
9
(0 32) 273.15
17.78 273.15 or 255.37
K(32)
5
9
(32 32) 273.15
0 273.15 or 273.15
K(212)
5
9
(212 32) 273.15
100 273.15 or 373.15
Pages 17–19 Exercises
11. f(x) g(x) x22xx9
x2x9
f(x) g(x) x22x(x9)
x23x9
f(x) g(x) (x22x)(x9)
x37x218x
g
f
(x)
x
x
2
2
9
x
, x9
12. f(x) g(x)
x
x
1
x21
x
x
1
(x2
x
1
)(x
1
1)
x
x
1
x3x
x
2
1
x1
x3
x
x2
1
1
, x1
5Chapter 1
f(x) g(x)
x
x
1
(x21)
x
x
1
(x2
x
1
)(x
1
1)
x
x
1
x3x
x
2
1
x1
, x1
f(x) g(x)
x
x
1
(x21)
x(x
x
1
)(x
1
1)
x2x, x 1
g
f
(x)
x
x
1
x21
1
x3x2x
x1
, x1 or 1
13. f(x) g(x)
x
3
7
x25x
x
3
7
(x2
x
5
x)(
7
x7)
x37x25x235x
x7
3
x7
x
x
1
x21
x3x22x1
x1
, x7
f(x) g(x)
x
3
7
(x25x)
x
3
7
(x2
x
5
x)(
7
x7)
x37x25x235x
x7
3
x7
x32x235x3
x7
x
3
7
x2
1
5x
x32x
3
235x
, x5, 0, 7
14. f(x) g(x) x3
x
2
x
5
(x
x
3
)(x
5
5)
x
2
x
5
x2
x
2
x
5
15
x
2
x
5
x
x
2
1
5
5
, x5
f(x) g(x) x3
x
2
x
5
(x
x
3
)(x
5
5)
x
2
x
5
x2
x
2
x
5
15
x
2
x
5
x2
x
4
x
5
15
, x5
f(x) g(x) (x3)
x
2
x
5
2x
x
2
5
6x
, x5
g
f
(x) x3
x
2
x
5
, x7
f(x) g(x)
x
3
7
(x25x)
3x
x
2
1
7
5x
, x7
g
f
(x)
x
3
7
x25x
x32x235x3
x7
x3
x
2
x
5
x22
2
x
x
15
, x0 or 5
15. [fg](x) f(g(x))
f(x4)
(x4)29
x28x16 9
x28x7
[gf](x) g(f(x))
g(x29)
x29 4
x25
16. [fg](x) f(g(x))
f(x6)
1
2
(x6) 7
1
2
x3 7
1
2
x4
[gf](x) g(f(x))
g(
1
2
x7)
1
2
x7 6
1
2
x1
17. [fg](x) f(g(x))
f(3x2)
3x24
[gf](x) g(f(x))
g(x4)
3(x4)2
3(x28x16)
3x224x48
18. [fg](x) f(g(x))
f(5x2)
(5x2)21
25x41
[gf](x) g(f(x))
g(x21)
5(x21)2
5(x42x21)
5x410x25
19. [fg](x) f(g(x))
f(x3x21)
2(x3x21)
2x32x22
[gf](x) g(f(x))
g(2x)
(2x)3(2x)21
8x34x21
20. [fg](x) f(g(x))
f(x25x6)
1x25x6
x25x7
[gf](x) g(f(x))
g(1 x)
(x1)25(x1) 6
x22x1 5x5 6
x27x12
21. [fg](x) f(g(x))
f
x
1
1
x
1
1
1
x
1
1
x
x
1
1
x
x
1
, x1
[gf](x) g(f(x))
g(x1)
x1
1
1
1
x
, x0
22. Domain of f(x): all reals
Domain of g(x): all reals
Domain of [fg](x): all reals
23. Domain of f(x): x0
Domain of g(x): all reals
g(x) 0
7 x0
7 x
Domain of [fg](x) is x7.
24. Domain of f(x): x2
Domain of g(x): x0
g(x) 2
1
4
x2
1 8x
1
8
x
Domain of [fg](x) is x
1
8
, x0.
25. x1f(x0) f(2)
9 2 or 7
x2f(x1) f(7)
9 7 or 2
x3f(x2) f(2)
9 2 or 7
7, 2, 7
26. x1f(x0) f(1)
(1)21 or 2
x2f(x1) f(2)
(2)21 or 5
x3f(x2) f(5)
(5)21 or 26
2, 5, 26
27. x1f(x0) f(1)
1(3 1) or 2
x2f(x1) f(2)
2(3 2) or 2
x3f(x2) f(2)
2(3 2) or 2
2, 2, 2
Chapter 1 6
28. $43.98 $38.59 $31.99 $114.56
Let xthe 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 [TS](x).
[TS](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) m1xb1and g(x)
m2xb2. Then [fg](x) m1(m2xb2) b1
m1m2xm1b2b1
Since m1and m2are constants, m1m2is a
constant. Similarly, m1, b2, and b1are constants,
so m1b2b1is a constant. Thus, [fg](x) is a
linear function if f(x) and g(x) are both linear.
30a. WnWpWf
FpdFfd
d(FpFf)
30b. Wnd(FpFf)
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. (fg)(x) f(g(x))
f(1 x2)
x2
1
(x
2
x2
1)
x2
(1 x2) 1
So, f(x) x1 and f
1
2
1
2
1
1
2
.
33a. v(p)
7
4
p
7
33b. r(v) 0.84v
33c. r(p) r(v(p))
r
7
4
p
7
0.84
7
4
p
7
5.
4
8
7
8p
or
1
1
4
17
7
5
p
33d. r(423.18)
147
1
(4
1
2
7
3
5
.18)
$52.94
r(225.64)
147
1
(2
1
2
7
5
5
.64)
$28.23
r(797.05)
147
1
(7
1
9
7
7
5
.05)
$99.72
7Chapter 1
34a. Iprt
5000(0.08)(1)
400
Iprt
5400(0.08)(1)
432
Iprt
5832(0.08)(1)
466.56
Iprt
6298.56(0.08)(1)
503.88
Iprt
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.
37. g(4)
(
4
4
(
)
3
4
)
5
6
4
1
6
5
5
1
9
6
or 3
1
1
6
1
38.
39. f(n1) 2(n1)2(n1) 9
2(n22n1) n1 9
2n25n12
The correct choice is C.
Graphing Linear Equations
Page 23 Check for Understanding
1. mrepresents 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.
1-3
xy
26
13
00
13
26
39
y
x
O
9.
1
2
x6 0
1
2
x6
x12
10. Since m0 and b19, this function has no
x-intercept, and therefore no zeros.
11a. (38.500, 173), (44.125, 188)
11b. m
44.
1
1
8
2
8
5
1
3
7
8
3
.500
5.
1
6
5
25
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.
Pages 24–25 Exercises
12. The y-intercept is 9. The slope is 4.
13. The y-intercept is 3. The slope is 0.
4. Sample answer: Both graphs are lines. Both lines
have a y-intercept of 8. The graph of y5x8
slopes upward as you move from left to right on
the graph and the graph of y5y8 slopes
downward as you move from left to right on the
graph.
5. 3x 4(0) 2 03(0) 4y2 0
3x2 04y2 0
3x24y2
x
2
3
y
1
2
6. x2(0) 5 00 2y5 0
x5 02y5 0
x52y5
y
5
2
7. The y-intercept is 7. Graph (0, 7).
The slope is 1.
8. The y-intercept is 5. Graph (0, 5).
The slope is 0.
Chapter 1 8
O
y
x
(1, 8)
(0, 7)
O
y
x
(0, 5)
O
y
x
4
4
4
812
8
12
O
y
x
2442
8
16
24
O
y
x
y
4
x
9
O
y
x
y
3
O
y
x
1
2
0,
)(
)(
2
3
,0
O
y
x
5
2
0,
(5, 0)
)(
14. 2x3y15 0
3y2x15
y
2
3
x5
The y-intercept is 5. The slope is
2
3
.
15. x4 0
x4
There is no slope. The x-intercept is 4.
16. The y-intercept is 1. The slope is 6.
17. The y-intercept is 5. The slope is 2.
18. y8 0
y8
The y-intercept is 8. The slope is 0.
19. 2xy0
y2x
The y-intercept is 0. The slope is 2.
20. The y-intercept is 4. The slope is
2
3
.
21. The y-intercept is 150. The slope is 25.
22. 2x5y8
5y2x8
y
2
5
x
8
5
The y-intercept is
8
5
. The slope is
2
5
.
23. 3xy7
y3x7
y3x7
The y-intercept is 7. The slope is 3.
9Chapter 1
O
y
x
2
x
3
y
15 0
O
y
x
x
4 0
O
y
x
y
6
x
1
O
y
x
y
5 2
x
O
y
x
y
8 0
O
y
x
2
x
y
0
x
O
y
x
4
2
3
y
O
y
x
50
50
100
2426
y
25
x
150
O
y
x
2
x
5
y
8
O
y
x
3
x
y
7
24. 9x5 0
9x5
x
5
9
The y-intercept is 5.
25. 4x12 0
4x12
x3
The y-intercept is 12.
26. 3x1 0
3x1
x
1
3
The y-intercept is 1.
27. 14x0
x0
The slope is 14.
28. None; since m0 and b12, this function has
no x-intercepts, and therefore no zeros.
29. 5x8 0
5x8
x
8
5
The y-intercept is 8.
30. 5x2 0
5x2
x
2
5
31. The y-intercept is 3. The slope is
3
2
.
3
2
x3 0
3
2
x3
x2
32. Sample answer: f(x) 5; f(x) 0
33a. (1.0, 12.0), (10.0, 8.4)
m
8
1
.
0
4
.0
12
1
.
.
0
0
3
9
.6
or 0.4
(0.4) 0.4 ohms
33b. 0.4
1.0
12
25
v
.0
0.4
1
2
2
4
v
9.6 12 v
v2.4 volts
34. m
9
4
7
3
2
7
a
1
(
9
4)
2
7
2
7
a
8
4
2(a4) 56
2a8 56
2a48
a24
35a. (86.85, 90), (126.85, 100)
m
1
4
0
0
or
1
4
35b. For each 1 degree increase in the temperature,
there is a
1
4
-pascal increase in the pressure.
35c.
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 290t0
290t10,440
t36
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.
37c.
100 – 90
126.85 – 86.85
Chapter 1 10
O
f
(
x
)
x
f
(
x
) 12
4
8
O
f
(
x
)
x
8
5
()
, 0
2
2
4
2
6
8
f
(
x
) 5
x
8
O
y
x
(2, 0)
3
2
y
x
3
O
P
T
20
20 40 60 80
40
60
80
100
O
v
(
t
)
t
2000
8162432
4000
6000
8000
10,000
(36, 0)
(0, 10,440)
Ox
5
9
(
, 0
)
f
(
x
) 9
x
5
f
(
x
)
Ox
(3, 0)
f
(
x
) 4
x
12
f
(
x
)
Ox
1
3
()
,0
f
(
x
) 3
x
1
f
(
x
)
Ox
(0, 0)
14
28
f
(
x
) 14
x
f
(
x
)
38. Afunction 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)
m
6
7
7
6
.2
5
5
6
0
1
2
7
0
.2
or 0.86
39b. 1805(0.86) $1552.30
39c. 1 MPC 1 0.86
0.14
39d. 1805(0.14) $252.70
40. (fg)(x) 2xx24 or x22x4
(fg)(x) 2x(x24)
x22x4
41a. 1 0.12 0.88
d(p) 0.88p
41b. r(d) d100
41c. r(d(p)) r(0.88p)
0.88p100
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. [fg](3) f(g(3))
f(3 2)
f(5)
(5)24(5) 5
25 20 5 or 50
[gf](3) g(f(3))
g((3)24(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.
{(3, 14), (2, 13), (1, 12),
(0, 11)}, yes
46. Let ssum.
4
s
15
s60
The correct choice is D.
Graphing Calculator Exploration:
Analyzing Families of Linear
Graphs
Page 26
1. See students’ graphs. All of the graphs are lines
with y-intercept at (0, 2). Each line has a
different slope.
2. Aline 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 mare
parallel. The graphs of lines with the same value
for bhave the same y-intercept.
Writing Linear Equations
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.
yy1m(xx1)ymx b
y(4)
1
4
(x3) 4
1
4
(3) b
y4
1
4
x
3
4
1
4
9
b
x4y19 0Substitute the slope
and intercept into the
general form.
y
1
4
x
1
4
9
Write in standard form.
x4y19 0
3. 55 represents the hourly rate and 49 represents
the fee for coming to the house.
4. m
0
3
6
0
y
1
2
x3
3
6
or
1
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 point-
slope form.
6. ymx b→y
1
4
x10
1-4
1-3B
11 Chapter 1
xy
314
213
112
011
7. y2 4(x3)
y2 4x12
y4x10
8. m
9
7
2
5
7
2
y2
7
2
(x5)
y2
7
2
x
3
2
5
y
7
2
x
3
2
1
9. y2 0(x(9))
y2 0
y2
10a. y5.9x2
10b. y5.9(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 Exercises
11. ymx b→y5x2
12. y5 8(x(7))
y5 8x56
y8x61
13. ymx b→y
3
4
x
14. ymx b→y12x
1
2
15. y5 6(x4) 16. x12
y5 6x24
y6x19
17. m
9
8
5
1
y5
4
9
(x1)
4
9
y5
4
9
x
4
9
y
4
9
x
4
9
9
18. (8, 0), (0, 5) y0
5
8
(x(8))
m
0
5
(
0
8)
y
5
8
x5
5
8
27b. Using sample answer from part a,
y
9
7
(19)
1
7
6
17
7
1
1
7
6
or
18
7
7
or about 26.7 mpg
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)
is
3
3
9
5
or
3
4
. The slope of the line through
(3, 3), and (1, 6) is
1
6
(
3
3)
or
3
4
. Since these two
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.
30. 3x2y5 0
2y3x5
y
3
2
x
5
2
31a. (1995, 70,583), (1997, 82,805)
m
82,
1
8
9
0
9
5
7
–1
7
9
0
9
,
9
583
12,
2
222
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. (fg)(x) f(x) g(x)
x3(x23x7)
x53x47x3
g
f
(x)
x2
x
3
3
x7
, where g(x) 0
34.
{(4, 16), (3, 9), (2, 4)}, yes
Chapter 1 12
O
y
x
3
x
2
y
5 0
xy
416
39
24
35. x
1
y
y
y
1
y
y
2
1
y
2
y2
y
1
2
y2
y
1
1
2
y2
2
y
1
The correct choice is A.
19. m
1
3
1
8
0
11
or 0
y1 0(x8)
y1 0
y1
20. x421. x0
22. y0 0.25(x24)
y0.25x6
23. y(4)
1
2
(x(2))
y4
1
2
x1
1
2
xy5 0
x2y10 0
x2y10
24. m
1
3
(
0
2)
3
3
1
y0 1(x2)
yx2
xy2
25a. t2
x
20
7
0
0
0
00
25b. t2
14,49
2
4
00
0
7000
2 3.747 or 5.747; about 5.7 weeks
26. Ax By C0
By Ax C
y
B
A
x
C
B
; m
B
A
27a. Sample answer: using (20, 28) and (27, 37),
m
3
2
7
7
2
2
8
0
y28
9
7
(x20)
9
7
y28
9
7
x
18
7
0
y
9
7
x
1
7
6
4. Let xoriginal 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 [TS](x).
[TS](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. [fg)(x) f(g(x)
f(x1)
x1
1
1
1
x
, x0
[gf](x) g(f(x))
g(
x
1
1
)
x
1
1
1
x
1
1
x
x
1
1
x
x
1
, x1
6. 2x4y8
4y2x8
y
1
2
x2
7. 3x2y
3
2
xy
8. 5x3 0
5x3
x
3
5
9. m
8
7
5
2
y5
3
5
(x2)
3
5
y5
3
5
x
6
5
3
5
xy
1
5
9
0
3x5y19 0
10a. (1990, 6,478,216), (2000, 8,186,453)
m8,186,453 6,478,216
2000 – 1990
Page 31 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.
10b. y – 6,478,216 170,823.7(x– 1990)
y – 6,478,216 170,823.7x – 339,939,163
y170,823.7x – 333,460,947
Writing Equations of Parallel and
Perpendicular Lines
Pages 35–36 Check for Understanding
1. If A, B, and Care the same or the ratios of the As
and the Bs and the Cs are proportional, then the
lines are coinciding. If Aand Bare the same and
Cis 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. 4x3y19 0
y
4
3
x
1
3
9
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. yx6xy8 0
yx8
parallel
8. y2x84x2y16 0
y2x8
coinciding
9. y9 5(x5)
y9 5x25
5xy16 0
10. 6x5y24
y
6
5
x
2
5
4
y(5)
5
6
(x(10))
y5
5
6
x
2
3
5
6y30 5x50
5x6y80 0
11. mof EF: m
8
4
4
3
mof EH: m
2
6
4
3
4
1
or 4
2
3
mof GH: m
2
6
6
7
mof FG: m
6
7
8
4
4
1
or 4
2
3
parallelogram
Pages 36–37 Exercises
12. y5x18 2x10y10 0
y
1
5
x1
slopes are opposite reciprocals; perpendicular
1-5
13 Chapter 1
O
y
x
2
x
4
y
8
or about 170,823.7
1,708,237
10
O
y
x
3
x
2
y
2. f(4) 7 42
7 16 or 9
3. g(n2)
n
3
21
n
3
1
13. y7x5 0y7x9 0
y7x5y7x9
same slopes, different y-intercepts; parallel
14. different slopes, not reciprocals; none of these
15. horizontal line, vertical line; perpendicular
16. y4x34.8x1.2y3.6
y4x3
same slopes, same y-intercepts; coinciding
17. 4x6y11 3x2y9
y
2
3
x
1
6
1
y
3
2
x
9
2
Slopes are opposite reciprocals; perpendicular.
18. y3x23xy2
y3x2
different slopes, not reciprocals; none of these
19. 5x9y14 y
5
9
x
1
9
4
y
5
9
x
1
9
4
same slopes, same y-intercepts; coinciding
20. y4x2 0y4x1 0
y4x2y4x1
same slopes, different y-intercepts; parallel
21. None of these; the slopes are not the same nor
opposite reciprocals.
22. y(8) 2(x0)
y8 2x
2xy8 0
23. m
(
4
9)
or
4
9
y(15)
4
9
(x12)
y15
4
9
x
1
3
6
9y135 4x48
4x9y183 0
24. y(11) 0(x4)
y11 0
25. y(3)
1
5
(x0)
y3
1
5
x
5y15 x
x5y15 0
26. m
(
6
1)
or 6; perpendicular slope is
1
6
y(2)
1
6
(x7)
y2
1
6
x
7
6
6y12 x7
x6y5 0
27. x12 is a vertical line; perpendicular slope is 0.
y(13) 0(x6)
y13 0
28a. 5y4x10
4x5y10 0m
(
4
5)
or
4
5
y8
4
5
(x(15))
y8
4
5
x12
5y40 4x60
4x5y100 0
28b. perpendicular slope:
5
4
y8
5
4
(x(15))
y8
5
4
x
7
4
5
4y32 5x75
5x4y43 0
29a. 8x14y3 0kx 7y10 0
m
(
8
14)
or
4
7
m
(
k
7)
or
k
7
4
7
k
7
→k4
29b.
k
7
7
4
4k49
k
4
4
9
30a. Sample answer: y1 0, x1 0
30b. Sample answer: x7 0, x9 0
31. altitude from Ato BC:
mof BC
5
10
(
4
5)
0
6
or 0
mof altitude is undefined; x7
altitude from Bto AC:
mof AC
4
5
7
10
1
3
5
or 5
mof altitude
1
5
y(5)
1
5
(x10)
y5
1
5
x2
5y25 x10
x5y15 0
altitude from Cto AB:
mof AB
1
5
0
1
7
0
3
15
or 5
mof altitude
1
5
y(5)
1
5
(x4)
y5
1
5
x
4
5
5y25 x4
x5y29 0
32. We are given ym1xb1and ym2xb2with
m1m2and b1b2. Assume that the lines
intersect at point (x1, y1). Then y1m1x1b1
and y1m2x1b2. Substitute m1x1b1for y1in
y1m2x1b2. Then m1x1b1m2x1b2.
Since m1m2, substitute m1for m2. The result is
m1x1b1m1x1b2. Subtract m1x1from each
side to find b1b2. However, this contradicts the
given information that b1b2. Thus, the
Chapter 1 14
assumption is incorrect and the lines do not share
any points.
33a. Let xregular espressos.
Let ylarge espressos.
216x162y783 248x186y914
y
4
3
x
2
6
9
y
4
3
x
4
9
5
3
7
No; the lines that represent the situation do not
coincide.
33b. Let xregular espressos.
Let ylarge espressos.
216x162y783 344x258y1247
y
4
3
x
2
6
9
y
4
3
x
2
6
9
Yes; the lines that represent the situation
coincide.
34a. (15, 1939.20), (16, 1943.09)
m 1943.09 – 1939.20
16 – 15
36a. (40, 295), (80, 565)
m
56
8
5
0
2
4
9
0
5
2
4
7
0
0
or 6.75
y565 6.75(x80)
y565 6.75x540
y6.75x25
36b. $6.75 36c. $25
37. 3x2y6 0
y
3
2
x3
38. [gh](x) g(h(x))
g(x2)
x21
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. 2xy12 x2y6
y2x12 x2(2x12) 6
x4x24 6
3x30
x10
2(10) y12
y8
2x2y2(10) 2(8)
20 (16) or 4
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.
1-6
15 Chapter 1
3.
1
89
or 3.89
y1943.09 3.89(x16)
y1943.09 3.89x62.24
y3.89x1880.85
(16, 1943.09), (17, 1976.76)
m
33
1
.67
or 33.67
1976.76 – 1943.09
17 – 16
y1976.76 33.67(x17)
y1976.76 33.67x572.39
y33.67x1404.37
(17, 1976.76), (18, 1962.44)
m
or –14.32
14.32
1
1962.44 1976.76
18 17
y
x
O
3
x
2
y
6 0
y1962.44 –14.32(x18)
y1962.44 –14.32x257.76
y–14.32 2220.2
(18, 1962.44), (19, 1940.47)
m
–21
1
.97
or –21.97
1940.47 1962.44
19 18
y1940.47 21.97(x19)
y1940.47 21.97x417.43
y21.97x2357.9
34b. parallel lines or the same line; no
34c. y3.89(22) 1880.85
1966.43
y33.67(22) 1404.37
2145.11
y14.32(22) 2220.2
1905.16
y21.97(22) 2357.9
1874.56
No; the equations take only one pair of days into
account.
35. y5 2(x1)
y5 2x2
y2x7
Personal Consumption on Durable Goods
3500
3000
2500
2000
1500
500
01995 1997 1999
Year
2001 2003
Dollars
1000
Car Weight and Mileage
80
60
0
Weight (hundreds of pounds)
15 3525 30200
Average
Mileage
20
40
All-Time NFL Coaching Victories
200
0
Years
010 203040
Wins
400
Students per Computer
25
20
5
10
0’99 ’01
Year
’97’93 ’95’89 ’91
Average 15
4b. Sample answer: using (1995, 2294) and
(2002, 3158)
m
86
7
4
or 123.4
y3158 123.4(x2002)
y123.4x243,888.8
4c. y132.8x262,621.2; r0.98
4d. y132.8(2010) 262,621.2
4306.8
$4306.80; yes, the correlation coefficient shows a
strong correlation.
5a.
5b. Sample answer: using (1997, 6.1) and (2001, 4.9)
m
20
4
0
.9
1
–
–
6
1
.
9
1
97
–1
4
.2
or 0.3
y6.1 0.3(x1997)
y0.3x605.2
5c. y1.61x3231.43; r0.97
5d. 1 1.61x3231.43
3230.43 1.61x
2006.47 x
2006; yes, the number of students per computer is
decreasing steadily.
Pages 42–44 Exercises
6a.
6b. Sample answer: using (18, 170) and (40, 324)
m
1
2
5
2
4
or 7
y324 7(x40)
y7x44
6c. y7.57x33.38; r0.88
6d. y7.57(16) 33.38
154.5
155; yes, ris fairly close to 1. (Actual data is
159.)
324 – 170
40 – 18
3158 – 2294
2002 – 1995
7a.
7b. Sample answer: using (1995, 23,255) and
(2002, 30,832)
m
75
7
77
or 1082.43
y23,255 1082.43(x1995)
y1082.43x2,136,192.85
7c. y1164.11x2,299,128.75; r0.99
7d. y1164.11(2010) – 2,299,128.75
40,732.35
$40,732.35; yes, rshows a strong relationship.
8a.
8b. Sample answer: using (17.5, 65.4) and
(35.0, 27.7)
m
2
3
7
5
.
.
7
0
6
1
5
7
.
.
4
5
1
3
7
7
.5
.7
or 2.15
y65.4 2.15(x17.5)
y2.15x103.0
8c. y1.72x87.59; r0.77
8d. y1.72(45.0) 87.59
y10.19
10.19; no, rdoesn’t show a particularly strong
relationship.
9a.
30,832 – 23,255
2002 – 1995
Chapter 1 16
Personal Income
35
30
25
0
Year
19971995 20011999 200
3
Dollars
(thousands)
15
20
Acorn Size and Range
30,000
20,000
10,000
0
Acorn Size (cm3)
012108642
Range
(hundreds
of km2)
O
y
5
10
15
20
25
12345
x
y
0.82
x
24
World Population
7000
6000
5000
4000
3000
2000
1000
0
Year
0 20001000
Millions
of People
9b. Sample answer: using (0.3, 233) and (3.4, 7900)
m
7
3
9
.
0
4
0
0
2
.
3
3
3
7
3
6
.
6
1
7
or 2473.23
y7900 2473.23(x3.4)
y2473.23x508.97
9c. y885.82 6973.14; r0.38
9d. The correlation value does not show a strong or
moderate relationship.
10a.
10b. Sample answer: using (1990, 26.2) and
(2003, 37.6)
m
1
1
1
3
.4
or 0.88
y26.2 0.88(x1990)
y0.88x1725
10c. y0.84x1648.27; r0.984
10d. y0.84(2008) 1648.27 or 38.45
38.45%; yes, ris very close to 1.
11a.
11b. Sample answer: using (1, 200) and (2000, 6050)
m
5
1
8
9
5
9
0
9
or 2.93
y200 2.93(x1)
y2.93x197.07
11c. y1.65x289.00; r0.56
11d. y1.65(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.
6050 – 200
2000 – 1
37.6 – 26.2
2003 – 1990
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
yax byax b
a885.2867133 a625.041958
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 xcomputers.
Let yprinters.
24x40y38,736
y0.6x968.4
30x50y51,470
y0.6x1029.4
No; the lines do not coincide.
14b. Let xcomputers.
Let yprinters.
24x40y38,736
y0.6x968.4
30x50y48,420
y0.6x968.4
Yes; the lines coincide.
15. y(4) 6(x(3))
y4 6x18
6xy22 0
16a.
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. [fg](x) f(g(x))
f(x1)
(x1)3
x33x23x1
[gf](x) g(f(x))
g(x3)
x31
18. Yes; each domain value is paired with exactly one
range value.
17 Chapter 1
Working Women
40
35
30
25
20
15
10
5
0’88 ’92 ’96
Year
’00 ’02 ’04’98’94’90’86
Percent in
Management
O
y
x
19. The y-intercept is 1.
The slope is 3 (move down 3 and right 1).
The correct choice is C.
Piecewise Functions
Pages 48–49 Check for Understanding
1. f(x) xif x0
xif x0
2. reals, even integers
3. f(x)
x2 if x0
xif 0 x4
x2 if x4
4. Alex is correct because he is applying the
definition of a function.
5.
6.
7.
1-7
O
y
x
9. greatest integer function; his hours, c(h) is the
cost, c(h) 50hif [[h]] h
50[[h1]] if [[h]] h
10. long term lot:
2(6) 3(1) 12 3 or 15
shuttle facility:
4(3) 12
shuttle facility
Pages 49–51 Exercises
11.
12.
13.
Chapter 1 18
O
y
x
xf(x)
3 x23
2 x12
1 x01
0 x10
1 x21
2 x32
3 x43
4 x54
xf(x)
14
03
12
21
30
41
O
y
x
Ox
y
O
2
100
200
300
400
61048
y
x
xf(x)
0 x150
1 x2 100
2 x3 150
3 x4 200
xf(x)
14
32
50
72
94
Ox
y
xf(x)
2 x10
1 x01
0 x12
1 x23
2 x34
O
y
x
8.
O
y
x
O
y
x
14.
15.
16. ART
17.
18.
19. 20.
21.
19 Chapter 1
xf(x)
57
33
1.5 0
03
27
Ox
y
xf(x)
2 x13
1 x02
0 x11
1 x20
2 x31
y
x
O
O
y
x
xf(x)
06
14
22
30
42
Ox
y
xf(x)
1 x
2
3
3
2
3
x
1
3
2
1
3
x01
0 x
1
3
0
1
3
x
2
3
1
2
3
x12
O
y
x
2
4
–4
–1 1
–2
xf(x)
5 x4
2
5
4 x3
1
2
3 x2
2
3
2 x11
1 x02
1 x22
2 x31
3 x4
2
3
4 x5
1
2
5 x6
2
5
O
1
2
2
1
y
x
22.
6 if t
1
2
10 if
1
2
t1
16 if 1 t2
24 if 2 t24
24. Greatest integer; wis the weight in ounces, c(w) is
the cost in dollars,
c(w) 0.37 0.23(w1) if [[w]] w
0.37 0.23[[w]] if [[w]] w
25. Absolute value; wis the weight in pounds, d(w) is
the discrepancy, d(w) 1 w
26a. step
3.50 if 0.00 v25.00
5.95 if 25.01 v75.00
7.95 if 75.01 v125.00
9.95 if 125.01 v
26c.
27. If nis any integer, then all ordered pairs (x, y)
where xand yare both in the interval [n, n1)
are solutions.
28a. absolute value 28b. d(t) 65 t
28c.
28d. d(63) 65 63or 2
d(28) 65 28or 37
2
2
37
19.5 heating degree days
29a. step
5% if x$10,000
7.5% if $10,000 x$30,000
9.3% if x$30,000
29c.
29d. 9.5%
30. No; the functions are the same if xis positive.
However, if xis negative, the functions yield
different values. For example, [gf](1.5) 1
and [fg](1.5) 1; [gf](1.5) 2 and
[fg](1.5) 1.
Chapter 1 20
xf(x)
56
30
16
09
16
30
56
O
y
x
23. Step; tis the time in hours, c(t) is
the cost in dollars,
c(t)
O
26104812 2218 242014 16
8
16
24
d(t)
t
26b. vis the value of the order, s(v) is the shipping,
s(v)
O
25 125 15075
Value of Order
(dollars)
Shipping
(dollars)
50 100
O
20
20
40
60
80
6040 80
t
d
(
t
) |65
t
|
d
(
t
)
xf(x)
0 x1 0.37
1 0.37
1 x2 0.60
2 0.60
2 x3 0.83
3 0.83
0.2
0.4
0.6
0.8
1.0
24
O
c(w)
w
xf(x)
01
10
21
32
Ow
d(w)
y
x
0
Income
(thousands of dollars)
Tax
Rate
(percent) 5
10
10
20 30
29b. t(x)
31a.
31b. Sample answer: using (3,183,088, 53.4) and
(362,777, 3.3)
m
2
,8
5
2
0
0
.
,
1
311
or 0.0000178
y53.4 0.0000178(x3,183,088)
y0.0000178x3.26
31c. y0.0000136x4.55, r0.68
31d. y0.0000136(307,679) 4.55
y8.73
8.73%; No, the actual value is 22%.
32. y2 2(x4)
y2 2x8
2xy6 0
33a. (39, 29), (32, 15) 33b. m
1
3
5
2
2
3
9
9
1
7
4
or 2
33c. The average number of points scored each
minute.
34. p(x) (rc)(x)
(400x0.2x2) (0.1x200)
399.9x0.2x2200
35. Let xthe 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)
[TS](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.
3.3 53.4
362,777 3,183,088
Graphing Linear Inequalities
Pages 54–55 Check for Understanding
1. y2x6
2. Graph the lines 3 2xyand 2xy7. The
graph of 3 2xyis solid and the graph of
2xy7 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.
5. ART
6.
7.
1-8
21 Chapter 1
60
50
40
30
20
10
0051015 20 25 30 35
Working Population
(hundreds of thousands)
Percent who use
public transportation
Public Transport
O
x
y
! 4
y
x
O
3
x
y
^ 6
y
x
O
y
x
7 !
x
y
^ 9
O
y
y
! |
x
3|
x
O
y
x
2 ^
x
2
y
^ 4
O
y
x
2
519
5
y 6
x
O
y
y
! 2
x
1
x
8a. c(m) 45 0.4m
8b.
8c. Sample answer: (0, 45), (10, 49), (20, 50)
Pages 55–56 Exercises
9.
10.
11.
12. 13.
Chapter 1 22
O
c(m)
c
(
m
) ^ 45 0.4
m
20 60 10040
10
20
30
40
50
60
70
80
m
O
y
x
–
y
> – 5
x
O
y
y
< 3
x
O
y
x
–4 ^
x
y
^ 5
O
y
x
y 6 |
x
|
O
y
x
y
6 |
x
| 4
O
y
x
y
! |2
x
3|
O
y
x
8 ^ 2
x
y
! 6
y
x
|
x
3| !
y
1
O
O
y
2
x
4
y
6 7
x
14.
15. 16.
17.
18.
19.
20.
1
55
110
165
220
275
3
Hours
Ox
y
245
O
y
x
20
20 6040 80
40
60
80
21.
22.
23a. 8x10y8(60)
8x10y480
23b.
23c. Sample answer: (0, 48) (60, 0), (45, 6)
23d. Sample answer: using complex computer
programs and systems of inequalities.
24.
25a. points in the first and third quadrants
25b. If xand ysatisfy the inequality, then either
x0 and y0 or x0 and y0. If x0 and
y0, then xxand yy. Thus,
xyxy. Since xyis positive,
xyxy.
If x0 and y0, then xxand yy.
Then xyx(y) or (xy). Since
both xand yare negative, (xy) is negative,
and xy(xy).
26a. 8 x
5
1
00
; 4
1
4
y
5
1
00
23 Chapter 1
O
y
x
x
6 0
y
6 0
O
y
x
1
1086422
2
4
4
3
6
8
10
2
3
4
O
y
x
|y| 6
x
O
y
x
4.252
4.25
4.248
7.998 8.0028
O
r
a
10
100
200
3020 40 6050 70
0.6 (220
a
) ^
r
^ 0.9 (220
a
)
26b.
27a. 0.6(220 a) r0.9(220 a)
27b.
28a. step
28b. Let c(h) represent the cost for hhours.
Then c(h) 55hif [[h]] h
55[[h1]] if [[h]] h
28c.
29a. 3xy10
y3x10
y(2) 3(x0)
3xy2 0
29b. perpendicular slope:
1
3
y(2)
1
3
(x0)
y2
1
3
x
3y6 x
x3y6 0
30. m
7
5
4
1
y7
3
4
(x5)
3
4
y
3
4
x3
1
4
31a. (0, 23), (16, 48);
m
4
1
8
6
2
0
3
2
1
5
6
or 1.5625
31b. the average change in the temperature per hour
32.
95
8
94
59,049
8
6561
52,
8
488
or 6561 or 94
The correct choice is E.
Chapter 1 Study Guide and Assessment
Page 57 Understanding the Vocabulary
1. c2. f
3. d4. g
5. i6. a
7. h8. j
9. e10. b
Pages 58–60 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)24(3) 9
36 12 9 or 57
14. h(0.2) 6 2(0.2)3
6 0.016 or 5.984
15. g
1
3
2
5
1
3
Chapter 1 24
20. (fg)(x) f(x) g(x)
4 x23xor 4 3xx2
(fg)(x) f(x) g(x)
4 x2(3x)
4 3xx2
(fg)(x) f(x) g(x)
(4 x2)(3x)
12x3x3
g
f
(x)
g
f(
(
x
x
)
)
4
3x
x2
, x0
21. (fg)(x) f(x) g(x)
x27x12 x4
x28x16
(fg)(x) f(x) g(x)
x27x12 (x4)
x26x8
(fg)(x) f(x) g(x)
(x27x12)(x4)
x311x240x48
g
f
(x)
g
f(
(
x
x
)
)
x2
x
7
x
4
12
(x
x
4
)(x
4
3)
x3, x4
22. (fg)(x) f(x) g(x)
x21 x1
x2x
(fg)(x) f(x) g(x)
x21 (x1)
x2x2
(fg)(x) f(x) g(x)
(x21)(x1)
x3x2x1
g
f
(x)
g
f(
(
x
x
)
)
x
x
2
1
1
(x
x
1
)(x
1
1)
x1, x1
23. (fg)(x) f(x) g(x)
x24x
x
4
4
x38x216x4
x4
or
6
5
16. k(4c) (4c)22(4c) 4
16c28c4
17. f(m1) (m1)23(m1)
m22m1 3m3
m25m4
18. (fg)(x) f(x) g(x)
6x4 2
6x2
(fg)(x) f(x) g(x)
6x4 (2)
6x6
(fg)(x) f(x) g(x)
(6x4)(2)
12x8
g
f
(x)
g
f(
(
x
x
)
)
6x
2
4
3x2
19. (fg)(x) f(x) g(x)
x24xx2
x25x2
(fg)(x) f(x) g(x)
x24x(x2)
x23x2
(fg)(x) f(x) g(x)
(x24x)(x2)
x32x28x
g
f
(x)
f
g
(
(
x
x
)
)
x
x
2
4
2
x
, x2
2
5
3
x24x
x
4
4
, x4
(fg)(x) f(x) g(x)
x24x
x
4
4
x38x216x4
x4
x24x
x
4
4
, x4
(fg)(x) f(x) g(x)
(x24x)
x
4
4
4x, x4
g
f
(x)
g
f(
(
x
x
)
)
or , x4
x38x216x
4
x24x
x
4
4
24. [fg](x) f(g(x))
f(2x)
(2x)24
4x24
[gf](x) g(f(x))
g(x24)
2(x24)
2x28
25. [fg](x) f(g(x))
f(3x2)
0.5(3x2) 5
1.5x25
[gf](x) g(f(x))
g(0.5x5)
3(0.5x5)2
0.75x215x75
26. [fg](x) f(g(x))
f(3x)
2(3x)26
18x26
[gf](x) g(f(x))
g(2x26)
3(2x26)
6x218
27. [fg](x) f(g(x))
f(x2x1)
6 (x2x1)
x2x7
[gf](x) g(f(x))
g(6 x)
(6 x)2(6 x) 1
x211x31
28. [fg](x) f(g(x))
f(x1)
(x1)25
x22x4
[gf](x) g(f(x))
g(x25)
x25 1
x24
29. [fg](x) f(g(x))
f(2x210)
3 (2x210)
2x27
[gf](x) g(f(x))
g(3 x)
2(3 x)210
2x212x28
30. Domain of f(x): x16
Domain of g(x): all reals
g(x) 16
5 x16
x11
Domain of [fg](x) is x11.
31. The y-intercept is 6. The slope is 3.
32. The y-intercept is 8. The slope is 5.
33. y15 0
y15
The y-intercept is 15. The slope is 0.
34. 0 2xy7
y2x7
The y-intercept is 7. The slope is 2.
35. The y-intercept is 0. The slope is 2.
36. The y-intercept is 2. The slope is 8.
25 Chapter 1
O
y
x
6
6612
12
y
3
x
6
18
O
y
x
4
4–4
8
y
= 8 – 5
x
O
y
x
10
1010
20
y
15 0
O
y
x
y
2
x
7
O
y
x
y
2
x
O
y
x
y
8
x
2
38. The y-intercept is 6. The slope is
1
4
.
39. y2x340. yx1
41. y2
1
2
(x(5))
y2
1
2
x
5
2
y
1
2
x
9
2
42. m
2
5
(
2
4)
3
6
or
1
2
y5
1
2
(x2)
y5
1
2
x1
y
1
2
x4
43. (1, 0), (0, 4)
m
0
4
1
0
4
1
or 4
y(4) 4(x0)
y4 4x
y4x4
44. y145. y0
46. y0 0.1(x1)
y0.1x0.1
47. y1 1(x1)
y1 x1
xy0
48. y6
1
3
(x(1))
y6
1
3
x
1
3
3y18 x1
x3y19 0
49. m
2
1
or 2
y2 2(x(3))
y2 2x6
2xy4 0
50. y(8)
1
2
(x4)
y8
1
2
x2
2y16 x4
x2y20 0
51. m
(
4
2)
or 2, perpendicular slope is
1
2
y4
1
2
(x1)
y4
1
2
x
1
2
2y8 x1
x2y9 0
52. x8 is a vertical line; perpendicular slope is 0.
y(6) 0(x4)
y6 0
53a.
53b. Sample answer: using (1994, 18,458) and
(2000, 25,975)
m
75
6
17
or 1252.8
y25,975 1252.8(x2000)
y1252.8x2,479,625
53c. y1115.9x2,205,568; r0.9441275744
53d. y1115.9(2008) 2,205,568
35,159.2
35,159,200 visitors; Sample answer: This is a
good prediction, because the rvalue indicates a
strong relationship.
54.
25,975 – 18,458
2000 – 1994
Chapter 1 26
O
y
x
7
x
2
y
5
O
y
x
424
1
4
y
x
6
Overseas Visitors
21
14
7
0
Year
1994 1996 20001998
Visitors
(millions)
28
O
f
(
x
)
x
37. 7x2y5
y
7
2
x
5
2
The y-intercept is
5
2
. The slope is
7
2
.
y
x
O
y
^ |
x
|
55.
56.
57.
58.
59.
60. ART
61.
62.
63. 64.
65.
66.
27 Chapter 1
h
(
x
)
x
O
O
f
(
x
)
x
f
(
x
)
x
1
xf(x)
2 x11
1 x00
0 x11
1 x22
2 x33
xf(x)
28
14
00
14
28
g
(
x
)
x
O
g
(
x
) |4
x
|
xf(x)
26
14
02
14
26
k
(
x
)
x
O
y
x
O
y
1 4
y
x
O
x
^ 5
y
x
O
x
y
^ 1
y
x
O
2
y
x
^ 4
y
x
O
y
3
x
1 2
y
x
O
y
! |
x
| 2
y
! |
x
2|
y
x
O
Page 61 Applications and Problem Solving
67a. d
1
2
(20)(1)2
10
d
1
2
(20)(2)2
40
d
1
2
(20)(3)2
90
d
1
2
(20)(4)2
160
d
1
2
(20)(5)2
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)
m
13
5
6
or 27.2; about $27.2 billion
68b. y500 27.2(x1999)
y27.2x53,872.8
69. y0.284x12.964; The correlation is
moderately negative, so the regression line is
somewhat representative of the data.
Page 61 Open-Ended Assessment
1. Possible answer: f(x) 4x4, g(x) x2;
[fg](x) f(g(x)) 4(x2) 4 4x24
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 x4
3a. y2x4 if x4
x1 if x1
3b. y3x1 if x1
Chapter 1
SAT & ACT Preparation
Page 65 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.
636 – 500
2004 – 1999
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.
4
1
3
1
3
3
2
3
5
1
5
3
Remember that dividing by a fraction is
equivalent to multiplying by its reciprocal
1
3
3
1
5
3
5
3
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.
77
77
75347 5 12 24
The correct choice is A.
6. Write each part of the expression without
exponents.
(4)216
(2)4
2
1
4
1
1
6
16
1
1
6
3
4
16
1
1
3
6
16
1
1
3
6
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.
1
3
3
1
5
3
4
1
3
2
3
5
Chapter 1 28
8. First combine the numbers inside the square root
symbol. Then find the square root of the result.
64
36
100
10
The correct choice is A.
9. 60 2 2 3 5
223 5
The number of distinct prime factors of 60 is 3.
The correct choice is C.
10. First find the number of fish that are not tetras.
1
8
(24) or 3 are tetras. 24 3 or 21 are not
tetras. Then
2
3
of these are guppies.
2
3
(21) 14
The answer is 14.
29 Chapter 1
y
x
(1, 3)
y
5
x
2
y
2
x
5
O
Solving Systems of Equations in
Two Variables
Page 69 Graphing Calculator Exploration
1. (1, 480)
2. 3x 4y320 →y
3
4
x80
5x2y340 y
5
2
x170
(76.923077, 22.30769)
3. accurate to a maximum of 8 digits
4. 5x7y70 →y
5
7
x10
10x14y120 y
5
7
x
6
7
0
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.
Page 70 Check for Understanding
1. Sample answer:
4x7x21
y2x1
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.
4. 2y3x6→y
3
2
x3
4y16 6xy
3
2
x4
Inconsistent; Sample answer: The graphs of the
equations are lines with slope
3
2
, but each
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)
2-1 6. no solution
7. 7xy97xy9
5xy15 7(2) y9
12x24 y5
x2(2, 5)
8. 3x4y1→3x4y1
2(6x2y) 2(3) 12x4y6
15x5
x
1
3
3x4y1
3
1
3
4y1
4y2
y
1
2
1
3
,
1
2
9. 30
1
3
x
3
2
y30(4) →10x45y120
2(5x4y) 2(14) 10x8y28
37y148
y4
5x4y14
5x4(4) 14
5x30
x6(6, 4)
10. Let brepresent the number of baseball racks and
krepresent the number of karate-belt racks.
b6kb6k
3b5k46,000 6(2000)
3(6k) 5k46,000 12,000
23k46,000
k2000
12,000 baseball, 2000 karate
Pages 71–72 Exercises
11. x3y18 →y
1
3
x6
x2y7y
1
2
x
7
2
consistent and independent
12. y0.5x→y0.5x
2yx4y
1
2
x2
inconsistent
13. 35x40y55 →y
7
8
x
1
8
1
7x8y11 y
7
8
x
1
8
1
consistent and dependent
Chapter 2 Systems of Linear Equations and Inequalities
Chapter 2 30
y
O
y
x
2
y
x
5
25. 2(3x10y5) →6x20y10
3(2x7y24) 6x21y72
41y82
2x7y24 y2
2x7(2) 24
2x10
x5(5, 2)
26. 2xy7x2y8
2(2y8) y7x2(3) 8
3y9x2
y3(2, 3)
27. 3(2x5y) 3(4) →6x15y12
2(3x6y) 2(5) 6x12y10
3y 2
2x5y4y
2
3
2x5
2
3
4
2x
2
3
x
1
3
1
3
,
2
3
28. 5
3
5
x
1
6
y5(1) →3x
5
6
y5
1
5
x
5
6
y1
1
5
x
5
6
y11
1
5
6
x16
x5
1
5
x
5
6
y11
1
5
(5)
5
6
y11
5
6
y10
y12 (5, 12)
29. 7(4x5y) 7(8) →28x35y56
5(3x7y) 5(10) 15x35y50
43x6
4x5y8x
4
6
3
4
4
6
3
5y8
5y
3
4
2
3
0
y
6
4
4
3
4
6
3
,
6
4
4
3
30. 2(3xy) 2(9) →6x2y18
4x2y84x2y8
2x10
3xy9x5
3(5) y9
y6
y6(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 ab, making
substitution very easy.
ab03a2b15
ab3(b) 2b15
5b15
b3
ab0
a(3) 0
a3(3, 3)
32a. B
31 Chapter 2
y
x
O
x
4
y
3(4, 3)
y
x
O
4
5
y
x
4
x
5
(5, 0)
y
x
O
y
3
x
10
y
x
2
(2, 4)
y
x
O
1
3
y
x
1
3
y
x
5
6
y
x
O
1
2
y
x
2
y
x
2
(0, 2)
y
x
O
1
4
y
x
3
3
2
y
x
3
(0, 3)
14. 15.
(5, 0) (4, 3)
16. 17.
(2, 4) no solution
18. 19.
(0, 2) (0, 3)
20. 3x8y10 →y
3
8
x
5
2
16x32y75 y
1
2
x
7
3
5
2
Consistent and independent; if each equation is
written in slope-intercept form, they have
different slopes, which means they will intersect
at some point.
21. 3(5xy) 3(16) →15x3y48
2x3y32x3y3
17x51
x3
5xy16
5(3) y16
y1
y1(3, 1)
22. 3x5y8→3x5y8
3(x2y) 3(1) 3x6y3
11y11
x2y1y1
x2(1) 1
x1(1, 1)
23. x4.5 yy6 x
x4.5 6 xy6 5.25
2x10.5 y0.75
x5.25 (5.25, 0.75)
24. 5(2x3y) 5(3) →10x15y15
12x15y412x15y4
22x11
2x3y3x
1
2
2
1
2
3y3
3y2
y
2
3
1
2
,
2
3
32b. S4V0SV30,000
S4V4VV30,000
5V30,000
V6000
S4V0
S4(6000) 0
S24,000 0
S24,000
Spartans: 24,000; visitors: 6000
33a. Let brepresent the base and represent the leg.
Perimeter of first triangle: b220
Perimeter of second triangle: 6 b20
b220
b20 2
6 b20 b220
6 20 220 b2(6) 20
6b8
6
6, 6, 8; 6, 6, 8
33b. isosceles
34. y(3) 4(x4)
y4x19
y(3)
1
4
(x4)
y
1
4
x2
35a. Let xrepresent the number of refills. Then,
x1 number of drinks purchased.
C2.95 0.50x
C0.85 0.85x
C2.95 0.50x
0.85 0.85x2.95 0.50x
0.35x2.1
x6
x1 7C2.95 0.50x
C2.95 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.
a
b
d
e
36b.
a
b
d
e
, cf
36c.
a
b
d
e
, cf
37. Let xrepresent the full incentive.
Let yrepresent the value of the computer.
x516 y
3
4
.5
x264 y
3
4
.5
x264 y
3
4
.5
(516 y) 264 y
451.5 0.875y264 y
y$1500
38. Let xrepresent the number of people in line
behind you. 200 xrepresents the number in
front of you. Let represent the whole line.
200 x1 x
3x
200 x1 x3x3x
201 x3(201)
603 people 603
39.
40.
41. y6 2(x0)
y2x6
42. $12,500
43. [fg](x) f(g(x))
f(x2)
3(x2) 5
3x1
44. {18}, {3, 3}; no, because there are two range
values paired with a single domain value.
45.
5
25
5
5
1
1
The correct choice is A.
Solving Systems of Equations in
Three Variables
Page 76 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.
2-2
Chapter 2 32
y
x
O
2
x
7
y
xf(x)
21
11
03
11
21
x
O
f
(
x
)
f
(
x
) 2|x| 3
2. The solution would be an equation in two
variables. Sample example: the system 2x4y
6z12, x2y3z6, and 3x5y6z
27 has a solution of all values of xand ythat
satisfy 5xy39.
2x4y6z12 2(x2y3z) 2(6)
3x5y6z27 2x4y6z12
5xy39 ↓
2x4y 6z12
2x4y6z12
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(4x2yz) 4(7) 2(4x2yz) 2(7)
2x2y4z4x3y2z8
↓↓
16x8y4z28 8x4y2z14
2x2y4z4x3y 2z8
18x10y24 9x7y6
18x10y24 →18x10y24
2(9x7y) 2(6) 18x14y12
4y12
y3
9x7y64x2yz7
9x7(3) 64(3) 2(3) z7
x3z1
(3, 3, 1)
5. 2(xyz) 2(7) x2y3z12
x2y3z12 3x2y7z30
↓2x4z18
2x2y2z14
x2y3z12
x5z2
2(x5z) 2(2) →2x10z4
2x4z18 2x14z18
14z14
z1
x5z2xyz7
x5(1) 27 y1 7
x7y1(7, 1, 1)
6. 2(2x2y3z)2(2x3y7z)
2(6) 2(1)
4x3y2z04x3y2z0
↓↓
4x4y6z12 4x6y14z2
4x3y2z04x3y2z0
y4z12 3y12z2
3(y4z) 3(12) →3y12z36
3y12z23y12z2
0 38
no solution
7. 75
1
2
a(1)2v0(1) s0
0.5av0s0
75
1
2
a(2.5)2v0(2.5)s0
3.125a2.5v0s0
3
1
2
a(4)2v0(4) s0
8a4v0s0
2.5(75) 2.5(0.5av0s0)
75 3.125a2.5v0s0
↓
187.5 1.25a2.5v02.5s0
75 3.125a2.5v0s0
112.5 1.875a1.5s0
4(75) 4(0.5av0s0)
3 8a4v0s0
↓
300 2a4v04s0
3 8a4v0s0
297 6a3s0
2(112.5) 2(1.875a1.5s0)
297 6a3s0
↓
225 3.75a3s0
297 6a 3s0
72 2.25a
32 a
297 6a3s03 8a4v0s0
297 6(32) 3s03 8(32) 4v0s0
35 s056 v0
acceleration 32 ft/s2, initial velocity: 56 ft/s,
initial height: 35 ft
Pages 76–77 Exercises
8. 3(5x3yz) 3(11)
x2y3z5
↓
15x9y3z33
x2y3z5
16x11y28
2(5x3yz) 2(11)
3x2y2z13
↓
10x6y2z22
3x2y2z13
7x4y9
4(16x11y)4(28) →64x44y112
11(7x4y) 11(9) 77x44y99
13x 13
x1
16x11y28 x2y3z5
16(1) 11y28 1 2(4) 3z5
y4z4
(1, 4, 4)
33 Chapter 2
9. 7(x3y2z) 7(16) x3y2z16
7x5yz0x6y2z18
↓3y2z 2
7x21y14z112
7x5yz0
26y15z112
15(3yz) 15(2) →45y15z30
26y15z112 26y15z112
71y142
y2
3yz2x6yz18
3(2) z2x6(2) 4 18
z4x2
(2, 2, 4)
10. 2(2xy2z) 2(11)
x2y9z13
↓
4x2y4z22
x 2y9z13
3x5z35
3(x3z) 3(7) →3x9z21
3x5z35 3x5z35
14z14
z1
x3z72xy2z11
x3(1) 72(10) y2(1) 11
x10 y7
(10, 7, 1)
11. 3(x3y2z) 3(8) 5(x3y2z)
5(8)
3x5yz95x6y3z15
↓↓
3x9y6z24 5x15y10z40
3x5yz95x6y3z15
4y7z33 9y13z55
9(4y7z) 9(33) →36y63z297
4(9y13z) 4(55) 36y52z220
11z77
z7
4y7z33 x3y2z8
4y7(7) 33 x3(4) 2(7) 8
y4x6
(6, 4, 7)
12. 8xyz4
yz5
8xy9
1(8xy) 1(9) →8xy9
11xy15 11xy15
3x6
x2
8xz4yz5
8(2) z4y12 5
z12 y7
(2, 7, 12)
13. 3(xyz) 3(3) 2(xyz) 2(3)
4x3y2z12 2x2y2z5
↓↓
3x3y3z92x2y2z6
4x3y2z12 2x2y2z5
7xz21 0 11
no solution
14. 3(36x15y50z) 3(10)
5(54x5y30z) 5(160)
↓
108x45y150z30
270x25y150z800
162x20y770
81(2x25y) 81(40) →162x2025y3240
162x20y770 162x20y770
2005y4010
y2
2x25y40 36x15y50z10
2x25(2) 40 36(5) 15(2) 50z10
x5z4
(5, 2, 4)
15. 4(x3yz) 4(54)
4x2y3z32
↓
4x12y4z216
4x2y3z32
10y3z184
5(2y8z) 5(78) →10y40z390
10yz184 10yz184
41z574
z14
2y8z78 x3yz54
2y8(14) 78 x3(17) 14 54
y17 x11
(11, 17, 14)
16. 1.8x1.2y z0.7
1.2yz0.7
1.8x1.2y0
3(1.8x1.2y) 3(0) →5.4x3.6y0
1.2(1.5x3y) 1.2(3) 1.8x3.6y3.6
7.2x3.6
x0.5
1.5x3y31.8xz0.7
1.5(0.5) 3y31.8(0.5) z0.7
y0.75 z0.2
(0.5, 0.75, 0.2)
17. yx2zz1 2x
y(y14) 2z7 1 2x
7 z4 x
xy14
4 y14
10 y(4, 10, 7)
Chapter 2 34
18.
5
2
3
4
x
1
6
y
1
3
z
5
2
(12)
1
8
x
2
3
y
5
6
z8
↓
1
8
5
x
1
5
2
y
5
6
z30
1
8
x
2
3
y
5
6
z8
2x
1
4
y38
7
4
3
4
x
1
6
y
1
3
z
7
4
(12)
1
3
6
x
5
8
y
1
7
2
z25
↓
2
1
1
6
x
2
7
4
y
1
7
2
z21
1
3
6
x
5
8
y
1
7
2
z25
9
8
x
1
1
2
1
y4
1
9
6
2x
1
4
y
1
9
6
(38) →
9
8
x
6
9
4
y
17
8
1
9
8
x
1
1
2
1
y4
9
8
x
1
1
2
1
y4
2
1
0
9
3
2
y
20
8
3
y24
2x
1
4
y38
3
4
x
1
6
y
1
3
z12
2x
1
4
(24) 38
3
4
(16)
1
6
(24)
1
3
z12
x16 z12
(16, 24, 12)
19. Let xrepresent amount in International Fund, y
represent amount in Fixed Assets Fund, and z
represent amount in company stock.
xyz2000
x2z
0.045x0.026y0.002z58
xyz2000
2zyz2000
y3z2000
y2000 3z
0.045x0.026y0.002z58
0.045(2z) 0.026y0.002z58
0.026y0.088z58
0.026y0.088z58
0.026(2000 3z) 0.088z58
z600
x2zxyz2000
x2(600) 1200 y600 2000
x1200 y200
International Fund $1200; Fixed Assets Fund
$200; company stock $600
20a. Sample answer: xyz15;
2xz1; 2yz7
20b. Sample answer: 4xyz12;
4xyz10; 5yz9
20c. Sample answer: xyz6;
2xy2z8; x2y3z2
21. 124
1
2
a(1)2v0(1) s0
272
1
2
a(3)2v0(3) s0
82
1
2
a(8)2v0(8) s0
↓
124
1
2
av0s0
272
9
2
a3v0s0
82 32a8v0s0
1(124) 1
1
2
av0s0
272
9
2
a3v0s0
↓
124
1
2
av0s0
272
9
2
a3v0s0
148 4a2v0
1(124) 1
1
2
av0s0
82 32a8v0s0
↓
124
1
2
av0s0
82 32a8v0s0
42 31.5a7v0
7(148) 7(4a2v0)
2(42) 2(31.5a7v0)
↓
1036 28a14v0
84 63a14v0
1120 35a
32 a
148 4a2v0124
1
2
av0s0
148 4(32) 2v0124
1
2
(32) 138 s0
138 v02 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 xy1 and
2x2y2.
35 Chapter 2
23. xyz 2
yxz 2
zxy 2
xyz 2→xyz 2
yxz 2yxz 2
(xy) (yz xz) 0
(xy) z(xy) 0
(1 z)(xy) 0
1 z0orxy0
z1xy
yxz 2→yxz 2
zxy 2zxy 2
(yz) (xz xy) 0
(yz) x(yz) 0
(1 x)(yz) 0
1 x0oryz0
x1yz
If z1 and x1, 1 1y2 and y1.
If z1 and yz, x1 1 2 and x1.
If xyand x1, 1 1z2 and z1.
If xyand yz, xyz.
xxx2
xx22
x2x2 0
(x2)(x1) 0
x2 0orx1 0
x22x1
If x2, y2 and z2.
The answers are (1, 1, 1) and (2, 2, 2).
24. 3x4y375 →3x4y375
2(5x2y) 2(345) 10x4y690
7x315
x45
5x2y345
5(45) 2y345
y60 (45, 60)
25.
26.
AB: d(1
3)2
(2
(1))2
5
BC: d(2 (
1))2
(6
2)2
5
CD: d(6 2
)2(3
6)2
5
AD: d(6 3
)2(3
(
1))2
5
AB: m
2
1
(
3
1)
B
C
: m
2
6
(
2
1)
4
3
3
4
AB BC CD AD 5 units; ABCD is
rhombus. Slope of A
B
4
3
and slope of B
C
3
4
,
so A
B
⊥B
C
. A rhombus with a right angle is a
square.
27a. (20, 3000), (60, 5000)
m
500
6
0
0
3
2
0
0
00
20
4
0
0
0
or 50
y5000 50(x60)
y50x2000
C(x) 50x2000
27b. $2000; $50
27c.
28. As2Ar2
2 s2(2
)2
2
s2
The correct choice is C.
Modeling Real-World Data with
Matrices
Pages 82–83 Check for Understanding
1. Sample answer:
film pain reliever blow dryer
(24 exp.) (100 ct)
Atlanta $4.03 $6.78 $18.98
Los Angeles $4.21 $7.41 $20.49
Mexico City $3.97 $7.43 $32.25
Tokyo $7.08 $36.57 $63.71
2. 2 4
3. The sum of two matrices exists if the matrices
have the same dimensions.
2-3
Chapter 2 36
y
x
O
1
3
y
x
2
x
O
c
(
x
)
c
(
x
) 50
x
2000
Cost
($1000)
Televisions Produced
1
2
3
4
123456789
y
x
D
(3, 6)
C
(6, 2)
B
(2, 1)
(1, 3)
A
O
4. Anthony is correct. A third order matrix has 3
rows and 3 columns. This matrix has 4 rows and 3
columns.
5. 2yx32yy5 3
xy5y2
xy5
x2 5
x7(7, 2)
6. 18 4xy24 12y
24 12y2 y
18 4xy
18 4x2
5 x(5, 2)
7. 16 4x16 4x
0 y4 x
2x8 y(4, 0)
8. XZ
1 3
6 (2)
4 (1)
2 0
18. x2yy2(2y) 6
y2x6y2
x2y
x2(2)
x4(4, 2)
19. 27 3y27 3y
8 5x3y9 y
8 5x3y
8 5x3(9)
7 x(7, 9)
20. 4x3y11
xy1
4x3y11 →4x3y11
3(xy) 13x3y3
7x14
x2
xy1
2 y1
y1(2, 1)
21. 2x10 2x10 y3x
y3xx5y3(5)
y15 y15
(5, 15)
22. 12 6x12 6x2 y1
2 y12 x1 y
12y10 x(2, 1)
23. xy0
yy2
3 2yx
6 4 2x
6 4 2xxy0
1 x1 y0
y1(1, 1)
24. x21 2
xy5
5 yx
y4 2
y4 2xy5
y6x6 5
x1(1, 6)
25. 3
6
3xy
15
6z
y1
3z
x
4
37 Chapter 2
9. impossible
10. ZX
3 1
2 6
1 4
0 (2)
4
4
3
2
2
8
5
2
11. 4X
4(1)
4(6)
4(4)
4(2)
12. impossible
13. YX [0 3]
[0(4) (3)(2) 0(1) (3)(6)]
or [6 18]
14.
Budget Viewers
($ million) (million)
soft-drink 40.1 78.6
package delivery 22.9 21.9
telecommunications 154.9 88.9
Pages 83–86 Exercises
15. y2x1y2(y5) 1
xy5y11
xy5
x11 5
x6 (6, 11)
16. 9 x2y13 4x1
13 4x13 x
9 x2y
9 3 2y
3 y(3, 3)
17. 4x15 x4x15 x
5 2yx5
5 2y
2.5 y(5, 2.5)
1
6
4
2
4
24
16
8
3x15
12 6z
3y3 6
9z3xy
3x15 12 6z3y3 6
x52 zy3
(5, 3, 2)
6
3xy
15
6z
3y3
9z
3x
12
26. 2
4
2x8z
16
6
xz
8
w5
3y37. 5A
5(7)
5(1)
5(5)
5(6)
Chapter 2 38
2w10 16 2w10 16 6y6
6y6w3y1
2x2z4
16 2x8z
2x2z42x2z4
2x8z16 2x2(2) 4
10z20 x0
z2(3, 0, 1, 2)
27. AB
7 5
1 8
5 3
6 (1)
4
2x8z
16
6
2x2z
16
2w10
6y
28. impossible
29. impossible
30. DC
2 3
0 (1)
2 1
1 (2)
3 0
4 0
0 4
2 5
4 9
12
9
8
7
31. BAB(A)
57
61
35
18
5
1
1
1
3
4
4
3
13
or
32. CDC(D)
2
0
2
1
3
4
0
2
4
3
1
1
2
0
0
4
5
9
22
57
3 (5) 5 (7)
1 68 (1)
3 (2)
1 0
1 2
2 (1)
0 (3)
0 (4)
4 0
5 2
9 (4)
or
1
1
3
3
3
4
4
7
5
33.
4D
4(2)
4(0)
4(2)
4(1)
4(3)
4(4)
4(0)
4(2)
4(4)
8
0
8
4
12
16
0
8
16
34. 2F
2(0)
2(0)
2(1)
2(4)
2(6)
2(1)
35. FEF(E)
2
5
4
1
8
3
0
0
1
4
6
1
0
0
2
8
12
2
36. EFE(F)
0
0
1
4
6
1
2
5
4
1
8
3
2
5
3
3
14
2
2
5
3
3
14
2
35
5
25
30
or
39. impossible
40. FC
3
1
1
2
0
0
4
5
9
0
0
1
4
6
1
26
1
15
53
38. BA
7
1
5
6
5
8
3
1
8(0) (4)(2) 2(4)
3(0) 1(2) (5)(4)
6(2) (1)(0) 0(0)
1(2) 4(0) 0(0)
8(1) (4)(3) 2(4)
3(1) 1(3) (5)(4)
8(2) (4)(0) 2(2)
3(2) 1(0) (5)(2)
12
16
4
14
16
22
42. AA
7
1
5
6
7
1
5
6
5(7) 7(1)
6(7) 1(1)
5(5) 7(6)
6(5) 1(6)
3(7) 5(1)
1(7) 8(1)
3(5) 5(6)
1(5) 8(6)
6(3) (1)(1) 0(1)
1(3) 4(1) 0(1)
6(4) (1)(5) 0(9)
1(4) 4(5) 0(9)
17
1
12
2
29
24
41. ED
2
0
2
1
3
4
0
2
4
2
5
4
1
8
3
or
42
41
17
36
43. FD
2
0
2
1
3
4
0
2
4
0
0
1
4
6
1
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)
12
2
9
13
2
8
EFD
2 (12)
5 2
4 (9)
1 13
8 2
3 (8)
10
3
13
14
10
5
5
8
3
1
3(7)
3(1)
(3)(5)
(3)(6)
6(6) 12(2)
24(6) (12)(2)
6(6) 18(2)
39 Chapter 2
44. 3AB 3
5
8
3
1
7
1
5
6
5
8
3
1
21
3
15
18
15(3) (21)(1)
18(3) (3)(1)
15(5) (21)(8)
18(5) (3)(8)
243
66
24
57
45. (BA)E
2
5
4
1
8
3
7
1
5
6
5
8
3
1
3(7) 5(1)
1(7) 8(1)
3(5) 5(6)
1(5) 8(6)
2
5
4
1
8
3
2
5
4
1
8
3
26
1
15
53
15(4) 26(1)
53(4) 1(1)
15(8) 26(3)
53(8) 1(3)
15(2) 26(5)
53(2) 1(5)
46. F2EC F(2EC)
2EC 2
3
1
1
2
0
0
4
5
9
2
5
4
1
8
3
160
111
86
213
42
421
3
1
1
2
0
0
4
5
9
4
10
8
2
16
6
2(2)
2(5)
2(4)
2(1)
2(8)
2(3)
3
1
1
2
0
0
4
5
9
16(4) 8(5) (4)(9)
6(4) (2)(5) 10(9)
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
6
32
12
60
56
F(2EC)
0 (60)
0 (6)
1 32
4 12
6 (60)
1 56
60
6
31
16
66
57
47. 3XY 3
6
2
3
4
3
5
4
4
6
2
8
2
6
2
3
4
3
5
3(4)
3(4)
3(6)
3(2)
3(8)
3(2)
6
2
3
4
3
5
12
12
18
6
24
6
6(3) 12(4)
24(3) (12)(4)
6(3) 18(4)
6(3) 12(5)
24(3) (12)(5)
6(3) 18(5)
12
168
72
30
120
90
78
12
72
48. 2K3J2
(3)
5
1
4
1
7
2
1
3
3(5)
3(1)
(3)(4)
(3)(1)
2(7)
2(2)
2(1)
2(3)
15
3
12
3
14
4
2
6
14 (15)
4 3
2 12
6 (3)
49. Sample answer:
1996 2000 2006
18 to 24 8485 8526 8695
25 to 34 10,102 9316 9078
35 to 44 8766 9036 8433
45 to 54 6045 6921 7900
55 to 64 2444 2741 3521
65 and older 2381 2440 2572
50a.
24
21
7
6
31
28
9
4
26
22
12
17
19
17
6
4
24
24
6
2
18
16
6
12
29
7
14
3
24
21
7
6
31
28
9
4
26
22
12
17
19
17
6
4
24
24
6
2
18
26
6
12
19 (24)
17 (21)
6 (7)
4 (6)
24 (31)
24 (28)
6 (9)
2 (4)
18 (26)
16 (22)
6(12)
12 (17)
or
TV Radio Recording
Classical 875
Jazz 644
Opera 631
Musicals 522
50b. classical performances on TV
5
4
1
2
7
4
3
2
8
6
6
5
51a.
3
5
2
4
b
d
a
c
3
5
2
455. 2x6y8z5
2x9y12z5
15y20z10
2(2x9y12z) 2(5) →4x18y24z10
4x6y4z34x6y4z3
24y20z13
15y20z10 →15y20z10
1(24y20z) 1(13) 24y20z13
9y3
y
1
3
15y20z10 2x6y8z5
15
1
3
20z10 2x6
1
3
8
1
4
5
z
1
4
x
1
2
1
2
,
1
3
,
1
4
56. 4x2y7→y2x
7
2
12x6y21 y2x
7
2
consistent and dependent
57.
58.
59. Sample answer: using (60, 83) and (10, 65),
m
6
1
5
0
8
6
3
0
1
5
8
0
or 0.36
y65 0.36(x10)
y0.36x61.4
60. m
7
5
4
1
y7
3
4
(x5)
3
4
y
3
4
x3
1
4
61. f(x) 5x3
0 5x3
3
5
x
62. [fg](x) f(x) g(x)
2
5
x
(40x10)
16x24x
63. f(x) 4 6xx3
f(14) 4 6(14) (14)3
2656
Chapter 2 40
xf(x)
28
15
02
15
28
2a3c22b3d3
4a5c44b5d5
2(2a3c) 2(2) →4a6c4
4a5c44a5c4
c0
4a5c4
4a5(0) 4
a1
2(2b3d) 2(3) →4b6d6
4b5d54b5d5
d1
4b5d5
4b5(1) 5
b0
51b. a matrix equal to the original matrix
52a. [42 592118]
52b. 33.81 30.94 27.25
[42 592118]
15.06 13.25 8.75
54 54 46.44
52.06 44.69 34.38
42
33.81
59
15.06
21(54) 18
52.06
42
30.94
59
13.25
21(54) 18
44.69
42
27.25
59
8.75
21
46.44
18
34.38
[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
nth triangular number is
n(n
2
1)
. Solve
n(n
2
1)
2001. The solution is 63. So the 63rd entry in the
first row is
63(63
2
1)
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.
ABCD
A0100
B1011
C0102
D0120
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.
3
5
2
4
2(b) 3(d)
4(b) (5)(d)
2(a) 3(c)
4(a) (5)(c)
y
x
O
6 3
x
y
12
x
O
f
(
x
)
f
(
x
) |3
x
| 2
64.
2x
x
3
3
2
x
2(2x3) x(3 x)
4x6 3xx2
x2x6 0
(x3)(x2) 0
x3 0orx2 0
x3x2
The correct choice is A.
Page 86 Graphing Calculator Exploration
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
and B
.
b12
b22
b11
b21
a12
a22
a11
a21
2d. Let A
, B
,
b12
b22
b11
b21
a12
a22
a11
a21
41 Chapter 2
AB
b12
b22
b11
b21
a12
a22
a11
a21
b12
b22
a12
a22
b11
b21
a11
a21
a12
a22
b12
b22
a11
a21
b11
b21
a12
a22
a11
a21
b21
b12
b11
b21
2b. Let A
, B
,
b12
b22
b11
b21
a12
a22
a11
a21
and C
c12
c22
c11
c21
(AB) C
c12
c22
c11
c21
b12
b22
a12
a22
b11
b21
a11
a21
(a12 b12)c12
(a22 b22) c22
(a11 b11) c11
(a21 b21)c21
a12 (b12 c12)
a22 (b22 c22)
a11 (b11 c11)
a21 (b21 c21)
b12 c12
b22 c22
b11 c11
b21 c21
a12
a22
a11
a21
A(BC)
2c. Let A
and B
.
b12
b22
b11
b21
a12
a23
a11
a21
AB
a11b12 a12b22
a21b12 a22b23
a11b11 a12b21
a21b11 a22b21
BA
Thus, AB BA, since a12b21 b12a21.
b11a12 b12a22
b21a12 b22a22
b11a11 b12a21
b21a11 b22a21
and C
.
c12
c22
c11
c21
(AB)C
c12
c22
c11
c21
a11b12 a12b22
a21b12 a22b22
a11b11 a12b21
a21b11 a22b21
a11b11c11 a11b12c21 a12b21c11 a21b22c21
a21b11c11 a21b12c21 a22b21c11 a22b22c21
a11b11c12 a11b12c22 a12b21c12 a12b22c22
a21b11c12 a21b12c22 a22b21c12 a22b22c22
A(BC)
b11c12 b12c22
b21c12 b22c22
b11c11 b12c21
b21c11 b22c21
a12
a22
a11
a21
a11b11c11 a11b12c21 a12b21c11 a21b22c21
a21b11c11 a21b12c21 a22b21c11 a22b22c21
Therefore (AB)CA(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.
Transformation Matrices
Page 87
1. The new figure is a 90° counterclockwise rotation
of LMN.
2. The new figure is an 180° counterclockwise
rotation of LMN.
3. The new figure is an 270° counterclockwise
rotation of LMN.
2-4A
a11b11c12 a11b12c22 a12b21c12 a12b22c22
a21b11c12 a21b12c22 a22b21c12 a22b22c22
7.
0
1
3
2
4
1
1
2
0
1
1
0
L(6, 4), M(3, 2), N(1, 2)
10a. 10b.
x3
y4
1
2
3
2
6
4
1
2
3
2
6
4
0
1
1
0
4. See students’ work for graphs. Multiplying a
vertex matrix by
results in a vertex
matrix for a figure that is a 90° counterclockwise
rotation of the original figure. Multiplying a
vertex matrix by
results in a vertex
matrix for a figure that is a 180° counterclockwise
rotation of the original figure. Multiplying a
vertex matrix by
results in a vertex
matrix for a figure that is a 270° counterclockwise
rotation of the original figure.
Modeling Motion with Matrices
Pages 92–93 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 xis 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)andareflection over the line
yx,which interchanges the values for xand y.
4a. 64b. 24c. 34d. 4
5. 1.5
0
3
1.5
4.5
3
7.5
0
2
1
3
2
5
2-4
1
0
0
1
0
1
1
0
1
0
0
1
A(1, 2), B(4, 1), C(3, 2), D(0, 1)
8.
2
1
4
1
2
3
1
2
1
4
3
2
1
0
0
1
0
1
3
2
4
1
1
2
Chapter 2 42
J(3, 7.5), K(1.5, 4.5), L(0, 3)
P(2, 3), Q(4, 1), R(2, 1)
9.
0
1
1
0
0
1
1
0
0
1
1
0
y
x
Q
P
RP
R
Q
O
Landing
Ball
4N
3E
y
x
K
K
L
L
J
J
y
x
A
A
D
B
C
D
B
C
O
y
x
A
A
D
B
C
D
B
C
O
y
x
N
M
N
M
L
L
O
6.
A(2, 1), B(2, 1), C(2, 3), D(2, 3)
2
3
2
3
2
1
2
1
1
2
1
2
1
2
1
2
1
1
3
1
3
3
1
3
Pages 93–96 Exercises
11. 3
A(3, 3), B(3, 12), C(15, 3)
12.
3
4
0
4
15
4
9
6
2
4
7
3
2
X(0, 6), Y
3
3
4
, 6
3
4
, Z
2
1
4
, 1
1
2
13. 2
P(6, 0), Q(4, 4), R(2, 6), S(8,4)
14a. 2
0
2
2
0
0
2
2
0
0
1
1
0
0
1
1
0
8
4
2
6
4
4
6
0
4
2
1
3
2
2
3
0
3
2
5
9
0
8
15
3
3
12
3
3
5
1
1
4
1
1
14b. 3
2
A(3, 0), B(0, 3), O(3, 0), D(0, 3); A(6, 0),
B(0, 6), C(6, 0), D(0, 6)
14c. The final results are the same image.
15.
W(1, 2), X(4, 3), Y(6, 3)
6
3
4
3
1
2
3
2
3
2
3
2
3
1
1
5
2
0
0
6
6
0
0
6
6
0
0
3
3
0
0
3
3
0
0
3
3
0
0
3
3
0
0
1
1
0
0
1
1
0
43 Chapter 2
y
x
O
S
S
R
PP
Q
Q
R
3
A(2, 0), B(0, 2), C(2, 0), D(0, 2); A(6, 0),
B(0, 6), C(6, 0), D(0, 6)
0
6
6
0
0
6
6
0
0
2
2
0
0
2
2
0
y
x
B
C
A
A
A
D
D
DC
C
B
B
y
x
O
W
W
Y
Y
X
X
16.
2
1
2
1
2
1
2
1
3
2
4
7
1
5
0
0
y
x
Q
Q
R
R
O
P
P
O
y
x
B
C
A
A
D
D
DC
C
B
B
A
y
x
A
A
B
B
C
C
2
2468101214
4
6
8
10
12
y
x
O
Z
Z
Y
Y
X
X
y
x
C
D
E
E
F
F
D
C
O
Q(2, 1), P(1, 4), Q(2, 6), R(1, 1)
17.
C(0, 5), D(4, 9), E(8, 5), F(4, 1)
4
1
8
5
4
9
0
5
3
4
3
4
3
4
3
4
1
3
5
1
1
5
3
1
1
1
2
6
1
4
2
1
a3b1
c3d3
2ab2
2cd1
a3b1
c3d3
Thus, a1, b0, c0, and d1. By
substitution, Rx-axis
.
0
1
1
0
Chapter 2 44
18a.
F(2, 1), G(6, 1), H(4, 3)
18a-b.
18b.
F(1, 4), G(5, 6), H(3, 2)
18c. translation of 5 units left and 3 units up
19.
A(1, 2), B(0, 4), C(2, 3)
20.
D(2, 4), E(6, 2), F(3, 4), G(1, 2)
21.
H(2, 1), I(1, 3), J(5, 1), K(4, 2)
4
2
5
1
1
3
2
1
2
4
1
5
3
1
1
2
1
0
0
1
1
2
3
4
6
2
2
4
1
2
3
4
6
2
2
4
0
1
1
0
2
3
0
4
1
2
2
3
0
4
1
2
0
1
1
0
3
2
5
6
1
4
1
5
1
5
1
5
4
3
6
1
2
1
4
3
6
1
2
1
6
2
6
2
6
2
2
1
0
3
4
1
1
3
2
1
1
3
1
3
2
1
1
3
b
d
a
c
1
3
2
1
1
3
a3b
c3d
2ab
2cd
a3b
c3d
22.
L(1, 1), M(2, 2), N(1, 3)
23.
O(0, 0), P(4, 0), Q(4, 4), R(0, 4)
24.
S(2, 1), T(1, 3), U(2, 5), V(4, 4),
W(4, 2)
25a. Let
Rx-axis
b
d
a
c
4
2
4
4
2
5
1
3
2
1
2
4
4
4
5
2
3
1
1
2
1
0
0
1
0
4
4
4
4
0
0
0
0
4
4
4
4
0
0
0
0
1
1
0
1
3
2
2
1
1
3
1
2
2
1
1
1
0
0
1
y
x
G
F
H
G
F
H
G
F
H
O
y
x
A
B
C
A
B
C
O
y
x
DD
EE
FF
GG
O
JK
H
H
J
I
K
I
O
y
x
N
L
M
M
NL
O
y
x
RQ
P
P
y
x
P
Q
R
O
O
T
US
S
T
U
V
W
VW
O
y
x
25b. Let
Ry-axis.
b
d
a
c
45 Chapter 2
1
3
2
1
1
3
1
3
2
1
1
3
b
d
a
c
1
3
2
1
1
3
a3b
c3d
2ab
2cd
a3b
c3d
Thus, a1, b0, c0, and d1. By
substitution, Ry-axis
.
25c. Let
Ryx.
b
d
a
c
0
1
1
0
a3b1
c3d3
2ab2
2cd1
a3b1
c3d3
3
1
1
2
3
1
1
3
2
1
1
3
b
d
a
c
3
1
1
2
3
1
a3b
c3d
2ab
2cd
a3b
c3d
Thus, a0, b1, c1, and d0. By
substitution, Ryx
.
25d. Let
Rot90
b
d
a
c
1
0
0
1
a3b3
c3d1
2ab1
2cd2
a3b3
c3d1
3
1
1
2
3
1
1
3
2
1
1
3
b
d
a
c
3
1
1
2
3
1
a3b
c3d
2ab
2cd
a3b
c3d
Thus, a0, b1, c1, and d0. By
substitution, Rot90
25e. Let
Rot180.
b
d
a
c
1
0
0
1
a3b3
c3d1
2ab1
2cd2
a3b3
c3d1
1
3
2
1
1
3
1
3
2
1
1
3
b
d
a
c
1
3
2
1
1
3
a3b
c3d
2ab
2cd
a3b
c3d
Thus, a1, b0, c0, and d1. By
substitution, Rot180
.
25f. Let
Rot270.
b
d
a
c
0
1
1
0
a3b1
c3d3
2ab2
2cd1
a3b1
c3d3
3
1
1
2
3
1
1
3
2
1
1
3
b
d
a
c
3
1
1
2
3
1
a3b
c3d
2ab
2cd
a3b
c3d
Thus, a0, b1, c1, and d0. By
substitution, Rot270
.
1
0
0
1
a3b3
c3d1
2ab1
2cd2
a3b3
c3d1
26.
1
2
3
2
6
4
1
2
3
2
6
4
0
1
1
0
J(4, 1), K(1, 3), L(1, 7)
27.
1
2
3
2
6
4
1
2
3
2
6
4
0
1
1
0
1
7
1
3
4
1
2
5
2
5
2
5
1
2
3
2
6
4
J(6, 4), K(3, 2), L(1, 2)
1
2
3
2
6
4
1
2
3
2
6
4
0
1
1
0
28.
1
2
3
2
6
4
1
2
3
2
6
4
0
1
1
0
J(4, 6), K(2, 3), L(2, 1)
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
c
, c
, c
, and
1
1
1
1
1
1
1
1
1
1
1
1
2
1
2
3
4
6
1
2
3
2
6
4
1
0
0
1
L
J
K
L
x
y
O
K
J
L
K
J
J
J
K
x
y
K
L
L
L
K
J
O
J
J
J
L
K
L
L
O
y
x
K
K
c
, where cis the number of squares
moved.
29b. The knight moves in combinations of 2 vertical-1
horizontal or 1 vertical-2 horizontal squares.
1
1
1
1
37. Let xrepresent hardback books and yrepresent
paperback books.
4x7y5.75
3x5y4.25
3(4x7y) 3(5.75) →12x21y17.25
4(3x5y) 4(4.25) 12x20y17
y0.25
4x7y5.75
4x7(0.25) 5.75
x1
hardbacks $1, paperbacks $0.25
38. xy3
yx3
(3, 2)
39. y1 4(x2))
y1 4x8
4xy9 0
40. y6 2(x1)
y2x4
41. (fg)(x) f(x) g(x)
x3(x23x7)
x53x47x3
g
f
(x)
g
f(
(
x
x
)
)
x2
x
3
3
x7
42. 2xy12 →2x y12
2(x2y) 2(6) 2x4y12
3y24
y8
2xy12
2x(8) 12
x10
2x2y2(10) 2(8)
4
The correct choice is B.
Page 96 Mid-Chapter Quiz
1.
1
2
x5y17 →y
1
1
0
x3
2
5
3x2y18 y
3
2
x9
Chapter 2 46
These can be either up or down, left or right.
Sample matrices are
,
1
2
1
2
,
,
,
,
,
, and
.
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
1
2
1
2
1
2
1
2
1
2
1
2
29c. The king can move 1 unit in any direction. The
matrices describing this are
,
1
0
1
0
,
,
,
,
,
1
1
1
1
1
1
1
1
0
1
0
1
0
1
0
1
1
0
1
0
, and
.
30. Consider
.
Dilation with scale factor 1
1
Rotation of 180°
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.
34.
34a. Sample answer: the figure would be enlarged
disproportionally.
34b.
34c. See students’ work; the figure appears as if
blown out of proportion.
6
6
12
0
3
4
3
2
2
3
4
0
1
2
1
1
0
2
3
0
2
3
4
0
1
2
1
1
b
d
a
c
b
d
a
c
0
1
1
0
b
d
a
c
b
d
a
c
b
d
a
c
1
1
1
1
1
1
1
1
35.
36. xyz1.8 x2y4.6
yz5.6 x2y3.8
x2y3.8 2x0.8
x0.4
x2y3.8 yz5.6
13
12
4
4
8 5
4 8
3 1
2 (2)
5
8
1
2
8
4
3
2
B
C
D
A
AD
C
2
4
6
24681012
4
6
B
y
x
O
y
x
(3, 2)
(0, 0)
(4, 2)
(2, 4)
x
y
3
x
O
y
x
5
y
17
1
2
3
x
2
y
18
(4, 3)
2. 4xy86x2y9
y8 4x6x2(8 4x) 9
x
1
2
4xy8
4
1
2
y8
y6
1
2
, 6
3. Let xrepresent trucks and yrepresent cars.
x4y
6x5y29,000 x4y
6(4y) 5y29,000 x4(1000)
y1000 4000
4000 trucks, 1000 cars
4. 2xy4z13
3xy2z1
5x2z12
2(3xy2z) 2(1) →6x2y4z2
4x2yz19 4x2yz19
10x3z17
2(5x2z) 2(12) →10x4z24
10x3z17 10x3z17
7z7
z1
5x2z12 2xy4z13
5x2(1) 12 2(2) y4(1) 13
x2y5
(2, 5, 1)
5. xy1xy1
2xy2
1
3
y1
3x1y1
1
3
x
1
3
4xyz8
4
1
3
1
1
3
z8
z8
1
3
, 1
1
3
, 8
6. y3 xy3 x
y2x2x3 x
x3
y2x
y2(3)
6(3, 6)
7. AB
7 6
4 10
5 8
0 (9)
3 (2)
1 5
10. The result is the original figure. The original
figure is represented by
. The
reflection over the x-axis is found by
. The
reflection of the image over the x-axis is found by
. The matrix
for the final image is the same as that of the
original figure.
Determinants and Multiplicative
Inverses of Matrices
Page 102 Check for Understanding
1. Sample answer: a matrix with a nonzero
determinant
2. Sample answer:
is not a square
0
5
2
3
3
4
2-5
c
f
b
e
a
d
c
f
b
e
a
d
0
1
1
0
c
f
b
e
a
d
c
f
b
e
a
d
0
1
1
0
c
f
b
e
a
d
47 Chapter 2
8. impossible
9. B3AB(3A)
3A3
7
4
5
0
3
1
1
14
13
9
1
4
3(7)
3(4)
3(5)
3(0)
3(3)
3(1)
or
21
12
15
0
9
3
B(3A)
21
12
15
0
9
3
6
10
8
9
2
5
6 21
10 (12)
8 (15)
9 0
2 (9)
5 3
27
2
7
9
11
8
matrix.
also has no determinant.
5
1
9
2
1
0
3. Sample answer:
4. Sample answer: The system has a solution if
ad bc 0, since you can use the inverse of the
matrix
to find the solution.
c
d
a
b
0
0
0
1
0
0
1
0
0
1
0
0
1
0
0
0
5.
4(3) (2)(1) or 10
1
3
4
2
6.
12(32) (15)(26) or 6
26
32
12
15
7.
0
1
7
1
15
10
4
5
2
4
1
0
4(95) 1(33) 0(20)
413
8.
6
4
(1)
6(0) 4(27) (1)(27)
135
9.
2(7) 5(3) or 29
3
7
2
5
3
0
0
9
3
0
0
9
3
0
3
0
1
3
0
4
3
0
6
0
9
15
10
5
2
1
7
5
2
1
7
15
10
2
1
9
3
2
7
5
10.
4(9) 6(6) or 0
does not exist
6
9
4
6
11.
3
24
x
y
4
5
5
320.
4
(1)
(2)
4(5) 1(2) 2(4)
26
21.
3
2
0
1
0
3
2
3
1
2
1
0
2
1
3
0
2
1
3
2
1
2
1
3
1
2
1
4
0
2
Chapter 2 48
1
1
3
4
5
5
3
4
5
5
3
1
54
35
1
1
3
x
y
4
5
5
3
4
5
5
3
1
1
3
3
24
4
5
5
3
1
1
1
3
1
1
1
2
3
9
x
y
1
1
1
3
1
,
1
1
2
3
9
12.
63
85
x
y
3
9
6
5
3
1
9
3
6
9
5
3
6
9
5
1
63
59
3
1
9
3
1
9
63
85
3
6
9
5
x
y
3
9
6
5
3
6
9
5
(8, 5)
13. Let xrepresent the amount of metal with 55%
aluminum content, and let yrepresent the amount
of metal with 80% aluminum content.
xy20
0.55x0.8y0.7(xy)
0.55x0.8y0.7x0.7y
0.15x0.1y0
15x10y0
→
2
1
5
2
1
5
x
y
1
10
1
15
1
1
10
15
1
1
10
15
1
1
10
15
1
11
15 10
20
0
x
y
1
10
1
15
xy20
15x10y0
8
5
x
y
2
1
5
20
0
1
1
10
15
8 kg of the metal with 55% aluminum and 12 kg of
the metal with 80% aluminum
Pages 102–105 Exercises
14.
3(5) 2(4) or 7
4
5
3
2
8
12
x
y
15.
4(1) 0(1) or 4
1
1
4
0
16.
9(16) 12(12) or 0
12
16
9
12
17.
2(1) (2)(3) or 4
3
1
2
2
18.
13(8) (5)(7) or 69
7
8
13
5
19.
6(8) 0(5) or 48
5
8
6
0
2
(1)
3
2(6) 1(2) 3(9)
37
22.
8
9
3
8(6) 9(19) 3(11)
5
2
3
1
7
4
3
1
7
4
5
2
3
7
4
9
5
2
8
3
1
0
3
3
1
2
0
3
1
2
0
0
3
90
23.
4
6
7
4(2) 6(7) 7(5)
2
1
3
1
4
1
3
1
4
1
2
1
7
4
1
6
2
1
4
3
1
1
24.
15
2
9
36
12
15
25
31
17
25
36
15
12
15
31
17
2
9
31
17
2
9
12
15
25(78) 36(313) 15(669)
3183
25.
2.3
2.2
6.6
3.6
0.5
8.2
1.5
4.3
1.6
1.5
(36)
2.2
6.6
4.3
1.6
2.2
6.6
0.5
8.2
0(17) 1(12) 4(25)
112
27.
2(2) (2)(3) or 10
3
2
2
2
1
1
0
3
2
2
2
28.
2(0) 1(0) or 0
does not exist
29.
4(2) 1(2) or 6
2
2
4
1
0
0
2
1
1
6
2
4
2
1
30.
6(7) (6)(7) or 84
7
7
6
6
8
1
4
7
6
7
6
23
1.5(14.74) 3.6(31.9) 2.3(36.06)
175.668
26.
0
1
(4)
2
3
3
8
3
4
3
8
3
4
2
3
4
3
4
1
2
3
0
3
8
0.5
8.2
4.3
1.6
31.
4(12) 8(6) or 0
does not exist
6
12
4
8
32.
9(36) 27(13) or 27
13
36
9
27 39.
1
1
x
y
3
1
9
5
49 Chapter 2
2
1
7
13
9
36
27
33.
3
4
1
2
5
1
8
or 1
3
4
1
8
5
1
2
1
2
1
8
5
3
4
34.
1
7
x
y
1
2
4
1
1
9
1
4
2
1
1
4
2
1
1
41
12
1
9
1
9
1
7
1
4
2
1
x
y
1
2
4
1
1
4
2
1
(1, 3)
35.
12
12
x
y
6
6
9
4
1
3
x
y
7
1
8
6
9
6
4
6
9
6
4
1
96
46
7
1
8
7
1
8
12
12
6
9
6
4
x
y
6
6
9
4
6
9
6
4
(0, 2)
36.
26
41
x
y
5
2
1
3
0
2
x
y
1
1
7
5
1
2
3
5
1
2
3
1
15
32
1
1
7
1
1
7
26
41
5
1
2
3
x
y
5
2
1
3
5
1
2
3
(9, 7)
37.
7
0
x
y
8
3
4
3
9
7
x
y
3
1
6
8
4
3
3
8
4
3
3
1
48
33
3
1
6
3
1
6
7
0
8
4
3
3
x
y
8
3
4
3
8
4
3
3
1
7
2
,
1
7
2
38.
24
3
x
y
5
4
3
5
1
7
2
1
7
2
x
y
3
1
7
5
3
4
5
5
3
4
5
1
35
54
3
1
7
3
1
7
24
3
5
3
4
5
x
y
5
4
3
5
5
3
4
5
(3, 3)
3
3
x
y
1
6
3
9
1
5
3
9
1
5
1
93
51
1
6
1
6
1
1
3
9
1
5
x
y
3
1
9
5
3
9
1
5
1
3
,
2
3
40.
4
0
1
x
y
z
3
2
1
2
2
1
3
1
2
1
3
2
3
x
y
1
9
x
y
z
3
2
1
2
2
1
3
1
2
10
3
8
1
3
1
4
3
5
1
9
4
0
1
10
3
8
1
3
1
4
3
5
1
9
4(4) 1(0) (10)(1)
3(4) 3(0) (3)(1)
5(4) 1(0) 8(1)
x
y
z
1
9
6
9
12
x
y
z
2
3
, 1,
4
3
41.
9
5
1
x
y
z
3
1
1
5
2
1
6
9
3
2
3
1
4
3
x
y
z
1
9
x
y
z
3
1
1
5
2
1
6
9
3
1
21
33
2
15
21
1
12
15
1
9
9
5
1
1
21
33
2
15
21
1
12
15
1
9
1(9) (2)(5) 1(1)
12(9) (15)(5) 21(1)
15(9) 21(5) (33)(1)
x
y
z
1
9
2
12
3
x
y
z
2
9
4
3
1
3
x
y
z
2
9
,
4
3
,
1
3
42. 216
43. 30,143
44.
4.74
1.2
x
y
0.5
6.5
0.3
12 48. Let xrepresent the number of gallons of 10%
alcohol solution, and let yrepresent the number of
gallons of 25% alcohol solution.
xy12
0.10x0.25y0.15(xy)
0.10x0.25y0.15x0.15y
0.05x0.10y0
xy 12 →
0.05x0.10y0
12
0
x
y
1
0.10
1
0.05
Chapter 2 50
7.
1
95
0.5
0.3
6.5
12
0.5
0.3
6.5
12
1
0.3 0.5
12 6.5
7.
1
95
x
y
0.5
65
0.3
12
0.5
0.3
6.5
12
7.
1
95
4.74
1.2
0.5
0.3
6.5
12
(3.8, 7.2)
45. 2(x2yz) 2(7) →2x4y2z14
6x2y2z46x2y2z4
8x2y18
2(6x2y2z) 2(4) →12x4y4z8
4x6y4z14 4x6y4z14
16x10y22
18
22
x
y
2
10
8
16
3.8
7.2
x
y
1
1
12
2
8
10
16
2
8
10
16
1
82
16 10
1
1
12
1
1
12
18
22
2
8
10
16
x
y
2
10
8
16
2
8
10
16
x2yz7
2 2(1) z7
z3
(2, 1, 3)
46. Let xrepresent the number of cars produced in
the first year, and let yrepresent the number of
cars in the second year.
xy390,000 →xy390,000
xy90,000 xy90,000
390,000
90,000
x
y
1
1
1
1
2
1
x
y
1
2
1
1
1
1
1
1
1
1
1
11
11
1
2
x
y
1
1
1
1
1
1
1
1
1
2
390,000
90,000
1
1
1
1
150,000 in the second year and 240,000 in the first year
47. Let A
and I
.
A1
a11a22
a
22
a21a12
a11a22
a1
a
2
21a12
a11a22
a2
a
1
21a12
a11a22
a
11
a21a12
0
1
1
0
a12
a22
a11
a21
240,000
150,000
x
y
AA1
a
a
1
1
1
1
a
a
2
2
2
2
a
a
2
2
1
1
a
a
1
1
2
2
a
a
2
1
1
1a
a
2
2
2
2
a
a
2
2
1
1
a
a
1
2
2
2
a
a
1
1
1
1
a
a
2
2
2
2
a
a
2
2
1
1
a
a
1
1
2
2
a11a12 a12a11
a11a22 a21a12
I
Thus, AA1I.
0
1
1
0
0.
1
15
1
1
0.10
0.05
1
1
0.10
0.05
1
11
0.05 0.10
0.
1
15
x
y
1
0.10
1
0.05
1
1
0.10
0.05
0.
1
15
12
0
1
1
0.10
0.05
8 gal of 10% and 4 gal of 25%
49. Yes
8
4
x
y
A
. Does (A2)1(A1)2
b
d
a
c
A2
ab bd
bc d2
a2bc
ac cd
(A2)1
ab bd
a2bc
bc d2
ac cd
1
a2d22abcd b2d2
A1
ad
1
bc
ad
d
bc
ad
b
bc
ad
c
bc
ad
a
bc
b
a
d
c
(A1)2
Thus, (A2)1(A1)2.
50. A
1
2
1
1
1
b
d
f
a
c
e
ab bd
a2bc
bc d2
ac cd
1
a2d22abcd b2d2
1
2
1
1
1
3
4
0
1
0
3
1
2
1
(3)
1
1
2
(1(4) 3(3) 1(12))
1
2
17or 8.5
8
1
2
or 8.5 square units
4
0
0
3
1
1
0
3
1
1
4
0
5
1
06
53
38
44
22
53
38
44
22
1
38 53
22 44
5
1
06
x
y
53
44
38
22
53
38
44
22
5
1
06
49,109
31,614
53
38
44
22
computer system: $959, printer: $239
959
239
x
y
51. Let xrepresent the cost of complete computer
systems, and let yrepresent the cost of printers.
day 1: 38x53y49,109
day 2: 22x44y31,614
day 3: 21x26y26,353
using day 1 and day 2:
49,109
31,614
x
y
53
44
38
22
52. Let xrepresent Jessi’s first test score, and let y
represent Jessi’s second test score.
xy179 →xy179
yx7xy7
179
7
x
y
1
1
1
1
60. [fg](x) f(g(x))
f(x1)
(x1)23(x1) 2
x22x1 3x3 2
x2x
[gf](x) g(f(x))
g(x23x2)
x23x2 1
x23x1
61. No, more than one element of the range is paired
with the same element of the domain.
62. The radius of circle Eis 3, so the circumference is
2(3) or about 18.85. The diagonal of the square
has length 6, so each side has length
6
2
6
22
32
. The perimeter of the square is 4(32
) or
122
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.
Graphing Calculator Exploration:
Augmented Matrices and
Reduced Row-Echelon Form
Page 106
1.
,
,
(1, 5, 2)
2.
,
,
(7, 1, 2)
3.
,
(1, 1, 2, 2)
4. Exercise 1: x1, y5, z2; Exercise 2:
x7, y1, z2; Exercise 3: w1, x1,
y2, z2; They are the solutions for the
system.
5. The calculator would show the first part of the
number and follow it by ....
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.
2-6
1000: 1
0100:1
0010: 2
0001:2
1111:0
2111:1
1111:0
0210:0
100: 7
010: 1
001:2
111: 6
234: 3
484:12
100:1
010: 5
001:2
212: 7
125:1
411:1
2-5B
51 Chapter 2
1
2
1
1
1
1
1
1
1
1
1
11
11
1
2
1
2
179
7
1
1
1
1
x
y
1
1
1
1
1
1
1
1
first test: 86, second test: 93
53.
3
4
3
4
3
4
3
4
4
9
0
5
4
1
8
5
86
93
x
y
4 (3)
9 4
0 (3)
5 4
4 (3)
1 4
8 (3)
5 4
H(5, 9), I(1, 5), J(3, 9), K(1, 13)
54.
3
4
55. x3y2z6→x3y2z6
2(4xyz) 2(8) 8x2y2z16
9xy22
4(4xyz) 4(8) →16x4y4z32
7x5y4z10 7x5y4z10
9xy22
1(9xy) 1(22) →9xy22
9xy22 9xy22
0 0
infinitely many solutions
56.
57. y5
1
2
(x2)
y5
1
2
x1
2y10 x2
x2y8 0
58. m
3
2
5
1
1
2
or 2
(y5) 2(x1) or (y3) 2(x2)
y5 2(x1)
y5 2x2
y2x7
59a.
1
1
2
or approximately 0.0833
59b.
1
1
2
1
x
8
18 12x
1.5 x; 1.5 ft
6
2
4
1
30
7
0
8
4
1
13
3
9
1
5
5
9
x
O
g
(
x
)
g
(
x
) 2
x
5
xf(x)
6 x52
5 x40
4 x32
3 x24
2 x16
Chapter 2 52
y
x
O
x
y
3
x
2
y
4
3
(
1
3,
)
1
3
(7, 0)
(7, 0.5)
(1, 0)
(1, 3) (0, 4)
y
x
O
y
x
O
2
(
(4, 3)
(0, 2)
1
3,
)
1
3
y
x
O
(3, 5) (5, 3)
(7, 7)
O
y
x
(1200, 2040)
2000
2000
4000
6000
4000 6000
y
1.70
x
y
0.45
x
1500
y
x
O
y
x
1
y
x
1
(1,0)
y
x
O
2
3
1
211
(0, 1)
(
2
3, 1
)
y
3
x
3
y
3
x
1
y
1
O
y
x
2
x
5
y
25
5
x
7
y
14
y
3
x
2
8. Let xrepresent the number of greeting cards sold,
and let yrepresent the income in dollars.
x0
y0
y0.45x1500
y1.70x
at least 1200 cards
Pages 110–111 Exercises
9.
10.
11.
12. yes, it is true for both inequalities:
y
1
3
x5y2x1
2
?
1
3
(3) 52
?
2(3) 1
2 4true 2 7true
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.
4.
5.
(1, 0), (1, 3), (0, 4), (7, 0.5), (7, 0)
6.
f(x, y) 4x3y
f(0, 2) 4(0) 3(2) or 6
f(4, 3) 4(4) 3(3) or 25
f
2
1
3
,
1
3
4
2
1
3
3
1
3
or 8
1
3
25, 6
7.
f(x, y) 3x4y
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
13.
3
5
, 3
1
5
,
2
5
, 1
1
5
,
1
3
5
,
1
5
14.
(0, 0), (0, 3), (3, 3)
15.
(2, 5), (7, 0), (4, 0), (1, 5)
16. f(x, y) 8xy
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
17.
f(x, y) 3xy
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
18.
f(x, y) yx
f(0, 0) 0 0 or 0
f(0, 4) 4 0 or 4
f(2, 0) 0 2 or 2
4, 2
19.
f(x, y) xy
f(0, 2) 0 2 or 2
f(5, 6) 5 6 or 11
f(10, 6) 10 6 or 16
16, 2
20.
f(x, y) 4x2y7
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
21.
f(x, y) 2xy
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
y
x
O
2
5,4
5
0, 2
(
2
5, 1
y
0.5
x
1
y
2
x
2
y
3
x
2
)()
()
1
5
y
x
O
(0, 0)
(3, 3)
(0, 3)
y
0
x
y
x
3 0
y
x
O
(2, 5)
(4, 0) (7, 0)
(1, 5)
5
x
3
y
20
y
x
7
y
5 0
y
x
O
(0, 2)(5, 2)
(5, 4)
2
x
5
y
10
x
5
y
2
y
x
O
(0, 4)
(2, 0)
(0, 0)
y
2
x
4
x
2 2
y
0
y
x
O
(10, 6)
(5, 6)
(0, 2)
4
x
5
y
10
2
x
5
y
10
y
6
O
y
x
(0, 4)
(3, 1)
(0, 1)
y
1
x
0
x
y
4
y
x
O
(1, 2)
(2, 2) (4, 1)
(3, 1)
x
4
y
7
x
6
y
10
y
4
x
6
2
x
y
7
22.
f(x, y) 2xy5
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. x4, x4, y4, y4
24. Sample answer: y3, x4, 4x3y12
25a. 3y2x11 3y2x11
y03(0) 2x11
5
1
2
x
5
1
2
, 0
y2x13 y2x13
y00 2x13
6
1
2
x
6
1
2
, 0
y16 xy16 x
y2x13 2x13 16 x
x
2
3
9
y16 x
y16
2
3
9
1
3
9
2
3
9
,
1
3
9
y16 x2y17
2y17 2(16 x) 17
x7
1
2
2y17
y8
1
2
7
1
2
, 8
1
2
2y17 2y17
y3x1y8
1
2
y3x1
8
1
2
3x1
2
1
2
x
2
1
2
, 8
1
2
y7 2xy7 2x
y3x13x1 7 2x
x
6
5
y7 2x
y7 2
6
5
2
5
3
6
5
,
2
5
3
3y2x11 3y2x11
y7 2x3(7 2x) 2x11
2
1
2
x
y7 2x
y7 2
2
1
2
2
2
1
2
, 2
25b. f(x, y) 5x6y
f
5
1
2
, 0
5
5
1
2
6(0) or 27
1
2
f
6
1
2
, 0
5
6
1
2
6(0) or 32
1
2
f
2
3
9
,
1
3
9
5
2
3
9
6
1
3
9
or 86
1
3
f
7
1
2
, 8
1
2
5
7
1
2
6
8
1
2
or 88
1
2
f
2
1
2
, 8
1
2
5
2
1
2
6
8
1
2
or 63
1
2
f
6
5
,
2
5
3
5
6
5
6
2
5
3
or 33
1
5
f
2
1
2
, 2
5
2
1
2
6(2) or 24
1
2
max at
7
1
2
, 8
1
2
88
1
2
; min at
2
1
2
, 2
24
1
2
26. xy200
2xy300
x0
y0
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 xrepresent the Main Street site, and let y
represent the High Street site.
x20
y20
10x20y1200
27b. f(x, y) 30x40y
27c. f(x, y) 30x40y
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 ft2at the Main St. site and 20 ft2at 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 ft2at Main St.
Chapter 2 54
y
x
O
(1, 4)
(1, 8) (3, 8)
(3, 2)
(6, 2)
y
5
x
y
8
y
2
2
x
y
2 16
O
y
x
(100, 100)
(0, 200)
100
200
300
200
(150, 0)
(0, 0)
300
2
x
y
300
x
y
200
x
0
y
0
y
x
O
(20, 20 )
(20, 50 )
(80, 20 )
10
x
20
y
1200
y
20
x
20
y
x
O
y
2
x
8
d
p
O
80
60
40
20
123123
20
40
60
80
1
7
30.
31.
32. {16}, {4, 4}; no, two y-values for one x-value
33.
wx
4
yz
15
wxyz60
Linear Programming
Pages 115–116 Check for Understanding
1. Sample answer: These inequalities are usually
included because in real life, you cannot make less
than 0 of something.
2-7
1
2
2
3
55 Chapter 2
O
y
x
(5, 0)
(3, 4)
(0, 6)
(0, 0)
2
x
y
10
2
x
3
y
18
y
0
x
0
y
x
O
1
2
3
4
5
6
7
8
22468101214
0.5
x
1.5
y
7
3
x
9
y
2
28a. 3 is $3 profit on each batch of garlic dressing and
2 is $2 profit on each batch of raspberry
dressing.
28b. 2x3y18
2xy10
x0
y0
f(x, y) 3x2y
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
29.
2(2) (3)(1)or7
1
2
2
3
14020
20
40
60
80
100
120
(120, 24)
140
160
60 100
y
x
O
3
x
5
y
480
25
x
50
y
4200
(0, 84)
(160,0)
(0,0)
O
y
x
(50, 225)
(100, 150)
(50, 150)
3
x
2
y
600
x
50
100 200 300
100
200
300
y
150
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.
4.
5a. 25x50y4200
5b. 3x5y480
5c.
5d. P(x, y) 5x8y
5e. P(x, y) 5x8y
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 xthe number of brochures.
Let ythe number of fliers.
3x2y600
x50
y150
C(x, y) 8x4y
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
7. Let xthe number of Explorers.
Let ythe number of Grande Expeditions.
xy375
2x3y450
x0
y0
R(x, y) 250x350y
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 xthe number of loaves of light whole wheat.
Let ythe number of loaves of regular whole
wheat.
2x3y90
x2y80
x0
y0
P(x, y) 1x1.50y
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
Pages 116–118 Exercises
9. infeasible
10. unbounded
11.
f(x, y) 3 3y
f
3
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 gthe number of cups of Good Start food
and sthe number of cups of Sirius food.
0.84g0.56s1.54
12b. 0.21g0.49s0.56
12c.
12d. C(g, s) 36g22s
12e. C(g, s) 36g22s
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 dthe number of day-shift workers
and nthe number of night-shift workers.
d5
n6
dn14
13b.
13c. $5.50 4 $7.50 4 $52
$7.50 8 $60
C(n, d) 52d60n
13d. C(n, d) 52d60n
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
O
y
x
(0, 150)
(225, 0)
(0, 0)
2
x
3
y
450
x
0
y
0
x
y
375
100 200 300
100
200
300
400
O
y
x
(45, 0)
(0, 30)
(0, 0)
x
2
y
80
x
0
y
0
2
x
3
y
90
20 40 60 80
20
20
40
60
y
x
O
5
x
3
y
15
y
6
y
x
O
2
x
y
48
x
2
y
42
8
8
16
24
32
40
48
56
162432404856
y
x
O
4
x
3
y
12
( , 3)
(4, 3)
(4, 0)(3, 0)
y
3
x
4
3
4
(1.5, 0.5)
1
12
2
3
s
g
O
0.81
y
0.56 s
1.54
0.21
g
0.49 s
0.56
(2.67, 0)
(0, 2.75)
d
n
O
d
n
14
(9, 5)
(6, 8)
d
5
n
6
13e. $776
14a. Let xthe number of acres of corn.
Let ythe number of acres of soybeans.
xy180
x40
y20
x2y
P(x, y) 150x250y
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 xthe questions from section I.
Let ythe questions from section II.
6x15y90
y2
x0
S(x, y) 10x15y
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 xthe number of food containers.
Let ythe number of drink containers.
xy1200
x300
y450
P(x, y) 17.50x20y
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
17. Let xamount to deposit at First Bank.
Let yamount to deposit at City Bank.
xy11,000
0 x7500
1000 y7000
I(x, y) 0.06x0.065y
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 xthe number of nurses.
Let ythe number of nurse’s aides.
xy50
xy20
y12
x2y
C(x, y) 35,000x18,000y
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 xunits of snack-size bags.
Let yunits of family-size bags.
xy2400
x600
y900
P(x, y) 12x18y
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
O
y
x
(40, 20) (120, 60)
(160, 20)
x
y
180
x
40
y
20
x
2
y
100 200
100
200
O
y
x
(0, 6)
(0, 2)
(10, 2)
6
x
15
y
90
x
0
y
2
812416
4
8
12
O
y
x
(300, 900)
(750, 450)
(300, 450)
x
y
1200
x
300
y
450
800 1200400
400
800
1200
O
y
x
(4000, 7000)
(7500, 3500)
(0, 7000)
(7500, 1000)
(0, 1000)
x
y
11,000
y
1000
x
7500
y
7000
8000 12,0004000
4000
8000
12,000
O
y
x
(24, 12)
(33.3, 16.7)
(38, 12)
x
y
50
x
y
20
x
2
y
y
12
40 6020
20
40
60
O
y
x
(600, 900)
(1500, 900)
(600, 1800)
x
y
2400
x
600
y
900
1600 2400800
800
1600
2400
23b. Sample answer: Spend more than 30 hours per
week on these services.
24.
f(x, y)
1
3
x
1
2
y
f(15, 10)
1
3
(15)
1
2
(10) or 0
f(0, 5)
1
3
(0)
1
2
(5) or 2
1
2
f(15, 10)
1
3
(15)
1
2
(10) or 10
minimum: 2
1
2
, maximum: 10
25. 4xy64xy6
x2y12 4(2y12) y6
8y48 y6
y6
x12y12
x2(6) 12 or 0 (0, 6)
26.
27. Sample answer: C$13.65 $0.15(n30);
C$13.65 $0.15(n30)
C$13.65 $0.15(42 30)
$15.45
28.
2x
x
3
3
2
x
2(2x3) x(3 x)
4x6 3xx2
x2x6 0
(x3)(x2) 0
x3 0 or x2 0
x3x2
The correct choice is A.
Chapter 2 Study Guide and Assessment
Page 119 Understanding and Using the Vocabulary
1. translation 2. added
3. determinant 4. inconsistent
5. scalar multiplication 6. equal matrices
7. polygonal convex set 8. reflections
9. element 10. multiplied
Chapter 2 58
O
y
x
(3.8, 0)
(3, 2)
(0, 0)
(0, 8) 20
x
8
y
76
12
x
6
y
48
x
0
y
0
O
y
x
(4, 5)
(9, 0)
(0, 0)
(0, 6)
x
y
9
x
2
y
16
x
4
y
24
x
0
y
0
8124
4
8
12
O
y
x
C
(12, 9)
D
(12, 10)
E
(0, 14)
A
(0, 1)
B
(3, 3)
F
(0, 11)
2
x
6
y
84
8
x
3
y
33 2
x
3
y
3
y
0
x
12
8124
4
8
12
O
y
x
(40, 10)
(60, 0)
(25, 10)
(25, 0)
30
x
60
y
1800
y
0
y
10
x
25
40 6020
20
40
60
O
y
x
(0, 5)
(15, 10)
(15, 10)
y
5 5
y
x
5
y
x
5
10 201020
10
20
10
20
20. Let xbatches of soap.
Let ybatches of shampoo.
12x6y48
20x8y76
x0
y0
3 batches of soap and 2 batches of shampoo
21. Let xthe number of small monitors.
Let ythe number of large monitors.
x2y16
xy9
x4y24
x0
y0
P(x, y) 40x40y
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
22.
Area of trapezoid ACDE
1
2
(12)(13 1)
84
Area of ABF
1
2
(10)(3)
15
Area of shaded origin 84 15
69 square units
23a. Let xoil changes.
Let ytune-ups.
x25
0 y10
30x60y30(60)
P(x, y) 12x20y
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
xy
19
06
13
20
33
x
O
y
y
3|
x
2|
Pages 120–122 Skills and Concepts
11. 2y4xyx2
2(x2) 4xy2 2
2x4 4xy4
x2
(2, 4)
12. 6yx0yx5
6(x5) x0y6 5
6x30 x0y1
x6
(6, 1)
13. 3yx12x5y
x1 3y2(1 3y) 5y
2 6y5y
1
2
1
y
2x5y
2x5
1
2
1
x
1
5
1
1
5
1
,
1
2
1
14. 2y15x4y6x1
2(6x1) 15x4y6(2) 1
12x2 15x4y13 (2, 13)
x2
15. 5(3x2y) 5(1) →15x10y5
2(2x5y) 2(12) 4x10y24
19x19
x1
2x5y12
2(1) 5y12
y2(1, 2)
16. x5y20.5 →x5y20.5
3yx13.5 x3y13.5
8y34
x5y20.5 y4.25
x5(4.25) 20.5
x0.75 (0.75, 4.25)
17. 3(x2y3z) 3(2) 3(x4y3z) 3(14)
3x5y4z03x5y4z0
↓↓
3x6y9z63x12y9z42
3x5y4z03x5y4z0
y5z6 7y13z42
7(y5z) 7(16) →7y35z42
7y13z42 7y13z42
48z0
z0
y5z6x2y3z2
y5(0) 6x2(6) 3(0) 2
y6x10
(10, 6, 0)
18. x2y6z42(x2y6z) 2(4)
xy2z32x3y4z5
3y4z7↓
2x4y12z18
2x3y4z15
7y16z13
4(3y4z) 4(7) →12y16z28
7y16z13 7y16z13
5y15
y3
3y4z7xy2z3
3(3) 4z7x3 2(0.5) 3
z0.5 x1
(1, 3, 0.5)
19. x2yz72(3xyz) 2(2)
3xyz22x3y2z7
4xy9↓
6x2y2z4
2x3y2z7
8x5y11
5(4xy) 5(9) →20x5y45
8x5y11 8x5y11
28x56
x2
4xy9x2yz7
4(2) y92 2(1) z7
y1z3
(2, 1, 3)
20. AB
8 (5)
4 (2)
7 (3)
0 2
59 Chapter 2
3
6
4
2
21. BA
5 8
2 (4)
3 7
2 0
13
2
10
2
22. 3B
3(5)
3(2)
3(3)
3(2)
15
6
9
6
23. 4C
4(2)
4(5)
8
20
24. AB
5
2
3
2
8
4
7
0
7(5) 8(2)
0(2) (4)(2)
7(3) 8(2)
0(3) (4)(2)
or
25. impossible
26. 4A4B4A(4B)
4A
4B
4(5)
4(2)
4(3)
4(2)
4(8)
4(4)
4(7)
4(0)
51
8
5
8
20
8
12
8
32
16
28
0
4A(4B)
32 20
16 8
28 12
0 (8)
52
8
40
8
27. impossible
28.
A(7, 1), B(1, 3), C(2, 7)
29.
W(2, 3), X(1, 2), Y(0, 4), Z(1, 2)
30.
D(2, 3), E(2, 5), F(1, 5), G(1, 3)
31. 0.5
P(6, 8), Q(2, 4), R(2, 2)
0.5
0.5
0.5
1
1.5
2
1
1
1
2
3
4
1
3
1
5
2
5
2
3
1
3
1
5
2
5
2
3
0
1
1
0
1
2
0
4
1
2
2
3
1
2
0
4
1
2
2
3
0
1
1
0
2
7
1
3
7
1
3
4
3
4
3
4
5
3
2
1
4
3
32.
5
3
2
1
4
3
5
3
2
1
4
3
0
1
1
0
Chapter 2 60
A
B
C
C
B
A
y
x
O
W
X
Y
Z
Y
W
Z
X
y
x
O
D
EF
G
F
G
D
E
y
x
O
P
Q
R
R
Q
P
2
4
4
2
2246
8
6
y
x
O
A(3, 4), B(1, 2), C(3, 5)
3
5
1
2
3
4
5
3
2
1
4
3
1
0
0
1
33.
1
3
1
3
1
3
5
3
5
3
5
3
4
6
4
6
4
6
34.
3(7) (4)(5) or 1
5
7
3
4
35.
8(3) (6)(4) or 0
4
3
8
6
36.
4
6
4
1
2
3
3
5
7
3
(1)
4
3(10) 1(62) 4(29)
24
37.
4
1
6
0
3
2
5
7
2
2
3
5
7
6
4
5
7
6
4
2
3
5
0
(4)
5(16) 0(44) 4(20)
160
38. no, not a square matrix
39.
3(5) (1)(8) or 23
8
5
3
1
3
2
7
2
1
6
7
2
1
6
3
2
2
1
3
8
3
5
1
40.
5(4) 10(2) or 0
no inverse
41.
3(4) 1(5) or 7
5
4
3
1
2
4
5
10
1
7
5
3
4
1
42.
3(7) 5(2) or 11
2
7
3
5
1
1
1
2
3
7
5
43.
2(1) 6(5) or 32
5
1
2
6
3
1
2
5
2
1
6
44.
2(2) (1)(4) or 0
no inverse
4
2
2
1
A
C
B
B
C
A
y
x
O
13
5
x
y
50.
f(x, y) 3x2y1
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 xgallons in the truck.
Let ygallons in the motorcycle.
xy28
0 x25
0 y6
m(x, y) 22x42y
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
Page 123 Applications and Problem Solving
52.
Broadman 30; Girard 49; Niles 43
30
49
43
2(5) 5(3) 5(1)
8(5) 2(3) 3(1)
6(5) 4(3) 1(1)
5
3
1
5
3
1
5
2
4
2
8
6
61 Chapter 2
45.
1
2
x
y
5
3
2
1
1
5
2
3
1
5
2
3
1
1
25
11
1
1
1
2
5
2
3
1
x
y
5
3
2
1
5
2
3
1
(13, 5)
46.
3
6
x
y
2
4
3
6
2
1
4
2
3
4
6
2
3
4
6
1
32
64
2
1
4
2
1
4
3
6
2
3
4
6
x
y
2
4
3
6
2
3
4
6
(1, 0)
47.
1
2
x
y
5
4
3
2
1
0
x
y
1
2
5
3
4
2
5
3
4
2
1
35
24
1
2
1
2
1
2
5
3
4
2
x
y
5
4
3
2
5
3
4
2
(7, 4)
48.
48
1
.31
2.7
4.6
8.8
2.9
2.7
4.6
8.8
2.9
1
4.6 2.7
2.9 8.8
7
4
x
y
48
1
.31
x
y
2.7
8.8
4.6
2.9
2.7
4.6
8.8
2.9
O
y
x
(1, 5)
(4, 2)
(6, 2)
(1, 2)
y
x
6
x
1
y
2
y
10 2
x
O
y
x
(22, 6)
(25, 3)
(25, 0)
(0, 0)
(0, 6)
x
y
28
x
25
y
0
y
6
20 3010
10
20
30
48
1
.31
8.4
74.61
2.7
4.6
8.8
2.9
(5.7, 6.6)
49.
f(x, y) 2x3y
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
5.7
6.6
x
y
O
y
x
(4, 7)
(0, 9)
(6, 4)
(0, 4) (6, 5)
x
y
11
x
6
2
y
x
18
y
4
8124
4
8
12
53. Let xthe shortest side.
Let ythe middle-length side.
Let zthe longest side.
xyz83 xyz83
z3xxy3x83
z
1
2
(xy) 17 4xy83
z
1
2
(xy) 17
3x
1
2
(xy) 17
5xy34
z3x
z3(13)
z39
13 in., 31 in., 39 in.
54a. Let xnumber of Voyagers.
Let ynumber of Explorers.
5x6y240
3x2y120
5x18y540
x0
y0
P(x, y) 2.40x5.00y
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
Page 123 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
. The coefficient
has a
determinant of 0, so it has no inverse.
4
8
2
4
12
24
x
y
4
8
2
4
83
34
x
y
1
1
4
5
Chapter 2 SAT & ACT Preparation
Page 125 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.
(1 2)(2 3)(3 4)
1
2
(20 x)
(3)(5)(7)
1
2
(20 x)
105
1
2
(20 x)
210 20 x
x190
The correct choice is D.
2. First convert the numbers to improper fractions.
5
1
3
6
1
4
1
3
6
2
4
5
Express both fractions with a common
denominator. Then subtract.
5
1
3
6
1
4
1
3
6
2
4
5
6
1
4
2
7
1
5
2
1
1
2
1
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: xyz
The correct choice is D.
4. 2x 2 1 5
2x 26
x 23
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.
Chapter 2 62
O
y
x
(30, 15)
(18, 25)
(40, 0)
(0, 0)
(0, 30)
3
x
2
y
120
5
x
6
y
240
5
x
18
y
540
y
0
40 60
20
20
40
60
1
9
1
4
1
5
1
4
1
5
1
41
51
1
9
1
9
83
34
1
4
1
5
x
y
1
1
4
5
1
4
1
5
13
31
x
y
6. Start by simplifying the fraction expression on the
right side of the equation.
2
5
x
x
2
5
2
5
2
5
x
x
4
5
To finish solving the equation, treat it as a
proportion and write the cross products.
2
5
x
x
4
5
5(2 x) 4(5 x)
10 5x20 4x
x10
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 nand n1. Represent
even integers by 2k, and odd integers by 2k1.
I. n(n1) 2n1
2n1 is odd for any value of n. So statement
I is always true.
II. n(n1) (n2) 3n3
If nis even, then 3n3 3(2k) 3 is odd.
If nis odd, then 3n3 3(2k1) 3
6k3 3 6k6 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(n1) (n2) 3n3 3(n1)
The correct choice is D.
8. Start by representing the relationships that are
given in the problem. Let Prepresent the number
of pennies; Nthe number of nickels; Dthe number
of dimes; and Qthe number of quarters. He has
twice as many pennies as nickels.
P2N
Similarly, N2Dand D2Q. 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 D2Q. 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.
9.
3
2
4
9
1
1
2
8
2
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 tbe the number of tapes sold.
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
costs
1
3
of $12 or $4. You can calculate the total
sales of CDs by subtracting $480 from $600 to get
$120.
Now you can find tusing an equation.
4t120
t30
Thirty cassette tapes were sold. The answer is 30.
3
2
9
4
3
2
3
2
2
63 Chapter 2
Price each Number Total
Sold Sales
CDs 40 $480
Tapes t
Total $600
Price each Number Total
Sold Sales
CDs $12 40 $480
Tapes $4 t$120
Total $600
Symmetry and Coordinate
Graphs
Page 133 Graphing Calculator Exploration
1. f(x) f(x)2. f(x) f(x)
3. even; odd
4. f(x) x83x42x22
f(x) (x)83(x)42(x)22
x83x42x22
f(x)
f(x) x74x5x3
f(x) (x)74(x)5(x)3
x74x5x3
(x74x5x3)
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.
Pages 133–134 Check for Understanding
1. The graph of yx212 is an even function.
The graph of xy 6isanoddfunction. The graphs
of xy24 and 17x216xy 17y2225 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: y0, x0, yx, yx
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.
3-1 Algebraically: Substituting (x, y) into the
equation followed by substituting (x, y) is the
same as substituting (x, y).
6. f(x) x69x
f(x) (x)69(x)f(x) (x69x)
f(x) x69xf(x) x69x
no
7. f(x)
5
1
x
x19
f(x)
5(
1
x)
(x)19 f(x)
5
1
x
x19
f(x)
5
1
x
x19 f(x)
5
1
x
x19
yes
8. 6x2y1→6a2b1
x-axis 6a2(b) 1
6a2b1no
y-axis 6(a)2b1
6a2b1yes
yx6(b)2a1
6b2a1no
yx6(b)2(a) 1
6b2a1no
y-axis
9. x3y34→a3b34
x-axis a3(b)34
a3b34no
y-axis (a)3b34
a3b34no
yx(b)3(a)34
a3b34yes
yx(b)3(a)34
a3b34no
yx
10.
11. y2 x2
→b2 a
2
x-axis b2 a
2
no
y-axis b2 (
a)2
b2 a
2
yes
Chapter 3 64
Chapter 3 The Nature of Graphs
y
x
(4, 2) (4, 2)
(1, 2) (1, 2)
(2.5, 3) (2.5, 3)
O
x
O
y
2
1
1
11
(1,1) (1,1)
(√2,0) (√2,0)
2
2
2
(0√2)
12. yx3→ba3
x-axis ba3
ba3yes
y-axis b(a)3
ba3no
13. x-intercept:
2
x
5
2
y
9
2
1
2
x
5
2
0
9
2
1
2
x
5
2
1
x225
x5(5, 0)
other points:
when x6when x6
3
2
6
5
y
9
2
1
3
2
6
5
y
9
2
1
y
9
2
2
1
5
1
y
9
2
2
1
5
1
y2
9
2
9
5
y2
9
2
9
5
y y
6,
,
6,
,
6,
Pages 134–136 Exercises
14. f(x) 3x
f(x) 3(x)f(x) (3x)
f(x) 3xf(x) 3x
yes
15. f(x) x31
f(x) (x)31f(x) (x31)
f(x) x31f(x) x31
no
16. f(x) 5x26x9
f(x) 5(x)26(x) 9
f(x) 5x26x9
f(x) (5x26x9)
f(x) 5x26x9
no
17. f(x)
4
1
x7
f(x)
4(
1
x)7
f(x)
4
1
x7
f(x)
4
1
x7
f(x)
4
1
x7
yes
311
5
311
5
311
5
311
5
311
5
18. f(x) 7x58x
f(x) 7(x)58(x)
f(x) 7x58x
f(x) (7x58x)
f(x) 7x58x
yes
19. f(x)
1
x
x100
f(x)
(
1
x)
(x)100 f(x)
1
x
x100
f(x)
1
x
x100 f(x)
1
x
x100
no
20. yes;
g(x)
x2
x
1
g(x)
(x
(
)2
x
)
1
Replace x with x.
g(x)
x2
x
1
(x)2x2
g(x)
x2
x
1
Determine the opposite
of the function.
g(x) g(x)
21. xy 5→ab 5
x-axis a(b) 5
ab 5
ab 5no
y-axis (a)b5
ab 5
ab 5no
yx(b)(a) 5
ab 5yes
yx(b)(a) 5
ab 5yes
yxand yx
22. xy21→ab21
x-axis a(b)21
ab21yes
y-axis (a) b21
ab21no
yx(b) (a)21
ba21no
yx(b) (a)21
ba21no
x-axis
23. y8x→b8a
x-axis (b) 8a
b8ano
y-axis b8(a)
b8ano
yx(a) 8(b)
a8bno
yx(a) 8(b)
a8bno
none of these
65 Chapter 3
(1,1)
(1,1)
y
x
O
Chapter 3 66
y
x
(4, 4)
(1, 2)
(2, 1)
(4, 4)
(1, 2)
(2, 1)
O
y
x
(4, 4)
(1, 2)
(2, 1)
(4, 4)
(1, 2)
(2, 1)
O
y
x
(4, 4)
(1, 2)
(2, 1)
(2, 3)
O
y
x
O
y
x
O
28.
29.
30. Sample answer:
31. y2x2→b2a2
x-axis (b)2a2
b2a2
y-axis b2(a)2
b2a2yes; both
32. x3y→a3b
x-axis a3(b)
a3bno
y-axis (a)3b
a3byes
y-axis
24. y
x
1
2
→b
a
1
2
x-axis (b)
a
1
2
b
a
1
2
no
y-axis b
(
1
a)2
b
a
1
2
yes
yx(a)
(b
1
)2
a
b
1
2
no
yx(a)
(
1
b)2
a
b
1
2
no; y-axis
25. x2y24→a2b24
x-axis a2(b)24
a2b24yes
y-axis (a)2b24
a2b24yes
yx(b)2(a)24
a2b24yes
yx(b)2(a)24
a2b24yes
all
26. y2
4
9
x2
4→b2
4
9
a2
4
x-axis (b)2
4
9
a2
4
b2
4
9
a2
4yes
y-axis b2
4(
9
a)2
4
b2
4
9
a2
4yes
yx(a)2
4(
9
b)2
4
a2
4
9
b2
4no
yx(a)2
4(
9
b)2
4
a2
4
9
b2
4no
x-axis and y-axis
27. x2
y
1
2
→a2
b
1
2
x-axis a2
(
1
b)2
a2
b
1
2
yes
y-axis (a)2
b
1
2
a2
b
1
2
yes
yx(b)2
(a
1
)2
b2
a
1
2
a2
b
1
2
yes
yx(b)2
(
1
a)2
b2
a
1
2
a2
b
1
2
yes
x-axis, y-axis, yx, yx
67 Chapter 3
33. y23x0→b23a0
x-axis (b)23a0
b23a0yes
y-axis b23(a) 0
b23a0no
x-axis
34. y2x2→b2a2
x-axis (b)2a2
b2a2yes
y-axis b2(a)2
b2a2
yes; both
35. x
12
8y2
→a12
8b2
x-axis a12
8(b)
2
a12
8b2
yes
y-axis (a)
12
8b2
a12
8b2
yes; both
36. yxy →bab
x-axis (b)a(b)
bab no
y-axis b(a)b
bab no
neither
37. yx3x→ba3a
x-axis (b)a3a
ba3ayes
y-axis b(a)3(a)
ba3ano
x-axis
The equation yx3xis symmetric about the
x-axis.
38a.
x
8
2
1
y
0
2
1→
a
8
2
1
b
0
2
1
x-axis
a
8
2
(
1
b
0
)2
1
a
8
2
1
b
0
2
1yes
y-axis
(
8
a)2
1
b
0
2
1
a
8
2
1
b
0
2
1yes
origin
(
8
a)2
(
1
b
0
)2
1
a
8
2
1
b
0
2
1yes
x- and y-axis symmetry
38b.
38c. (2, 5
), (2, 5
), (2, 5
)
39. Sample answer: y0
40. Sample answer:
yx1yx1yx1
yx1y2x4y2x4
y2x4y2x4
y
x
O
y
x
O
y
x
O
y
x
(1, 0)
1
1
(1, 0)
O
y
x
O
1
1
1
1
y
x
O
y
x
1
1
1212
2
2
O
Chapter 3 68
46. 4x2y7→y2x
7
2
12x6y21 y2x
7
2
consistent and dependent
47.
48. m
16
2
2
0
14
2
or 7
y2 7(x0)
y7x2
49. [fg](x) f(g(x))
f(x6)
2(x6) 11
2x23
[gf](x) g(f(x))
g(2x11)
(2x11) 6
2x5
50. 7537577537
7510
The correct choice is B.
Families of Graphs
Page 142 Check for Understanding
1. y(x4)37
2. The graph of y(x3)2is a translation of yx2
three units to the left. The graph of yx23 is a
translation of yx2three units up.
3. reflections and translations
4. When c1, the graph of yf(x) is compressed
horizontally by a factor of c.
When c1, the graph of yf(x) is unchanged.
When 0 c1, the graph is expanded
horizontally by a factor of
1
c
.
5a. g(x)
3x
1
5b. h(x)
3x1
5c. k(x)
3x2
1
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
horizontally by a factor of
1
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
3-2
y
x
x
1
y
O
O
y
x
(80, 0)
(50, 75)
(0, 0)
(0, 112.5)
3
x
4
y
450
5
x
2
y
400
x
0
y
0
100 15050
50
100
150
45. 3(2xyz) 3(0) →6x3y3z0
3x2y3z21 3x2y3z21
9xy21
3x2y3z21
4x5y3z2
7x3y23
3(9xy) 3(21) →27x3y63
7x3y23 7x3y23
20x40
x2
9xy21 2xyz0
9(2) y21 2(2) (3) z0
y3z7
(2, 3, 7)
38
47
59
74
y
x
O
y
x
y
x
2
41.
1
y
2
2
1
x
6
2
1
(
1
6
2
)2
1
x
6
2
1
3
1
x
6
2
1
1
x
6
2
2
x232
x42
(42
, 6) or (42
, 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.
Consider the graph of y
1
x
.
43. Let xnumber of bicycles.
Let ynumber of tricycles.
3x4y450
5x2y400
x 0
y 0
P(x, y) 6x4y
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
44.
4(5) 3(6)
7(5) 2(6)
4(8) 3(9)
7(8) 2(9)
5
6
8
9
3
2
4
7
69 Chapter 3
y
x
O
y
x
O
xf(x)
0 x150
1 x2 100
2 x3 150
3 x4 200
4 x5 250
$50
0
$100
$150
$200
$250
10543
Time (h)
2
$50
00
$100
$150
$200
$250
1643
Time (h)
52
y
x
O
y
x
O
y
x
O
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
20c. compressed vertically by a factor of
1
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
22c. compressed horizontally by a factor of
3
5
,
translated up 2 units
23a. translated left 2 units, compressed vertically by
a factor of
1
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
24b. compressed horizontally by a factor of
1
6
,
translated 8 units up
24c. The portion of parent graph on the left of the y-
axis is replaced by a reflection of the portion on
the right of the y-axis.
25a. compressed horizontally by a factor of
2
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.
26. yx2
27. y
x
0
.25
4
3
28. 29.
9a. translated up 3 units, portion of graph below
x-axis reflected over the x-axis
9b. reflected over the x-axis, compressed
horizontally by a factor of
1
2
9c. translated left 1 unit, compressed vertically by a
factor of 0.75
10. 11.
12a.
12b.
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
vertically by a factor of
3
4
.
15. The graph of g(x) is the graph of f(x) compressed
horizontally by a factor of
1
5
.
16. The graph of g(x) is a translation of f(x) right 5
units.
17. The graph of g(x) is the graph of f(x) expanded
vertically by a factor of 3.
18. The graph of g(x) is the graph of f(x) reflected over
the x-axis.
19. 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.
Chapter 3 70
36a. 0
[7.6, 7.6] scl:1 by [5, 5] scl:1
36b. 0.66
[7.6, 7.6] scl:1 by [5, 5] scl:1
36c. 0.25
[7.6, 7.6] scl:1 by [5, 5] scl:1
37a. 0
[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
y
x
O
y
x
O
y
x
O
y
x
O
448
4
4
8
12
y
x
O
y
f
(|
x
|)
y
f
(
x
)
4
4
48
8
4
8
30. 31.
32. 33.
34.
35a. 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
35c. 1.5
[7.6, 7.6] scl:1 by [5, 5] scl:1
37c. 0.6
[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.
39. The x-intercept will be
a
b
.
40a. y
40b.
41a. A
1
2
bh
1
2
(10)(5)
25 units2
41b. The area of the triangle is A
1
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 ycf(x) would be 25cunits2.
41c. The area of the triangle is A
1
2
(10)(5)
25 units2. Its area is the same as that of the
original triangle. The area of the triangle formed
by yf(xc) would be 25 units2.
0.25[[x1]] 1.50 if [[x]] x
0.25[[x]] 1.50 if [[x]] x
71 Chapter 3
xy
0 x11.50
1 x21.75
2 x32.00
3 x42.25
4 x52.50
1
2
3
4
5
32
Fare Units
5410
Price
(dollars)
0
8
O
412
4
8
x
y
8
O
412
4
8
x
y
8
O
412
4
8
x
y
y
x
48
4
4
4
O
y
x
48
4
4
8
O
y
x
O
48
4
4
8
12
y
x
O
48
4
4
8
42a. (1) yx2(2) yx3
(3) yx2(4) yx3
42b. (1) (2)
(3) (4)
42c. (1) y(x3)25(2)y(x3)35
(3) y(x3)25(4)y(x3)35
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 xnumber of preschoolers.
Let ynumber of school-age children.
xy50
x 3(10)
y 5(10)
x 0
y 0
I(x, y) 18x6y
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
46.
A(4, 5), B(3, 1), C(1, 2)
47. x225 9 y12 2z
x56 z
48. 5(6x5y) 5(14)→30x25y70
6(5x2y) 6(3) 30x12y18
13y52
y4
5x2y3
5x2(4) 3
x1(1, 4)
49. The graph implies a negative linear relationship.
50. 3x4y0→y
3
4
x
perpendicular slope:
4
3
1
2
3
1
4
5
2
1
1
3
5
4
1
0
0
1
O
y
x
(30, 0)
(30, 20)
(0, 0)
(0, 50)
x
y
50
y
50
y
0
x
30
40 60
20
20
40
60
Chapter 3 72
51. 5d2p500 →p
5
2
d250
250
52. [fg](x) f(g(x))
f(x26x9)
2
3
(x26x9) 2
2
3
x24x4
[gf](x) g(f(x))
g
2
3
x2
2
3
x2
26
2
3
x2
9
4
9
x2
8
3
x4 4x12 9
4
9
x2
2
3
0
x25
53. If m1; d1
5
1
0
or 49.
If m10; d10
5
1
0
0
or 5.
If m50; d50
5
5
0
0
or 49.
If m100; d100
1
5
0
0
0
or 99.5
If m1000; d1000
1
5
0
0
00
or 999.95.
The correct choice is A.
Graphs of Nonlinear Inequalities
Page 149 Check for Understanding
1. Aknowledge of transformations can help
determine the graph of the boundary of the
shaded region, y5 x2
.
2. When solving a one variable inequality
algebraically, you must consider the case where
the quantity inside the absolute value is non-
negative 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
5. y5x47x38
3
?5(1)47(1)38
3 4; yes
6. y3x41
3
?3(0) 41
3 3; no
3-3
y
x
O
y
x
O
y
x
O
y
x
O
7. 8.
9.
10. Case 1 Case 2
x64x64
(x6) 4x6 4
x6 4x2
x10
x10
{xx10 or x2}
11. Case 1 Case 2
3x4x3x4x
(3x4) x3x4 x
3x4 x2x4
4x4x2
x1
{x1 x2}
12a. x120.005
12b. Case 1 Case 2
x120.005 x120.005
(x12) 0.005 x12 0.005
x12 0.005 x12.005
x11.995
x11.995
12.005 cm, 11.995 cm
Pages 150–151 Exercises
13. yx34x2214. yx27
0
?(1)34(1)228
?3 27
0 1; no 8 8; no
15. yx11
1
1
?(2)
11
1
1 2; yes
16. y0.2x29x7
63
?0.2(10)29(10) 7
63 63; no
17. y
x2
x
6
9
?
(6
)2
6
6
9 5; yes
18. y2x37
0
?2037
0 7; yes
19. yx
2yx
2
0
? 0
24
? 1
2
0 2; yes 4 3; no
yx
2yx
2
1
? 1
20
? 1
2
1 3; yes 0 i2; no
yx
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.
20. 21.
22. 23.
24. 25.
26. 27.
28. 29.
73 Chapter 3
y
x
O
y
x
O
4812
4
8
12
Ox
y
y
x
O
y
x
O
y
x
O
y
x
O
y
x
O
y
x
O
y
x
O
y
x
O
y
x
O
y
x
O
30. 31.
32.
33. Case 1 Case 2
x45x45
(x4) 5x4 5
x4 5x1
x9
x9
{xx9 or x1}
34. Case 1 Case 2
3x1242 3x1242
(3x12) 42 3x12 42
3x12 42 3x30
3x54 x10
x18
{xx18 or x10}
35. Case 1 Case 2
7 2x8 37 2x8 3
(7 2x) 8 37 2x8 3
7 2x8 32x4
2x18 x2
x9
{x2 x9}
36. Case 1 Case 2
5 xx5 xx
(5 x) x5 xx
5 xx2x5
5 0; true x2.5
{xx2.5}
37. Case 1 Case 2
5x805x80
(5x8) 05x8 0
5x8 05x8
5x8x
8
5
x
8
5
no solution
38. Case 1 Case 2
2x92x02x92x0
(2x9) 2x02x9 2x0
2x9 2x09 0; true
4x9
x
9
4
all real numbers
Chapter 3 74
39. Case 1 Case 2
2
3
x58
2
3
x58
2
3
((x5)) 8
2
3
(x5) 8
2
3
(x5) 8
2
3
x
1
3
0
8
2
3
x
1
3
0
8
2
3
x
1
3
4
2
3
x
3
3
4
x7
x17
{x17 x7}
40. x37.51.2
Case 1 Case 2
x37.51.2 x37.51.2
(x37.5) 1.2 x37.5 1.2
x37.5 1.2 x38.7
x36.3
x36.3
36.3 x38.7
41. Case 1 Case 2
3x7 x13x7x1
3((x7)) x13(x7) x1
3(x7) x13x21 x1
3x21 x12x20
4x22 x10
x5.5
{x5.5 x10}
42. 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
35
. The length of side AC is 45
the area of
the triangle is 0.5(35
)(45
) or 30.
43. 0.10(90) 0.15(75) 0.20(76)
0.40(80) 0.15(x) 80
0.15x12.55
x83
2
3
44.
[1, 8] scl:1 by [1, 8] scl:1
44a. b0
44b. none
44c. b0 or b4
44d. b4
44e. 0 b4
45a.
45b. The shaded region shows all points (x, y) where
xrepresents 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.
47. y
a
1
4
→b
a
1
4
x-axis (b)
a
1
4
b
a
1
4
no
y-axis b
(
1
a)4
b
a
1
4
yes
yx(a)
(b
1
)4
a
(b
1
)4
no
yx(a)
(
1
b)4
a
(b
1
)4
no
y-axis
48.
2
1
8
3
8
5
4
3
8
5
4
1
83
45
x
P
(
x
)
900
100
200
300
400
1000 1100 1200 1300 1400 1500
O
xf(x)
211
18
05
18
211
f
(
x
)
x
O
States
with
Teen
Courts
50
45
40
35
30
25
20
15
10
5
019800 1990 2000 2010
Year
49.
3
4
50.
51.
52. [fg](4) f(g(4))
f(0.5(4) 1)
f(1)
5(1) 9
14
[gf](4) g(f(4))
g(5(4) 9)
g(29)
0.5(29) 1
13.5
6
4
21
30
3
4
(8)
3
4
(7)
3
4
(4)
3
4
(0)
7
0
8
4
75 Chapter 3
53. Student A 15
Student B
1
3
(15) 15 or 20
Let xnumber of years past.
20 x2(15 x)
20 x30 2x
x10
Page 151 Mid-Chapter Quiz
1. x2y29 0→a2b29 0
x-axis a2(b)29 0
a2b29 0yes
y-axis (a)2b29 0
a2b29 0yes
yx(b)2(a)29 0
a2b29 0yes
yx(b)2(a)29 0
a2b29 0yes
x2y29 0 →f(x) x2
9
f(x) (x)
29
f(x) (x2
9
)
f(x)
x2
9
f(x) x2
9
yes
x-axis, y-axis, yx, yx, origin
2. 5x26x9 y→5a26a9 b
x-axis 5a26a9 (b)
5a26a9 bno
y-axis 5(a)26(a) 9 b
5a26a9 bno
yx5(b)26(b) 9 (a)
5b26b9 ano
yx5(b)26(b) 9 (a)
5b26b9 ano
5x26x9 y→f(x) 5x26x9
f(x) 5(x)26(x) 9
5x26x9
f(x) (5x26x9)
f(x) 5x26x9 no
none of these
3. x
7
y
→a
7
b
x-axis a
(
7
b)
a
7
b
no
y-axis (a)
7
b
a
7
b
no
yx(b)
(a
7
)
a
7
b
yes
yx(b)
(
7
a)
a
7
b
yes
x
7
y
→f(x)
7
x
f(x)
(
7
x)
f(x)
7
x
f(x)
7
x
f(x)
7
x
yes
yx, yx, origin
y
x
O
y
x
O
4. yx1→ba1
x-axis (b) a1
ba1no
y-axis b(a)1
ba1yes
yx(a) (b)1
ab1no
yx(a) (b)1
ab1no
yx1→f(x) x1
f(x) x1f(x) (x1)
f(x) x1f(x) x1 no
y-axis
5a. translated down 2 units
5b. reflected over the x-axis, translated right 3 units
5c. compressed vertically by a factor of
1
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
7. 8.
9. Case 1 Case 2
2x715 2x715
(2x7) 15 2x7 15
2x7 15 2x22
2x8x11
x4
4 x11
10. x643
Case 1 Case 2
x643x643
(x64) 3x64 3
x64 3x67
x61
x61
61 x67
Inverse Functions and Relations
Pages 155–156 Check for Understanding
1. Sample answer: First, let yf(x). Then
interchange xand y. Finally, solve the resulting
equation for y.
2. nis 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.
3-4
Chapter 3 76
9. f(x) 3x2
y3x2
x3y2
x2 3y
y
1
3
x
2
3
f1(x)
1
3
x
2
3
; f1(x) is a function.
10. f(x)
x
1
3
y
x
1
3
x
y
1
3
y3
1
x
y
3
1
x
or
f1(x) ; f1(x) is a function.
11. f(x) (x2)26
y(x2)26
x(y2)26
x6 (y2)2
x6
y2
y2 x6
f1(x) 2 x6
; f1(x) is not a function.
12. Reflect the graph of yx2over the line yx.
Then, translate the new graph 1 unit to the left
and up 3 units.
13. f(x)
1
2
x5
y
1
2
x5
x
1
2
y5
x5
1
2
y
y2x10
f1(x) 2x10
[ff1](x) f(2x10)
1
2
(2x10)5
x
[f1f](x) f1
1
2
x5
2
1
2
x5
10
x
Since [ff1](x) [f1f](x) x, fand f1are
inverse functions.
14a. B(r) 1000(1 r)3
B1000(1 r)3
10
B
00
(1 r)3
3
10
B
00
1 r
r1
3B
10
1
3x
1
3x
f(x) x1
xf(x)
23
12
01
12
23
f1(x)
xf
1(x)
32
21
10
21
32
f
(
x
)
f
(
x
)
f
1(
x
)
x
O
5. She is wrong. The inverse is f1(x) (x3)22,
which is a function.
6.
7.
8.
f(x) x31
xf(x)
27
10
01
12
29
f1(x)
xf
1(x)
72
01
10
21
92
x
O
f
1(
x
)
f
(
x
)
f
(
x
)
f(x) (x3)21
xf(x)
13
20
31
40
53
f1(x)
xf
1(x)
31
02
13
04
35
x
O
f
1(
x
)
f
(
x
)
f
(
x
)
x
O
y
14b. r1
1 or 0.0323; 3.23%
Pages 156–158 Exercises
15.
16.
17.
31100
10
3B
10
77 Chapter 3
18.
19.
20.
f(x) x2
xf(x)
24
13
02
13
24
f1(x)
xf
1(x)
42
31
20
31
42
f
(
x
)
f
(
x
)
f
1(
x
)
x
O
f(x) 2x
xf(x)
24
12
00
12
24
f1(x)
xf
1(x)
42
21
00
21
42
f
(
x
)
f
(
x
)
f
1(
x
)
x
O
f(x) x32
xf(x)
210
13
02
11
26
f1(x)
xf
1(x)
10 2
31
20
11
62
x
O
f
1(
x
)
f
(
x
)
f
(
x
)
f(x) x510
xf(x)
242
111
010
19
222
f1(x)
xf
1(x)
42 2
11 1
10 0
91
22 2
x
O
f
1(
x
)
f
(
x
)
f
(
x
)
10
10
10
10
f(x) [x]
xf(x)
2 x12
1 x01
0 x10
1 x21
2 x32
f1(x)
xf
1(x)
22 x1
11 x0
00 x1
11 x2
22 x3
f
(
x
)
f
(
x
)
f
1(
x
)
x
O
f(x) 3
xf(x)
23
13
03
13
23
f1(x)
xf
1(x)
32
31
30
31
32
f
(
x
)
f
(
x
)
f
1(
x
)
x
O
21.
22.
23.
24. f(x) x24
yx24
xy24
x4 y2
yx4
; f1(x) x4
25. f(x) 2x7
y2x7
x2y7
x7 2y
y
x
2
7
f1(x)
x
2
7
; f1(x) is a function.
26. f(x) x2
yx2
xy2
yx2
f1(x) x2; f1(x) is a function.
27. f(x)
1
x
y
1
x
x
1
y
y
1
x
f1(x)
1
x
; f1(x) is a function.
28. f(x)
x
1
2
y
x
1
2
x
y
1
2
y2
1
x
y
1
x
f1(x)
1
x
; f1(x) is not a function.
29. f(x) (x3)27
y(x3)27
x(y3)27
x7 (y3)2
x7
y3
y3 x7
f1(x) 3 x7
; f1(x) is not a function.
Chapter 3 78
f(x) x22x4
xf(x)
37
24
13
04
17
f1(x)
xf
1(x)
73
42
31
40
71
x
O
f
1(
x
)
f
(
x
)
f
(
x
)
4
8
4
8
4488
f(x) (x2)25
xf(x)
49
36
25
16
09
f1(x)
xf
1(x)
94
63
52
61
90
f
(
x
)
f
(
x
)
f
1(
x
)
x
O
f(x) (x1)24
xf(x)
45
23
14
03
25
f1(x)
xf
1(x)
54
32
41
30
52
O
f
1(
x
)
f
(
x
)
f
(
x
)
f(x) x24
xf(x)
28
15
04
15
28
f1(x) x4
xf
1(x)
82
51
40
x
O
f
1(
x
)
f
(
x
)
f
(
x
)
4
4
4
4
35. Reflect the part of the graph of x2that lies in the
first quadrant about yx. Then, translate 5 units
to the left.
36. Reflect the graph of x2about yx. Then,
translate 2 units to the right and up 1 unit.
37. Reflect the graph of x3about yxto obtain the
graph of
3x
. Reflect the graph of
3x
about the
x-axis. Then, translate 3 units to the left and
down 2 units.
38. Reflect the graph of x5about yx. Then,
translate 4 units to the right. Finally, stretch the
translated graph vertically by a factor of 2.
79 Chapter 3
30. f(x) x24x3
yx24x3
xy24y3
x1 y24y4
x1 (y2)2
x1
y2
y2 x1
f1(x) 2 x1
; f1(x) is not a function.
31. f(x)
x
1
2
y
x
1
2
x
y
1
2
y2
1
x
y
1
x
2
f1(x)
1
x
2; f1(x) is not a function.
32. f(x)
(x
1
1)2
y
(x
1
1)2
x
(y
1
1)2
(y1)2
1
x
y1
1
x
y1
f1(x) 1 ; f1(x) is not a function.
33. f(x)
(x
2
2)3
y
(x
2
2)3
x
(y
2
2)3
(y2)3
2
x
y2
3
2
x
y2
3
2
x
f1(x) 2
3
2
x
; f1(x) is not a function.
34. g(x)
x2
3
2x
y
x2
3
2x
x
y2
3
2y
y22y
3
x
y22y1
3
x
1
(y1)2
3
x
1
y1
3
x
1
y1
3
x
1
g1(x) 1
3
x
1
1
x
1
x
x
O
f
(
x
)
x
O
f
(
x
)
f
(
x
)
x
O
x
O
f
(
x
)
39. f(x)
2
3
x
1
6
y
2
3
x
1
6
x
2
3
y
1
6
x
1
6
2
3
y
y
3
2
x
1
4
f1(x)
3
2
x
1
4
[ff1](x) f
3
2
x
1
4
2
3
3
2
x
1
4
1
6
x
1
6
1
6
x
[f1f](x) f1
2
3
x
1
6
3
2
2
3
x
1
6
1
4
x
1
4
1
4
x
Since [ff1](x) [f1f](x) x, fand f1are
inverse functions.
40. f(x) (x3)34
y(x3)34
x(y3)34
x4 (y3)3
3x4
y3
y3
3x4
f1(x) 3
3x4
[ff1](x) f(3
3x4
)
[(3
3x4
) 3]34
x4 4
x
[f1f](x) f1[(x3)34]
3
3[(x
3)3
4] 4
3 x3
x
Since [ff1](x) [f1f](x) x, fand f1are
inverse functions.
41a.
41b. No; the graph of d(x) fails the horizontal line test.
41c. d1(x) gives the numbers that are 4 units from x
on the number line. There are always two such
numbers, so d1associates two values with each
x-value. Hence, d1(x) is not a function.
42a. v2gh
v22gh
h
2
v
g
2
h
2(
v
3
2
2)
h
6
v
4
2
43a. Sample answer: yx
43b. The graph of the function must be symmetric
about the line yx.
43c. Yes, because the line yxis the axis of
symmetry and the reflection line.
44a.
44b. positive real numbers; positive multiples of 10
44c.
44d. positive multiples of 10; positive real numbers
44e. C1(x) gives the possible lengths of phone calls
that cost x.
45. It must be translated up 6 units and 5 units to the
left; y(x6)25; y6 x5
.
46a. KE
1
2
mv2
2KE mv2
2K
m
E
v2
v
2K
m
E
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 x3
y2 x3
x2 y3
x2 y3
(x2)2y3
y(x2)23
C1(x) (x2)23
Chapter 3 80
d(x) x4
xd(x)
62
51
40
31
22
d1(x)
xd
1(x)
26
15
04
13
22
d
1(
x
)
x
O
xC(x)
0 x1 $0.10
1 x2 $0.20
2 x3 $0.30
3 x4 $0.40
4 x5 $0.50
C
(
x
)
x
1
$0.10
$0.20
$0.30
$0.40
$0.50
$0.60
$0.70
$0.80
2345678
O
xC
1(x)
$0.10 0 x1
$0.20 1 x2
$0.30 2 x3
$0.40 3 x4
$0.50 4 x5
C
1(
x
)
x
20¢
1
2
3
4
5
6
7
8
40¢ 60¢ 80¢
O
46b. v
2K
m
E
v
2(1
1
5)
v5.477; 5.5 m/sec
42b. h
6
v
4
2
h
(7
6
5
4
)2
h87.89
Yes. The pump can propel
water to a height of about
88 ft.
81 Chapter 3
O
b
a
(6, 8)
(0, 0)
(0, 9)
(8, 0)
4
a
b
32
a
6
b
54
b
0
a
0
1
2
2
4
1
1
2
4
1
1
1
42
11
1
2
1
2
10
6
2
4
1
1
x
y
2
1
4
1
2
4
1
1
(1, 7)
52.
1
2
1
2
(3)
1
2
(6)
1
2
(9)
1
2
(6)
3
6
9
6
1
7
x
y
3
2
3
9
2
3
y
x
O
annx→p(x) →
positive even
positive even
positive odd
positive odd
annx→p(x) →
negative even
negative even
negative odd
negative odd
53.
54.
1
4
4;
1
4
4 1; neither
55. m
2
5
7
0
yy1m(xx1)
5
5
or 1y7 1(x0)
yx7
56. bc180
If P
Q
is perpendicular to Q
R
, then m∠PQR 90.
Since the angles of a triangle total 180,
ad90 180.
ad90
abcd180 90 or 270
The correct choice is C.
Continuity and End Behavior
Page 165 Check for Understanding
1. Sample answer: The function approaches 1 as x
approaches 2 from the left, but the function
approaches 4 as xapproaches 2 from the right.
This means the function fails the second condition
in the continuity test.
2.
3. Infinite discontinuity; f(x) →as x→,
f(x) →as x→.
4. f(x) x2is decreasing for x0 and increasing for
x0, g(x) x2is increasing for x0 and
decreasing for x0. Reflecting a graph switches
the monotonicity. In other words, if f(x) is
increasing, the reflection will be decreasing and
vice versa.
5. No; yis undefined when x3.
6. No; f(x) approaches 6 as xapproaches 2 from the
left but f(x) approaches 6 as xapproaches 2
from the right.
3-5
47d. C1(x) (x2)23
C1(1) (1 2)23 or 6, F
C1(2.899) (2.899 2)23 or 21, U
C1(2.123) (2.123 2)23 or 14, N
C1(0.449) (0.449 2)23 or 3, C
C1(2.796) (2.796 2)23 or 20, T
C1(1.464) (1.464 2)23 or 9, I
C1(2.243) (2.243 2)23 or 15, O
C1(2.123) (2.123 2)23 or 14, N
C1(2.690) (2.690 2)23 or 19, S
C1(0) (0 2)23 or 1, A
C1(2.583) (2.583 2)23 or 18, R
C1(0.828) (0.828 2)23 or 5, E
C1(1) (1 2)23 or 6, F
C1(2.899) (2.899 2)23 or 21, U
C1(2.123) (2.123 2)23 or 14, N
FUNCTIONS ARE FUN
48. Case 1 Case 2
2x 462x46
(2x4) 62x4 6
2x4 62x2
2x10 x1
x5
{x5 x1}
49. both
50a. a0, b0, 4ab32, a6b54
50b.
G(a, b) ab
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. 4x2y10 →4x2y10
y6 xxy6
10
6
x
y
2
1
4
1
24.
y→0 as x→, y→0 as x→.
25.
f(x) →2 as x→, f(x) →2 as x→.
26.
[6, 6] scl:1 by [30, 30] scl:5
increasing for x3 and x1; decreasing for
3 x1
27.
[7.6, 7.6] scl:1 by [5, 5] scl:1
decreasing for all x
Chapter 3 82
x
O
f
(
x
)
x
O
y
y
x
1
2
xy
10,000 1 108
1000 1 106
100 1 104
10 0.01
0undefined
10 0.01
100 1 104
1000 1 106
10,000 1 108
f(x)
x
1
3
2
xf(x)
10,000 2
1000 2.000000001
100 2.000001
10 2.001
0undefined
10 1.999
100 1.999999
1000 1.999999999
10,000 2
7. an: positive, n: odd
y→as x→, y→as x→.
8. an: negative, n: even
y→as x→, y→as x→.
9.
decreasing for x3; increasing for x3
10.
decreasing for x1 and x1; increasing for
1 x1
11a. t411b. when t411c. 10 amps
Pages 166–168 Exercises
12. Yes; the function is defined when x1; the
function approaches 3 as xapproaches 1 from
both sides; and y3 when x1.
13. No; the function is undefined when x2.
14. Yes; the function is defined when x3; the
function approaches 0 as xapproaches 3 from
both sides; and f(3) 0.
15. Yes; the function is defined when x3; the
function approaches 1 (in fact is equal to 1) as x
approaches 3 from both sides; and y1 when
x3.
16. No; f(x) approaches 7 as xapproaches 4 from
the left, but f(x) approaches 6 as xapproaches 4
from the right.
17. Yes; the function is defined when x1; f(x)
approaches 3 as xapproaches 1 from both sides;
and f(1) 3.
18. jump discontinuity
19. Sample answer: x0; g(x) is undefined when
x0.
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→.
33b. Since fis odd, its graph must be symmetric with
respect to the origin. Therefore, fis increasing
for 2 x0 and decreasing for x2. fmust
have a jump discontinuity when x3 and
f(x) →as x→.
34a. polynomial
34b.
[5, 80] scl:10 by [500, 12000] scl:1000
0.5 t39.5
34c. 0 t0.5 and t39.5
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.
[1, 8] scl:1 by [10, 1] scl:1
x4
36b. Answers will vary.
36c. 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.
36d. See graph in 36a. x4
36e. The slope is negative; see students’ work.
37a. 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.
83 Chapter 3
x
O
f
(
x
)
28.
[7.6, 7.6] scl:1 by [8, 2] scl:1
decreasing for x1 and x1
29.
[25, 25] scl:5 by [25, 25] scl:5
increasing for x1 and x5; decreasing for
1 x2 and 2 x5
30.
[7.6, 7.6] scl:1 by [1, 9] scl:1
decreasing for x2 and 0 x2; increasing
for 2 x0 and x2
31.
[7.6, 7.6] scl:1 by [1, 9] scl:1
decreasing for x
3
2
and 0 x
3
2
; increasing
for
3
2
x0 and x
3
2
32. As the denominator, r, gets larger, the value of
U(r) gets smaller. U(r) approaches 0.
33a. Since fis even, its graph must be symmetric
with respect to the y-axis. Therefore, fis
decreasing for 2 x0 and increasing for
x2. fmust have a jump discontinuity when
x3 and f(x) →as x→.
x
O
f
(
x
)
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.
38a.
38b. 0 x1, 1 x2, 2 x4, 4 x6,
6 x8, x8
39. For the function to be continuous at 2, bx aand
x2amust approach the same value as x
approaches 2 from the left and right, respectively.
Plugging in x2 to find that common value gives
2ba4 a. Solving for bgives b2. For the
function to be continuous at 2, b
x
and
bx amust approach the same value as x
approaches 2 from the left and right,
respectively. Plugging in x2 gives b
2
2ba. We already know b2, so the
equation becomes 0 4 a. Hence, a4.
40. f(x) (x5)2
y(x5)2
x(y5)2
x
y5
y5 x
f1(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) x2y
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
43.
5(2) 8(4) or 42
44a. c47.5h35
44b. c47.5h35
c47.52
1
4
35
c$141.875
45. f(x) 2x22x8
f(2) 2(2)22(2) 8
8 4 8 or 20
46. The volume of the cube is x3.
The volume of the other box is
x(x1)(x1) x(x21) or x3x.
The difference between the volumes of the two
boxes is x3(x3x) or x.
The correct choice is A.
4
2
5
8
Gap Discontinuities
Page 170
1. {all real numbers xx3}
[10, 10] scl:1 by [6, 50] scl:10
2. {all real numbers x2 x4}
[9.4, 9.4] scl:1 by [6.2, 6.2] scl:1
3. {all real numbers xx3 or x1}
[18.8, 18.8] scl:1 by [12.4, 12.4] scl:1
4. {all real numbers xx3 or x2}
[4.7, 4.7] scl:1 by [25, 25] scl:10
5. {all real numbers xx1 or x1}
[4.7, 4.7] scl:1 by [3.1, 3.1] scl:1
3-5B
Chapter 3 84
5
020 46810121416
10
15
20
25
30
35
40
Percent
with
Similar
Computer
Usage
function, then the graph of y
is like the graph of f(x), but with an infinite
number of “interval bites” removed.
13. Yes; sample justification: the equation y
is a possible equation for
the function described.
14a.
[15, 15] scl:2 by [10, 20] scl:2
14b.
[9.1, 9.1] scl:1 by [6, 6] scl:1
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 xapproaches 1 from the left and
increasing as xmoves away from 1 to the right.
Therefore, on the interval 0.99 x1.01, (1, 4) is
a relative minimum.
3-6
x2(x2) (2x4)(x4)
((x2) or (x4))
f(x)
(x[[x]]0.25)
12. Yes; sample justification: if f(x) is a polynomial
85 Chapter 3
6. {all real numbers xx6 or x2}
[9.4, 9.4] scl:1 by [6.2, 6.2] scl:1
7. {all real numbers xx3 or x4}
[9.4, 9.4] scl:1 by [3, 9.4] scl:1
8. {all real numbers xx2 or 1 x1
or x2}
[4.7, 4.7] scl:1 by [2, 8] scl:1
9. {all real numbers xx1 or 2 x3 or x4}
[3, 6.4] scl:1 by [2, 8] scl:1
10. {all real numbers xx1 or x2}
[9.4, 9.4] scl:1 by [6.2, 6.2] scl:1
11. Sample answer: y
x2
((x2) or ((x5) and (x7)) or x8))
3. Sample answer:
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)
[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)
12b. 2 weeks
12c. $58.80 per acre
12d. Rain or other bad weather could delay harvest
and/or destroy part of the crop.
Pages 177–179 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)
[5, 5] scl:1 by [8, 2] scl:1
20. rel. max.: (1.53, 1.13);
rel. min.: (1.53, 13.13)
[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)
[5, 5] scl:1 by [8, 2] scl:1
Chapter 3 86
y
x
O
(3, 1)
(0, 4)
(4, 6)
x
Time (weeks)
024681012141618
Profit
(dollars)
10
0
20
30
40
50
60
70
80
P
(
x
)
23. rel. max.: (1, 1); rel. min.: (0.25, 3.25)
[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)
[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
f
2
3
2.85
f(0.77) 2.86 min.
34. The point of inflection is now at x6 and there
is now a minimum at x3.
35a. V(x) 2x(12.5 2x)(17 2x)
35b.
[1, 12] scl:1 by [200, 500] scl:100
2.37 cm by 2.37 cm
35c. See students’ work.
36a. Psd 25d
s(200s15,000) 25(200s15,000)
200s220,000s375,000
[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 2MB2AB2
AM 2x222
AM x24
f(x) 5000(x24
) 3500(10 x)
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
38. equations of the form yxnor y
nx
, where nis
odd
39.
[3, 7] scl:1 by [50, 10] scl:10
The particle is at rest when t0.14 and when
t3.52. Its positions at these times are
s(0.14) 8.79 and s(3.52) 47.51.
87 Chapter 3
Chapter 3 88
x
O
y
O
y
x
(10, 190)
(120, 80)
(10, 80)
x
y
200
100
100
200
200
x
10
y
80
O
y
x
C
(2, 1)
D
(3, 1)
B
(2, 4)
A
(3, 4)
46. 3A3
42
57
or
12 6
15 21
3(4) 3(2)
3(5) 3(7)
2B2
35
43
or
610
86
2(3) 2(5)
2(4) 2(3)
3A2B
610
86
12 6
15 21
12 (6) 6 10
15 (8) 21 6
64
727
y
x
O
Q
T
X
P
RYS
47. Let xnumber of 1-point free throws.
Let ynumber of 2-point field goals.
Let znumber of 3-point field goals.
1x2y3z32
xyz17
y0.50(18)
1x2y3z32 →1x2y3z32
1(xyz17) xyz17
y2z15
y0.50(18) y2z15
y99 2z15
z3
xyz17
x9 3 17
x55 free throws, 9 2-point field
goals, 3 3-point field goals
48. y6 4
y2
49. 2x3y15 →y
2
3
x5
6x4y16 →y
3
2
x4
2
3
3
2
1; perpendicular
50. Arelation 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.
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 x5.
42.
43. Let xunits of notebook paper.
Let yunits of newsprint.
xy200
x10
y80
P(x, y) 400x350y
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
44.
3 x2, 1 y4
45.
1(5) 2(3) or 1; yes
13
25
89 Chapter 3
x
O
f
(
x
)
y
x
O
y
x
O
y
x
O
x
O
y
4
8
4
48
8
48
x
O
y
x
O
yy
x
O
V
O
P
6. x2, x1y
(x2
x
)(
3
x1)
y
x2
x
x
3
2
y
y
no horizontal
asymptotes
7. f(x)
x
1
1
2
8. The parent graph is translated 4 units right. The
vertical asymptote is now at x4. The horizontal
asymptote, y0, is unchanged.
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 y1.
10. 3x5
x33x2
4x
5
→3x5
x
2
0
3
3x29x
5x5
5x15
20
y3x5
11. 12.
13a.
1
1
x
x
1
2
x
2
3
x
x
3
3
x
x
2
3
x
x
3
x
2
3
Graphs of Rational Functions
Pages 185–186 Check for Understanding
1.
1a. x2, y6
1b. y
x
1
2
6
2. Sample graphs:
Vertical Asymptote Horizontal Asymptote
Slant Asymptote
3. Sample answer: f(x)
x(
x
x
1
1)
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. x5
f(x)
x
x
5
y
x
x
5
y(x5) x
xy 5yx
xy x5y
x(y1) 5y
x
y
5
y
1
; y1
3-7
13b. P0, V0
13c. The pressure approaches 0.
Pages 186–188 Exercises
14. x4f(x)
x
2
x
4
y
x
2
x
4
y(x4) 2x
xy 4y2x
xy 2x4y
x(y2) 4y
x
y
4y
2
; y2
15. x6y
x
x
2
6
y
x
x
2
2
x
x
2
2
x
6
2
Chapter 3 90
y
no horizontal
asymptote
16. x
1
2
, x5y
(2x
x
1
)(x
1
5)
y
2x2x
9x
1
5
y
y; y0
17. x1, x3y
x2
x
4x
2
3
y
y; y0
18. no vertical asymptote, y
x2
x
2
1
y
x
x
2
2
x
x
2
2
x
1
2
1
x
x
2
2
1
4
x
x
3
2
x
x
2
x
2
2
x
x
2
2
4
x
x
2
x
3
2
1
x
x
1
2
2
9
x
x
5
2
x
x
2
x
1
2
2
x
x
2
2
9
x
x
2
x
5
2
1
1
x
x
6
2
y; y1
1
1
x
1
2
x
O
y
x
O
y
4
8
48
4
8
48
19. x1y
(x
x2
1
1
)2
y
x2
x2
2
x
1
1
y
y; y1
20. x2
y
(x
x3
2)4
y
y
y; y0
21. f(x)
x
1
3
1
22. f(x)
x
1
2
3
23. f(x)
1
x
1
24. The parent graph is translated 3 units up. The
vertical asymptote, x0, is unchanged. The
horizontal asymptote is now y3.
25. The parent graph is translated 4 units right and
expanded vertically by a factor of 2. The vertical
asymptote is now x4. The horizontal asymptote,
y0, is unchanged.
1
x
1
8
x
2
x
4
2
3
x
2
3
1
x
6
4
x
x
3
4
x
x
4
4
8
x
x
4
3
24
x4
x2
3
x
2
4
x
1
x
6
4
x3
x48x324x232x16
1
2
x
x
1
2
1
x
1
2
x
x
2
2
2
x
x
2
x
1
2
x
x
2
2
x
1
2
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 x3 and the
horizontal asymptote is now y1.
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,
x0, is unchanged. The horizontal asymptote is
now y2.
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 x3 and the horizontal asymptote
is y3.
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 x5 and the horizontal
asymptote is y4.
91 Chapter 3
x
O
y
4
4
44
x
O
y
4
8
44
4
8
88
48124
4
4
8
12
8
12
x
O
y
x
O
y
4
8
12
16
4
8
12
16
4
8
4
x
O
yy
x
(2, 4)
O
x
2, 1
4
)(
O
y
x
(2, 0)
O
y
x
(3, 0)
O
y
x
O
y
30. x1
x4x23
x3
→x1
x
1
4
x24x
x3
x4
1yx1
31. x3
xx23
x4
→x3
4
x
x2
3x4
3x
4yx3
32. x2
x21x32
x2x
4
→x2
x22
1
x3x
2x24
2x22
2yx2
33.
1
2
x
5
4
2x3x24
x1
→
1
2
x
5
4
x2
3
2
x
5
2
x1
5
2
x
1
4
5
1
4
1
y
1
2
x
5
4
34. No; the degree of the numerator is 2 more than
that of the denominator.
35. 36.
37. 38.
39. 40.
1
4
1
2x3
41a. C(t)
48
4
0
0
3
t
t
41b. C(t)
48
4
0
0
3
t
t
10
48
4
0
0
3
t
t
400 10t480 3t
7t80
t11.43 L
42. Sample answer: The circuit melts or one of the
components burns up.
43. To get the proper x-intercepts, x2 and x3
should be factors of the numerator. The vertical
asymptote indicates that x4 should be a factor
of the denominator. To get point discontinuity
at (5, 0), make x5 a factor of both the
numerator and denominator with a bigger
exponent in the numerator. Thus, a sample
answer is f(x) .
44a. Vx2hA(x) 4xh2x2
120 x2hA(x) 4x
1
x
2
2
0
2x2
1
x
2
2
0
hA(x)
48
x
0
2x2
44b.
44c. The surface area approaches infinity.
45. If the degree of the denominator is larger than
that of the numerator, then y0 will be a
horizontal asymptote. To make the graph intersect
the x-axis, the simplest numerator to use is x.
Thus, a sample answer is f(x)
x2x
1
.
46a. Avertical asymptote at r0 and a horizontal
asymptote at F0.
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.
a
a
2
3
9
47b.
The slope approaches 6.
(x2)(x3)(x5)2
(x4)(x5)
48. abs. max.: (2, 1)
[1, 6] scl:1 by [5, 2] scl:1
49. x29 y
y29 x
y2x9
yx9
50. f(x, y) yx
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
51. 4
4(5)
4(4)
4(6)
4(8)
5
4
6
8
Chapter 3 92
A
(
x
)
280
260
240
320
300
220
200
180
160
140
24681012
14 1618
x
O
x2.9 2.99 3 3.01 3.1
m5.9 5.99 — 6.01 6.1
52. Let xprice of film and yprice of sunscreen.
8x2y35.10
3xy14.30 8x2y35.10
y14.30 3x8x2(14.30 3x) 35.10
2x28.60 35.10
x3.25
y14.30 3x
y14.30 3(3.25)
y4.55
$3.25; $4.55
53. xy3xy3
0 0
?33 2
?3
0 3no 5 3yes
xy3xy3
4 2
?32 4
?3
2 3no 2 3no
(3, 2)
54. 15yx1 →y
1
1
5
x
1
1
5
55. [fg](x) f(g(x))
f(2 x2)
8(2 x2)
16 8x2
[gf](x) g(f(x))
g(8x)
2 (8x)2
2 64x2
20
16
24
32
56. Let xthe width of each card and ythe height
of each card. The rectangle has a base of 4xor 5y.
The rectangle has a height of xy.
Abh
180 4x(xy)4x5y
180 4x
x
4
5
x
y
4
5
x
180
36
5
x2
y
4(
5
5)
25 x2y4
5 x
Perimeter 2(4x) 2(xy)
P2(4 5) 2(5 4)
P58 in.
The correct choice is B.
Direct, Inverse, and Joint Variation
Pages 193–194 Check for Understanding
1a. inverse
1b. neither
1c. direct
2. Sample answer:
Suppose yvaries directly as xn.
Then y1kx1nand y2kx2n
y1kx1n
y
y
1
2
k
k
x
x
1
2
Division property of equality.
y
y
1
2
x
x
1
2
Simplify.
3. The line does not go through the origin, therefore
its equation is not of the form ykxn.
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 kxy12
4(3) k15y12
12 ky
4
5
6. ykx2y
2
3
x2
54 k(9)2y
2
3
(6)2
2
3
ky24
7. ykxz3y0.5xz3
16 k(4)(2)3y(0.5)(8)(3)3
0.5 ky108
8. y
k
w
x2
z
y
0.
w
42
xz
3
k(3
2
)(
2
10)
y
0.4(
4
42
)(20)
0.4 ky2
9. yvaries directly as x4;
1
7
.
10. Avaries jointly as and w; 1
11. yvaries inversely as x; 3.
3-8
12a. Vkhg2
288 k(40)(1.5)2
3.2 k
V3.2hg2
12b. V3.2hg2
V3.2(75)(2)2
V960
50 960 48,000 m3
Pages 194–196 Exercises
13. ykx y 0.2x
0.3 k(1.5) y0.2(6)
0.2 ky1.2
14. xy kxy50
25(2) kx(40) 50
50 kx1.25
15. ykxz y 15xz
36 k(1.2)(2) y15(0.4)(3)
15 ky18
16. x2ykx
2y36
(2)2(9) k32y36
36 ky4
17. rkt2r16t2
4 k
1
2
2r16
1
4
2
16 kr1
18. x
yk
1.21
(0.44) k
0.484 k
x
y0.484 or y0.484 1
x
93 Chapter 3
y0.484
y1.21
19. ykx3z2y
1
1
2
x3z2
9 k(3)3(2)2y
1
1
2
(4)3(3)2
1
1
2
ky48
20. y
k
z2
x
y
0
z
.3
2x
1
6
k(
6
2
2
0)
y
0.3
5
(2
14)
0.3 ky0.168
21. y
k
w
xz
y
2
w
xz
3
k(2)
4
(3)
y
2(4
)(
4
7)
2 ky14
22. y
k
x
z
3
2
y
x
2
2
z2
6
k(
3
9
3
)2
y
2(
6
34)2
2 ky
2
4
7
23. a
k
c
b2
a
15
c
b2
45
k
1
(6
2
)2
96
1
1
5
0
b2
15 k8 b
24. x2ykyx
232
(4)2(2) k8x232
32 kx2
1
0.16
25. Cvaries directly as d; .
26. yvaries directly as x;
1
4
.
27. yvaries jointly as xand the square of z;
4
3
.
28. Vvaries directly as the cube of r;
4
3
.
29. yvaries inversely as the square of x;
5
4
.
30. yvaries inversely as the square root of x; 2.
31. Avaries jointly as hand the quantity b1b2; 0.5.
32. yvaries directly as xand inversely as the square
of z;
1
3
.
33. yvaries directly as x2and inversely as the cube of
z; 7.
34. yvaries jointly as the product of the cube of xand
zand 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. T1km1d1and T2km2d2
T1T2
km1d1km2d2Substitution property
m1d1m2d2of equality
35c. m1d1m2d2
75(3.3) (125)d2
1.98 d2; 1.98 meters
36a. tr k
36b. tr ktr36,000
45(800) kt(1000) 36,000
36,000 kt36 minutes
37. If yvaries directly as xthen there is a nonzero
constant ksuch that ykx. Solving for x, we find
x
1
k
y
1
k
is a nonzero constant, so xvaries
directly as y.
38a. I
d
k
2
38b. I
d
k
2
a2b2c2I
5
d
7
2
6
16
6
k
2
(6)2(25)2c2I
(6
5
.
7
5
6
)2
576 k6.5 cI13.6 lux
39. ais doubled
a
k
c
b
3
2
a
a
a2
k
c
b
3
2
40a. FG
m1
d
2
m2
1
4
kb2
1
8
c3
k
1
2
b
2
1
2
c
3
Chapter 3 94
86422468
2
4
6
8
2
4
6
8
y
x
O
y
x
O
40b. FG
m1
d
2
m2
1.99 1020 G
6.67 1011 G; 6.67 1011
N
k
g
m
2
2
40c. FG
m1
d
2
m2
(6.67 1011)
3.53 1022 N
40d. 3.53 1022 (1.99 1020)x
178 x; about 178 times greater
41. R
k
r
L
2
R
1.6
8
(
0.0
1
0
0
3
)2
8(3)
1.07 102
(0
k
.0
0
2
1)2
R1.78 103
1.68 108k
42.
43. f(x) (x3)36
y(x3)36
x(y3)36
x6 (y3)3
3x6
y3
y
3x6
3
f1(x)
3x6
3; f1(x) is a function.
44.
A(1, 2), B(
2), C(1, 4), D(3, 0)
45. 4x2y7→y2x
7
2
12x6y21 →y2x
7
2
consistent and dependent
46.
47. m
2
1
0
8
0
.6
0
2
1
3
9
.
9
2
5
y18.6 0.92(x2000)
4
5
.6
or 0.92 y0.92x1858.60
48. 144 429 or 1221
12 is divisible by 3, 4, 6, and 12.
3
0
1
4
3
2
1
2
3
0
1
4
3
2
1
2
0
1
1
0
(5.98 1024)(1.99 1030)
(1.50 1011)2
(5.98 1024)(7.36 1022)
(3.84 108)2
The correct choice is D.
Chapter 3 Study Guide and Assessment
Page 197 Understanding and Using the
Vocabulary
1. even 2. continuous 3. point
4. decreasing 5. maximum 6. rational
7. inverse 8. monotonic 9. slant
10. Joint
Pages 198–200 Skills and Concepts
11. f(x) 2(x)f(x) (2x)
f(x) 2xf(x) 2xyes
12. f(x) (x)22f(x) (x22)
f(x) x22f(x) x22no
13. f(x) (x)2(x) 3
f(x) x2x3
f(x) (x2x3)
f(x) x2x3no
14. f(x) (x)36(x) 1
f(x) x36x1
f(x) (x36x1)
f(x) x36x1no
15. xy 4→ab 4
x-axis a(b) 4
ab 4no
y-axis (a)b4
ab 4no
yx(b)(a) 4
ab 4yes
yx(b)(a) 4
ab 4yes
yxand yx
16. xy24→ab24
x-axis a(b)24
ab24yes
y-axis (a) b24
ab24no
yx(b) (a)24
a2b4no
yx(b) (a)24
95 Chapter 3
y
x
O
y
x
O
y
x
O
y
x
O
a2b4no
x-axis
17. x2y→a2b
x-axis a2(b)
a2bno
y-axis (a) 2b
a2bno
yx(b) 2(a)
b2ano
yx(b) 2(a)
b2ano; none
18. x2
1
y
→a2
1
b
x-axis a2
(
1
b)
a2
1
b
no
y-axis (a)2
1
b
a2
1
b
yes
yx(b)2
(a
1
)
b2
a
1
no
yx(b)2
(
1
a)
b2
a
1
no; y-axis
19. The graph of g(x) is a translation of the graph of
f(x) up 5 units.
20. The graph of g(x) is a translation of the graph of
f(x) left 2 units.
21. The graph of g(x) is the graph of f(x) expanded
vertically by a factor of 6.
22. The graph of g(x) is the graph of f(x) expanded
horizontally by a factor of
4
3
and translated down
4 units.
23. 24.
25. 26.
27. Case 1 Case 2
4x574x57
(4x5) 74x5 7
4x5 74x2
Chapter 3 96
f(x) 3x1
xf(x)
27
14
01
12
f1(x)
xf
1(x)
72
41
10
21
f
(
x
)
f
(
x
)
f
1(
x
)
x
O
f(x)
1
4
x5
xf(x)
25.5
15.25
05
14.75
24.5
f1(x)
xf
1(x)
5.5 2
5.25 1
50
4.75 1
4.5 2
f
(
x
)
f
(
x
)
f
1(
x
)
x
24
4
6
2
4
6
646
O
y
x
O
f
1(
x
)
f
(
x
)
f(x) (x1)24
xf(x)
30
23
14
03
10
f1(x)
xf
1(x)
03
32
41
30
01
x
O
f
1(
x
)
f
(
x
)
f
(
x
)
y
x
1
2
1
xy
1000 1.000001
100 1.0001
10 1.01
12
10 1.01
100 1.0001
1000 1.000001
4x12 x0.5
x3
{xx3 or x0.5}
28. Case 1 Case 2
x32 11 x32 11
(x3) 2 11 x3 2 11
x5 11 x12
x6
x6
{x6 x12}
29.
30.
31.
32.
33. f(x) (x2)38
y(x2)38
x(y2)38
x8 (y2)3
3x8
y2
y
3x8
2
f1(x)
3x8
2; yes
34. f(x) 3(x7)4
y3(x7)4
x3(y7)4
3
x
(y7)4
4
3
x
y7
y7
4
3
x
f1(x) 7
4
3
x
; no
35. Yes; the function is defined when x2; the
function approaches 6 as xapproaches 2 from both
sides; and y6 when x2.
36. No; the function is undefined when x1.
37. Yes; the function is defined when x1; the
function approaches 2 as xapproaches 1 from both
sides; and y2 when x1.
38. an: negative, n: odd
y→as x→, y→as x→.
39. an: positive, n: odd
y→as x→, y→as x→.
40.
y→1 as x→, y→1 as x→.
41. an: positive, n: odd
y→as x→, y→as x→.
f(x)
2
x
3
xf(x)
32.3
22
11
1
2
1
0—
1
2
7
15
24
33.7
f1(x)
xf
1(x)
2.3 3
22
11
1
1
2
7
1
2
51
42
3.7 3
42.
[5, 5] scl:1 by [20, 10] scl:5
decreasing for x2 and x 1;
increasing for 2 x1
43.
[6, 6] scl:1 by [5, 20] scl:5
decreasing for x3 and 0 x3;
increasing for 3 x0 and x3
44. abs. max.: (2, 1)
45. rel. max.: (0, 4), rel. min.: (2, 0)
46. f(2.9) 0.029
f(3) 0
f(3.1) 0.031 min.
47. f(0.1) 6.996
f(0) 7
f(0.1) 7.004 pt of inflection
48. f(x)
1
x
1
49. f(x)
2
x
50. The parent graph is translated 2 units left and
expanded vertically by a factor of 3. The vertical
asymptote is now x3. The horizontal
asymptote, f(x) 0, is unchanged.
51. The parent graph is translated 3 units right and
then translated 2 units up. The vertical asymptote
is now x3 and the horizontal asymptote is
f(x) 2.
52. x1y
x
x
1
y
x
x
x
x
1
x
97 Chapter 3
x
O
f
(
x
)
x
O
f
(
x
)
y;y1
53. x2y
x
x
2
2
1
y
x
x
2
2
x
1
2
x
x
2
x
2
2
1
1
1
x
y
no horizontal asymptotes
54. x3, y
(x
x2
3
9
)2
y
x2
x26
x
9
9
y
x
x
2
2
6
x
x
2
x
9
2
x
x
2
2
x
9
2
1
x
1
2
1
x
x
2
2
y; y1
55. x2
xx22
x1
→x2
1
x
x2
2x
2x
1yes; yx2
56. ykxz y 0.625xz
5 k(4)(2) y0.625(6)(3)
0.625 ky11.25
57. yy
20 10
140 kx
14
x196
58. y
k
z
x2
y
32
z
0x2
7.2
k(0
4
.3)2
y
32
4
0
0
(1)2
320 ky8
Page 201 Applications and Problem Solving
59. x6.50.2;
Case 1 Case 2
x6.50.2 x6.50.2
(x6.5) 0.2 x6.5 0.2
x6.5 0.2 x6.7
x6.3
x6.3 6.3 x6.7
140
x
k
49
140
x
k
x
1
6
x
x
9
2
1
x
9
2
y
x
O
Page 201 Open-Ended Assessment
1a. Sample answer: xy2
1b. Sample answer: yx2
1c. Sample answer: xy 1
1d. Sample answer: xy 1
1e. Sample answer: yx3
2. Sample answer: 2(x4)21
Chapter 3 98
xC(x)
0 x10.40
1 x20.80
2 x31.20
3 x41.60
4 x52.00
5 x62.40
xC
1(x)
0.40 0 x1
0.80 1 x2
1.20 2 x3
1.60 3 x4
2.00 4 x5
2.40 5 x6
0.40
0
0.80
1.20
1.60
2.00
2.40
2.80
3.20
3.60
Cost
(dollars)
10643
Time (min)
59782
1
0
2
3
4
5
6
7
8
9
Time
(min)
0.400 1.20
Cost (dollars)
2.00 3.602.80
t
O
h
(
t
)
1
0.5
0.5 1
60a.
60b. positive real numbers;
positive multiples of $0.40
60c.
60d. positive multiples of $0.40;
positive real numbers
60e. C1(x) gives the possible number of minutes
spent using the scanner that cost xdollars.
61a.
61b. 1.08 m
y
x
O
y
x
O
y
x
O
y
x
O
3a. Sample answer:
3b. abs. min.: (2, 3); rel. max.: (0, 0); rel. min.:
(2, 1)
Chapter 3 SAT & ACT Preparation
Page 203 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, a2b2.
Factor it.
3
y
y
2
9
9
(y
3y
3)
(y
9
3)
Factor the denominator. Both the numerator and
denominator contain the factor (y3). Simplify
the fraction.
(y
3y
3)
(y
9
3)
(y
3(y
3)
(y
3
)
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 xyz, except for
the numbers. Multiply both sides of the equation
xyzby 2.
2(xy) 2zor 2x2y2z
So choice A is true. For choice B, start with xy
and subtract yfrom each side.
xyyy0
So choice B is true. For choice C, start with xy
and subtract zfrom each side.
xzyz
So choice C is true. For choice D, substitute yfor x
and xyfor z.
x
2
z
y
x
2
y
2
2
y
y
So choice D is also true. For choice E, write each
side of the equation in terms of y.
zy(xy) xy
2x2y
y2y
So choice E is not true. The correct choice is E.
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.
miles
m
pe
il
r
es
gallon
gallons
4
2
5
5
0
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.
Afaster 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.
4.
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.
9
1
0
0
0
1
9
0
0
0
10
9
00
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.
(10x4x22x8) (3x43x32x9)
(10x43x4) (3x3) (x2) (2x2x)
(8 9)
7x43x3x20 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 nin the answer choices.
Find the expression that has a remainder of 7.
Choice A:
n
8
1
21
8
1
2
8
2
2R6
The remainder is 6.
Choice B:
n
8
2
21
8
2
2
8
3
2R7
You could also reason that since ndivided by 8 has
a remainder of 5, then (n2) divided by 8 will
have a remainder of (5 2) or 7. The correct
choice is B.
99 Chapter 3
y
x
(0, 0)
(2, 1)
(2, 3)
O
O
y
x
8.Simplify the expression inside the square root
symbol. Factor 100 from each term. Then factor
the trinomial.
10
x
(x
3
3)
10
The correct choice is B.
9. Since a b c, substitute abfor cin
ac 5. So, a(ab) 5. Then b5 or
b5. Substitute 5 for bin bc3. So,
5 c3. Then c8 or c8.
The correct choice is B.
10(x3
)2
x3
10(x3
)(x
3)
x3
10x26
x9
x3
100(x
26x
9)
x3
100x2
60
0x9
00
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.
4x2y24
2xy12
Now simplify the second equation. Multiply both
sides by 2x.
7
2
y
x
7
7y7(2x)
7y14x
Divide both sides by 7.
y2x
You need to find the value of x. Substitute 2xfor y
in the first equation.
2xy12
2x(2x) 12
4x12
x3
The answer is 3.
Chapter 3 100
y
x
O
Polynomial Functions
Pages 209–210 Check for Understanding
1. Azero 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. Acomplex number is any number in the form
abi, where aand bare real numbers and iis
the imaginary unit. In a pure imaginary number,
a0 and b0. Examples: 2i, 3i;
Nonexamples: 5, 1 i
4.
5. 3; 1 6. 5; 8
7. no; f(x) x35x23x18
f(5) (5)35(5)23(5) 18
f(5) 125 125 15 18
f(5) 33
8. yes; f(x) x35x23x18
f(6) (6)35(6)23(6) 18
f(6) 216 180 18 18
f(6) 0
9. (x(5))(x7) 0
(x5)(x7) 0
x22x35 0; even; 2
10. (x6)(x2i)(x(2i))(xi) (x(i)) 0
(x6)(x2i)(x2i)(xi)(xi) 0
(x6)(x24i2)(x2i2) 0
(x6)(x24)(x21) 0
(x36x24x24)(x21) 0
x56x45x330x24x24 0;
odd; 1
4-1 11. 2; x214x49 0
(x7)(x7) 0
x7 0x7 0
x7x7
12. 3; a32a28a0
a(a22a8) 0
a(a4)(a2) 0
a0a4 0a2 0
a4a2
13. 4; t41 0
(t21)(t21) 0
(t1)(t1)(t21) 0
t1 0t1 0t21 0
t1t1t21
ti
14a. x2r262V(x) Bh
r236 x2V(x) (36 x2)(2x)
14b. V(x) (36 x2)(2x)
V(x) (36x2)(2x)
V(x) 72x2x3
14c. V(x) 72x2x3
V(4) 72(4) 2(4)3
V(4) 502.65 units3
Pages 210–212 Exercises
15. 4; 5 16. 7; 3 17. 3; 5 18. 5; 25
19. 6; 120. 2; 1
21. Yes; the coefficients are complex numbers and the
exponents of the variable are nonnegative
integers.
101 Chapter 4
Chapter 4 Polynomial and Rational Functions
f
(
x
)
x
O
60
20
40
f
(
x
)
x
2 14
x
49
(0, 49)
(7, 0)
1248
f
(
a
)
a
O
30
10
10
20
f
(
a
)
a
2 2
a
2 8
a
(0, 0)(4, 0) (2, 0)
2424
f
(
t
)
t
O
f
(
t
)
t
4 1
(1, 0)(1, 0)
22. No;
a
1
a1, which is a negative exponent.
23. yes; f(a) a413a212a
f(0) (0)413(0)212(0)
f(0) 0
24. no; f(a) a413a212a
f(1) (1)413(1)212(1)
f(1) 1 13 12
f(1) 24
25. yes; f(a) a413a212a
f(1) (1)413(1)212(1)
f(1) 1 13 12
f(1) 0
26. yes; f(a) a413a212a
f(4) (4)413(4)212(4)
f(4) 256 208 48
f(4) 0
27. no; f(a) a413a212a
f(3) (3)413(3)212(3)
f(3) 81 117 36
f(3) 72
28. yes; f(a) a413a212a
f(3) (3)413(3)212(3)
f(3) 81 117 36
f(3) 0
29. f(b) b43b22b4
f(2) (2)43(2)22(2) 4
f(2) 16 12 4 4
f(2) 12; no
30. f(x) x44x3x24x
f(1) (1)44(1)3(1)24(1)
f(1) 1 4 1 4
f(1) 0; yes
31a. 3; 1 31b. 2; 2 31c. 4; 2
32. (x(2))(x3) 0
(x2)(x3) 0
x2x6 0; even; 2
33. (x(1))(x1)(x5) 0
(x1)(x1)(x5) 0
(x21)(x5) 0
x35x2x5 0; odd; 3
34. (x(2))(x(0.5))(x4) 0
(x2)(x0.5)(x4) 0
(x22.5x1)(x4) 0
x34x22.5x210xx4 0
x31.5x29x4 0
2x33x218x8 0; odd; 3
35. (x(3))(x(2i))(x2i) 0
(x3)(x2i)(x2i) 0
(x3)(x24i2) 0
(x3)(x24) 0
x33x24x12 0; odd; 1
36. (x(5i))(x(i))(xi)(x5i) 0
(x5i)(xi)(xi)(x5i) 0
(x5i)(x5i)(xi)(xi) 0
(x225)(x21) 0
x426x225 0; even; 0
Chapter 4 102
f
(
x
)
x
O
f
(
x
)
x
8
(8, 0)
(0, 8)
f
(
a
)
a
O
10 2010
20
40
60
80
20
f
(
a
)
a
2
81
(0, 81)
(9, 0)(9, 0)
f
(
b
)
b
O
242
60
80
40
20
4
f
(
b
)
b
2
36 (0, 36)
f
(
t
)
t
O
4
12
4
f
(
t
)
t
3
2
t
2 4
t
8
(2, 0) (2, 0)
(0, 8)
2424
37. (x(1))(x1)(x4)(x(4)(x5) 0
(x1)(x1)(x4)(x4)(x5) 0
(x21)(x216)(x5) 0
(x417x216)(x5) 0
x55x417x385x216x80 0; odd; 5
38. (x(1))(x1)(x3)(x(3)) 0
(x1)(x1)(x3)(x3) 0
(x21)(x29) 0
x410x29 0
39. 1; x8 0
x8
40. 2; a281 0
(a9)(a9) 0
a9 0a9 0
a9a9
41. 2; b236 0
b236
b6i
42. 3; t32t24t8 0
t2(t2) 4(t2) 0
(t2)(t24) 0
(t2)(t2)(t2) 0
t2 0t2 0t2 0
t2t2t2
103 Chapter 4
f
(
n
)
n
O
20
10
20
10
f
(
n
)
n
3 9
n
(3, 0)(0, 0)(3, 0)
2424
f
(
c
)
c
O
100
50
100
50
f
(
c
) 6
c
3
3
c
2 45
c
(3, 0)(0, 0)
(2.5, 0)
2424
f
(
a
)
a
O
f
(
a
)
a
4
a
2 2
(1, 0)(1, 0)
(0, 2)
f
(
x
)
x
O
20
10
20
10
f
(
x
)
x
4 10
x
2 9
(3, 0)(1, 0) (1, 0) (3, 0)
(0, 9)
2424
f
(
m
)
m
O
f
(
m
) 4
m
4 17
m
2 4
(0, 4)
f
(
u
)
u
O
f
(
u
)
(
u
1)(
u
2 1)
(1, 0) (1, 0)
(0, 1)
y
x
O
y
x
O
y
x
O
y
x
O
y
x
O
47. 4; 4m417m24 0
(4m21)(m24) 0
4m1 0m24 0
m
4
1
m4
m0.5im2i
48. (u1)(u21) 0
(u1)(u1)(u1) 0
u1 0u1 0u1 0
u1u1u1
49a. 49b.
49c. 49d.
49e. 49f. not possible
43. 3; n39n0
n(n29) 0
n(n3)(n3) 0
n0n3 0n3 0
n3n3
44. 3; 6c33c245c0
c(6c23c45) 0
c(c3)(6c15) 0
c0c3 06c15 0
c3c
1
6
5
c2.5
45. 4; a4a22 0
(a22)(a21) 0
(a22)(a1)(a1) 0
a22 0a1 0a1 0
a22a1a1
a2
i
46. 4; x410x29 0
(x29)(x21) 0
(x3)(x3)(x1)(x1) 0
x3 0x3 0x1 0x1 0
x3x3x1x1
50.
[5, 5] sc11 by [2, 8] sc11
50a. 450b. 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) 99,000x355,000x265,000x
51b. r0.15 x1 r
x1 0.15
x1.15
V(x) 99,000x355,000x265,000x
V(1.15) 99,000(1.15)355,000(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. d(t)
1
2
at2d(t)
1
2
at2
d(30)
1
2
(16.4)(30)2d(60)
1
2
(16.4)(60)2
d(30) 7380 ft d(60) 29,520 ft
d(t)
1
2
at2
d(120)
1
2
(16.4)(120)2
d(120) 118,080 ft
53b. It quadruples; (2t)24t2.
54. Let xthe width of the sidewalk.
The length of the pool would be 70 2xfeet.
The width of the pool would be 50 2xfeet.
Aw
2400 (70 2x)(50 2x)
2400 3500 240x4x2
0 4x2240x1100
0 x260x275
0 (x55)(x5)
x55 0x5 0
x55 x5
Use x5 since 55 is an unreasonable solution.
5 ft
55. Let xthe number of pizzas.
(160 16x)(16 0.40x) 4000
6.4x2192x2560 4000
6.4x2192x1440 0
x230x225 0
(x15)(x15) 0
x15 0x15 0
x15 x15
16 0.40x16 0.40(15)
$10
56. (xB)(xC) 0
x2Cx Bx BC 0
x2(CB)xBC 0
from x2Bx C0
B1
C1 1
C2
C2
Sample answer: 1; 2
57. y
x(x
x
2
)(x
2
2)
58a. Let xthe width. The length
1
2
(52 2x) or
26 x. A(x) x(26 x)
58b. A(x) x(26 x)
A(x) 26xx2
[5, 30] sc15 by [2, 200] sc120
x13
26 x26 13
13
13 yd by 13 yd
59. The graph of y2x31 is the graph of y2x3
shifted 1 unit up.
60. (6, 9)
61.
15(3) (9)(5)
45 45 or 0; no
5
3
15
9
CBB
BC C
Chapter 4 104
62. AB
2
6
9
7
3
5
1
4
2
3
2(9) (1)(7)
3(9) 4(7)
2(3) (1)(5)
3(3) 4(5)
63. x4y9
4yx9
y
1
4
x
9
4
64. Parallel; the lines have the same slope.
10
18
25
1
1
29
2(2) (1)(6)
3(2) 4(6)
y
x
O
x
4
y
9
65. [f g](x) f(g(x))
f
1
2
x6
1
2
x6
24
1
4
x26x36 4
1
4
x26x32
[g f](x) g(f(x))
g(x24)
1
2
(x24) 6
1
2
x22 6
1
2
x24
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
small car symbol must represent
27
20
or 135 real
cars.
The correct choice is A.
Quadratic Equations
Pages 218–219 Check for Understanding
1. Add 4 to each side of the equation to get t26t
4. Determine the value needed to make t26t
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.
t3 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.
por
3a. equals 0 3b. negative number
3c. positive number
4. Graphing
Factoring
x24x5 0
(x5)(x1) 0
x5 0x1 0
x5x1
13 29
10
13 (13)
24(
5)(7)
2(5)
4-2
Completing the Square
x24x5 0
x24x5
x24x4 5 4
(x2)29
x2 3
x2 3x2 3
x1x5
Quadratic Formula
x4 424
(1)(5
)
2(1)
105 Chapter 4
f
(
x
)
x
O
(5, 0) (1, 0)
2624
4
8
4
f
(
x
)
x
2 4
x
5
x
4
2
36
x
4
2
6
x2 3
x2 3x2 3
x1x5
See students’ work.
5. x28x20 0
x28x20
x28x16 20 16
(x4)236
x4 6
x4 6x4 6
x2x10
6. 2a211a21 0
a2
1
2
1
a
2
2
1
0
a2
1
2
1
a
2
2
1
a2
1
2
1
a
1
1
2
6
1
2
2
1
1
1
2
6
1
a
1
4
1
2
2
1
8
6
9
a
1
4
1
1
4
7
a
1
4
1
1
4
7
a
1
4
1
1
4
7
a
3
2
a7
7. b24ac 1224(1)(36)
0; 1 real
m12
m
1
2
2
or 6
8. b24ac (6)24(1)(13)
16; 2 imaginary
t
6
2
4i
3 2i
9. p26p5 0
(p5)(p1) 0
p5 0p1 0
p5p1
10. r24r10 0
r(4) (4)2
4(1
)(10)
2(1)
(6) 16
2(1)
0
2(1)
r4 24
2
r4 2i6
2
r2 i6
11. P12I0.02I2
1600 12I0.02I2
0.02I212I1600 0
I2600I80,000 0
(I200)(I400) 0
I200 0I400 0
I200 amps I400 amps
Pages 219–221 Exercises
12. z22z24 0
z22z24
z22z1 24 1
(z1)225
z1 5
z1 5z1 5
z6z4
13. p23p88 0
p23p88
p23p
9
4
88
9
4
p
3
2
2
36
4
1
p
3
2
1
2
9
p
3
2
1
2
9
p11 p8
14. x210x21 0
x210x21
x210x25 21 25
(x5)24
x5 2
x5 2x5 2
x7x3
15. d2
3
4
d
1
8
0
d2
3
4
d
1
8
d2
3
4
d
6
9
4
1
8
6
9
4
d
3
8
2
6
1
4
d
3
8
1
8
d
3
8
1
8
d
3
8
1
8
d
1
2
d
1
4
16. 3g212g4
g24g
4
3
g24g4
4
3
4
(g2)2
8
3
g2 26
3
17. t23t7 0
t23t7
t23t
9
4
7
9
4
t
3
2
2
3
4
7
t
3
2
37
2
Chapter 4 106
t
3
2
18.
1
2
2
1
4
19. b24ac (6)24(4)(25)
364
2 imaginary; the discriminant is negative.
20. b24ac 724(6)(3) or 121; 2 real
m
m
7
1
2
11
m
3
2
,
1
3
21. b24ac (5)24(1)(9) or 11; 2 imaginary
s5 11
2(1)
7 121
2(6)
37
2
s
22. b24ac (84)24(36)(49) or 0; 1 real
d
d
8
7
4
2
or
7
6
23. b24ac (2)24(4)(9) or 140; 2 imaginary
x2 140
2(4)
84 0
2(36)
5 i11
2
x2 2i35
8
x
24. 3p24p8
3p24p8 0
b24ac 424(3)(8) or 112; 2 real
p4 112
2(3)
1 i35
4
p4 47
6
p
25. 2k25k9
2k25k9 0
b24ac 524(2)(9) or 97; 2 real
k5 97
2(2)
2 27
3
k
26. 7 i5
27. 5 2i
28. sbb24
ac
2a
5 97
4
s5 (5)2
4(3
)(9)
2(3)
s5 83
6
g2 26
3
s5 i83
6
29. x23x28 0
(x7)(x4) 0
x7 0x4 0
x7x4
30. 4w219w5 0
(4w1)(w5) 0
4w1 0w5 0
4w1w5
w
1
4
31. 4r2r5
4r2r5 0
(4r5)(r1) 0
4r5 0r1 0
4r5r1
r
5
4
32. pbb24
ac
2a
35. b24ac 0
824(1)(c) 0
64 4c0
4c64
c16
36a. Abh
A12(16)
A192
(12 2x)(16 2x)
1
2
(192)
(12 2x)(16 2x) 96
36b. (12 2x)(16 2x) 96
192 56x4x296
4x256x96 0
x214x24 0
36c. roots: 2, 12
12 2x12 2(2) or 8
16 2x16 2(2) or 12
8 ft by 12 ft
12 2x12 2(12) or 12
16 2x16 2(12) or 8
∅
37a. d(t) v0t
1
2
gt2
d(t) 5t
1
2
(32)t2
d(t) 5t16t2
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) 5t16t2
50 5t16t2
37e. 50 5t16t2
16t25t50 0
tbb24
ac
2a
107 Chapter 4
p2 224
(1)(8)
2(1)
p2 28
2
p2 2i7
2
p1 i7
33. x26
(26
)
24(
1)(2
)
2(1)
x26
32
2
x
x6
22
34a. P0.01A20.05A107
P0.01(25)20.05(25) 107
P6.25 1.25 107
P114.5 mm Hg
34b. P0.01A20.05A107
125 0.01A20.05A107
0 0.01A20.05A18
Abb24
ac
2a
26
42
2
A0.05 0.052
4(0
.01)(
18)
2(0.01)
A0.05 0.722
5
0.02
P
A
O
150
125
100
75
50
25
25 7550 100
P
0.01
A
2 0.05
A
107
f
(
x
)
x
O
10
20
20
10
f
(
x
)
x
2 14
x
24
10 20
d
(
t
)
t
O
d
(
t
) 5
t
16
t
2
t5 (5)2
4(1
6)(5
0)
2(16)
t5 3225
32
tt
t1.93 s t1.62
about 1.93 s
5 3225
32
5 3225
32
Aor A
A40 A45
40 years old
34c.
As a woman gets older, the normal systolic
pressure increases.
0.05 0.722
5
0.02
0.05 0.722
5
0.02
38. ax2bx c0
x2
a
b
x
a
c
0
x2
a
b
x
a
c
x2
a
b
x
2
b
a
2
a
c
2
b
a
2
x
2
b
a
2
a
c
4
b
a
2
2
x
2
b
a
2
4a
4
c
a
2b2
The Remainder and Factor
Theorems
Page 226 Check for Understanding
1. The Remainder Theorem states that if a
polynomial P(x) is divided by xr, 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 xris a factor if and only if P(r) 0.
2. (x34x27x8) (x5); x2x2; 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 (x3) is a
factor.
5. 6.
7. f(x) x22x15
f(3) (3)22(3) 15
9 6 15 or 0; yes
8. f(x) x4x22
f(3) (3)4(3)22
81 9 2 or 92; no
9. f(x) x35x2x5
f(1) (1)35(1)21 5
1 5 1 5 or 0
x1 is a factor x24x5(x5)(x1)
(x5), (x1), (x1)
10. f(x) x36x211x6
f(1) (1)36(1)211(1) 6
1 6 11 6 or 0
116116
156
1560
x1 is a factor x25x6(x2)(x3)
(x1), (x2), (x3)
11. 6 k2
k4
12a. 12 12b. 12 12c. 11
12d. f(x) x7x9x12 2x2
x12 x9x72x2
x(x11 x8x62x)
x2(x10 x7x52)
x, x2, x11 x8x62x, or x10 x7x52
13. hr4Vr2h
Vr2(r4)
5r2(r4)
5r34r2
0 r34r25
0 (r34r25)
r1 0hr4
r1 in. h1 4 or 5 in.
4-3
Chapter 4 108
xy
20
11
02
11
20
x
2
b
a
b24
ac
2a
x
2
b
a
b24
ac
2a
x
39. 2; 18a23a1 0
(3a1)(6a1) 0
3a1 06a1 0
3a16a1
a
1
3
a
1
6
40.
41. f(x) (x9)2
y(x9)2
x(y9)2
x
y9
yx
9
f1(x) x
9
42. 3x4y375 →3x4y375
2(5x2y) 2(345) 10x4y690
7x315
x45
3x4y375
3(45) 4y375
y60 (45, 60)
43. m
6
2
1
.
9
8
–
–
5
2
9
.4
5
60
m
0
2
.
4
4
619 x24
m60 x$643
44. 3y8x12
3y8x12
y3x4;
8
3
45. x2x20 (x5)(x4)
The correct choice is A.
619 – x
2.8 – 3.2
bb24
ac
2a
f
(
a
)
a
O
f
(
a
)
18
a
2 3
a
1
y
x
O
y
|
x
| 2
2114
22
116
x1, R6
51117 15
52015
1430
x24x3
11515
145
1450
1107k
11 6
1166 k
11405
15 5
155 0
Pages 226–228 Exercises
14. 15.
x13 x26x9, R 1
16.
x33x26x12, R23
17.
x32x24x8
18.
3x35x210x14, R12
19.
2x22x, R 3
20. f(x) x2221. f(x) x532
f(1) (1)22f(2) (2)532
1 2 or 1; no 32 32 or 0;
yes
22. f(x) x46x28
f(2
) (2
)46(2
)28
4 12 8 or 0; yes
23. f(x) x3x6
f(2) (2)32 6
8 2 6 or 12; no
24. f(x) 4x34x22x3
f(1) 4(1)34(1)22(1) 3
4 4 2 3 or 13; no
25. f(x) 2x33x2x
f(1) 2(1)33(1)21
2 3 1 or 0; yes
26a-d.
d
27. (6
)436 36 36 or 0
28.
x26x7 (x1)(x7)
(x1)(x1)(x7)
29.
x23x2 (x1)(x2)
(x2), (x1), (x2)
30.
x249 (x7)(x7)
(x1), (x7), (x7)
31.
x2x2 (x2)(x1)
(x4), (x2), (x1)
32.
x24 (x2)(x2)
(x2), (x2), (x2)
33.
x25x4 (x1)(x4)
(x1), (x1), (x4)
34.
x22x1 (x1)(x1)
(x1), (x1), (x1)
35.
x52x45x310x24x8
x44x33x24x4
2 times
36.
x2x2 (x2)(x1)
1 time; 2, 1
37. f(x) 2x3x2xk
f(1) 2(1)3(1)21 k
0 2 1 1 k
2 k
38. f(x) x3kx22x4
f(2) (2)3k(2)22(2) 4
0 8 4k4 4
0 4k8
2 k
39. f(x) x318x2kx 4
f(2) (2)318(2)2k(2) 4
0 8 72 2k4
0 2k68
34 k
40. f(x) x34x2kx 1
f(1) (1)34(1)2k(1) 1
0 1 4 k1
0 k4
4 k
109 Chapter 4
7120 91
791
113 0
3192728
318 27
1691
2110 01
2612 24
13612 23
21080 16
24816
1248 0
132542
3510 14
351014 12
12023
220
22 03
r1328
114 26
11244
215 8 8
21140
11717
167
1670
21144
264
1320
11149 49
1049
1049 0
41528
448
1120
21248
208
1040
11414
154
1540
11331
121
1210
2109024016
2410 20 8 16
12510 4 8 0
212510 4 8
28 688
14 3 440
214 3 44
212 3052
16 1526 56
11212
112
1120
41. d(t) v0t
1
2
at2
25 4t
1
2
(0.4)t2
0 0.2t24t25
50.2 4 25
125
0.2 5 0
t5 0
t5 s
42. 1117ab
12 55 a
1255 a5 ab
21255 a5 ab
20 10 10 2a
1055 a15 ab
5 ab05 ab0
15 ab05 a10 0
20 2b0a5 0
2b20 a5
b10
43a. V(x) (3 x)(4 x)(5 x)
V(x) (12 7xx2)(5 x)
V(x) x312x247x60
43b.
43c. Vwh
V3 4 5 or 60
3
5
V
3
5
(60)
36
V(x) x312x247x60
36 x312x247x60
43d. 36 x312x247x60
0 x312x247x24
[3, 10] sc11 by [200, 100] sc125
about 0.60 ft
44a.
1
2
(20 2x) or 10 x
w18 2x
hx
V(x) (10 x)(18 2x)(x)
V(x) (180 38x2x2)(x)
V(x) 2x338x2180x
44b.
44c. V(x) 2x338x2180x
224 2x338x2180x
44d. 224 2x338x2180x
0 2x338x2180x224
0 x319x290x112
2119 90 112
234 112
117 5602 in.
45. P(3 4i) 0 and P(3 4i) 0 implies that these
are both roots of ax2bx c. Since this
polynomial is of degree 2 it has only these two
roots.
x3 4i
x3 4i
(x3)216
x26x9 16
x26x25 0
a1, b6, c25
46. r25r8 0
r25r8
r25r
2
4
5
8
2
4
5
r
5
2
2
5
4
7
r
5
2
r
5
2
2
57
47a. f(x) x44x3x24x
f(2) (2)44(2)3(2)24(2)
f(2) 16 32 4 8 or 12; no
47b. f(0) (0)44(0)3(0)24(0)
f(0) 0 0 0 0 or 0; yes
47c. f(2) (2)44(2)3(2)24(2)
f(2) 16 32 4 8 or 36; no
47d. f(4) (4)44(4)3(4)24(4)
f(4) 256 256 16 16 or 0; yes
48. f(x) x532
(0, 32); point of inflection
[4, 4] sc11 by [50, 10] sc110
49. wider than parent graph and moved 1 unit left
57
2
Chapter 4 110
V
(
x
)
x
O
V
(
x
)
x
3 12
x
2 47
x
60
V
(
x
)
x
O
50
10O
4812
15O
20O
25O
V
(
x
)
2
x
3 38
x
2 180
x
50. Let xnumber of 100 foot units of Pipe A and y
number of 100 foot units of Pipe B.
4x6y48
2x2y18
2xy 16
x0
y 0
P(x, y) 34x40y
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. 4x2y3z64x2y3z6
2x7y3z→2x7y3z0
3x9y13 2z3x9y2z13
4x2y3z6
2x7y3z0
6x9y6
2(2x7y3z) 2(0)
3(3x9y2z) 3(13)
↓
4x14y6z0
9x27y6z39
5x 13y39
5(6x9y) 5(6)
6(5x13y) 6(39)
↓
30x45y30
30x78y234
33y204
y
6
1
8
1
6x9y64x2y3z6
6x9
6
11
8
64(
9
1
1
1
) 2
6
1
8
1
3z6
6x
5
1
4
1
6
3z
2
1
9
1
4
x
9
1
1
1
z
9
1
8
1
9
1
1
1
,
6
1
8
1
,
9
11
8
52. M
7
2
2
,
2
2
6
or (4.5, 4)
N
2
2
2
,
6
2
3
or (2, 1.5)
slope of M
N
4
4
.5
1
(
.5
2)
or 1
slope of R
I
2
7
(
(
3
2
)
)
or 1
Since the slopes are the same, M
N
LR
I
.
53. ababab
ac bc a cbcacbc
I. true II. true III. false
The correct choice is D.
The Rational Root Theorem
Page 232 Graphing Calculator Exploration
1. 3; 1, 1, 2
2. 2; 1, 2
3. (1) 1 positive;
f(x) (x)44(x)33(x)24(x) 4
f(x) x44x33x24x4; 3 or 1
(2) 1 positive; f(x) (x)33(x) 2
f(x) x33x2; 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.
Page 233 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 qmust equal 1.
Therefore,
p
q
becomes
p
1
or p, and pis defined as a
factor of an.
3. Sample answer: f(x) x3x2x3;
f(x) (x)3(x)2(x) 3
f(x) x3x2x3; 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.
5.
p
q
1, 2
rational root1
4-4
111 Chapter 4
O
y
x
(8, 0)
(3, 6)
(7, 2)
(0, 0)
(0, 8) 2
x
y
16
2
x
2
y
18
4
x
6
y
48
y
0
x
0
r1412
11320
11564
21234
2161324
6. p: 1, 3
q: 1, 2
P
q
: 1, 3,
1
2
,
3
2
rational roots: 3,
1
2
, 1
7. 2 or 0; f(x) 8(x)36(x)223(x) 6
f(x) 8x36x223x6; 1
p
q
: 1, 2, 3, 6,
1
2
,
3
2
,
1
4
,
3
4
,
1
8
,
3
8
8x210x3 0
(4x1)(2x3) 0
4x1 02x3 0
4x12x3
x
1
4
x
3
2
or 1
1
2
1
1
2
,
1
4
, 2
8. 1; f(x) (x)37(x)27(x) 15
f(x) x37x27x15; 2 or 0
p
q
: 1, 3, 5, 15
5, 3, 1
9. r2152x2
V
1
3
r2h
1152
1
3
(152x2)(15 x)
3456 (152x2)(15 x)
3456 3375 225x15x2x3
x315x2225x81 0
Possible rational roots: 1, 9, 81
f(x) x315x2225x81 0
f(1) 128 f(1) 320
f(9) 0f(9) 2592
f(81) 611,712 f(81) 414,720
xrepresents 9 cm.
Pages 234–235 Exercises
10.
q
p
:
1, 2, 3, 6
x24x3 0
(x3)(x1) 0
x3, x1
rational roots: 3, 1, 2
11.
p
q
: 1, 2, 3, 6, 9, 18
x24x9 0
does not factor
rational root: 2
12.
p
q
: 1, 2
x34x25x2
x22x1 0
(x1)(x1) 0
x1, x1
rational roots: 1, 2
13.
p
q
: 1, 2, 4, 5, 10, 20
x23x10 0
(x5)(x2) 0
x5, x2
rational roots: 2, 2, 5
14. p: 1, 3
q: 1, 2
p
q
: 1, 3,
1
2
,
3
2
2x36 0
x33
x
33
rational root:
1
2
Chapter 4 112
r8623 6
18 221 15
281030
r17 7 15
118 15 0
116 116
314 50
512 30
r1256
11328
214 3 0
r12118
113414
2149 0
r2383
12 530
12 1912
1
2
2460
1
2
2261.5
32 9 19 60
32310
r15972
114520
r1452
2121 0
r15420
114812
116218
21310 0
r21063
1
2
2 0060
15. p: 1
q: 1, 2, 3, 6
p
q
: 1,
1
2
,
1
3
,
1
6
6x338x218x2
6x236x6 0
x26x1 0
does not factor
rational roots:
1
3
,
1
2
16. 4; 3 or 1;
f(x) (x)42(x)37(x) 4(x) 15
f(x) x42x37x4x15; 1 negative
1 positive
17. f(x) x37x6
0 or 2 negative
x2x6 0
(x3)(x2) 0
x3, x2
rational zeros: 2, 1, 3
18. 1 positive
f(x) x32x28
1 negative
f(x) x32x28x
0 x(x22x8)
x(x4)(x2)
x0, x4, x2
rational zeros: 2, 0, 4
19. 1 positive
f(x) x33x210x24
2 or 0 negative
x26x8 0
(x4)(x2) 0
x4, x2
rational zeros: 4, 2, 3
20. 2 or 0 positive
f(x) 10x317x27x2
1 negative
10x222x4 0
5x211x2 0
(5x1)(x2) 0
x
1
5
, x2
rational zeros:
1
2
,
1
5
, 2
21. 2 or 0 positive
f(x) x42x39x22x8
2 or 0 negative
x33x26x8
x22x8 0
(x4)(x2) 0
x4, x2
rational zeros: 4, 1, 1, 2
22. 2 or 0 positive
f(x) x45x24
2 or 0 negative
x3x24x4
x24 0
(x2)(x2) 0
x2, x2
rational zeros: 2, 1, 1, 2
23a. f(x) (x2)(x2)(x1)2
0 (x2)(x2)(x1)2
x2 0x2 0(x1)20
x2x2x1 0
x1
23b. f(x) (x2)(x2)(x1)2
f(x) (x24)(x22x1)
f(x) x42x33x28x4
23c. 1 positive
f(x) x42x33x28x4
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.
w4
h21
V() wh
V() (4)(21)
V() (24)(21)
V() 23924
113 Chapter 4
r635171
1
2
638 18 2 0
r638182
1
3
636 60
r1310 24
316 8 14
r10 17 72
1
2
10 22 4 0
r1076
11 1612
11160
r12928
113680
r1368
11280
r10504
111440
r1144
11040
24b. V() 23924
2208 23924
24c. 2208 23924
0 239242208
12 w4h21
w12 4 or 8 h2(12) 1 or 23
12 in. 8 in. 23 in.
25a. Sample answer: x4x3x2x3 0
25b. Sample answer: x3x22 0
25c. Sample answer: x3x0
26a. Let the length.
h9
V()
1
3
Bh
V()
1
3
(2)(9)
V()
1
3
332
26b. V()
1
3
332
6300
1
3
332
26c. 6300
1
3
332
0
1
3
3326300
0 39218,900
30 h9
h30 9 or 21
base: 30 in. by 30 in., height: 21 in.
27. d0.0000008x2(200 x)
0.8 0.0000008x2(200 x)
0 0.00016x20.0000008x30.8
0 8x31600x28,000,000
0 x3200x21,000,000
x100 ft
28. The graphs are reflections of each other over the
x-axis. The zeros are the same.
29.
x8
30. b24ac 624(4)(25)
364; 2 imaginary
31. (x1)(x(1))(x2)(x(2)) 0
(x1)(x1)(x2)(x2) 0
(x21)(x24) 0
x45x24 0
32. y4.3x8424.3
y4.3(2008) 8424.3
y$210.10
33.
2x
x
3
3
2
x
2(2x3) x(3 x)
4x6 3xx2
x2x6 0
(x3)(x2) 0
x3 0x2 0
x3x2
The correct choice is A.
Page 235 Mid-Chapter Quiz
1. (x1)(x(1))(x2i)(x(2i)) 0
(x1)(x1)(x2i)(x2i) 0
(x21)(x24) 0
x43x24 0
2. 3; x311x230x0
x(x211x30) 0
x(x6)(x5) 0
x0x6 0x5 0
x6x5
3. x25x150
x25x
2
4
5
150
2
4
5
x
5
2
2
62
4
5
x
5
2
2
2
5
x
5
2
25
x
5
2
25
x
5
2
25
x10 x15
4. b24ac (39)24(6)(45)
441; 2 real roots
b
b
39
1
2
21
b
39
1
2
21
or b
39
1
2
21
b5b
3
2
5.
x2x4
6.
2; no
7.
x2x6 (x3)(x2)
(x3)(x1)(x2)
8.
p
q
: 1, 3
x23x1 0
does not factor
rational root: 3
39 441
2(6)
Chapter 4 114
r294 2208
12 2 15 184 0
r190 18,900
30 1 21 630 0
r1200 0 1,000,000
100 1 100 10,000 0
71156
756
180
21328
228
1140
41426
40 8
1022
11256
116
1160
r1610 3
31310
9. 1 positive
F(x) x44x33x24x4
3 or 1 negative
x35x28x4 0
x24x4 0
(x2)(x2) 0
x2, x2
rational zeros: 2, 1, 1
10. Let rradius.
hr6
V
1
3
r2h
27
1
3
r2(r6)
0
1
3
r32r227
0 r36r281
r3hr6
h3 6 or 9
r3 cm, h9 cm
Locating Zeros of a Polynomial
Function
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.
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.
4-5
4. Nikki; the sign changes between 2 and 1.
5.
4 and 5, 1 and 0
6.
2 and 1, at 1, 3 and 4
7.
approximate zero: 2.3
8.
zeros: 2, 1
9. Sample answer:
upper bound: 2
f(x) x48x2
lower bound: 0
10. Sample answer:
upper bound: 2
f(x) x4x23
lower bound: 2
11a. Let xamount of increase.
V(x) (25 x)(30 x)(5 x)
V(x) (750 55xx2)(5 x)
V(x) x360x21025x3750
115 Chapter 4
r14344
1158 4 0
r158 4
1144 0
r16 0 81
31927 0
f
(
x
)
x
O
ba
f
(
b
)
f
(
a
)
(
b
,
f
(
b
))
(
a
,
f
(
a
))
r142
21610
1153
0142
1135
2126
3115
41 0 2
51 1 3
r1324
215812
1142 2
01324
11240
21144
31 022
411212
r2403
02403
12225
22003
322615
r132
2110
1120
r10082
111175
2124 0 2
r10082
010082
r101 03
1112 21
212510 17
r101 03
1112 21
212510 17
←
11b. Vwh1.5V1.5(3750)
V25(30)(5) 1.5V5625
V3750
V(x) x360x21025x3750
5625 x360x21025x3750
11c. 5625 x360x21025x3750
0 x360x21025x1875
x1.7
25 x25 1.7 30 x30 1.7
26.7 31.7
5 x5 1.7
6.7
about 26.7 cm by 31.7 cm by 6.7 cm
Pages 240–242 Exercises
12.
1 and 2
13.
0 and 1, 2 and 3
14.
at 1, at 2
15.
3 and 2, 2 and 1, 1 and 2, 2 and 3
16.
2 and 1, 0 and 1, 1 and 2
17.
no real zeros
18.
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:
upper bound: 1
f(x) 3x32x25x1
lower bound: 0
27. Sample answers:
upper bound: 2
f(x) x2x1
lower bound: 1
28. Sample answers:
upper bound: 6
f(x) x46x32x26x13
lower bound: 2
Chapter 4 116
r1601025 1875
11611086 789
21621149 423
r1 002
11113
01 002
11 111
21 24 6
31 39 25
r251
1278
0251
1232
2211
3214
r120 12
214815 28
113320
0120 12
111101
2100 10
r108010
3131319
212486
111773
01 08010
11 1773
21 2486
3131319
r1031
21211
11123
01 031
11 121
21213
r2 0133
3261960 183
224921 45
122369
02 0 1 33
12 2 3 0 3
22 4 9 16 35
r62454 3
6612 18 111
56 624 117
r3251
13 16 5
r3251
03251
r16 2613
117 9 310
21818 30 47
r162 613
1153310
2146625
313715 58
412618 85
5113958
61021895
r111
11 01
2111
r111
112 1
29. Sample answers:
upper bound: 2
f(x) x35x23x20
lower bound: 6
30. Sample answers:
upper bound: 4
f(x) x43x32x23x5
lower bound: 2
31. Sample answers:
upper bound: 1
f(x) x55x43x320x215
lower bound: 7
32a. 432b. 1, 5
32c. 3 or 1; f(x) x43x32x23x5
1 negative real zero
32d.
2 and 1, 3 and 4
32e. Sample answers:
upper bound: 4 (See table in 32d.)
f(x) x43x32x23x5
lower bound: 2
32f. 1.4, 3.4 (Use TABLE feature of a graphing
calculator.)
33a. 1890: P(0) 0.78(0)4133(0)37500(0)2
147,500(0) 1,440,000
1,440,000
1910: P(20) 0.78(20)4133(20)37500(20)2
147,500(20) 1,440,000
2,329,200
1930: P(40) 0.78(40)4133(40)37500(40)2
147,500(40) 1,440,000
1,855,200
1950: P(60) 0.78(60)4133(60)37500(60)2
147,500(60) 1,440,000
1,909,200
1970: P(80) 0.78(80)4133(80)37500(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)4133(90)37500(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) (x2
)(x2
)(x1)
f(x) (x22)(x1)
f(x) x3x22x2; 2
, 1
35a. 37.44 60x360x260x
35b. f(x) 60x360x260x37.44
35c.
about
1
2
35d. 0.4 (Use TABLE feature of a graphing
calculator.)
36. Sample answer: f(x) x21
117 Chapter 4
r15320
116 317
217 11 2
r15320
114713
21392
3129 7
4117 8
51035
611338
r13235
114 216
215 8 13 21
r13235
112416
2114515
31 02314
41121139
r15320 015
116 32323 8
r15 320 0 15
114 727 27 8
213 938 76 137
312 947141 408
411 748192 753
510 335175 860
6113212 57
71211 57 399 2808
r13235
512843210
41121139
31 02314
2114515
112416
013235
1142 16
215813 21
3161645 130
r13235
114 216
215 8 13 21
f
(
x
)
x
O
f
(
x
)
x
3
x
2 2
x
2
f
(
x
)
x
O
2O
4O
40
20
1221
f
(
x
)
60
x
3 60
x
2 60
x
37.44
42.
7(6) 3(9) or 15
43. (x, y)
3
2
8
,
2
2
4
(2.5, 1)
44. x2y4 0
y
1
2
x2;
1
2
; 2
45.
yxcannot be true.
The correct choice is B.
Rational Equations and Partial
Fractions
Page 247 Check for Understanding
1. Multiply by the LCD, 6(b2). 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.
4. x
x
2
2
2
x
2
2
x
x
2
2
(x2)
2
x
2
2
(x2)
x(x2) 2 2(x2) 2
x22x2 2x4 2
x24x4 0
(x2)(x2) 0
x2 0x2 0
x2x2
If you solve the equation, you will get x2.
However, if x2, the denominators will equal 0.
5. b
5
b
4
b
5
b
b(4)b
b25 4b
b24b5 0
(b5)(b1) 0
b5 0b1 0
b5b1
6.
b
9
5
b
3
3
b
9
5
(b5)(b3)
b
3
3
(b5)(b3)
9(b3) 3(b5)
9b27 3b15
6b42 0
b7
4-6
9
6
7
3
Chapter 4 118
f
(
x
)
x
O
2O,000
40,000
60,000
80,000
100,000
1612842420
f
(
x
)
0.125
x
5 3.125
x
4 4000
r2528 15
3211 5 0
y
x
O
4
x
x
1
y
37a.
37b. f(0) 0.125(0)53.125(0)44000
f(0) 4000 deer
37c. 1920 1905 15
f(15) 0.125(15)53.125(15)44000
f(15) 67,281.25
about 67,281 deer
37d. in 1930
38a. 81.58 6x418x324x218x
38b. 81.58 6x418x324x218x
0 6x418x324x218x81.58
about 1.1 (Use TABLE feature of a graphing
calculator.)
38c. xrate 1
1.1 rate 1
0.1 rate
about 10%
39. 2 or 0; f(x) 2x35x228x15
1 negative zero
2x211x5 0
(2x1)(x5) 0
x0.5 x5
rational zeros: 3, 0.5, 5
40. d(t) v0t
1
2
gt2
1750 4t
1
2
(9.8)t2
4.9t24t1750 0
tbb24
ac
2a
A
B
yx
CD
t4 (4)2
4(4
.9)(1
750)
2(4.9)
t4 34,31
6
9.8
tt
t19.3 t18.5
about 19.3 s
41.
4 34,31
6
9.8
4 34,31
6
9.8
7.
t
t
4
t
3
4
t2
16
4t
t
t
4
t
3
4
(t)(t4)
t2
16
4t
(t)(t4)
(t4)(t4) 3(t) 16
t216 3t16
t23t0
t(t3) 0
t0t3 0
t3
But t0, so t3.
8.
3
p
p
2
1
1
(p
3p
1)
(p
1
1)
3
p
p
2
1
1
p
A
1
p
B
1
3p1 A(p1) B(p1)
Let p1.
3(1) 1 A(1 1) B(1 1)
2 2B
1 B
Let p1.
3(1) 1 A(1 1) B(1 1)
4 2A
2 A
3
p
p
2
1
1
p
2
1
p
1
1
9. 5
1
x
1
x
6
; exclude: 0
5x1 16
5x15
x3
Test x1: 5
(
1
1)
(
16
1)
4 16 true
Test x1: 5
1
1
1
1
6
6 16 false
Test x4: 5
1
4
1
4
6
5
1
4
4true
Solution: x0, x3
10. 1
a
5
1
7
6
; exclude: 1
6(a1) 30 7(a1)
6a6 30 7a7
31 a
Test a1: 1
1
5
1
7
6
3
2
7
6
true
Test a2: 1
2
5
1
7
6
6
7
6
false
Test a36: 1
36
5
1
7
6
1
1
7
7
6
4
4
8
2
4
4
9
2
true
Solution: a1, a31
11a.
3
3
6
0
x
20
57.14
11b.
3
3
6
0
x
20
57.14
3 60 20 57.14(3 x)
200 171.42 57.14x
0.50 x; 0.50 h
Pages 247–250 Exercises
12.
1
t
2
t8 0
1
t
2
t8
t(0)t
12 t28t0
t28t12 0
(t6)(t2) 0
t6 0t2 0
t6t2
13.
m
1
m
2
m2
34
m
1
2m2
m
2
m2
34
2m2
m234m2m2
0 m234m
0 m(m34)
m0m34 0
m34
But m0, so m34.
14.
y
2
2
3
y
y
y
2
y
2
2
3
y
(y)(y2)
y
y
2
(y)(y2)
2y3(y2) y2
5y6 y2
y25y6 0
(y3)(y2) 0
y3 0y2 0
y3y2
But y2, so y3.
15.
n21
0
1
2
n
n
1
5
2
n
n
1
5
n21
0
1
2
n
n
1
5
(n1)(n1)
2
n
n
1
5
(n1)(n1)
10 (2n5)(n1) (2n5)(n1)
2n23n5 2n23n5
6n10
n
5
3
16.
b
1
2
b
1
2
b
3
1
b
1
2
b
1
2
(b2)(b1)
b
3
1
(b2)(b1)
b1 b1 3(b2)
2b2 3b6
4 b
17.
3a
7
a
3
4a
5
4
2a
3
a
2
3(a
7
a
1)
4(a
5
1)
2(a
3
a
1)
3(a
7
a
1)
4(a
5
1)
(12)(a1)(a1)
2(a
3
a
1)
(12)(a1)(a1)
4(a1)7a3(a1)5 6(a1)3a
28a228a15a15 18a218a
10a225a15 0
2a25a3 0
(2a1)(a3) 0
2a1 0a3 0
a
1
2
a3
119 Chapter 4
18. 1
1
1
a
a
a
1
1(1 a)(a1)
1
1
a
a
a
1
(1 a)(a1)
a1 a2aa1 a(1 a)
a22a1 a22a1
0 0
all reals except 1
19.
2q
2
q
3
2q
2
q
3
1
2q
2
q
3
2q
2
q
3
(2q3)(2q3) 1(2q3)(2q3)
2q(2q3) 2q(2q3) (2q3)(2q3)
4q26q4q26q4q29
0 4q212q9
23.
x
x
2
2
6
x
x(
x
x
6
2)
x
x
2
2
6
x
A
x
x
B
2
x6 A(x2) B(x)
Let x2.
2 6 A(2 2) B(2)
4 2B
2 B
Let x0.
0 6 A(0 2) B(0)
6 2A
3 A
x
x
2
2
6
x
3
x
x
2
2
24.
5
m
m
2
4
4
(m
5m
2)(
m
4
2)
5
m
m
2
4
4
m
A
2
m
B
2
5m4 A(m2) B(m2)
Let m2.
5(2) 4 A(2 2) B(2 2)
6 4B
1.5 B
Let m2.
5(2) 4 A(2 2) B(2 2)
14 4A
3.5 A
5
m
m
2
4
4
m
3
.5
2
m
1
.5
2
25.
3y2
4
4
y
y1
(3y
1
4
)(
y
y1)
3y2
4
4
y
y1
3y
A
1
y
B
1
4yA(y1) B(3y1)
Let y1.
4(1) A(1 1) B(3(1) 1)
4 2B
2 B
Let y
1
3
.
4
1
3
A
1
3
1
B
3
1
3
1
4
3
2
3
A
2A
3y2
4
4
y
y1
3y
2
1
y
2
1
26.
9
x2
9
9
x
(x
9
3
)(x
9x
3)
9
x2
9
9
x
(x
A
3)
(x
B
3)
Let x3.
9 9(3) A(3 3) B(3 3)
18 6B
3 B
Let x3.
9 9(3) A(3 3) B(3 3)
36 6A
6 A
9
x2
9
9
x
x
6
3
x
3
3
x
6
3
,
x
3
3
27a. a(a6)
Chapter 4 120
q12 144
4(4)(
9)
2 4
12 288
8
12 122
8
20.
3
1
m
6m
3m
9
3m
4m
3
3
1
m
6m
3m
9
(12m)
3m
4m
3
(12m)
4 4(6m9) 3(3m3)
4 24m36 9m9
15m23
m
2
1
3
5
21.
x
4
1
2
7
x
x
3
1
x
4
1
(x1)(2 x)(x1)
2
7
x
x
3
1
(x1)(2 x)(x1)
4(2 x)(x1) 7(x1)(x1)
3(x1)(2 x)
4(x2x2) 7(x21)
3(x23x2)
4x24x8 4x29x13
5 13x
1
5
3
x
22a. (n1)(n2) 22b. 1, 2
22c. 1
n
n
6
1
n
4
2
1
n
n
6
1
(n1)(n2)
n
4
2
(n1)(n2)
(n1)(n2) (n2)(n6) 4(n1)
n2n2 n24n12 4n4
2n2n18 0
n1 1 4
(2)(1
8)
2 2
3 32
2
1 145
4
27b.
a
a
2
a
a
4
6
a
a
2
(a)(a6)
a
a
4
6
(a)(a6)
(a2)(a6) (a4)(a)
a28a12 a24a
12 4a
3 a
27c. 0, 6
27d. Test a1:
1
1
2
1
1
4
6
3
5
7
false
Test a1:
1
1
2
1
1
4
6
1
3
5
true
Test a4:
4
4
2
4
4
4
6
1
2
0false
Test a7:
7
7
2
7
7
4
6
5
7
3true
Solution: 0 a3, 6 a
28.
w
2
3
2
w
9
; exclude: 0
2 3w29
w9
Test w1:
2
1
3
29
1
1 29 true
Test w1:
2
1
3
2
1
9
5 29 false
Test w10:
(1
2
0)
3
2
1
9
0
3
1
2
0
2
1
9
0
true
Solution: w0, w9
29.
(
(
x
x
5
3
)
)
(
(
x
x
6
4
)
)
2
0; exclude 5, 6
(x3)(x4) 0
x3 0x4 0
x3x4
Test x0:
(
(
0
0
5
3
)
)
(
(
0
0
6
4
)
)
2
0
1
1
2
80
0true
Test x3.5: 0
(3.5 3)(3.5 4)
(3.5 5)(3.5 6)2
30.
x2
x
2
4x
1
6
5
0
(x
x2
5
)(x
1
6
1)
0; exclude 5, 1
x216 0
x216
x4
Test x5: 0
(5)216
(5 5)(5 1)
121 Chapter 4
9
0
.
.
3
2
7
5
5
0false
Test x4.5: 0
(4.5 3)(4.5 4)
(4.5 5)(4.5 6)2
0
1
.
.
7
1
5
25
0true
Test x5.5: 0
(5.5 3)(5.5 4)
(5.5 5)(5.5 6)2
0
3
.
.
1
7
2
5
5
0false
Test x6.5:( 0
6.5 3)(6.5 4)
(6.5 5)(6.5 6)2
0
8
.
.
3
7
7
5
5
0false
Solution: x3, 4 x5
4
9
0
0true
Test x2: 0
(2)216
(2 5)(2 1)
7
12
0false
Test x0:
02
0
4(0
1
)
6
5
0
1
5
6
0true
Test x4.5: 0
4.5216
(4.5 5)(4.5 1)
4
2
.2
.7
5
5
0false
Test x6:
62
6
2
24
1
6
5
0
2
7
0
0true
Solution: x4, 1 x4, x5
31.
4
1
a
8
5
a
1
2
; exclude: 0
4
1
a
8
5
a
1
2
2 5 4a
7
4
a
Test a1:
4(
1
1)
8(
5
1)
1
2
7
8
1
2
false
Test a1:
4(
1
1)
8(
5
1)
1
2
7
8
1
2
true
Test a2:
4(
1
2)
8(
5
2)
1
2
1
7
6
1
2
false
Solution: 0 a
7
4
32.
2b
1
1
b
1
1
1
8
5
; exclude:
1
2
, 1
2b
1
1
b
1
1
1
8
5
15(b1) 15(2b1) 8(2b1)(b1)
45b30 16b224b8
0 16b221b22
0 (16b11)(b2)
16b11 0b2 0
b
1
1
6
1
b2
Test b2:
2(2
1
)1
2
1
1
1
8
5
4
3
1
8
5
false
Test b0.8:
2(0.
1
8) 1
(0.8
1
)1
1
8
5
1
3
0
1
8
5
true
Test b0.6:
2(0.
1
6) 1
(0.6
1
)1
1
8
5
5
2
1
8
5
false
Test b0:
2(0)
1
1
0
1
1
1
8
5
2
1
8
5
true
Test b3:
2(3)
1
1
3
1
1
1
8
5
2
1
8
1
1
8
5
false
Solution: 1 b
1
1
6
1
,
1
2
b2
33.
y
7
1
7; exclude 1
7 7(y1)
1 y1
0 y
Test y2:
2
7
1
7
7 7false
Test y0.5:
0.5
7
1
7
14 7true
Test y1:
1
7
1
7
7
2
7 false
Solution: 1 y0
34. Let xthe number.
4
1
x
x10
2
5
20 5x252x
5x252x20 0
(5x2)(x10) 0
5x2 0x10 0
x
2
5
x10
35.
x
x
2
5
0.30
x
x
2
5
0.30; exclude 5
x2 0.30(x5)
x2 0.30x1.5
0.7x3.5
x5
Test x6:
6
6
2
5
0.30
0.36 0.30 true
Test x0:
0
0
2
5
0.30
0.4 0.30 false
Test x6:
6
6
2
5
0.30
8 0.30 true
Solution: x5 or x5
36a.
1
8
d
1
i
3
1
2
36b.
1
8
d
1
i
3
1
2
1
8
(32di)
d
1
i
3
1
2
(32di)
4di32 di
3di32
di10
2
3
cm
37. Sample answer:
x
x
3
x
1
2
38. Let xcapacity of larger truck.
5
2
x
x
3
5(x3) 2x
5x15 2x
3x15
x5 tons
39a.
1
1
0
2
1
r
1
r
2
1
0
39b.
1
1
0
2
1
r
1
r
2
1
0
1
1
0
(20r)
2
1
r
1
r
2
1
0
(20r)
2r10 20 r
r30
2r2(30) or 60; 60 ohms, 30 ohms
40. Let xthe number of quiz questions to be
answered.
2
1
0
1
x
x
0.70
11 x0.70(20 x)
11 x14 0.70x
0.3x3
x10 questions
41. Let xthe speed of the wind.
20
1
0
06
2
x
20
7
0
3
8
x
1062(200 x) 738(200 x)
212,400 1062x147,600 738x
64,800 1800x
36 x; 36 mph
Chapter 4 122
42.
a
1
1
b
1
c
a
1
1
b
(a)(b)(c)
1
c
(a)(b)(c)
bc ac ab
bc ab ac
bc a(bc)
b
b
c
c
a
43a.
1
x
1
2
1
y
1
z
1
x
1
2
3
1
0
4
1
5
43b.
1
x
1
2
3
1
0
4
1
5
1
x
6
1
0
9
1
0
1
x
(360x)
6
1
0
9
1
0
(360x)
360 6x4x
360 10x
36 x
44. Let xnumber of gallons of gasoline.
20
g
m
15,0
x
0
g
0m
20x15,000
x750 gallons
750 $1.20 $900
x$1.20 $900 $200
x583
1
3
gallons
Let ynumber of miles per gallon.
y
g
m
583
1
3
y15,000
y25.7; about 25.7 mpg
45. T
d
s
10
2
3
s
2
6
5
s
2
6
5
10
2
3
(3)(s5)(s5)
s
2
6
5
s
2
6
5
(3)(s5)(s5)
32(s5)(s5) 26(3)(s5) 26(3)(s5)
32s2800 78s390 78s390
32s2156s800 0
8s239s200 0
(8s25)(s8) 0
8s25 0s8 0
s
2
8
5
s8
8 mph
46.
3x
5
y
5y
3
5
x
y
5
5
y
y
11 1
10
47.
3 and 2, 2 and 1, 1 and 2
15,000 m
583
1
3
g
48. 51030 0
52525
15525
no
49. 2; 12x28x15 0
(6x5)(2x3) 0
6x5 02x3 0
x
5
6
x
3
2
50. 3x12 0
3x12
3x12 3x12
x4x4
51. y
2x
x
3
3
?
2(6)
6
3
3
5
2
false
no
52. y2121x2→b2121a2
52a. (b)2121a252b. b2121(a)2
b2121a2yes b2121a2yes
52c. (a)2121(b)252d. (a)2121(b)2
a2121b2no a2121b2no
53a. Let xshort answer questions and y
essay questions.
xy20
2x12y60
x0
y0
S(x, y) 5x15y
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
123 Chapter 4
r1235
31105
21031
11141
01235
113 05
214 5 5
O
y
x
4
4
8
12
16
20
8121620
(18, 2)
(20, 0)
(0, 0)
(0, 5)
2
x
12
y
60
x
y
20
y
0
x
0
53b. Let xshort answer questions and y
essay questions.
xy 20
2x12y120
x 0
y 0
S(x, y) 5x15y
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
54.
x
35
35
11
11
Radical Equations and
Inequalities
Pages 254–255 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
t
2
1 4t4
4t3
t
3
4
4-7
Chapter 4 124
O
y
x
4
4
8
12
16
20
8121620
(20, 0)
(12, 8)
(0, 0)
(0, 10) 2
x
12
y
120
x
y
20
y
0
x
0
x
1(3) 1(3) 1(5) 1(5)
1(3) 1(3) 1(5) 1(5)
x
55. yy1m(xx1)
y1 2(x(3))
y1 2x6
2xy7 0
56a. m
300
2
0
0
5
6
0
0
00
56b. $2000; $50
m50
y3000 50(x20)
y50x2000
C(x) 50x2000
56c.
57. Aof JKL
1
2
(9)(7) or 31.5
Aof small triangle
1
2
(5)(3) or 7.5
Aof shaded region 31.5 7.5 or 24
The answer is 24.
00
00
C
(
x
)
x
0
$1000
$2000
$3000
Cost
Televisions Produced
$4000
86420
C
(
x
) 50
x
2000
5.
3x4
12 3Check:
3x4
12 3
3x4
9
3733
4
12 3
x4 729
3729
12 3
x733 9 12 3
3 3
6. 5 x4
2Check: 5 x4
2
x4
35 13
4
2
x4 95 9
2
x13 5 3 2
no real solution
7. 6x4
2x1
0
6x4 2x10
4x14
x3.5
Check: 6x4
2x1
0
6
7
2
4
2
7
2
10
21
4
7 1
0
17
17
8. a4
a3
7
a4
7 a3
a4 49 14a3
a3
42 14a3
1764 196(a3)
9 a3
12 a
Check: a4
a3
7
12
4
12
3
7
16
9
7
4 3 7
Check: 1 4
t
2
1 4
3
4
2
1 3
2
2 2
9. 5x4
85x4 0
5x4 64 5x4
5x60 x0.8
x12
Test x1: 5(1)
4
8
1
8meaningless
Test x0: 5(0)
4
8
4
8true
Test x13: 5(13)
4
8
69
8false
Solution: 0.8 x12
10. 3 4a
5
10 4a5 0
4a
5
74a5
4a5 49 a1.25
4a54
a13.5
Test a0: 3 4(0)
5
10
3 5
10 meaningless
Test a2: 3 4(2)
5
10
4 3
10 true
Test a14: 3 4(14)
5
10
3 51
10 false
Solution: 1.25 a13.5
11a. vv02
64h
90 102
64h
90 100
64h
11b. 90 100
64h
Check: 90 100
64h
8100 100 64h90 100
64(12
5)
8000 64h90 8100
1125 h; 125 ft 90 90
Pages 255–257 Exercises
12. x8
5Check: x8
5
x8
25 17
8
5
x17 25
5
5 5
13.
3y7
4Check:
3y7
4
y7 64
371
7 4
y71
364
4
4 4
14. 8n
5
1 2Check: 8n
5
1 2
8n
5
38
7
4
5
1 2
8n5 914
5
1 2
8n14 3 1 2
n
7
4
2 2
15. x16
x
4
x16 x8x
16
0 8x
0 x
0 x
Check: x16
x
4
0 1
6
0
4
16
4
4 4
16. 43m2
15
4
3m2
15
1
3m215 1
3m216
m2
1
3
6
m
4
3
3
Check: 43m2
15
4
43
4
3
3
2
15
4
41
4
4 4
Check: 43m2
15
4
43
4
3
3
2
15
4
41
4
4 4
17. 9u
4
7u
20
9u4 7u 20
2u16
u8
Check: 9u
4
7u
20
9(8)
4
7(8)
20
76
76
no real solution
18.
36u
5
2 3
36u
5
5
6u5 125
6u120
u20
Check:
36u
5
2 3
36(20
) 5
2 3
3125
2 3
5 2 3
3 3
19. 4m2
3m
2
2m5 0
4m2
3m
2
2m5
4m23m2 4m220m25
23 23m
1 m
Check: 4m2
3m
2
2m5 0
4(1)
23(
1)
2
2(1) 5 0
9
2 5 0
3 3 0
0 0
20. k9
k
3
k9
3
k
k9 3 23k
k
6 23k
36 4(3k)
36 12k
3 k
Check: k9
k
3
3 9
3
3
12
3
3
23
3
3
3
3
125 Chapter 4
21. a2
1
1 a1
2
a21 2a2
1
1 a12
2a2
1
10
4(a21) 100
a21 25
a4
Check: a2
1
1 a1
2
4 2
1
1 4 1
2
25
1 16
5 1 4
4 4
22. 3x4
2x7
3
3x4
3 2x7
3x4 9 62x7
2x7
x2 62x7
x24x4 36(2x7)
x268x256 0
(x4)(x64) 0
x4 0x64 0
x4x64
Check: 3x4
2x7
3
3(4)
4
2(4)
7
3
16
1
3
4 1 3
3 3
Check: 3x4
2x7
3
3(64)
4
2(64)
7
3
196
121
3
14 11 3
3 3
23. 2
37b
1
4 0
2
37b
1
4
8(7b1) 64
7b1 8
7b9
b
9
7
Check: 2
37b
1
4 0
2
37
9
7
1
4 0
2
38
4 0
4 4 0
0 0
24.
43t
2 0Check:
43t
2 0
43t
2
43
1
3
6
2 0
3t16
416
2 0
t
1
3
6
2 2 0
0 0
25. x2
7 x9
x2 14x2
49 x9
14x2
42
196(x2) 1764
x2 9
x7
Check: x2
7 x9
7 2
7 7 9
3 7 4
4 4
no real solution
26. 2x1
2x6
5
2x1
5 2x6
2x1 25 102x6
2x6
30 102x6
3 2x6
9 2x6
3 2x
3
2
x
Check: 2x1
2x6
5
2
3
2
1
2
3
2
6
5
4
9
5
2 3 5
5 5
27. 3x1
0
x11
1
3x10 x11 2x11
1
2x2 2x11
x1 x11
x22x1 x11
x23x10 0
(x5)(x2) 0
x5 0x2 0
x5x2
Check: 3x1
0
x11
1
3(5)
10
5 1
1
1
25
16
1
5 4 1
5 3
Check: 3x1
0
x11
1
3(2)
10
2
11
1
4
9
1
2 3 1
2 2
Solution: x2
28a. 3t1
4
t6
3t1
4
6 t
3t14 36 12tt2
0 t215t50
0 (t5)(t10)
t5 0t10 0
t5t10
Check: 3t1
4
t6
3(5)
14
5 6
1
5 6
1 5 6
6 6
Check: 3t1
4
t6
3(10)
14
10 6
16
10 6
4 10 6
14 6
10
28b. 5
Chapter 4 126
29. 2x7
5
2x7 25
2x32
x16
2x7 0
2x7
x
7
2
34. m2
3m
4
m2 3m4
2m2
m1
m2 0
m2
3m4 0
3m4
m
4
3
Test m3: 3
2
3(3)
4
1
5
meaningless
Test m1.6: 1.6
2
3(1.
6) 4
0.4
0.8
meaningless
Test m1.2: 1.2
2
3(1.
2) 4
0.8
0.4
false
Test m0: 0 2
3(0)
4
2
4
true
Solution: m1
35. 2c5
7
2c5 49
2c54
c27
2c5 0
2c5
c2.5
127 Chapter 4
Test x0: 2(0)
7
5
7
5
meaningless
Test x4: 2(4)
7
5
8 7
5
1 5
false
Test x17: 2(17)
7
5
27
5
true
Solution: x 16
30. b4
6
b4 36
b32
b4 0
b4
Test b5: 5
4
6
1
6
meaningless
Test b0: 0 4
6
4
6
2 6
true
Test b33: 33
4
6
37
6
false
Solution: 4 b32
31. a5
4
a5 16
a21
a5 0
a5
Test a0: 0 5
4
5
4
meaningless
Test a6: 6 5
4
1
4
1 4
true
Test a22: 22
5
4
17
4
false
Solution: 5 a21
32. 2x5
6
2x5 36
2x41
x20.5
2x5 0
2x5
x2.5
Test x0: 2(0)
5
6
5
6
meaningless
Test x5: 2(5)
5
6
5
6
true
Test x22: 2(22)
5
6
39
6
false
Solution: 2.5 x20.5
33.
45y9
2
5y9 16
5y25
y5
5y9 0
5y9
y1.8
Test y0:
45(0)
9
2
49
2
meaningless
Test y2:
45(0)
9
2
41
2
true
Test y6:
45(6)
9
2
421
2
false
Solution: 1.8 y5
Test c0: 2(0)
5
7
5
7
meaningless
Test c5: 2(5)
5
7
5
7
false
Test c28: 2(28)
5
7
51
7
true
Solution: c27
36a. t
2
g
s
3
2(7
g
.2)
3
14
g
.4
36b. 3
14
g
.4
9
14
g
.4
9g14.4
g1.6 m/s2
37. x5
3x3
x5
3(x3
)2
x5
3x26
x9
(x5)3x26x9
x315x275x 125 x26x9
x316x281x134 0
Use a graphing calculator to find the zero.
[2, 10] sc11 by [10, 10] sc11
about 7.88
38a. s30fd
s30(0.6
)(25)
s450
s21.2 mph
38b. s30fd
35 30(0.6
)d
1225 18d
68.06 d; about 68 ft
r16116
115 60
38c. No; it is not a linear function.
39a. T2
g
39b. t2
g
T2
9
1
.8
T2
8
1
.9
T2.01 s T2.11 s
39c. Let xthe new length of the pendulum.
2
2
g
2
g
x
4
g
2
g
x
2
g
g
x
4
g
g
x
4x
It must be multiplied by 4.
40.
T
T
a
b
r
r
a
b
3
2
6
2
8
5
7
67,20
r
0
b
,000
3
4
5
7
0
1
,
,
6
9
2
6
5
9
3.03
r
b31023
50,625rb31.43 1029
rb32.83 1024
r141,433,433.8; about 141,433,434 mi
41. 2x9
ab
2x9
ab
2x9 0, so ab0
no real solution when ab0
42. T
t
2
c
t
2
c
2p2
108
t(
2
200)
t(
2
200)
2
50
2
108
t
2
200
t
2
20
0
2
2500
t
2
416
25
00
t2400t40,000
4
44.
p
q
: 6, 3, 2, 1
x36x211x6 0
x25x6 0
(x3)(x2) 0
x3 0x2 0
x3x2
3, 2, 1, 1
45a. point discontinuity
45b. jump discontinuity
45c. infinite discontinuity
46a. p
w
v
p
10
w
56
46b. x- and y-axes 46c. It increases.
46d. It is halved.
Chapter 4 128
t2832t173,056 t2400t50,000
123,056 1232t
99.88 t; about 99.88 psi
43.
2
a
a
2
1
a
3
4a
3
2
; exclude:
1
2
2
a
a
2
1
(6)(2a1)
a
3
2(2a
3
1)
(6)(2a1)
6(a2) a(2)(2a1) 3(3)
6a12 4a22a9
0 4a24a3
0 (2a3)(2a1)
2a3 02a1 0
a
3
2
a
1
2
3
2
t2400t50,000
4
t2832t173,056
4
p
w
O
10O
10O
10O
10O
1056
w
p
47.
4(0) (1)(2) 6(5)
4(0) 0(2) 2(5)
3
2
1
0
2
5
6
2
1
0
4
4
4(3)(1)(2) 6(1)
4(3) 0(2) 2(1)
r15 556
11611 6 0
48. 4(abc) 4(6) →4a4b4c24
2a3b4c32a3b4c3
2a7b21
4(abc) 4(6) →4a4b4c24
4a8b4c12 4a8b4c12
12b12
b1
2a7b21 abc6
2a7(1) 21 7 1 c6
a7c2
(7, 1, 2)
49. y3.54x7125.4
y3.54(2010) 7125.4
y10 students
50. 7y4x3 0
y
4
7
x
3
7
perpendicular slope:
7
4
y5
7
4
(x2)
y
7
4
x
3
2
51. Ar2Ar2
1
2
2(1)2
1
4
1
1
4
1
5
4
The correct choice is C.
20
14
28
10
Modeling Real-World Data with
Po lynomial Functions
Pages 261-262 Check for Understanding
1a. Sample answer:
1b. Sample answer:
1c. Sample answer:
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) 1.98x42.95x35.91x20.22x4.89
6. Sample answer: f(x) 3.007x20.001x7.896
7a. Sample answer: f(x) 0.48x58.0
7b. Sample answer: 2010 1950 60
f(x) 0.48x58.0
0.48(60) 58.0
86.8%
7c. Sample answer: f(x) 0.48x58.0
89 0.48x58.0
64 x
1950 64 2014
Pages 262-264 Exercises
8. cubic 9. quadratic
10. linear 11. quadratic
4-8 12. f(x) 1.25x5
13. f(x) 8x23x9
14. Sample answer:
f(x) 1.03x45.16x36.08x20.23x0.94
15. Sample answer:
f(x) 0.09x32.70x224.63x65.21
16. Sample answer:
f(x) 4.05x40.09x36.69x2222.03x
2697.74
17. Sample answer:
f(x) 0.02x38.79x23.35x27.43
18a. Sample answer: f(x) 1.99x21.74x2.76
18b. Sample answer:
f(x) 0.96x30.56x20.36x4.05
18c. Sample answer: Cubic; the value of r2for the
cubic function is closer to 1.
19a. Sample answer: f(x) 0.126x22.732
19b. Sample answer:
2010 1900 110
f(x) 0.126x22.732
f(110) 0.126(110) 22.732
f(110) 36.592
37
19c. Sample answer:
2025 1900 125
f(x) 0.126x22.732
f(125) 0.126(125) 22.732
f(125) 38.482
38
20. Sample answer:
21a. Sample answer:
f(x) 0.008x40.138x30.621x20.097x
18.961
21b. Sample answer: 1994 1992 2
f(x) 0.008x40.138x30.621x20.097x
18.961
f(2) 0.008(2)40.138(2)30.621(2)2
0.097(2) 18.961
f(2) 20.663
about 21%
22. Asixth-degree polynomial; there are 5 changes in
direction.
23a. Sample answer:
f(x) 0.109x20.001x48.696
129 Chapter 4
y
x
O
y
x
O
y
x
O
x12345678 9
f(x)1363–13–49–112–209 –347
23b. Sample answer:
f(x) 0.109x20.001x48.696
100 0.109x20.001x48.696
0 0.109x20.001x51.304
[5, 25] sc11 by [50, 10] sc15
root: (21.7, 0)
1985 22 2007
23c. Sample answer: 1998 1985 13
f(x) 0.109x20.001x48.696
f(13) 0.109(13)20.001(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.
24a. Sample answer:
f(x) 0.033x31.471x21.368x5.563
24b. Sample answer: 1996 1990 6
f(x) 0.033x31.471x21.368x5.563
f(6) 0.033(6)31.471(6)21.368(6) 5.563
f(6) 43.183
about 43.18 million
24c. Sample answer:
f(x) 0.033x31.471x21.368x5.563
200 0.033x31.471x21.368x5.563
0 0.033x31.471x21.368x194.437
[5, 25] sc15 by [300, 50] sc150
root: (14.8, 0)
14 1990 2004
about 2004
25. 5 b2
0Check: 5 b2
0
5 b2
5 23
2
0
25 b25 25
0
23 b5 5 0
0 0
26.
p
6
3
p
p
3
1
p
6
3
p
p
3
(p3)(p3) 1(p3)(p3)
6(p3) p(p3) (p3)(p3)
6p18 p23pp29
9p9
p1
27.
1, 1
28a. Let xnumber of weeks.
P(120 10x)(0.48 0.03x)
P57.6 1.2x0.3x2
[20, 20] sc12 by [40, 60] sc15
maximum: (2, 58.8)
2 weeks
28b. $58.80 per tree
29. x0.10x0.90x
0.90x0.10(0.90x) 0.90x0.09x
0.99x
The correct choice is B.
Fitting a Polynomial Function to a
Set of Points
Page 266
1. y7x34x217x15
2. y7x34x217x15; yes
3. Sample answer: y5x62x540x42x3
x28x4
4. Infinitely many; suppose that you are given a set
of npoints 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 x-
coordinates. You could find polynomial functions
that went through the original npoints 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.
Chapter 4 Study Guide and Assessment
Page 267 Understanding and Using the
Vocabulary
1. Quadratic Formula 2. Integral Root Theorem
3. zero 4. Factor Theorem
5. polynomial function 6. lower bound
4-8B
Chapter 4 130
r210 1 2
123320
0210 1 2
12 11 2 0
7. Extraneous 8. complex roots
9. complex numbers 10. quadratic equation
Pages 268-270 Skills and Concepts
11. no; f(a) a33a23a4
f(0) (0)33(0)23(0) 4
f(0) 4
12. yes; f(a) a33a23a4
f(4) (4)33(4)23(4) 4
f(4) 0
13. no; f(a) a33a23a4
f(2) (2)33(2)23(2) 4
f(2) 18
14. f(t) t42t23t1
f(3) (3)42(3)23(3) 1
f(3) 73
no
15. 3; x32x23x0
x(x22x3) 0
x(x3)(x1) 0
x0x3 0x1 0
x3x1
16. b24ac (7)24(2)(4)
81; 2 real
x
x
7
4
9
x
7
4
9
x
7
4
9
x4x
1
2
17. b24ac (10)24(3)(5)
40; 2 real
m10 40
2(3)
781
2(2)
a
4
2
a2
21. b24ac (1)24(5)(10)
199; 2 imaginary
r1 199
2(5)
131 Chapter 4
O
f
(
x
)
f
(
x
)
x
3 2
x
2 3
x
x
m10 2 10
6
m
18. b24ac (1)24(1)(6)
23; 2 imaginary
x1 23
2(1)
5 10
3
x
19. b24ac 324(2)(8)
73; 2 real
y3 73
2(2)
1 i23
2
y3 73
4
20. b24ac 424(1)(4)
0; 1 real
a4 0
2(1)
r
22. f(x) x3x210x8
f(2) (2)3(2)210(2) 8
8 4 20 8 or 0; yes
23. f(x) 2x35x27x1
f(5) 2(5)35(5)27(5) 1
250 125 35 1 or 161; no
24. f(x) 4x37x1
f
1
2
4
1
2
37
1
2
1
4
8
7
2
1 or 4; no
25. f(x) x410x29
f(3) (3)410(3)29
81 90 9 or 0; yes
26.
p
q
: 1, 2
x2x2 0
(x2)(x1) 0
x2 0x1 0
x2x1
rational roots: 1, 1, 2
27.
p
q
: 1
rational root: 1
28. p: 1, 2, 4
q: 1, 2
p
q
: 1, 2, 4;
1
2
2x22x2 0
x2x1 0
does not factor
rational root: 2
1 i199
10
r1212
11120
r10111
111 012
111010
r2224
12 026
22220
29. p: 1, 3
q: 1, 2
p
q
: 1, 3,
1
2
,
3
2
rational root:
3
2
30.
p
q
: 1, 2, 4
x42x33x25x2 0
x33x1 0
rational roots: 2, 2
31. p: 1, 2, 4, 8
q: 1, 3
p
q
: 1, 2, 4, 8,
1
3
,
2
3
,
4
3
,
8
3
3x210x8 0
(3x4)(x2) 0
3x4 0x2 0
x
4
3
x2
rational roots: 2,
4
3
, 1
32. p: 1, 2
q: 1, 2, 4
p
q
: 1, 2,
1
4
,
1
2
4x28 0
x22 0
does not factor
rational root:
1
4
33.
p
q
: 1, 5
x3x25x5 0
x25 0
does not factor
rational roots: 1, 1
34. 1 positive
f(x) x3x234x56
2 or 0 negative
x26x8 0
(x4)(x2) 0
x4 0x2 0
x4x2
rational zeros: 4, 2, 7
35. 2 or 0 positive
f(x) 2x311x212x9
1 negative
2x212x18 0
x26x9 0
(x3)(x3) 0
x3 0x3 0
x3x3
rational zeros:
1
2
, 3
36. 2 or 0 positive
f(x) x413x236
2 or 0 negative
x32x29x18 0
x25x6 0
(x3)(x2) 0
x3 0x2 0
x3x2
rational zeros: 3, 2, 2, 3
Chapter 4 132
r3728
13 10 8 0
r10405
11155 0
r4182
1
4
4080
r1013 0 36
212 918 0
r12 918
315 6 0
r1134 56
71 6 8 0
r211 12 9
1
2
212 18 0
r1155
11050
r23611 3
12 5112 15
12 1741
32 9 21 52 153
323320 57
1
2
24413
1
2
9
1
2
227
1
2
5
3
4
3
2
263
3
2
1
10
4
5
3
2
20620
r1071124
111657 3
212352 0
r12352
210310
r1031
21211
11123
41 4 13 53
4141351
37.
1 and 0
38.
0 and 1, 3 and 4
39.
1 and 0, 3 and 4
40.
1 and 0
41.
2 and 1, 0 and 1, 1 and 2
42.
0 and 1
43. Use the TABLE feature of a graphing calculator.
4.9, 1.8, 2.2
44. n
n
6
5 0
n
n
6
5
(n) 0(n)
n25n6 0
(n6)(n1) 0
n6 0n1 0
n6n1
45.
1
x
x
2
x2
3
1
x
(2x2)
x
2
x2
3
(2x2)
2xx3
x3
46.
5
6
2m
2m
2
3m
1
3
5
6
6(m1)(m1)
2(m
2m
1)
3(m
1
1)
6(m1)(m1)
5(m1)(m1) (2m)(3)(m1) 2(m1)
5m25 6m26m2m2
0 m28m3
m8 (8)2
4(1
)
(3)
2(1)
133 Chapter 4
r30 0 1
2361223
13332
030 0 1
r142
0142
1131
2122
3111
41 0 2
r133
1141
0133
1125
2115
31 03
41 1 1
r110 1
213611
1122 1
0110 1
1100 1
r4211 3
24733
143811
04111 3
14563
249 7 17
r92524 6
193458 64
092524 6
1916 82
29710 14
m
m4 13
47.
3
y
2
5
y
; exclude: 0
3
y
2
y
5
y
y
3 2y5
y1
Test y2:
3
2
2
5
2
7
2
5
2
true
Test y0.5:
0
3
.5
2
0
5
.5
8 10 false
Test y1:
3
1
2
5
1
1 5true
Solution: y1, y0
48.
x
2
1
1
x
1
1
; exclude 1, 1
x
2
1
(x1)(x1)
1
x
1
1
(x1)(x1)
2(x1) (x1)(x1) (x1)
2x2 x2x2
0 x23x
0 x(x3)
x0x3 0
x3
Test x2:
2
2
1
1
2
1
1
2
4
3
true
Test x0.5:
0.5
2
1
1
0.5
1
1
4 1false
Test x0.5:
0.5
2
1
1
0.5
1
1
4
3
3true
Test x2:
2
2
1
1
2
1
1
2
3
0false
Test x4:
4
2
1
1
4
1
1
2
5
2
3
true
Solution: x1, 0 x1, x3
49. 5 x2
0Check: 5 x2
0
5 x2
5 23
2
0
25 x25 25
0
23 x5 5 0
0 0
8 52
2
50.
34a
1
8 5Check:
34a
1
8 5
34a
1
3
34(6.
5) 1
8 5
4a1 27
327
8 5
4a26 3 8 5
a6.5 5 5
51. 3 x8
x35
9 6x8
x8 x35
6x8
18
x8
3
x8 9
x1
Check: 3 x8
x35
3 1 8
1 3
5
3 9
36
3 3 6
6 6
52. x5
7x5 0
x5 49 x5
x54
Test x0: 0 5
7
5
7meaningless
Test x10: 10
5
7
5
7true
Test x60: 60
5
7
55
7false
Solution: 5 x54
53. 4 2a
7
62a7 0
2a
7
22a7
2a7 4a3.5
2a3
a1.5
Test a5: 4 2(5)
7
6
4 3
6meaningless
Test a2: 4 2(2)
7
6
4 3
6false
Test a0: 4 2(0)
7
6
4 7
6true
Solution: a1.5
54. cubic
55. f(x) 2x2x3
Page 271 Applications and Problem Solving
56. Let xwidth of window.
Let x6 height of window.
Aw
315 x(x6)
315 x26x
0 x26x315
0 (x21)(x15)
x21 0x15 0
x21 x15
Since distance cannot be negative, x15 and
x6 21. the window should be 15 in. by 21 in.
57. Let xwidth.
Let x6 length.
(x12)(x6) (x6)(x) 288
x218x72 x26x288
12x72 288
x18
x6 24
18 ft by 24 ft
58a.
58b. g(x) 0.006x40.140x30.053x21.79x
x(0.006x30.140x20.053x1.79)
x(x323.3
x28.83
x298.3
)
rational zeros: 0, about 23.5
59. T2
g
1.6 2
9
.8
0.25
9
.8
0.06
9
.8
0.64 ; about 0.64 m
Page 271 Open-Ended Assessment
1. Sample answer:
x
x
3
2x
2
1
x
x
3
(x3)(2x1)
2x
2
1
(x3)(2x1)
x(2x1) 2(x3)
2x2x2x6
2x2x6 0
(2x3)(x2) 0
2x3 0x2 0
x
3
2
x2
2a. Sample answer: x4 x2
2b. Sample answer: x4 x2
(x4)2x2
x28x16 x2
x29x18 0
(x6)(x3) 0
x6 0x3 0
x6x3
Check: x4 x2
x4 x2
6 4 6 2
3 4 3 2
2 4
1 1
2 2 1 1
The solution is 6. Since 1 1, 3 is an
extraneous root.
3a. Sample answer:
3b. Sample answer: f(x) x3x22x
3c. Sample answer: 2, 0, 1
Chapter 4 134
x3210.5
f(x)12 0 2 1.125
x00.512
f(x)00.625 0 8
r123.333 8.833 298.333
1122.333 13.503 311.836
5118.333 82.835 712.508
23.5 1 0.167 12.758 0
g
(
x
)
x
10
100
200
20
O
g
(
x
)
0.006
x
4
0.140
x
3 0.053
x
2
1.79
x
Chapter 4 SAT & ACT Preparation
Page 273 SAT and ACT Practice
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
(1 (
2)2
(3
(1))2
324
2
9 1
6
25
or 5
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.
524232The 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 Cmust 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 Bis aunits from the y-axis, so point Cmust
be adunits from the y-axis. The x-coordinate of
point Cis ad. So point Chas coordinates
(ad, b). The correct choice is E.
4.You may want to draw a diagram.
Use the formula for the perimeter of a rectangle,
where represents the length and wrepresents
the width.
22wP
Replace wwith s. Replace with s6.
2(s6) 2s60
The correct choice is E.
5. First find the slope of the given line. Write the
equation in the form ymx b.
3x6y12
6y3x12
y
1
2
x2The slope is
1
2
.
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: xis 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
327
1012
3 104or 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 xor y. You need to find the value of
6x6y.
Notice that the first equation contains 5yand the
second contains 1y. If you subtract the second
from the first, you have 6y. Similarly, subtraction
of the xvalues gives a result of 6x. Use the same
strategy that you would for solving a system.
Subtract the second equation from the first.
10x5y14
4x5y2
6x6y12
The correct choice is C.
135 Chapter 4
O
y
x
4
3
(1, 3)
(2, 1)
s
6
s
8. You can solve this problem using the midpoint
formula or by sketching a graph.
The midpoint formula:
x1
2
x2
,
y1
2
y2
3
2
(4)
,
5
2
3
2
1
,
8
2
1
2
,4
The correct choice is B.
9. The expression x22ax a2is a perfect square
trinomial and can be factored as (xa)2. The
square of a real quantity is never negative.
The correct choice is A.
(xa2x22ax a2
(x2a2) 2ax
So the quantity in Column A equals the quantity
in Column B plus the sum of the squares of xand
a. Since neither xnor aequal 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.
One method of solving this problem is to “plug-in”
numbers for the segment lengths. Since
EG
5
3
EF, let EF 3. Then EG 5. This means
that FG must equal 2, since EF FG EG.
HF 5FG 5(2) 10
HG HF FG 10 2 8
H
EF
G
3
8
The answer is .375 or 3/8.
32 8
EFG H
Chapter 4 136
O
y
x
1
2(3, 5)
(4, 3)
, 4
()
Angles and Degree Measure
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.
2. Add 29,
4
6
5
0
, and
3
2
6
6
00
.
3. 270° 360k°where kis an integer
4.
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° 4630)
72° 4630
7. 128° 3045
128° 30
6
1
0
°
45
36
1
0
°
0
128.513°
8. 29° 6629° 6
6
1
0
°
6
36
1
0
°
0
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°
13.
4
3
5
6
3
0
1.26
360(1)° 453°
360 453°
93°; II
5-1 Pages 281–283 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° 2812)
183° 2812
21. 286.88° 286° (0.88 60)
286° 52.8
286° 52(0.8 60)
286° 5248
286° 5248
22. 27.465° 27° (0.465 60)
27° 27.9
27° 27(0.9 60)
27° 2754
27° 2754
23. 246.876° 246° (0.876 60)
246° 52.56
246° 52(0.56 60)
246° 5233.6
246° 5233.6
24. 23° 143023° 14
6
1
0
°
30
36
1
0
°
0
23.242°
25. 14° 520
14° 5
6
1
0
°
20
36
1
0
°
0
14.089°
26. 233° 2515 233° 25
6
1
0
°
15
36
1
0
°
0
233.421°
27. 173° 2435173° 24
6
1
0
°
35
36
1
0
°
0
173.410°
28. 405° 1618
405° 16
6
1
0
°
18
36
1
0
°
0
405.272°
29. 1002° 3030 1002° 30
6
1
0
°
30
36
1
0
°
0
1002.508°
30. 3 360° 1080° 31. 2 360° 720°
32. 1.5 360° 540° 33. 7.5 (360°) 2700°
34. 2.25 360° 810° 35. 5.75 (360°) 2070°
36. 4 360° 1440°
37. 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°
137 Chapter 5
Chapter 5 The Trigonometric Functions
O
y
x
14.
7
3
9
6
8
0
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.
2
1
4
(360°) 15°
6
1
0
2
1
4
360°
0.25°, or 0.25(60) 15
6
1
0
6
1
0
2
1
4
360°
0.0042°,
or 0.0042(60)(60) 15
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.
4
3
0
6
0
0
°
°
1.11 46.
3
2
6
8
0
0
°
°
0.78
360(1)° 400° 360(1)° 280°
360° 400° 360° 280°
40°; I 80°; I
47.
9
3
4
6
0
0
°
°
2.61 48.
1
3
0
6
5
0
9
°
°
2.94
360(2)° 940° 360(2)° 1059°
720° 940° 720° 1059°
220°; III 339°; IV
49.
3
6
6
2
0
4
°
°
1.73
1.73 1 0.73
0.73 360° 264°
360° 264° 96°; II
50.
3
9
6
8
0
9
°
°
2.75
2.75 2 0.75
0.75 360° 269°
360° 269° 91°; II
51.
1
3
2
6
7
0
5
°
°
3.54
360(3)° 1275°
1080° 1275°
195°; III
52. 360° 360° 327° or 33°
53. 180° 180° 148° or 32°
54. 563° 360° 203°
180° 203° 180° or 23°
55. 420° 360° 60° 56. 360° 197° 163°
360° 60° 300° 180° 163° 17°
360° 300° 60°
57.
1
3
0
6
4
0
5
°
°
2.90
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.
90 r
s
e
e
v
c
o
o
l
n
u
d
tions
rev
3
o
6
lu
0
t
°
ion
32,400°/second
90 r
s
e
e
v
c
o
o
l
n
u
d
tions
6
1
0
m
se
i
c
n
o
u
n
t
d
e
s
rev
3
o
6
lu
0
t
°
ion
1,944,000°/minute
60. 90k, where kis an integer
61.
95 r
m
ev
in
ol
u
u
t
t
e
ions
rev
3
o
6
lu
0
t
°
ion
34,200°/minute
30 seconds
1
2
minute
34,200°
1
2
17,100°
62.
rev
3
o
6
lu
0
t
°
ion
10,800,000 or 1.08 107
rev
3
o
6
lu
0
t
°
ion
36,000,000 or 3.6 107
1.08 107to 3.6 107degrees
63.
62
s
r
e
o
c
t
o
a
n
ti
d
ons
ro
3
ta
6
t
0
i
°
on
22,320° second
62
s
r
e
o
c
t
o
a
n
ti
d
ons
ro
3
ta
6
t
0
i
°
on
60
m
s
i
e
n
c
u
o
t
n
e
ds
1,339,200°/minute
62
s
r
e
o
c
t
o
a
n
ti
d
ons
ro
3
ta
6
t
0
i
°
on
60
m
s
i
e
n
c
u
o
t
n
e
ds
60 m
ho
in
u
u
r
tes
80,352,000°/hour
62
s
r
e
o
c
t
o
a
n
ti
d
ons
ro
3
ta
6
t
0
i
°
on
60
m
s
i
e
n
c
u
o
t
n
e
ds
60 m
ho
in
u
u
r
tes
24
h
d
o
a
u
y
rs
1,928,448,000°/day
64. 25° 120k°, where kis an integer
65a. 44.4499° 44° (0.4499 60)
44° 26.994
44° 26(0.994 60)
44° 2659.64
44° 2659.64
68.2616° 68° (0.2616 60)
68° 15.696
68° 15(0.696 60)
68° 1541.76
68° 1541.76
65b. 24° 333224° 33
6
1
0
°
32
36
1
0
°
0
24.559°
81° 4534.481° 45
6
1
0
°
34.4
36
1
0
°
0
81.760°
66a. Sydney:
7
1
0
r
m
ot
i
a
n
t
u
io
te
n
s
60 m
ho
in
u
u
r
tes
24
d
h
a
o
y
urs
20.6 r
d
o
a
t
y
ations
San Antonio:
1r
1
ev
h
o
o
lu
u
t
r
ion
24
d
h
a
o
y
urs
24 rev
d
o
a
l
y
utions
24 20.6 3.4 revolutions
about 3.4 revolutions
66b. Sydney:
20.6 r
d
o
a
t
y
ations
7
w
d
e
a
e
y
k
s
ro
3
ta
6
t
0
i
°
on
51,840°
San Antonio:
24 rev
d
o
a
l
y
utions
7
w
d
e
a
e
y
k
s
rev
3
o
6
lu
0
t
°
ion
60,480°
60,480° 51,840° 8640°
67a. Use graphing calculator to find cubic regression.
Sample answer: f(x) 0.00055x30.0797x2
3.7242x76.2147
67b. 2010 1950 60
f(x) 0.00055x30.0797x23.7242x
76.2147
f(60) 0.00055(60)30.0797(60)2
3.7242(60) 76.2147
20.8827
Sample answer: about 20.9%
100,000 revolutions
minute
30,000 revolutions
minute
Chapter 5 138
139 Chapter 5
O
f
(
x
)
f
(
x
) |(
x
1)2 2|
x
O
y
y
x
2 1
x
1
x
O
y
(3, 5)
(1, 5)
(0, 3)
x
20
48
52
24
10
26
68.
36n
5
15 10
36n
5
5
6n5 125
6n120
n20
Check:
36n
5
15 10
36(20)
5
15 10
3125
15 10
5 15 10
10 10
69.
x
x
3
2
2
x25
3
x6
x
x
3
2
2
(x2)
3
(x3)
(x2)(x3)
x
x
3
2
(x2)(x3)(2)
(x2)(x3)
(x2)
3
(x3)
(x3)(x3) (x2)(x3)(2) 3
x26x9 2x210x12 3
0 x24x
0 x(x4)
x0 or x4 0
x4
70. 2 1108 1
2424
1212 25
25
71. (x(5))(x(6))(x10) 0
(x5)(x6)(x10) 0
(x211x30)(x10) 0
x3x280x300 0
72. r1t1r2t2
18(3) r2(11)
18
(
11
3)
r2
4.91 r2
about 4.91
73. x1 0
x1
Point discontinuity
74.
decreasing for x1, increasing for x1
75. expanded vertically by a factor of 3, translated
down 2 units
76.
77. [fg](x) f(g(x))
f(x0.3x)
(x0.3x) 0.2(x0.3x)
x0.3x0.2x0.06x
0.56x
78. m∠EOD 180° m∠EOA m∠BOD
180° 85° 15°
80°
m∠OED m∠EDO
m∠OED
1
2
(180° m∠EOD)
1
2
(180° 80°)
50°
m∠ECA 180° m∠EOC m∠OED
180° (80° 15°) 50°
35°
The correct choice is D.
Trigonometric Ratios in Right
Triangles
Page 284 Graphing Calculator Exploration
1. Sample answers:
2. R1
1
5
3
or about 0.3846
R1
1
3
5
9
or about 0.3846
R2
1
1
2
3
or about 0.9231
R2
3
3
6
9
or about 0.9231
R3
1
5
2
or about 0.4167
R3
1
3
5
6
or about 0.4167
3. R1
1
1
2
3
or about 0.9231
R1
3
3
6
9
or about 0.9231
R2
1
5
3
or about 0.3846
R2
1
3
5
9
or about 0.3846
R3
1
5
2
or 2.4
R3
3
1
6
5
or 2.4
4. Each ratio has the same value for all 22.6° angles.
5. yes 6. Yes; the triangles are similar.
5-2
Pages 287–288 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
3. sin A
a
c
, cos A
b
c
, tan A
a
b
,
csc A
a
c
, sec A
b
c
, cot A
a
b
4. sin Acos B, csc Asec B, tan Acot B
5. (TV)2(VU)2(TU )2
172152(TU )2
514 (TU )2
514
TU
sin T
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos T
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin Tor cos Tor 17514
514
17
514
15514
514
15
514
11. (AC)2(CB)2(AB)2
8252(AB)2
89 (AB)2
89
AB
sin A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin Aor cos Aor 889
89
8
89
589
89
5
89
Chapter 5 140
tan T
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
tan T
1
1
5
7
6. csc v
si
1
nv
7. tan v
co
1
tv
csc vor
5
2
tan v
1
1
.5
or about 0.6667
1
2
5
8. (PS)2(QS)2(QP)2
(PS)262202
(PS)2364
PS 364
or 291
sin P
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos P
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin P
2
6
0
or
1
3
0
cos Por 91
10
291
20
tan P
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc P
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan Por csc P
2
6
0
or
1
3
0
391
91
6
291
tan A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
tan A
5
8
12. (AC)2(BC)2(AB)2
(AC)2122402
(AC)21456
AC 1456
or 491
sin A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin A
1
4
2
0
or
1
3
0
cos Aor
tan A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
tan Aor
13. tangent
14. cot v
ta
1
nv
15. csc v
si
1
nv
cot vor 3 csc vor
7
3
1
3
7
1
1
3
391
91
12
491
91
10
491
40
16. cos v
se
1
cv
17. sin v
cs
1
cv
cos vor
9
5
sin v
2
1
.5
or 0.4
1
5
9
18. tan v
co
1
tv
19. sec v
co
1
sv
tan v
0.
1
75
or about 1.3333 sec v
0.1
1
25
or 8
20. (RT )2(TS)2(RS)2
142(TS)2482
(TS)22108
TS 2108
or 2527
sin R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin Ror cos R
1
4
4
8
or
2
7
4
527
24
2527
48
tan R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan Ror csc Ror 24527
527
48
2527
527
7
2527
14
sec P
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot P
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec Por cot Por 91
3
291
6
1091
91
20
291
9. cos v
I
I
o
t
cos 45°
2
4
I
I
o
t
0.5I0It
Pages 288–290 Exercises
10. (AC)2(CB)2(AB)2
802602(AB)2
10,000 (AB)2
100 AB
sin A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin A
1
6
0
0
0
or
3
5
cos A
1
8
0
0
0
or
4
5
tan A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
tan A
6
8
0
0
or
3
4
It
Io
2
2
It
Io
sec Rcot Rside adjacent
side opposite
hypotenuse
side adjacent
sec R
4
1
8
4
or
2
7
4
cot Ror 7527
527
14
2527
21. (ST)2(TR)2(SR)2
382(TR)2402
(TR)2156
TR 156
or 239
sin R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin R
3
4
8
0
or
1
2
9
0
cos Ror 39
20
239
40
28. sin R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
sin R
3
7
a2b2c2
32b272
b240
b40
or 210
cos R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
tan R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
cos Rtan Ror
csc R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
sec R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
csc R
7
3
sec Ror
cot R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
cot R
29a. tan v
g
v2
r
29b. tan v
g
v2
r
tan 11°
9.8(
v
1
2
5.5)
tan 13°
9.8(
v
1
2
5.5)
29.53 v235.07 v2
5.4 v5.9 v
about 5.4 m/s about 5.9 m/s
29c. tan v
g
v2
r
29d. increase
tan 15°
9.8(
v
1
2
5.5)
40.70 v2
6.4 v
about 6.4 m/s
30. sin v
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos v
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
c
s
o
in
s
v
v
c
s
o
in
s
v
v
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
c
s
o
in
s
v
v
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
c
s
o
in
s
v
v
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
c
s
o
in
s
v
v
tan v
31a. ∠90° L23.5° cos
(N
3
1
6
0
5
)360
∠90° 26° 23.5° cos
(172
36
1
5
0)360
∠90° 26° 23.5° (0.99997)
∠87.5°
∠90° L23.5° cos
(N
3
1
6
0
5
)360
∠90° 26° 23.5° cos
(355
36
1
5
0)360
∠90° 26° 23.5° 1
∠40.5°
31b. ∠90° L23.5° cos
(N
3
1
6
0
5
)360
∠90° 64° 23.5° cos
(172
36
1
5
0)360
∠90° 64° 23.5° 0.99997
∠49.5°
∠90° L23.5° cos
(N
3
1
6
0
5
)360
∠90° 64° 23.5° cos[
(355
36
1
5
0)360
]
∠90° 64° 23.5° 1
∠2.5°
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
——
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
210
3
710
20
7
210
310
20
3
210
210
7
141 Chapter 5
v82° 84° 86° 88°
sin 0.990 0.995 0.998 0.999
cos 0.139 0.105 0.070 0.035
v72° 74° 76° 78° 80°
sin 0.951 0.961 0.970 0.978 0.985
cos 0.309 0.276 0.242 0.208 0.174
v18° 16° 14° 12° 10°
sin 0.309 0.276 0.242 0.208 0.174
cos 0.951 0.961 0.970 0.978 0.985
tan 0.325 0.287 0.249 0.213 0.176
tan R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan Ror csc R
4
3
0
8
or
2
1
0
9
1939
39
38
239
sec R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec Ror cot Ror
22. (ST )2(TR)2(SR)2
(7
)292(SR)2
88 (SR)2
88
SR; 88
or 222
sin R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin Ror cos Ror 922
44
9
222
154
44
7
222
39
19
239
38
2039
39
40
239
tan R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan Rcsc Ror 2154
7
222
7
7
9
sec R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec Rcot Ror
23. cot (90° v) tan v24a. 0.7963540136
cot (90° v) 1.3
24b. 0.186524036 24c. 35.34015106
24d. 1.37638192
25.
25a. 125b. 0
26.
26a. 026b. 1
26c. 0
27.
s
s
i
i
n
n
v
v
r
i
n
sin
sin
27
4
°
5
5
°
5
n
1.5103 n
97
7
9
7
222
9
v8° 6° 4° 2°
sin 0.139 0.105 0.070 0.035
cos 0.990 0.995 0.998 0.999
tan 0.141 0.105 0.070 0.035
31c. 87.5° 40.5° 47°
49.5° 25° 47°
neither
32. xt
sin
c
(
o
B
s
A
A)
x10
sin(
c
6
o
0
s
°
4
1°
41°)
x10(0.4314)
x4.31; about 4.31 cm
33. 88.37° 88° (0.37 60)
88° 22.2
88° 22(0.2 60)
88° 2212
88° 2212
34. positive: 1
f(x) x42x36x1
negative: 3 or 1
35. 35a. 23 employees
35b. $1076
[10, 50] scl:10 by [10, 1200] scl:100
36.
7
(3)
5
7(2) (3)(8) 5(8)
78
37. m
3
6
5
2
m
4
2
or
1
2
yy1m(xx1)
y3
1
2
(x6)
y
1
2
x6
38. A
1
2
bh 2x2(2) or 4
12
1
2
(2x)(3x)3x3(2) or 6
12 3x2a2b2c2
4 x24262c2
2 x52 c2
52
c; 52
or 213
The correct choice is C.
Trigonometric Functions on the
Unit Circle
Page 296 Check for Understanding
1. Terminal side of a 180° angle in standard position
is the negative x-axis which intersects the unit
circle at (1, 0). Since csc v
1
y
, csc 180°
1
0
which is undefined.
5-3
0
2
4
8
1
0
4
8
1
0
0
2
5
1
0
3
0
2
7
4
8
2.
As vgoes from 0° to 90°, the y-coordinate
increases. As vgoes from 90° to 180°, the
y-coordinate decreases.
3. cot v
x
y
c
s
o
in
s
v
v
4.
5. (1, 0); tan 180°
y
x
or
0
1
; 0
6. (0, 1); sec(90°)
1
x
or
1
0
; undefined
7.
,
sin 30° ycos 30° x
sin 30°
1
2
cos 30°
tan 30°
y
x
csc 30°
1
y
tan 30° csc 30°
csc 30° 2
tan 30°
tan 30°
sec 30°
1
x
cot 30°
x
y
sec 30° cot 30°
2
3
1
2
1
2
3
3
3
1
3
1
1
2
1
2
2
3
3
2
1
2
3
2
Chapter 5 142
O
y
x
O
1
1
1
1
y
x
Function Quadrant
IIIIII IV
sin or cos
cos or sec
tan or cot
sec 30° cot 30° 3
sec 30° 23
3
2
3
8. terminal side — Quadrant III
reference angle: 225° 180° 45°
,
sin 225° ycos 225° x
sin 225° cos 225° 2
2
2
2
2
2
2
2
12. cos v
x
r
r2x2y2
cos v
1
2
22(1)2y2
x1, r23 y2
3
y
Quadrant II, so y 3
sin v
y
r
tan v
y
x
csc v
y
r
sin vtan vor 3
csc v2
3
3
1
3
2
143 Chapter 5
tan 225°
2
2
2
2
csc 225° 2
sec 225°
1
x
cot 225°
x
y
sec 225° 2
9. rx2y
2
r324
2
t25
or 5
sin v
y
r
cos v
x
r
tan v
y
x
sin v
4
5
cos v
3
5
tan v
4
3
csc v
y
r
sec v
x
r
cot v
x
y
csc v
5
4
sec v
5
3
cot v
3
4
10. rx2y
2
r(6)2
62
r72
or 62
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
6
6
or 1
6
62
6
62
sin vcos v2
2
2
2
csc v
y
r
sec v
x
r
csc vor 2
sec vor 2
62
6
62
6
cot v
x
y
cot v
6
6
or 1
11. tan v
y
x
r2x2y2
tan v1r212(1)2
x1, y1r22
r2
sin v
y
r
cos v
x
r
sin vcos v1
2
1
2
sin v cos v2
2
2
2
csc v
y
r
sec v
x
r
csc vor 2
sec vor 2
2
1
2
1
cot v
x
y
cot v
1
1
or 1
csc v
sec v
x
r
cot v
x
y
23
3
sec v
2
1
or 2cot v1
3
cot v
13. C2rcos LC2rcos L
C2(3960) cos 0° C2(3960) cos 90°
C24,881.41 C0
The circumference goes from about 24,881 miles to
0 miles.
Pages 296–298 Exercises
14. (0, 1); sin 90° yor 1
15. (1, 0); tan 360°
y
x
or
0
1
; 0
16. (1, 0); cot(180°)
x
y
or
0
1
; undefined
17. (0, 1); csc 270°
1
y
or
1
1
; 1
18. (0, 1); cos(270°) xor 0
19. (1, 0); sec 180°
1
x
or
1
1
; 1
20. Sample answers: 0°, 180° 21. undefined
22.
,
sin 45° ycos 45° x
sin 45° cos 45° 2
2
2
2
2
2
2
2
3
3
tan 45°
y
x
csc 45°
1
y
tan 45° or 1
2
2
2
2
csc 45°
csc 45° 2
sec 45°
1
x
cot 45°
x
y
2
2
sec 45° 1
2
2
tan 225°
y
x
csc 225°
1
y
csc 225° 1
2
2
tan 225° 1csc 225° 2
2
sec 225° 1
2
2
cot 225°
2
2
2
2
sec 225° 2
2
cot 225° 1
csc 45° 1
2
2
cot 45° or 1
2
2
2
2
sec 45°
sec 45° 2
2
2
23. terminal side — Quadrant II
reference angle: 180° 150° 30°
,
1
2
sin 150° ycos 150° x
sin 150°
1
2
cos 150°
tan 150°
y
x
csc 150°
1
y
3
2
3
2
sec 210°
1
x
cot 210°
x
y
Chapter 5 144
tan 150° csc 150° 2
1
3
tan 150° 3
3
sec 150°
1
x
cot 150°
x
y
csc 315° 2
2
csc 315° 2
sec 315°
1
x
cot 315°
x
y
sec 315° cot 315° or 1
2
2
2
2
1
2
2
sec 315° 2
2
sec 315° 2
25. terminal side — Quadrant III
reference angle: 210° 180° 30°
,
1
2
sin 210° ycos 210° x
sin 210°
1
2
cos 210°
tan 210°
y
x
csc 210°
1
y
3
2
3
2
sec 210° 1
2
3
cot 210°
2
3
1
2
sec 210° cot 210° 3
2
3
sec 210°
26. terminal side — Quadrant IV
reference angle: 360° 330° 30°
,
1
2
sin 330° ycos 330° x
sin 330°
1
2
cos 330° 3
2
3
2
23
3
tan 330°
y
x
csc 330°
1
y
tan 330°
1
2
2
3
csc 330° 1
1
2
tan 330° csc 330° 2
1
3
tan 330° 3
3
sec 330°
1
x
cot 330°
x
y
tan 150°
1
2
2
3
csc150° 1
1
2
sec 150° 1
2
3
cot 150°
2
3
1
2
sec 150° 2
3
cot 150° 3
sec 150°
24. terminal side — Quadrant IV
reference angle: 360° 315° 45°
,
sin 315° ycos 315° x
sin 315° cos 315°
tan 315°
y
x
csc 315°
1
y
tan 315° or 1csc 315° 1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
23
3
tan 210°
1
2
2
3
tan 210° 1
3
csc 210° 1
1
2
csc 210° 2
tan 210° 3
3
sec 330° 1
2
3
cot 330°
2
3
1
2
sec 330° 2
3
cot 330° 3
sec 330° 23
3
tan 420°
y
x
csc 420°
1
y
sin v cos v65
65
865
65
145 Chapter 5
27. terminal side — Quadrant I
reference angle: 420° 360° 60°
1
2
,
sin 420° ycos 420° x
sin 420° cos 420°
1
2
3
2
3
2
sec 420° 1
1
2
cot 420°
1
2
2
3
sec 420° 2
cot 420° 1
3
cot 420°
28. terminal side — Quadrant IV
reference angle: 45°
,
2
2
2
2
3
3
cot (45°)
x
y
cot (45°) or 1
29. terminal side — Quadrant 1
reference angle: 390° 360° 30°
,
1
2
3
2
2
2
2
2
31. rx2y
2
r(6)2
62
r72
or 62
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
6
6
or 1
6
62
6
62
sin vcos v2
2
2
2
csc v
y
r
sec v
x
r
cot v
x
y
csc vsec vcot v
6
6
or 2
or 2
or 1
32. rx2y
2
r220
2
r4
or 2
sin v
y
r
cos v
x
r
tan v
y
x
sin v
0
2
or 0 cos v
2
2
or 1 tan v
0
2
or 0
csc v
y
r
sec v
x
r
cot v
x
y
csc v
2
0
sec v
2
2
or 1 cot v
2
0
undefined undefined
33. rx2y
2
r12(
8)2
r65
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
1
8
or 8
1
65
8
65
62
6
62
6
csc v
y
r
sec v
x
r
csc vor sec vor 65
65
1
65
8
65
8
cot v
x
y
cot v
1
8
or
1
8
34. rx2y
2
r52(
3)2
r34
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
5
3
or
3
5
5
34
3
34
sin v cos v534
34
334
34
csc v
y
r
sec v
x
r
csc vor sec v34
5
34
3
34
3
cot v
x
y
cot v
5
3
or
5
3
35. rx2y
2
r(8)2
152
r289
or 17
sin v
y
r
cos v
x
r
tan v
y
x
sin v
1
1
5
7
cos v
17
8
or
1
8
7
tan v
15
8
or
1
8
5
csc v
y
r
sec v
x
r
cot v
x
y
csc v
1
1
7
5
sec v
17
8
or
1
8
7
cot v
15
8
or
1
8
5
tan 420°
2
3
1
2
csc 420° 1
2
3
csc 420° 2
3
tan 420° 3
csc 420° 23
3
sec 420°
1
x
cot 420°
x
y
csc 390°
1
y
csc 390° or 2
1
1
2
30. rx2y
2
r(4)2
(3
)2
r25
or 5
sin v
y
r
cos v
x
r
tan v
y
x
sin v
3
5
cos v
4
5
tan v
3
4
or
3
4
csc v
y
r
sec v
x
r
cot v
x
y
csc v
5
3
sec v
5
4
cot v
4
3
or
4
3
36. rx2y
2
rx2y
2
r52(
6)2
r(5)2
62
r61
r61
sin v
y
r
sin v
y
r
sin vsin v6
61
6
61
41. tan v
y
x
r2x2y2
tan v2r21222
y2, x1r25
r5
sin v
y
r
cos v
x
r
csc v
y
r
sin vcos vcsc v5
2
1
5
2
5
Chapter 5 146
sin v sin v
The sine of one angle is the negative of the sine of
the other angle.
37. If sin v0, ymust be negative, so the terminal
side is located in Quadrant III or IV
38. cos v
x
r
r2x2y2
cos v
1
1
2
3
132(12)2y2
x12, r13 25 y2
5 y
Quadrant III, so y5
sin v
y
r
tan v
y
x
sin v
13
5
or
1
5
3
tan v
1
5
2
or
1
5
2
csc v
y
r
sec v
x
r
csc v
13
5
or
1
5
3
sec v
1
1
3
2
or
1
1
3
2
cot v
y
r
cot v
1
5
2
or
1
5
2
39. csc v
y
r
r2x2y2
csc v22
2x212
r2, y13 x2
3
x
Quadrant II, so x3
sin v
y
r
cos v
x
r
tan v
y
x
sin v
1
2
cos vor tan v1
3
3
2
3
2
661
61
661
61
tan v
sec v
x
r
cot v
x
y
sec vcot vor 3
3
1
2
3
3
3
sec v23
3
40. sin v
y
r
r2x2y2
sin v
1
5
52x2(1)2
y1, r5 24 x2
26
x
Quadrant IV, so x26
cos v
x
r
tan v
y
x
csc v
y
r
cos vtan vcsc v
5
1
or 5
1
26
26
5
tan v6
12
sec v
x
r
cot v
x
y
sec vcot vor 26
26
1
5
26
sec v56
12
sin vcos v5
5
25
5
sec v
x
r
cot v
x
y
sec vor 5
cot v
1
2
42. sec v
x
r
r2x2y2
sec v3
(3
)212y2
r3
, x12 y2
2
y
Quadrant IV, so y2
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan vor 2
2
1
1
3
2
3
5
1
sin v cos v3
3
6
3
csc v cot v2
2
6
2
43. cot v
x
y
r2x2y2
cot v1 (Quadrant III) r2(1)2(1)2
x1, y1r22
r2
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
1
1
or 1
1
2
1
2
sin v cos v2
2
2
2
csc v
y
r
sec v
x
r
csc vor 2
sec vor 2
44. csc v
y
r
r2x2y2
csc v22
2x2(1)2
r2, y13 x2
3
x
Quadrant III, so x3
tan v
y
x
tan v1
3
2
1
2
1
csc v
y
r
cot v
x
y
csc vcot v1
2
3
2
tan v
45. gsin vcos v0
sin v0 or cos v0
v0° v90°
46a. kis an even integer. 46b. kis an odd integer.
47. cos v
I
I
o
t
cos v1
ItIo
cos v1
v0°
3
3
48. Let x1. y3(1)
y3
r2x2y2
r2(1)2(3)2
r210
r10
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
3
1
or 3
1
10
3
10
55. f(x) x216
yx216
xy216
x16 y2
x16
y
56.
2(2) (3)1
7
1
7
57. 3(8m3n4p) 3(6) →24m9n12p 18
4m9n2p44m 9n 2p4
28m14p 14
4(8m3n4p) 4(6) →32m12n16p 24
6m12n5p16m12n5p1
38m 11p 23
11(28m14p) 11(14)
14(38m11p) 14(23)
↓
308m154p154
532m154p322
224m 168
m
3
4
38m11p23 4m9n2p4
38
3
4
11p23 4
3
4
9n2
1
2
4
p
1
2
n
2
3
3
4
,
2
3
,
1
2
58. 2x4y72x4y7
2(9) 4(3)
?72(1) 4(2)
?7
6 7; yes 10 7; yes
2x4y7
2(2) 4(2)
?7
12 7; no
59. absolute value; f(x) 2
1
2
x
60. Aof square Aof circle A
s2r2A
22(1)2A
0.86 A
The correct choice is C.
Applying Trigonometric Functions
Pages 301–302 Check for Understanding
1a. cos or sec 1b. tan or cot
1c. sin or csc
2. Sample answer: Find a.
5-4
21
32
21
32
147 Chapter 5
sin vcos v10
10
310
10
csc v
y
r
sec v
x
r
csc vsec vor 10
10
1
10
3
sin v
5
7
51.
3
8
6
4
0
0
2.33 360(2)° 840°
720° 840°
120°
360° 120° 240°; III
52. 5 b2
0
5 b2
25 b2
23 b
53. xbb24
ac
2a
x9 (9)2
4(4
)(5)
2(4)
A
DC
B
4
36
36 60˚
300˚
cot v
x
y
cot v
3
1
or
1
3
49a. 4 2(36) 76 ft
49b. ABC is equilateral.
m∠BCA 60°
m∠ACD m∠BCA 90°
m∠ACD 60° 90°
m∠ACD 30°
Since AC 36, AD 18.
18 4 22 ft
49c. Refer to 49b for diagram and reasoning.
Since AC 30, AD 15.
154 19 ft
49d.
1
2
r4
50. sin v
cs
1
cv
sin v1
7
5
x9 1
8
x
9
8
1
or x
9
8
1
x
1
8
0
or 1.25 x
8
8
or 1
54. k
y
x
ykx
k
9
15
y(0.6)(21)
k0.6 y12.6
O
f
(
x
)
f
(
x
)
x
2 16
15 10 52
2
4
6
4
6
5
x
B
10
38˚
C
a
A
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. tan A
a
b
6. sin B
b
c
tan 76°
1
a
13
sin 26°
1
c
8
13 tan 76° acsin 26° 18
52.1 ac
sin
18
26°
c41.1
7. cos B
a
c
cos 16° 45
—
1
a
3
—
13 cos 16° 45a
12.4 a
16. tan B
—
a
b
—
tan 49° 13
—
1
a
0
—
atan 49° 1310
a
—
tan 4
1
9
0
°13
—
a8.6
17. sin A
—
a
c
—
sin 16° 55
—
13
a
.7
—
13.7 sin 16° 55a
4.0 a
18. cos B
—
a
c
—
cos 47° 18
—
22
c
.3
—
ccos 47° 1822.3
c
—
cos
2
4
2
7
.
°
3
18
—
c 32.9
19. sin 30°
—
1
h
2
—
cos 30°
—
1
n
2
—
12 sin 30° h12 cos 30° n
6 h10.4 n
tan 45°
m
6
sin 45°
—
p
6
—
mtan 45° 6psin 45° 6
m
tan
6
45°
p
sin
6
45°
m6p8.5
20a. cos 36°
—
10
x
.8
—
xcos 36° 10.8
x
—
co
1
s
0
3
.8
6°
—
x13.3 cm
20b. tan 36°
10.8 tan 36°
1
2
s
2 10.8 tan 36° s
15.7 s
about 15.7 cm
20c. P5s
P5(15.7)
P78.5 cm
21a. cos 42° 30
xcos 42° 30
—
1
2
—
(14.6)
x
x9.9 m
21b. tan 42° 30
—
1
2
—
(14.6) tan 42° 30x
6.7 x
about 6.7 m
x
1
2
(14.6)
1
2
(14.6)
cos 42° 30
1
2
(14.6)
x
1
2
s
10.8
Chapter 5 148
8a. Let xaltitude.
sin 55° 30
1
x
0
10 sin 55° 30x
8.2 x
about 8.2 cm
8b. Let x
1
2
of the base.
cos 55° 30
1
x
0
10 cos 55° 30x
5.66 x
base 2x
base 2(5.66)
base 11.3 cm
9. tan 13° 15
—
17
x
5
—
xtan 13° 15175
x
tan
1
1
7
3
5
°15
x743.2 ft
Pages 302–304 Exercises
10. tan A
—
a
b
—
11. cos B
—
a
c
—
tan 37°
a
6
cos 67°
—
1
a
6
—
6 tan 37° a16 cos 67° a
4.5 a6.3 a
12. sin B
—
b
c
—
13. sin A
—
a
c
—
sin 62°
—
2
b
4
—
sin 29°
—
4
c
.6
—
24 sin 62° bcsin 29° 4.6
21.2 bc
—
sin
4.
2
6
9°
—
c9.5
14. cos B
—
a
c
—
15. tan B
—
a
b
—
cos 77°
—
17
c
.3
—
tan 61°
—
33
a
.2
—
ccos 77° 17.3 atan 61° 33.2
c
—
co
1
s
7
7
.3
7°
—
a
—
ta
3
n
3.
6
2
1°
—
c76.9 a18.4
8c. A
—
1
2
—
bh
A
—
1
2
—
(11.3)(8.2)
A46.7 cm2
21c. A
—
1
2
—
bh
A
—
1
2
—
(14.6)(6.7)
A48.8 m2
28. Let Mrepresent the point of intersection of the
altitude and E
F
. Since GEF is isosceles, the
altitude bisects E
F
. EMG is a right triangle.
Therefore, sin v
—
a
s
—
or ssin vaand tan v
—
0.
a
5b
—
or 0.5btan va.
29. Latasha: Markisha:
sin 35°
—
25
x
0
—
sin 42°
—
22
x
5
—
250 sin 35° x225 sin 42° x
143.4 x150.6 x
1506 143.4 7.2
Markisha’s; about 7.2 ft
30. Let xthe height of the building.
Let ythe distance between the buildings.
tan 47° 30
—
x
y
—
tan 54° 54
40
y
x
ytan 47° 30xy tan 54° 5440 x
y
—
tan 4
x
7° 30
—
y
—
tan
40
54
°
x
54
—
tan 4
x
7° 30
—
tan
40
54
°
x
54
—
tan 54° 54(x) tan 47° 30(40 x)
xtan 54° 5440 tan 47° 30
xtan 47° 30
x(tan 54° 54tan 47° 30) 40 tan 47° 30
x40 tan 47° 30
———
tan 54° 54tan 47° 30
149 Chapter 5
22a. r
—
1
2
—
(6.4) or 3.2
cos 30°
—
3
a
.2
—
3.2 cos 30° a
2.771281292 a
about 2.8 cm
22b. Let xside of hexagon.
sin 30°
32 sin 30°
—
1
2
—
x
2 3.2 sin 30° x
3.2 x; 32 cm
1
2
x
3.2
22c. P6s
P6(3.2)
P19.2 cm
22d. A
—
1
2
—
pa
A
—
1
2
—
(19.2)(2.771281292)
A26.6 cm2
23. sin 10° 2136
—
19
x
5.8
—
xsin 10° 2136195.8
x
—
sin 1
1
0
9
°
5
2
.8
136
—
x1088.8 ft
24. height: V
—
1
3
—
area of base height
tan V
—
1
3
—
(s2)
—
1
2
—
stan
V
—
1
6
—
s3tan
x
1
2
s
—
1
2
—
stan x
25a.
25b. 84 8 76
tan 60°
—
7
x
6
—
xtan 60° 76
x
—
tan
76
60°
—
x43.9 ft
60˚
8 ft
84 ft
25c. sin 60°
—
7
x
6
—
xsin 60° 76
x
—
sin
76
60°
—
x87.8 ft
26a. tan 6°
—
39
x
00
—
xtan 6° 3900
x
—
t
3
a
9
n
0
6
0
°
—
x37,106.0 ft
26b. sin 6°
—
39
x
00
—
xsin 6° 3900
x
—
s
3
i
9
n
0
6
0
°
—
x37,310.4 ft
27. Yacht: Barge:
tan 20°
—
20
x
8
—
tan 12° 30
—
20
x
8
—
xtan 20° 208 xtan 12° 30208
x
—
ta
2
n
0
2
8
0°
—
x
—
tan
2
1
0
2
8
°30
—
x571.5 x938.2
938.2 571.5 366.8 ft; no
x131.7 ft
31. terminal side — Quadrant II
reference angle: 180°120° 60°
—
1
2
—
,
sin 120° ycos 120° x
sin 120° cos 120°
—
1
2
—
3
2
3
2
tan 120°
y
x
csc 120°
1
y
tan 120°
2
3
1
2
tan 120° 3
csc 120°
sec 120°
—
1
x
—
cot 120°
—
x
y
—
cot 120°
1
2
2
3
23
3
sec 120° 2cot 120°
32. (PR)2(RQ)2(PQ)2
7222(PQ)2
53 (PQ)2
53
PQ
sin P
—
P
r
—
cos P
—
q
r
—
sin Pcos P7
53
2
53
3
3
sin Pcos P
tan P
—
P
q
—
tan P
—
2
7
—
33. 43° 153543° 15°
6
1
0
°
35
36
1
0
°
0
43.260°
753
53
253
53
csc 120° 1
2
3
sec 120° 1
1
2
34.
35. Let xthe cost of notebooks and ythe cost of
pencils.
3x2y5.89
4xy6.20
3x2y5.80 →3x2y5.80
2(4xy) 2(6.20) 8x2y12.40
5x6.60
4xy6.20 x$1.32
4(1.32) y6.20
y$0.92
36.
—
m
hh
m
o
i
u
l
r
e
s
s
—
xhours
—
m
h
x
—
miles
The correct choice is E.
Page 304 Mid-Chapter Quiz
1. 34.605 ° 34° (0.605 60)
34° 36.3
34° 36(0.3 60)
34° 3618
34° 3618
2.
—
3
4
6
0
0
0
°
°
—
1.11
1.11 1 0.11
0.11 360° 40°
360° 40° 320°; IV
3. (GI )2(IH )2(GH )2
(GI )2102122
(GI )244
GI 44
or 211
sin G
—
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
—
cos G
—
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
—
sin G
—
1
1
0
2
—
or
5
6
cos Gor
tan G
—
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
—
csc G
—
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
—
tan Gcsc G
1
1
2
0
or
—
6
5
—
10
211
11
6
211
12
4. rx2y
2
r22(
5)2
r29
sin v
—
y
r
—
cos v
x
r
tan v
y
x
sin vcos vtan v
—
2
5
—
or
5
2
2
29
5
29
Chapter 5 150
O
y
y
|
x
2|
x
tan G
sec G
—
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
—
cot G
—
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
—
sec Gcot Gor 11
5
211
10
12
211
511
11
sec G611
11
sin v cos v
csc v
—
y
r
—
sec v
x
r
csc vor sec v
cot v
—
x
y
—
cot v
—
2
5
—
or
2
5
5. tan 27.8°
—
55
x
0
—
xtan 27.8° 550
x
tan
55
2
0
7.8°
x1043.2 ft
Solving Right Triangles
Pages 308–309 Check for Understanding
1a. linear 1b. angle
2. They are complementary.
3. Sample answer:
4. Marta; they need to find the inverse of the cosine,
not
—
c
1
os
—
.
5. 60°, 300° 6. 150°, 330°
7. sin
sin1
3
2
5-5
29
2
29
5
29
5
229
29
529
29
Let Asin1.Then sin A.
3
2
3
2
sin
sin1
8. Let Acos1
—
3
5
—
. Then cos A
—
3
5
—
r2x2y2
5232y2
16 y2
4 y
tan A
—
4
3
—
; tan
cos1
—
3
5
—
—
4
3
—
3
2
3
2
30˚
60˚
11. A78° 90°
A12°
Find b.Find c.
tan B
—
a
b
—
cos B
—
a
c
—
tan 78°
—
4
b
1
—
cos 78°
—
4
c
1
—
41 tan 78° bccos 78° 41
192.9 bc
—
cos
41
78°
—
c197.2
A12°, b192.9, c197.2
12. a2b2c2Find B.
112212c2tan B
—
a
b
—
562
ctan B
—
2
1
1
1
—
23.7 cBtan1
—
2
1
1
1
—
B62.4°
Find A.
A62.35402464 90
A27.64597536
c23.7, A27.6°, B62.4°
13. 3.2° B90°
B58°
Find a.Find b.
sin A
—
a
c
—
cos A
—
b
c
—
sin 32°
—
1
a
3
—
cos 32°
—
1
b
3
—
13 sin 32° a13 cos 32° b
6.9 a11.0 b
B58°, a6.9, b11.0
14a. tan x
—
1
2
2
1
8
0
0
0
—
xtan1
—
1
2
2
1
8
0
0
0
—
x31.4°
14b. tan 38°
—
12
x
80
—
xtan 38° 1280
x
—
ta
1
n
28
3
0
8°
—
x1638.3 ft
14c. tan 65°
—
12
x
80
—
xtan 65° 1280
x
—
ta
1
n
28
6
0
5°
—
x596.9 ft
Pages 309–312 Exercises
15. 90° 16. 120°, 300°
17. 30°, 330° 18. 90°, 270°
19. 225°, 315° 20. 135°, 315°
21. Sample answers: 30°, 150°, 390°, 510°
22. Let Aarccos
—
4
5
—
. Then cos A
—
4
5
—
.
cos
arccos
—
4
5
—
—
4
5
—
23. Let Atan1
—
2
3
—
. Then tan A
—
2
3
—
.
tan
tan1
—
2
3
—
—
2
3
—
24. Let Acos1
—
2
5
—
. Then cos A
—
2
5
—
.
sec A
—
co
1
sA
—
sec A
sec A
—
5
2
—
sec
cos1
—
2
5
—
—
5
2
—
25. Let Aarcsin 1. Then sin A1.
csc A
—
sin
1
A
—
csc A
—
1
1
—
or 1
csc (arcsin 1) 1
26. Let Acos1
—
1
5
3
—
. Then cos A
—
1
5
3
—
r2x2y2
13252y2
144 y2
12 y
tan A
—
1
5
2
—
; tan
cos1
—
1
5
3
—
—
1
5
2
—
27. Let Asin1
—
2
5
—
. Then sin A
—
2
5
—
r2x2y2
52x222
21 x2
21
x
cos A; cos
sin1
—
2
5
—
28. tan N
—
m
n
—
tan N
—
1
9
5
—
Ntan1
—
1
9
5
—
N59.0°
21
5
21
5
1
2
5
151 Chapter 5
9. tan R
—
r
s
—
tan R
—
1
7
0
—
Rtan1
—
1
7
0
—
R35.0°
10. cos S
—
r
t
—
cos S
—
1
2
2
0
—
Scos1
—
1
2
2
0
—
S53.1°
29. sin M
—
m
p
—
sin M
—
1
8
4
—
Msin1
—
1
8
4
—
M34.8°
30. cos M
—
n
p
—
cos M
—
2
3
2
0
—
Mcos1
—
2
3
2
0
—
M42.8°
31. tan N
—
m
n
—
tan N
—
1
1
8
4
.
.
8
3
—
Ntan1
—
1
1
8
4
.
.
8
3
—
N52.7°
32. cos N
—
m
p
—
cos N
—
1
7
7
.
.
2
1
—
Ncos1
—
1
7
7
.
.
2
1
—
N65.1°
33. sin M
—
m
p
—
sin M
—
3
5
2
4
.
.
5
7
—
Msin1
—
3
5
2
4
.
.
5
7
—
M36.5°
34. tan A
—
1
2
8
4
—
tan B
2
1
4
8
Atan1
—
1
2
8
4
—
Btan1
—
2
1
4
8
—
A36.9° B53.1°
35.
—
1
2
—
(14) 7
base angles: vertex angle 2m∠B
tan A
—
8
7
—
tan B
—
7
8
—
Atan1
—
8
7
—
Btan1
—
7
8
—
A48.8° B41.2°
2m∠B82.4°
about 48.8°, 48.8°, and 82.4°
42. A33° 90°
A57°
sin B
—
b
c
—
cos B
—
a
c
—
sin 33°
—
15
b
.2
—
cos 33°
—
15
a
.2
—
15.2 sin 33° b15.2 cos 33° a
8.3 b12.7 a
43. 14° B90°
B76°
sin A
—
a
c
—
cos A
—
b
c
—
sin 14°
—
9
a
.8
—
cos 14°
—
9
b
.8
—
9.8 sin 14° a9.8 cos 14° b
2.4 a9.5 b
44a. sin v
—
1
6
0
4
2
7
0
—
vsin1
—
1
6
0
4
2
7
0
—
v39.4°
44b. tan 39.4°
—
64
x
7
—
xtan 39.4° 647
x
—
tan
64
3
7
9.4°
—
x788.5 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
tan v
—
1
4
5
—
vtan1
—
1
4
5
—
v14.9°
47a. tan v
—
1
8
00
—
47b. tan v
—
1
5
00
—
vtan1
—
1
8
00
—
vtan1
—
1
5
00
—
v4.6° v2.9°
48. tan v
—
2
4
2
5
00
—
vtan1
—
2
4
2
5
00
—
v1.2°
49.
—
65
h
m
ou
i
r
les
—
—
52
m
80
il
f
e
eet
—
—
360
1
0
h
se
o
c
u
o
r
nds
—
95.3
—
se
f
c
e
o
e
n
t
d
—
tan v
—
g
v2
r
—
tan v
—
32
9
(
5
1
.
2
3
0
2
0)
—
vtan1
—
32
9
(
5
1
.
2
3
0
2
0)
—
v13.3°
Chapter 5 152
36. a2b2c2
212b2302
b459
b21.4
44.427004 B90
B45.6°
sin A
a
c
sin A
—
2
3
1
0
—
Asin1
—
2
3
1
0
—
A44.4°
37. 35° B90°
B55°
tan A
—
a
b
—
cos A
—
b
c
—
tan 35°
—
a
8
—
cos 35°
—
8
c
—
8 tan 35° accos 35° 8
5.6 ac
—
cos
8
35°
—
c9.8
38. A47° 90°
A43°
tan B
—
a
b
—
sin B
—
b
c
—
tan 47°
—
12
a
.5
—
sin 47°
—
12
c
.5
—
atan 47° 12.5 csin 47° 12.5
a
—
ta
1
n
2.
4
5
7°
—
c
—
si
1
m
2.
4
5
7°
—
a11.7 c17.1
39. a2b2c2tan A
—
a
b
—
3.824.22c2tan A
—
3
4
.
.
8
2
—
32.08
cAtan1
—
3
4
.
.
8
2
—
5.7 cA42.1°
42.13759477 B90
B47.9°
40. a2b2c2
a23.729.52
a76.56
a8.7
A22.92175446 90
A67.1°
sin B
—
b
c
—
sin B
—
3
9
.
.
7
5
—
Bsin1
—
3
9
.
.
7
5
—
B22.9°
41. 51.5° B90°
B38.5°
tan A
—
a
b
—
sin A
—
a
c
—
tan 51.5°
—
13
b
.3
—
sin 51.5°
—
13
c
.3
—
btan 51.5° 13.3 csin 51.5° 13.3
b
—
tan
13
5
.
1
3
.5°
—
c
—
sin
13
5
.
1
3
.5°
—
b10.6 c17.0
50.
—
s
s
i
i
n
n
v
v
r
i
—
n
—
s
s
in
in
6
v
0
r
°
—
2.42
—
si
2
n
.4
6
2
0°
—
sin vr
0.3579 sin vr
sin10.3579 vr
21.0° vr
51. Draw the altitude from Yto XZ. Call the point of
intersection W.
m∠Xm∠XYW 90°
30° m∠XYW 90°
m∠XYW 60°
In XYW:
cos 30°
—
X
1
W
6
—
sin 30°
—
W
16
Y
—
16 cos 30° XW 16 sin 30° WY
13.9 XW 8 WY
In ZYW:
sin Z
—
2
8
4
—
tan 19.5°
—
W
8
Z
—
Zsin1
—
2
8
4
—
WZ tan 19.5° 8
Z19.5° WZ
—
tan
8
19.5°
—
WZ 22.6
cos m∠WYZ
—
2
8
4
—
m∠WYZ cos1
—
2
8
4
—
m∠WYZ 70.5°
Ym∠XYW m∠WYZ y XW WZ
Y60° 70.5° y13.9 22.6
Y 130.5° y36.5
52. baseball stadium: football stadium:
tan 63°
—
10
x
00
—
tan 18°
—
10
y
00
—
xtan 63° 1000 ytan 18° 1000
x
—
ta
1
n
00
6
0
3°
—
y
—
ta
1
n
00
1
0
8°
—
x509.5 y3077.7
distance xy
distance 509.5 3077.7
distance 3587.2 ft
53. (FD)2(DE)2(FE)2
72(DE)2152
(DE)2176
DE 176
or 411
sin F
—
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
—
cos F
—
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
—
sin Fcos F
tan F
—
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
—
csc F
—
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
—
tan Fcsc For
sec F
—
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
—
cot F
—
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
—
sec F
1
7
5
cot For
54. Use TABLE feature of a graphing calculator.
0.3, 1.4, 4.3
711
44
7
411
1511
44
15
411
411
7
7
15
411
15
55. x-axis y3x22
(y)3x22
y3x22; no
y-axis y3x22
y3(x)22
y3x22; yes
yxy
3x22
(x)3(y)22
x3y22; no
yxy
3x22
(x)3(y)22
x3y22; no
y-axis
56.
2
2
1
3
3
6
5
4
5
3
0
1
1
0
153 Chapter 5
1(5) 0(4)
0(5) 1(4)
1(5) 0(3)
0(5) 1(3)
1(1) 0(3)
0(1) 1(3)
1(3) 0(6)
0(3) 1(6)
1(2) 0(2)
0(2) 1(2)
(5, 3), (5, 4), (3, 6), (1, 3), (2, 2)
2
2
1
3
3
6
5
4
5
3
57.
2
1
2
2
1
2
2
5
7
2
0
3
3
2
6
4
8
9
2 (2)
0 1
3 (2)
3 2
2 1
6 2
4 (2)
8 (5)
9 (7)
58. m
1
2
9
2
5
.2
0
4
1
2
8
.
8
5
0
m
—
2
7
0
0
.3
—
or 0.29
y22.2 0.29(x1950)
y0.29x587.7
59. 2x5y10 0
5y2x10
y
—
2
5
—
x2
—
2
5
—
; 2
60.
—
a
a
c
—
b
—
a
c
c
—
d10
—
ab
a
c
c
d
—
10
The correct choice is A.
The Law of Sines
Page 316 Check for Understanding
1.
—
sin
x
30°
—
2x
sin 90°
x3
sin 60°
5-6
0
1
5
1
1
8
2
3
2
2x2x2x
2x
1
x3
2
3
x
1
2
x
2
x
60˚
30˚
3
x
2. Sample answer:
3. Area of WXYZ Area of triangle ZWY Area of
triangle XYW.
m∠Xm∠Z
triangle ZWY:triangle XYW:
K
—
2a
1
b
—
sin Zk
—
1
2
—
ab sin X
K
—
2a
1
b
—
sin X
K
—
1
2
—
ab sin X
1
2
ab sin X
Kab sin X
4. Both; if the measures of two angles and a non-
included side are known or if the measures of two
angles and the included side are known, the
triangle is unqiue.
5. C180° (40° 59°) or 81°
—
sin
a
A
—
—
sin
c
C
——
sin
b
B
—
—
sin
c
C
—
—
sin
a
40°
—
—
sin
14
81°
——
sin
b
59°
—
—
sin
14
81°
—
a
—
14
si
s
n
in
81
4
°
0°
—
b
—
14
si
s
n
in
81
5
°
9°
—
a9.111200533 b12.14992798
C81°, a9.1, b12.1
6. C180° (27.3° 55.9°) or 96.8°
—
sin
b
B
—
—
sin
a
A
—
—
sin
c
C
—
—
sin
a
A
—
—
sin 5
b
5.9°
—
—
sin
8
2
.6
7.3°
——
sin 9
c
6.8°
—
—
sin
8
2
.6
7.3°
—
b
—
8.6
sin
sin
27
5
.
5
3
.
°
9°
—
c
—
8.6
sin
sin
27
9
.
6
3
.
°
8°
—
b15.52671055 c18.61879792
C96.8°, b15.5, c18.6
7. A180° (17° 5598° 15) or 63° 50
—
sin
c
C
—
—
sin
a
A
—
—
sin 98
c
°15
—
—
sin 6
1
3
7
°50
—
c
—
17
si
s
n
in
63
9
°
8
5
°
0
1
5
—
c18.7
8. K
—
1
2
—
bc sin A
K
—
1
2
—
(14)(12) sin 78°
K82.2 units2
9. C180° (22° 105°) or 53°
K
—
1
2
—
b2
sin
s
A
in
s
B
in C
K
—
1
2
—
(14)2
—
sin
s
2
i
2
n
°
1
s
0
i
5
n
°
53°
—
K30.4 units2
10. Let dthe distance from the fan to the pitcher’s
mound.
v180° (24° 125° 42) or 150° 6
—
sin 1
d
50° 6
—
—
sin
6
5
0
°
.5
42
—
d
—
60.5
sin
sin
5°
1
4
5
2
0
°6
—
d303.7 ft
Pages 316–318 Exercises
11. B180° (40° 70°) or 70°
—
sin
b
B
—
—
sin
a
A
—
—
sin
c
C
—
—
sin
a
A
—
—
sin
b
70°
—
—
sin
20
40°
——
sin
c
70°
—
—
sin
20
40°
—
b
—
20
si
s
n
in
40
7
°
0°
—
c
—
20
si
s
n
in
40
7
°
0°
—
b29.238044 c29.238044
B70°, b29.2, c29.2
12. A180° (100° 50°) or 30°
—
sin
a
A
—
—
sin
c
C
——
sin
b
B
—
—
sin
c
C
—
—
sin
a
30°
—
—
sin
30
50°
—
—
sin
b
100°
—
—
sin
30
50
—
a
—
30
si
s
n
in
50
3
°
0°
—
b
—
30
s
s
in
in
5
1
0
0
°
0°
—
a19.58110934 b38.56725658
A30°, a19.6, b38.6
13. C180° (25° 35°) or 120°
—
sin
a
A
—
—
sin
b
B
——
sin
c
C
—
—
sin
b
B
—
—
sin
a
25°
—
—
sin
12
35°
—
—
sin
c
120°
—
—
sin
12
35°
—
a
—
12
si
s
n
in
35
2
°
5°
—
c
—
12
s
s
in
in
3
1
5
2
°
0°
—
a8.84174945 c18.11843058
C120°, a8.8, c18.1
14. C180° (65° 50°) or 65°
—
sin
a
A
—
—
sin
c
C
——
sin
b
B
—
—
sin
c
C
—
—
sin
a
65°
—
—
sin
12
65°
——
sin
b
50°
—
—
sin
12
65°
—
a
—
12
si
s
n
in
65
6
°
5°
—
b
—
12
si
s
n
in
65
5
°
0°
—
a12 b10.14283828
C65°, a12, b10.1
15. A180° (24.8° 61.3°) or 93.9°
—
sin
b
B
—
—
sin
a
A
—
—
sin
c
C
—
—
sin
a
A
—
—
sin 2
b
4.8°
—
—
sin
8
9
.2
3.9°
——
sin 6
c
1.3°
—
—
sin
8
9
.2
3.9°
—
b
—
8.2
sin
sin
93
2
.
4
9
.
°
8°
—
c
—
8.2
sin
sin
93
6
.
1
9
.
°
3°
—
b3.447490503 c7.209293255
A93.9°, b3.4, c7.2
16. B180° (39° 1564° 45) or 76°
—
sin
a
A
—
—
sin
c
C
——
sin
b
B
—
—
sin
c
C
—
—
sin 3
a
951
—
—
sin
1
6
9
4
.
°
3
45
—
—
sin
b
76°
—
—
sin
1
6
9
4
.
°
3
45
—
a
—
19.
s
3
in
si
6
n
4°
39
4
°
5
15
—
b
—
1
s
9
in
.3
6
s
4
in
°4
7
5
6
°
—
a13.50118124 b20.7049599
B76°, a13.5, b20.7
17. C180° (37° 2051° 30) or 91° 10
—
sin
a
B
—
—
sin
c
C
—
—
sin 5
b
1° 30
—
—
sin
1
9
2
1
5
°10
—
b
—
12
s
5
in
si
9
n
1
5
°
1
1
°
0
3
1
—
b97.8
18. A180° (29° 3423° 48) or 126° 38
—
sin
a
A
—
—
sin
b
B
—
—
sin 12
a
6° 38
—
—
sin 2
1
9
1
°34
—
a
—
11
s
s
in
in
2
1
9
2
°
6
3
°
4
3
8
—
a17.9
Chapter 5 154
C
A
bc
B
10
35˚40˚
5
5
45˚
99
72˚
19. K
—
1
2
—
bc sin A
K
—
1
2
—
(14)(9) sin 28°
K29.6 units2
20. A180° (37° 84°) or 59°
K
—
1
2
—
a2
—
sin
s
B
in
s
A
in C
—
K
—
1
2
—
(5)2
—
sin 3
si
7
n
°
5
si
9
n
°
84°
—
K8.7 units2
21. C180° (15° 113°) or 52°
K
—
1
2
—
b2
—
sin
s
A
in
s
B
in C
—
K
—
1
2
—
(7)2
—
sin
s
1
i
5
n
°
1
s
1
i
3
n
°
52°
—
K5.4 units2
22. K
—
1
2
—
bc sin A
K
—
1
2
—
(146.2)(209.3) sin 62.2°
K13,533.9 units2
23. K
—
1
2
—
ac sin B
K
—
1
2
—
(12.7)(5.8) sin 42.8°
K25.0 units2
24. B180° (53.8° 65.4°) or 60.8°
K
—
1
2
—
a2
—
sin
s
B
in
s
A
in C
—
K(19.2)2
K181.3 units2
25. Kab sin X(formula from Exercise 3)
K(14)(20) sin 57°
K234.8 cm2
26. Area of pentagon
5 Area of triangle
360° 5 72°
K
—
1
2
—
(9)(9) sin 72° 5K5(38.51778891)
K38.51778891 5K192.6 in2
27. Area of octagon
8 Area of triangle
360° 8 45°
K
—
1
2
—
(5)(5) sin 45° 8K8(8.838834765)
K8.838834765 8K70.7 ft2
28a. 180° (95° 40°) 45°
28b.
—
sin
x
95°
—
—
sin
80
45°
——
sin
y
40°
—
—
sin
80
45°
—
x
—
80
si
s
n
in
45
9
°
5°
—
y
—
80
si
s
n
in
45
4
°
0°
—
x112.7065642 y72.72311643
about 112.7 ft and 72.7 ft
sin 60.8° sin 65.4°
———
sin 53.8°
1
—
2
28c. P112.7 72.7 80
P265.4 ft
29. Applying the Law of Sines,
—
sin
m
M
—
—
sin
n
N
—
and
—
sin
r
R
—
—
sin
s
S
—
. Thus sin M
—
nsi
n
nN
—
and sin R
—
rsi
s
nS
—
. Since ∠M∠R, sin Msin Rand
—
ms
n
in N
—
—
rsi
s
nS
—
. However, ∠N∠Sand
sin Nsin S, so
—
m
n
—
—
r
s
—
and
—
m
r
—
—
n
s
—
. Similar
proportions can be derived for pand t. Therefore,
MNP RST.
30. 360° 5 72°
triangle: K
—
1
2
—
(300)(300)sin 72°
K42,797.54323
pentagon: 5K5(42,797.54323)
5K213,987.7 ft
31a. v180° (20° 1562° 30) or 97° 15
Let xthe distance from the balloon to the
soccer fields.
—
sin 6
x
2° 30
—
—
sin 9
4
7° 15
—
x
—
4
s
s
in
in
9
6
7
2
°
°
1
3
5
0
—
x3.6 mi
31b. v180° (20° 1562° 30) or 97° 15
Let ythe distance from the balloon to the
football field.
—
sin 2
4
0° 15
—
—
sin 9
4
7° 15
—
y
—
4
s
s
in
in
9
2
7
0
°
°
1
1
5
5
—
y1.4 mi
32. 180° 30° 150°
v180° (26.8° 150°) or 3.2°
Let xthe length of the track.
—
sin 2
x
6.8°
—
—
sin
10
3
0
.2°
—
x
—
100
si
s
n
in
3.
2
2
6
°
.8°
—
x807.7 ft
33a. Let xthe distance of the second part of the
flight.
v180° (13° 160°) or 7°
—
sin
x
13°
—
—
si
8
n
0
7°
—
x
—
80
s
s
in
in
7
1
°
3°
—
x147.6670329
distance of flight 80 147.7 or about 227.7 mi
33b. Let ythe distance of a direct flight.
—
sin
y
160°
—
—
si
8
n
0
7°
—
y
—
80
s
s
i
i
n
n
7
1
°
60°
—
y224.5 mi
155 Chapter 5
y
x
63˚
27˚
62˚
13 ft
55˚63˚
37. sin v
—
y
r
—
r2x2y2
sin v
—
1
6
—
62x2(1)2
y1, r6 35 x2
35
x
Quadrant IV, so x35
cos v
—
x
r
—
tan v
—
y
x
—
csc v
—
y
r
—
cos vtan vcsc v
—
6
1
—
or 6
1
35
35
6
Chapter 5 156
34. 90° 63° 27°
180° (55° 63°) 62°
Let xthe vertical
distance.
Let ythe length of the
overhang.
—
sin
x
27°
—
—
sin
13
63°
——
sin
y
63°
—
—
sin
66
62°
—
x
—
13
si
s
n
in
63
2
°
7°
—
y
—
6.6
sin
sin
62
6
°
3°
—
x6.623830843 y6.684288563
about 6.7 ft
35a.
—
sin
a
A
—
—
sin
b
B
—
—
a
b
—
—
s
s
i
i
n
n
B
A
—
35b.
—
sin
a
A
—
—
sin
c
C
—
—
a
c
—
—
s
s
i
i
n
n
C
A
—
—
a
c
—
1
—
s
s
i
i
n
n
C
A
—
1
—
a
c
—
—
c
c
—
—
s
s
i
i
n
n
C
A
—
—
s
s
i
i
n
n
C
C
—
—
a
c
c
—
—
sin A
si
nC
sin C
—
35c. From Exercise 34b,
—
a
c
c
—
—
sin A
si
nC
sin C
—
or
—
sin A
a
s
c
in C
—
—
sin
c
C
—
.
—
sin
a
A
—
—
sin
c
C
—
—
a
c
—
—
s
s
i
i
n
n
C
A
—
—
a
c
—
1
—
s
s
i
i
n
n
C
A
—
1
—
a
c
—
—
c
c
—
—
s
s
i
i
n
n
C
A
—
—
s
s
i
i
n
n
C
C
—
—
a
c
c
—
—
sin A
si
nC
sin C
—
—
sin
c
C
—
—
sin A
a
s
c
in C
—
Therefore,
—
sin A
a
s
c
in C
—
—
sin A
a
s
c
in C
—
or
—
a
a
c
c
—
—
s
s
i
i
n
n
A
A
s
s
i
i
n
n
C
C
—
.
35d.
—
sin
a
A
—
—
sin
b
B
—
—
a
b
—
—
s
s
i
i
n
n
B
A
—
—
a
b
—
1
—
s
s
i
i
n
n
B
A
—
1
—
a
b
—
—
b
b
—
—
s
s
i
i
n
n
B
A
—
—
s
s
i
i
n
n
B
B
—
—
a
b
b
—
—
sin A
si
nB
sin B
—
—
a
b
b
—
—
sin A
si
nB
sin B
—
36. tan v
—
4
2
5
0
—
vtan1
—
4
2
5
0
—
v66.0°
O
y
x
4
8
12
16
12 16
(4, 11)
(8, 7)
(2, 4)
(2, 11)
(8, 4)
x
y
15
x
2
x
8
y
11
y
4
O
y
6
3
x
y
12
x
tan v
sec v
—
x
r
—
cot v
—
x
y
—
sec vcot vor 35
35
1
6
35
35
35
sec v
38. 83° 360k°
39. Let xstandard carts and let ydeluxe carts.
2 x8
4 y11
xy15
M(x, y) 100x250y
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. 4xy2z0
3x4y2z20
7x5y2z20
3(3x4y2z) 3(20)
2(2x5y3z) 2(14)
↓
9x12y6z60
4x10y6z28
5x22y88
5(7x5y) 5(20) →35x25y100
7(5x22y) 7(88) 35x154y616
129y516
y4
7x5y20 4xy2z0
7x5(4) 20 4(0) 4 2z0
x0z2
(0, 4, 2)
41. 6 3xy3xy12
y3x6y3x12
635
35
42. Area of one face of the small cube 12or 1 in2.
Surface area of the small cube 6 1 or 6 in2.
Area of one face of large cube 22or 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 in224 in2
or 24 in2.
The correct choice is A.
Page 319 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.
The Ambiguous Case for the Law
of Sines
Page 323 Graphing Calculator Exploration
1. B44.1°, C23.9°, c1.8
2. B52.7°, C76.3°, b41.0; B25.3°,
C103.7°, b22.0
3. The answers are slightly different.
4. Answers will vary if rounded numbers are used to
find some values.
Page 324 Check for Understanding
1. Atriangle cannot exist if m∠A90° and
absin Aor if m∠A90° and ab.
2.
5-7
5. Since 44° 90°, consider Case I.
asin B23 sin 44°
asin B23 (0.6947)
asin B15.97714252
12 16.0; 0 solutions
6. Since 17° 90°, consider Case I.
asin C10 sin 17°
2.923717047
2.9 10 11; 1 solution
sin
c
C
sin
a
A
sin
11
17°
si
1
n
0
A
sin A
10 s
1
in
1
17°
Asin1
10 s
1
in
1
17°
A15.41404614
B180° (15.4° 17°) or about 147.6°
sin
c
C
sin
b
B
sin
11
17°
sin 1
b
47.6°
b
11 s
s
i
i
n
n
1
1
4
7
7
°
.6°
b20.16738057
A15.4°, B147.6°, b20.2
7. Since 140° 90°, consider Case II.
3 10; no solutions
8. Since 38° 90°, consider Case I.
bsin A10 sin 38°
bsin A6.156614753
6.2 8 10; 2 solutions
sin
a
A
sin
b
B
sin
8
38°
si
1
n
0
B
sin B
10 si
8
n38°
Bsin1
10 si
8
n38°
B50.31590502
180° 180° 50.3° or 129.7°
Solution 1
C180° (50.3° 38°) or 91.7°
sin
a
A
sin
c
C
sin
8
38°
sin 9
c
1.7°
c
8s
s
i
i
n
n
9
3
1
8
.
°
7°
c12.98843472
Solution 2
C180° (129.7° 38°) or 12.3°
sin
a
A
sin
c
C
sin
8
38°
sin 1
c
2.3°
c
8s
s
i
i
n
n
1
3
2
8
.
°
3°
c2.768149638
B50.3°, C91.7°, c13.0; B129.7°,
C12.3°, c2.8
157 Chapter 5
A
B
C
10
6
30˚
56.4˚
93.6˚
A
10
6
C
B
30˚26.4˚
123.6˚
180° 180° 56.4°
123.6°
C180° (30° 123.6°)
26.4°
sin
6
30°
si
1
n
0
B
sin B
10 si
6
n30°
Bsin1
10 si
6
n30°
B56.44269024
C180° (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 Bfrom 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
80˚
100˚
10˚
45 70
39.3˚
y
x
˚
80˚
100˚
10˚
45 70
9. Since 130° 90°, consider Case II.
17 5; 1 solution
sin
c
C
sin
b
B
sin
1
1
7
30°
sin
5
B
sin B
5sin
17
130°
Bsin1
5sin
17
130°
B13.02094264
A180° (13.0 130°) or 37.0°
sin
c
C
sin
a
A
sin
1
1
7
30°
sin 3
a
7.0°
a
17
si
s
n
in
1
3
3
7
0
.
°
0°
a13.35543321
A37.0°, B13.0°, a13.4
10a.
10b. 90° 10° 80°
180° 80° 100°
sin
7
1
0
00°
si
4
n
5
x
sin x
45 si
7
n
0
100°
xsin1
45 si
7
n
0
100°
x39.3°
10c. v180° (100° 39.3°)
or about 40.7°
sin 4
y
0.7°
sin
7
1
0
00°
y
70
si
s
n
in
1
4
0
0
0
.
°
7°
y46.4 ft
Pages 324–326 Exercises
11. Since 57° 90°, consider Case I.
bsin A19 sin 57°
bsin A15.93474079
11 15.9; 0 solutions
12. Since 30° 90°, consider Case I.
csin A26 sin 30°
csin A13
13 13; 1 solution.
13. Since 61° 90°, consider Case I.
asin B12 sin 61°
asin B10.49543649
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.
asin B32 sin 37°
asin B19.25808074
27 19.3; 2 solutions
17. Since 65° 90°, consider Case I.
bsin A57 sin 65°
bsin A51.65954386
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 A29 sin 58°
b sin A24.59339479
26 24.6; 2 solutions
sin
a
A
sin
b
B
sin
26
58°
si
2
n
9
B
sin B
29 s
2
in
6
58°
Bsin1
29 s
2
in
6
58°
B71.06720496
180° 180° 71.1° or 108.9°
Solution 1
C180° (58° 71.1°) or 50.9°
sin
a
A
sin
c
C
sin
26
58°
sin 5
c
0.9°
c
26
s
s
i
i
n
n
5
5
8
0
°
.9°
c23.80359004
Solution 2
C180° (58° 108.9)° or 13.1°
sin
a
A
sin
c
C
sin
26
58°
sin 1
c
3.1°
c
26
s
s
i
i
n
n
5
1
8
3
°
.1°
c6.931727606
B71.1°, C50.9°, c23.8; B108.9°,
C13.1°, c6.9
Chapter 5 158
70 ft
45 ft
10˚
20. Since 30° 90°, consider Case I.
bsin A8 sin 30°
bsin A4
4 4; 1 solution
sin
a
A
sin
b
B
sin
4
30°
sin
8
B
sin B
8sin
4
30°
Bsin1
8sin
4
30°
B90
C180° (30° 90°) or 60°
sin
a
A
sin
c
C
sin
4
30°
sin
c
60°
C
4
s
s
in
in
3
6
0
0
°
°
C6.92820323
B90°; C60°, c6.9
21. Since 70° 90°, consider Case I.
asin C25 sin 70°
asin C23.49231552
24 23.5; 2 solutions
sin
c
C
sin
a
A
sin
24
70°
si
2
n
5
A
sin A
25 s
2
in
4
70°
Asin1
25 s
2
in
4
70°
A78.1941432
180° 180° 79.2° or 101.8°
Solution 1
B180° (70° 79.2°) or 31.8°
sin
c
C
sin
b
B
sin
24
70°
sin 3
b
1.8°
b
24
s
s
i
i
n
n
7
3
0
1
°
.8°
b13.46081025
Solution 2
B180° (70° 101.8°) or 8.2°
sin
c
C
sin
b
B
sin
24
70°
sin
b
8.2°
b
24
si
s
n
in
7
8
0
.
°
2°
b3.640196918
A78.2°, B31.8°, b13.5; A101.8°,
B8.2°, b3.6
22. C180° (40° 60°) or 80°
sin
c
C
sin
a
A
sin
c
C
sin
b
B
sin
20
80°
sin
a
40°
sin
20
8
°
0°
sin
b
60°
a
20
si
s
n
in
80
4
°
0°
b
20
si
s
n
in
80
6
°
0°
a13.05407289 b17.58770483
C80°, a13.1, b17.6
23. Since 90° 90°; consider Case II.
12 14; no solution
24. Since 36° 90°, consider Case I.
csin B30 sin 36°
csin B17.63355757
19 17.6; 2 solutions
sin
b
B
sin
c
C
sin
19
36°
si
3
n
0
C
sin C
30 s
1
in
9
36°
Csin1
30 s
1
in
9
36°
C68.1377773
180° 180° 68.1° or 111.9°
Solution 1
A180° (36° 68.1°) or 75.9°
sin
b
B
sin
a
A
sin
19
36°
sin 7
a
5.9°
a
19
s
s
i
i
n
n
3
7
6
5
°
.9°
a31.34565276
Solution 2
A180° (36° 111.9°) or 32.1°
sin
b
B
sin
a
A
sin
19
36°
sin 3
a
2.1°
a
19
s
s
i
i
n
n
3
3
6
2.1°
a17.1953669
A75.9°, C68.1°, a31.3; A32.1°,
C111.9°, a17.2
25. Since 107.2° 90°, consider Case II.
17.2 12.2; 1 solution
sin
a
A
sin
c
C
sin
1
1
7
0
.2
7.2°
s
1
i
2
n
.2
C
sin C
12.2 s
1
i
7
n
.2
107.2°
Csin1
12.2 s
1
i
7
n
.2
107.2°
C42.65491459
B180° (107.2° 42.7°) or 30.1°
sin
a
A
sin
b
B
sin
1
1
7
0
.2
7.2°
sin 3
b
0.1°
b
17
s
.
i
2
n
s
1
in
07
3
.
0
2
.
°
1°
b9.042067456
B30.1°, C42.7°, b9.0
26. Since 76° 90°, consider Case I.
bsin A20 sin 76°
bsin A19.40591453
5 19.4; no solution
159 Chapter 5
27. Since 47° 90°, consider Case I.
16 10; 1 solution
sin
c
C
sin
a
A
sin
16
47°
si
1
n
0
A
sin A
10 s
1
in
6
47°
Asin1
10 s
1
in
6
47°
A27.19987995
B180° (47° 27.2) or 105.8°
sin
c
C
sin
b
B
sin
16
47°
sin 1
b
05.8°
b
16 s
s
i
i
n
n
1
4
0
7
5
°
.8°
b21,0506609
A27.2°, B105.8°, b21.1
28. Since 40° 90°, consider Case I.
csin B60 sin 40°
csin B38.56725658
42 38.6; 2 solutions
sin
b
B
sin
c
C
sin
42
40°
si
6
n
0
C
sin C
60 s
4
in
2
40°
Csin1
60 s
4
in
2
40°
C66.67417652
180° 180° 66.7° or 113.3°
Solution 1
A180° (40° 66.7°) or 73.3°
sin
b
B
sin
a
A
sin
42
40°
sin 7
a
3.3°
a
42
s
s
i
i
n
n
4
7
0
3
°
.3°
a62.58450564
Solution 2
A180° (40° 113.3°) or 26.7°
sin
b
B
sin
a
A
sin
42
40°
sin 2
a
6.7°
a
42
s
s
i
i
n
n
4
2
0
6
°
.7°
a29.33237132
A73.3°, C66.7°, a62.6; A26.7°,
C113.3°, a29.3
29. Since 125.3° 90°, consider Case II.
32 40; no solution
30.
sin
21
5
.
7
7
.4°
s
1
i
9
n
.3
x
sin x
19.3
2
si
1
n
.7
57.4°
xsin1
19.3
2
si
1
n
.7
57.4°
x48.52786934
v180 (57.4° 48.5°) or 74.1°
sin
21
5
.
7
7
.4°
sin 7
y
4.1°
y
21.
s
7
in
si
5
n
7.
7
4
4
°
.1°
y24.76922417
31.
sin
a
A
sin
b
B
sin
15
29°
si
2
n
0
B
sin B
20 s
1
in
5
29°
Bsin1
20 s
1
in
5
29°
B40.27168721
180° 180° 40.3° or 139.7°
Solution 1
C180° (29° 40.3°) or 110.7°
sin
a
A
sin
c
C
sin
15
29°
sin 1
c
10.7°
c
15
s
s
i
i
n
n
2
1
9
10
°
.7°
c28.93721187
Perimeter abc
15 20 28.9 or about 63.9 units
Solution 2
C180° (29° 139.7°) or 11.3°
sin
a
A
sin
c
C
sin
15
29°
sin 1
c
1.3°
c
15
s
s
i
i
n
n
2
1
9
1
°
.3°
c6.047576406
Perimeter abc
15 20 6.0 or about 41.0 units
32.
sin
b
B
sin
a
A
sin
13
55°
si
1
n
5
A
sin A
15 s
1
in
3
55°
Asin1
15 s
1
in
3
55°
A70.93970395
180° 180° 70.9° or 109.1°
Solution 1
C180° (70.9° 55°) or 54.1°
sin
b
B
sin
c
C
sin
13
55°
sin 5
c
4.1°
c
13
s
s
i
i
n
n
5
5
5
4
°
.1°
c12.8489656
Perimeter abc
15 13 12.8 or about 40.8
Solution 2
c180° (109.1° 55°) or 15.9°
sin
b
B
sin
c
C
sin
13
55°
sin 1
c
5.9°
c
13
s
s
i
i
n
n
5
1
5
5
°
.9°
c4.35832749
Perimeter abc
15 13 4.4 or about 32.4
A70.9°, B55°, C54.1°
Chapter 5 160
24.8 cm
19.3 cm 21.7 cm
48.5˚
57.4˚
74.1˚
33. side opposite 37° 15 18 or 33
side between vand 37° 15 22 or 37
Let xthe measure of the third angle.
sin
33
37°
si
3
n
7
x
sin x
37 s
3
in
3
37°
xsin1
37 s
3
in
3
37°
x42.43569405
v180 (37° 42.4°) or about 100.6°
34a. absin A34b. absin A
a14 sin 30° a14 sin 30°
a7a7 or a14
34c. absin A
a14 sin 30°
a7 and a14
7 a14
35.
s
1
in
84
5
.
9
5
°
s
1
in
40
x
sin x
140
18
si
4
n
.5
59°
xsin1
140
18
si
4
n
.5
59°
x40.57365664
v180° (59° 40.6°) or about 80.4°
90° 80.4° 9.6°
36a. 12° 90° and 316 450 sin 12°; 2 solutions
si
3
n
1
1
6
2°
s
4
i
5
n
0
v
sin v
450
3
s
1
in
6
12°
vsin1
450
3
s
1
in
6
12°
v17.22211674
180° 180° 17.2° or 162.8°
turn angle 180° 162.8° or 17.2°
about 17.2° east of north
36b. v180° (162.8° 12°) or about 5.2°
sin
x
5.2°
si
3
n
1
1
6
2°
x
316
sin
sin
12
5
°
.2°
x138.3094714
drt
138.3 23t
6.013455278 t; about 6 hr
36c. 180° 20° 160°
180° (160° 12°) 8°
si
2
n
00
8°
sin
c
160°
c
200
s
s
in
in
8
1
°
60°
c491.5032301
Since 491.5 450, the ship
will not reach port.
37. Distance from satellite to center of earth is 3960
1240 or 5200 miles.
angle across from 5200 mi side 45° 90° or 135°
si
5
n
2
1
0
3
0
5°
3
si
9
n
60
x
sin x
3960
5
s
2
i
0
n
0
135°
xsin1
3960
5
s
2
i
0
n
0
135°
x32.58083835
v180° (135° 32.6°) or about 21.4°
2
3
1
6
.
0
4
°
°
(2 hours) 0.0689953425 hours or about 4.1
minutes
38. Pturns 20(360°) or 7200° every second which
equals 72° every 0.01 second.
si
P
n
Q
O
si
O
n
P
Q
sin
15
72°
sin
5
Q
sin Q
5si
1
n
5
72°
Qsin1
5si
1
n
5
72°
Q18.48273235
m∠P180° (72° 18.5°) or about 89.5°
s
Q
in
O
P
si
P
n
Q
O
sin
Q
8
O
9.5°
sin
15
72°
QO
15
s
s
i
i
n
n
7
8
2
9
°
.5°
QO 15.77133282
QO 5 15.8 5 or about 10.8 cm
39a. bcsin B
12 17 sin B
1
1
2
7
sin B
sin1
1
1
2
7
B
44.90087216 B
B44.9°
39b. bcsin B
12 17 sin B
1
1
2
7
sin B
sin1
1
1
2
7
B
44.90087216 B
B44.9°
39c. bcsin B
12 17 sin B
1
1
2
7
sin B
sin1
1
1
2
7
B
44.90087216 B
B44.9°
40. Area of rhombus 2(Area of triangle)
triangle:
K
1
2
bc sin A
K
1
2
(24)(24) sin 32°
K152.6167481
rhombus:
A2(152.6) or about 305.2 in2
161 Chapter 5
20˚
160˚
c
12˚
200 mi.
Port
C
10
8c
B
A
50˚
41. tan 22°
7
x
5
x
tan
75
22°
x185.6 m
42. 3;
1
2
44136
216
4212
0
4x22x12 0
x2 (2)2
4(4
)(12)
2(4)
4. Sample answers:
A180° (90° 53°) or 37°
sin B
b
c
sin 53°
1
c
0
c
sin
10
53°
c12.52135658
tan B
a
b
tan 53°
1
a
0
a
tan
10
53°
a7.535540501
A37°, a7.5, c12.5
C180° (5.5° 45°) or 80°
sin
b
B
sin
a
A
sin
b
45°
sin
10
55°
b
10
si
s
n
in
55
4
°
5°
b8.6321799
sin
c
C
sin
a
A
sin
c
80°
sin
10
55°
c
10
si
s
n
in
55
8
°
0°
c12.0222828
C80°, b8.6, c12.0
c2a2b22ab cos C
c2102822(10)(8) cos 50°
c261.15398245
c7.820101179
sin
a
A
sin
c
C
si
1
n
0
A
sin
7.
5
8
0°
sin A
10 s
7
i
.
n
8
50°
Asin1
10 s
7
i
.
n
8
50°
A78.4024367
B180° (78.4° 50°) or 51.6°
A78.4°, B51.6°, c7.8
5. a2b2c22bc cos A
3223824622(38)(46) cos A
322
2(
3
3
8
8
2
)(4
6)
462
cos A
cos1
A
43.49782861 A
sin
a
A
sin
b
B
sin
3
4
2
3.5°
si
3
n
8
B
sin B
38 si
3
n
2
43.5°
Bsin1
38 si
3
n
2
43.5°
B54.8
C180° (43.5° 54.8°) or 81.7°
A43.5°, B54.8°, C81.7°
322382462
2(38)(46)
Chapter 5 162
x2 188
8
x22i47
8
x
43. no
4x
9
x
1
x
44. 5x2y9y3x1
5x2(3x1) 9y3(7) 1
5x6x2 9y22
x7
(7, 22)
45. 2x5y7
y
2
5
x
7
5
perpendicular slope:
5
2
y4
5
2
(x(6))
y4
5
2
x15
2y8 5x30
5x2y22
46. Perimeter of XYZ 4 8 9 or 21
length of A
B
1
3
of perimeter
1
3
(21) or 7
The answer is 7.
The Law of Cosines
Pages 330–331 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 c2a2b22ab cos C. Since
cos 90° 0, c2a2b22ab(0) or c2a2b2.
5-8
4
x
x
1
1
x
9
x
1
x
3
x
1
x
x
1
1
x
9
x
1
x
3
x
1
1
3
x
3
x
1
1i47
4
C
10
B
A
a
c
53˚
C
10
bc
B
A
45˚
55˚
50 ft
65 ft
45.0˚
c
65 ft
65 ft
c
6. c2a2b22ab cos C
c22523022(25)(30) cos 160°
c22934.538931
c54.1713848
sin
c
C
sin
a
A
sin
54
1
.
6
2
0°
si
2
n
5
A
sin A
25 s
5
i
4
n
.2
160°
Asin1
25 s
5
i
4
n
.2
160°
A9.1
B180° (9.1° 160°) or 10.9°
A9.1°, B10.9°, c54.2
7. The angle with greatest measure is across from
the longest side.
2121821422(18)(14) cos v
212
2(
1
1
8
8
2
)(1
4)
142
cos v
cos1
212
2(
1
1
8
8
2
)(1
4)
142
v
81.0 v
about 81.0°
8. s
1
2
(2 7 8) 8.5
K8.5(8.
5 2
)(8.5
7)(8.5
8)
6.4 units2
9. s
1
2
(25 13 17) 27.5
K27.5(2
7.5
25)(2
7.5
13)(2
7.5
17)
102.3 units2
10. a2b2c2
652652c2
8450 c2
91.92388155 c
a2b2c22bc cos v
65265291.922(65)(91.9) cos v
cos v
cos1
v
45.01488334 v
c2a2b22ab cos C
c26525022(65)(50) cos 45.0°
c22128.805922
c46.1 ft
65265291.92
2(65)(91.9)
65265291.92
2(65)(91.9)
Pages 331–332 Exercises
11. a2b2c22bc cos A
a2721022(7)(10) cos 51°
a260.89514525
a7.803534151
sin
a
A
sin
b
B
sin
7.
5
8
1°
sin
7
B
sin B
7si
7
n
.8
51°
Bsin1
7si
7
n
.8
51°
B44.22186872
C180° (51° 44.2°) or 84.8°
B44.2°, C84.8°, a7.8
12. c2a2b22ab cos C
7252622(5)(6) cos C
72
2(
5
5
2
)(
6)
62
cos C
cos1
72
2(
5
5
2
)(
6)
62
C
78.46304097 C
sin
a
A
sin
c
C
sin
5
A
sin 7
7
8.5°
sin A
5sin
7
78.5°
Asin1
5sin
7
78.5°
A44.42268919
B180° (44.4° 78.5°) or 57.1°
A44.4°, B57.1°, C78.5°
13. c2a2b22ab cos C
7242522(4)(5) cos C
72
2(
4
4
2
)(
5)
52
cos C
cos1
72
2(
4
4
2
)(
5)
52
C
101.536959 C
sin
a
A
sin
c
C
sin
4
A
sin 1
7
01.5°
sin A
4sin
7
101.5°
Asin1
4sin
7
101.5°
A34.05282227
B180° (34.1° 101.5°) or 44.4°
A34.1°, B44.4°, C101.5°
14. b2a2c22ac cos B
b21621222(16)(12) cos 63°
b2225.6676481
b15.02223845
sin
a
A
sin
b
B
si
1
n
6
A
si
1
n
5
6
.0
3°
sin A
16
1
si
5
n
.0
63°
Asin1
16
1
si
5
n
.0
63°
A71.62084388
C180° (71.6° 63°) or 45.4°
A71.6°, C45.4°, b15.0
163 Chapter 5
74 ft 38 ft
88 ft
x
15. b2a2c22ac cos B
13.7211.4212.22
2(11.4)(12.2) cos B
cos B
cos1
B
70.8801474 B
sin
a
A
sin
b
B
s
1
in
1.4
A
sin
13
7
.
0
7
.9°
sin A
11.4
1
si
3
n
.7
70.9°
Asin1
11.4
1
si
3
n
.7
70.9°
A51.84180107
C180° (51.8° 70.9°) or 57.3°
A51.8°, B70.9°, C57.3°
16. c2a2b22ab cos C
c221.521322(21.5)(13) cos 79.3°
c2527.462362
c22.96654876
sin
a
A
sin
c
C
s
2
i
1
n
.5
A
sin
23
7
.
9
0
.3°
sin A
21.5
2
si
3
n
.0
79.3°
Asin1
21.5
2
si
3
n
.0
79.3°
A66.90667662
B180° (66.9° 79.3°) or 33.8°
A66.9°, B33.8°, c23.0
17. 14.9223.8236.92
2(23.8)(36.9) cos v
cos v
cos1
v
13.75878964 v
about 13.8°
18. d124026022(40)(60) cos 132°
d128411.826911
d191.71601229
180° 132° 48°
d224026022(40)(60) cos 48°
d221988.173089
d244.58893461
about 91.7 cm and 44.6 cm
19. s
1
2
(4 6 8) 9
K9(9
4)(9
6)(9
8)
11.6 units2
20. s
1
2
(17 13 19) 24.5
K24.5(2
4.5
17)(2
4.5
13)(2
4.5
19)
107.8 units2
21. s
1
2
(20 30 40) 45
K45(45
20
)(45
30)(4
5 4
0)
290.5 units2
22. s
1
2
(33 51 42) 63
K63(63
33
)(63
51)(6
3 4
2)
690.1 units2
14.9223.8236.92
2(23.8)(36.9)
14.9223.8236.92
2(23.8)(36.9)
13.7211.4212.22
2(11.4)(12.2)
13.7211.4212.22
2(11.4)(12.2)
23. s
1
2
(174 138 188) 250
K250
76 1
12 6
2
11,486.3 units2
24. s
1
2
(11.5 13.7 12.2) 18.7
K187(1
8.7
11.5)
(18.7
13.
7)(18.
7 1
2.2)
66.1 units2
25a. d23024822(30)(48) cos 120°
d24644
d68.1 in.
25b. Area of parallelogram 2(Area of triangle)
K
1
2
(30)(48) sin 120°
K623.5382907
2K2(623.5382907) or about 1247.1 in2
26a. s
1
2
(15 15 24.6) 27.3
K27.3(2
7.3
15)(2
7.3
15)(2
7.3
24.6)
105.6
Area of rhombus 2(105.6) 211.2 cm2
26b. 24.621521522(15)(15) cos v
cos v
cos1
v
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°.
x240029022(400)(90) cos 45°
x2117,188.3118
x342.3 ft
28.
3827428822(74)(88) cos v
cos v
cos1
382
2(
7
7
4
4
2
)(8
8)
882
v
25.28734695 v
sin v
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
sin 25.3°
7
x
4
31.60970664 x
about 31.6 ft
29a. x2100222022(100)(220) cos 10°
x215,068.45887
x122.7536511
about 122.8 mi
29b. (100 122.7536511) 220 2.7536511
about 2.8 mi
382742882
2(74)(88)
24.62152152
2(15)(15)
24.62152152
2(15)(15)
Chapter 5 164
125 ft
II
I
III
201.5 ft
202 ft
180.25 ft
158 ft
y
x
75˚
82.5˚
30.
I: K
1
2
(201.5)(202) sin 82.5°
K20,177.3901
II: x2201.5220222(201.5)(202) cos 82.5°
x270,780.6348
x266.046302
y21582180.2522(158)(180.25) cos 75°
y242.711.98851
y206.6687894
206.72266.021252
2(266.0)(125) cos v
cos v
cos1
v
48.93361962 v
K
1
2
(266.0)(125) sin 48.9°
K12,536.58384
III: K
1
2
(180.25)(158) sin 75°
K13,754.54228
Area of pentagon I II III
20,177.4 12,536.6 13,754.5
46,468.5 ft
31. I: 2423524022(35)(40) cos v
242
2(
3
3
5
5
2
)(4
0)
402
cos v
cos1
242
2(
3
3
5
5
2
)(4
0)
402
v
36.56185036 v
II: 2423022022(30)(20) cos v
242
2(
3
3
0
0
2
)(2
0)
202
cos v
cos1
242
2(
3
3
0
0
2
)(2
0)
202
v
52.89099505 v
the player 30 ft and 20 ft from the posts
32a. sin 6°
20,
x
000
x
2
s
0
in
,0
6
0
°
0
x191,335.4 ft
32b. sin 3°
15,
y
000
y
1
s
5
in
,0
3
0
°
0
y286,609.8 ft
32c. 6° 3° 3°
d2191,335.42286,609.82
2(191,335.4)(286,609.8) cos 3°
d29,227,519,077
d96,060.0 ft
33. Since 63.2° 90°, consider Case I.
bsin A18 sin 63.2°
bsin A16.06654473
17 16.1; 2 solutions
206.72266.021252
2(266.0)(125)
206.72266.021252
2(266.0)(125)
165 Chapter 5
O
50
40
C
A
(0, 0)
B
(50, 0)
51˚
y
x
O
50
40
A
(0, 0)
C
B
(50, 0)
51˚
y
x
34. tan v
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
tan v
5
7
7
0
0
0
vtan1
5
7
7
0
0
0
v39.2°
35. 775° 2(360°) 55°
reference angle 55°or 55°
36. 317k6
312 36 3k
1412 k30 3k
30 3k0
k10
37. m
5
5
t
t
2
t
t
m
4
3
t
t
or
4
3
38.
2
y
x2
3
2
y
3x
3
6
8
y
x
3
6
The correct choice is A.
Graphing Calculator Exploration:
Solving Triangles
Page 334
1. AB 12.1, B25.5°, C119.5°
2.
Find yusing A
C
.
y(tan 51°)x
Find yusing B
C
.
sin
40
51°
si
5
n
0
C
sin C
50 s
4
in
0
51°
Csin1
50 s
4
in
0
51°
C76.27180414
B80 51 76.27180414
52.72819586
tan(180 52.72819586)
x
y
50
(x50) tan(127.2718041) y
Set the two values of y equal to each other.
(tan 51°)x(x50), tan 127.2718041°
(tan 51°)xx(tan 127.2718041°
50(tan 127.2718041°)
x
x25.77612538
y(tan 51°) (25.77612538) 31.83086394
50(tan 127.2718041°)
tan 51° tan 127.271804°
5-8B
Ccould also equal 180 76.27180414 or
103.728195°
B180 51 103.7281959°
25.2718041
tan (180 25.2718041)
x
y
50
(x50) tan 154.7281959° y
Set the two values of yequal to each other.
(tan 51°) x(x50)tan 154.7281959°
(tan 51°) xx(tan 154.7281959°) 50(tan
154.7281959°)
x
x13.82829048
y(tan 51°) (13.82828048)
17.07651659
B52.7°, C76.3°, b40.9; B25.3°,
C 103.7°, b220
3. Law of Cosines
4. Sample answer: put vertex Aat the origin and
vertex Cat (3, 0).
Chapter 5 Study Guide and Assessment
Page 335 Understanding and Using the
Vocabulary
1. false; depression 2. false; arcosine
3. true 4. false; adjacent to
5. true 6. false; coterminal
7. true 8. false; Law of Cosines
9. false; terminal side 10. true
Pages 336–338 Skills and Concepts
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° 730)
17° 730
50(tan 154.7281959°)
tan 51° tan 154.7281959°
Chapter 5 166
13.
8
3
6
6
0
0
°
°
2.39
360(2)° 860°
720° 860°
140°; II
14.
1
3
1
6
4
0
6
°
°
3.18
360(3)° 1146°
1080° 1146°
66°; I
15.
3
1
6
5
0
6
°
°
0.43
360(1)° 156°
360° 156°
204°; III
16.
9
3
9
6
8
0
°
°
2.77
360(2)° 998°
720° 998°
278°; IV
17.
3
3
6
0
0
0
°
°
0.83
360(1)° 300°
360° 300°
60°; I
18.
1
3
0
6
7
0
2
°
°
2.98
360(2)° 1072°
720° 1072°
352°; IV
19.
6
3
5
6
4
0
°
°
1.82
360(1)° 654°
360° 654°
294°; IV
20.
3
8
6
3
0
2
°
°
2.31
360(2)° 832°
720° 832°
112°
360° 112° 248°; III
21. 284° has terminal side in first quadrant.
360° 284° 76°
22.
5
3
9
6
2
0
°
°
1.64
360(1)° 592°
360° 592°
232°
terminal side in third quadrant
232° 180° 52°
23. (BC)2(AC)2(AB)2
15292(AB)2
306 (AB)2
306
AB
334
AB
sin A
o
h
p
y
p
p
o
o
s
t
it
e
e
nu
si
s
d
e
e
cos A
a
h
d
y
ja
p
c
o
e
t
n
en
t
u
si
s
d
e
e
sin Aor cos Aor
tan A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
tan A
1
9
5
or
5
3
334
34
9
334
534
34
15
334
24. (PM)2(PN)2(MN)2
82122(MN)2
208 (MN)2
208
MN
413
MN
sin M
o
h
p
y
p
p
o
o
s
t
it
e
e
nu
si
s
d
e
e
cos M
a
h
d
y
ja
p
c
o
e
t
n
en
t
u
si
s
d
e
e
sin Mor cos Mor
tan M
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc M
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan M
1
8
2
or
3
2
csc Mor
sec M
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot M
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec Mor cot M
1
8
2
or
2
3
25. (MP)2(PN)2(MN)2
(MP)2102122
(MP)244
MP 44
or 211
sin M
o
h
p
y
p
p
o
o
s
t
it
e
e
nu
si
s
d
e
e
cos M
a
h
d
y
ja
p
c
o
e
t
n
en
t
u
si
s
d
e
e
sin M
1
1
0
2
or
5
6
cos Mor
tan M
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc M
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan Mor csc M
1
1
2
0
or
6
5
sec M
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot M
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec Mor cot Mor
26. sec v
co
1
sv
cos v
se
1
cv
cos vor
5
7
27. rx2y
2
r323
2
r18
or 32
sin v
y
r
cos v
x
r
tan v
y
x
1
7
5
11
5
211
10
611
11
12
211
511
11
10
211
11
6
211
12
13
2
413
8
13
3
413
12
213
13
8
413
313
13
12
413
29. r x2y
2
r82(
2)2
r68
or 217
sin v
y
x
cos v
x
y
tan v
y
x
sin vcos vtan v
8
2
or
1
4
8
217
2
217
167 Chapter 5
sin vcos vtan v
3
3
or 1
sin vcos v
csc v
y
r
sec v
x
r
csc vor 2
sec vor 2
cot v
x
y
cot v
3
3
or 1
28. rx2y
2
r(5)2
122
r169
or 13
sin v
y
r
cos v
x
r
tan v
y
x
sin v
1
1
2
3
cos v
13
5
or
1
5
3
tan v
12
5
or
1
5
2
csc v
y
r
sec v
x
r
cot v
x
y
csc v
1
1
3
2
sec v
13
5
or
1
5
3
cot v
12
5
or
1
5
2
32
3
32
3
2
2
2
2
3
32
3
32
sin v cos v
csc v
y
r
sec v
x
r
417
17
17
17
csc vor 17
sec vor 17
4
217
8
217
2
cot v
x
y
cot v
8
2
or 4
30. rx2y
2
r(2)2
02
r4
or 2
sin v
y
r
cos v
x
r
tan v
y
x
sin v
0
2
or 0 cos v
2
2
or 1tan v
0
2
or 0
csc v
y
r
sec v
x
r
cot v
x
y
csc v
2
0
sec v
2
2
or 1cot v
0
2
undefined undefined
31. rx2y
2
r425
2
r41
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
5
4
4
41
5
41
sin vcos v
csc v
y
r
sec v
x
r
cot v
x
y
csc vsec vcot v
4
5
32. rx2y
2
r(5)2
(9
)2
r106
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
9
5
or
9
5
5
106
9
106
41
4
41
5
441
41
541
41
sin v cos v
csc v
y
r
sec v
x
r
csc vor sec vor
cot v
x
y
cot v
5
9
or
5
9
33. rx2y
2
r(4)2
42
r32
or 42
sin v
y
r
cos v
x
r
tan v
y
x
sin vcos vtan v
4
4
or 1
sin vcos v
csc v
y
r
sec v
x
r
cot v
x
y
csc vsec vcot v
4
4
or 1
csc v2
sec v2
42
4
42
4
2
2
2
2
4
42
4
42
106
5
106
5
106
9
106
9
5106
106
9106
106
34. rx2y
2
r520
2
r25
or 5
sin v
y
r
cos v
x
r
tan v
y
x
sin v
0
5
or 0 cos v
5
5
or 1 tan v
0
5
or 0
csc v
y
r
sec v
x
r
cot v
x
y
csc v
5
0
sec v
5
5
or 1 cot v
5
0
undefined undefined
35. cos v
x
r
r2x2y2
cos v
3
8
82(3)2y2
x3, r8 55 y2
55
y
Quadrant 11, so y55
sin v
y
r
tan v
y
x
sin vtan vor
csc v
y
r
sec v
x
r
cot v
x
y
csc v
8
55
sec v
8
3
or
8
3
cot v
5
3
5
csc v
8
55
55
cot v
3
55
55
36. tan v
y
x
r2x2y2
tan v3; Quadrant III r2(1)2(3)2
y3, x1r210
r10
sin v
—
y
r
—
cos v
—
x
r
—
csc v
—
x
r
—
sin v
1
3
0
cos v
1
1
0
csc v
1
3
0
or
3
10
sin v
3
10
10
cos v
1
1
0
0
sec v
—
x
r
—
cot v
—
x
y
—
sec v
1
1
0
or 10
cot v
1
3
or
1
3
55
3
55
3
55
8
41.40962211° B90°
B48.6°
a13.2, A41.4°, B48.6°
44. 64° B90°
B26°
sin A
—
a
c
—
cos A
—
b
c
—
sin 64°
—
2
a
8
—
cos 64°
—
2
b
8
—
28 sin 64° a28 cos 64° b
25.2 a12.3 b
B26°, a25.2, b12.3
45. A180° (70° 58°) or 52°
—
sin
b
B
—
—
sin
a
A
—
—
sin
c
C
—
—
sin
a
A
—
—
sin
b
70°
—
—
sin
84
52°
——
sin
c
58°
—
—
sin
84
52°
—
b
—
84
si
s
n
in
52
7
°
0°
—
c
—
84
si
s
n
in
52
5
°
8°
—
b100.1689124 c90.39983243
A52°, b100.2, c90.4
46. A180° (57° 49°) or 74°
—
sin
a
A
—
—
sin
c
C
——
sin
b
B
—
—
sin
c
C
—
—
sin
a
74°
—
—
sin
8
49°
——
sin
b
57°
—
—
sin
8
49°
—
a
—
8
s
s
in
in
4
7
9
4
°
°
—
b
—
8
s
s
in
in
4
5
9
7
°
°
—
a10.1891739 b8.889995197
A74°, a10.2, b8.9
47. B180° (20° 64°) or 96°
K
—
1
2
—
a2
—
sin
s
B
in
s
A
in C
—
K
—
1
2
—
(19)2
—
sin 9
si
6
n
°
2
si
0
n
°
64°
—
K471.7 units2
48. C180° (56° 78°) or 46°
K
—
1
2
—
b2
—
sin
s
A
in
s
B
in C
—
K
—
1
2
—
(24)2
—
sin 5
si
6
n
°
7
si
8
n
°
46°
—
K175.6 units2
49. K
—
1
2
—
bc sin A
K
—
1
2
—
(65.5)(89.4) sin 58.2°
K2488.4 units2
50. K
—
1
2
—
ac sin B
K
—
1
2
—
(18.4)(6.7) sin 22.6°
K23.7 units2
Chapter 5 168
37. sin B
—
b
c
—
sin 42°
—
1
b
5
—
15 sin 42° b
10.0 b
38. sin A
—
a
c
—
sin 38°
—
2
c
4
—
csin 38° 24
c
—
sin
24
38°
—
c39.0
39. tan B
—
a
b
—
tan 67°
—
2
a
4
—
atan 67° 24
a
tan
24
67°
a10.2
40. 30°, 210°
41. 180°
42. A49° 90°
A41°
tan B
—
a
b
—
cos B
—
a
c
—
tan 49°
—
1
b
6
—
cos 49°
—
1
c
6
—
16 tan 49° bccos 49° 16
18.4 bc
—
cos
16
49°
—
c24.4
A41°, b18.4, c24.4
43. a2b2c2
a2152202
a175
a13.2
cos A
—
b
c
—
cos A
—
1
2
5
0
—
Acos1
—
1
2
5
0
—
A41.4°
51. Since 38.7° 90°, consider Case I.
csin A203 sin 38.7°
c sin A126.9242592
172 126.9; 2 solutions
—
sin
a
A
—
—
sin
c
C
—
sin
1
3
7
8
2
.7°
—
s
2
in
03
C
—
sin C
—
203 s
1
i
7
n
2
38.7°
—
Csin1
—
203 s
1
i
7
n
2
38.7°
—
C47.55552829
180° 180° 47.6° or 132.4°
Solution 1
B180° (38.7° 47.6°) or 93.7°
—
sin
b
B
—
—
sin
a
A
—
—
sin
b
937°
—
—
sin
1
3
7
8
2
.7°
—
b
—
17
s
2
in
si
3
n
8
9
.7
3
°
.7°
—
b274.5059341
Solution 2
B(180°(38.7° 132.4°) or 8.9°
sin
b
B
sin
a
A
—
sin
b
8.9°
—
—
sin
1
3
7
8
2
.7°
—
b
—
17
s
2
in
s
3
in
8.
8
7
.
°
9°
—
b42.34881128
B93.7°, C47.6°, b274.5; B8.9°,
C132.4°, b42.3
52. Since 57° 90°, consider Case I.
bsin A19 sin 57°
bsin A15.93474074
12 15.9; no solution
53. Since 29° 90°, consider Case I.
csin A15 sin 29°
csin A7.272144304
12 7.3; 2 solutions
—
sin
a
A
—
—
sin
c
C
—
sin
12
29°
si
1
n
5
C
sin C
—
15 s
1
in
2
29°
—
Csin1
—
15 s
1
in
2
29°
—
C37.30170167
180 180° 37.3° or 142.7°
Solution 1
B180° (29° 37.3°) or 113.7°
—
sin
a
A
—
—
sin
b
B
—
sin
12
29°
sin 1
b
13.7°
b
—
12
s
s
i
i
n
n
2
1
9
13
°
.7°
—
b22.6647614
Solution 2
B180° (29° 142.7°) or 8.3°
—
sin
a
A
—
—
sin
b
B
—
—
sin
12
29°
—
—
sin
b
8.3°
—
b
—
12
si
s
n
in
2
8
9
.
°
3°
—
b3.573829815
B113.7°, C37.3°, b22.7;
B8.3°, C142.7°, b3.6
54. Since 45° 90°, consider Case I.
83 79; 1 solution
—
sin
a
A
—
—
sin
b
B
—
—
sin
83
45°
—
—
si
7
n
9
B
—
sin B
—
79 s
8
in
3
45°
—
Bsin1
—
79 s
8
in
3
45°
—
B42.30130394
C180° (45° 42.3°) or 92.7°
—
sin
a
A
—
—
sin
c
C
—
—
sin
83
45°
—
—
sin 9
c
2.7°
—
c
—
83
s
s
i
i
n
n
4
9
5
2
°
.7°
—
c117.2495453
B42.3°, C92.7°, c117.2
55. a2b2c22bc cos A
a24024522(40)(45) cos 51°
a21359.446592
a36.87067388
—
sin
a
A
—
—
sin
b
B
—
—
si
3
n
6
5
.9
1°
—
—
si
4
n
0
B
—
sin B
—
40
3
si
6
n
.9
51°
—
Bsin1
—
40
3
si
6
n
.9
51°
—
B57.39811237
C180° (51° 57.4) or 71.6°
a36.9, B57.4°, C71.6°
56. b2a2c22ac cos B
b25126122(51)(61) cos 19°
b2438.9834226
b20.95193124
—
sin
b
B
—
—
sin
a
A
—
—
si
2
n
1
1
.0
9°
—
—
si
5
n
1
A
—
sin A
—
51
2
si
1
n
.0
19°
—
Asin1
—
51
2
si
1
n
.0
19°
—
A52.4178316
C180° (52.4° 19°) or 108.6°
b21.0, A52.4°, C108.6°
57. c2a2b22ab cos C
2021121322(11)(13) cos C
—
202
2(
1
1
1
1
2
)(1
3)
132
—
cos C
cos1
—
202
2(
1
1
1
1
2
)(1
3)
132
—
C
112.6198649 C
—
sin
a
A
—
—
sin
c
C
—
—
si
1
n
1
A
—
—
sin 1
2
1
0
2.6°
—
sin A
—
11 sin
20
112.6°
—
Asin1
—
11 sin
20
112.6°
—
A30.51023741
B180° (30.5° 112.6°) or 36.9°
A30.5, B36.9°, C112.6°
169 Chapter 5
58. b2a2c22ac cos B
b24226.522(42)(6.5) cos 24°
b21307.45418
b36.15873588
—
sin
b
B
—
—
sin
c
C
—
—
si
3
n
6
2
.2
4°
—
—
si
6
n
.5
C
—
sin C
—
6.5
3
s
6
in
.2
24°
—
Csin1
—
6.5
3
s
6
in
.2
24°
—
C4.192989407
A180° (24° 4.2°) or 151.8°
b36.2, A151.8°, C4.2°
Page 339 Applications and Problem Solving
59a. sin v
—
1
8
2
—
vsin1
—
1
8
2
—
v41.8°
59b. cos v
—
1
x
2
—
cos 41.8°
—
1
x
2
—
12 cos 41.8° x
8.94427191 x
about 8.9 ft
60a. x24.528.222(4.5)(8.2) cos 32°
x224.9040505
x5.0 mi
60b.
—
s
8
in
.2
v
—
—
sin
5.
3
0
2°
—
sin v
—
8.2 s
5
i
.
n
0
32°
—
vsin1
—
8.2 s
5
i
.
n
0
32°
—
v60.54476292
180 v180 60.5 or about 119.5°
Page 339 Open-Ended Assessment
1. K
—
1
2
—
ab sin C
125
—
1
2
—
ab sin 35°
435.86 ab
Sample answer: about 40 cm and 10.9 cm
2a. Sample answer: a10, b24, A30°;
10 24, 10 24 sin 30°
2b. Sample answer: b18; 10 18, 10 18 sin 30°
Chapter 5 SAT & ACT Preparation
Page 341 SAT and ACT Practice
1. There are several ways to solve this problem. Use
the Pythagorean Theorem on the large triangle.
(2y3y)24232
(5y)216 9
25y225
—
2
2
5
5
y2
—
—
2
2
5
5
—
y21
y1
Since yis 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.
5y3 4
5y7
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.
(xh)2(yk)2r2
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 xand
yvariables 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)23 3 n2.
So nmust equal 111. The correct choice is C.
4.
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 72
. The correct choice is B.
Chapter 5 170
C
D
A
B
45˚
45˚
7
7
7
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 ais as
small as possible and bis as large as possible.
Since the smallest value of amust be slightly
greater than 4, and the largest value of bmust be
slightly less than 9, this minimum value of the
fraction must be larger than
—
4
9
—
. The maximum
value of the fraction occurs when ais as large as
possible and bis as small as possible. This
maximum must be smaller than
—
7
7
—
or 1. The
correct choice is A.
6. Start by making a sketch of the situation.
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 yrepresent the length
of side AC.
In the smaller right triangle,
y24262
y216 36
x252
You do not need to solve for y.
In the larger triangle,
102x2y2
100 x252
x248
x48
x43
The correct choice is B.
8. Factor the polynomial in the numerator of the
fraction. Simplify the fraction. Solve for x.
—
x2
x
7
x
4
12
—
5
—
(x
x
3
)(x
4
4)
—
5
x3 5
x2
The correct choice is B.
9.
T is on the circle, so O
T
is a radius of the circle.
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.
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.
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
36,000
1:00 P. M .
10:00 A.M.
9000
9000
9000
9000
C
X
D
AB
4
4
5
5
3
30
TA
C
O
Angles and Radian Measure
Pages 347–348 Check for Understanding
1. 2. 90°;
—
4
—
3. Divide 10 by 8.
4. Let R2r. For the circle with radius R, sRvor
2rvwhich is 2(rv). Thus, s2s. For the circle
with radius R, A
1
2
R2vor
1
2
(2r)2vwhich is
1
2
(4r2)vor 4
1
2
r2v
. Thus, A4A.
5. 240° 240°
18
0°
6. 570° 570°
18
0°
4
3
19
6
7.
3
2
3
2
18
0°
270°
8. 1.75 1.75
18
0°
100.3°
9. reference angle:
3
4
or
4
; Quadrant 2
sin
3
4
2
2
10. reference angle:
11
6
or
5
6
; Quadrant 3
tan
11
6
3
3
11. srv12. 77° 77°
18
0°
s15
5
6
7
1
7
8
0
s39.3 in. srv
s15
7
1
7
8
0
s20.2 in.
6-1 Pages 348–351 Exercises
16. 135° 135°
18
0°
17. 210° 210°
18
0°
3
4
7
6
18. 300° 300°
18
0°
19. 450° 450°
18
0°
5
3
5
2
20. 75° 75°
18
0°
21. 1250° 1250°
18
0°
5
1
2
12
1
5
8
22.
7
1
2
7
1
2
18
0°
23.
11
3
11
3
18
0°
105° 660°
24. 17 17
18
0°
25. 3.5 3.5
18
0°
974.0° 200.5°
26.
6
.2
6
.2
18
0°
27. 17.5 17.5
18
0°
29.0° 1002.7°
28. reference angle: 2
5
3
or
3
; Quadrant 4
sin
5
3
2
3
29. reference angle:
7
6
or
6
; Quadrant 3
tan
7
6
3
3
30. reference angle:
5
4
or
4
; Quadrant 3
cos
5
4
2
2
31. reference angle:
7
6
or
6
; Quadrant 3
sin
7
6
1
2
32.
14
3
is coterminal with
2
3
reference angle:
2
3
or
3
; Quadrant 2
tan
14
3
3
33.
19
6
is coterminal with
5
6
reference angle:
5
6
or
6
; Quadrant 2
cos
19
6
2
3
34. srv35. srv
s14
2
3
s14
5
1
2
s29.3 cm s18.3 cm
36. 150° 150°
18
0°
37. 282° 282°
18
0°
5
6
4
3
7
0
srvsrv
s14
5
6
s14
4
3
7
0
s 36.7 cm s68.9 cm
38. srv39. 320° 320°
18
0°
s14
3
1
1
16
9
s12.0 cm srv
s14
16
9
s78.2 cm
Chapter 6 172
Chapter 6 Graphs of Trigonometric Functions
13. A
1
2
r2v
A
1
2
(1.42)
2
3
A2.1 units2
14. 54° 54°
18
0°
—
3
1
0
—
A
1
2
r2v
A
1
2
(62)
3
1
0
A17.0 units2
15. 30° 30°
18
0°
6
srv
s1.4
6
s0.7 m
y
x
1
4
3
O
40. r
1
2
d78° 78°
18
0°
r
1
2
(22)
1
3
3
0
r11 srv
s11
1
3
3
0
s15.0 in.
41. srvd2r
70.7 r
5
4
d2(18.0)
18.00360716 rd36.0 m
42. 60° 60°
18
0°
srv
3
14.2 r
3
13.56 r; about 13.6 cm
53b. srv2.5 2.5
18
0°
5 2v143.2°
2.5 v
54. 330° 330°
18
0°
11
6
srvsrv
s2
11
6
11.5 8v
s11.5 in. 1.44 v; about 1.4 radians
55. srv0.5 0.5
18
0°
10.5 22.9v26.3°
0.46 v; about 0.5
56a. 45° 34° 11° 56b. 45° 31° 14°
11° 11°
18
0°
14° 14°
18
0°
1
1
1
8
0
—
7
9
0
—
srvsrv
s3960
1
1
1
8
0
s3960
7
9
0
s760.3 mi s967.6 mi
56c. 34° 31° 3°
3° 3°
18
0°
6
0
srv
s3960
—
6
0
—
s207.3 mi
57. 84.5° 84.5°
18
0°
80° 80°
18
0°
1
3
6
6
9
0
4
9
srvsrv
s0.70
1
3
6
6
9
0
s0.67
4
9
s1.03 mi s0.94 mi
1.03 1.46 0.94 1.8 5.23 mi
58a. r
1
2
d1.5 rotations 1.5 2radians
r
1
2
2
1
2
3radians
r1.25
srv
s1.25(3)
s11.8 ft
58b. srv3.6 3.6
18
0°
4
1
2
1.25v206.3°
3.6 v
59a. v2
2
or
3
2
A
1
2
r2v
A
1
2
(152)
—
3
2
—
A530.1 ft2
59b. A
1
2
r2v
750
1
2
r2
3
2
318.3098862 r2
17.84124116 r; about 17.8 ft
173 Chapter 6
43. A
1
2
r2v
A
1
2
(102)
—
5
1
2
—
A65.4 units2
45. A
1
2
r2v
A
1
2
(72)
—
8
—
A9.6 units2
46. A
1
2
r2v
A
1
2
(12.52)
—
4
7
—
A140.2 units2
47. 225° 225°
—
18
0°
—
48. 82° 82°
—
18
0°
—
—
5
4
—
—
4
9
1
0
—
A
—
1
2
—
r2vA
—
1
2
—
r2v
A
1
2
(62)
5
4
A
1
2
(7.32)
4
9
1
0
A70.7 units2A38.1 units2
44. 90° 90°
18
0°
2
A
1
2
r2v
A
1
2
(222)
2
A380.1 units2
49a. srv
6 r(1.2)
5 r; 5 ft
49b. A
1
2
r2v
A
1
2
(52)(1.2)
A15 ft2
50a. 135° 135°
18
0°
50b. A
1
2
r2v
—
3
4
—
A
1
2
(48.42)
3
4
srvA2757.8 mm2
114 r
3
4
48.38 r; about 48.4 mm
51a. A
1
2
r2v
15
1
2
r2(0.2)
150 r2
12.247 r
about 12.2 in.
52a. A
1
2
r2v
15.3
1
2
(32)v
3.4 v; 3.4 radians
51b. srv
s12.2(0.2)
s2.4 in.
52c. srv
s3(3.4)
s10.2 m
52b. 3.4 3.4
18
0°
194.8°
53a. 225° 225°
18
0°
5
4
srv
s2
5
4
s7.9 ft
60. 3.5 km 350,000 cm
srv
350,000 32v
10,937.5 v; 10,937.5 radians
61. Area of segment Area of sector Area of triangle
A
1
2
r2a
—
1
2
—
rrsin a
A
1
2
r2(asin a)
62. s
1
2
(6 8 12)
13
Ks(s
a)(s
b)(s
c)
K13(13
6)(
13
8)(13
12)
K455
K21.3 in2
63. Since 152° 90°, consider Case II.
10.2 12, so there is no solution.
64. C180° 38° 27° 115°
sin
56
1
0
15°
sin
a
27°
a280.52
sin 38°
280
x
.52
x172.7 yd
65. I, III
66a. Find a quadratic regression line using a
graphing calculator. Sample answer: y102x2
505x18,430
66b. 2020 1970 50
y102x2505x18,430
y102(50)2505(50) 18,430
y248,180
Sample answer: about 248,180
67.
f(x) (x)43(x32(x)26(x) 10
f(x) x43x32x26x10
Sample answers: 4; 2
68. 2 161212
288
14 4
4
No; there is a remainder of 4.
69. x2y216 →a2b216
x-axis a2b216
a2(b)216
a2b216; yes
y-axis a2b216
(a)2b216
a2b216; yes
yxa
2b216
(b)2(a)216
a2b216; yes
yxa
2b216
(b)2(a)216
a2b216; yes
all
70. 4x2y3z6
5x4y3z75
9x6y81
2(4x2y3z) 2(6) →8x4y6z12
3(3x3y2z) 3(2) 9x9y6z6
17x5y6
5(9x6y) 5(81) →45x30y405
6(17x5y) 6(6) 102x30y36
147x441
x3
9x6y81 4x2y3z6
9(3) 6y81 4(3) 2(9) 3z6
y9z8
(3, 9, 8)
71. b
72. Since q0, q0. Given that p0,
pqpqand pq0. So the
expression pqis nonnegative.
The correct choice is B.
Linear and Angular Velocity
Page 355 Check for Understanding
1.
2.
1
5
m
re
i
v
n
2
1
ra
r
d
e
i
v
ans
1
6
m
0
i
s
n
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.
5. Since angular velocity is
—
v
t
—
, the radius has no
effect on the angular velocity. Let R2r. For
a circle with radius R, vR
v
t
or (2r)
v
t
which is
2
r
v
t
. Thus v2v.
6. 5.8 211.6or about 36.4 radians
7. 710 21420 or about 4461.1 radians
8. 3.2 26.49. 700 21400
q
v
t
q
v
t
q
6.
7
4
q
14
1
0
5
0
q2.9 radians/s q293.2 radians/min
6-2
Chapter 6 174
r132610
1124212
211426
3102010
41121466
r132610
114246
21581030
10. vrq11. vrq
v12(36) v7(5)
v432 in./s v110.0 m/min
12a. r3960 22,300 or 26,260 mi
srv
s26,260(2)
s164,996.4 mi
12b. vr
v
t
v26,260
2
2
4
v6874.9 mph
Pages 355–358 Exercises
13. 3 26or about 18.8 radians
14. 2.7 25.4or about 17.0 radians
15. 13.2 226.4or about 82.9 radians
16. 15.4 230.8or about 96.8 radians
17. 60.7 2121.4or about 381.4 radians
18. 3900 27800or about 24,504.4 radians
19. 1.8 23.620. 3.5 27
q
v
t
q
v
t
q
3.
9
6
q
7
3
q1.3 radians/s q7.3 radians/min
21. 17.2 234.422. 28.4 256.8
q
v
t
q
v
t
q
34
1
.
2
4
q
56
1
.
9
8
q9.0 radians/s q9.4 radians/s
23. 100 220024. 122.6 2245.2
q
v
t
q
v
t
q
20
1
0
6
q
24
2
5
7
.2
q39.3 radians/min q28.5 radians/min
25.
1
5
r
0
e
s
v
e
o
c
l
o
u
n
ti
d
o
s
n
1
2
re
r
v
a
o
d
lu
ia
ti
n
o
s
n
0.1 radian/s
26.
—
500
1
r
m
ev
i
o
n
l
u
u
t
t
e
ions
—
—
6
1
0
m
se
i
c
n
o
u
n
t
d
e
s
—
—
1
2
re
r
v
a
o
d
lu
ia
ti
n
o
s
n
—
52.4
radians/s
27.
8
1
5
s
r
e
a
c
d
o
i
n
an
d
s
6
1
0
m
se
i
c
n
o
u
n
t
d
e
s
1
2
re
r
v
a
o
d
lu
ia
ti
n
o
s
n
811.7 rpm
28. vrq29. vrq
v8(16.6) v4(27.4)
v132.8 cm/s v109.6 ft/s
30. vrq31. vrq
v1.8(6.1)v17(75.3)
v34.5 m/min v4021.6 in./s
32. vrq33. vrq
v39(805.6) v88.9(64.5)
v31,418.4 in./min v18,014.0 mm/min
34a.
1s
1
e
2
c
0
o
°
nd
6
1
0
m
se
i
c
n
o
u
n
t
d
e
s
1re
3
v
6
ol
0
u
°
tion
20 rpm
34b. 120° 120°
18
0°
2
3
q
v
t
vrq
qv10.5 in./s
q
2
3
2
3
1
35a. In 1 second, the second hand moves
6
1
0
(360°)
or 6°.
6° 6°
18
0°
or
3
0
vrv
v30
3
0
v3.1 mm/s
35b. In 1 second, the minute hand moves
6
1
0
6
1
0
(360°) or 0.1°.
0.1° 0.1°
18
0°
or
0
1
.
8
1
0
vrv
v27
0
1
.
8
1
0
v0.05 mm/s
35c. In 1 second, the hour hand moves
1
1
2
6
1
0
6
1
0
(360°) or about 0.008°.
0.008° 0.008°
18
0°
or
0.
1
0
8
0
0
8
vrv
v18
0.
1
0
8
0
0
8
v0.003 mm/s
36a. r
1
2
dvr
v
t
r
1
2
(80) or 40 v40
2
4
5
v5.6 ft/s
36b. vr
v
t
8 40
2
t
t31 s
37a. 3 26radians 1 minute 60 seconds
vr
v
t
v22
—
1
2
—
6
6
0
v7.1 ft/s
37b. vr
v
t
37c. 7.1 3.1 4 ft/s
3.1 r
6
6
0
9.87 r; about 9.9 ft
38a. 35° 35°
18
0°
7
3
6
lighter child: q
v
t
q
q1.2 radians/s
heavier child: q
v
t
q
q1.2 radians/s
7
3
6
1
2
7
3
6
1
2
175 Chapter 6
v5
2
3
38b. lighter child: vrq
v9(1.2)
v11.0 ft/s
heavier child: vrq
v6(1.2)
v7.3 ft/s
39a. 3 miles 190,080 inches
r
1
2
dsrv
r
1
2
(30) 190,080 15v
r15 12,672 v
12,672
1rev
2
o
lution
2017 revolutions
39b.
2.75
s
r
e
e
c
v
o
o
n
lu
d
tions
6
1
0
m
se
i
c
n
o
u
n
t
d
e
s
60
1
m
h
i
o
n
u
u
r
tes
1
2
re
r
v
a
o
d
lu
ia
ti
n
o
s
n
19,800radians/hour
vrq
v15(19,800)
v933,053.0181
14.7 mph
40a. Mercury: Venus:
vr
v
t
vr
v
t
v2440
14
2
0
7.6
v6052
58
2
3
2.5
v10.9 km/h v6.5 km/h
Earth: Mars:
vr
v
t
vr
v
t
v6356
23
2
.9
35
v3375
24
2
.6
23
v1668.5 km/h v861.2 km/h
40b. The linear velocity of Earth is about twice that
of Mars.
41a. vvmcos qt
v
4
cos t
41b. v
4
cost
0
4
cos t
0 cos t
t
2
or t
3
2
t
1
2
or 0.5 s t
3
2
or 1.5 s
42a. 3960 200 4160 miles
C2rtCspeed
C2(4160) t26,138.05088 17,000
C26138.05088 t1.537532405
q
v
t
q
1
2
.5
4
q4.1 radians/h
42b. q
v
t
tCspeed
4
2
t
2
2r17,000
—
2
—
t
—
2
—
(17,000) 2r
r
4250 r
4250 3960 290; about 290 mi
2
(17,000)
2
1mile
933,053.0181 inches
42c. 3960 500 4460; C2(4460) or
28023.00647
t28,023.00647 17,000 or 1.648412145
q
v
t
q
1
2
.6
5
q3.8
Its angular velocity is between 3.8 radians/h and
4.1 radians/h.
43a. Bclockwise; Ccounterclockwise
43b. vArA
v
t
A
vA3.0
12
1
0
vA360
The linear velocity of each of the three rollers is
the same.
vBrB
v
t
BvCrC
v
t
C
360 2.0
v
1
B
360 4.8
v
1
C
180 vB75 vC
180 rpm 75 rpm
44. 105° 105°
18
0°
7
1
2
A
1
2
r2v
A
1
2
(7.22)
7
1
2
A47.5 cm2
45. r
1
2
d
r
1
2
(7.3)
r3.65
v360° 10 or 36°
sin v
x
r
cos v
y
r
sin 36°
3.
x
65
cos 36°
3.
y
65
x2.145416171 y2.952912029
A
1
2
bh
A
1
2
(2.15)(2.95)
A3.16761261
Area of pentagon 10(3.17) or about 31.68 cm2
46. 35°205535° 20
6
1
0
°
55
36
1
0
°
0
35.349°
47. 10 k5
8 Check: 10 k5
8
k5
2 10 9 5
8
k5
4 10 4
8
k9 10 2 8
12 8
no real solution
48. (x(4))(x3i)(x(3i)) 0
(x4)(x3i)(x3i) 0
(x4)(x29) 0
x34x29x36 0
Chapter 6 176
x
y
r
49.
50. m
0
6
5
8
m
1
5
4
or
1
5
4
yy1m(xx1)
y0
1
5
4
(x(6))
y
1
5
4
x
1
7
5
51. P2a2b
P2
3
4
b
2b
P
3
2
b2b
P
7
2
b
2
7
P
bThe correct choice is D.
Graphing Sine and
Cosine Functions
Page 363 Check for Understanding
1. Sample answer:
period: 6
2. Sample answers:
3
2
,
2
,
5
2
3. cos xcos(x2)
4.
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
2nand its minimum value occurs
when x
3
2
2n, where nis an integer. The
6-3
maximum value of the cosine function occurs
when xn, where nis an even integer, and its
minimum value occurs when xn, where nis
an odd integer.
5. yes; 4 6. 07. 1
8.
3
2
2n, where nis an integer
9.
10.
11. Neither; the period is not 2.
12. April (month 4):
y49 28 sin
6
(t4)
y49 28 sin
6
(4 4)
y49
October (month 10):
y49 28 sin
6
(t4)
y49 28 sin
6
(10 4)
y49
The average temperatures are the same.
Pages 363–366 Exercises
13. yes; 6 14. no 15. yes; 20 16. no
17. no 18. no 19. 120. 0
21. 022. 123. 124. 1
25. sin cos 0 (1)
1
26. sin 2cos 20 1
1
27. 2n, where nis an integer
28.
2
2n, where nis an integer
29.
2
n, where nis an integer
30. v2n, where nis an integer
31.
177 Chapter 6
y
x
3 1
x
y
O
y
x
O
y
sin
x
y
cos
x
2
1
1
345
x
y
O
y
cos
x
1
1
675
x
y
O
y
sin
x
1
1
435
x
y
O
y
sin
x
1
1
324
x
y
O
32.
33.
34.
35.
36.
37. ycos x; the maximum value of 1 occurs when
x4, the minimum value of 1 occurs when
x5, and the x-intercepts are
7
2
,
9
2
, and
11
2
.
38. Neither; the graph does not cross the x-axis.
39. ysin x; the maximum value of 1 occurs when
x
11
2
, the minimum value of 1 occurs
when x
13
2
, and the x-intercepts are 7,
6, and 5.
40. Sample answer: a shift of
2
to the left
41. x
2
n, where nis an integer
42. xn, where nis an integer
43a. csc v
si1
nv
43b. csc v
si1
nv
1
—
si
1
nv
—
1
si1
nv
sin v1sin v1
2
2n, where n
3
2
2n, where n
is an integer is an integer
43c. csc vis undefined when sin v0.
n, where nis an integer
44a. sec v
co
1
sv
44b. sec v
co
1
sv
1
co
1
sv
1
co
1
sv
cos v1cos v1
2n, where n2n, where n
is an integer is an integer
44c. sec vis undefined when cos v0.
2
n, where nis an integer
45.
[0, 2] sc1
—
2
—
by [2, 2] sc11
x
3
4
,
7
4
46.
[0, 2] sc1
—
2
—
by [2, 2] sc11
0 x
4
,
5
4
x2
47.
[0, 2] sc1
2
by [2, 2] sc11
none
48.
[0, 2] sc1
2
by [2, 2] sc11
x0,
2
x,
3
2
x2
Chapter 6 178
y
cos
x
1
1
9108
x
y
O
y
cos
x
1
1
435
x
y
O
y
sin
x
1
1
65
x
y
O
y
cos
x
1
1
23
x
y
O
y
sin
x
1
1
45
x
y
O
49.
[0, 2] sc1
2
by [2, 2] sc11
x0,
2
, 2
50.
[0, 2] sc1
2
by [2, 2] sc11
x
4
,
5
4
51a. July (month 7):
y43 31 sin
6
(t4)
y43 31 sin
6
(7 4)
y74
January (month 1):
y43 31 sin
6
(t4)
y43 31 sin
6
(1 4)
y12
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 nis an integer
52b. 252c. 252d. 2
52e.
52f. It expands the graph vertically.
53a. P100 20 sin 2t
P100 20 sin 2(0) or 100
P100 20 sin 2(0.25) or 120
P100 20 sin 2(0.5) or 100
P100 20 sin 2(0.75) or 80
P100 20 sin 2(1) or 100
53b. 0.25 s 53c. 0.75 s
54a. v3.5 cos
t
—
m
k
—
v3.5 cos
0.9
—
1
1
9
.9
.6
9
—
v3.3 cm
v3.5 cos
t
m
k
v3.5 cos
1.7
—
1
1
9
.9
.6
9
—
v2.0 cm
54b. v3.5 cos
t
—
m
k
—
0 3.5 cos
t
—
1
1
9
.9
.6
9
—
0 cos
t
—
1
1
9
.9
.6
9
—
cos10 t
—
1
1
9
.9
.6
9
—
1.570796327 t
—
1
1
9
.9
.6
9
—
0.5005164776 t; about 0.5 s
54c. v3.5 cos
t
—
m
k
—
3.5 3.5 cos
t
—
1
1
9
.9
.6
9
—
1 cos
t
—
1
1
9
.9
.6
9
—
cos11 t
—
1
1
9
.9
.6
9
—
2t
—
1
1
9
.9
.6
9
—
2.00206591 t; about 2.0 s
55a.
4
2
n
, where nis an integer
55b. 155c. 155d.
55e.
56a. P500 200 sin [0.4(t2)]
P500 200 sin [0.4(0 2)] or about 357
pumas
D1500 400 sin (0.4t)
D1500 400 sin (0.4(0)) or 1500 deer
56b. P500 200 sin [0.4(t2)]
P500 200 sin [0.4(10 2)] or about 488
pumas
D1500 400 sin (0.4t)
D1500 400 sin (0.4(10)) or about 1197 deer
56c. P500 200 sin [0.4(t2)]
P500 200 sin [0.4(25 2)] or about 545
pumas
D1500 400 sin (0.4t)
D1500 400 sin (0.4(25)) or about 1282 deer
57.
500
1rm
ev
io
n
lu
ut
te
ions
6
1
0m
se
ic
n
o
u
n
td
es
1
2
rer
va
od
lu
ia
ti
n
os
n
52.4 radians
per second
58. 1.5 1.5
18
0°
85.9°
59. 45°, 135°
179 Chapter 6
y
2 sin
x
1
2
2
1
22
x
y
O
y
cos 2
x
1
1
22
x
y
O
60.
x
2
2
2
x
x
x
x
2
2
4
4
1(x2)(x2)
x
2
2
1(x2)(x2)
2
x
x
(1)(x2)(x2)
x
x
2
2
4
4
1(x2)(2) (x2)(x)(1)(x24)
2x4 x22xx24
x2
But, x2, so there is no solution.
61. 1 positive real zero
f(x) 2x33x211x6
2 or 0 negative real zeros
22311 6
4146
27 30
2x27x3 0
(2x1)(x3) 0
2x1 0or x3 0
x
1
2
x3
3,
1
2
, 2
62. 112918
136
13612
12; no
63. vertical:
x2x0
x(x1) 0
x0 or x1 0
x1
horizontal:
y1
64. reflected over the x-axis, expanded vertically by a
factor of 3
65.
1
0
5
4
1
4
2
1
3
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.
Page 367 History of Mathematics
1.
[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.
Chapter 6 180
nn
2n3
12
212
336
480
5 150
6 252
7 392
8 576
9 810
10 1100
11 1452
12 1872
13 2366
14 2940
15 3600
16 4352
17 5202
18 6156
19 7220
20 8400
21 9702
22 11,132
23 12,696
24 14,400
25 16,250
26 18,252
27 20,412
28 22,736
29 25,230
30 27,900
O
g
(
x
)
x
g
(
x
)
x
2
x
2
x
2
4
(1)
1
4
1
3
0
5
1
3
0
5
1
4
2(5) 4(5) 1(1)
11
66.
1
6
2
4
3
2
0
1
1
0
1(1) 0(6)
0(1) 1(6)
1(2) 0(4)
0(2) 1(4)
1(3) 0(2)
0(3) 1(2)
A(3, 2), B(2, 4), C(1, 6)
67. x
3
2
y
y
2
3
xART
1
6
2
4
3
2
y
x
O
3
x
y
2
181 Chapter 6
Amplitude and Period of Sine and
Cosine Functions
Page 369 Graphing Calculator Exploration
1.
2. The graph is shrunk horizontally.
3. The graph of f(x) sin kx for k0 is the graph of
f(x) sin kxreflected over the y-axis.
Pages 372–373 Check for Understanding
1. Sample answer: y5 sin 2v
2. The graphs are a reflection of each other over the
v-axis.
3. A: period
2
2
or
B: period
2
5
C: period or 4
D: period 2
C has the greatest period.
4. Period and frequency are reciprocals of each other.
5.
All three graphs are periodic and curve above and
below the x-axis. The amplitude of y3 cos vis 3,
while the amplitude of ycos vand ycos 3vis
1. The period of ycos 3vis
2
3
, while the period
of ycos vand y3 cos vis 2.
6. 2.52.5
2
1
2
6-4 7.
2
4
2
8. 1010;
2
2
9. 33;
2
2
10. 0.50.5; 12
11.
1
5
1
5
; 8
12. A0.8
2
k
A0.8 k
2
or 2
y0.8 sin 2v
2
1
4
2
1
6
181 Chapter 6
y
3 cos
y
cos 3
y
cos
2
2
32
y
O
y
2.5 cos
2
2
234
y
O
y
sin 4
1
1
23
2
y
O
y
10 sin 2
8
4
8
4
32
y
O
y
3 cos 2
3
1
2
3
2
1
32
y
O
y
0.5 sin
6
1
1
428106
y
O
y
cos
1
5
0.2
0.2
4286
4
y
O
13. A7
2
k
3
A7kor 6
y7 sin 6v
14. A1.5
2
k
5
A1.5 k
2
5
or
2
5
y1.5 cos
2
5
v
15. A
3
4
2
k
6
A
3
4
k
2
6
or
3
y
3
4
cos
3
v
16. A0.25
2
k
2
1
94
A0.25 k588
y0.25 sin (588t)
Pages 373–377 Exercises
17. 22
18.
3
4
3
4
19. 1.51.5
2
3
20.
2
2
21. 8
22.
2
6
3
23. 55;
2
1
2
24. 22;
0
2
.
5
4
2
1
4
Chapter 6 182
y
2 sin
2
2
2 3
y
O
y
cos
3
4
0.5
0.5
2 3
y
O
y
1.5 sin
1
2
2
1
2 3
y
O
y
cos 2
1
1
2 3
y
O
y
cos
4
1
1
2468
y
O
y
sin 6
1
1
2
y
O
y
5 cos
4
4
432
y
O
y
2 cos 0.5
2
2
432
y
O
25.
2
5
2
5
;
2
9
26. 88;
0
2
.
5
4
27. 33; 4
28.
2
3
2
3
;
1
3
4
29. 33;
2
2
2
3
7
2
2
30. 33;
0
2
.
5
4
31.
1
3
1
3
;
2
3
32.
1
3
1
3
; 6
33. 44; 4
34. 2.52.5; 10
35. 0.50.5;
6
2
9
8
3
1
49
2
1
5
2
1
2
2
1
3
183 Chapter 6
y
sin 9
4
0.4
0.4
2
5
2
y
O
y
8 sin 0.5
4
8
4
8
2 345
y
O
y
3 sin
2
4
4
2
2
2
23
25
2
2
y
O
y
cos
0.4
0.8
0.8
0.4
2
33
7
23
25
23
2
y
O
y
3 sin 2
2
2
45
32
x
O
y
3 cos 0.5
2
2
45
32
y
O
1
1
45
32
y
cos 3
1
3
y
O
1
1
45
32
y
sin
1
3
3
y
O
4
2
4
2
45
32
y
4 sin
2
y
O
2
2
45
32
y
2.5 cos
5
y
O
36. A0.4
2
k
10
A0.4 k
1
5
y0.4 sin
5
v
37. A35.7
2
k
4
A35.7 k8
y35.7 sin 8v
38. A
1
4
2
k
3
A
1
4
k6
y
1
4
sin 6v
39. A0.34
2
k
0.75
A0.34 k
8
3
y0.34 sin
8
3
v
40. A4.5
2
k
5
4
A4.5 k
8
5
y4.5 sin
8
5
v
41. A16
2
k
30
A16 k
1
5
y16 sin
1
5
v
42. A5
2
k
2
A5k1
y5 cos v
43. A
5
8
2
k
7
A
5
8
k14
y
5
8
cos 14v
44. A7.5
2
k
6
A7.5 k
1
3
y7.5 cos
3
v
45. A0.5
2
k
0.3
A0.5 k
2
3
0
y0.5 cos
2
3
0
v
46. A
2
5
2
k
3
5
A
2
5
k
1
3
0
y
2
5
cos
1
3
0
v
47. A17.9
2
k
16
A17.9 k
8
y17.9 cos
8
v
48. A1.5
2
k
2
A1.5 k4
y1.5 sin 4v, y1.5 cos 4v
49. cosine curve A2
2
k
4
k
1
2
y2 cos
2
v
50. sine curve A0.5
2
k
k2
y0.5 sin 2v
51. cosine curve A3
2
k
2
k1
y3 cos v
52. sine curve A1.5
2
k
4
k
1
2
y1.5 sin
2
v
53. A3.8
2
k
1
1
20
A3.8 k240
y3.8 sin(240t)
54. A15
2
k
3
1
6
A15 k72
y15 cos (72t)
55.
All the graphs have the same shape, but have
been translated vertically.
56a. A
A
2
(A)
2
k
8
k
4
A
3
2
A1.5; down first, so A1.5
y1.5 sin
4
t
56b. y1.5 sin
4
t
y1.5 sin
4
(3)
y1.1 ft
Chapter 6 184
56c. y1.5 sin
4
t
y1.5 sin
4
(12)
y0 ft
57a. Maximum value of sin v1.
Maximum value of 2 sin v2 1 or 3
57b. Minimum value of sin v1
Minimum value of 2 sin v2 (1) or 1
57c.
2
1
2
37d.
58a. A0.2
2
k
2
1
62
A0.2 k524
y0.2 sin (524t)
2
45
32
y
2 sin
y
O
58b. A
1
2
(0.2)
2
k
5
1
24
A0.1 k1048
y0.1 sin (1048t)
58c. A2(0.2)
2
k
1
1
31
A0.4 k262
y0.4 sin (262t)
59a. yAcos
t
g
y1.5 cos
t
9
6
.8
59b. y1.5 cos
t
9
6
.8
y1.5 cos
4
9
6
.8
y0.6
about 0.6 m to the right
59c. y1.5 cos
t
9
6
.8
y1.5 cos
7.9
9
6
.8
y1.2
about 1.2 m to the left
60a.
2
n, where nis an integer
60b. 1
60c.
2
1
2
60d.
61a. y1.5 cos
t
m
k
2
k
6.8
y1.5 cos
t
1
0
8
.
.
4
5
k0.9 s/cycle
y1.5 cos 6.8tfrequency:
0
1
.9
1.1 hertz
61b. y1.5 cos
t
m
k
2
k
5.6
y1.5 cos
t
1
0
8
.
.
6
5
k1.1 s/cycle
y1.5 cos 5.6tfrequency:
1
1
.1
0.9 hertz
61c. y1.5 cos
t
m
k
2
k
4.8
y1.5 cos
t
1
0
8
.
.
8
5
k1.3 s/cycle
y1.5 cos 4.8tfrequency:
1
1
.3
0.8 hertz
61d. It increases. 61e. It decreases.
62. 0
63. 84 2168radians
q
v
t
q
16
6
8
q88.0 radians/s
64. 73° 73°
18
0°
srv
7
1
3
8
0
s9
7
1
3
8
0
s11.5 in.
65. a2b2c2tan A
a
b
15.1219.52c2tan A
1
1
5
9
.
.
1
5
24.66292764 cAtan1
1
1
5
9
.
.
1
5
A37.75273111
B180° (90° 37.8°) or 52.2°
c24.7, A37.8°, B52.2°
66. T2
g
4.1 2
9
.8
0.6525352667
9
.8
0.458022743
9
.8
4.17 ; about 4.17 m
67. b24ac 524(3)(10)
95
2 imaginary roots
68a. Let xthe number of Model 28 cards and let
ythe number of Model 74 cards.
30x15y240
20x30y360
12x10y480
x0
y0
P(x, y) 100x60y
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
69.
3
2
3
4
1
1
2
1
0
1
1
0
185 Chapter 6
1
1
45
32
y
cos( )
y
O
O
y
x
8
8
16
24
32
40
16 24 32 40
(3, 10)
(0, 12)
(8, 0)
(0, 0)
12
x
10
y
480
30
x
15
y
240
20
x
30
y
360
x
0
y
0
1(1) 0(1)
0(1) (1)(1)
1(2) 0(1)
0(2) (1)(1)
1(3) 0(2)
0(3) (1)(2)
1(3) 0(4)
0(3) (1)(4)
g
(
x
)
x
O
(2, 1), (1, 1), (3, 4), (3, 2)
70.
3
2
3
4
1
1
2
1
71. y14.7x140.1
y14.7(20) 140.1
y$434.10
72.
{(4, 16), (3, 9), (2, 4)}; yes
73. As2radius
1
2
(10) or 5
100 s2Ar2
10 sA(5)2or 25
4(25) 100
The correct choice is C.
Page 377 Mid-Chapter Quiz
1.
5
6
5
6
18
0°
150°
2. r
1
2
dsrv
r
1
2
(0.5) or 0.25 s0.25
5
3
s1.3 m
3. A
1
2
r2v
A
1
2
(82)
2
5
A40.2 ft2
4. 7.8 215.6or about 49.0 radians
5. 8.6 217.2
q
v
t
q
17
7
.2
q7.7 radians/s
Chapter 6 186
xx
2y
4(4)216
3(3)29
2(2)24
6. vrq
v3(8)
v75.4 meters/s
1
1
89
7
y
cos
x
x
y
O
8
4
8
4
681012
42
y
7 cos
3
y
O
7. 1
8.
9. 77; 6
10. A5
2
k
3
A5k6
y5 sin 6v
2
1
3
2
2
345
2
()
y
3 cos
2
y
O
2
4
345
2
y
sin 2 3
y
O
Translations of Sine
and Cosine Functions
Page 378 Graphing Calculator Exploration
1.
2. The graph shifts farther to the left.
3. The graph shifts farther to the right.
Page 383 Check for Understanding
1. Both graphs are the sine curve. The graph of y
sin x1 has a vertical shift of 1 unit upward,
while the graph of ysin (x1) 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 ysin xand ycos x, and find the sum of
their ordinates.
5. Jamal;
k
c
or 3
6.
2
7. 3; y3
A1;
2
2
2
—
6
6-5
8. 22;
2
2
;
2
; 5
9.
1
2
1
2
; 4;
2
; 3
10. A20
2
k
1c0
A20 k2h100
y20 sin 2v100
11. A0.6
2
k
12.4
A0.6 k
6
.2
k
c
2.13 h7
2.13
c
6
.2
4
1
2
2
1
2
13c.
Pages 383–386 Exercises
14.
k
c
1
2
or 2; A1;
2
1
2
15.
k
c
2
; A1;
2
2
16.
k
c
or 2; A2; 8
17.
1
2
; y
1
2
A1; 4
2
1
2
2
1
4
2
—
1
4
187 Chapter 6
xsin xcos xy
0sin 0 cos 0 1
2
sin
2
cos
2
1
sin cos 1
3
2
sin
3
2
cos
3
2
1
2sin 2cos 21
2
4
6
345
2
y
2 sin (2 ) 5
y
O
2
4
345
2
()
y
3 cos
4
2
1
2
y
O
c
2.
6
1
.
3
2
y0.6 cos
6
.2
v
2.
6
1
.
3
2
7
12.
13a.
130
2
70
100; P100
13b. A
130
2
70
2
k
1
A30 k2
A30 P30 sin 2t100
1
1
2
y
sin
x
cos
x
x
y
O
40
80
120
3465
21
P
30 sin 2
t
100
P
30 sin 2
t
100
t
P
O
1
1
45
32
y
sin( 2)
y
O
1
1
45
32
y
sin(2 )
y
O
1
1
2
2
234567
()
y
2 cos
4
2
y
O
1
2
1
345
2
y
sin 1
2
2
y
O
18. 4; y4
A5;
2
1
2
19. 7; y7
A1;
2
2
20.
k
c
2
4
or 2; 3
21. 33;
2
1
2;
2
; 0
22. 66;
2
1
2;
3
; 2
23. 11; 6;
4
; 2
1
2
1
3
2
1
3
3
1
2
1
24. 55;
2
3
;
3
; 20
25.
1
4
1
4
; 4; 0; 3
26. 1010; 8; 16; 5
27. A
2
2
(6)
4;
2
to the left or
2
;
A4down 2, or 2
A4; 4
28. A7
2
k
3h7
c
2
3
4
1
4
2
1
4
0
1
2
2
1
2
Chapter 6 188
2
4
8
6
345
2
y
5 cos 4
y
O
2
4
6
8
345
2
y
O
y
7 cos 2
2
2
345
2
()
y
3 cos
2
y
O
2
4
2
4
6
8
345
2
()
y
6 sin 2
3
y
O
2
4
345
2
()
y
2 sin
3
12
y
O
10
20
30
345
2
y
O
y
20 5 cos (3 )
2
4
345
2
y
3cos
1
4
4
y
O
5
5
15
10
345
2
()
y
10 sin 5 4
4
y
O
A7k
2
3c
2
3
y7 sin
2
3
v
2
3
7
29. A50
2
k
3
4
2
h25
c
8
3
A50 k
8
3
c
4
3
y50 sin
8
3
v
4
3
25
30. A
3
4
2
k
5
1
c
0
h
1
4
A
3
4
k10 c10
y
3
4
sin (10v10)
1
4
31. A3.5
2
k
2
4
c
4
h7
A3.5 k4c
y3.5 cos (4v) 7
32. A
4
5
2
k
6
1
c
2
3
h
7
5
A
4
5
k12 c4
y
4
5
cos (12v4)
7
5
38.
39.
40.
41a. 2000 1000 3000
2000 1000 1000
41b. 10,000 5000 15,000
10,000 5000 5000
41c.
[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.
33. A100
2
k
45 0h110
c
2
4
5
189 Chapter 6
xsin xsin xx
00 0
2
1
2
1 2.57
03.14
3
2
1
3
2
1 3.71
2026.28
2
4
6
8
3
2
y
x
sin
x
x
y
O
xcos xsin xcos xsin x
010 1
2
01 1
10 1
3
2
011
210 1
xsin x2xsin 2xsin xsin 2x
0000 0
4
0.71
2
11.71
2
101
0200
3
2
1301
20400
y
cos
x
sin
x
1
2
2
1
2
x
y
O
y
sin
x
sin 2
x
1
2
2
1
2
x
y
O
y
2 sin
x
3 cos
x
y
3 cos
xy
2 sin
x
2
4
2
4
2 3
x
y
O
y
cos 2
x
cos 3
x
y
cos 3
xy
cos 2
x
2
4
2
2
3
2
2
x
y
O
A100 k
2
4
5
c0
y100 cos
2
4
5
v
110
34. A
1
2
(3)
2
k
4h1
A2k
1
2
A2; 2
y2 cos
2
v
1
35. A
3.5
2
(2.5)
2
k
2
c
0h3
A0.5 k2c0
A0.5; 0.5
y0.5 sin 2v3
36.
37.
X
51. 7(3x5y) 7(4) →21x35y28
14x35y21 14x35y21
35x49
x1.4
3x5y4
3(1.4) 5y4
y0.04 (1.4, 0.04)
52.
53. 3xy7 0yy1m(xx1)
y3x7y(2) 3(x3)
slope: 3 y2 3x9
3xy11 0
54. 4 inches
1
3
foot
75 42
1
3
1050 cubic feet
1050 7.48 7854 gal
The correct answer is 7854.
Modeling Real-World Data
with Sinusoidal Functions
Pages 390–391 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
6-6
0
0
0
0
42.
43a. 46 42 4 ft
43b. r
1
2
dt21 4
r
1
2
(42) t25
r21
43c.
3
6
r
0
ev
s
o
e
l
c
u
o
t
n
io
d
n
s
s
1
x
re
s
v
e
o
co
lu
n
t
d
io
s
n
x20 s
43d. A21
2
k
20 h25
A21; 21 k
1
0
h25 21 sin
1
0
t
43e. h25 21 sin
1
0
t
46 25 21 sin
1
0
t
1 sin
1
0
t
sin1sin
1
0
t
5 t; 5 s
43f. h25 21 sin
1
0
t
h25 21 sin
1
0
10
h25 ft
44.
k
c
0
2
or 0
k
c
or
4
There is a
4
phase difference.
45a. ysin x
45b. y
co
x
sx
45c. ycos x245d. ysin x
46.
2
k
2
1
94
k588y0.25 sin 588t
47. vrq
v7(19.2)
v134.4 cm/s
48. asymptote: x2
y
x
x
3
2
y(x2) x3
yx 2yx3
2y3 xyx
2y3 x(1 y)
1
2y
y
3
x
asymptote: y1
2
2
49. f(x)
x
3
1
y
x
3
1
x
y
3
1
x(y1) 3
y1
3
x
y
3
x
1
50.
X
5
5
3
3
1
1
1
1
Chapter 6 190
xx 4y
66 42
44 40
22 42
00 44
y
x
y
x
cos
xy
cos
x
3
2
2
3
24 3
x
y
O
y
x
3
x
2
x
y
O
X
1(5) 1(5)
1(5) 1(5)
1(3) 1(3)
1(3) 1(3)
y
x
O
y
|
x
4|
4a. y5 cos
6
t
y5 cos
6
0
y 5
5 units below equilibrium
8d. h3 cos
5
3
t
3.5
h3 cos
5
3
(25)
3.5
h2 units
9a. R1200 300 sin
2
t
R1200 300 sin
2
0
R1200
9b. H250 25 sin
2
t
4
H250 25 sin
2
0
4
H232
9c. R: 1200 300 1500
H: 250 25 275 no
9d. R1200 300 sin
2
t
1500 1200 300 sin
2
t
300 300 sin
2
t
1 sin
2
t
sin11
2
t
sin11
2
t
1 t
January 1, 1971
9e. 250 25 225
H250 25 sin
2
t
4
225 250 25 sin
2
t
4
25 25 sin
2
t
4
1 sin
2
t
4
sin11
2
t
4
2
sin11
4
t
0.5 t
July 1, 1969;
2
k
k4
next minimum: July 1, 1973
9f. See students’ work.
10. A
4
2
2
k
10
A2k
5
y2 cos
5
t
11. h4.25; A3.55;
2
k
12.40; 4.68
c
2.
3
3
.
4
1
y3.55 sin
6
.2
t
2.
3
3
.
4
1
4.24
12a. h47.5; A23.5;
2
k
12; 4
c
2
3
y23.5 sin
6
t
2
3
47.5
c
6
c
6
.2
2
2
191 Chapter 6
4b. 5 units above
equilibrium
4c. y5 cos
6
t
y5 cos
6
7
y4.33
about 4.33 units above equilibrium
5. A
140
2
80
h
140
2
80
A30 h110
2
k
1
k2P30 sin 2t110
6a. A
66°
2
41°
6b. h
66°
2
41°
A12.5° h53.5°
6c. 12 months
6d. A12.5
2
k
12 h53.5
k
6
y12.5 cos
6
tc
53.5
41 12.5 cos
6
1 c
53.5
12.5 12.5 cos
6
c
1 cos
6
c
cos11
6
c
cos11
6
c
0.5 c
Sample answer: y12.5 cos
6
t0.5
53.5
6e. y12.5 cos
6
t0.5
53.5
y12.5 cos
6
(2) 0.5
53.5
y42.82517529
Sample answer: About 42.8°; it is somewhat
close to the actual average.
6f. y12.5 cos
6
t0.5
53.5
y12.5 cos
6
(10) 0.5
53.5
y53.20504268
Sample answer: About 53.2°; it is close to the
actual average.
Pages 391–394 Exercises
7a. 0.5
7b.
2
k
660
k
3
1
30
7c. 330 hertz
8a. 3.5 36.5 units
8b. 3.5 3 0.5 units
8c.
2
k
or
6
5
2
5
3
1
3
1
30
k
6
.2
k
6
12b. y23.5 sin
6
t
2
3
47.5
y23.5 sin
6
3
2
3
47.5
y35.75
about 35.8°
12c. y23.5 sin
6
t
2
3
47.5
y23.5 sin
6
8
2
3
47.5
y67.9°
13a. A
81°
2
73°
13b. h
81°
2
73°
A4° h77°
13c. 12 months
13d. A4
2
k
12 h77
k
6
y4 cos
6
tc
77
73 4 cos
6
1 c
77
4 4 cos
6
c
1 cos
6
c
cos11
6
c
cos11
6
c
0.5235987756 c
Sample answer: y4 cos
6
t0.5
77
13e. y4 cos
6
t0.5
77
y4 cos
6
8 0.5
77
y80.41594391
Sample answer: About 80.4°; it is very close to
the actual average.
13f. y4 cos
6
t0.5
77
y4 cos
6
5 0.5
77
y79.08118409
Sample answer: About 79.1°; it is close to the
actual average.
14.
k
c
1
or
increase shift by
2
;
2
3
2
k
c
3
2
1
c
3
2
c
3
2
Sample answer: y3 cos
x
3
2
5
15a. A
13.25
2
1.88
15b. h
13.25
2
1.88
A5.685 ft h7.565 ft
15c. 4:53 P.M.4:30 A.M.12:23 or about 12.4 h
15d. A5.685
2
k
12.4 h7.565
k
6
.2
4:30 A.M.4.5 hrs
h5.685 sin
6
.2
tc
7.565
13.25 5.685 sin
6
.2
4.5 c
5.685 5.685 sin
4
6
.5
.2
c
1 sin
4
6
.5
.2
c
sin11
4
6
.5
.2
c
sin11
4
6
.5
.2
c
0.7093918895 c
Sample answer: h5.685 sin
6
.2
t0.71
7.565
15e. 7:30 P.M.19.5 hrs
h5.685 sin
6
.2
t0.71
7.565
h5.685 sin
6
.2
19.5 0.71
7.565
h8.993306129
Sample answer: about 8.99 ft
16a. Table at bottom of page.
Chapter 6 192
Sunrise A.M.Time Sunset P.M.Time Daylight Hours
Month A.M.in Decimals P.M.in Decimals (P.M.-A.M.)
January 7:19 7.317 4:47 16.783 9.47 h
February 6:56 6.933 5:24 17.4 10.47 h
March 6:16 6.267 5:57 17.95 11.68 h
April 5:25 5.416 6:29 18.483 13.07 h
May 4:44 4.733 7:01 19.017 14.28 h
June 4:24 4.4 7:26 19.433 15.03 h
July 4:33 4.55 7:28 19.467 14.92 h
August 5:01 5.017 7:01 19.017 14 h
September 5:31 5.517 6:14 18.233 12.72 h
October 6:01 6.017 5:24 17.4 11.38 h
November 6:36 6.6 4:43 16.717 10.12 h
December 7:08 7.133 4:28 16.467 9.33 h
16b. A
15.03
2
9.33
16c. h
15.03
2
9.33
A2.85 h h12.18 h
16d. 12 months
16e. A2.85
2
k
12 h12.18
k
6
y2.85 cos
6
tc
12.18
9.47 2.85 cos
6
1c
12.18
2.71 2.85 cos
6
c
0.950877193 cos
6
c
cos10.950877193
6
c
cos10.950877193
6
c
0.2088597251 c
Sample answer: y2.85 cos
6
t0.21
12.18
17. 70.5 19.5 51
y70.5 19.5 sin
6
tc
51 70.5 19.5 sin
6
1 c
19.5 19.5 sin
6
c
1 sin
6
c
sin11
6
c
sin11
6
c
2.094395102 c
Sample answer: about 2.09
18a.
14
1
re
m
vo
in
lu
u
t
t
i
e
ons
6
1
0
m
se
i
c
n
o
u
n
t
d
e
s
1
2
re
r
v
a
o
d
lu
ia
ti
n
o
s
n
7
1
5
rad/s
y3.5 cos
7
1
5
t
18b. y3.5 cos
7
1
5
t
y3.5 cos
7
1
5
4
y3.197409102
about (4, 3.20)
19. A
120
2
(120)
2
k
60
A120 k
3
0
VR120 sin
3
0
t
20. See students’ work.
21. 33;
2
2
;
2
; 5
22. 2nwhere nis an integer
23. 800°
18
0°
40
9
24. 4023222022(32)(20) cos v
cos v
402
2(
3
3
2
2
2
)(2
0)
202
vcos1
402
2(
3
3
2
2
2
)(2
0)
202
v97.9°
180° 97.9° 82.1°
about 97.9°, 82.1°, 97.9°, 82.1°
25.
2
m
m
2
1
1
6
6
(m
2m
4)
(m
16
4)
(m
2m
4)
(m
16
4)
m
A
4
m
B
4
2m16 A(m4) B(m4)
Let m4.
2(4) 16 A(4 4) B(4 4)
8 8B
1 B
Let m4.
2(4) 16 A(4 4) B(4 4)
24 8A
3 A
m
A
4
m
B
4
m
3
4
m
1
4
26. 22k16
42k84k14
2k42k74k20
4k20 0
k5
27.
increasing: x1; decreasing: x1
28. The correct choice is E.
Graphing Other Trigonometric
Functions
Page 400 Check for Understanding
1. Sample answers: , , 2
2. The asymptotes of ytan vand ysec vare the
same. The period of ytan vis and the period
of ysec vis 2.
3. Sample answers:
2
,
3
2
4. 05. 1
6. n, where nis an odd integer
7.
4
n, where nis an integer
6-7
193 Chapter 6
2
8
6
4
345
2
y
3 cos(2 ) 5
y
O
O
f
(
x
)
x
f
(
x
)
2
x
1
5
8.
1
;
4
9.
2
2
;
2
; h1
10. k:
2
k
3c:
3
h: h4
c
2
9
ycsc
2
3
v
2
9
4
11. k:
k
2c:
4
h: h0
c
1
2
c
2
3
4
1
24.
4
n, where nis an integer
25.
4
n, where nis an integer
26.
3
4
n, where nis an integer
27.
2
n, where nis an odd integer
28. n, where nis an integer
29.
1
;
2
30. 6; no phase shift; no vertical shift
31.
2
1
2; no phase shift; h5
32. 2;
2
; h1
4
1
2
1
2
2
1
3
2
1
Chapter 6 194
2
4
2
4
22
()
y
tan
4
y
O
2
4
2
4
22
y
sec(2 ) 1
y
O
c
8
ycot
1
2
v
8
12a. fma 12b. Ffsec v
f73(9.8) F715.4 sec v
f715.4 N
12c.
2
1
2; no phase shift, no vertical shift
12d. 715.4 N
12e. The tension becomes greater.
Pages 400–403 Exercises
13. 014. 0
15. undefined 16. 1
17. 118. undefined
19. undefined 20. 0
21. n, where nis an integer
22. n, where nis an even integer
23.
3
2
2n, where nis an integer
800
400
2
F
715.4 sec
2
y
O
4
8
4
8
22
()
y
cot
2
y
O
2
4
2
4
22
y
sec
3
y
O
2
4
6
8
22
y
csc 5
y
O
2
4
2
4
22
()
y
tan 1
4
2
y
O
k
2
3
k
1
2
33.
2
2
;
2
; h3
34. 6;
2
; h2
35.
2, , 0, , 2
36. k:
k
2c: 0h: h6
c
1
2
6
1
3
2
1
3
41. k:
2
k
3
c:
6
c
2
h: h5
k6c3
ycsc (6v3) 5
42. k:
2
k
3c: h: h8
c
2
3
ysec
2
3
v
2
3
8
43. k:
k
2
c:
2
c
4
h: h7
k2c
2
ytan
2v
2
7
44a. 40
no phase shift
no vertical shift
44b. d10 tan
4
0
t
d10 tan
4
0
(3)
d2.4 ft from the center
44c. d10 tan
4
0
t
d10 tan
4
0
(15)
d24.1 ft from the center
45.
The graph of ycsc vhas no range values
between 1 and 1, while the graphs of y3 csc v
and y3 csc vhave no range values between
3 and 3. The graphs of y3 csc vand y
3 csc vare reflections of each other.
4
0
c
2
3
195 Chapter 6
6
4
8
12
2
22
y
csc(2 ) 3
y
O
6
4
8
2
2
22
()
y
sec 2
6
3
y
O
2
4
2
4
22
y
sec
y
cos
y
O
c0
ytan
2
v
6
37. k:
k
2
c:
2
c
8
h: h7
k2c
—
4
—
ycot
2v
4
7
38. k:
2
k
c:
2
c
4
h: h10
k2c
2
ysec
2v
2
10
39. k:
2
k
3c: h: h1
c
2
3
ycsc
2
3
v
2
3
1
40. k:
k
5c: h: h12
c
5
ycot
5
v
5
12
c
1
5
c
2
3
20
20
40
40
101020 20
d
10 tan
40
t
d
t
O
2
4
2
4
22
y
O
46a. fm9.8
f7 9.8
f68.6 N
46b. F
1
2
fsec
2
v
F
1
2
(68.6) sec
2
v
F34.3 sec
2
v
46c. 4
2
1
2
40
20
F
34.3 sec
2
F
O
k
—
1
2
—
k
2
3
k
1
5
k
2
3
46d. 34.3 N
46e. The tension becomes greater.
47a. 220 A
47b.
6
2
0
3
1
0
s
47c.
3
1
60
6
60
52. C180° (62°3175°18) or 42°11
sin
a
A
sin
b
B
sin
5
6
7
2
.3
°31
sin 7
b
5°18
b
57.
s
3
in
si
6
n
2°
7
3
5
1
°
18
b62.47505783
sin
a
A
sin
c
C
sin
5
6
7
2
.3
°31
sin 4
c
2°11
c
57
s
.3
in
si
6
n
2°
4
3
2
1
°
11
c43.37198044
C42°11, b62.5, c43.4
Chapter 6 196
47d. I220 sin
60t
6
I220 sin
6060
6
I110 A
48. y1 tan
v
2
49a. A
3.99
2
0.55
49b. h
3.99
2
0.55
A1.72 ft h2.27 ft
49c. 12:19 P.M.12:03 A.M.12:16 or about 12.3 hr
49d. A1.72
2
k
12.3 h2.27
k
1
2
2
.3
12:03 0.05 hr since midnight
h1.72 sin
1
2
2
.3
tc
2.27
3.99 1.72 sin
1
2
2
.3
0.05 c
2.27
1.72 1.72 sin
0
1
.
2
1
.
3
c
1 sin
0
1
.
2
1
.
3
c
sin11
0
1
.
2
1
.
3
c
sin11
0
1
.
2
1
.
3
c
1.545254923 c
Sample answer: h1.72 sin
1
2
2
.3
t1.55
2.27
49e. noon 12 hrs since midnight
h1.72 sin
1
2
2
.3
t1.55
2.27
h1.72 sin
1
2
2
.3
12 1.55
2.27
h3.964014939
Sample answer: 3.96 ft
50. 22; 4
51. srv
s18
3
s6cm
2
1
2
1
2
1
2
22
y
2 cos
2
y
O
73˚
4 m
53a.
53b. tan 73°
4
x
x4 tan 73°
x13.1 m
53c. cos 73°
4
y
y
cos
4
73°
y13.7 m
54. a2b2c2sin A
a
c
7242c2sin A
7
65
65
csin A
7
65
65
cos A
b
c
tan A
a
b
cos A
4
65
tan A
7
4
cos A
4
65
65
55.
—
x2
x2
3
x
4
10
—
0
—
(
(
x
x
2
5
)
)
(
(
x
x
2
2
)
)
—
0zeros: 2, 2
excluded values: 5, 2
Test 3: 0
5
5
0false
Test 0:
02
0
3
2
(
0)
4
10
0
1
4
0
0false
Test 3:
32
3
3
2
(
3)
4
10
0
5
10
0true
Test 6:
62
6
3
2
(
6)
4
10
0
3
8
2
0false
2 x5
56. k
0
6
.5
tkr
k12 10 12r
r0.83
(3)24
(3)23(3) 10
57. 3x2y8
y
3
2
x4
y2x1
2yx4
y
1
2
x2
58. y17.98x35.47; 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.
Sound Beats
Page 404
1.
the third graph
2.
Sample answer: The graph seems to stay above
the x-axis for an interval of xvalues, and then
stay below the x-axis for another interval of x
values.
3. 0.38623583
4. no
5. 1.78043; yes; the value for f(x) is negative and
corresponds to a point not graphed by the
calculator.
6. 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.
7. See students’ work.
8. Yes; no, they only provide plausible visual
evidence.
6-7B
Trigonometric Inverses and Their
Graphs
Page 410 Check for Understanding
1. ysin1xis the inverse relation of ysin x, y
(sin x)1is the function y
sin
1
x
, and ysin(x1)
is the function ysin
1
x
.
2. For every yvalue there are more than one xvalue.
The graph of ycos1xfails the vertical line
test.
3. The domain of ySin xis the set of real numbers
between
2
and
2
, inclusive, while the domain of
ysin xis 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. yArcsin x
xArcsin y
Sin xyor ySin x
7. yCos
x
2
xCos
y
2
Cos1xy
2
yCos1x
2
8. Let vArctan 1. 9. Let vTan11.
Tan v1Tan v1
v
4
v
4
cos (Tan11) cos v
cos
4
2
2
6-8
197 Chapter 6
y
x
O
11
y
Arcsin
x
2
2
x
y
O
1
1
y
Sin
x
2
2
x
y
O
1
1
y
Cos
2
2
()
x
2
x
y
O
11
y
Cos1
x
2
2
2
x
y
O
10. Let vCos1
2
2
.
Cos v
2
2
v
—
4
—
cos
Cos1
2
2
2
cos
v
2
cos
4
2
cos
—
4
—
2
2
11. true
12. false; sample answer: x1; when x1,
Cos1(1) , Cos1(1) 0
13a. C2r13b. C40,212 cos v
C2(6400)
C40,212 km
13c. C40,212 cos v
3593 40,212 cos v
4
3
0
5
,2
9
1
3
2
cos v
cos1
4
3
0
5
,2
9
1
3
2
v
1.48 v; about 1.48 radians
13d. C40,212 cos v
C40,212 cos 0
C40,212 km
Pages 410–412 Exercises
14. yarccos x
xarccos y
cos xyor ycos x
15. ySin x
xSin y
Sin1xyor ySin1x
16. yarctan x
xarctan y
tan xyor ytan x
17. yArccos 2x
xArccos 2y
Cos x2y
1
2
Cos xyor y
1
2
Cos x
18. y
2
Arcsin x
x
2
Arcsin y
x
—
2
—
Arcsin y
Sin
x
2
yor ySin
x
2
19. ytan
2
x
xtan
2
y
tan1x
2
y
2 tan 1xy; y2 tan1x
Chapter 6 198
11
2
y
arccos
x
x
y
O
1
1
2
y
cos
x
x
y
O
1
1
y
Sin
x
2
2
x
y
O
11
y
Sin1
x
2
2
x
y
O
2
2
y
arctan
x
x
y
O
2
2
y
tan
x
x
y
O
11
y
Arccos 2
x
x
y
O
1
1
y
Cos
x
1
2
x
y
O
11
2
y
Arcsin
x
2
x
y
O
1
1
y
Sin
()
x
2
2
x
y
O
2
2
y
tan
x
2
x
y
O
2
2
y
2 tan1
x
x
y
O
20. yTan
x
2
xTan
y
2
Tan1xy
2
Tan1x
2
y
No; the inverse is yTan1x
2
.
21.
30. Let aSin11 and Cos1
1
2
.
Sin a1Cos
1
2
a
2
3
sin
Sin11 Cos1
1
2
sin (a)
sin
2
3
sin
6
1
2
31. No; there is no angle with the sine of 2.
32. false; sample answer: x2; when x2,
Cos1(cos 2) Cos11, or 0, not 2.
33. true
34. false; sample answer: x1; when x1,
Arccos (1) and Arccos ((1)) 0.
35. true 36. true
37. false; sample answer: x
2
; when x
2
, cos1
2
is undefined.
38.
39. y54.5 23.5 sin
6
t
2
3
54.5 54.5 23.5 sin
6
t
2
3
0 23.5 sin
6
t
2
3
0 sin
6
t
2
3
sin1 0
6
t
2
3
0
6
t
2
3
or
6
t
2
3
2
3
6
t
5
3
6
t
4 t10 t
April and October
40. PVI Cos v
7.3 122(0.62) Cos v
0.0965097832 Cos v
Cos10.0965097832 v
1.47 v; about 1.47 radians
41.
4
n, where nis an integer
42. II0cos2v
1 8 cos2v
1
8
cos2v
1
8
cos v
cos1
1
8
v
1.21 v; about 1.21 radians
43a. 6:18 12:24 18:42 or 6:42 P.M.
43b. 12.4 h
43c. A
7.05
2
(0.30)
A3.675 ft
199 Chapter 6
4
2
4
2
y
Cot
x
2
x
y
O
44
22
y
Arccot
x
2
x
y
O
22. Let vSin10.
Sin v0
v0
23. Let vArccos 0.
Cos v0
v
—
2
—
25. If ytan
4
, then
y1.
Sin1
tan
4
Sin1y
Sin11
—
2
—
24. Let vTan1
3
3
.
Tan v
3
3
v
6
26. If yCos1
2
2
, then y
—
4
—
.
sin
2 Cos1
2
2
sin (2y)
sin
2
4
sin
2
1
27. If yTan13
, then y
3
.
cos (Tan13
) cos y
cos
3
1
2
28. Let aTan11 and Sin11.
Tan a1Sin 1
a
4
2
cos (Tan11 Sin11) cos (a)
cos
4
2
cos
4
2
2
29. Let aCos10 and Sin1
1
2
.
Cos a0Sin
1
2
a
2
6
cos
Cos10 Sin1
1
2
cos (a)
cos
2
6
cos
2
3
1
2
y
x
O
y
tan(Tan1
x
)
43d. A3.675
2
k
12.4 h
7.05
2
(0.30)
k
6
.2
h3.375
6:18 6.3 h
y3.675 sin
6
.2
tc
3.375
7.05 3.675 sin
6
.2
6.3 c
3.375
3.675 3.675 sin
6
6
.3
.2
c
1 sin
6
6
.3
.2
c
sin11
6
6
.3
.2
c
sin11
6
6
.3
.2
c
1.621467176 c
Sample answer:
y3.375 3.675 sin
6
.2
t1.62
43e. y3.375 3.675
sin
6
.2
t1.62
6 3.375 3.675
sin
6
.2
t1.62
2.625 3.675 sin
6
.2
t1.62
2
3
.
.
6
6
2
7
5
5
sin
6
.2
t1.62
sin1
2
3
.
.
6
6
2
7
5
5
—
6
.2
—
t1.62
sin 1
2
3
.
.
6
6
2
7
5
5
1.62
6
.2
t
6
.2
sin1
2
3
.
.
6
6
2
7
5
5
1.62
t
4.767243867 t
0.767243867 60 46.03463204;
Sample answer: about 4:46 A.M.
44.
45a. vcos1
D
2
c
d
vcos1
6
2(
10
4
)
v1.47 radians
45b. LD(dD)v2Csin v
L(6) (4 6)1.47 2(10) sin 1.47
L35.81 in.
46. n, where nis an integer
47. A5
2
k
3h8
c
2
3
Chapter 6 200
1
1
22
y
sin (Tan1
x
)
x
y
O
c
2
3
y5 sin
2
3
v
2
3
8
1
1
11910
y
cos
x
x
y
O
30 30
30 30
x
25˚
48.
49. v25°
a180° (25° 25°)
or 130°
sin
30
25°
sin
x
130°
x
30
s
s
in
in
2
1
5
3
°
0°
x54.4 units
v2(25°) or 50°
a
1
2
(180° 50°) or 65°
sin
30
65°
sin
y
50°
y
30
si
s
n
in
65
5
°
0°
y25.4 units
50. 210° 180° 30°
51. p: 1, 2, 3, 6
q: 1, 2
—
p
q
—
: 1, 2, 3, 6,
1
2
,
3
2
52.
decreasing for x2 and x2
53. [f g](x) f(g(x))
f(3x)
(3x)31
27x31
[g f](x) g(f(x))
g(x31)
3(x31)
3x33
54. D4, F6, G7, H8
value: (4 6 7 8)4 (25)4 or 100
The correct choice is D.
30 30
30 30
y
g
(
x
)
g
(
x
)
3
1
x
2
x
O
A5k
—
2
3
—
Chapter 6 Study Guide and Assessment
Page 413 Understanding and Using the
Vocabulary
1. radian 2. angular
3. the same 4. amplitude
5. angle 6. phase
7. radian 8. frequency
9. sunusoidal 10. domain
Pages 414–416 Skills and Concepts
11. 60° 60°
18
0°
12. 75° 75°
18
0°
3
5
1
2
13. 240° 240°
18
0°
14.
5
6
5
6
18
0°
4
3
150°
15.
7
4
7
4
18
0°
16. 2.4 2.4
18
0°
315° 137.5°
17. srv18. 75° 75°
18
0°
s15
3
4
5
1
2
s35.3 cm srv
s15
5
1
2
s19.6 cm
19. 150° 150°
18
0°
20. srv
5
6
s15
5
srvs9.4 cm
s15
5
6
s39.3 cm
21. 5 210or about 31.4 radians
22. 3.8 27.6or about 23.9 radians
23. 50.4 2100.8or about 316.7 radians
24. 350 2700or about 2199.1 radians
25. 1.8 23.626. 3.6 27.2
q
v
t
q
v
t
q
3.
5
6
q
7.
2
2
q2.3 radians/s q11.3 radians/min
27. 15.4 230.828. 50 2100
q
v
t
q
v
t
q
30
1
.
5
8
q
10
1
0
2
q6.5 radians/s q26.2 radians/min
29. 130. 031. 132. 0
33. 44;
2
2
34. 0.50.5;
2
4
2
35.
1
3
1
3
; 4
36. A4
2
k
2
4
c
2h1
A4k4c8
y4 sin (4v8) 1
37. A0.5
2
k
2
c
3
h3
A0.5 k2c
2
3
y0.5 sin
2v
2
3
3
38. A
—
3
4
—
2
k
4
8
c
0h5
A
3
4
k8c0
y
3
4
cos 8v5
39. A
120
2
80
2
k
1h
120
2
80
A20 k2h100
y20 sin 2t100
40. A
130
2
100
—
2
k
—
1h
130
2
100
A15 k2h115
y15 sin 2t115
41. period:
2
1
or 2, no phase shift, no vertical shift
2
1
2
201 Chapter 6
2
4
4
2
3
2
y
4 cos 2
y
O
1
1
3
2
y
0.5 sin 4
y
O
1
1
6
42
y
cos
1
3
2
y
O
1
1
4
32
y
csc
1
3
y
O
42.
3
;
6
; no vertical shift
43. vertical shift: 4
44. vertical shift: 2
2
3
Page 417 Applications and Problem Solving
50a. A11.5 2
k
12 3h64
c
6
Chapter 6 202
8
4
8
4
2
y
2 tan 3
()
2
y
O
2
6
4
2
32
y
sec 4
y
O
6
2
2
32
y
tan 2
y
O
45. Let vArctan 1.
Tan v1
v
4
46. Let vSin11.
Sin v1
v
2
47. If ytan
4
, then y1.
Cos1
tan
4
Cos1y
Cos11
Let vCos11.
Cos v1
v0
48. If ySin1
2
3
, then y
—
3
—
.
sin
Sin1
2
3
sin y
sin
—
3
—
2
3
49. Let aArctan 3
and Arcsin
1
2
.
Tan a3
Sin
1
2
a
3
6
cos (Arctan 3
Arcsin
1
2
cos (a)
cos
3
6
cos
2
0
c
2
y11.5 sin
6
t
2
64
50b. April: month 4
y11.5 sin
6
t
2
64
y11.5 sin
6
4
2
64
y69.75; about 69.8°
50c. July: month 7
y11.5 sin
6
t
2
64
y11.5 sin
6
7
2
64
y74.0°
51. B
IL s
F
in v
0.04
5.0(1
0
)
.2
sin v
0.04(5.0(1) sin v) 0.2
sin v
0.04
0
(5
.2
.0)(1)
sin v1
v
2
Page 417 Open-Ended Assessment
1. A
1
2
r2v
26.2
1
2
r2v
Sample answer: r5 in., v
2
3
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.
2b. Sample answer: If the equation does not have
the form yAcos kv, the graph has been
translated. The graph of y2 cos 2vhas not
been translated. The graph of y2 cos (2v)
3 has been translated vertically and
horizontally.
1
1
2
x
y
O
1
1
2
x
y
O
k
6
Chapter 6 SAT & ACT Preparation
Page 419 SAT and ACT Practice
1. Since there is no diagram, draw one. Sketch a
right triangle and mark the information given.
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).
sin v
4
5
The correct choice is B.
2. Let xbe 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 x2. Write an equation that says that
the sum of these two integers is 56. Then solve for x.
x(x2) 56
2x2 56
2x54
x27
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.
a2b2(ab)(ab)
sin vcos 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 Bare (9, 12).
Since point Ahas 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).
Aslightly different way of solving this problem is
to write the equation of the line containing points
Aand B.
y
1
9
2
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.
x22x8 0
(x4)(x2) 0
(sin vcos v)(sin vcos v)
sin vcos v
If either of the two factors equals 0, then the
statement is true. Set each factor equal to 0 and
solve for x.
x4 0or x2 0
x4x2
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.
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.
cos v
hy
a
p
d
o
ja
te
ce
n
n
u
t
se
4
5
The correct choice is C.
7. Look at the powers of the variables in the
equation. There is an x2term, an xterm, and a y
term, but no y2term. It cannot represent a line,
because of the x2term. 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
ya(xh)2k. 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.
I
16
n
n
1
16
16
n
(n
1
1)
16; an integer
II
16n
16
n
16
16(
1
n
6
n
1)
n
n
1
; not an integer
III
16n
1
2
6n
n
n(16
1
n
6n
1)
16n
16
1
; not an integer
Only expression I is an integer.
The correct choice is A.
9. Since x1, 1 x0. So x1 x
xx1
1
.
Since x1, xx 11. So
xx1
1
1.
The correct choice is D.
203 Chapter 6
3
4
hypotenuse
opposite
adjacent
3
5
CB
A
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.
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.
Chapter 6 204
Basic Trigonometric Identities
Page 427 Check for Understanding
1. Sample answer: x45°
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. tan v
co
1
tv
, cot v
ta
1
nv
,
c
s
o
in
s
v
v
cot v,
1 cot2vcsc2v
4. tan(A)
c
s
o
in
s
(
(
A
A
)
)
c
s
o
i
s
n
A
A
c
s
o
in
s
A
A
tan A
5. Rosalinda is correct; there may be other values for
which the equation is not true.
6. Sample answer: v0°
sin vcos vtan v
sin 0° cos 0° tan 0°
0 1 0
1 0
7. Sample answer: x45°
sec2xcsc2x1
sec245° csc245° 1
(2
)2(2
)21
2 2 1
4 1
8. sec v
co
1
sv
9. tan v
co
1
tv
sec vtan v1
2
5
1
2
3
7-1 12.
7
3
2
3
cos
7
3
cos 2
3
cos
3
13. 330° 360° 30°
csc (330°)
sin (
1
330°)
205 Chapter 7
Chapter 7 Trigonometric Identities and Equations
sec v
3
2
tan v 2
5
tan v
2
5
5
10. sin2vcos2v1
1
5
2cos2v1
2
1
5
cos2v1
cos2v
2
2
4
5
cos v
2
5
6
Quadrant III, so
2
5
6
11. tan2v1 sec2v
4
7
21sec2v
1
4
6
9
1sec2v
6
4
5
9
sec2v
7
65
sec v
Quadrant IV, so
7
65
1
sin (360° 30°)
sin
1
30°
csc 30°
14.
c
c
s
o
c
tv
v
si
1
nv
c
s
o
in
s
v
v
co
1
sv
sec v
15. cos xcsc xtan xcos x
sin
1
x
c
s
o
in
s
x
x
1
16. cos xcot xsin xcos x
c
s
o
in
s
x
x
sin x
c
s
o
i
s
n
2
x
x
sin x
cos2x
si
nx
sin2x
sin
1
x
csc x
17. B
Fc
I
s
cv
BIFcsc v
F
c
B
sc
I
v
FBI
cs
1
cv
FBIsin v
Pages 427–430 Exercises
18. Sample answer: 45°
sin vcos vcot v
sin 45° cos 45° cot 45°
1
2
2
2
2
si
1
n v
—
c
s
o
in
s
v
v
1
2
1
19. Sample answer: 45°
t
s
a
e
n
cv
v
sin v
t
s
a
e
n
c4
4
5
5
°
°
sin 45°
1
2
2
2
2
2
2
20. Sample answer: 30°
sec2x1
c
c
o
sc
s
x
x
sec230° 1
c
c
o
sc
s
3
3
0
0
°
°
21
2
3
2
23
3
29. 1 cot2vcsc2v
1 cot2v
3
11
2
1 cot2v
1
9
1
cot2v
2
9
cot v
32
Quadrant II, so
32
30. tan2v1 sec2v
tan2v1
5
4
2
tan2v1
2
15
6
tan2v
1
9
6
tan v
3
4
Quadrant II, so
3
4
31. sin2vcos2v 1
1
3
2cos2v 1
1
9
cos2v 1
cos2v
8
9
cos v
2
32
Quadrant III, so cos v
2
32
tan v
c
so
in
sv
v
tan v
1
3
—
2
32
Chapter 7 206
1
9
2
13
4
1
3
21. Sample answer: 30°
sin xcos x 1
sin 30° cos 30° 1
1
2
2
3
1
3
4
1
2
3
1
22. Sample answer: 0°
sin ytan ycos y
sin 0° tan 0° cos 0°
0 01
01
23. Sample answer: 45°
tan2Acot2A1
tan245° cot245° 1
1 1 1
2 1
24. Sample answer: 0
cos v
2
cos vcos
2
cos 0
2
cos 0 cos
2
cos
2
cos 0 cos
2
01 0
01
25. csc v
si
1
nv
26. cot v
ta
1
nv
csc vcot v1
4
3
1
2
5
csc v
5
2
cot v4
3
cot v
27. sin2vcos2v1
1
4
2cos2v1
1
1
6
cos2v1
cos2v
1
1
5
6
4
3
3
cos v
15
4
Quadrant I, so
28. sin2vcos2v1
15
4
sin2v
2
3
21
sin2v
4
9
1
sin2v
5
9
sin v
5
3
Quadrant II, so
5
3
tan v
2
1
2
or
42
32. tan2v1 sec2v
2
3
21 sec2v
4
9
1 sec2v
1
9
3
sec2v
3
13
sec v
Quadrant III, so sec v
3
13
cos v
se
1
cv
cos v1
3
13
cos v or
33. cos v
se
1
cv
cos v1
7
5
3
1
3
13
3
13
sin2vcos2v 1
sin2v
5
7
2 1
sin2v
2
45
9
1
sin2v
2
44
9
sin v
2
76
Quadrant III, so
2
76
cos v
5
7
34. sec v
co
1
sv
sec v
sec v8
Quadrant IV, so 3
7
35. 1 cot2vcsc2vsin v
cs
1
cv
1
4
3
2csc2vsin v
1
1
9
6
csc2vsin v
3
5
2
9
5
csc2v
5
3
csc v
Quadrant IV, so
5
3
36. 1 cot2vcsc2v
1 (8)2csc2v
1 64 csc2v
65 csc2v
65
csc v
Quadrant IV, so 65
37. sec v
co
1
sv
sec v1
43
1
5
3
1
1
8
40.
19
5
2(2)
5
207 Chapter 7
Quadrant II, so
4
13
tan v
c
so
in
sv
v
tan v
4
13
—
43
tan v
1
3
3
or
3
39
4
33
2
3
39
2
2
4
13
2
2
43
2
sec2 Atan2 A
2sin2 A2cos2A
tan
19
5
sin
19
5
—
cos
19
5
sin 2(2)
5
——
cos 2(2)
5
tan
5
41.
—
10
3
—
3
3
sin
5
—
cos
5
csc
—
10
3
—
1
sin
10
3
1
sin 3
3
1
sin
3
csc
3
42. 1290° 7(180°) 30°
sec (1290°)
cos (
1
1290°)
co
1
s30°
sec 30°
43. 660° 2(360°) 60°
cot (660°)
c
s
o
in
s
(
(
6
6
6
6
0
0
°
°
)
)
c
s
o
in
s
6
6
0
0
°
°
cot 60°
44.
—
t
s
a
e
n
cx
x
—
sin
1
x
csc x
45.
—
c
c
o
o
s
tv
v
—
si
1
nv
csc v
46.
—
c
s
o
in
s
(
(
v
v
)
)
—
—
c
s
o
in
s
v
v
—
tan v
47. (sin xcos x)2(sin xcos x)2
sin2x2sin xcos xcos2xsin2x
2sin xcos xcos2x
2sin2x2 cos2x
2(sin2xcos2x)
2
c
s
o
in
s
v
v
cos v
co
1
s x
—
c
s
o
in
s
x
x
cos (2(360°) 60°)
sin (2(360°) 60°)
1
cos (7(180°) 30°)
tan2v1 sec2v
tan2v1 82
tan2v1 64
tan2v 63
tan v 37
sin2vcos2v1
sin2v
43
21
sin2v
1
3
6
1
sin2v
1
1
3
6
sin v
4
13
9
9
3
1
2
6
4
9
8
3
9
9
2
1
13
6
2
1
3
6
1
2
38. 390° 360° 30°
sin 390° sin (360° 30°)
sin 30°
39.
27
8
3
3
8
cos
27
8
cos 3
3
8
cos
3
8
sec v
4
3
or
4
33
48. sin xcos xsec xcot xsin xcos x
—
co
1
sx
—
—
c
s
o
in
s
x
x
—
cos x
49. cos xtan xsin xcot xcos x
—
c
s
o
in
s
x
x
—
sin x
—
c
s
o
in
s
x
x
—
sin xcos x
50. (1 cos v)(csc vcot v) (1 cos v)
—
si
1
nv
—
—
c
s
o
in
s
v
v
—
(1 cos v)
—
1
sin
co
v
sv
—
—
1
si
c
n
o
v
s2v
—
s
s
i
i
n
n
2
v
v
sin v
51. 1 cot2vcos2vcos2vcot2v
1 cot2vcos2v(1 cot2v)
csc2vcos2v(csc2v)
csc2v(1 cos2v)
csc2v(sin2v)
—
sin
1
2v
—
(sin2v)
1
52.
1
sin
co
x
sx
1
sin
co
x
sx
sin xsin xcos x
1 cos2x
sin xsin xcos x
1 cos2x
56a. e
W
A
se
s
cv
eAs Wsec v
s
e
e
A
c
s
v
W
W eAs cos v
56b. WeAs cos v
W0.80(0.75)(1000) cos 40°
W459.6266659
459.63 W
57. FNmg cos v0
FNmg cos v
mg sin vmkFN0
mg sin vmk(mg cos v)0
mk(mg cos v)mg sin v
mk
m
m
g
g
c
s
o
in
s
v
v
mk
c
s
o
in
s
v
v
mktan v
58.
v
3
2
6
n
0°
18
n
0°
, tan v, so h
2ta
a
nv
a
2
cot v.
a
2
h
Chapter 7 208
1
2
s
c
i
o
n
s
x
2x
2
si
s
n
i
2
n
x
x
sin
2
x
2csc x
53. cos4a2cos2asin2asin4a(cos2asin2a)2
12or 1
54. II0cos2v
0I0cos2v
0cos2v
0cos v
cos10v
90° v
55. Let (x, y) be the point where the terminal side of
Aintersects the unit circle when Ais in standard
position. When Ais reflected about the x-axis to
obtain A, the y-coordinate is multiplied by 1,
but the x-coordinate is unchanged. So,
sin (A) ysin Aand
cos (A) xcos A.
y
x
O
(
x
,
y
)
(
x
,
y
)
A
A
ha
The area of the isosceles triangle is
1
2
(a)
a
2
cot
18
n
0°
a
4
2
cot
18
n
0°
. There are nsuch triangles, so
A
1
4
na2cot
18
n
0°
.
59.
sin vEF and cos vOF since the circle is a unit
circle. tan v
O
CD
D
C
1
D
CD.
sec v
O
CO
D
C
1
O
CO. EOF OBA, so
O
EF
F
O
BA
A
B
1
A
BA. Then cot v
c
s
o
in
s
v
v
O
EF
F
BA.
Also by similar triangles,
E
E
O
F
O
O
B
A
, or
E
1
F
O
1
B
.
Then csc v
si
1
nv
E
1
F
O
1
B
OB.
60. Cos1135°
2
2
y
x
O
AC
DF
E
B
61.
62. 2(3° 30) 7° 7° 7°
18
0°
1
7
8
0
srv
s20
1
7
8
0
s2.44 cm
63. B180° (90° 20°) or 70°
sin A
a
c
cos A
b
c
sin 20°
3
a
5
cos 20°
3
b
5
35 sin 20° a35 cos 20° b
11.97070502 a32.88924173 b
a12.0, B70°, b32.9
64. 2
2184
4
1
0
4
25 2
0
2x25x20
(2x1)(x2) 0
2x1 0 or x2 0
x
1
2
x 2
2,
1
2
, 2
65. 2x27x4 0
x2
7
2
x20
x2
7
2
x2
x2
7
2
x
4
1
9
6
2
4
1
9
6
x
7
4
2
8
1
1
6
x
7
4
9
4
x
7
4
9
4
x0.5 or 4
66. continuous
67. 4(xy2z) 4(3) 4x4y8z12
4xyz0→4xy z0
3y9z12
xy2z3
x5y4z11
4y2z14
4(3y9z) 4(12) →12y36z48
3(4y2z) 3(14) 12y6z42
30z90
z3
3y9z12 xy2z3
3y9(3) 12 x(5) 2(3) 3
y5x2
(2, 5, 3)
68. m
4
4
2
5
yy1m(xx1)
2
9
or
2
9
y4
2
9
(x(4))
y
2
9
x
2
9
8
69. m∠BCD 40°
40
1
2
m-
(BC)
80 m-
(BC)
m∠BAC
1
2
m-
BC
m∠BAC
1
2
(80)
m∠BAC 40°
The correct choice is C.
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.
5.
[2, 2] sc1
2
by [2, 2] sc11
sin x
Pages 433–434 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)212.
3. Sample answer: They are the trigonometric
functions with which most people are most
familiar.
4. Answers will vary.
5. cos x
c
c
s
o
c
tx
x
cos x
cos x
co
1
sx
cos xcos x
c
s
o
in
s
x
x
—
sin
1
x
7-2
209 Chapter 7
y
x
1
67
6
2
35
313
6
1
O
y
cos
x
()
6
6.
tan x
1
sec x
sin
co
x
s
x
1
Pages 434–436 Exercises
13. tan A
s
c
e
sc
c
A
A
tan A
co
1
s A
—
sin
1
A
Chapter 7 210
sin
co
x
s
x
1
1
c
s
o
in
s
x
x
co
1
sx
sin
co
x
s
x
1
1
sin
co
x
s
x
1
sin
co
x
s
x
1
sin
co
x
s
x
1
7. csc vcot v
csc v
1
cot v
csc vcot v
csc v
1
cot v
c
c
s
s
c
c
v
v
c
c
o
o
t
t
v
v
csc vcot v
c
c
s
s
c
c
2v
v
c
c
o
o
t
t2
v
v
csc vcot v
csc vcot v
csc v
1
cot v
csc vcot vcsc vcot v
8. sin vtan vsec vcos v
sin vtan v
co
1
sv
cos v
sin vtan v
co
1
sv
c
c
o
o
s
s
2
v
v
sin vtan v
1
co
c
s
o
v
s2v
sin vtan v
s
c
i
o
n
s
2
v
v
sin vtan vsin v
c
s
o
in
s
v
v
sin vtan vsin vtan v
9. (sin Acos A)21 2 sin2Acot A
sin2A2 sin Acos Acos2A1 2 sin2Acot A
1 2 sin Acos A1 2 sin2Acot A
1 2 sin Acos A
s
s
i
i
n
n
A
A
1 2 sin2Acot A
1 2 sin2A
c
s
o
in
s
A
A
1 2 sin2Acot A
1 2 sin2Acot A1 2 sin2Acot A
10. Sample answer: sin x
1
4
tan x
1
4
sec x
t
s
a
e
n
cx
x
1
4
1
4
c
s
o
in
s
x
x
—
co
1
s x
csc vcot v
(1 cot2v) cot2v
sin x
1
4
11. Sample answer: cos x1
cot xsin xcos xcot x
c
s
o
in
s
x
x
sin xcos x
c
s
o
in
s
x
x
cos xsin2xcos2x
cos2xsin2xcos x
1cos x
cos x1
12.
Ic
R
o2
sv
R
I2c
c
o
s
t
c
v
v
Ic
R
o2
sv
I
c
s
o
in
s
v
v
—
R2
si
1
n v
Ic
R
o2
sv
s
s
i
i
n
n
v
v
Ic
R
o2
sv
Ic
R
o2
sv
I
c
s
o
in
s
v
v
—
R2
si
1
n v
tan A
c
s
o
in
s
A
A
tan Atan A
14. cos vsin vcot v
cos vsin v
c
s
o
in
s
v
v
cos vcos v
15. sec xtan x
1
cos
si
x
nx
sec xtan x
co
1
sx
c
s
o
in
s
x
x
sec xtan xsec xtan x
16.
si
1
n
x
ta
c
n
os
x
x
sec x
sec x
1
c
s
o
in
s
x
x
sin xcos x
c
c
o
o
s
s
x
x
sec x
1
c
s
o
in
s
x
x
sin xcos x
sec x
cos xsin x
cos x(sin xcos x)
co
1
sx
sec x
sec xsec x
17. sec xcsc xtan xcot x
sec xcsc x
c
s
o
in
s
x
x
c
s
o
in
s
x
x
sec xcsc x
c
s
o
in
s
x
x
s
s
i
i
n
n
x
x
c
s
o
in
s
x
x
c
c
o
o
s
s
x
x
sec xcsc x
cos
si
x
n2
si
x
nx
sin
co
x
s2
co
x
sx
sec xcsc x
sin
c
2
os
x
x
si
c
n
os
x
2x
sec xcsc x
cos x
1
sin x
sec xcsc x
co
1
sx
sin
1
x
sec xcsc xsec xcsc x
18. sin vcos v
s
2
in
si
v
n2
v
c
os
1
v
sin vcos v2 sin2v(sin2vcos2v)
sin vcos v
sin vcos v
s
s
in
in
2v
v
c
c
o
o
s
s
2
v
v
sin vcos v(sin vcos v)(sin vcos v)
sin vcos v
sin vcos vsin vcos v
19. (sin Acos A)2
2
sec
se
A
cA
csc
cs
A
cA
(sin Acos A)2
sec A
2
csc A
s
s
e
e
c
c
A
A
c
c
s
s
c
c
A
A
(sin Acos A)22
se
1
cA
cs
1
cA
1
(sin Acos A)22 cos Asin A1
(sin Acos A)22 cos Asin Asin2Acos2A
(sin Acos A)2(sin Acos A)2
20. (sin v1)(tan vsec v)cos v
sin vtan vtan vsin vsec vsec vcos v
sin v
c
s
o
in
s
v
v
c
s
o
in
s
v
v
sin v
co
1
sv
co
1
sv
cos v
cos v
sin2vsin vsin v1
cos v
27. sin xcos x
1
co
t
s
a
x
nx
1
sin
co
x
tx
sin xcos xsin x
1
c
s
o
in
s
x
x
cos x
1
c
s
o
in
s
x
x
211 Chapter 7
sin
c
2
o
v
s
v
1
cos v
c
c
o
o
s
s2
v
v
cos v
cos vcos v
21.
1
cos
si
y
ny
1
cos
si
y
ny
1
cos
si
y
ny
1
1
s
s
i
i
n
n
y
y
1
cos
si
y
ny
cos
1
y
(1
s
in
s
2in
y
y)
1
cos
si
y
ny
cos y
c
(1
os
2y
sin y)
1
cos
si
y
ny
1
cos
si
y
ny
1
cos
si
y
ny
22. cos vcos (v) sin vsin (v) 1
cos vcos vsin v(sin v)1
cos2vsin2v1
11
23. csc x1
cs
c
c
o
x
t2
x
1
csc x1
c
c
s
s
c
c
2
x
x
1
1
csc x1
csc x1 csc x1
24. cos Bcot Bcsc Bsin B
cos Bcot B
sin
1
B
sin B
cos Bcot B
sin
1
B
s
s
i
i
n
n
2
B
B
cos Bcot B
1
si
s
n
in
B
2B
cos Bcot B
c
s
o
i
s
n
2
B
B
cos Bcot Bcos B
c
s
o
in
s
B
B
cos Bcot Bcos Bcot B
25. sin vcos vtan vcos2v1
sin vcos v
c
s
o
in
s
v
v
cos2v1
sin2vcos2v1
11
26. (csc xcot x)2
1
1
c
c
o
o
s
s
x
x
csc2x2 csc xcot x cot2x
1
1
c
c
o
o
s
s
x
x
sin
1
2x
2
sin
1
x
c
s
o
in
s
x
x
c
s
o
in
s2
2
x
x
1
1
c
c
o
o
s
s
x
x
1
1
c
c
o
o
s
s
x
x
1 2 cos xcos2x
sin2x
(csc x1)(csc x1)
csc x1
(1
1
c
c
o
o
s
s2
x
x
)2
1
1
c
c
o
o
s
s
x
x
1
1
c
c
o
o
s
s
x
x
(1 cos x)2
(1 cos x)(1 cos x)
1
1
c
c
o
o
s
s
x
x
1
1
c
c
o
o
s
s
x
x
sin xcos x
c
c
o
o
s
s
x
x
s
s
i
i
n
n
x
x
sin x
1
c
s
o
in
s
x
x
cos x
1
c
s
o
in
s
x
x
sin xcos x
cos
c
x
o
s2
s
x
in x
sin
s
x
in
2
c
x
os x
sin xcos x
sin
c
x
o
s2
c
x
os x
sin
s
x
in
2
c
x
os x
sin xcos x
s
s
in
in
2x
x
c
c
o
o
s
s
2
x
x
sin xcos x(sin xcos x)(sin xcos x)
sin xcos x
sin xcos xsin xcos x
28. sin vcos vtan vsin vsec vcos vtan v
sin vcos v
c
s
o
in
s
v
v
sin vsec vcos vtan v
sin vcos v
s
c
i
o
n
s
2
v
v
sec vcos vtan v
sin v
c
c
o
o
s
s
2
v
v
s
c
i
o
n
s
2
v
v
sec vcos vtan v
sin v
cos2v
co
sv
sin2v
sec vcos vtan v
sin v
co
1
sv
sec vcos vtan v
sin vsec vsec vcos vtan v
sin v
c
c
o
o
s
s
v
v
sec vsec vcos vtan v
cos v
c
s
o
in
s
v
v
sec vsec vcos vtan v
cos vtan vsec vsec vcos vtan v
sec vcos vtan vsec vcos vtan v
29. Sample answer: sec x2
c
c
s
o
c
tx
x
2
2
si
1
n v
—
c
s
o
in
s
v
v
co
1
sx
2
sec x2
30. Sample answer: tan x2
1
1
t
c
a
o
n
tx
x
2
2
1
c
s
o
in
s
x
x
——
1
c
s
o
in
s
x
x
2
cos x
co
sx
sin x
——
sin x
si
nx
cos x
c
s
o
in
s
x
x
2
tan x2
31. Sample answer: cos x0
co
1
tx
s
c
e
sc
c
x
x
cos x
tan xcos x
co
1
s x
—
sin
1
x
tan x
c
s
o
in
s
x
x
cos x
tan xtan xcos x
0cos x
32. Sample answer: sin x
1
2
1
sin
co
x
sx
1
sin
co
x
sx
4
sin x(
s
1
in
2x
cos x)
4
1 2 cos xcos2x
sin x(1 cos x)
38. yes
[2, 2] sc1
2
by [4, 4] sc11
39. no
[2, 2] sc1
2
by [4, 4] sc11
40a. PI02Rsin22ft
PI02R(1 cos22pft)
40b. PI02Rsin22ft
P
cs
I
c2
02
2
R
ft
41. f(x) x
1 4
x2
Chapter 7 212
4
1 2 cos xcos2xsin2x
sin x(1 cos x)
sin
2
x
(1
2
co
c
s
os
x
x)
4
sin
2(
x
1
(1
co
c
s
o
x
s
)
x)
4
sin
2
x
4
24 sin x
1
2
sin x
33. Sample answer: sin x1
cos2x2 sin x20
1 sin2x2 sin x2 0
0sin2x2 sin x1
0(sin x1)2
0sin x1
sin x1
34. Sample answer: cot x1
csc xsin xtan xcos x
csc xsin x
c
s
o
in
s
x
x
cos x
csc x
s
c
i
o
n
s
2
x
x
c
c
o
o
s
s
2
x
x
csc x
co
1
sx
sin
1
x
co
1
sx
c
s
o
in
s
x
x
1
cot x1
35.
t
t
a
a
n
n
3
v
v
1
1
sec2v1 0
(tan2v1) 1 0
(tan v1)(tan2vtan v1)
tan v1
tan2vtan v1 tan2v1 1 0
tan v1 0
tan v1
cot v
ta
1
nv
cot v
1
1
cot v1
36. no
[2 , 2 ]sc1
2
by [2, 8] sc11
37. yes
[2 , 2 ]sc1
2
by [4, 4] sc11
f(x)
1
2
tan v
14
1
2
tan
v
2
f(x)
1
2
tan v
1 ta
n2v
f(x)
1
2
tan v
sec2v
f(x)
1
2
tan v
sec v
f(x)
f(x)
1
2
sin v
42. sin asin asin c⇒sin a
s
s
i
i
n
n
a
c
cos b
s
co
in
s
a
b
⇒cos bsin acos b
cos ccos acos b⇒cos b
c
c
o
o
s
s
a
c
Then cos bsin acos b
s
s
i
i
n
n
a
c
c
c
o
o
s
s
a
c
c
s
o
in
s
a
a
c
s
o
in
s
c
c
tan acot c
43. y
2v0
2
g
c
v
o
2
s2v
x
c
s
o
i
s
n
v
v
y
2v
g
0
v
2
2
sec2vxtan v
y
2
g
v
x
0
2
2
(1 tan2v) xtan v
1
2
c
s
o
in
s
v
v
co
1
s v
44. We find the area of ABTP by subtracting the area
of OAP from the area of OBT.
1
2
OB BT
1
2
OA AP
1
2
1 tan v
1
2
cos vsin v
1
2
c
s
o
in
s
v
v
cos vsin v
1
2
sin v
co
1
sv
cos v
1
2
sin v
co
1
sv
c
c
o
o
s
s
2
v
v
1
2
sin v
1
co
c
s
os
v
2v
1
2
sin v
s
c
i
o
n
s
2
v
v
1
2
c
s
o
in
s
v
v
sin2v
1
2
tan vsin2v
45. By the Law of Sines,
sin
b
b
sin
a
a
, so b
a
s
s
in
in
a
b
.
Then
A
1
2
ab sin
A
1
2
a
a
s
s
in
in
a
b
sin
A
a2s
2
in
si
n
s
a
in
Aa2sin bsin
2 sin (180° (b))
51. Let xthe number of shirts and ythe number
of pants.
x1.5y100
2.5x2y180
1.5x3y195
x0
y0
P(x, y)5x4.5y
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
53.
a
a
b
b
b
b
a
a
a
a
b
b
b
b
a
a
a
a
b
b
1
a
(a
b
b)
1
The correct choice is D.
Sum and Difference Identities
Pages 441–442 Check for Understanding
1. Find a counterexample, such as x30° and
y60°.
2. Find the cosine, sine, or tangent, respectively, of
the sum or difference, then take the reciprocal.
3. The opposite side for 90° Ais the adjacent side
for A, so the right-triangle ratio for sin (90° A) is
the same as that for cos A.
4. cot (ab)
tan (a
1
b)
1
1
ta
n
t
a
an
a
ta
ta
n
n
b
b
7-3
213 Chapter 7
A
a
2
2
s
s
i
i
n
n
(
b
b
s
in
)
46.
c
s
o
in
s
x
x
cos xsin x
c
s
o
in
s
x
x
co
1
s x
c
s
o
in
s
x
x
tan xcos xsin xtan x
sec xtan x
1
cos
si
x
nx
sin
co
x
s
x
1
1
cos
si
x
nx
1
47. A2
36
k
0°
180°
2
c
45°
A2k2c90°
y2sin (2x90°)
48.
1
1
5
6
1
1
5
6
18
0°
168.75°
168.75° 168° 0.75°
6
1
0
°
168° 45
168° 45
49.
33y1
2 0Check:
33y1
2 0
33y1
2
33(3)
1
20
3y18
38
20
y32 20
50. x1 0
x1
f(x)
x3
x1
y
x
3
x
1
y(x1) 3x
yx y3x
y3xyx
yx(3 y)
3
y
y
x
3 y0
y3
sin xcos2x sin2x
cos x
O
y
x
(72, 0)
(40, 40)
(0, 65)
(0, 0)
x
1.5
y
100
2.5
x
2
y
180
1.5
x
3
y
195
20
20
40
60
80
100
40 60 80 100
y
0
x
0
1
ta
n
t
a
an
a
ta
ta
n
n
b
b
c
c
o
o
t
t
a
a
c
c
o
o
t
t
b
b
1
co
1
t a
co
1
t b
co
1
t a
co
1
t b
A
90˚ A
c
c
o
o
t
t
a
a
c
ot
c
b
o
tb
1
5. cos 165° cos (45° 120°)
cos 45° cos 120° sin 45° sin 120°
1
2
3
2
2
2
2
2
10. sin (90° A)cos A
sin 90° cos Acos 90° sin Acos A
1 cos A0 sin Acos A
cos Acos A
11. tan v
2
cot v
Chapter 7 214
2
6
4
6. tan
1
2
tan
3
4
tan
3
tan
4
——
1 tan
3
tan
4
3
1
1 3
1
2 3
7. 795° 2(360°) 75°
sec 795° sec 75°
cos 75° cos (30° 45°)
cos 30° cos 45° sin 30° sin 45°
1
2
2
2
2
2
3
2
4 23
2
6
2
4
sec 795° 4
6
2
6
2
8. cos x 1 si
n2x
cos y1 si
n2y
1
4
9
2
1
1
4
2
6
8
5
1
or
1
1
5
6
or
sin (xy) sin xcos ycos xsin y
4
9
1
4
4
15
3
6
65
9. csc x
sin
1
x
cos x
1 si
n2x
5
3
sin
1
x
1
3
5
2
sin x
3
5
1
2
6
5
or
4
5
tan x
c
s
o
in
s
x
x
or
3
4
3
5
4
5
65
9
15
4
15
4
65
9
sin y
1 co
s2y
1
1
5
3
2
1
1
4
6
4
9
or
1
1
2
3
tan (xy)
1
ta
n
t
x
an
x
ta
ta
n
n
y
y
3
4
1
5
2
——
1
3
4
1
5
2
6
1
3
6
6
2
3
0
—
4
5
cot v
sin v
2
——
cos v
2
cot v
sin vcos
2
cos vsin
2
———
cos vcos
2
sin vsin
2
cot v
c
s
o
in
s
v
v
cot v
cot vcot v
12. sin (xy)
1
cs
c
c
o
x
t
s
x
e
t
c
a
y
ny
sin (xy)
1
c
s
o
in
s
x
x
c
s
o
in
s
y
y
——
sin
1
x
co
1
s y
(sin v) 0 (cos v) 1
(cos v) 0 (sin v) 1
sin (xy)
s
s
i
i
n
n
x
x
c
c
o
o
s
s
y
y
1
c
s
o
in
s
x
x
c
s
o
in
s
y
y
——
sin
1
x
co
1
s y
sin (xy)
sin (xy) sin (xy)
13. sin (nq0t90°) sin nq0tcos 90° cos nq0tsin 90°
sin nq0t0 cos nq0t1
cos nq0t
Pages 442–445 Exercises
14. cos 105° cos (45° 60°)
cos 45° cos 60° sin 45° sin 60°
1
2
3
2
2
2
2
2
sin xcos ycos xsin y
1
15. sin 165° sin (120° 45°)
sin 120° cos 45° cos 120° sin 45°
1
2
2
2
2
2
3
2
2
6
4
16. cos
7
1
2
cos
4
3
cos
4
cos
3
sin
4
sin
3
1
2
3
2
2
2
2
2
6
2
4
2
6
4
17. sin
1
2
sin
3
4
sin
3
cos
4
cos
3
sin
4
1
2
2
2
2
2
3
2
6
2
4
tan y
c
s
o
in
s
y
y
or
1
5
2
1
1
2
3
—
1
5
3
18. tan 195° tan (45° 150°)
tan 45° tan 150°
1 tan 45° tan 150°
25.
11
1
3
2
4(2)
1
1
7
2
cot
11
1
3
2
cot
1
1
7
2
tan
1
1
7
2
tan
7
6
4
3
2
cot
11
1
3
2
3
1
2
2 3
26. sin x1 co
s2x
cos y1 si
n2y
1
1
8
7
2
1
1
3
2
7
2
2
2
2
8
5
9
or
1
1
5
7
1
1
2
3
2
6
5
9
or
3
3
5
7
sin (xy) sin xcos ycos xsin y
1
1
5
7
3
3
5
7
1
8
7
1
3
2
7
6
6
2
2
1
9
27. sin x1 co
s2x
sin y1 co
s2y
1
3
5
2
1
4
5
2
1
2
6
5
or
4
5
2
9
5
or
3
5
cos (xy) cos xcos ysin xsin y
3
5
4
5
4
5
3
5
2
2
4
5
28. cos x1 si
n2x
sin y1 co
s2y
1
1
8
7
2
1
3
5
2
2
2
2
8
5
9
or
1
1
5
7
1
2
6
5
or
4
5
tan x
c
s
o
in
s
x
x
tan y
c
s
o
in
s
y
y
1
8
5
4
3
tan (xy)
1
ta
n
t
x
an
x
ta
ta
n
n
y
y
3
7
6
7
1
1
2
5
—
7
4
7
5
1
8
5
4
3
1
1
8
5
4
3
4
5
—
3
5
1
8
7
—
1
1
5
7
3
3
3
—
3
3
3
3
3
1
——
1
3
3
1
tan
6
tan
4
——
1 tan
p
6
tan
4
215 Chapter 7
1
3
3
1 1
3
3
3
3
3
—
3
3
3
or 2 3
12 63
6
19. cos
1
2
cos
4
3
cos
4
cos
3
sin
4
sin
p
3
1
2
3
2
2
2
2
2
20. tan 165° tan (45° 120°)
tan 45° tan 120°
1 tan 45° tan 120°
2
6
4
1 (3
)
1 1 (3
)
4
2
2
3
or 2 3
21. tan
2
1
3
2
tan
4
5
3
tan
4
tan
5
3
——
1 tan
4
tan
5
3
1 3
1 3
4
2
2
3
or 2 3
22. 735° 2(360°) 15°
sin 735° sin 15°
sin 15° sin (45° 30°)
sin 45° cos 30° cos 45° sin 30°
2
2
2
3
2
2
1
2
6
4
2
23. 1275° 3(360°) 195°
sec 1275° sec 195°
cos 195° cos (150° 45°)
cos 150° cos 45° sin 150° sin 45°
2
3
2
2
1
2
2
2
6
4
2
sec 1275°
6
4
2
2
6
24. sin
5
1
2
sin
6
4
sin
6
cos
4
cos
6
sin
4
1
2
2
2
2
3
2
2
2
4
6
csc
5
2
2
4
6
6
2
1 (3
)
1 1 (3
)
29. sec x
tan2x
1
cos y
1 si
n2y
5
3
2
1
1
1
3
2
3
9
4
or
3
34
8
9
or
2
3
2
cos x
3
34
or
3
34
34
tan x
c
s
o
in
s
x
x
5
3
sin x
—
3
34
34
3
5
1
2
6
5
9
or
1
5
3
cos x1 si
n2x
cos y
se
1
cy
1
3
5
2
1
—
1
5
3
Chapter 7 216
sin x
5
34
34
cos (xy) cos xcos ysin xsin y
3
34
34
2
3
2
5
34
34
1
3
6
10
6
2
8
5
10
3
2
4
1217
534
102
30. tan x
co
1
tx
cos y
se
1
cy
1
—
3
2
1
—
6
5
5
6
2
3
sin y 1 co
s2y
1
2
3
2
5
9
or
3
5
tan y
c
s
o
in
s
y
y
or
2
5
tan (xy)
1
ta
n
t
x
an
x
ta
ta
n
n
y
y
5
6
2
5
——
1
5
6
2
5
3
5
—
2
3
10
1
6
2
5
——
12
1
5
2
5
1
1
0
2
6
5
5
5
270
1
1
9
22
5
31. sin x
cs
1
cx
sec ytan2y
1
1
5
2
2
1
1
—
5
3
1
2
6
5
or
4
5
1
5
3
sin y1 co
s2y
1
1
5
3
2
1
1
4
6
4
9
or
1
1
2
3
cos(xy) cos xcos ysin xsin y
4
5
1
5
3
3
5
1
1
2
3
5
6
6
5
sec (xy)
cos (x
1
y)
6
5
5
6
32. cos a1 si
n2a
sin b1 co
s2b
1
1
5
2
1
2
7
2
2
2
4
5
or
2
5
6
4
4
5
9
or
3
7
5
sin (ab) sin acos bcos asin b
1
5
2
7
2
5
6
3
7
5
26
3
5
30
33. sin x1 co
s2x
sin y1 co
s2y
1
1
3
2
1
3
4
2
8
9
or
2
3
2
1
7
6
or
4
7
cos (xy) cos xcos ysin xsin y
1
3
3
4
2
3
2
4
7
32
1
2
14
34. cos
2
xsin x
cos
2
cos xsin
2
sin xsin x
0 cos x1 sin xsin x
sin xsin x
35. cos (60° A)sin (30° A)
cos 60° cos Asin 60° sin Asin 30° cos A
cos 30° sin A
1
2
cos A
2
3
sin A
1
2
cos A
2
3
sin A
36. sin (A)sin A
sin Acos cos Asin sin A
(sin A)(1) (cos A)(0) sin A
sin Asin A
1
—
5
5
6
5
37. cos (180° x) cos x
cos 180° cos xsin 180° sin xcos x
1 cos x0 sin xcos x
cos xcos x
38. tan (x45°)
1
1
t
t
a
a
n
n
x
x
1
1
t
t
a
a
n
n
x
x
tan xtan 45°
1 tan xtan 45°
217 Chapter 7
1
tan
(ta
x
n
x)
1
(1)
1
1
t
t
a
a
n
n
x
x
1
1
t
t
a
a
n
n
x
x
1
1
t
t
a
a
n
n
x
x
39. sin (AB)
ta
s
n
ec
A
A
se
ta
c
n
B
B
sin (AB)
c
s
o
in
s
A
A
+
c
s
o
in
s
B
B
——
co
1
s A
co
1
s B
sin (AB)
c
c
o
o
s
s
A
A
c
c
o
o
s
s
B
B
c
s
o
in
s
A
A
+
c
s
o
in
s
B
B
——
co
1
s A
co
1
s B
sin (AB)
sin (AB) sin (AB)
40. cos (AB)
1
se
t
c
a
A
nA
sec
ta
B
nB
cos (AB)
1
c
s
o
in
s
A
A
c
s
o
in
s
B
B
——
co
1
s A
co
1
s B
sin Acos Bcos Asin B
1
cos (AB)
c
c
o
o
s
s
A
A
c
c
o
o
s
s
B
B
1
c
s
o
in
s
A
A
c
s
o
in
s
B
B
——
co
1
s A
co
1
s B
cos (AB)
cos (AB) cos (AB)
41. sec (AB)
1
se
t
c
a
A
nA
sec
ta
B
nB
sec (AB)
co
1
s A
co
1
s B
——
1
c
s
o
in
s
A
A
c
s
o
in
s
B
B
cos Acos Bsin Asin B
1
sec (AB)
c
c
o
o
s
s
A
A
c
c
o
o
s
s
B
B
co
1
s A
co
1
s B
——
1
c
s
o
in
s
A
A
c
s
o
in
s
B
B
sec (AB) 1
cos Acos Bsin Asin B
sec (AB)
cos (A
1
B)
sec (AB) sec (AB)
42. sin (xy) sin (xy) sin2xsin2y
(sin xcos ycos xsin y)(sin xcos ycos xsin y)
sin2xsin2y
(sin xcos y)2(cos xsin y)2sin2xsin2y
sin2xcos2ycos2xsin2ysin2xsin2y
sin2xcos2ysin2xsin2ysin2xsin2y
cos2xsin2ysin2xsin2y
sin2x(cos2ysin2y) sin2y(sin2xcos2x)
sin2xsin2y
(sin2x)(1) (sin2y)(1) sin2xsin2y
sin2xsin2ysin2xsin2y
n
sin
a
2
30°
sin 30°
n
sin
a
2
cos 30° cos
a
2
sin 30°
1
2
n2sin
a
2
2
3
cos
a
2
1
2
n3
sin
a
2
cos
a
2
45. The given expression is the expanded form of the
sine of the difference of
3
Aand
3
A. We have
sin
3
A
3
A sin (2A)
sin 2A
46a.
f(xh
h
)f(x)
sin (xh) sin x
h
46b.
46c. cos x
47. tan (ab)
c
s
o
in
s
(
(
a
a
b
b
)
)
tan (ab) sin acos bcos asin b
cos acos bsin asin b
sin xcos hcos xsin hsin x
h
y
x
1
0.5
54 678321
1
0.5
O
y
sin
x
cos 0.1 cos
x
sin 0.1 sin
x
0.1
tan (ab)
c
s
o
in
s
a
a
c
c
o
o
s
s
b
b
c
c
o
o
s
s
a
a
c
s
o
in
s
b
b
———
c
c
o
o
s
s
a
a
c
c
o
o
s
s
b
b
c
s
o
in
s
a
a
s
co
in
s
b
b
tan (ab)
1
ta
n
t
a
an
a
ta
ta
n
n
b
b
Replace bwith bto find tan(ab).
tan (a(b)) tan atan (b)
1 tan atan (b)
tan (ab)
1
ta
n
t
a
an
a
ta
ta
n
n
b
b
48a. Answers will vary.
43. VLI0qLcos qt
2
VLI0qLcos qtcos
2
sin qtsin
2
VLI0qL(cos qt0 sin qt1)
VLI0qL(sin qt)
VLI0qLsin qt
44. n
sin
1
2
(ab)
sin
b
2
n
sin
1
2
(a60°)
sin
6
2
0°
x
37˚12′
6˚40′
35 ft
48b. tan Atan Btan Ctan Atan Btan C
tan Atan Btan (180° (AB))
tan Atan Btan(180° (AB))
tan Atan Btan 180° tan (AB)
1 tan 180° tan (AB)
56. srvA
1
2
r2v
18 r(2.9) A
1
2
(6.2)2(2.9)
6.2 r; 6.2 ft A55.7 ft2
57. c270213022(70)(130) cos 130°
c233498.7345
c183 miles
58. 120° 90°, consider Case 2.
4 12, 0 solutions
59.
v37° 126° 40or 30° 32
a90° 6° 40or 96° 40
b180° (30° 3296° 40) or 52° 48
sin 3
3
0
5
°32
sin 5
x
2° 48
x
35
si
s
n
in
30
5
°
2
3
°
2
4
8
x54.87 ft
60. 4x33x2x0
x(4x23x1) 0
x(4x1)(x1) 0
x0or4x1 0orx1 0
x
1
4
x1
61. Case 1 Case 2
x14x14
(x1) 4x1 4
x1 4x3
x5
x5{xx5 or x3}
62.
1(6) 3(2)
2
6
1
3
Chapter 7 218
tan Atan Btan 180° tan (AB)
1 tan 180° tan (AB)
tan Atan B0 tan (AB)
1 0 tan (AB)
tan Atan B
tan Atan Btan (AB)
tan Atan Btan (AB)
(tan Atan B)
1
1
t
t
a
a
n
n
A
A
t
t
a
a
n
n
B
B
tan (AB)
tan Atan B (AB)
tan (AB)(1 tan Atan B) tan (AB)
tan Atan B (AB)
(1 tan Atan B1) tan (AB)
tan Atan B (AB)
tan Atan Btan (AB)tan Atan B(AB)
49. sec2x
1
1
c
s
o
in
s2
2
x
x
csc2xcot2x
sec2x
1
cos
c2
os
x
2x
1 cot2xcot2x
sec2x
cos
1
2x
c
c
o
o
s
s
2
2x
x
1
sec2xsec2x1 1
sec2xsec 2x
50. sin2vcos2v1
1
8
2cos2v1
cos2v
6
6
3
4
cos v
3
8
7
Quadrant III, so
3
8
7
0 tan (AB)
1 0 tan (AB)
51. Arctan 3
3
sin (Arctan 3
) sin
3
2
3
52. k, where kis an integer
53. A
86
2
50
2
4
2
h
86
2
50
A18 68 68
y18 sin
2
tc68
50 18 sin
2
1 c68
18 18 sin
2
c
1 sin
2
c
sin1(1)
2
c
3
2
2
c
c
y18sin
2
t68
54. 88;
36
1
0
360;
3
1
0°
30°
55. sin (540°) sin (360° 180°)
0
6 6 or 12
63. fg(4) f(g(4))
f(5(4) 1)
f(21)
3(21)24
1319
gf(4) g(f(4))
g(3(4)24)
g(44)
5(44) 1
221
64. (8)62 862
(
8
8
6
)
2
62
88
62 (1)62 1
The correct choice is A.
tan v
c
s
o
in
s
v
v
1
8
—
3
8
7
3
1
7
21
7
Page 445 Mid-Chapter Quiz
1. csc v
si
1
nv
7
2
Quadrant 1, so
3
2
5
2. tan2v1 sec2vcos v
se
1
cv
4
3
21sec2v
1
9
6
1sec2v
3
5
2
9
5
sec2v
5
3
sec v
Quadrant II, so
5
3
3.
19
4
5
4
cos
19
4
cos 5
4
cos
4
4.
1t
1
an2x
1c
1
ot2x
1
sec
1
2x
csc
1
2x
1
cos2xsin2x1
11
5.
csc2
s
v
e
c2s
v
ec2v
csc2v
s
c
e
sc
c
2
2v
v
s
s
e
e
c
c
2
2v
v
csc2v
1csc2v
c
s
o
in
s2
2
v
v
1csc2v
cot2v1csc2v
csc2vcsc2v
6. cot xsec xsin x2 tan xcos xcsc x
c
s
o
in
s
x
x
co
1
sx
sin x2
c
s
o
in
s
x
x
cos x
sin
1
x
12 1
11
7. tan (ab)
1
c
ot
c
a
ot
a
ta
ta
n
n
b
b
tan (ab)
1
ta
1
na
tan b
——
ta
1
na
tan b
sin
1
2v
—
cos
1
2v
1
5
3
1
—
2
7
9. cos x1 si
n2x
cos y1 si
n2y
1
2
3
2
1
3
4
2
5
9
or
3
5
1
7
6
or
4
7
cos (xy) cos xcos ysin xsin y
3
5
4
7
2
3
3
4
3
1
5
2
6
10. tan x
5
4
tan ysec2y
1
221
3
tan (xy)
1
ta
n
t
x
an
x
ta
ta
n
n
y
y
219 Chapter 7
tan (ab)
t
t
a
a
n
n
a
a
1
ta
1
na
tan b
——
ta
1
na
tan b
tan (ab)
1
ta
n
t
a
an
a
ta
ta
n
n
b
b
tan (ab) tan(ab)
8. cos 75° cos(30° 45°)
cos 30° cos 45° sin 30° sin 45°
2
3
2
2
1
2
2
2
6
4
2
5
4
3
1
5
4
3
5 4
4
3
4 5
4
3
80
4
5
1
9
3
or
80
59
41
3
Reduction Identities
Page 447
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 awere 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 aare 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
7-3B
543
453
1 cot2vcsc2v
1 cot2v
7
2
2
1 cot2v
4
4
9
cot2v
4
4
5
cot v
3
2
5
Double-Angle and Half-Angle
Identities
Page 453 Check for Understanding
1. If you are only given the value of cos v, then
cos 2v2 cos2v1 is the best identity to use.
If you are only given the value of sin v, then
cos 2v1 2 sin2vis the best identity to use. If
you are given the values of both cos vand sin v,
then cos 2vcos2vsin2vis just as good as the
other two.
2. cos 21 2 sin2v
cos 2v12 sin2v
cos
2v
2
1
sin2v
1c
2
os 2v
sin2v
1c
2
o
s2v
sin v
Letting v
a
2
yields sin
a
2
,
or sin
a
2
1
2
c
os
.
3a. III or IV 3b. I or II 3c. I, II, III or IV
4. sin 2v2 sin v
sin 2
2
2 sin
2
sin 2 sin
2
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.
6. sin
8
sin
4
2
1 cos 2
2
2
7-4 8. sin2vcos2v1tan v
c
s
o
in
s
v
v
2
5
2cos2v1
2
5
—
5
21
Chapter 7 220
(Quadrant I)
1 cos
4
2
cos2v
2
2
1
5
or
2
21
21
cos v
5
21
(Quadrant I)
sin 2v2 sin vcos v
2
2
5
5
21
4
25
21
cos 2vcos2vsin2v
5
21
2
2
5
2
1
2
7
5
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
2
21
21
——
1
2
21
21
2
2
21
or
4
17
21
9. tan2v1 sec2vsin2vcos2v1
4
3
21sec2vsin2v
3
5
21
2
9
5
sec2vsin2v
1
2
6
5
5
3
sec v(Quadrant III) sin v
4
5
cos v
se
1
cv
(Quadrant III)
or
3
5
sin 2v2 sin vcos v
2
4
5
3
5
2
2
4
5
cos 2vcos2vsin2v
3
5
2
4
5
2
2
7
5
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
4
3
—
1
4
3
2
1
3
5
4
21
21
—
1
2
7
1
or
2
7
4
10. tan 2v
cot v
2
tan v
tan 2v
cot v
2
tan v
t
t
a
a
n
n
v
v
tan 2v
tan 2v
1
2
t
t
a
a
n
n
v
2v
tan 2vtan 2v
2 tan v
cot vtan vtan2v
8
3
—
7
9
2
2
2
7. tan 165° tan
33
2
0°
1
1
c
c
o
o
s
s
3
3
3
3
0
0
°
°
(Quadrant II)
(2 3
)
3
2
1
3
3
—
1
3
3
1
2
2
2
11. 1
1
2
sin 2A
sec A
se
cA
sin A
1
1
2
sin 2A
1
1
2
sin 2A
c
c
o
o
s
s
A
A
1
1
2
sin 2A1 sin Acos A
1
1
2
sin 2A1
1
2
2 sin Acos A
1
1
2
sin 2A1
1
2
sin 2A
12. sin
2
x
cos
2
x
sin
2
x
sin
2
x
2 sin
2
x
cos
2
x
——
2
co
1
sA
sin A
——
co
1
sA
co
1
s A
sin A
——
co
1
s A
16. tan
5
1
2
tan
5
6
2
221 Chapter 7
sin
2
x
sin
2
x
sin
2
x
13. cos 2v2 cos2v1
cos 2v1 2 cos2v
1
2
cos 2v
1
2
cos2v
PI02Rsin2qt
PI02R(1 cos2qt)
PI02R1
1
2
cos 2qt
1
2
PI02R
1
2
1
2
cos 2qt
P
1
2
I02R
1
2
I02Rcos 2qt
Pages 454–455 Exercises
14. cos 15° cos
3
2
0°
1c
2
os 30°
(Quadrant I)
1
2
3
—
2
sin 2
2
x
—
2
15. sin 75° sin
15
2
0°
1c
o
2
s150
°
(Quadrant I)
1
2
3
2
2 3
2
2
2
3
(Quadrant I)
1 cos
5
6
——
1 cos
5
6
1
2
3
1
2
3
2
2
3
—
2
2
3
(2 3
)(2 3
)
(2 3
)(2 3
)
(Quadrant I)
1 cos
3
4
——
2
2 3
17. sin
3
8
sin
3
4
2
(2 3
)2
43
1
2
2
—
2
18. cos
7
1
2
cos
7
6
2
2
2
2
(Quadrant II)
1 cos
7
6
π
——
2
2
2
2
—
2
2
2
(2 2
)(2 2
)
(2 2
)(2 2
)
2
2
2
2
2
1
(2 2
)2
42
22
2
1
2
2
—
1
2
2
1
2
3
2
2
2
3
19. tan 22.5° tan
4
2
5°
1
1
c
c
o
o
s
s
4
4
5
5
°
°
(Quadrant I)
20. tan
2
v
1
1
c
c
o
o
s
s
v
v
23. tan2v1 sec2v
(2)21sec2v
5sec2v
5
sec v(Quadrant II)
cos vsec v
or
5
5
sin2vcos2v1
sin2v
5
5
21
sin2v
2
2
0
5
sin v
2
5
5
(Quadrant II)
sin 2v2 sin vcos v
2
2
5
5
5
5
4
5
cos 2vcos2vsin2v
5
5
2
2
5
5
2
3
5
tan 2v
1
2
t
t
a
a
n
n
v
2v
1
2(
(
2
2
)
)2
4
3
or
4
3
24. cos v
se
1
cv
1
5
Chapter 7 222
1
1
4
1
1
4
3
4
—
5
4
3
5
or
5
15
sin 2v2 sin vcos v
2
3
5
4
5
2
2
4
5
cos 2vcos2vsin 2v
4
5
2
3
5
2
2
7
5
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
3
4
1
3
4
2
or
2
7
4
3
2
—
1
7
6
22. sin2vcos2v1
1
3
2cos2v1
cos2v
8
9
cos v
2
3
2
(Quadrant I)
sin 2v2 sin vcos v
2
1
3
2
3
2
4
9
2
cos 2vcos2vsin2v
2
3
2
2
1
3
2
7
9
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
4
2
1
4
2
2
1
4
3
3
4
sin2vcos2v1
sin2v
3
4
21
sin2v
1
7
6
sin v
4
7
(Quadrant II)
tan v
c
s
o
in
s
v
v
4
7
—
3
4
3
7
sin 2v2 sin vcos v
2
4
7
3
4
3
8
7
cos 2vcos2vsin2v
3
4
2
4
7
2
1
2
6
or
1
8
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
3
7
1
3
7
2
or 37
2
3
7
—
2
9
21. sin2vcos2v1
sin2v
4
5
21
sin2v
2
9
5
sin v
3
5
(Quadrant I)
tan v
c
s
o
in
s
v
v
3
4
3
5
—
4
5
tan v
c
s
o
in
s
v
v
1
3
—
2
3
2
or
4
2
1
22
or
4
7
2
2
2
—
1
1
4
6
25. 1 cot2vcsc2vtan v
co
1
tv
1
3
2
2csc2v
1
4
3
csc2v
2
3
2
13
csc v(Quadrant III)
sin v
cs
1
cv
sin2vcos2v1
2
13
13
2cos2v1
1
2
13
1
3
2
tan 2v
1
2
ta
ta
n
n
v
2v
223 Chapter 7
(Quadrant III)
sin 2v2 sin vcos v
2
2
13
13
3
13
13
1
1
2
3
cos 2vcos2vsin2v
3
13
13
2
2
13
13
2
1
5
3
tan 2v
1
2
ta
ta
n
n
v
2v
or
1
5
2
4
3
5
9
2
2
3
1
2
3
2
26. sin v
cs
1
cv
2
5
1
—
5
2
(Quadrant IV)
tan v
c
s
o
in
s
v
v
2
5
—
5
21
2
2
21
21
1
2
21
21
2
or
4
17
21
4
21
21
——
1
2
7
1
27. sin2acos2a1tan a
c
s
o
in
s
a
a
sin2a
3
2
21
sin2a
7
9
or
2
14
7
2
3
7
3
2
or
2
5
14
28. csc 2v
1
2
sec vcsc v
sin
1
2v
1
2
sec vcsc v
2sin
1
vcos v
1
2
sec vcsc v
1
2
si
1
nv
co
1
sv
1
2
sec vcsc v
1
2
csc vsec v
1
2
sec vcsc v
1
2
sec vcsc v
1
2
sec vcsc v
29. cos Asin A
cos
c
A
os
2
s
A
in A
cos Asin A
c
c
o
o
s
s
2A
A
s
s
i
i
n
n
2
A
A
cos Asin A
cos Asin Acos Asin A
30. (sin vcos v)21sin 2v
sin2v2 sin vcos vcos2v1 sin 2v
2 sin vcos v1 1sin 2v
2 sin vcos vsin 2v
sin 2vsin 2v
31. cos x1
2
c
(
o
c
s
os
2x
x
1
1)
cos x1
2c
2
o
(
s
c
2
os
x
x
1
1
)
1
cos x1
2
2(
c
c
o
o
s
s
2
x
x
1
2
)
cos x1
2
2
(
(
c
c
o
o
s
s
2
x
x
1
1
)
)
cos x1
cos x1 cos x1
32. sec 2v
c
c
o
o
s
s
2
2
v
v
s
s
i
i
n
n
2
2
v
v
sec 2v
cos
1
2v
sec 2vsec 2v
33. tan
A
2
1
sin
co
A
sA
2(cos x1)(cos x1)
2(cos x1)
(cos Asin A)(cos Asin A)
cos Asin A
14
5
2
or
2
13
13
2
13
cos2v
1
1
6
17
9
cos v
3
13
13
sin2vcos2v1
2
5
2cos2v1
cos2v
2
2
1
5
cos v
5
21
or
2
21
21
sin 2v2 sin vcos v
2
2
5
5
21
4
25
21
cos 2vcos2vsin2v
5
21
2
2
5
2
1
2
7
5
2
21
sin a
3
7
(Quadrant II)
tan 2a
1
2
ta
ta
n
n
a
2a
2
2
14
1
2
14
2
39a. tan45°
1
2
tan 45° tan
L
2
1 tan 45° tan
L
2
Chapter 7 224
tan
A
2
sin 2
A
2
1 cos 2
A
2
tan
A
2
2 sin
A
2
cos
A
2
1 2 cos2
A
2
1
tan
A
2
2 sin
A
2
cos
A
2
2 cos2
A
2
tan
A
2
tan
A
2
tan
A
2
34. sin 3x3 sin x4 sin3x
sin(2xx) 3 sin x4 sin3x
sin 2xcos xcos 2xsin x3 sin x4 sin3x
2 sin xcos2x(1 2 sin2x) sin x3 sin x4 sin3x
2sinx(1 sin2x)(1 2sin
2x)sinx3 sin x4 sin3x
2sinx2sin
3xsin x2sin
3x3 sin x4 sin3x
3 sin x4 sin3x3 sin x4 sin3x
35. cos 3x4 cos3x3 cos x
cos(2xx) 4 cos3x3 cos x
(2 cos2x1)cos x2 sin2xcos x4 cos3x3 cos x
(2 cos2x1)cos x2(1 cos2x)cos x4 cos3x3cos x
2 cos3xcos x2 cos x2 cos3x4 cos3x3 cos x
4 cos3x3 cos x4 cos3x3 cos x
36.
s
s
i
i
n
n
2
2
2
v
v
2
v
g
2
sin22v
2
v
g
2
sin2v
sin
A
2
cos
A
2
R2v2cos vsin(v45°)
gcos245°
(2 sin
sin
v
2
co
v
sv)2
4sin
s
2
in
v
2
c
v
os2v
4 cos2v
37. ∠PBD is an inscribed angle that subtends the
same arc as the central angle ∠POD, so m∠PBD
1
2
v. By right triangle trigonometry, tan
1
2
v
B
P
A
A
1
PA
OA
1
sin
co
v
sv
.
38. R2v2cos vsin(va)
gcos2a
R2v2cos v(sin vcos 45° cos vsin 45°)
gcos245°
R
2v2cos v(sin v)
2
2
(cos v)
2
2
g
2
2
2
R
2
2
2
v2cos v(sin vcos v)
g
1
2
R
R
v2
g
2
(2 cos vsin v(2 cos2v1)1)
R
v2
g
2
(sin 2vcos 2v1)
2
v2(2 cos vsin v2 cos2v)
g
1 tan
L
2
1 1 tan
L
2
1
1
1
c
c
o
o
s
s
L
L
1
1
1
c
c
o
o
s
s
L
L
39b.
1
1
1
c
c
o
o
s
s
6
6
0
0
˚
˚
1
1
1
c
c
o
o
s
s
6
6
0
0
˚
˚
1
1
1
c
c
o
o
s
s
L
L
1
1
1
c
c
o
o
s
s
L
L
1
1
1
2
1
1
2
1
1
1
2
1
1
2
1
1
3
1
1
3
3
3
3
3
3
3
12
6
6
3
or 2 3
40. tan (a30°)
2
7
1
tan (a30°) 3
3
tan a
3
3
3 3
tan a
tan a3
tan a3
3
3
(1 3
) tan a3
3
3
tan a3
3
3
1 3
tan atan 30°
1 tan atan 30°
tan a9 3
333
tan a
6
3
5
3
41. cos
1
2
cos
3
4
cos
3
cos
4
sin
3
sin
4
1
2
2
2
2
3
2
2
2
4
6
sec
1
2
1
cos
1
2
1
2
4
6
4
2
4
4
6
or 6
2
42. Sample answer:
sin(
)2cos(
)2sin cos
0 (1)
1
1
43. srv
1
1
7
0
1
1
7
0
18
0°
17 10 v97.4°
1
1
7
0
v
44. Let xthe distance from Ato the point beneath
the mountain peak.
tan 21°10
570
h
x
h(570 x) tan 21°10
tan 36°40
h
x
hxtan 36°40
(570 x) tan 21°10xtan 36°40
570 tan 21°10xtan 36°40xtan 21°10
570 tan 21°10x(tan 36°40tan 21°10)
x
617.7646751 x
tan 36°40
h
x
tan 36°40
61
h
7.8
h460 ft
45. (x(3))(x0.5)(x6)(x2) 0
(x3)(x0.5)(x6)(x2) 0
(x22.5x1.5)(x28x12) 0
x45.5x39.5x242x18 0
2x411x319x284x36 0
46. y2x5
x2y5
x52y
x
2
5
y
47. x2y11 3x5y11
x11 2y3(11 2y) 5y11
33 6y5y11
11y22
y2
x2y11
x2(2) 11
x7(7, 2)
48. ab 3
b
a
3
(ab)264
a22ab b264
a22a
a
3
a
3
264
a26
a
3
264
a2
a
3
270
a2b270
The correct answer is 70.
570 tan 21°10
tan 36°40tan 21°10
225 Chapter 7
y
x
y
2
x
5
y
x
5
2
O
Solving Trigonometric Equations
Page 458 Graphing Calculator Exploration
1.
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.
5.
[0, 2] sc1
4
by [3, 3] sc11
5a. The x-intercepts of the graph are the solutions of
the equation sin x2 cos x. They are the same.
5b. ytan 0.5xcos xor ycos xtan 0.5x
Page 459 Check for Understanding
1. Atrigonometric 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 xis any
integer
7-5
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 x1 06. 2 cos x3
0
2 sin x12 cos x3
sin x
1
2
cos x
2
3
x30° x30°
7. sin xcot x
2
3
sin x
c
s
o
in
s
x
x
2
3
cos x
2
3
x30° or x330°
8. cos 2xsin2x2
2 cos2x1 (1 cos2x) 2
2 cos2x1 cos2x1
3 cos2x0
cos2x0
cos x0
x90° or x270°
9. 3 tan2x1 0
3 tan2x1
tan2x
1
3
tan x
3
3
x
6
or x
5
6
or x
7
6
or x
11
6
10. 2 sin2x5 sin x3
2 sin2x5 sin x3 0
(2 sin x1)(sin x3) 0
2 sin x1 0orsin x3 0
sin x
1
2
sin x3
x
7
6
or x
11
6
no solutions
11. sin22xcos2x0
1 cos22xcos2x0
1 (2 cos2x1)2cos2x0
1 (4 cos4x4 cos2x1) cos2x0
4 cos4x5 cos2x0
cos2x(4 cos2x5) 0
cos2x0or 4 cos2x5 0
cos x0cos2x
5
4
x
2
kcos x
2
5
no solutions
12. tan2x2 tan x1 0
(tan x1)(tan x1) 0
tan x1 0
tan x1
x
3
4
k
13. cos2x3 cos x2
cos2x3 cos x2 0
(cos x1)(cos x2) 0
cos x1 0orcos x2 0
cos x1cos x2
x(2k1)no solutions
14. sin 2xcos x0
2 sin xcos xcos x0
cos x(2 sin x1) 0
cos x0or2 sin x1 0
x
2
ksin x
1
2
x
6
2k
or x
5
6
2k
15. 2 cos v1 0
2 cos v1
cos v
1
2
cos v
1
2
at
2
3
and
4
3
2
3
v
4
3
16. WFd cos v
1500 100 20 cos v
0.75 cos v
v41.41°
Pages 459–461 Exercises
17. 2
sin x1 018. 2 cos x1 0
2
sin x12 cos x1
sin x
1
2
cos x
1
2
sin x
2
2
x120°
x45°
19. sin 2x1 0
2 sin xcos x1 0
sin2xcos2x
1
4
sin2x(1 sin2x)
1
4
sin2xsin4x
1
4
0
sin4xsin2x
1
4
0
sin2x
1
2
sin2x
1
2
0
sin2x
1
2
0
sin2x
1
2
sin x
1
2
or
2
2
x45°
Chapter 7 226
20. tan 2x3
0
tan 2x3
1
2
t
t
a
a
n
n
x
2x
3
2 tan x3
(1 tan2x)
2 tan x3
3
tan2x
3
tan2x2 tan x3
0
(3
tan x1)(tan x3
) 0
3
tan x1 0tan x3
0
tan x
1
3
tan x3
tan x
3
3
x60°
x30°
21. cos2xcos x
cos2xcos x0
cos x(cos x1) 0
cos x0or cos x1 0
x90° cos x1
x0°
22. sin x1 cos2x
sin x1 1 sin2x
sin2xsin x2 0
(sin x1)(sin x2) 0
sin x1 0orsin x2 0
sin x1sin x2
x90° no solution
23. 2
cos x1 0
2
cos x1
cos x
2
2
x135° or x225°
24. cos xtan x
1
2
cos x
c
s
o
in
s
x
x
1
2
sin x
1
2
x30° or x150°
25. sin xtan xsin x0
sin x(tan x1) 0
sin x0ortan x1 0
x0° or x180° tan x1
x45° or x225°
26. 2 cos2x3 cos x2 0
(2 cos x1)(cos x2) 0
2 cos x1 0orcos x2 0
2 cos x1cos x2
cos x
1
2
no solution
x60° or x300°
27. sin 2xsin x
2 sin xcos x–sin x
2 sin xcos xsin x0
sin x(2 cos x1) 0
sin x0or2 cos x1 0
x0° or x180° 2 cos x1
cos x
1
2
x120°
or x240°
28. cos (x45°) cos (x45°) 2
cos xcos 45° sin xsin 45°
cos xcos 45° sin xsin 45° 2
cos x
2
2
sin x
2
2
cos x
2
2
sin x
2
2
2
2
cos x2
cos x1
x0°
29. 2 sin vcos v3
sin v0
sin v(2 cos v3
) 0
sin v0or2 cos v+ 3
0
v0° or v180° 2 cos v3
cos v
2
3
v150°
or v210°
30. (2 sin x1)(2 cos2x1) 0
2 sin x1 0or 2 cos2x1 0
2 sin x12 cos2x1
sin x
1
2
cos2x
1
2
x
6
cos x
2
2
or x
5
6
x
4
or x
3
4
or x
5
4
or x
7
4
31. 4 sin2x1 4 sin x
4 sin2x4 sin x1 0
(2 sin x1)(2 sin x1) 0
2 sin x1 0
2 sin x1
sin x
1
2
x
7
6
or x
11
6
32. 2
tan x2 sin x
2
c
s
o
in
s
x
x
2 sin x
2
2 cos x
2
2
cos x
x
4
or x
7
4
2
tan x2 sin xwould also be true if both tan x
and sin xequal 0. Since tan x
c
s
o
in
s
x
x
, tan xequals
0 when sin x0. Therefore xcan also equal 0 and
.
0,
4
, ,
7
4
33. sin xcos 2x1
sin x1 2 sin2x1
2 sin2xsin x0
sin x(2 sin x1) 0
sin x0or2 sin x1 0
x0 or x2 sin x1
sin x
1
2
x
7
6
or
x
11
6
227 Chapter 7
34. cot2xcsc x1
csc2x1 csc x1
csc2xcsc x2 0
(csc x2)(csc x1) 0
csc x2 0orcsc x1 0
csc x2or csc x1
sin x
1
2
sin x1
x
6
or x
5
6
x
3
2
35. sin xcos x0
sin xcos x
sin2xcos2x
sin2xcos2x0
sin2x1 sin2x0
2 sin2x1 0
sin2x
1
2
sin x
1
2
or
2
2
sin xand cos xmust be opposites, so x
3
4
or x
7
4
.
36. 1 3 sin vcos 2v
1 3 sin v1 2 sin2v
2 sin2v3 sin v2 0
(2 sin v1)(sin v2) 0
2 sin v1 0orsin v2 0
2 sin v1sin v2
sin v
1
2
no solution
v
7
6
or v
11
6
37. sin x
1
2
x
7
6
2 kor x
11
6
2k
38. cos xtan x2 cos2x1
cos x
c
s
o
in
s
x
x
2 cos2x1
sin x2(1 sin2x) 1
2 sin2xsin x1 0
(2 sin x1)(sin x1) 0
2 sin x1 0or sin x1 0
2 sin x1sin x1
sin x
1
2
x
3
2
2k
x
6
2kor x
5
6
2k
39. 3 tan2x3
tan x
3 tan2x3
tan x0
tan x(3 tan x3
) 0
tan x0or3 tan x3
0
xk3 tan x3
tan x
3
3
x
6
k
40. 2(1 sin2x) 3 sin x
2 2 sin2 x3 sin x
2 sin2x3 sin x2 0
(2 sin x1)(sin x2) 0
2 sin x1 0or sin x2 0
2 sin x1sin x2
sin x
1
2
no solution
x
6
2kor
x
5
6
2k
41.
cos x
1
sin x
cos xsin x
(cos xsin x)(cos xsin x) 1
cos2xsin2x1
cos2x(1 cos2x) 1
2 cos2x1 1
2 cos2x2
cos2x1
cos x1
xk
42. 2 tan2x3 sec x0
2(sec2x1) 3 sec x0
(2 sec x1)(sec x2) 0
2 sec2x3 sec x2 0
2 sec x1 0orsec x2 0
2 sec x1sec x2
sec x
1
2
cos x
1
2
cos x2x
3
2kor
no solution x
5
3
2k
43. sin xcos x
1
2
sin2xcos2x
1
4
sin2x(1 sin2x)
1
4
sin2xsin4x)
1
4
sin4xsin2x
1
4
0
sin2x
1
2
sin2x
1
2
0
sin2x
1
2
0
sin2x
1
2
sin x
2
2
x
4
k
44. cos2xsin2x
2
3
cos2x(1 cos2x)
2
3
2 cos2x1
2
3
2 cos2x
2
2
3
cos2x
2
4
3
cos x
x
1
2
kor x
1
1
1
2
k
45. sin4x1 0
(sin2x1)(sin2x1) 0
sin2x1 0orsin
2x1 0
sin2x1sin
2x1
sin x1no solutions
x
2
k
46. sec2x2 sec x0
sec x(sec x2) 0
sec x0orsec x2 0
cos x
1
0
sec x2
no solution cos x
1
2
x
2
3
2kor
x
4
3
2k
2
3
2
Chapter 7 228
47. sin xcos x1
sin2x2 sin xcos xcos2x1
sin2x2 sin xcos x1 sin2x1
2 sin xcos x0
sin xcos x0
sin2xcos2x0
sin2x(1 sin2x) 0
sin2x0or1 sin2x0
sin x0sin
2x1
x2ksin x1
x
2
2k
48. 2 sin xcsc x3
2 sin2x1 3 sin x
2 sin2x3 sin x1 0
(2 sin x1)(sin x1) 0
2 sin x1 0or sin x1 0
2 sin x1sin x1
sin x
1
2
x
2
2k
x
6
2kor
x
5
6
2k
49. cos v
2
3
cos v
2
3
at
5
6
and
7
6
5
6
v
7
6
50. cos v
1
2
0
cos v
1
2
cos v
1
2
at
3
and
5
3
0 v
3
or
5
3
v2
51. 2
sin v1 0
2
sin v1
sin v
2
2
sin v
2
2
at
4
and
3
4
0 v
4
or
3
4
v2
52. 0.4636, 3.6052 53. 0, 1.8955
54. 0.3218, 3.4633
55. sin v
D
sin v
5.5
0
.00
1
3
07
sin v0.0001833333333
v0.01°
56. sin 2xsin x
2 sin xcos xsin x
2 sin xcos x – sin x0
sin x(2 cos x1) 0
The product on the left side of the inequality is
equal to 0 when xis 0,
3
, , or
5
3
. For the product
to be negative, one factor must be positive and the
other negative. This occurs if
3
xor
5
3
x
2.
57. R
v
g
2
sin 2v
20
1
9
5
.8
2
sin 2v
0.8711111111 sin 2v
2v60.5880156 or 2v119.4119844
v30.29° v59.71°
58a. n1sin i n2sin r
1.00 sin 35° 2.42 sin r
sin r
1.00
2
s
.4
in
2
35°
sin r0.2370150563
r13.71°
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. D0.5 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.
t
3.32
2
6
5
4
0
7
0
7773
t0.0013 s
60. asin(bx c) dd
a
2
asin(bx c)
a
2
sin(bx c)
1
2
The period of the function sin(bx c) is
36
b
0°
, so
the given interval consists of bperiods.
360°
36
b
0°
229 Chapter 7
The equation sin (bx c)
1
2
has two solutions
per period, so the total number of solutions is 2b.
61.
3
4
x
y
17
22
3
4
cos vsin v
sin vcos v
17
22
3 cos v4 sin v
3 sin v4 cos v
3 cos v4 sin v17
3 sin v4 cos v22
↓
9 cos v12 sin v317
16 cos v12 sin v82
25 cos v82
317
cos v
8
2
25
3
17
v18.68020037
360 v341.32°
y
360˚
270˚
90˚180˚
O
y
cos
2
3
2
3
2
3
62. cot 67.5° cot
13
2
5°
cot v
ta
1
nv
tan
13
2
5°
1
1
–c
c
o
o
s
s
1
1
3
3
5
5
°
°
(Quadrant 1)
67. 2
1032
24 2
12 1
0
x22x1 0
(x1)(x1) 0
x1 0x1 0
x1x1
(x2)(x1(x1)
68.
[5, 5] sc11 by [2, 8] sc11
max: (1, 7), min: (1, 3)
69. 3x4 16 6 2y
x4y3(4, 3)
70. xyz1xyz1
2xy3z5xyz11
3x4z62x12
x6
3x4z6xyz11
3(6) 4z66 y(3) 11
4z12 y2
z3
(6, 2, 3)
71.
72. A
1
2
bh
A
1
2
(6)(1)
A3
The correct choice is C.
Chapter 7 230
1
2
2
1
2
2
2
2
2
2 –
2
2
(2 2
)(2 2
)
(2 2
)(2 2
)
(2
4
2
2
)2
22
2
cot 67.5 1
2
2
2
2
2 2
2
(2 2
)
(2 2
)(2 2
)
2
1
63.
t
s
a
e
n
cx
x
5
2
22
2
4 2
5
2
sin x
5
2
Sample answer: sin x
5
2
64. A
2
3
, 2
65.
45
h
m
ou
i
r
les
52
m
8
i
0
le
ft
12 in
ft
ches
3
1
60
h
0
ou
se
r
c
792 in/sec
vr
v
t
792 7
1
v
79
7
2
79
7
2
radians 218 rps
66. undefined
c
s
o
in
s
x
x
—
co
1
s x
xg(x)
74
52
30
12
14
g
(
x
)
x
g
(
x
) |
x
3|
O
Page 462 History of Mathematics
1. x252522(5)(5) cos 10°
x0.87
x252522(5)(5) cos 20°
x1.74
x252522(5)(5) cos 30°
x2.59
x252522(5)(5) cos 40°
x3.42
x252522(5)(5) cos 50°
x4.23
x252522(5)(5) cos 60°
x5
x252522(5)(5) cos 70°
x5.74
x252522(5)(5) cos 80°
x6.43
x252522(5)(5) cos 90°
x7.07
Normal Form of a Linear Equation
Page 467 Check for Understanding
1. Normal means perpendicular
2. Compute cos 30° and sin 30°. Use these as the
coefficients of xand y, respectively, in the normal
form. The normal form is
2
3
x
1
2
y10 0.
3. The statement is true. The given line is tangent to
the circle centered at the origin with radius p.
7-6
4.
See students’ work for sample problems.
5. xcos fysin fp0
xcos 30° ysin 30° 10 0
2
3
x
1
2
y10 0
3
xy20 0
6. xcos fysin fp0
xcos 150° ysin 150° 3
0
2
3
x
1
2
y3
0
3
xy23
0
7. xcos fysin fp0
xcos
7
4
ysin
7
4
52
0
2
2
x
2
2
y52
0
2
x2
y102
0
xy10 0
8. 4x3y10 A2
B2
423
2
or 5
4x3y10 0
4
5
x
3
5
y
10
5
0
4
5
x
3
5
y2 0
sin f
3
5
, cos f
4
5
, p2; Quadrant III
tan for
3
4
f
9. y3x2
3xy2 0
A2
B2
321
2
or 10
3
10
x
1
10
y
2
10
0
3
10
10
x
1
1
0
0
y
5
10
0
sin f
1
1
0
0
, cos f
3
10
10
, p
5
10
; Quadrant I
tan for
1
3
1
1
0
0
—
3
10
10
3
5
4
5
231 Chapter 7
Angle Length of
Measure Chord (cm)
10° 0.87
20° 1.74
30° 2.59
40° 3.42
50° 4.23
60° 5.00
70° 5.74
80° 6.43
90° 7.07
Slope-Intercept Form: ymx b, displays
slope and y-intercept
Point-Slope Form: yy1m(xx1),
displays slope and a point on the line
Standard Form: Ax by C0, displays
no information
Normal Form: xcos fysin fp0,
displays length of the normal and the angle
the normal makes with the x-axis
f18°
10. 2
x2
y6
2
x2
y6 0
A2
B2
2
2
(2
)2
or 2
2
2
x
2
2
y
6
2
0
2
2
x
2
2
y3 0
sin f
2
2
, cos f
2
2
, p3; Quadrant IV
tan for 1
f315°
11a. 3x4y8
y
3
4
x2
11b. 3x4y8
3x4y8 0
A2
B2
32(
4)2
or 5
3
5
x
4
5
y
8
5
0
p
8
5
or 1.6 miles
Pages 467–469 Exercises
12. xcos fysin fp0
xcos 60° ysin 60° 15 0
1
2
x
2
3
y15 0
x3
y30 0
13. xcos fysin vp0
xcos
4
ysin
4
12 0
2
2
x
2
2
y12 0
2
x2
y24 0
14. xcos fysin fp0
xcos 135° ysin 135° 32
0
2
2
x
2
2
y32
0
2
x2
y62
0
xy6 0
15. xcos fysin fp0
xcos
5
6
ysin
5
6
23
0
2
3
x
1
2
y23
0
3
xy43
0
16. xcos fysin fp0
xcos
2
ysin
2
2 0
0x1y2 0
y2 0
2
2
—
2
2
17. xcos fysin fp0
xcos 210° ysin 210° 5 0
2
3
x
1
2
y 5 0
3
xy10 0
18. xcos fysin fp0
xcos
4
3
ysin
4
3
5 0
1
2
x
2
3
y5 0
x3
y10 0
19. xcos fysin fp0
xcos 300° ysin 300°
3
2
0
1
2
x
2
3
y
3
2
0
x3
y3 0
20. xcos fysin fp0
xcos
11
6
ysin
11
6
43
0
2
3
x
1
2
y43
0
3
xy83
0
21. A2
B2
521
22
or 13
5
13
x
1
1
2
3
y
6
1
5
3
0
1
5
3
x
1
1
2
3
y5 0
sinf
1
1
2
3
, cos f
1
5
3
, p5; Quadrant III
tan for
1
5
2
–
1
1
2
3
—
–
1
5
3
Chapter 7 232
y
x
3
x
4
y
8
O
f247°
22. xy1
xy1 0
A2
B2
121
2
or 2
1
2
x
1
2
y
1
2
0
2
2
x
2
2
y
2
2
0
sin f
2
2
, cos f
2
2
, p
2
2
; Quadrant I
tan for 1
f45°
23. 3x4y15
3x4y15 0
A2
B2
32(
4)2
or 5
3
5
x
4
5
y
1
5
5
0
3
5
x
4
5
y3 0
sin f
4
5
, cos f
3
5
, p3; Quadrant IV
2
2
—
2
2
tan for
4
3
f307°
4
5
—
3
5
24. y2x – 4
2xy4 0
A2
B2
(2)2
12
or 5
2
5
x
1
5
y
4
5
0
2
5
5
x
5
5
y
4
5
5
0
sin f
5
5
, cos f
2
5
5
, p
4
5
5
; Quadrant IV
tan for
1
2
–
5
5
—
2
5
5
28.
3
x
y4
3
x
y4 0
x3y12 0
A2
B2
12(
3)2
or 10
1
10
x
3
10
y
12
10
0
1
1
0
0
x
3
10
10
y
6
5
10
0
sin f
3
10
10
, cos f
1
1
0
0
, p
6
5
10
; Quadrant II
tan for 3
3
10
10
—
–
1
1
0
0
233 Chapter 7
f333°
25. x3
x3 0
A2
B2
120
2
or 1
1
1
x
3
1
0
x3 0
sin f0, sin f1, p3
tan f
0
1
or 0
f0°
26. 3
xy2
3
xy2 0
A2
B2
(3
)2(
1)2
or 2
2
3
x
1
2
y
2
2
0
2
3
x
1
2
y1 0
sin f
1
2
, cos f
2
3
, p1; Quadrant III
tan for
3
3
–
1
2
—
–
2
3
f210°
27. y2
1
4
(x20)
y2
1
4
x5
x4y28 0
A2
B2
(1)2
42
or 17
1
17
x
4
17
y
2
1
8
7
0
1
1
7
7
x
4
17
17
y
28
17
17
0
sin f
4
17
17
, cos f
1
1
7
7
, p
28
17
17
; Quadrant II
tan for 4
4
17
17
—
–
1
1
7
7
f104°
f108°
29.
2
x
0
2
y
4
1
2
x
0
2
y
4
1 0
6x5y124 0
A2
B2
625
2
or 61
6
61
x
5
61
y
12
6
4
1
0
6
61
61
x
5
61
61
y
120
6
1
61
0
sin f
5
61
61
, cos f
6
61
61
, p
120
6
1
61
; Quadrant I
tan for
5
6
5
61
61
—
6
61
61
f40°
30.
A2
B2
628
2
or 10; p10
cos f
1
6
0
or
3
5
, sin f
1
8
0
or
4
5
xcos fysin fp0
3
5
x
4
5
y10 0
3x4y50 0
31.
A2
B2
(4)2
42
or 42
; p42
cos f
4
4
2
or
2
2
, sin f
4
4
2
or
2
2
xcos fysin fp0
2
2
x
2
2
y42
0
xy8 0
32. 22
xy18
22
xy18 0
A2
B2
2(2)2
(1)2
9
3
2
3
2
x
1
3
y
1
3
8
0
p
1
3
8
6 units
33a.
y
x
O
45˚
1.25 ft
33b. p1.25, f45°
xcos (45°) ysin (45°) 1.25 0
2
2
x
2
2
y1.25 0
2
x2
y2.5 0
34a.
fand the supplement of vare complementary
angles of a right triangle, so f180° v90°.
Simplifying this equation gives vf90°.
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
reciprocal of the slope of . That is,
ta
1
nv
cot v.
34d. The slope of is the negative reciprocal of the
slope of the normal, or
ta
1
nf
cot f.
35a. A2
B2
521
22
or 13
1
5
3
x
1
1
2
3
y
3
1
9
3
0
1
5
3
x
1
1
2
3
y3 0
35b. sin f
1
1
2
3
, cos f
1
5
3
; Quadrant I
tan for
1
5
2
1
1
2
3
—
1
5
3
36a.
The angles of the quadrilateral are 180° a,
90°, f2f1, and 90°. Then 180° a90°
f2f190° 360°, which simplifies to
f2f1a. If the lines intersect so that ais an
interior angle of the quadrilateral, the equation
works out to be f2180° f1a.
36b. tan f2tan(f1a)
1
ta
n
t
f
a1
n
f1
ta
ta
n
n
a
a
If the lines intersect so that ais an interior
angle of the quadrilateral, the equation works
out to be tan f2
1
ta
n
t
f
a1
n
f1
ta
ta
n
n
a
a
.
37. 5xy15
5xy15 0
A2
B2
52(
1)2
or 26
5
26
x
1
26
y
1
2
5
6
0
5
26
26
x
2
2
6
6
y
15
26
26
0, p
15
26
26
3x4y36
3x4y – 36 0
A2B
2
324
2
or 5
3
5
x
4
5
y
3
5
6
0, p
3
5
6
5x2y20
5x2y20 0
52(
2)2
25
4
or 29
5
29
x
2
29
y
2
2
0
9
0
5
29
29
x
2
29
29
y
20
29
29
0, p
20
29
29
15
26
26
3
5
6
20
29
29
13.85564879
13.85564879 500 6927.824395; $6927.82
38. 2 cos2x7 cos x4 0
(2 cos x1)(cos x4) 0
2 cos x1 0orcos x4 0
2 cos x1cos x4
cos x
1
2
no solution
x
3
or x
5
3
39. sin x
1 co
s2
xsin y
1 co
s2y
Chapter 7 234
y
x
O
f67°
f90° 67° 90° or 157°
xcos 157° ysin 157° 3 0
1
1
2
3
x
1
5
3
y3 0
35c. See students’ work.
35d. The line with normal form xcos fysin f
p0 makes an angle of fwith 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 fdwith 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.
y
x
O
x
y
3 0
12
135
13
y
x
O
1
2
1
1
6
2
1
2
3
2
3
3
6
6
or
6
35
5
9
or
3
5
sin(xy) sin xcos ycos xsin y
6
35
2
3
1
6
3
5
2
35
18
5
O
y
x
(2, 6)
(2, 3)
(5, 3)
x
y
8
y
3
x
2
40. A1,
2
4
2
or 90°
41. r
d
2
r
13
2
.4
or 6.7
x26.726.722(6.7)(6.7) cos 26°20
x29.316604344
x3.05 cm
42.
x
x
5
25
1
7
x2
x
1
5
x
x
5
x2
17
25
x
1
5
(x5)(x5)
x
x
5
(x5)(x5)
(x
5)
1
(x
7
5)
(x5)(x5)
x
1
5
x(x5) 17 x5
x25x17 x5
x24x12 0
(x6)(x2) 0
x6 0orx2 0
x6x2
43. original box: Vwh
4 6 2
48
new box: Vwh
1.5(48) (4 x)(6 x)(2 x)
72 x312x244x48
0 x312x244x24
V(0.5) is closer to zero, so x0.5.
4 x 4 0.5 or 4.5
6 x6 0.5 or 6.5
2 x2 0.5 or 2.5
4.5 in. by 6.5 in. by 2.5 in.
44.
f(x, y) 3xy4
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
45.
1
5
2
1
3
4
2
1
3
4
1
12
43
235 Chapter 7
y
90˚
45˚
1
1
O
y
sin 4
xV(x)
0.4 4.416
0.5 1.125
0
15
x
y
2
3
1
4
1
5
1
5
0
15
2
1
3
4
x
y
2
3
1
4
2
1
3
4
1
5
30
15
x
y
(6, 3)
46. The value of 2abcannot be determined from
the given information. The correct choice is E.
Distance From a Point to a Line
Page 474 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 C0 is the standard form of
the equation of the line.
3. In the figure, Pand Qare 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.
4. The formula is valid in either case. Examples will
vary. For a vertical line, xa, the formula
subtracts afrom the x-coordinate of the point. For
a horizontal line, yb, the formula subtracts b
from the y-coordinate of the point.
5. 2x3y2 →2x3y2 0
dAx1By1C
A2
B2
7-7
6
3
x
y
d2(1) (3)(2) 2
22(
3)2
d
2
13
or
2
13
13
6. 6xy3 →6xy3 0
dAx1By1C
A2
B2
P
Q
d6(2) (1)(3) 3
62(
1)2
d
1
3
2
7
or
12
37
37
7. 3x5y1When x2, y1. Use (2, 1).
3x5y3 →3x5y3 0
dAx1By1C
A2
B2
14. y4
2
3
x→2x3y12 0
dAx1By1C
A2
B2
Chapter 7 236
d3(2) (5)(1) 3
32(
5)2
d
4
34
or
2
17
34
2
17
34
8. y
1
3
x3Use (0, 3).
y
1
3
x7 →x3y21 0
dAx1By1C
A2
B2
d
d
30
10
or 310
310
9. d1d22x13y14
22(
3)2
6x18y15
628
2
1(0) 3(3) 21
123
2
6x1
1
8
0
y15
2x1
3
1
y
3
14
613
x813
y513
20x30y40
(20 613
)x
(30 813
)y40 513
0;
6x1
1
8
0
y15
2x1
3
1
y
3
14
613
x813
y513
20x30y40
(20 613
)x(813
30)y 40 513
0
10. (2000, 0)
dAx1By1C
A2
B2
d5(2000) (3)(0) 0
52(
3)2
d
1
0,0
3
0
4
0
or about 1715 ft
Pages 475–476 Exercises
11. dAx1By1C
A2
B2
d3(2) (4)(0) 15
32(
4)2
d
21
5
2
5
1
12. dAx1By1C
A2
B2
d
dor
5
17
34
10
34
5(3) (3)(5) 10
52(
3)2
5
17
34
13. 2xy3 → 2xy3 0
dAx1By1C
A2
B2
d2(0) (1)(0) 3
(2)2
(1
)2
d
3
5
or
3
5
5
d2(2) 3(3) (12)
223
2
d
2
1
5
3
or
25
1
3
13
25
13
13
15. y2x5 → 2xy5 0
dAx1By1C
A2
B2
d2(3) 3(1)(1) (5)
22(
1)2
d
0
5
or 0
16. y
4
3
x6 →4x3y18 0
dAx1By1C
A2
B2
d
d
5
16
or
1
5
6
1
5
6
17. dAx1By1C
A2
B2
4(1) 3(2) (18)
423
2
d
d
1
10
or
1
1
0
0
1
1
0
0
18. 6x8y3When x0, y
3
8
. Use 0,
3
8
.
6x8y5 →6x8y5 0
dAx1By1C
A2
B2
3(0) (1)(0) 1
32(
1)2
d
d
8
10
or
4
5
4
5
19. 4x5y12 When x3, y0. Use (3, 0).
4x5y6 →4x5y6 0
dAx1By1C
A2
B2
6(0) (8)
3
8
1
62(
8)2
d
d
6
41
or
6
41
41
20. y2x1Use (0, 1).
2xy2 →2xy2 0
dAx1By1C
A2
B2
4(3) (5)(0) (6)
42(
5)2
d2(0) (1)(1) (2)
22(
1)2
dor
3
5
5
35
5
3
5
21. y3x6Use (0, 6).
3xy4 →3xy4 0
dAx1By1C
A2
B2
27. y
2
3
x1 →2x3y3 0
y3x2 →3xy2 0
237 Chapter 7
d
d
2
10
or
5
10
22. y
8
5
x1Use (0, 1).
8x15 5y→8x5y15 0
dAx1By1C
A2
B2
3(0) 1(6)(1) (4)
321
2
d
d
20
89
or
20
89
89
20
89
89
23. y
3
2
xUse (0, 0).
y
3
2
x4 →3x2y8 0
dAx1By1C
A2
B2
8(0) (5)(1) 15
82(
5)2
d
d
8
13
or
8
13
13
8
13
13
24. yx6Use (0, 6).
xy1 0
dAx1By1C
A2
B2
3(0) 2(0) 8
322
2
d
d
5
2
or
5
2
2
25. d1d25x112y126
52(
12)2
3x14y110
324
2
1(0) 1(6) (1)
121
2
3x14
5
y110
5x11
1
2
3
y126
39x52y130 25x60y130
14x112y0
x8y0
3x14
5
y110
5x11
1
2
3
y126
39x52y130 25x60y130
64x8y260 0
16x2y65 0
26. d1d2
15x18y168
(15)
282
4x1y16
421
2
68x17y102 1512
x817
y6817
(68 1517
)x(17 817
)y102 6817
0
15x18y168
17
4x1y16
17
68x17y102 1512
x817
y6817
(68 1517
)x(17 817
)y102 6817
0
15x18y168
17
4x1y16
17
d1d2
3x1y12
32 1
2
2x13y13
22 (
3)2
210
x310
y310
313
x13
y213
(210
313
)x(13
310
)y310
213
0
210
x310
y310
313
x13
y213
(210
313
)x(13
310
)y310
213
0
28a. Linda: (19, 112)
dAx1By1C
A2
B2
3x1y12
10
2x13y13
13
3x1y12
10
2x13y13
13
d
d
3
5
2
or 6.4
Father: (45, 120)
dAx1By1C
A2
B2
4(19) (3)(112) 228
42(
3)2
d4(45) (3)(120) 228
42(
3)2
d
48
5
or 9.6
Linda
28b. 4x3y228 0
4x3(140) 228 0
4x192
x48
29. Let x1.
tan v
y
x
tan 40°
1
y
y0.8390996312
m
0.8
1
3
9
0
0
yy1m(xx1)
m0.839 y0.839 0.839(x1)
y0.839x
0.839xy0
dAx1By1C
A2
B2
d0.839(16) 1(12) 0
0.839
212
d1.092068438
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.
dAx1By1C
A2
B2
d
d
6
1
5
3
d5
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.
5(5) (12)(6) 32
52(
12)2
31. m1
4
3
7
1
or
3
4
y7
3
4
(x1)
3x4y25 0
a1Ax1By1C
A2
B2
33. 2x7y5
2x7y5 0
A2
B2
22(
7)2
or 53
2
53
x
7
53
y
5
53
0
2
53
53
x
7
53
53
y
5
53
53
0
34. cos 2A1 2 sin2A
1 2
6
3
2
5
6
35.
2
1
2,
6
1
0°
60°
36. 110 3 330 180° (60° 40°) 80°
x2330233022(330)(330) cos 80°
x2179979.4269
x424.24 miles
37. T2
g
T2
9
2
.8
T2.8 s
38. 2
18 k
220
11020 k
20 k0
k20
39. 2xyz92xyz9
2(x3y2z) 2(10) →2x6y4z20
7y5z11
x2yz7
x3y2z10
y2z3
5(yz) 5(3) →5y5z15
7y5z11 7y5z11
2y4
y2
yz3x2yz7
2 z3x2(2) (5) 7
5 zx6
(6, 2, 5)
40. square: As2triangle: A
1
2
bh
16 s26
1
2
(4)h
4 s3 h
AE sh
AE 4 3 or 7
EF AE
EF 7
The correct choice is C.
Chapter 7 238
a1
a1
3
5
4
m2
1
3
(
4
3)
or
7
2
y4
7
2
(x(3))
7x2y13 0
a2Ax1By1C
A2
B2
3(1) (4)(3) 25
32(
4)2
a2
a2
34
53
or
34
53
53
m3
7
1
(
(
3
1
)
)
or 5
y7 5(x1)
5xy2 0
a3Ax1By1C
A2
B2
7(1) 2(7) 13
722
2
a3
a3
1
2
7
6
or
17
26
26
3
5
4
,
34
53
53
,
17
26
26
32.
The standard form of the equation of the line
through (0, 0) and (4, 12) is 3xy0. The
standard form of the equation of the line through
(4, 12) and (10, 0) is 2xy20 0. The
standard form for the x-axis is y0. To find the
bisector of the angle at the origin, set
3
x
10
y
y
and solve to obtain y
1
3
10
x. To find the
bisector of the angle of the triangle at (10, 0), set
2x
y
5
20
yand 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
system of equations, substitute y
1
3
10
xinto
the equation of the other bisector and solve for xto
get x.Then y
20(1 10
)
5 3 5
210
20(1 10
)
5 3 5
210
5(3) (1)(4) 2
52(
1)2
y
x
O
2
4
6
8
10
2468
1
3
10
.This y-coordinate is the
inradius of the triangle. The approximate value is
3.33.
60
5 35
210
y
120˚300˚480˚
1
1
O
y
csc ( 60˚)
Chapter 7 Study Guide and Assessment
Page 477 Understanding and Using the
Vocabulary
1. b2. g3. d4. a
5. i6. j7. h8. f
9. e10. c
Pages 478–480 Skills and Concepts
11. csc v
si
1
nv
2
12. tan2v1 sec2v
421 sec2v
17 sec2v
17
sec v
13. sin v
cs
1
cv
sin2vcos2v1
3
5
2cos2v1
1
5
3
1
1
2
19.
sin4
s
x
in
2c
x
os4x
1 cot2x
1 cot2x
sin2
s
x
in
2
c
x
os2x
1 cot2x
1
c
s
o
in
s
2
2
x
x
1 cot2x
1 cot2x1 cot2x
20. cos 195° cos (150° 45°)
cos 150° cos 45° sin 150° sin 45°
2
3
2
2
1
2
2
2
6
4
2
21. cos 15° cos (45° 30°)
cos 45° cos 30° sin 45° sin 30°
2
2
2
3
2
2
1
2
6
4
2
22. sin
1
1
7
2
sin
1
1
7
2
sin
4
7
6
sin
4
cos
7
6
cos
4
sin
7
6
2
2
2
3
2
2
1
2
6
4
2
6
4
2
23. tan
1
1
1
2
tan
2
3
4
tan
2
3
tan
4
1 tan
2
3
tan
4
(sin2xcos2x)(sin2xcos2x)
sin2x
239 Chapter 7
3
5
14. sec v
co
1
sv
tan2v1 sec2v
tan2v1
5
4
2
1
4
5
1
(
3
3
)
1
(1)
1
1
3
3
4
2
2
3
or 2 3
24. cos x
1 si
n2x
sin y
1 co
s2x
1
2
7
5
2
1
2
3
2
5
6
7
2
6
5
or
2
2
4
5
5
9
or
3
5
cos (xy) cos xcos ysin xsin y
2
2
4
5
2
3
2
7
5
3
5
48
7
7
5
5
cos2v
1
2
6
5
cos v
4
5
5
4
tan2v
1
9
6
tan v
3
4
15. csc xcos2xcsc x
sin
1
x
(1 sin2x)
sin
1
x
sin
1
x
sin
1
x
sin x
sin x
16. cos2xtan2xcos2x1
cos2x
c
s
o
in
s
2
2
x
x
cos2x1
cos2xsin2x1
1 1
17.
1
1
c
c
o
o
s
s
v
v
(csc vcot v)2
1
1
c
c
o
o
s
s
v
v
si
1
nv
c
s
o
in
s
v
v
2
1
1
c
c
o
o
s
s
v
v
(1
sin
c
2
os
v
v)2
1
1
c
c
o
o
s
s
v
v
(1
1
c
c
o
o
s
s2
v
v
)2
1
1
c
c
o
o
s
s
v
v
1
1
c
c
o
o
s
s
v
v
1
1
c
c
o
o
s
s
v
v
18.
se
t
c
a
v
n
v
1
se
t
c
a
v
n
v
1
se
t
c
a
v
n
v
1
tan
se
v
c
(s
2
e
v
c
v
1
1)
se
t
c
a
v
n
v
1
tan v
t
(
a
se
n
c
2
v
v
1)
se
t
c
a
v
n
v
1
se
t
c
a
v
n
v
1
(1 cos v)2
(1 cos v)(1 cos v)
25. cos y
se
1
cy
sin y
1 co
s2y
1
2
3
2
2
3
5
9
or
3
5
tan y
c
s
o
in
s
y
y
or
2
5
3
5
2
3
1
3
2
2
7
4
33. sin 4vsin 2(2v)
2 sin 2vcos 2v
2
2
2
4
5
2
7
5
3
6
3
2
6
5
34. tan x1 sec x
(tan x1)2sec2x
tan2x2 tan x1 tan2x1
2 tan x0
tan x0
x0°
35. sin2xcos 2xcos x0
1 cos2x2 cos2x1 cos x0
cos2xcos x0
cos x(cos x1) 0
cos x0orcos x1 0
x90° or x270° cos x1
x0°
Chapter 7 240
tan (xy)
1
ta
n
t
x
an
x
ta
ta
n
n
y
y
5
4
2
5
1
5
4
2
5
5 2
4
5
8 5
8
5
1
8
0
5
4
5
5
180
6
8
1
25
or
180
61
255
26. cos 75° cos
15
2
0°
1c
o
2
s150
°
(Quadrant I)
1
2
3
2
2
3
2
27. sin
7
8
sin
7
4
2
(Quadrant II)
1 cos
7
4
2
2
2
2
28. sin 22.5° sin
4
2
5°
1c
2
os 45°
(Quadrant I)
1
2
2
2
2
2
2
1
2
2
2
29. tan
1
2
tan
6
2
(Quadrant I)
1 cos
6
1 cos
6
1
2
3
1
2
3
2
2
3
3
(2 3
)(2 3
)
(2 3
)(2 3
)
2 3
30. sin2vcos2v1
sin2v
3
5
21
sin2v
1
2
6
5
sin v
4
5
sin 2v2 sin vcos v
2
4
5
3
5
2
2
4
5
31. cos 2v2 cos2v1
2
3
5
21
2
7
5
32. tan v
c
s
o
in
s
v
v
tan 2v
1
2
t
t
a
a
n
n
v
2v
or
4
3
2
4
3
1
4
3
2
4
5
3
5
(2 3
)2
4 3
36. cos 2xsin x1
1 2 sin2xsin x1
2 sin2xsin x0
sin x(2 sin x1) 0
sin x0or2 sin x1 0
x0° or x180° sin x
1
2
x30° or
x150°
37. sin xtan x
2
2
tan x0
tan xsin x
2
2
0
tan x0orsin x
2
2
0
xksin x
2
2
x
4
2kor
3
4
2k
38. sin 2xsin x0
2 sin xcos xsin x0
sin x(2 cosx1) 0
sin x0or2 cos x1 0
xkcos x
1
2
x
2
3
2k
or x
4
3
2k
39. cos2x2 cos x
cos2xcos x2 0
(cos x1)(cos x2) 0
cos x1 0orcos x2 0
cos x1cos x2
x2kno solution
40. xcos fysin fp0
xcos
3
ysin
3
23
0
1
2
x
2
3
y23
0
x3
y43
0
41. xcos fysin fp0
xcos 90° ysin 90° 5 0
0x1y5 0
y5 0
42. xcos fysin fp0
xcos
2
3
ysin
2
3
3 0
1
2
x
2
3
y3 0
x3
y6 0
43. xcos fysin fp0
xcos 225° ysin 225° 42
0
2
2
x
2
2
y42
0
xy8 0
44.
A2
B2
723
2
or 58
7
58
x
3
58
y
8
58
0
7
58
58
x
3
58
58
y
4
29
58
0
sin f
3
58
58
, cosf
7
58
58
, p
4
29
58
; Quadrant I
tan for
3
7
3
58
58
7
58
58
45. 6x4y5
6x4y5 0
A2
B2
62(
4)2
or 213
2
6
13
x
2
4
13
y
2
5
13
0
3
13
13
x
2
13
13
y
5
26
13
0
sin f
2
13
13
, cos f
3
13
13
, p
5
26
13
; Quadrant II
tan for
2
3
f146°
46. 9x5y3
9x5y3 0
A2
B2
925
2
or 106
1
9
06
x
1
5
06
y
1
3
06
0
9
10
1
6
06
x
5
10
1
6
06
y
3
10
1
6
06
0
sin f
5
10
1
6
06
,cos f
9
10
1
6
06
,p
3
10
1
6
06
;Quadrant I
tan for
5
9
106
9
10
1
6
06
2
13
13
——
3
13
13
241 Chapter 7
f23°
f29°
47. x7y5
x7y5 0
A2
B2
12(
7)2
or 52
5
1
2
x
5
7
2
y
5
5
2
0
10
2
x
7
1
0
2
y
2
2
0
sin f
7
1
0
2
, cos f
10
2
, p
2
2
; Quadrant II
tan for 7
7
1
0
2
—
10
2
f98°
48. d
A
x1
A
B
2
y
1
B2
C
d2(5) (3)(6) 2
22(
3)2
d
6
13
or
6
13
13
49. 2y3x6 → 3x2y6 0
d
Ax
1
A2
B
y1
B
2
C
d3(3) 2(4) (6)
322
2
d
2
1
3
3
or
23
1
3
13
23
13
13
50. 4y3x1 →3x4y1 0
d
Ax
1
A2
B
y1
B
2
C
d
d
5
23
or
2
5
3
2
5
3
3(2) (4)(4) (1)
32(
4)2
51. y
1
3
x6 →x3y18 0
d
A
x1
A
B
2
y
1
B2
C
d1(21) (3)(20) 18
12(
3)2
57. x3y2 0
y
3
5
x3 →3x5y15 0
d1d23x15y115
32(
5)2
x13y12
(1)2
32
Chapter 7 242
d
2
1
1
0
or
21
10
10
52. y
3
x
6Use (0, 6).
y
3
x
2 →x3y6 0
d
A
x1
A
B
2
y
1
B2
C
d
d
24
10
or
12
5
10
d
12
5
10
53. y
3
4
x3Use (0, 3).
y
3
4
x
1
2
→3x4y2 0
d
Ax
1
A2
B
y1
B
2
C
d
d
5
14
or
1
5
4
d
1
5
4
54. xy1Use (0, 1).
xy5 →xy5 0
d
Ax
1
A2
B
y1
B
2
C
d1(0) 1(1) (5)
121
2
3(0) (4)(3) (2)
32(
4)2
1(0) (3)(6) 6
12(
3)2
d
4
2
or 22
d22
55. y
2
3
x2Use (0, 2).
d
Ax
1
A2
B
y1
B
2
C
d
d
9
13
or
9
13
13
56. y3x2 →3xy2 0
y
2
x
3
2
→x2y3 0
d1d2x12y13
122
2
3x1y12
321
2
2(0) (3)(2) 3
22(
3)2
3x1
y
1
1
0
2
x1
2y
5
13
35
x5
y25
10
x210
y310
(35
10
)x(5
210
)y25
310
0
3x1
y
1
1
0
2
x1
2y
5
13
35
x5
y25
10
x210
y310
(35
10
)x(5
210
)y25
310
0
3x15y115
34
x13y12
10
34
x334
y234
310
x510
y
1510
(34
310
)x(334
510
)y
234
1510
0
310
x510
y1510
34
x334
y
234
(34
310
)x(334
510
)y234
1510
0
Page 481 Applications and Problem Solving
58. The formulas are equivalent.
v
2
0
g
2
s
ta
ec
n
2
2
v
v
v02
c
s
o
in
s
2
2
v
v
2g
cos
1
2v
3x15y115
34
x13y12
10
c
c
o
o
s
s
2
2
v
v
v02
c
s
o
in
s
2
2
v
v
2g
cos
1
2v
v02
2
s
g
in2v
59. d
Ax
1
A2
B
y1
B
2
C
d
d
64
2
0
0
0
d1431 ft
60. sin 30°
10
x
0
30° 45° v90°
100 sin 30° xv15°
50 x
cos v
x
y
cos 15°
5
y
0
y
cos
50
15°
y51.76 yd
Page 481 Open-Ended Assessment
1. Sample answer: 15°; 15°
3
2
0°
sin
3
2
0°
1c
2
os 30°
2
2
3
cos
3
2
0°
1c
2
os 30°
2
2
3
1
2
3
2
1
2
3
2
4(1600) (2)(0) 0
42(
2)2
tan
3
2
0°
1
1
c
c
o
o
s
s
3
3
0
0
°
°
1
2
3
1
2
3
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.
abc180
xa180
yb180
zc180
Add the last three equations.
xaybzc180 180 180
xyzabc180 180 180
Replace abcwith 180.
xyz180 180 180 180
xyz180 180 or 360
The correct choice is D.
4. Since xy90°, x90° y.
Then sin xsin (90° y).
sin (90° y) cos y
c
s
o
in
s
x
y
sin(
c
9
o
0
s
°
y
y)
c
c
o
o
s
s
y
y
1
The correct choice is D.
Another solution is to draw a diagram and notice
that sin x
b
c
and cos y
b
c
.
c
s
o
in
s
x
y
1
5. In order to represent the slopes, you need the
coordinates of point A. Since Alies on the y-axis,
let its coordinates be (0, y). Then calculate the two
slopes. The slope of A
B
is
0
y
(
0
3)
3
y
. The slope
of A
D
is
0
y
0
3
3
y
. The sum of the slopes is
3
y
3
y
0.
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 abx
ab90 x
The correct choice is A.
b
c
—
b
c
243 Chapter 7
2
2
3
2
2
3
(2
4
3
3
)2
(2 3
)(2 3
)
(2 3
)(2 3
)
2 3
2. Sample answer: sin xtan x
1
co
c
s
o
x
s2x
sin xtan x
1
co
c
s
o
x
s2x
sin x
c
s
o
in
s
x
x
s
c
i
o
n
s
2
x
x
s
c
i
o
n
s
2
x
x
s
c
i
o
n
s
2
x
x
SAT & ACT Preparation
Page 483 SAT and ACT Practice
1. The problem states that the measure of ∠Ais 80°.
Since the measure of ∠Bis half the measure of
∠A, the measure of ∠Bmust be 40°. Because ∠A,
∠B, and ∠Care interior angles of a triangle, the
sum of their measures must equal 180°.
m∠Am∠Bm∠C180
80 40 m∠C180
120 m∠C180
m∠C60
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
2x2 for yin the second equation.
7x3y11
7x3(2x2) 11
7x6x6 11
x5
Then use this value for xto calculate the value
for y.
y2x2
y2(5) 2 or 8
The point of intersection is (5, 8). The correct
choice is A.
a
b
c
x
˚
y
˚
7. Simplify the fraction. One method is to multiply
both numerator and denominator by
y
y
2
2
.
y
y
2
2
y
1
y
1
2
y
y
1
2
y
1
y
1
2
y
y
1
2
9. Since the volume Vvaries directly with the
temperature T, the volume and temperature
satisfy the equation V kT, where kis a constant.
When V12, T60. So 12 60k, or k
1
5
.
The relationship is V
1
5
T.
To find the volume when the temperature is 70°,
substitute 70 for Tin the equation V
1
5
T.
V
1
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 sbe the length of the other two
sides. Write and solve an inequality.
2s13
s6.5
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.
Chapter 7 244
y2
y3
2
y
y
1
(y
y
(y
1
2
)(
y
1)
1)
y
(
(
y
y
1
1
)
)
(
(
y
y
1
1
)
)
y
y
2
1
y
Another method is to write both the numerator
and denominator as fractions, and then simplify.
y
2
y
1
——
y2
y
2
2
y1
y
1
y
1
2
y
y
1
2
y2
y
1
y2
y
2
2
y1
y
(
(
y
y
1
1
)
)
(
(
y
y
1
1
)
)
y
y
2
1
y
The correct choice is A.
8. Since the triangles are similar, use a proportion
with corresponding sides of the two triangles.
B
AC
C
B
A
D
E
2
2
3
A
4
E
2AE 4(2 3)
AE 10
The correct choice is E.
Geometric Vectors
Page 490 Check for Understanding
1. Sample answer:
Draw a
u
. Then draw b
u
so that its initial point (tip)
is on the terminal point (tail) of a
u
. Draw a dashed
line from the initial point of a
u
to the terminal
point of b
u
. 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°
8.
9.
10.
8-1
245 Chapter 8
Chapter 8 Vectors and Parametric Equations
a
a
b
b
x y
3.5 cm
x
y
70˚
x y
2.6 cm
x
y
210˚
4y z
12.9 cm
z
4y
51˚
3.5 cm, 70°
2.6 cm, 210°
12.9 cm, 51°
11.
2.9 cm, 12°
12. h2.9 cos 55° v2.9 sin 55°
h1.66 cm v2.38 cm
13a.
13b. Use the Pythagorean Theorem.
c2a2b2
c2(100)2(5)2
c210,025
c10,02
5 or about 100.12 m/s
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.
18.
3 cm, 101°
19.
3.4 cm, 25°
20.
3.8 cm, 359°
x
1
3
2.9 cm
2z
x
1
3
2z
12˚
5100
3 cm
r
s
r
s
101˚
3.4 cm s
t
s
t
25˚
3.8 cm s
u
su
359˚
21.
5.5 cm, 324°
22.
3.9 cm, 155°
23.
5.2 cm, 128°
24.
4.2 cm, 45°
25.
8.2 cm, 322°
26.
3.5 cm, 22°
27.
5.4 cm, 133°
28.
5.5 cm, 358°
29. Draw to scale:
3.4 cm, 301°
30. Draw to scale:
11.7 cm, 357°
31. h2.6 cos 128°v2.6 sin 128°
h1.60 cm v2.05 cm
32. h1.4 cos 45°v1.4 sin 45°
h0.99 cm v0.99 cm
33. h2.1 cos 14°v2.1 sin 14°
h2.04 cm v0.51 cm
Chapter 8 246
5.5 cm
u
r
u
r
324˚
3.9 cm
r
t
r
t
155˚
5.2 cm
r
2r
128˚
4.2 cm
3s
s
45˚
8.2 cm
3u
2s
2s
3u
322˚
3.5 cm
r
t
u
r
t
u
22˚
5.4 cm
r
s
u
r
s
u
133˚
5.5 cm
2s
u
r
1
2
2su
r
1
2
358˚
3.4 cm
r
2t
s
3u
2t
3u
s
r
301˚
11.7 cm
3t
2u
2u
3t
357˚
34. h3.0 cos 340°v3.0 sin 340°
h2.82 cm v1.03 cm
35. c2a2b2
c2(29.2)2(35.2)2
c22091.68
c2091.
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:
38.
61 N, 23° north of east
39. Sometimes;
40a. v1.5 sin 52° h1.5 cos 52°
v1.18 N h0.92 N
40b. h1.5 cos 78° v1.5 sin 78°
h0.31 N v1.47 N
41. h47 cos 40° v47 sin 40°
h36 mph v30 mph
42. It is true when k1 or when a
u
is the zero vector.
43. c2a2b2
c2(50)2(50)2
c5000
or about 71 lb
44.
a
u
b
u
24
equilateral triangle a
u
b
u
24 lb
45. The origin is not in the interior of the acute angle.
d1d2
d1
x
1
2
y
(
2
1)2
or
x
y
2
2
d2
y
02
5
1
2
or y5
x
y
2
2
(y5)
xy2 2
(y5)
xy2 2
y52
xy2 2
y52
0
x(1 2
)y2 52
0
46. csc vcos vtan v
si
1
nv
cos v
c
s
o
in
s
v
v
s
s
i
i
n
n
v
v
c
c
o
o
s
s
v
v
1
47.
4
nwhere nis an integer
48.
2
,
5
2
scl
2
by [3, 3] scl1
x,
3
2
for 0 x2
49. tan 18°29
0.
5
5b
b
0.5 tan
5
18°29
b29.9 cm
sin 18°29
h
5
h
sin 1
5
8°29
h15.8 cm
247 Chapter 8
s
s
r
r
s r
r s
6.1
60 N
35 N
40 N
23˚
a
a
b
5
5
b
a
b
2.3
2.7
aa
b
1.5
5
b
a
b
a
b
2.3
2.7
ab
60˚
60˚60˚
60˚
60˚
60˚
2.4 cm
50. vovolume of original box
vnvolume of new box
voowoho
(w1) w2w
(w1)2w2
2w32w2
vnnwnhn
(w2) (w1) (2w2)
(w23w2)(2w2)
2w38w210w4
2w38w210w4 160
wo3
o2w
2 3 or 6
how1
3 1 or 4
So, the dimensions of the original box are
3 ft 4 ft 6 ft
51. g(x)
(x
x
1
)(x
2
3)
vertical: As xapproaches 1 and 3, the expression
approaches or . So, x1 and x3 are
vertical asymptotes.
horizontal: y
x2
x
2x
2
3
y
x
x
2
x
2
2
x
x
2
2
2
x
x
2
x
3
2
2. Use XY
u
(x2
x1)2
(y2
y1)2
and replace
the values for xand y.
x(5, 6), y(3, 4)
XY
u
[3 (
5)]2
[4
(
6)]2
(8)2
(2)2
64
4
or 68
3. Jacqui is correct. The representation is incorrect.
2, 00, 5is not equal to 51, 0(2)0, 1.
The correct expression is 2i
u
5j
u
.
4. MP
u
3 2, 4 (1)or 5, 5
MP
u
(5)2
(5)2
50
or 52
units
5. MP
u
0 5, 5 6or 5, 1
MP
u
(5)2
(1
)2
26
units
6. MP
u
4 (19), 0 4or 23, 4
MP
u
(23)2
(4)2
545
units
7. t
u
u
u
v
u
1, 43, 2
1 3, 4 (2)or 2, 2
8. t
u
1
2
u
u
v
u
1
2
1, 43, 2
1
2
, 23, 2
1
2
3, 2 (2)or 3
1
2
, 4
9. t
u
4u
u
6v
u
4 1, 463, 2
4, 1618, 12
4 18, 16 (12)or 14, 4
10. t
u
8u
u
81, 4
8(1), 8(4)or 8, 32
11. 8, 682(
6)2
100
or 10
8i
u
6j
u
12. 7, 5(7)2
(5
)2
74
7i
u
5j
u
13. Let T
u
represent the force Terrell exerts.
Let W
u
represent the force Mr. Walker exerts.
T
u
x400 cos 65° W
u
x600 cos 110°
169.05 205.21
T
u
y400 sin 65° W
u
y600 sin 110°
362.52 563.82
T
u
169.05, 362.52, W
u
205.21, 563.82
T
u
W
u
36.16, 926.34
T
u
W
u
(36.
16)2
(926.3
4)2
927 N
Pages 497–499 Exercises
14. YZ
u
2 4, 8 2or 2, 6
YZ
u
(2)2
62
40
or 210
Chapter 8 248
w2810156
1264160
121020136
221234 88
321452 0
y
As xincreases positively or negatively, the
expression approaches 0. So, y0 is a horizontal
asymptote.
52. Let x, x2, and x4 be 3 consecutive odd
intergers.
3x2(x4) 3x4 15
3x2x8 3
3x2x11
x11
The correct answer 15.
Algebraic Vectors
Pages 496–497 Check for Understanding
1. Sample answer: a
u
8, 6, b
u
6, 8; equal
vectors have the same magnitude and direction.
8-2
1
x
x
2
2
1
2
x
x
3
2
15. YZ
u
1 (5), 2 7or 4, 5
YZ
u
42(
5)2
41
16. YZ
u
1 (2), 3 5or 3, 2
YZ
u
32(
2)2
13
17. YZ
u
0 5, 3 4or 5, 7
YZ
u
(5)2
(7
)2
74
18. YZ
u
0 3, 4 1or 3, 3
YZ
u
(3)2
32
18
or 32
19. YZ
u
1 (4), 19 12or 5, 7
YZ
u
527
2
74
20. YZ
u
7 5, 6 0or 2, 6
YZ
u
226
2
40
or 210
21. YZ
u
23 14, 14 (23)or 9, 9
YZ
u
929
2
162
or 92
22. AB
u
36 31, 45 (33)or 5, 12
AB
u
52(
12)2
169
or 13
23. a
u
b
u
c
u
6, 34, 8
6 (4), 3 8or 2, 11
24. a
u
2b
u
c
u
26, 34, 8
12, 64, 8
12 (4), 6 8or 8, 14
25. a
u
b
u
2c
u
6, 324, 8
6, 38, 16
6 (8), 3 16or 2, 19
26. a
u
2b
u
3c
u
26, 334, 8
12, 612, 24
12 (12), 6 24or 0, 30
27. a
u
b
u
4c
u
6, 344, 8
6, 316, 32
6 (16), 3 32or 22, 29
28. a
u
b
u
2c
u
6, 3 24, 8
6, 38, 16
6 (8), 3 (16)or 14, 13
29. a
u
3b
u
36, 3
3 6, 3 3or 18, 9
30. a
u
2
1
c
u
1
2
4, 8
1
2
(4),
1
2
8or 2, 4
31. a
u
6b
u
4c
u
66, 344, 8
36, 1816, 32
36 (16), 18 32or 20, 50
249 Chapter 8
32. a
u
0.4b
u
1.2c
u
0.46, 31.24, 8
2.4, 1.24.8, 9.6
2.4 (4.8), 1.2 9.6or 7.2, 8.4
33. a
u
1
3
(2b
u
5c
u
)
1
3
(26, 354, 8)
1
3
(12, 620, 40)
1
3
(12 (20), 6 40)
1
3
32, 34or
3
3
2
,
3
3
4
34. a
u
(3b
u
c
u
) 5b
u
36, 34, 856, 3
18, 9(4, 830, 15
18 (4) 30, 9 8 15or 44, 32
35. 3m
u
2.5n
u
35, 62.56 9
[15, 1815, 22.5
[15 (15), 18 (22.5
30, 4.5
36. 3, 4324
2
25
or 5
3i
u
4j
u
37. 2, 322(
3)2
13
2i
u
3j
u
38. 6, 11(6)2
(1
1)2
157
6i
u
11j
u
39. 3.5, 12(3.5)2
122
156.2
5
or 12.5
3.5i
u
12j
u
40. 4, 1(4)2
12
17
4i
u
j
u
41. 16, 34(16)
2(
34)2
1412
or 2353
16i
u
34j
u
42. ST
u
4 (9), 3 2or 5, 5
5i
u
5j
u
43. Student needs to show that
(v
u
1v
u
2) v
u
3v
u
1(v
u
2v
u
3)
(v
u
1v
u
2) v
u
3[a, bc, d] e, f
ac, bde, f
ace, bdf
ace, bdf
a, bce, df
a, b[c, de, f]
v
u
1(v
u
2v
u
3)
44a.
44b. sin 20°
1
F
u
0
y
0
F
u
y100 sin 20°
34 N
Fx
Fy
100 N
20˚
53. Let a400, b600, C46.3
°
c2400260022(400)(600) cos 46.3
°
c218.8578.39
c434
Pabc
400 600 434
1434 ft
s
1
2
(abc)
s
1
2
(1434) or 717
ks(s
a)(s
b)(s
c)
k717(7
17
400)(7
17
600)(7
17
434)
k7,525
,766,0
79
k86,751 sq ft
54. Sample answer:
An upper bound is 2.
Alower bound
is 1.
55.
[4, 4] scl1 by [4, 4] scl1
max: (0, 3), min: (0.67, 2.85)
56.
y→as x→, y→as x→
57. 7x1 7x1
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.
45a.
45b. sin 30°
V
u
15
k
V
u
k
sin
15
30°
30 mph
46a. Since QR
u
ST
u
0, QR
u
ST
u
.
So, they are opposites.
46b. QR
u
and ST
u
have the same magnitude, but
opposite direction. So, they are parallel.
Quadrilateral QRST is a parallelogram.
47a. t
d
r
1
5
5
m
0
/
m
s
or 30 s
47b. drt
(1.0 m/s)(30 s) or 30 m
47c. V
u
BV
u
C0.51.0
125
2
26
or about 5.1 m/s
48. cos v
(x2
v
u
x1)
→(x2x1) v
u
cos v
sin v
(y2
v
u
y1)
→(y2y1) v
u
sin v
49. PQ
u
2 8, 5 (7)
10, 12
PQ
u
(10)
212
2
244
RS
u
7 8, 0 (7)
1, 7
RS
u
(1)2
72
50
none
50. d
Ax1By1C
A2
B2
Chapter 8 250
V
x
V
k
Shore
Surfer
V
s
15
30˚
d
d
3
5
2
8
or about 4.2
51. sin 255° sin (225° 30°)
sin 225° cos 30° cos 225° sin 30°
2
2
2
3
2
2
1
2
52. yAsin (kx c)
A: A17
A17 or 17
k:
2
k
4
k8
c:
k
c
60°
8
c
60°
c480°
y17 sin (8x480°)
6
2
4
3(1) 7(4) 1
32(
7)2
f(x) 3x22x1
r321
13 12
23 49
f(x) 3x22x1
r321
13 56
23 817
f(x) x23x1
xf(x)
10,000 99,970,001
1000 997,001
100 9701
10 71
01
10 131
100 10,301
1000 1,003,001
10,000 100,030,001
32
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
pointing up. Then, vector 2i
u
is two units along the
x-axis, vector 3j
u
is three units along the y-axis,
and vector 4k
u
is four units along the z-axis. Draw
broken lines to represent three planes.
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
the j
u
-term must be the same (), and the
coefficient for the k
u
-term is 0, so the correct way
to express the vector as a sum of unit vectors is
i
u
4j
u
.
4.
OG
u
42(
1)2
72
66
5. RS
u
3 (2), 9 5, 3 8or 5, 4, 11
RS
u
524
2(
11)2
162
or 92
6. RS
u
10 3, 4 7, 0 (1)or 7, 11, 1
RS
u
72(
11)2
12
171
or 319
7. a
u
3f
u
g
u
31, 3, 83, 9, 1
3, 9, 243, 9, 1
3 3, 9 9, 24 (1)or 6, 0, 25
8. a
u
2g
u
5f
u
23, 9, 151, 3, 8
6, 18, 25, 15, 40
6 5, 18 (15), 2 (40)or 1, 33, 38
9. EF
u
6 (5), 6 (2), 6 4
11, 4, 2
11i
u
4j
u
2k
u
10. EF
u
12 (12), 17 15, 22 (9)
0, 2, 13
2j
u
13k
u
8-3
z
y
x
B
(10,3,15)
16
4
6
2
8
12
14
10
42
6
8
42
4
2
462
4
81012
G(4,1, 7)
251 Chapter 8
z
y
x
z
y
x
B
(4,1,3)
11. 132, 3454, 01322
3454
202
11,947
,540
3457 N
Pages 503–504 Exercises
12.
OB
u
421
2(
3)2
26
13.
OB
u
722
242
69
14.
OB
u
102
(3)2
15
2
334
15. TM
u
3 2, 1 5, 4 4or 1, 4, 8
TM
u
12(
4)2
(8)2
81
or 9
16. TM
u
3 (2), 5 4, 2 7or 1, 1, 5
TM
u
(1)2
12
(5)2
27
or 33
17. TM
u
3 2, 1 5, 0 4or 1, 4, 4
TM
u
12(
4)2
(4)2
33
18. TM
u
1 3, 1 (5), 2 6or 4, 6, 4
TM
u
(4)2
62
(4)2
68
or 217
19. TM
u
2 (5), 1 8, 6 3or 3, 9, 9
TM
u
32(
9)2
(9)2
171
or 319
z
y
x
B(7, 2, 4)
35. G1G
u
2(x2
x1)2
(y2
y1)2
(z2
z1)2
(x1
x2)2
(y1
y2)2
(z1
z2)2
G1G
u
2
because (xy)2(yx)2for all real numbers x
and y.
36. If m
u
m1, m2, m3, then
m
u
(m1)2
(m2
)2(m
3)2
. If m
u
m1, m2, m3, then m
u
(m1)
2(
m2)2
(m3
)2
.
Since m12(m1)2, m22(m2)2, and m32
(m3)2, m
u
m
u
.
37. 3, 2, 46, 2, 5F
u
O
u
9, 0, 9F
u
O
u
F
u
9, 0, 9or
9, 0, 9
38. m
1
2
(x1x2, y1y2, z1z2)
1
2
(2 4, 3 5, 6 2)
1
2
(6, 8, 8)
(3, 4, 4)
39a. OK
u
1 0, 4 0, 0 0or 1, 4, 0
i
u
4j
u
39b. TK
u
1 2, 4 4, 0 0or 1, 0, 0
i
u
40. c
u
b
u
a
u
c
u
3, 1, 51, 3, 1
c
u
2, 2, 4
41a.
41b. Find distance between (0, 0, 0) and (15, 15, 15).
d(15
0)2
(15
0)2
(15
0)2
675
or about 26 feet
41c. sin v
1
2
5
6
vsin1
1
6
5
75
v35.25°
42. AB
u
(1 2
)2(
3
0
)2(0
0)2
4
or 2
BC
u
(1 1
)2
1
3
3
2
2
3
2
0
2
36
3
63
or 1.69
AC
u
(1 2
)2
1
3
0
2
2
3
2
0
2
2
or 1.41
No, the distances between the points are not
equal. Aand Bare 2 units apart, Band Care 1.69
units apart, and Aand Care 1.41 units apart.
43. 3, 51, 23 (1), 5 2
2, 7
20. TM
u
1 0, 4 6, 3 3or 1, 2, 6
TM
u
12(
2)2
(6)2
41
21. CJ
u
3 (1), 5 3, 4 10
or 4, 8, 14)
CJ
u
42(
8)2
(14
)2
276
or 269
22. u
u
6w
u
2z
u
62, 6, 123, 0, 4
12, 36, 66, 0, 8
18, 36, 2
23. u
u
1
2
v
u
w
u
2z
u
u
u
1
2
4, 3, 52, 6, 123, 0, 4
2,
3
2
,
5
2
2, 6, 16, 0, 8
6, 7
1
2
, 11
1
2
24. u
u
3
4
v
u
w
u
3
4
4, 3, 52, 6, 1
3,
9
4
,
1
4
5
2, 6, 1
1, 8
1
4
, 4
3
4
25. u
u
3v
u
2
3
w
u
2z
u
34, 3, 5
2
3
2, 6, 123, 0, 4
12, 9, 15
4
3
, 4,
2
3
6, 0, 8
16
2
3
, 13, 23
2
3
26. u
u
0.75v
u
0.25w
u
0.754, 3, 50.252, 6, 1
3, 2.25, 3.750.5, 1.5, 0.25
3.5, 0.75, 3.5
27. u
u
4w
u
z
u
42, 6, 13, 0, 4
8, 24, 4(3, 0, 4
5, 24, 8
28.
2
3
f
u
3g
u
2
5
h
u
2
3
3, 4.5, 132, 1, 6
2
5
6, 3, 3
2, 3,
2
3
6, 3, 18
1
5
2
,
6
5
,
6
5
5
5
2
,
3
5
6
,
2
1
7
5
8
29. LB
u
5 2, 6 2, 2 7or 3, 8, 5
3i
u
8j
u
5k
u
30. LB
u
4 (6), 5 1, 1 0or 2, 4, 1
2i
u
4j
u
k
u
31. LB
u
7 9, 3 7, 2 (11)or 2, 4, 9
2i
u
4j
u
9k
u
32. LB
u
8 12, 7 2, 5 6or 20, 5, 11
20i
u
5j
u
11k
u
33. LB
u
8 (1), 5 2, 10 (4)
or 7, 3, 6
7i
u
3j
u
6 k
u
34. LB
u
6 (9), 5 12, 5 (5)
or 15, 7, 0
15i
u
7j
u
Chapter 8 252
12
12
12
6
6
6
z
y
x
O
44. AB
u
3 5, 3 2or 8, 1
CD
u
d10, d20or d1, d2
AB
u
CD
u
8, 1d1, d2
D(8, 1)
45. cot X
sin 2X
1 cos 2X
i
u
j
u
k
u
2. a
u
a
u
axayaz
axayaz
i
u
j
u
k
u
(ayazayaz)i
u
(axazaxaz)j
u
(axayaxay) k
u
0i
u
0j
u
0k
u
0, 0, 0
0
u
3. Sample answer: No, because a vector cannot be
perpendicular to itself.
4. 5, 23, 75(3) 2(7)
15 14
1, no
5. 8, 24.5, 188(4.5) 2(18)
36 36
0, yes
6. 4, 9, 83, 2, 24(3) 9(2) 8(2)
12 18 16
10, no
i
u
j
u
k
u
7. 1, 3, 22, 1, 5
132
215
i
u
j
u
k
u
13i
u
j
u
5k
u
or 13, 1, 5, yes
13, 1, 51, 3, 2
13(1) 1(3) (5)(2)
13 3 10 0
13, 1, 52, 1, 5
13(2) 1(1) (5)(5)
26 1 25 0
i
u
j
u
k
u
8. 6, 2, 104, 1, 9
6210
41 9
i
u
j
u
k
u
8i
u
14j
u
2k
u
or 8, 14, 2, yes
8, 14, 26, 2, 10
8(6) (14)(2) (2)(10)
48 28 20 0
8, 14, 24, 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)
TU
u
2, 1, 2
UV
u
1, 3, 5
TU
u
UV
u
i
u
j
u
k
u
212
135
i
u
j
u
k
u
i
u
8j
u
5k
u
or 1, 8, 5
1
3
2
1
2
5
2
1
2
5
1
3
2
1
6
4
10
9
6
4
10
9
2
1
3
1
1
2
2
5
1
2
2
5
3
1
ay
ay
ax
ax
az
az
ax
ax
az
az
ay
ay
253 Chapter 8
cot X
2 sin Xcos X
1 cos2Xsin2X
cot X
2 sin Xcos X
2 sin2X
c
s
o
in
s
X
X
cot X
cot Xcot X
46. cos v
2
3
sin2v1 cos2v
sin2v1
4
9
sin2v
5
9
sin v
3
5
47. y6 sin
2
v
amplitude 6or 6
period
2
k
or 4
48. 16
m
re
i
v
n
2
1
r
r
e
a
v
d
6
1
0
m
s
i
e
n
c
8
1
5
radians per second
49. Yes, because substituting 7 for xand 2 for y
results in the inequality 2 180 which is true.
y4x23x5
2 4(7)23(7) 5
2 180
50.
3
2
3
2
1
1
4
3
3
2
4
3
So, A, C, and D are not correct.
2
3
2
3
1
1
3
4
2
3
3
4
So, B is not correct.
The correct choice is E.
Perpendicular Vectors
Pages 508–509 Check for Understanding
1. Sample answer: Vector v
w
is the negative of
vector w
v
.
i
u
j
u
k
u
v
u
w
u
103
124
i
u
j
u
k
u
6i
u
7j
u
3k
u
i
u
j
u
k
u
v
u
w
u
124
103
i
u
j
u
k
u
6i
u
7j
u
3k
u
2
0
1
1
4
3
1
1
4
3
2
0
0
2
1
0
3
4
1
1
3
4
0
2
8-4
2
1
2
10. AB
u
(0.65, 0, 0.3) (0, 0, 0)
0.65, 0, 0.3
F
u
0, 0, 32
T
u
AB
u
F
u
i
u
j
u
k
u
0.65 0 0.3
0032
i
u
j
u
k
u
0i
u
20.8j
u
0k
u
T
u
02(
20.8
)20
2
20.8 foot-pounds
Pages 509–511 Exercises
11. 4.86, 34(6) 8(3)
24 24
0, yes
12. 3, 54, 23(4) 5(2)
12 10
2, no
13. 5, 13, 65(3) (1)(6)
15 6
21, no
14. 7, 20, 27(0) 2(2)
0 4
4, no
15. 8, 4(2, 48(2) 4(4)
16 16
32, no
16. 4, 9, 36, 7, 54(6) 9(7) (3)(5)
24 63 15
24, no
17. 3, 1, 42, 8, 23(2) 1(8) 4(2)
6 8 8
6, no
18. 2, 4, 816, 4, 22(16) 4(4) 8(2)
32 16 16
0, yes
19. 7, 2, 43, 8, 17(3) (2)(8) 4(1)
21 16 4
9, no
20. a
u
b
u
3, 128, 2
24 24
0, yes
b
u
c
u
8, 23, 2
24 4
28, no
a
u
c
u
3, 123, 2
9 24
15, no
0.3
0
0.65
0
0.3
32
0.65
0
0.3
32
0
0
i
u
j
u
k
u
21. 0, 1, 21, 1, 4
012
114
i
u
j
u
k
u
2i
u
2j
u
k
u
or 2, 2, 1,yes
2, 2, 10, 1, 2
2(0) 2(1) (1)(2)
2 2 2 0
2, 2, 11, 1, 4
2(1) 2(1) (1)(4)
2 2 4 0
i
u
j
u
k
u
22. 5, 2, 32, 5, 0
523
250
i
u
j
u
k
u
15i
u
6j
u
29k
u
or 15, 6, 29,yes
15, 6, 295, 2, 3
(15)(5) (6)(2) 29(3)
75 12 87 0
15, 6, 292, 5, 0
(15)(2) (6)(5) 29(0)
30 30 0 0
i
u
j
u
k
u
23. 3, 2, 01, 4, 0
320
140
i
u
j
u
k
u
0i
u
0j
u
10k
u
or 0, 0, 10,yes
0, 0, 103, 2, 0
0(3) 0(2) 10(0)
0 0 0 0
0, 0, 101, 4, 0
0(1) 0(4) 10(0)
0 0 0 0
i
u
j
u
k
u
24. 1, 3, 25, 1, 2
132
512
i
u
j
u
k
u
4i
u
12j
u
16k
u
or 4, 12, 16, yes
4, 12, 161, 3, 2
4(1) 12(3) 16(2)
4 36 32 0
4, 12, 165, 1, 2
4(5) 12(1) 16(2)
20 12 32 0
i
u
j
u
k
u
25. 3, 1, 24, 4, 0
312
440
i
u
j
u
k
u
8i
u
8j
u
16k
u
or 8, 8, 16, yes
8, 8, 163, 1, 2
8(3) 8(1) 16(2)
24 8 32 0
8, 8, 164, 4, 0
8(4) 8(4) 16(0)
32 32 0 0
1
4
3
4
2
0
3
4
2
0
1
4
3
1
1
5
2
2
1
5
2
2
3
1
2
4
3
1
0
0
3
1
0
0
2
4
2
5
5
2
3
0
5
2
3
0
2
5
1
1
0
1
2
4
0
1
2
4
1
1
Chapter 8 254
i
u
j
u
k
u
26. 4, 0, 27, 1, 0
40 2
71 0
i
u
j
u
k
u
2i
u
14j
u
4k
u
or 2, 14, 4, yes
2, 14, 44, 0, 2
2(4) 14(0) 4(2)
8 0 8 0
2, 14, 47, 1, 0
2(7) 14(1) 4(0)
14 14 0 0
27. Sample answer:
Let v
u
v1, v2, v3and v
u
v1, v2, v3
i
u
j
u
k
u
v
u
(v
u
)
v1v2v3
v1v2v3
i
u
j
u
k
u
0i
u
0j
u
0k
u
0
i
u
j
u
k
u
28. a
u
(bc
u
)
a1a2a3
(b1c1)(b2c2)(b3c3)
i
u
j
u
a3
(b3c3)
a1
(b1c1)
a3
(b3c3)
a2
(b2c2)
v2
v2
v1
v1
v3
v3
v1
v1
v3
v3
v2
v2
0
1
4
7
2
0
4
7
2
0
0
1
30. Sample answer:
Let T(2, 1, 0), U(3, 0, 0), and V(5, 2, 0).
TU
u
1, 1, 0
UV
u
8, 2, 0
i
u
j
u
k
u
TU
u
UV
u
110
820
i
u
j
u
k
u
0i
u
0j
u
6k
u
or 0, 0, 6
31. Sample answer:
Let T(0, 0, 1), U(1, 0, 1), and V(1, 1, 1).
TU
u
1, 0, 0
UV
u
2, 1, 2
i
u
j
u
k
u
TU
u
UV
u
100
212
i
u
j
u
k
u
0i
u
2j
u
k
u
or 0, 2, 1
32. The expression is false. m
u
n
u
and n
u
m
u
have
the same magnitude but are opposite in direction.
33a.
33b. T
u
AB
u
F
u
AB
u
0.04 cos (30°), 0, 0.04 sin (30°)
0.02(3
), 0, 0.02
F
u
0, 0, 600
i
u
j
u
k
u
AB
u
F
u
0.023
00.02
00600
0i
u
123
j
u
0k
u
T
u
AB
u
F
u
123
or about 21 N-m
i
u
j
u
k
u
34. x
u
y
u
230
114
i
u
j
u
k
u
12i
u
8j
u
5k
u
A
1
2
x
u
y
u
1
2
122
(8)2
(5)
2
1
2
233
35a. o
u
120, 310, 60
c
u
29, 18, 21
35b. o
u
c
u
120(29) 310(18) 60(21)
$10,320
3
1
2
1
0
4
2
1
0
4
3
1
0
1
1
2
0
2
1
2
0
2
0
1
1
2
1
8
0
0
1
8
0
0
1
2
255 Chapter 8
k
u
a2
(b2c2)
a1
(b1c1)
i
u
j
u
k
u
i
u
j
u
k
u
( a
u
b
u
) ( a
u
c
u
)
29. Sample answer:
Let T(0, 2, 2), U(1, 2, 3), and V(4, 0, 1)
TU
u
1, 4, 5
UV
u
3, 2, 2
i
u
j
u
k
u
TU
u
UV
u
145
322
i
u
j
u
k
u
2i
u
17j
u
14k
u
or 2, 17, 14
4
2
1
3
5
2
1
3
5
2
4
2
a2
c2
a1
c1
a3
c3
a1
c1
a3
c3
a2
c2
a2
b2
a1
b1
a3
b3
a1
b1
a3
b3
a2
b2
30˚
elbow
forearm
600 N
0.04 m
[a2(b3c3) a3(b2c2)]i
u
[a1(b3c3) a3(b1c1)]j
u
[a1(b2c2) a2(b1c1)]k
u
[(a2b3a2c3)(a3b2a3c2)]i
u
[(a1b3a1c3)(a3b1a3c1)]j
u
[(a1b2a1c2) (a2b1a2c1)]k
u
[(a2b3a3b2)(a2c3a3c2)]i
u
[(a1b3a3b1)(a1c3a3c1)]j
u
[(a1b2a2b1) (a1c2a2c1)]k
u
(a2b3a3b2)i
u
(a2c3a3c2)i
u
(a1b3a3b1)j
u
(a1c3a3c1)j
u
(a1b2a2b1)k
u
(a1c2a2c1)k
u
[(a2b3a3b2)i
u
(a1b3a3b1)j
u
(a1b2a2b1)k
u
][(a2c3a3c2)i
u
(a1c3a3c1)j
u
(a1c2a2c1)k
u
]
i
u
j
u
k
u
a2
b2
a1
b1
a3
b3
a1
b1
a3
b3
a2
b2
F
45˚
36a.
36b. WF
u
d
u
cos v
W120 4 cos 45°
W339 ft-lb
37a. X
u
2 1, 5 0, 0 3or 1, 5, 3)
Y
u
3 2, 1 5, 4 0or 1, 4, 4
X
u
Y
u
k
3
4
j
5
4
i
1
1
40. BA
u
2a
u
2b
u
22a
u
b
u
cos v
(a1
b1)2
(a2
b2)2
2
a12
a22
2
b12
b22
2
2
a12
a22
b12
b22
cos v
(a1b1)2(a2b2)2
a12a22b12b222
a12
a22
b12
b22
cos va122a1b1b12
a222a2b2b22
a12a22b12b22
2
a12
a22
b12
b22
cos v
2a1b12a2b2
2
a12
a22
b12
b22
cos v
a1b1a2b2
a12
a22
b12
b22
cos v
a1b1a2b2
a
u
b
u
cos va
u
b
u
a
u
b
u
cos v
41. AB
u
5 3, 3 3, 2 (1)or 2, 0, 3
42. D(8, 3)
E(0, 2)
DE
u
0 8, 2 3) or 8, 5
DE
u
(8)2
(5
)2
89
43. 4xy6 0
A2
B2
421
2
or 17
4
17
17
x
1
1
7
7
y
6
17
17
0
p
6
17
17
1.46 units
sin f
1
1
7
7
cos f
4
17
17
tan f
1
4
f14°
44. A36°, b13, and c6
a2b2c22 bc cos A
a2132622(13)(6) cos 36°
a8.9
sin
8.
3
9
6°
si
1
n
3
B
Bsin1
13 s
8
i
.
n
9
36°
B59.41° or 59°25
C180° 36° 59°25
C84.59° or 84°35
45. tan 73°
h
4
cos 73°
4
4 tan 73° h
cos
4
73°
13.1 h; 13.1 m 13.7 m
46. 3 3x4
10
3x4
7
3x4 49
x17.67
47. 81 34
64 26(22)3or (23)2
4 22
2 21
9 32
So 64 4382
The correct choice is B.
Chapter 8 256
i
u
j
u
k
u
8i
u
7j
u
9k
u
or 8, 7, 9
37b. The cross product of two vectors is always a
vector perpendicular to the two vectors and the
plane in which they lie.
38a. vp
u
(q
u
r
u
)
q
u
r
u
k
4
5
j
1
1
i
2
3
5
4
1
1
3
4
1
1
3
4
5
4
i
u
j
u
k
u
i
u
22j
u
5k
u
or 1, 22, 5
p
u
(q
u
r
u
) 0, 0, 11, 22, 5
0(1) 0(22) (1)(5)
5 or 5 units3
38b.
1
4
5
0
1
1
0
2
3
1
1
2
3
4
5
2
3
4
5
1
1
0
0
(1)
1
1
2
3
4
5
2
3
4
5
1
1
5 or 5 units3
They are the same.
39. Need (kv
u
w
u
) u
u
0.
[k1, 21, 2] 5, 120
[k, 2k1, 2] 5, 120
k1, 2k25, 120
(k1)5 (2k2)12 0
5k5 24k24 0
29k19 0
k
1
2
9
9
u
u
u
u
u
u
2.7 cm
115˚
46˚
2.3 cm
Page 511 Mid-Chapter Quiz
1.
Fx2.3 cos 46° Fy2.3 sin 46°
1.6 cm 1.7 cm
2.
Fx27 cos 245° Fy27 sin 245°
11.4 mm 24.5 mm
3. CD
u
4 (9), 3 2or 5, 5
CD
u
52(
5)2
50 or 52
4. CD
u
5 3, 7 7, 2 (1)or 2, 0, 3
CD
u
220
232
13
5. r
u
t
u
2s
u
6, 224, 3
6, 28, 6
6 8, 2 6or 14, 8
6. r
u
3u
u
v
u
31, 3, 83, 9, 1
3, 9, 243, 9, 1
3 3, 9 9, 24 (1)or 6, 0, 25
7. 3, 64, 23(4) 6(2)
12 12
0; yes
8. 3, 2, 41, 4, 03(1) (2)(4) 4(0)
3 8
11; no
9. 1, 3, 22, 1, 1u
u
j
u
k
u
13 2
211
i
u
j
u
k
u
3
1
1
2
2
1
1
2
2
1
3
1
Graphing Calculator Exploration:
Finding Cross Products
Page 512
1. 49, 32 552. 168, 96, 76
3. 0, 0, 04. 11, 15, 3
5. 0, 0, 76. 0, 40, 0
7. u
u
x
u
6, 6, 12
u
u
v
u
626
2(
12)2
216
8. u
u
v
u
1, 13, 20
u
u
v
u
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)
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
resultant force is O
u
.
4.
5. F
u
1300i
u
F
u
2(170 cos 55°) i
u
(170 sin 55°)j
u
F
u
1F
u
2(300
170 c
os 55°
)2(1
70 sin
55°)2
421.19 N
tan v170 sin 55°
300 170 cos 55°
8-5
8-4B
257 Chapter 8
i
u
5j
u
7k
u
or 1, 5, 7, yes
1, 5, 71, 3, 2
(1)(1) 5(3) (7)(2)
1 15 14 0
1, 5, 72, 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
Current
23 knots 17˚
vtan1
6. F
u
150i
u
F
u
2100j
u
F
u
1F
u
2502
1002
111.8 N
tan v
1
5
0
0
0
or 2
vtan12
63.43°
7. horizontal 18 cos 40°
13.79 N
vertical 18 sin 40°
11.57 N
170 sin 55°
300 170 cos 55°
94˚
454 lb
342 lb
8. F
u
1(33 cos 90°)i
u
(33 sin 90°)j
u
or 33j
u
F
u
2(44 cos 60°)i
u
(44 sin 60°)j
u
or 22i
u
22 3
j
u
F
u
1F
u
2222
(33
223
)2
74 N
tan v
33
2
2
2
23
or
3
2
23
vtan1
3
2
23
73°
Aforce with magnitude 74 N and direction 73°
180° or 253° will produce equilibrium.
9a.
9b. If vis the angle between the resultant path of
the ferry and the line between the landings,
then sin v
1
4
2
or
1
3
. So vsin1
1
3
, or about
19.5°.
Pages 517–519 Exercises
10. 11.
12.
13. F
u
1425i
u
F
u
2390j
u
F
u
1F
u
24252
3902
576.82 N
tan v
3
4
9
2
0
5
or
7
8
8
5
vtan1
7
8
8
5
42.5°
14. v
u
165i
u
v
u
2(50 cos 300°)i
u
(50 sin 300°)j
u
or 25i
u
253
j
u
v
u
1v
u
2902
(253
)2
99.87 mph
tan v
25
9
0
3
or
5
1
8
3
vtan1
5
1
8
3
Apositive value for vis about 334.3°.
15. v
u
1(115 cos 60°)i
u
(115 sin 60°)j
u
or 57.5i
u
57.53
j
u
v
u
2(115 cos 120°)i
u
(115 sin 120°)j
u
or 57.5i
u
57.53
j
u
v
u
1v
u
2
02
(1153
)2
1153
199.19 km/h
Since tan vis undefined and the vertical
component is positive, v90°.
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.
17. F
u
1105i
u
F
u
2(110 cos 50°)i
u
(110 sin 50°)j
u
F
u
1 F
u
2
(105
110 c
os 50°
)2(1
10 sin
50°)2
194.87 N
tan v110 sin 50°
105 110 cos 50°
Chapter 8 258
vtan1
16.7°
20. F
u
1(70 cos 330°)i
u
(70 sin 330°)j
u
or 353
i
u
35j
u
F
u
2(40 cos 45°)i
u
(40 sin 45°)j
u
or 202
i
u
202
j
u
F
u
3(60 cos 135°)i
u
(60 sin 135°)j
u
or 30 2
302
j
u
tan v35 502
353
102
250 sin 25° 45 sin 250°
250 cos 25° 45 cos 250°
vtan1
35 502
353
102
37.5°
F
u
1F
u
2F
u
3(353
10
2
)2
(35
502
)2
58.6 lb
21. F
u
1(23 cos 60°)i
u
(23 sin 60°)j
u
or 11.5i
u
11.53j
u
F
u
2(23 cos 120°)i
u
(23 sin 120°)j
u
or 11.5i
u
11.53j
u
F
u
1F
u
2
02(
233
)2
233
39.8 N
Since tan vis undefined and the vertical
component is positive , v90°. A force with
magnitude 39.8 N and direction 90° 180° or
270° will produce equilibrium.
vtan1
25.62°
18. Fwsin v
52.1 75 sin v
5
7
2
5
.1
sin v
sin1
5
7
2
5
.1
v
44° v
19. F
u
1(250 cos 25°)i
u
(250 sin 25°)j
u
F
u
2(45 cos 250°)i
u
(45 sin 250°)j
u
F
u
1F
u
2
(250 c
os 25°
45
cos 25
0°)2
(250 s
in 25°
45
sin 25
0°)2
220.5 lb
tan v250 sin 25° 45 sin 250°
250 cos 25° 45 cos 250°
110 sin 50°
105 110 cos 50°
4 mph
12 mph
42 N
53˚
Wind
256 mph
27˚
22. agsin 40°
32 sin 40°
20.6 ft/s2
23. F
u
1(36 cos 20°)i
u
(36 sin 20°)j
u
F
u
2(48 cos 222° i
u
(48 sin 222°)j
u
F
u
1F
u
2
(36 co
s 20°
48 co
s 222°
)2(3
6 sin 2
0° 4
8 sin 2
22°)2
19.9 N
tan v36 sin 20° 48 sin 222°
36 cos 20° 48 cos 222°
F1
760 lb
F2
co
c
s
os
17
6
4
.2
.5
°
°
F1
761 lb
30. Sample answer: Method b is better. Let Fbe the
force exerted by the tractor, Tbe the tension in
the two halves of the rope, and vbe the angle
between the original line of the rope and half of
the rope after it is pulled. At equilibrium,
2Tsin vF0, or T
2s
F
in v
. So, if 0° v30°,
the force applied to the stump using method b is
greater than the force exerted by the tractor.
31. Let Tbe the tension in each towline and suppose
the axis of the ship is the vertical direction.
2Tsin 70° 6000 0
T
2s
6
i
0
n
0
7
0
0°
3192.5 tons
32. Let Tbe the tension in each wire. The halves of
the wire make angles of 30° and 150° with the
horizontal.
Tsin 30° Tsin 150° 25 0
1
2
T
1
2
T 25 0
T25 lb
33. u
u
v
u
9(3) 5(2) 3(5)
2
The vectors are not perpendicular since u
u
v
u
0.
34. AB
u
0 12, 11 (5), 21 18
12, 6, 3
35. d
2v
g
02
sin vcos v
2
3
1
2
002
sin 65° cos 65°
239.4 ft
36. Sample answer: A plot of the data suggests a
quadratic function. Performing a quadratic
regression and rounding the coefficients gives
y1.4x22x3.9.
37. b0.3
p0.2
bp0.66
bp
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)90p140b32(1 pb)
32 58p108b
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.
155
sin 174.5° cos 174.5° tan 6.2°
259 Chapter 8
vtan1
264.7° or 5.3° west of south
24a.
24b. F
u
170i
u
F
u
2(135 cos 165°)i
u
(135 sin 165°)j
u
F
u
1F
u
2F
u
3
(70
135 c
os 165
°11
5 cos 2
40°)2
(135 s
in 165
°11
5 sin 2
40°)2
134.5 lb
tan v135 sin 165° 115 sin 240°
70 135 cos 165° 115 cos 240°
36 sin 20° 48 sin 222°
36 cos 20° 48 cos 222°
O
b
p
0.1
0.1
0.2
0.3
0.4
0.5
0.2 0.3 0.4
(0.2, 0.46)
(0.33, 0.33)
(0.3, 0.3)
(0.2, 0.3)
vtan1
208.7° or 28.7° south of west
Since F
u
1F
u
2F
u
30, the vectors are not in
equilibrium.
25. WF
u
d
u
[(1600 cos 50°)i
u
(1600 sin 50°)j
u
] 1500i
u
(1600 cos 50°)(1500) (1600 sin 50°)(0)
1,542,690 N-m
26a. Sample answer: The horizontal forward force is
F
u
cos v. You can increase the horizontal forward
force by decreasing the angle vbetween the
handle and the lawn.
26b. Sample answer: Pushing the lawnmower at a
lower angle may cause back pain.
27a. tan v
1
3
8
or
1
6
vtan1
1
6
9.5° south of east
27b. s182
32
18.2 mph
28. Fcos v100 cos 25°
90.63 N
29. F1cos 174.5° F2cos 6.2° 0
F1sin 174.5° F2sin 6.2° 155 0
The first equation gives F2
co
c
s
os
17
6
4
.2
.5
°
°
F1.
Substitute into the second equation.
F1sin 174.5° F1 155 0
cos 174.5° sin 6.2°
cos 6.2°
135 sin 165° 115 sin 240°
70 135 cos 165° 115 cos 240°
135 lb 165˚
75˚120˚
115 lb
70 lb
F1(sin 174.5° cos 174.5° tan 6.2°) 155
38. *4 *(3) (434) [(3)3(3)]
60 (24)
84
The correct choice is A.
Vectors and Parametric Equations
Pages 523–524 Check for Understanding
1. When t0, x3 and y1. When t1, x7
and y1. The graph is a line through (3, 1) and
(7, 1).
2. Sample answer: For every single unit increment of
t, xincreases 1 unit and yincreases 2 units. Then,
the parametric equations of the line are x3 t,
y6 2t.
3. When t0, x1 and y0, so the line passes
through (1, 0). When t1, x0 and y1, so
the line passes through (0, 1), its y-intercept. The
slope of the line is
1
0
0
1
or 1.
4. x(4), y11t3, 8
x4, y11t3, 8
x4 3ty11 8t
x3t4y8t11
5. x1, y5t7, 2
x1 7ty5 2t
x1 7ty5 2t
6. 3x2y57. 4x6y12
2y3x56y4x12
y
3
2
x
5
2
y
2
3
x2
xtxt
y
3
2
t
5
2
y
2
3
t2
8. x4t39. x9t
x3 4t
9
x
t
1
4
x
3
4
ty4t2
y5t3y4
9
x
2
y5
1
4
x
3
4
3y
4
9
x2
y
5
4
x
3
4
10.
11a. receiver:
x5 0ty50 10t
x5y50 10t
defensive player:
x10 0.9ty54 10.72t
x10 0.9ty54 10.72t
8-6
11b. 50 10t0
50 10t
5 t
When t5, 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.
524–525 Exercises
12. x5, y7t2, 0
x5 2ty7 0t
x5 2ty7
13. x(1), y4t6, 10
x1, y4t6, 10
x1 6ty4 10t
x1 6ty4 10t
14. x(6), y10t3, 2
x6, y10t3, 2
x6 3ty10 2t
x6 3ty10 2t
15. x1, y5t7, 2
x1 7ty5 2t
x1 7ty5 2t
16. x1, y0t2, 4
x1, yt2, 4
x1 2ty4t
x1 2t
17. x3, y(5)t2, 5
x3, y5t2, 5
x3 2ty5 5t
x3 2ty5 5t
18. xt
y4t5
Chapter 8 260
tx y
122
021
160
210 1
O
y
x
19. 3x4y7
4y3x7
y
3
4
x
7
4
xt
y
3
4
t
7
4
20. 2xy3
y2x3
y2x3
xt
y2t3
21. 9xy1
y9x1
xt
y9t1
22. 2x3y11
3y2x11
y
2
3
x
1
3
1
xt
y
2
3
t
1
3
1
23. 4xy2
y4x2
xt
y4t2
24. 3x6y8
6y3x8
y
1
2
x
4
3
The slope is
1
2
.
y5
1
2
(x2)
y
1
2
x6
xt
y
1
2
t6
y
x
1
1
1
21
2
(, )
29. x3 2t
x3 2t
1
2
x
3
2
t
y1 5t
y1 5
1
2
x
3
2
y
5
2
x
1
2
7
30. Regardless of the value of t, xis always 8, so the
parametric equations represent the vertical line
with equation x8.
31a. x11, y(4)t3, 7
x11, y4t3, 7
31b. x11 3ty4 7t
x3t11 y7t4
31c. x3t11
x11 3t
1
3
x
1
3
1
t
y7t4
y7
1
3
x
1
3
1
4
y
7
3
x
8
3
9
32.
[5, 5] Tstep1
[10, 10] Xscl1
[10, 10] Yscl1
33.
[10, 10] Tstep1
[20, 20] Xscl2
[20, 20] Yscl2
34.
[10, 10] Tstep1
[10, 10] Xscl1
[10, 10] Yscl1
35a. x2 3tand y4 7t
If t0, then x2 and y4, so the part of the
line to the right of point (2, 4) is obtained.
35b. x0
2 3t0
3t2
t
2
3
36. xycos2tsin2t
1
0 cos2t1 and 0 sin2t1, so the graph is
the segment of the line with equation xy1
from (1, 0) to (0, 1).
261 Chapter 8
25. x2t
2
x
t
y1 t
y1
2
x
y
1
2
x1
26. x7
1
2
t
x7
1
2
t
2x14 t
y3t
y3(2x14)
y6x42
27. x4t11
x11 4t
1
4
x
1
4
1
t
yt3
y
1
4
x
1
4
1
3
y
1
4
x
2
4
3
28. x4t8
x8 4t
1
4
x2 t
y3 t
y3
1
4
x2
y
1
4
x5
37a. target drone:
x3 (1)ty4 0t
x3 ty4
missile:
x2 ty2 2t
37b. 3 t2 t
1 2t
1
2
t
When t
1
2
, the missile has a y-coordinate of 3,
not 4, so it does not intercept the drone.
38a. Ceres: x1 t, y4 t, z1 2t
Pallas: x7 2t, y6 2t, z1 t
38b. Adding the equations for xand yfor Ceres gives
xy3. Subtracting the equations for xand y
for Pallas results in xy1. The solution of
this system is x1 and y2. Eliminating t
from the equations for yand zresults in the
system 2yz7, y2z4 which has
solution y2 and z3. Hence, the paths cross
at (1, 2, 3).
38c. 1 t1 vt2
7 2t1 vt4
Ceres is at (1, 2, 3) when t2 but Pallas is at
(1, 2, 3) when t4. The asteroids will not
collide.
39. The line is parallel to the vector 0
1
3
, 5 1,
8 1or
1
3
, 4, 9. The vector equation of the
line is x
1
3
, y1, z1t
1
3
, 4, 9or
x
1
3
, y1, z1t
1
3
, 4, 9.
x
1
3
1
3
ty1 4t
x
1
3
1
3
ty1 4t
z1 9t
z1 9t
40. v
u
1(150 cos 330°)i
u
(150 sin 330°)j
u
v
u
2(50 cos 245°)i
u
(50 sin 245°)j
u
v
u
1v
u
2
(150 c
os 330
°50
cos 2
45°)2
(15
0 sin
330°
50 sin
245°)
2
162.2 km/h
tan v150 sin 330° 50 sin 245°
150 cos 330° 50 cos 245°
45. The slope is 1.
y1 1[x(3)]
y1 x3
xy4 0
46. The linear velocity of the belt around the larger
pulley is (120 rpm)
2
9
2
in./rev
1080
in./min. The linear velocity around the smaller
pulley must be the same, so its angular velocity is
(1080in./min)
2
1
r
3
ev
in.
180 rpm. The correct
choice is D.
Graphing Calculator Exploration:
Modeling with Parametric
Equations
Page 526
1. 408.7t418.3(t0.0083)
408.7t418.3t3.47189
9.6t3.47189
t
3.4
9
7
.
1
6
89
0.362 hr or 21.7 min
2. drt
408.7
3.4
9
7
.
1
6
89
147.8 mi
3. The time for plane 1 to fly 500 miles is
4
5
0
0
8
0
.7
. The
time for plane 2 is
4
5
1
0
8
0
.3
0.0083. Suppose the
speed of plane 1 is increased by amph.
408
5
.7
00
a
4
5
1
0
8
0
.3
0.0083
8-6B
Chapter 8 262
vtan1
47°534or 47°534south of east
41. 1, 33, 21(3) 3(2)
3
Since the inner product is not 0, the vectors are
not perpendicular.
42. Since A90°, ab, and absin A, no solution
exists.
43. Agraphing calculator indicates that there is one
real zero and that it is close to 1. f(1) 0, so the
zero is exactly 1.
44. x
3
2
y2
x2
3
2
y
2
3
x
4
3
y
y
2
3
x
4
3
150 sin 330° 50 sin 245°
150 cos 330° 50 cos 245°
408
5
.7
00
a
1
4
5
1
0
8
0
.3
0.0083
a408.7
500
4
5
1
0
8
0
.3
0.0083
6.7 mph
Modeling Motion Using
Parametric Equations
Page 531 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.
4. v
u
yv
u
sin v5. v
u
xv
u
cos v
50 sin 40° 20 cos 50°
32.14 ft/s 12.86 m/s
6. v
u
xv
u
cos vv
u
yv
u
sin v
45 cos 32° 45 sin 32°
38.16 ft/s 23.85 ft/s
7. v
u
xv
u
cos vv
u
yv
u
sin v
7.5 cos 20° 7.5 sin 20°
7.05 m/s 2.57 m/s
8-7
8a. 300 mph
52
m
8
i
0
le
ft
360
h
0s
440 ft/s
xtv
u
cos v
xt(440) cos 0°
x440t
ytv
u
sin v
1
2
gt2h
yt(440) sin 0°
1
2
(32)t23500
y16t23500
8b. Sample graph:
8c. 16t23500 0
16t23500
t2
35
1
0
6
0
t
35
1
0
6
0
t14.8 s
8d. x440t
440(14.8)
6512 ft
Pages 531–533 Exercises
9. v
u
xv
u
cos vv
u
yv
u
sin v
65 cos 60° 65 sin 60°
32.5 ft/s 56.29 ft/s
10. v
u
xv
u
cos vv
u
yv
u
sin v
47 cos 10.7° 47 sin 10.7°
46.18 m/s 8.73 m/s
11. v
u
xv
u
cos vv
u
yv
u
sin v
1200 cos 42° 1200 sin 42°
891.77 ft/s 802.96 ft/s
12. v
u
xv
u
cos vv
u
yv
u
sin v
17 cos 28° 17 sin 28°
15.01 ft/s 7.98 ft/s
13. v
u
xv
u
cos vv
u
yv
u
sin v
69 cos 37° 69 sin 37°
55.11 yd/s 41.53 yd/s
14. v
u
xv
u
cos vv
u
yv
u
sin v
46 cos 19° 46 sin 19°
43.49 km/h 14.98 km/h
15a. xtv
u
cos vytv
u
sin v
1
2
gt2
x175tcos 35° y175tsin 35° 16t2
15b. y0
175tsin 35° 16t20
t(175 sin 35° 16t) 0
175 sin 35° 16t0
175 sin 35° 16t
175 s
1
i
6
n35°
t
x175tcos 35°
175
175 s
1
i
6
n35°
cos 35°
899.32 ft or 299.77 yd
16. To find the time the projectile stays in the air, set
y0 and solve for t.
tv
u
sin v
1
2
gt20
t(v
u
sin v
1
2
gt) 0
v
u
sin v
1
2
gt 0
v
u
sin v
1
2
gt
2v
u
g
sin v
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 tin
the equation for x.
xtv
u
cos v
2v
u
g
sin v
v
u
cos v
v
u
2g
g
sin 2v
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. y300 when t7
7v
u
sin 78°
1
2
(32)72300
7v
u
sin 78° 784 300
7v
u
sin 78° 1084
v
u
7s
1
i
0
n
8
7
4
8°
v
u
158.32 ft/s
17b. x
1
3
tv
u
cos v50 yd
1
3
(7)(158.32) cos 78° 50
127 yd
18. xtv
u
cos v
v
u
x
cos v
t
ytv
u
sin v
1
2
gt2
y
v
u
x
cos v
v
u
sin v
1
2
g
v
u
x
cos v
2
yxtan v
2v
u
2
g
cos2v
The presence of the x2-term (due to the force of
gravity) means that yis 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 is v
u
, set y0 and solve for t.
tv
u
sin v
1
2
gt20
t
v
u
sin v
1
2
gt
0
v
u
sin v
1
2
gt 0
v
u
sin v
1
2
gt
2v
u
g
s
t
in v
t
To find the range, substitute this expression for t
in the equation for x.
xtv
u
cos v
2v
u
g
sin v
v
u
cos v
v
u
2
g
sin 2v
263 Chapter 8
y
x
4000
3000
2000
1000
Horizontal Distance
(feet)
2000 4000 6000
Height
(feet)
If the magnitude of the initial velocity is doubled
to 2v
u
, the range becomes
(2v
u
)
g
2sin 2v
or
4
v
u
2
g
sin 2v
. The projectile will travel four times as
far.
20a. 800 km/h
10
k
0
m
0m
360
h
0s
222.2 m/s
xtv
u
cos v
x222.2 tcos 45°
ytv
u
sin v
1
2
gt2
y222.2tsin 45°
1
2
(9.8)t2
y222.2tsin 45° 4.9t2
The negative coefficient in the t-term in the
equation for yindicates that the aircraft is
descending. The negative coefficient in the
equation for xis arbitrary.
20b. y222.2tsin 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.
42
2
3
.5
.4
s
m
169 m/s
or
169 m/s
10
k
0
m
0m
360
h
0s
608.4 km/h
21a. 70 mph
52
m
80
i
ft
360
h
0s
30
3
8
ft/s
y0
t
30
3
8
sin 35° 16t210 0
t
t
3.84 s
xtv
u
cos v
323.2 ft
21b. y8
t
30
3
8
sin 35° 16t210 8
16t2t
30
3
8
sin 35° 2 0
t
t3.71 s
xtv
u
cos v
312.4 ft
21c. From the calculations in part b, the time is
about 3.71s.
22a.
30
3
8
sin 35°
30
3
8
si
n 35°
2
4(1
6)2
3
3
08
sin 35°
30
3
8
s
in 35°
24
(16)
10
22b.
23a. y300 when t4.8
4.8v
u
sin 82°
1
2
(32)(4.8)2300
4.8v
u
sin 82° 368.64 668.64
v
u
4.8
66
s
8
in
.6
8
4
2°
v
u
140.7 ft/s
23b. x
1
3
tv
u
cos v100
131.3 yd
24a. x tv
u
cos vytv
u
sin v
1
2
gt2h
x155tcos 22° y155tsin 22° 16t23
24b. x420
155tcos 22° 420
t
155
4
c
2
o
0
s22°
y155tsin 22° 16t23
155
155
4
c
2
o
0
s22°
sin 22° 16
155
4
c
2
o
0
s22°
23
36.04 ft
Since 36.04 15, the ball will clear the fence.
24c. y0
155tsin 22° 16t23 0
t
t
3.68 s
xtv
u
cos v
528.86 ft
25. x11 t
x11 t
x11 t
y8 6t
y8 6(x11)
y6x58
26a. mg sin v300(9.8) sin 22°
1101.3 N
26b. mg cos v300(9.8) cos 22°
2725.9 N
27. cos A
1
2
7
1
.
.
4
9
Acos1
1
2
7
1
.
.
4
9
A37°
155 sin 22° (155 s
in 22°
)24
(16)
3
Chapter 8 264
y
x
O
t
28. 2(2xyz) 2(2)
x3y2z3.25
5xy0.75
1(2xyz) 1(2)
4x5yz2.5
6x 4y0.5
4(5xy) 4(0.75)
6x4y0.5
14x3.5
x
3
1
.
4
5
x0.25
5xy0.75 2xyz2
5(0.25) y0.75 2(0.25) 0.5 z2
y0.5 z1
29. 5232259
16
The correct choice is B.
Page 534 History of Mathematics
1.
7
3
°
6
1
0
2
°
3
7
6
.2
0
°
°
5
1
0
5
1
0
5000
x
stadia
x50(5000)
x250,000 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.
Transformation Matrices in
Three-Dimensional Space
Pages 539–540 Check for Understanding
1. Matrix Tmultiplies 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.
2. CC
u
8 6, 8 7, 2 3or 2, 1, 1
The matrix is
.
2
1
1
2
1
1
2
1
1
2
1
1
2
1
1
2
1
1
8-8
5a. BE
u
0 5, 2 5, 4 0or 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)
The matrix is
.
0
5
4
5
5
4
5
2
4
0
2
4
0
2
0
0
5
0
5
5
0
5
2
0
265 Chapter 8
3. VU
T, so the transformations
are the same.
4a-c.
0
0
1
0
1
0
1
0
0
Transformation Orientation Site Shape
Reflection yes no no
Translation no no no
Dilation no yes no
5b.
0 4
2 (1)
0 2
0 4
5 (1)
0 2
5 4
5 (1)
0 2
5 4
2 (1)
0 2
0 4
5 (1)
4 2
5 4
5 (1)
4 2
5 4
2 (1)
4 2
0 4
2 (1)
4 2
4
4
6
9
4
6
9
1
6
4
1
6
4
1
2
4
4
2
9
4
2
9
1
2
5c.
0
5
4
5
5
4
5
2
4
0
2
4
0
2
0
0
5
0
5
5
0
5
2
0
0
0
1
0
1
0
1
0
0
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
D
and
0
0
4
0
4
0
4
0
0
0
5
4
5
5
4
5
2
4
0
2
4
0
2
0
0
5
0
5
5
0
5
2
0
O
z
y
xA
C
B
D
E
F
G
H
T
.
2
0
0
2
0
0
2
0
0
2
0
0
2
0
0
2
0
0
2
0
0
2
0
0
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).
Pages 540–542 Exercises
7. FB
u
3 3, 1 7, 4 4or 0, 6, 0
A(2, 3, (6), 2) A(2, 3, 2
C(4, 7 (6), 1) C(4, 1, 1)
The matrix is
.
8. AH
u
4 (3), 1 (2), 2 2or 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)
The matrix is
9. CF
u
6 4, 0 (1), 0 2or 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)
The matrix is
.
6
0
0
3
1
2
4
1
1
4
1
2
1
0
4
2
2
3
4
1
2
4
1
2
4
2
2
4
2
2
3
2
2
3
1
2
3
1
2
3
2
2
3
7
4
2
3
2
4
7
1
4
1
1
3
1
4
2
3
2
The result is a translation of 2 units along the y-
axis and 4 units along the z-axis.
11.
0 1
0 (2)
4 (2)
0 1
3 (2)
5 (2)
0 1
3 (2)
2 (2)
0 1
0 (2)
1 (2)
Chapter 8 266
A
B
G
H
C
D
E
F
z
x
A
B
G
H
C
D
E
F
10.
0 0
0 (2)
4 4
0 0
3 (2)
5 4
0 0
3 (2)
2 4
0 0
0 (2)
1 4
2 0
3 (2)
2 4
2 0
3 (2)
5 4
2 0
0 (2)
4 4
2 0
0 (2)
1 4
2
1
6
2
1
9
2
2
8
2
2
5
0
2
8
0
1
9
0
1
6
0
2
5
O
z
y
x
A
C
B
D
E
F
G
H
2 1
3 (2)
2 (2)
2 1
3 (2)
5 (2)
2 1
0 (2)
4 (2)
2 1
0 (2)
1 (2)
The result is a translation of 1 unit along the x-
axis, 2 units along the y-axis, and 2 units
along the z-axis.
12.
0 1
0 5
4 (3)
0 1
3 5
5 (3)
0 1
3 5
2 (3)
0 1
0 5
1 (3)
3
1
0
3
1
3
3
2
2
3
2
1
1
2
2
1
1
3
1
1
0
1
2
1
2 1
3 5
2 (3)
2 1
3 5
5 (3)
2 1
0 5
4 (3)
2 1
0 5
1 (3)
The results is a translation of 1 unit along the x-
axis, 5 units along the y-axis, and 3 units along
the z-axis.
3
8
1
3
8
2
3
5
1
3
5
2
1
5
1
1
8
2
1
8
1
1
5
2
z
y
x
A
E
F
D
C
G
H
B
z
y
A
E
F
D
C
G
H
B
x
13.
2
3
2
2
3
5
2
0
4
2
0
1
0
0
4
0
3
5
0
3
2
0
0
1
0
0
1
0
1
0
1
0
0
18.
0
0
1
0
1
0
1
0
0
0
0
0.75
0
0.75
0
0.75
0
0
267 Chapter 8
The transformation does not change the figure.
14.
2
3
2
2
3
5
2
0
4
2
0
1
0
0
4
0
3
5
0
3
2
0
0
1
0
0
1
0
1
0
1
0
0
2
3
2
2
3
5
2
0
4
2
0
1
0
0
4
0
3
5
0
3
2
0
0
1
The transformation results in reflections over the
xy and xz-planes.
15.
2
3
2
2
3
5
2
0
4
2
0
1
0
0
4
0
3
5
0
3
2
0
0
1
0
0
1
0
1
0
1
0
0
2
3
2
2
3
5
2
0
4
2
0
1
0
0
4
0
3
5
0
3
2
0
0
1
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.
17.
, so the
figure is three times the original size and reflected
over the xy-plane.
0
0
3
0
3
0
3
0
0
0
0
1
0
1
0
1
0
0
0
0
3
0
3
0
3
0
0
2
3
2
2
3
5
2
0
4
2
0
1
0
0
4
0
3
5
0
3
2
0
0
1
z
y
x
A
E
F
D
C
G
H
B
z
y
x
A
O
F
E
D
H
B
G
C
33
3
6
9
12
69
3
6
69
z
y
x
A
E
F
D
C
G
H
B
, so the figure is three-fourths
the original site and reflected over all three
coordinate planes.
19a.
, so the transformation can
x
y
z
0
0
5
0
2
0
2
0
0
2x
2y
5z
0
0
0.75
0
0.75
0
0.75
0
0
be represented by the matrix
.
19b. The transformation will magnify the x- and
y-dimensions two-fold, and the z-dimension
5-fold.
20a.
23.6
72
0
23.6
72
0
23.6
72
0
23.6
72
0
23.6
72
0
0
0
5
0
2
0
2
0
0
20b.
247 23.6
74 72
59 0
136 23.6
71 72
53 0
20 23.6
58 72
27 0
351 23.6
62 72
52 0
302 23.6
83 72
37 0
20c. The result is a translation 23.6 units along the
x-axis and 72 units along the y-axis.
21. The matrix
would reflect the prism
0
0
1
0
1
0
1
0
0
374.6
10
52
325.6
11
37
270.6
2
59
159.6
1
53
43.6
14
27
over the yz-plane. The matrix
would reduce its dimensions by half.
0
0
0.5
0
0.5
0
0.5
0
0
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.
Sample answer:
22b. Sample graphs:
1
0
1
0
1
0
1
0
0
0
0
0.5
0
0.5
0
0.5
0
0
0
0
1
0
1
0
1
0
0
0
0
0.5
0
0.5
0
0.5
0
0
A
D
C
GF
E
B
H
z
y
x
A
D
B
H
G
C
F
E
z
y
x
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 Tchanges the x-coordinate of
(x, y, z) to x2z.)
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 x4z, use a third row of 1 0 4.
The matrix is
.
25a. The x-coordinates are unchanged, the
y-coordinates increase, and the z-coordinates
decrease, so the movement is dip-slip.
25b.
36.4
123.9
85.3
129.4
97.1
166.4
73.8
82.6
212.0
201.7
28.3
261.5
41.3
145.8
246.6
123.9
88.0
205.3
0
0
4
0
2
0
3
0
1
29. 80x380x280x24.2
80x380x280x24.2 0
Agraphing calculator indicates that there is a
solution between 0 and 1. By Descartes’ Rule of
Signs, it is the only solution. When x0.2, 80x3
80x280x24.2 4.36 and when x0.3,
80x380x280x24.2 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 x2y.
I. x2y4II. y4
III. x2y2IV.x2y
8
3
Only I and II are equivalent, so the correct choice
is A.
Chapter 8 Study Guide and Assessment
Page 543 Understanding and Using the
Vocabulary
1. resultant 2. unit
3. magnitude 4. cross
5. inner 6. vector
7. parallel 8. standard
9. direction 10. components
544–546 Skills and Concepts
11. 1.3 cm, 50° 12. 2.9 cm, 10°
13.
4.1 cm, 23°
14.
5.3 cm, 25°
15.
2.5 cm, 98°
Chapter 8 268
36.4
125.5
84.1
129.4
95.5
165.2
73.8
84.2
213.2
201.7
29.9
262.7
41.3
144.2
247.8
123.9
86.4
206.5
26a. La Shawna Jaimie
x0x35t
y16t2150 y16t2150
16t2150 0
150 16t2
1
1
5
6
0
t2
1
1
5
6
0
t
3.06 t
x35t
35
1
1
5
6
0
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. x5t1
x1 5t
x
5
1
t
y2t10
y2
x
5
1
10
y
2
5
x
4
5
8
28. sec
cos1
2
5
1
cos
cos1
2
5
0
1.6
1.2
0
1.6
1.2
0
1.6
1.2
0
1.6
1.2
0
1.6
1.2
0
1.6
1.2
1
2
5
5
2
4.1 cm
p
q
p
q
23˚
2.5 cm
3p
q
3p
q
98˚
5.3 cm
2p
q
2p
q
25˚
16.
3.5 cm, 82°
17. h1.3 cos 50° v1.3 sin 50°
h0.8 cm v1 cm
18. h2.9 cos 10° v2.9 sin 10°
h2.9 cm v0.5 cm
19. CD
u
7 2, 15 3or 5, 12
CD
u
521
22
169
or 13
20. CD
u
4 (2), 12 8or 6, 4
CD
u
624
2
52
or 213
21. CD
u
0 2, 9 (3)or 2, 12
CD
u
(2)2
122
148
or 237
22. CD
u
5 (6), 4 4or 1, 8
CD
u
12(
8)2
65
23. u
u
v
u
w
u
u
u
2, 53, 1
u
u
2 3, 5 (1)or 5, 6
24. u
u
v
u
w
u
u
u
2, 53, 1
u
u
2 3, 5 (1)or 1, 4
25. u
u
3v
u
2w
u
u
u
32, 523, 1
u
u
6, 156, 2
u
u
6 6, 15 (2)or 12, 17
26. u
u
3v
u
2w
u
u
u
32, 523, 1
u
u
6, 156, 2
u
u
6 6, 15 (2)or 0, 13
27. EF
u
6 2, 2 (1), 1 4or 4, 1, 3
EF
u
42(
1)2
(3)2
26
28. EF
u
1 9, 5 8, 11 5or 10, 3, 6
EF
u
(10)
2(
3)2
62
145
29. EF
u
2 (4), 1 (3), 7 0) or 6, 2, 7
EF
u
622
272
89
30. EF
u
4 3, 0 7, 5 (8)or 7, 7, 13
EF
u
(7)2
(7
)21
32
267
31. u
u
2w
u
5v
u
u
u
24, 1, 551, 7, 4
u
u
8, 2, 10(5, 35, 20
u
u
8 (5), 2 35, 10 (20)
u
u
13, 37, 30
32. u
u
0.25v
u
0.4w
u
u
u
0.251, 7, 40.44, 1, 5
u
u
0.25, 1.75, 11.6, 0.4, 2
u
u
0.25 1.6, 1.75 (0.4), 1 2
u
u
1.35, 1.35, 1
33. 5, 12, 65(2) (1)6
10 6
16; no
34. 2, 63, 42(3) 6(4)
6 24
18; no
35. 4, 1, 23, 4, 44(3) 1(4) (2)4
12 4 8
0; yes
36. 2, 1, 46, 2, 12(6) (1)(2) 4(1)
12 2 4
18; no
37. 5, 2, 102, 4, 4
5(2) 2(4) (10)(4)
10 8 40
42; no
i
u
j
u
k
u
38. 5, 2, 51, 0, 3
525
103
u
u
j
u
k
u
6i
u
10j
u
2k
u
or 6, 10, 2
6, 10, 25, 2, 5
6(5) 10(2) (2)(5)
30 20 10 0
6, 10, 21, 0, 3
6(1) 10(0) (2)(3)
6 0 6 0
39. 2, 3, 12, 3, 4
i
u
j
u
k
u
231
234
i
u
j
u
k
u
9i
u
6j
u
0k
u
or 9, 6, 0
9, 6, 02, 3, 1
9(2) (6)(3) 0(1)
18 18 0 0
9, 6, 02, 3, 4
9(2) (6)(3) 0(4)
18 18 0 0
3
3
2
2
1
4
2
2
1
4
3
3
2
0
5
1
5
3
5
1
25
03
269 Chapter 8
3.5 cm
4p
q
4p
q
82˚
40. 1, 0, 45, 2, 1
i
u
j
u
k
u
10 4
52 1
i
u
j
u
k
u
8i
u
19j
u
2k
u
or 8, 19, 2
8, 19, 21, 0, 4
(8)(1) 19(0)(2)(4)
8 0 8 0
8, 19, 25, 2, 1
(8)(5) 19(2) (2)(1)
40 38 2 0
41. 7, 2, 12, 5, 3
i
u
j
u
k
u
721
253
i
u
j
u
k
u
2
5
7
2
1
3
7
2
1
3
2
5
0
2
1
5
4
1
1
5
4
1
0
2
47. x4, y0t3, 6
x4, yt3, 6
x4 3ty6t
x4 3t
Chapter 8 270
i
u
19j
u
31k
u
or 1, 19, 31
1, 19, 317, 2, 1
1(7) (19)2 31(1)
7 (38) 31 0
1, 19, 312, 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)
xy
u
4 1, 2 2, 1 3or 5, 0, 4
yz
u
5 (4), 3 2, 0 (1)or 9, 5, 1
5, 0, 49, 5, 1
i
u
j
u
k
u
504
951
i
u
j
u
k
u
20i
u
31j
u
25k
u
or 20, 31, 25
43. F
u
1320i
u
F
u
2260j
u
F
u
1F
u
23202
2602
412.31 N
tan v
2
3
6
2
0
0
or
1
1
3
6
vtan1
1
1
3
6
39.09°
44. v
u
112j
u
v
u
2(30 cos 116°)i
u
(30 sin 116°)j
u
v
u
1v
u
2(30 co
s 116°
)2(1
2 3
0 sin
116°)2
41 m/s
tan v
12
3
0
3
c
0
os
s
1
in
16
1
°
16°
vtan1
12
3
0
3
c
0
os
s
1
in
16
1
°
16°
108.65°
45. x3, y(5)t4, 2
x3, y5t4, 2
x3 4ty5 2t
x3 4ty5 2t
46. x(1), y9t7, 5
x1, y9t7, 5
x1 7ty9 5t
x1 7ty9 5t
0
5
5
9
4
1
5
9
4
1
0
5
48. xt
y8t7
7 8t
49. xt
y
1
2
t
5
2
50. v
u
xv
u
cos vv
u
yv
u
sin v
15 cos 55° 15 sin 55°
8.60 ft/s 12.29 ft/s
51. v
u
xv
u
cos vv
u
yv
u
sin v
13.2 cos 66° 13.2 sin 66°
5.37 ft/s 12.06 ft/s
52. v
u
xv
u
cos vv
u
yv
u
sin v
18 cos 28° 18 sin 28°
15.89 m/s 8.45 m/s
53. CH
u
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
.
The matrix for the translated figure is
.
The figure moves 2 units along the x-axis and 3
units along the z-axis.
2
2
5
2
2
2
2
4
2
2
4
5
5
4
5
5
4
2
5
2
2
5
2
5
4
2
2
4
2
1
4
4
1
4
4
2
3
4
2
3
4
1
3
2
1
3
2
2
z
y
x
F
E
H
A
B
C
D
G
54.
0
0
1
0
1
0
1
0
0
58. F
u
190i
F
u
2(70 cos 30°)i
u
(70 sin 30°)j
u
or
353
i
u
35j
u
F
u
1F
u
2(90
353
)23
52
154.6 N
tan vor 7
18 73
35
90 353
271 Chapter 8
4
2
2
4
2
1
4
4
1
4
4
2
3
4
2
3
4
1
3
2
1
3
2
2
3 km/h
16 km/h
250 m
35˚
vtan1
13.1°
Page 547 Open-Ended Assessment
1a. Sample answer: X(4, 1), Y(1, 1)
XY
u
1 4, 1 (1)or 3, 2
1b. XY
u
(3)2
22
or 13
The magnitude of XY
u
only depends on the
differences of the coordinates of Xand Y, not the
actual coordinates.
2a. Sample answer: P(1, 1), Q(3, 3), R(3, 1), S(5, 3)
PQ
u
3 1, 3 1or 2, 2
RS
u
5 3, 3 1or 2, 2
PQ
u
and RS
u
are parallel because they have the
same direction. In fact, they are the same vector.
2b. Sample answer: a
u
8, 4, b
u
3, 6
a
u
b
u
8(3) (4)6 or 0
a
u
and b
u
are perpendicular because their inner
product is 0.
Chapter 8 SAT & ACT Preparation
Page 549 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.
5232b2
25 9 b2
16 b2
4 b
The height is 4.
Use the formula for the area of a triangle.
A
1
2
bh
A
1
2
(4)(3) or 6
Since the parallelogram consists of two triangles,
the area of the parallelogram is 2 6 or 12. The
correct choice is A.
7
18 73
The figure is reflected over the xz-plane.
Page 547 Applications and Problem Solving
55. AB
u
1 cos 120°, 0, 1 sin 120°or
1
2
, 0,
2
3
F
u
0, 0, 50
T
u
AB
u
F
u
uu u
ij k
1
2
0
2
3
0050
i
u
j
u
k
u
0i
u
25j
u
0k
u
or 0, 25, 0
T
u
02(
25)2
02
25 lb-ft
56. ytv
u
sin v
1
2
gt2h
0.5(38) sin 40°
1
2
(32)(0.5)22
10.2 ft
57a.
b
u
(16 cos 55°)c
u
(16 sin 55°)j
u
c
u
3j
u
b
u
c
u
(16 co
s 55°)
2(1
6 sin
55°
3)2
13.7 km/h
57b.
2
u
50
16
1
s
6
in
co
5
s
5
5
°
5
°
3
u250
16
1
s
6
in
co
5
s
5
5
°
5
°
3
u275.3 m
0
0
1
2
0
2
3
50
1
2
0
2
3
50
0
0
4
2
2
4
2
1
4
4
1
4
4
2
3
4
2
3
4
1
3
2
1
3
2
2
1 ft
50 lb 60˚
z
y
x
A
C
B
D
E
F
G
H
O
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 (xh)2(yk)2r2, where the center
is (h, k) and the radius is r. From the coordinates
of points Aand 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
.
(x4)2(y0)2(22
)2
(x4)2y24(2)
(x4)2y28
The correct choice is B.
3. Write the equation for the perimeter of a
rectangle. then replace xwith its value in terms
of y. Solve the equation for y.
p2x2y
p2
2
3
y
2y
p
4
3
y2y
p
1
3
0
y
3
1
p
0
y
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 x80
x40
Since xmust be greater than 40, xcannot 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.
3x2
9x3
x
2
3
x
3
9
x
2
3
x
1
3
x(
2
3
1
3
)
x1or x
The correct choice is E.
6. This figure looks more complex than it is. A semi-
circle 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.
Large semi-circle area
1
2
32
9
2
Medium semi-circle area
1
2
22
4
2
Small semi-circle area
1
2
12
1
2
Shaded area
9
2
4
2
1
2
6
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) ,
the denominator is only 0 when x– 1 0 or x1.
The correct choice is D.
8. Start by sketching a diagram of the counter
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
m
–
6
9
or –
2
3
The point-slope form of the line is
y– 6 –
3
2
(x– –2).
y– 6 –
3
2
x– 3
y–
3
2
x3
So the y-intercept of the line is 3.
The correct choice is B.
–3 – 6
4 – –2
x2–3x2
x– 1
Chapter 8 272
40
28 38
30
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
2
(AC)(AB)
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
2
(AC)(AB)
1
2
(CE)(ED)
1
2
(AC)2
1
2
(CE)2
1
2
[(AC)2(CE)2]
Using the Pythagorean Theorem for ACE, you
know that (AC)2(CE)2(AE)2or 1.
So the sum of the two areas is
1
2
(1)
1
2
. You can
grid the answer either as .5 or as 12.
273 Chapter 8
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
C
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
D
Chapter 9 274
Chapter 9 Polar Coordinates and Complex Numbers
1234
0˚
330
˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
A
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
B
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
5101520
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
E
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
F
11. 12.
13.
14. P1P22.52
(3)2
2(2
.5)(3
) cos
4
6
6.25
9 1
5 cos
5
1
2
5.25
15 co
s
5
1
2
4.37
15a.
15b. 210 (30) 240
A
3
N
60
(r2)
2
3
4
6
0
0
(202)
838 ft2
Pages 558–560 Exercises
16. 17.
Polar Coordinates
Pages 557–558 Check for Understanding
1. There are infinitely many ways to represent the
angle v. Also, rcan be positive or negative.
2. Draw the angle vin standard position. Extend the
terminal side of the angle in the opposite
direction. Locate the point that is runits from
the pole along this extension.
3. Sample answer: 60° and 300°
Plot (4, 120) such that vis in standard position
and ris 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.
r4
v120 180 or v120 180
60 300
4. The points 3 units from the origin in the opposite
direction are on the circle where r3.
5. All ordered pairs of the form (r, v) where r0.
6. 7.
8. 9.
10. Sample answer:
2,
13
7
,
2,
25
6
,
2,
7
6
,
2,
19
6
(r, v2k)
→
2,
6
2(1)
→
2,
13
6
→
2,
6
2(2)) →
2,
25
6
(r, v(2k1))
→
2,
6
(1)
→
2,
7
6
→
2,
6
(3)
→
2,
19
6
9-1
275 Chapter 9
2468
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
G
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
H
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
J
12340˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
K
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
L
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
M
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
N
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
P
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
Q
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
R
18. 19.
20. 21.
22. 23.
24. 25.
26. 27.
28. Sample answer:
2,
3
,
2,
7
3
, (2, 240°),
(2, 600°)
(r, v2k)
→
2,
3
2(0)
→
2,
3
→
2,
3
2(1)
→
2,
7
3
(r, v(2k1)180°)
→(2, 60° (1)180°) →(2, 240°)
→(2, 60° (3)180°) →(2, 600°)
12340˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234 1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
29. Sample answer: (1.5, 540°), (1.5, 900°), (1.5, 0°),
(1.5, 360°)
(r, v360k°)
→(1.5, 180° 360(1)°) →(1.5, 540°)
→(1.5, 180° 360(2)°) →(1.5, 900°)
(r, v(2k1)180°)
→(1.5, 180° (1)180°) →(1.5, 0°)
→(1.5, 180° (1)180°) →(1.5, 360°)
30. Sample answer:
1,
7
3
,
1,
13
3
,
1,
4
3
,
1,
10
3
(r, v2k)
→
1,
3
2(1)
→
1,
7
3
→
1,
3
2(2)→
1,
13
3
(r, v(2k1))
→
1,
3
(1)
→
1,
4
3
→
1,
3
(3)
→
1,
10
3
31. Sample answer: (4, 675°), (4, 1035°), (4, 135°),
(4, 495°)
(r, v360k°)
→(4, 315 360(1)°) →(4, 675°)
→(4, 315 360(2)°) →(4, 1035°)
(r, v(2k1)180°)
→(4, 315 (1)180°) →(4, 135°)
→(4, 315 (1)180°) →(4, 495°)
32. 33.
34. 35.
36. 37.
Boat
3 mph
8 mph
Chapter 9 276
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
r
0
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
2468
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
49a. When v120°, r17. The maximum speed at
120° is 17 knots.
49b. When v150°, r13. The maximum speed at
150° is 13 knots.
50a.
50b.
3
3
2
3
or 120°
Let R3 100 or 300 and let r0.25 100
or 25.
A
3
N
60
(R2)
3
N
60
(r2)
1
3
2
6
0
0
((3002)
1
3
2
6
0
0
((25)2)
1
3
2
6
0
0
(90,000 625)
93,593 ft2
If each person’s seat requires 6 ft2of space,
there are
93,
6
593
or 15,599 seats.
51. The distance formula is symmetric with respect to
(r1, v1) and (r2, v2). That is,
r22
r12
2r2r1
cos (v1
v2)
r12
r22
2r1r2
cos[
(v2
v1)]
r12
r22
2r1r2
cos(v2
v1)
52a.
52b. P1P2526
22(
5)(6) c
os (34
5° 3
10°)
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
3
8
sin1(sin v) sin1
3
8
v22.0°
54. 3, 2, 41, 4, 0(3)(1) (2)(4) (4)(0)
3 8 0
11
No, the vectors are not perpendicular because
their inner product is not 0.
38. 39.
40.
41. r2
or r2
for any v.
42. P1P2426
22(
4)(6) c
os (10
5° 1
70°)
16
36
48 cos
(65
°)
52
48 cos
(65
)
5.63
43. P1P2125
2 2(
1)(5) c
os
3
4
6
1 2
5 1
0 cos
7
1
2
26
10 cos
7
1
2
5.35
44. P1P2
(2.5)2
(1
.75)2
2(2.5
)(1.75
)cos
2
5
8
6.25
30.06
25
8.75 c
os
2
4
1
0
9.3125
8.75
cos
2
4
1
0
3.16
45. P1P21.32
(3.6
)22
(1.3)(
3.6) co
s (62
°(
47°))
1.69
12.96
9.3
6 cos
(62°
47
°)
14.65
9.3
6 cos
(15°
)
4.87
46. r
(3)2
42
5
sin v
4
5
, v53°
180° 53° 127°
Sample answer: (5, 127°)
47. There are 360° in a circle. If the circle is cut into 6
equal pieces, each slice measures
36
6
0
or 60°.
Beginning at the origin, the equation of the first
line is v0°. The equation of the next line,
rotating counterclockwise, is v0 60 or 60°.
The equation of the last line is v60 60 or
120°. Note that lines extend through the origin, so
3 lines create 6 pieces.
48. P1P2 r12
r22
2r1r2
cos (v
v)
r12
r22
2r1r2
cos 0
r12
r22
2r1r2
(r1r
2)2
r1 r2
55. Rewrite y9x3 as 9xy3 0.
d
Ax1By1C
A2
B2
63.
2
4
(1)
2(5) 4(5) 1(1)
1
4
1
3
0
5
4
3
0
5
1
4
1
0
5
4
1
4
2
1
3
277 Chapter 9
9(3) (1)(2) (3)
92(
1)2
32
82
82
82
32
82
3282
82
12
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
4628
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
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. A
B
is a
diameter.
Cd
50d
50 d
The correct choice is E.
Graphs of Polar Equations
Page 565 Check for Understanding
1. Sample answer: rsin 2v
The graph of a polar equation whose form is
racos nvor asin nv, where nis a positive
integer, is a rose.
2. 1 sin v1 for any value of v. Therefore, the
maximum value of r3 5 sin vis r3 5(1)
or 8. Likewise, the minimum value of r3 5
sin vis r3 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 vis not
always sufficient. For example, the interval 0 v
only generates two of the four petals for the
rose rsin 2v. rsin
2
v
is an example where
values of vfrom 0 to 4would have to be
considered.
5. 6.
cardioid limaçon
9-2
Distance is always positive.
56.
1
sin
s
2
in
a
2a
sin
1
2a
1
csc2a1
cot2a
57. Arc cos 30°
In a 30°-60°-90° right
triangle, the angle opposite
the smallest leg is 30°.
58. y5 cos 4v
Amplitude 5; Period
2
4
or
2
59. bsin A18.6 sin 30°
9.3
Since absin A, there is one solution.
Find B.Find C.
s
1
i8
n.6
B
sin
9.3
30°
C180° 90° 30°
18.6 sin 30° 9.3 sin B60°
18.69
s.
i
3
n30°
sin B
90° B
Find c.
sinc60°
sin
9.3
30°
csin 30° 9.3 sin 60°
c
9.3
sin
sin
30
6
°0°
c16.1
60. 3 or 1 positive
f(x) x34x24x1
0 negative
Q
P
1
Since there are only positive real zeros, the only
rational real zero is 1.
61. x3
x 5x
2
2
x
3
x
2
5
x
3x3
3
x
1
5
10
As x→,
x1
05
→0. Therefore, the slant
asymptote is yx3.
62. y-axis:
For x: f(x) x43x22
For x: f(x) (x)43(x)22
x43x22
So, in general, point (x, y) is on the graph if and
only if (x, y) is on the graph.
3
2
1682
41
2
1
3
Chapter 9 278
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
10
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
12963
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2
1
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
812164
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
2468 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
10 15 2051234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
12
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
12
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
81216446820˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
23410
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2341
7. 8.
rose spiral of Archimedes
9.
2 sin 2 cos 2
sin cos 2
sin 1 2 sin2
2 sin2sin 1 0
(2 sin 1)(sin 1) 0
2 sin 1 0orsin 1 0
sin
1
2
sin 1
6
or
5
6
or
3
2
If
6
or
5
6
is substituted in either original
equation, r1. If
3
2
is substituted in either
original equation, r2. The solutions are
1,
6
,
1,
5
6
, and
2,
3
2
.
10a.
10b. Sample answer: 0 v
14
3
Begin at the origin and “spiral” twice around it,
or through 4radians. Move straight up
through 4
2
or
9
2
radians. Now move to the
left slightly, through approximately
9
2
6
or
14
3
radians.
Pages 565–567 Exercises
11. 12.
circle cardioid
13. 14.
spiral of Archimedes lemniscate
15. 16.
rose rose
17. 18.
Spiral of Archimedes limaçon
19. 20.
cardioid lemniscate
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)
[2, 2] scl1 by [2, 2] scl1
29. (2, 3.5), (2, 5.9)
[6, 6] scl:1 by [6, 6] scl1
30. (3.6, 0.6), (2.0, 4.7)
[4, 4] scl1 by [4, 4] scl1
31a. If the lemniscate is 6 units from end to end, then
a
1
2
(6) or 3.
r29 cos 2or r29 sin 2
31b. If the lemniscate is 8 units from end to end, then
a
1
2
(8) or 4.
r216 cos 2or r216 sin 2
32.
This microphone will pick up more sounds from
behind than the cardioid microphone.
33. 0 v4: Begin at the origin and curl around
once, or through 2radians. Curl around a second
time and go through 22or 4radians.
34. All screens are [1, 1] scl1 by [1, 1] scl1
279 Chapter 9
4321 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2341
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
12
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
12
21. 22.
rose cardioid
23. Sample answer: rsin 3v
The graph of a polar equation of the form racos
3vor rasin 3vis a rose with 3 petals.
24. Sample answer: r
2
v
4
a
2
r
1
2
1
2
ar
2
25. 3 2 cos
1 cos
0
The solution is (3, 0)
26. 1 cos 1 cos
2 cos 0
cos 0
2
or
3
2
Substituting each angle
into either of the original
equations gives r1, so
the solutions of the system
are
1,
2
and
1,
3
2
.
27.
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
cos
1
2
0 or or
3
or
5
3
If 0 or is substituted in either original
equation, r0. If
3
or
5
3
is substituted in
either original equation, r3
or r3
,
respectively. The solutions are (0, 0), (0, ),
3
,
3
, and
3
,
5
3
.
2468
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
10˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
34a. rcos
2
rcos
4
rcos
6
rcos
8
When n10, two more outer rings will appear.
34b. rcos
3
rcos
5
rcos
7
rcos
9
When n11, the innermost loop will be on the
left and there will be an additional outer ring.
35. Sample answer: r1 sin v
Aheart resembles the shape of a cardioid. The
sine function orients the heart so that the axis of
symmetry is along the y-axis. If a1, the heart
points in the right direction.
36a. For a limaçon to go back on itself and have an
inner loop, rmust change sign. This will happen
if ba.
36b. For the other two cases, ab.
Experimentation shows that the dimple
disappears when a2b, so there is a
dimple if ba2b.
36c. For this remaining case, there is neither an
inner loop nor a dimple if a2b.
37a. Subtracting afrom vrotates the graph
counterclockwise by an angle of a.
37b. Multiplying vby 1 reflects the graph about the
polar axis or x-axis.
37c. Multiplying the function by 1 changes rto its
opposite, so the graph is reflected about the
origin.
37d. Multiplying the function by cresults in a
dilation by a factor of c. (Points on the graph
move closer to the origin if c1, or farther
away from the origin if c1.)
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, (2k1)180°)
→(4, 45° (1)180°) →(4, 135°)
→(4, 45° (1)180°) →(4, 225°)
Chapter 9 280
39. v
w
k
0
4
j
3
2
i
2
1
3. x2
rcos 2
r
co
2
s
r2 sec
4. To convert from polar coordinates to rectangular
coordinates, substitute rand vinto the equations
xrcos vand yrsin v. To convert from
rectangular coordinates to polar coordinates, use
the equation r
x2y
2
to find r. If x0, v
Arctan
y
x
. If x0, vArctan
y
x
. If x0, you
can use
2
or any coterminal angle for v.
5. r
(2
)2(
2
)2
vArctan
2
2
4
or 2
3
4
2,
3
4
6. r (2)2
(5
)2
vArctan
5
2
29
5.39 4.33
(5.39, 4.33)
7. x2 cos
4
3
y2 sin
4
3
13
(1, 3
)
8. x2.5 cos 250° y2.5 sin 250°
0.86 2.35
(0.86, 2.35)
9. y2
rsin v2
r
si
2
nv
r2 csc v
10. x2y216
(rcos v)2(rsin v)216
r2(cos2vsin2v) 16
r216
r4 or r4
11. r6
x2y
2
6
x2 y236
12. rsec v
r
r
rc
o
1
sv
1
x1
x1
281 Chapter 9
i
j
k
12i
8j
7k
12, 8, 7
2, 3, 012, 8, 724 (24) 0 or 0
1, 2, 412, 8, 712 (16) 28 or 0
40. 3.5 cm, 87°
41. tan2x
sin2x
cos4xcos2xsin2x
3
2
2
1
0
4
2
1
0
4
3
2
tan2x
sin2x
cos2x(cos2xsin2x)
tan2x
sin2 x
(cos2x)(1)
tan2x
tan2xtan2x
42. Find C.
C180° 21°1549°40
109°5
Find b.
sin 4
b
9°40
sin2
1
8
0
.9
9°5
bsin 109°528.9 sin 49°40
b
b23.3
Find a.
sin 2
a
1°15
sin2
1
8
0
.9
9°5
asin 109°528.9 sin 21°15
a
a11.1
43. NY LA Miami
Bus
Train
44.
1
8
6
4
1
8
1
8
2
1
8
3
So
1
8
3
1
3
6
2
3
6
The correct choice is A.
Polar and Rectangular
Coordinates
Page 571 Check for Understanding
1. Sample answer: (22
, 45°)
r
222
2
Arctan
2
2
8
45°
22
2. The quadrant that the point lies in determines
whether vis given by Arctan
y
x
or Arctan
y
x
.
9-3
1
8
3
1
3
6
1
8
6
4
1
3
6
$260
$426
$199
$322
$240
$254
28.9 sin 21°15
sin 109°5
28.9 sin 49°40
sin 109°5
sin2x
cos2x
y
x
y
r
sin
x
r
cos
r
O
13a.
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.
Pages 572–573 Exercises
14. r
22(
2)2
vArctan
2
2
8
or 22
4
Add 2to obtain v
7
4
.
22
,
7
4
15. r
021
2
1
or 1
Since x0 when y1, v
2
.
1,
2
16. r
12(
3
)2
vArctan
4
or 2
3
2,
3
17. r
1
4
2
4
3
2
vArctan
43
1
4
3
1
22. x1 cos
6
y1 sin
6
1
23
1
1
2
2
3
1
2
2
3
,
1
2
23. x2 cos 270° y2 sin 270°
02
(0, 2)
24. x4 cos 210° y4 sin 210°
4
2
3
4
1
2
23
2
(23
, 2)
25. x14 cos 130° y14 sin 130°
9.00 10.72
(9.00, 10.72)
26. x7
rcos v7
r
c
os
7
v
r7 sec v
27. y5
rsin v5
r
si
5
nv
r5 csc v
28. x2y225
(rcos v)2(rsin v)225
r2(cos2vsin2v) 25
r225
r5 or r5
29. x2y22y
(rcos v)2(rsin v)22rsin v
r2(cos2vsin2v) 2rsin v
r22rsin v
r2 sin v
30. x2y21
(rcos v)2(rsin v)21
r2(cos2vsin2v) 1
r2(cos 2v) 1
r2
cos
1
2v
r2sec 2v
31. x2(y2)24
x2y24y4 4
(r cos v)2(r sin v)24r sin v0
r2(cos2vsin2v) 4r sin v0
r24r sin v0
r24r sin v
r4 sin v
32. r2
x2y
2
2
x2y24
33. r3
x2y
2
3
x2y29
Chapter 9 282
4321 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
1
4
6
Arctan
1
3
or
4
3
2
4
or
1
2
1
2
,
4
3
18. r
328
2
vArctan
8
3
73
8.54 1.21
(8.54, 1.21)
19. r
42(
7)2
vArctan
47
65
8.06 1.05
Add 2to obtain v5.23.
(8.06, 5.23)
20. x3 cos
2
y3 sin
2
03
(0, 3)
21. x
1
2
cos
3
4
y
1
2
sin
3
4
1
2
2
2
1
2
2
2
4
2
42
42
,
42
34. v
3
Arctan
y
x
3
y
x
13
y3
x
35. r2 csc v
r
r
rs
2
in v
1
2
y
y2
36. r3 cos v
r23r cos v
x2y23x
37. r2sin 2v8
r22 sin v cos v8
2r sin v r cos v8
2yx 8
xy 4
38. yx
y
x
1
Arctan
y
x
Arctan 1
v
4
39. rsin v
r2rsin v
x2y2y
40. x325 cos 70° y325 sin 70°
111.16 305.40
(111.16, 305.40)
41.
5
2
4
2
4
4
—
6
5
4
—
6
283 Chapter 9
y
x
3
2
1
3
6
4
2
4
0.52 unit
42. Drop a perpendicular from the point with polar
coordinates (r, v) to the x-axis. ris the length of
the hypotenuse in the resulting right triangle.
xis the length of the side adjacent to angle v, so
cos v
x
r
. Solving for xgives xrcos v. yis the
length of the side opposite angle v, so sin v
y
r
.
Solving for ygives yrsin 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.)
y
x
(
r
, )
r
O
O
y
x
2120˚
43. horizontal distance:
25(4 2 cos 120°) 75 m east
vertical distance:
25(3 2 sin 120°) 118.30 m north
44a. x4 cos 20° y4 sin 20°
3.76 1.37
3.76, 1.37
x5 cos 70° y5 sin 70°
1.71 4.70
1.71, 4.70
44b. 3.76, 1.371.71, 4.70
3.76 1.71, 1.37 4.70
5.47, 6.07
44c. 5.47 rcos v; 6.07 rsin v
6
5.
.0
47
7
r
r
c
s
o
in
s
v
v
6
5.
.0
47
7
tan v
47.98 v; 47.98°
5.47 rcos 47.98°
cos5
4
.4
77
.98°
r
8.17 r
8.17∠47.98°
44d. 8.17 sin (3.14t47.98°)
45. r2asin v2acos v
r22ar sin var cos v
x2y22ay 2ax
x22ax y22ay 0
(xa)2(ya)22a2
The graph of the equation is the circle centered at
(a, a) with radius 2
a.
46.
47. Sample answer: (2, 405°), (2, 765°), (2, 225°),
(2, 585°)
(r, v360k°)
→(2, 45° 360(1)°) →(2, 405°)
→(2, 45° 360(2)°) →(2, 765°)
(r, v(2k1)180°)
→(2, 45° (1)180°) →(2, 225°)
→(2, 45° (3)180°) →(2, 585°)
48. r
250242522 50 425 cos 30°
r
382.52 mph
si
5
n
0
v
s
3
i
8
n
23
.5
0
2
°
50 sin 30° 382.52 sin v
50
38
si
2
n
.5
3
2
0°
sin v
3°45v
The direction is 3°45west of south.
4321 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
425 mph
50 mph
30˚
r
3. The graph of the equation xkis a vertical line.
Since the line is vertical, the x-axis is the normal
line through the origin. Therefore, f0° or f
180°, depending on whether kis positive or
negative, respectively. The origin is kunits
from the given vertical line, so pk. The polar
form of the given line is krcos (v0°) if kis
positive or krcos (v180°) if kis negative.
Both equations simplify to krcos 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.
5. A2
B2
32(
4)2
5
Since Cis negative, use 5.
3
5
x
4
5
y2 0
cos f
3
5
, sin f
4
5
, p2
fArctan
4
3
53° or 307°
prcos (vf)
2 rcos (v307°)
6.
A2
B2
(2)2
42
25
Since Cis negative, use 25
.
2
2
5
x
2
4
5
y
2
9
5
0
cos f
55
, sin f
2
55
, p
9
10
5
fArctan(2)
63°
Since cos f0, but sin f0, the normal lies in
the second quadrant.
f180° 63° or 117°
prcos (vf)
9
10
5
rcos (v117°)
7. 3 rcos (v60°)
0 rcos (v60°) 3
0 r(cos vcos 60° sin vsin 60°) 3
0
1
2
rcos v
23
rsin v3
0
1
2
x
23
y3
0 x3
y6 or
x3
y6 0
8. r2 sec
v
4
rcos (v
4
) 2
r
cos vcos
4
sin vsin
4
2 0
22
rcos v
22
rsin v2 0
22
x
22
y2 0
2
x2
y4 0
Chapter 9 284
90˚
2
1
1
2
180˚360˚
270˚
y
y
2 cos
O
y
x
30˚
,
3
2
1
2
()
49. sin2Acos A1
1 cos2Acos A1
0 cos2Acos A2
0 (cos A2)(cos A1)
cos A2 0orcos A1 0
cos A2cos A1
A0°
50.
51. The terminal side is in the
third quadrant and the
reference angle is
210 180 or 30°.
cos 210°
2
3
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.
a0.07, b0.73, c1.36
Sample answer:
y0.07x20.73x1.36
53. 21003020
2
4
8
1
0
2
0
124 5100
x42x34x25x10
54. m
62
25
5
1
14
75
(y145) 60(x17)
60 y60x875
55. xyand yz, so xz.
If xz, then 0
x
z
1.
The correct choice is C.
Polar Form of a Linear Equation
Pages 577–578 Check for Understanding
1. The polar equation of a line is prcos (vf).
rand vare the variables. pis the length of the
normal segment from the line to the origin and
fis the angle the normal makes with the positive
x-axis.
2. For rto be equal to p, we must have cos (vf)
1. The first positive value of vfor which this is
true is vf.
9-4
14.
A2
B2
62(
8)2
10
Since Cis negative, use 10.
1
6
0
x
1
8
0
y
2
11
0
0
cos f
3
5
, sin f
4
5
, p2.1
fArctan
4
3
53°
Since cos f0, but sin f0, the normal lies in
the fourth quadrant.
f360° 53° or 307°
prcos (vf)
2.1 rcos (v307°)
15.
A2
B2
322
2
13
Sinc Cis negative, use 13
.
3
13
x
2
13
y
5
13
0
cos f
3
13
13
, sin f
2
13
13
, p
5
13
13
fArctan
2
3
34°
prcos (vf)
5
13
13
rcos (v34°)
16.
A2
B2
42(
5)2
14
Since Cis negative, use 41
.
4
41
x
5
41
y
1
4
0
1
0
cos f
4
41
41
, sin f
5
41
41
, p
10
41
41
fArctan
5
4
51°
Since cos f0, but sin f0, the normal lies in
the fourth quadrant.
f360° 51° or 309°
prcos (vf)
10
41
41
rcos (v309°)
17.
A2
B2
(–1)2
32
10
Since Cis negative, use 10
.
1
10
x
3
10
y
7
10
0
cos f
1
1
0
0
sin f
3
10
10
, p
7
10
10
fArctan (3)
72°
Since cos f< 0, but sin f> 0, the normal lies in
the second quadrant.
f180° 72° or 108°
pr cos (vf)
7
10
10
r cos (v108°)
18. 6 r cos (v120°)
0 r (cos vcos 120° sin vsin 120°) 6
0
1
2
r cos v
23
r sin v6
0
1
2
x
23
y6
0 x3
y12 or
x3
y12 0
285 Chapter 9
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2468 1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2468
9. 10.
11a. prcos (vf) →5 rcos
v
5
6
Since the shortest distance is along the normal,
the answer is (p, f) or
5,
5
6
.
11b.
Pages 578–579 Exercises
12.
A2
B2
72(
24)2
25
Since Cis positive, use 25.
2
7
5
x
2
2
4
5
y4 0
cos f
2
7
5
, sin f
2
24
5
, p4
fArctan
2
7
4
74°
Since cos f0, but sin f0, the normal lies in
the second quadrant.
f180° 74° or 106°
prcos (vf)
4 rcos (v106°)
13.
A2
B2
212
202
29
Since Cis negative, use 29.
2
21
9
x
2
20
9
y
8
27
9
0
cos f
2
21
9
, sin f
2
20
9
, p3
fArctan
2
20
1
44°
prcos (vf)
3 rcos (v44°)
369120˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
24680˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
19. 4 r cos
v
4
0 r
cos vcos
4
sin vsin
4
4
0
22
r cos v
22
r sin v4
0
22
x
22
y4
0 2
x2
y8 or
2
x2
y8 0
20. 2 r cos (v)
0 r (cos vcos sin vsin ) 2
0 r cos v 2
0 x 2
x2
21. 1 rcos (v330°)
0 r(cos vcos 330° sin vsin 330°) 1
0
23
rcos v
1
2
rsin v1
0
23
x
1
2
y1
0 3
xy2 or
3
xy2 0
22. r11 sec
v
7
6
rcos
v
7
6
11
r
cos vcos
7
6
sin vsin
7
6
11 0
23
rcos v
1
2
rsin v11 0
23
x
1
2
y11 0
3
xy22 0
23. r5sec (v60°)
rcos (v60°) 5
r(cos vcos 60° sin vsin 60°) 5 0
1
2
rcos v
23
rsin v5 0
1
2
x
23
y5 0
x3
y10 0
24. 25.
26. 27.
Chapter 9 286
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2468 0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2468
28. 29.
30. m
46
or
2
3
(y1)
2
3
(x4) →2x3y5 0
A2
B2
223
2
13
Since Cis negative, use 13
.
0
cos f
2
13
13
, sin f
3
13
13
, p
5
13
13
fArctan
3
2
56°
prcos (vf)
5
13
13
rcos (v56°)
31. prcos (vf)
→p3 cos
4
f
→p2 cos
7
6
f
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 f2.25 and p
0.31.
0.31 rcos (v2.25)
32a. prcos (vf) →6 rcos (v16°)
Since the shortest distance is along the normal,
the closest the fly came was por 6 cm.
32b. (p, f) or (6, 15°)
33. Since both normal segments have length 2, pmust
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 rcos (v45°) and 2 rcos (v
135°).
34. m0
(y4) 0(x5) →y4 0
cos f0, sin f1, p4
Since cos f0 when sin f1, f90°.
prcos (vf)
4 rcos (v90°)
5
13
3
13
2
13
35a.
35b. prcos (vf)
→p125 cos (130 f)
→p300 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 p124.43.
124.43 rcos (v135°)
36. krsin (va)
kr [sin v cos acos v sin a]
kr sin v cos ar cos v sin a
ky cos ax sin a
This is the equation of a line in rectangular
coordinates. Solving the last equation for yyields
y(tan a)x
cok
sa
. The slope of the line shows
that ais 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 Pin polar coordinates is 180° (90° a)
a90°. Substitute into the original polar
equation to find the r-coordinate of P:
kr sin (a90° a)
kr sin 90°
kr
Therefore, kis the length of the normal segment.
37. prcos (vf)
→p40 cos (0° f)
→p40 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 f36° and
p32.36.
32.36 rcos (v36°)
38. r6
x2y
2
6
x2y236
39. The graph of a polar equation of the form
rasin nvis a rose.
287 Chapter 9
125
250
375
500
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
y
x
P
O
y
x
(1, 0)
2468
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
8642 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1
40. x3y6
y
x
36
y
1
3
x2
xt, y
1
3
t2
41. A
3N
60
r2
3
6
6
5
0
62
20.42 ft2
42. Since 360° lies on the x-axis
of the unit circle at (1, 0),
sin 360° yor 0.
43. 2x35x212x0
x(2x25x12) 0
x(2x3)(x4) 0
x0 or x
3
2
or x4
44. c2d248
(cd)(cd) 48
12(cd) 48
cd4
Page 579 Mid-Chapter Quiz
1. 2.
3. 4.
5. r(2
)2(
2
)2
vArctan
4
or 2
4
Since (2
, 2
) is in the third quadrant,
v
4
or
5
4
.
2,
5
4
6. r
02(
4)2
16
or 4
Since x0 when y4, v
3
2
.
4,
3
2
2
4. Sample answer: x21 0
(xi)(xi) 0, where the solutions are xi.
x2xixii20
x2(1) 0
x21 0
5. i6(i4)2i2
12(1)
1
6. i10 i2(i4)2i2i2
(1)2i2i2
1 (1) or 2
7. (2 3i) (6 i) (2 (6)) (3ii)
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)) (4i5i)
1 9i
10. (2 i)2(2 i)(2 i)
4 4ii2
3 4i
11.
1
i2i
1
i2i
1
1
2
2i
i
1
i
2
4
i
i
2
2
i
52
2
5
1
5
i
12. (2.5 3.1i) (6.2 4.3i)
(2.5 (6.2)) (3.1i4.3i)
3.7 7.4iN
Pages 583–585 Exercises
13. i6i4i2
1 1
1
14. i19 (i4)4i3
14i
i
15. i1776 (i4)444
1444
1
16. i9i5(i4)2i(i4)2i3
12i12i
i(i) or 0
17. (3 2i) (4 6i) (3 (4)) (2i6i)
1 8i
18. (7 4i) (2 3i) (7 2) (4i3i)
9 7i
19.
1
2
i
(2 i)
1
2
(2)
(i(i))
3
2
2i
20. (3 i) (4 5i) (3 (4)) (i(5i))
7 4i
21. (2 i)(4 3i) 8 10i3i2
5 10i
22. (1 4i)2(1 4i)(1 4i)
1 8i16i2
15 8i
Chapter 9 288
Complex Numbers (
a
bi
)
Reals
(
b
0)
Pure
Imaginary
(
a
0)
Imaginary
(
b
0)
7. x2y236
x2y
2
36
r6 or r6
8. r2 csc v
rsin v2
y2
9.
A2
B2
52(
12)2
13
Since Cis positive, use 13.
1
5
3
x
1
12
3
y
1
3
3
0
cos f
1
5
3
, sin f
1
12
3
, p
1
3
3
fArctan
1
5
2
67°
Since cos f0, but sin f0, the normal lies in
the second quadrant.
f180° 67° or 113°
prcos (f)
1
3
3
rcos (v113°)
10.
A2
B2
(2)2
(6
)2
210
Since Cis negative, use 210
.
2
6
10
y
2
2
10
0
cos f
1
1
0
0
, sin f
3
10
10
, p
1
1
0
0
fArctan
3
1
72°
Since cos f0 and sin f0, the normal lies in
the third quadrant.
f180° 72° or 252°
prcos (vf)
1
1
0
0
rcos (v252°)
Simplifying Complex Numbers
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.
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 abi.
9-5
2
210
x
23. (1 7
i)(2 5
i)
2 5
i27
i35
i2
(2 35
) (27
5
)i
24. (2 3
)(1 12
) (2 3
i)(1 12
i)
2 212
i3
i
36
i2
2 43
i3
i6
8 33
i
25.
1
2
2
ii
1
2
2
ii
1
1
2
2i
i
2
1
3
i
4
i2
2i2
4
53i
4
5
3
5
i
26.
34
2i
i
3
4
2i
i
4
4
i
i
12
1
6
1
1i
i2
2i2
101
711i
1
10
7
1
1
7
1
i
27.
5
5
i
i
5
5
i
i
5
5
i
i
25
25
1
0i
i
2
i2
24
2610i
1
12
3
1
5
3
i
28. (xi)(xi) 0
x2i20
x21 0
29. (x(2 i))(x(2 i)) 0
(x2 i)(x2 i) 0
x22xxi2x4 2ixi2ii20
x24x4 1 0
x24x5 0
30. (2 i)(3 2i)(1 4i) (6 i2i2)(1 4i)
(8 i)(1 4i)
8 31i4i2
12 31i
31. (1 3i)(2 2i)(1 2i) (2 8i6i2)(1 2i)
(4 8i)(1 2i)
4 16i16i2
12 16i
32.
1
2
i
1
2
i
1
2
3
i
1 2
i
1
2
3
i
1 2
i
34.
(2
3
i
i
)2
(2
3i)
(2i
i)
4
3
4
i
i
i2
3
3
4
ii
3
3
4
ii
3
3
4
4i
i
9
9
9i
1
6i
4
2
i2
132
59i
1
23
5
2
9
5
i
35. (1 i)(1 i)
(3 2i)(3 2i)
(1 i)2
(3 2i)2
289 Chapter 9
1
2
22
i3i6i2
1 2i2
1
2
6
2
2
3
i
3
1
6
36
33
62
i
33.
2
3
6
2
i
i
2
3
2
6
i
i
3
3
6
6
i
i
6 2
6
i3
2
i
12
i2
9 6i2
(6 2
3
) (2
6
3
2
)i
15
2
5
2
15
3
52
2
15
6
i
5
2i
12i
5
2i
12i
5
5
1
12
2i
i
2
1
5
0i
1
2
4
4
4
i
i
2
2
24
16
910i
1
2
6
4
9
1
1
6
0
9
i
36a. ZR(XLXC)j
→ Z10 (1 2)j→Z10 johms
→Z3 (1 1)j→Z3 0johms
36b. (10 j) (3 0j) (10 3) (1j0j)
13 johms
36c. S
Z
1
→ S
6
13j
6
13j
6
6
3
3j
j
3
6
6
3
9
j
j2
6
453j
0.13 0.07jsiemens
37a. x
8i
2
36
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)28i(3 4i) 25 0
7 24i24i32 25 0
0 0
(3 4i)28i(3 4i) 25 0
7 24i24i32 25 0
0 0
38. f(xyi) (xyi)2
x22xyiy2
(x2y2) 2xyi
39a. z02 i
z1i(2 i) i2or 1 2i
z2i(2i1) 2i2ior 2 i
z3i(2 i) 2ii2or 1 2i
z4i(1 2i) i2i2or 2 i
z5i(2 i) 2ii2or 1 2i
8i
(8i)2
4(1)(
25)
2(1)
1 2ii2
9 12i4i2
39b. z01 0i
z1(0.5 0.866i)(1 0i) 0.5 0.866i
z2(0.5 0.866i)(0.5 0.866i)
0.25 0.866i0.75
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 0.866i0.75
0.500 0.866i
40. (1 2i)3
(1
1
2i)3
111
2i
111
2i
1
11
1
2
2i
i
1
1
12
52i
1
1
2
1
5
12
25
i
41. c1(cos 2tisin 2t) c2(cos 2tisin 2t)
c1cos 2tc1isin 2tc2cos 2tc2isin 2t
(c1c2)(cos 2t) (c1c2)(isin 2t)
(c1c2)(cos 2t) only if c1c2
42. A2
b2
62(
2)2
210
Since Cis positive, use 210
.
2
6
10
x
2
2
10
y
2
3
10
0
cos f
3
10
10
, sin f
11
00
, p
3
20
10
fArctan
1
3
18°
Since cos f0, but sin f0, the normal lies in
the second quadrant.
f180° 18° or 162°
prcos (vf)
3
20
10
rcos (v162°)
43.
44. x(3), y6t1, 4
x3, y6t1, 4
45. u
1
4
8, 6, 422, 6, 3
2,
3
2
, 14, 12, 6
6,
2
2
7
, 5
1
(3 4i)(1 2i)
46. tan a
4
3
cot B
1
5
2
tan2a1 sec2a1 cot2Bcsc2B
4
3
21 sec2a1
1
5
2
2csc2B
2
9
5
sec2a
1
1
6
4
9
4
csc2B
2
9
5
cos2a
1
1
4
6
4
9
sin2B
3
5
cos a
1
1
2
3
sin B
sin2acos2a1sin
2Bcos2B1
sin2a
3
5
21
1
1
2
3
2cos2B1
sin2a
1
2
6
5
cos2B
1
2
6
5
9
sin a
4
5
cos B
1
5
3
cos (aB) cos acos Bsin asin B
3
5
1
5
3
4
5
1
1
2
3
3
6
3
5
47. amplitude
1
2
(7) or 3.5
period
2
1
2
or
6
y3.5 cos
6
t
48. hx3
tan 52°
x
x
4
3
5
xtan 52° 45 tan 52° x3
xtan 52° x3
45 tan 52°
x
t
an
45
52
t
°
an
5
2°
3
x127.40
hx3
127.40(3
) 221 ft
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 ddepth of the original pool.
The second pool’s width 5d4, the length
10d6, and the depth d2.
(5d4)(10d6)(d2) 3420
(50d270d24)(d2) 3420
50d3100d270d2140d24d48 3420
50d3170d2164d3372 0
25d385d282d1686 0
Use a graphing calculator to find the solution
d3.
The dimensions of the original pool are 15 ft by
30 ft by 3 ft.
51. 80 k(5)(8)
2 k
y2(16)(2)
64
Chapter 9 290
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2821147
h
8˚
30˚
60˚
45 ft
x
120˚
52˚
291 Chapter 9
y
x
O
(1, 1)
(6, 1)
(6, 8)
i
(2, 1)
2
2
1
1
2
2
1
1
O
i
(1, 2)
O
2
2
1
1
2
2
1
1
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2468
i
(4, )
3
6
11
6
3
3
4
6
5
3
2
6
7
3
5
1234
0
2
2
3
i
(2, 3)
52. y7 x2
x7 y2
x7 y2
x7 y2
x
7
y
f1(x) 7 x
53.
f(x, y) 2xy
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. x2y7z14
x 3y5z21
y2z7
x3y5z21 →5x15y25z105
5xy2z75xy 2z7
16y27z112
y2z7→16y32z112
16y27z112
1
6
y
2
7
z
1
1
2
59z0
z0
y2(0) 7→y7
x2(7) 7(0) 14 →x0
(0, 7, 0)
55. Since BC BD, m∠BDC m∠DCB x
m∠DBC 180 120 or 60.
xx60 180
2x120
x60
x40 60 40 or 100
The correct choice is A.
The Complex and Polar Form
of Complex Numbers
Pages 589–590 Check for Understanding
1. To find the absolute value of abi, square aand
b, add the squares, then take the square root of
the sum.
2. i0 i; cos
2
0 and sin
2
1
icos
2
isin
2
3. Sample answer: z1i, z2i
z1z2z1z2
i(i)ii
0ii
0 2i
9-6
4. The conjugate of abiis abi.
(ab
i)(a
bi)
a2b
2
, so the friend’s
method gives the same answer.
Sample answer: The absolute value of 2 3iis
223
2
13
. Using the friend’s method, the
absolute value is (2 3
i)(2
3i)
4 9
13
.
5. 2xy(xy)i5 4i
2xy5xy4
2x(x4) 5yx4
x1
y(1) 4 or 3
6. 7.
z(22
(1)2
z12(
2
)2
5
3
8. r22(
2)2
vArctan
2
2
2
8
or 22
7
4
vis in the fourth quadrant.
2 2i22
cos
7
4
isin
7
4
9. r425
2
vArctan
5
4
41
0.90
4 5i41
(cos 0.90 isin 0.90)
10. r(2)2
02
vArctan
0
2
4
or 2
is on the x-axis at 2.
2 2 (cos isin )
11. 12.
4
cos
3
isin
3
2(cos 3 isin 3)
4
1
2
i
—
2
3
—
2(0.99 i(0.14))
2 23
i1.98 0.28i
13.
3
2
(cos 2isin 2)
3
2
(1 i(0))
3
2
14.
15a. magnitude 102
152
325
18.03 N
15b. Arctan
1
1
5
0
56.31°
Pages 590–591 Exercises
16. 2x5yi12 15i
2x12 5y15
x6y3
17. 1 (xy)iy3xi
1 yxy3x
x(1) 3x
1 2x
1
2
x
18. 4x(y5)i2xy(x7)i
y5 x74x2xy
yx12 4x2x(x12)
3x12
x4
y(4) 12 or 8
19. 20.
z223
2
z32(
4)2
13
25
or 5
21. 22.
z(1)2
(5
)2
z02(
3)2
26
9
or 3
23. 24.
z(1)2
(5
)2
z42(
2
)2
6
18
or 32
25. r(4)2
62
52
or 213
26. r323
2
vArctan
3
3
18
or 32
4
3 3i32
cos
4
isin
4
27. r(1)2
(
3
)2
vArctan
1
3
4
or 2
4
3
vis in the third quadrant.
1 3
i2
cos
4
3
isin
4
3
28. r62(
8)2
vArctan
6
8
2
100
or 10 5.36
vis in the fourth quadrant.
6 8i10(cos 5.36 isin 5.36)
Chapter 9 292
6
11
6
3
3
4
6
5
3
2
6
7
3
5
1234
0
2
2
3
i
(, 2)
2
3
i
0.38
i
1
1
0.36
0.90
1.30
0.63
1
1
O
i
(2, 3)
O
i
(3, 4)
O
i
(1, 5)
O
i
(0, 3)
O
i
O
(1, 5)
2
2
1
1
2
2
1
1
i
O
(4, 2)
293 Chapter 9
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
i
(3, )
4
6
11
6
3
3
4
6
5
3
2
6
7
3
5
1234
0
2
2
3
i
(1,)
6
6
11
6
3
3
4
6
5
3
2
6
7
3
5
1234
0
2
2
3
i
(2, )
3
46
11
6
3
3
4
6
5
3
2
6
7
3
5
36912
0
2
2
3
i
(10, 6)
29. r(4)2
12
vArctan
1
4
17
2.90
vis in the second quadrant.
4 i17
(cos 2.90 isin 2.90)
30. r202
(21
)2
vArctan
2
2
0
1
2
841
or 29 5.47
vis in the fourth quadrant.
20 21i29(cos 5.47 isin 5.47)
31. r(2)2
42
vArctan
4
2
20
or 25
2.03
vis in the second quadrant.
2 4i25
(cos 2.03 isin 2.03)
32. r320
2
vArctan
0
3
9
or 3 0
vis on the x-axis at 3.
3 3(cos 0 isin 0)
33. r(4
2
)2
02
vArctan
4
0
2
32
or 42
vis on the x-axis at 42
.
42
42
(cos isin )
34. r02(
2)2
4
or 2
Since x0 when y2, v
3
2
.
2i2
cos
3
2
isin
3
2
35. 36.
3
cos
4
isin
4
cos
6
isin
6
3
2
2
i
2
2
2
3
1
2
i
3
2
2
3
2
2
i
37. 38.
2
cos
4
3
isin
4
3
10(cos 6 isin 6)
2
1
2
i
2
3
10(0.960 i(0.279))
1 3
i9.60 2.79i
6
11
6
3
3
4
6
5
3
2
6
7
3
5
1234
0
2
2
3
i
(2, )
4
5
6
11
6
3
3
4
6
5
3
2
6
7
3
5
1234
0
2
2
3
i
(2.5, 1)
6
11
6
3
3
4
6
5
3
2
6
7
3
5
2468
0
2
2
3
i
(5, 0)
6
11
6
3
3
4
6
5
3
2
6
7
3
5
1234
0
2
2
3
i
(3, )
i
O
0.25
i
0.44 0.5
i
0.44 0.44
i
0.50 0.39
i
0.60 0.39
i
1
1
1
1
i
O
1
1
1
1
i
i
O
1
1
1
1
0.5 0.5
i
0.5 0.5
i
39. 40.
2
cos
5
4
isin
5
4
2.5(cos 1 isin 1)
2
2
2
i
2
2
2.5(0.54 i(0.84))
2
2
i1.35 2.10i
41. 42.
5(cos 0 isin 0) 3(cos isin )
5(1 0) 3(1 0)
53
43.
44.
45.
46a. 40∠30° 40(cos 30° jsin 30°)
40
2
3
j
1
2
34.64 20j
60∠60° 60(cos 60° jsin 60°)
60
1
2
j
2
3
30 51.96j
46b. (34.64 20j) (30 51.96j)
(34.64 30) (20j51.96j)
64.64 71.96j
46c. v(t) rsin (250tv°)
r64.64
271
.962
vArctan
7
6
1
4
.
.
9
6
6
4
96.73 48°
v(t) 96.73 sin (250t48°)
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 visin v) can be written as
r(cos (v) isin (v)), or r(cos visin v).
48a. 10(cos 0.7 jsin 0.7) 7.65 6.44j
16(cos 0.5 jsin 0.5) 14.04 7.67j
48b. (7.65 6.44j) (14.04 7.67j)
(7.65 14.04) (6.44j7.67j)
21.69 14.11johms
48c. r21.69
214
.112
Arctan
2
1
1
4
.
.
6
1
9
1
25.88 0.58
21.69 14.11j25.88 (cos 0.58 jsin 0.58) ohms
49a. Translate 2 units to the right and down 3 units.
49b. Rotate 90° counterclockwise about the origin.
49c. Dilate by a factor of 3.
49d. Reflect about the real axis.
50a. Sample answer: let z11 iand z23 4i.
z1z2(1 i)(3 4i)
1 7i
50b. z12
cos
4
isin
4
1.41(cos 0.79 isin 0.79)
z25(cos 0.93 isin 0.93)
z1z252
(cos 1.71 isin 1.71)
7.07(cos 1.71 isin 1.71)
50c. Sample answer: Let z12 4iand
z21 3i. Then
z125
(cos 5.18 isin 5.18)
4.47(cos 5.18 isin 5.18),
z210
(cos 1.89 isin 1.89)
3.16(cos 1.89 isin 1.89), and
z1z2(2 4i)(1 3i)
10 10i
102
cos
4
isin
4
14.14(cos 0.79 isin 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.)
51. (6 2i)(2 3i) 12 22i6i2
6 22i
52. x3 cos 135° y3 sin 135°
3
2
2
3
2
2
3
2
2
3
2
2
3
2
2
,
3
2
2
53. magnitude (3)2
72
58
3, 73i
u
7j
u
54. tan 105° tan (60° 45°)
tan 60° tan 45°
1 tan 60° tan 45°
Chapter 9 294
1
3
(
3
)
1
(1)
1
3
1
3
1
1
3
3
4
2
2
3
2 3
55. w
v
t
12(
1
2)
or 24
vrq
18(24) or 432cm/s
432cm/s 4.32m/s
13.57 m/s
56. sin A
1
1
2
8
Asin1
2
3
A41.8°
47. 2a
1
3a
5
2a1 3a5
4 a
58. as x→, y→; as x→, y→
59. In the fourth month, the person will have received
3 pay raises.
$500(1.10)3$665.50
The correct choice is D.
3
31 3
y2x22
xy
1000 2 106
10 202
02
10 202
1000 2 106
Graphing Calculator Exploration:
Geometry in the Complex Plane
Page 592
1. They are collinear.
2. Yes. Mis the point obtained when T0, and Nis
the point obtained when T1.
3. The points are again collinear, but closer together.
4. The points are on the line through Mand N.
5. If one of a, b, or cequals 0, then aK bM cN is
on KMN. If none of a, b, or cequals 0, then
aK bM cN lies on or inside KMN.
6. Mis the point obtained when T0 and Nis the
point obtained when T1. Thus, a point between
Mand Nis obtained when 0 T1.
7. The distance between zand 1 iis 5. This
defines a circle of radius 5 centered at 1 i.
8. The distance between a point zand a point at
2 3iis 2.
z(2 3i)2
Products and Quotients of
Complex Numbers in Polar Form
Pages 596 Check for Understanding
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. r2 2 or 4 v
2
3
2
4
2
or 2
4(cos 2isin 2) 4(1 i(0))
4
9-7
9-6B 5. r
3
4
v
6
2
3
3
6
or
2
3
4
cos
2
isin
2
3
4
(0 (1)i)
3
4
i
6. r
4
2
or 2 v
9
4
2
9
4
2
4
or
11
4
2
cos
11
4
isin
11
4
2
2
2
i
2
2
2
2
i
7. r
1
2
(6) or 3 v
3
5
6
2
6
5
6
or
7
6
3
cos
7
6
isin
7
6
3
2
3
i
1
2
3
2
3
3
2
i
8. r122(
23
)2
r2(3)2
(3
)2
16
or 4 12
or 23
r4(23
) or 83
v1Arctan
2
2
3
v2Arctan
3
3
3
5
6
v
3
5
6
2
6
5
6
or
7
6
83
cos
7
6
isin
7
6
83
2
3
i
1
2
12 43
i
9. EIZ
2
cos
11
6
jsin
11
6
3
cos
3
jsin
3
r2(3) or 6 v
11
6
3
11
6
2
6
13
6
or
6
V6
cos
6
jsin
6
volts
Pages 596–598 Exercises
10. r4(7) or 28 v
3
2
3
3
3
or
28(cos isin ) 28(1 i(0))
28
11. r
6
2
or 3 v
3
4
4
2
4
or
2
3
cos
2
isin
2
3(0 i(1))
3i
12. ror
1
6
v
3
6
1
2
3
295 Chapter 9
2
6
6
or
6
1
6
cos
6
isin
6
1
6
2
3
i
1
2
12
3
1
1
2
i
13. r5(2) or 10 v
3
4
4
4
3
4
or
7
4
10
cos
7
4
isin
7
4
10
2
2
i
2
2
52
52
i
14. r6(3) or 18 v
3
5
6
2
6
5
6
3
6
or
2
18
cos
2
isin
2
18(0 i(1))
18i
15. r
3
1
or 3 v
7
3
2
14
6
3
6
or
11
6
3
cos
11
6
isin
11
6
3
2
3
i
1
2
3
2
3
3
2
i
16. r2(3) or 6 v240° 60°
300°
6(cos 300° isin 300°) 6
1
2
i
2
3
3 33
i
17. rv
7
4
3
4
2
2
2
or 2
2(cos isin ) 2(1 i(0))
2
18. r3(0.5) or 1.5 v4 2.5 or 6.5
1.5(cos 6.5 isin 6.5) 1.46 0.32i
19. r
4
1
or 4 v2 3.6 or 5.6
4[cos (5.6) isin (5.6)] 3.10 2.53i
20. r
2
1
0
5
or
4
3
v
7
6
11
3
7
6
22
6
15
6
or
2
4
3
cos
2
isin
2
4
3
(0 i(1))
4
3
i
21. r2(2
) or 22
v
3
4
2
3
4
2
4
or
5
4
22
cos
5
4
isin
5
4
22
2
2
i
2
2
2 2i
22. r2(6) or 12 v
3
6
2
6
6
or
6
12
cos
6
isin
6
12
2
3
i
1
2
63
6i
23. ror 8 v
5
3
3
4
1
2
2
2
2
24. r122(
2)2
r2(3)2
32
8
or 22
18
or 32
r22
(32
) or 12
v1Arctan
2
2
2v2Arctan
3
3
7
4
3
4
v
7
4
3
4
10
4
or
2
12
cos
2
isin
2
12(0 i(1))
12i
25. r1(2
)2
(
2
)2
r2(3
2
)2(
32
)2
4
or 2 36
or 6
r2 6 or 12
v1Arctan
2
2
2v2Arctan
3
3
2
2
7
4
5
4
v
7
4
5
4
12
4
or
12(cos isin ) 12(1 i(0))
12
26. r1(3
)2
(
1)2
r222(
23
)2
4
or 2 16
or 4
r
2
4
or
1
2
v1Arctan
1
3
2v2Arctan
2
11
6
5
3
v
11
6
5
3
11
6
10
6
or
6
1
2
cos
6
isin
6
1
2
2
3
i
1
2
4
3
1
4
i
27. r1(4
2
)2
(42
)
2
r2626
2
64
or 8 72
or 62
r
6
8
2
3
4
2
2
2
4
6
2
or
2
3
2
v1Arctan
4
4
2
2
v2Arctan
6
6
3
4
4
v
3
4
4
2
4
or
2
2
3
2
cos
2
isin
2
2
3
2
(0 i(1))
2
3
2
i
23
Chapter 9 296
4
3
8
cos
4
3
isin
4
3
8
1
2
i
2
3
4 43
i
4
4
or
34. r5 sec
v
5
6
rcos
v
5
6
5
r
cos vcos
5
6
sin vsin
5
6
5 0
2
3
rcos v
1
2
rsin v5 0
2
3
x
1
2
y5 0
3
xy10 0
35.
Since the triangle is isosceles, the base angles are
congruent. Each measures
180
2
50
or 65°.
sin
23
50°
sin
x
65°
23 sin 65° xsin 50°
23
si
s
n
in
50
6
°
5°
x
27.21 x; 27.21 lb
36. cos 2xsin x 1
1 2 sin2xsin x1
2 sin2xsin x 0
sin x(2 sin x1) 0
sin x0or2 sin x1 0
x0° sin x
1
2
x30°
37. y cos x
x cos y
arccos xy
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.
Powers and Roots of Complex
Numbers
Page 602 Graphing Calculator Exploration
1. Rewrite 1 in polar form as 1(cos 0 isin 0).
Follow the keystrokes to find the roots at 1,
0.5 0.87i, and 0.5 0.87i.
2. Rewrite iin polar form as 1
cos
2
isin
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.
9-8
297 Chapter 9
6
11
6
3
3
4
6
5
3
2
6
7
3
5
1234
0
2
2
3
i
z
1
z
2
z
2
z
1
28. I
E
Z
3
13
2j
r113 r232(
2)2
13
r
1
1
3
3
or 13
v10v2Arctan
2
3
0.59
v0 (0.59) or 0.59
I13
(cos 0.59 jsin 0.59) 3 2jamps
29. Z
E
I
4
1
00
3j
r1100 r242(
3)2
25
or 5
r
10
5
0
or 20
v10v2Arctan
4
3
0.64
v0 (0.64) or 0.64
z20(cos 0.64 jsin 0.64)
16 12johms
30. Start at z1in the complex
plane. Since the modulus
of z2is 1, z1z2and
z
z1
2
will
both have the same
modulus as z1. Then z1z2
and
z
z1
2
can be located by
rotating z1by
6
counterclockwise and
clockwise, respectively.
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 a1, the equation will be the form z2bz
c0. The coefficient cis the product of the
solutions, which is 6
cos
7
6
isin
7
6
, or
33
3iin 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.
b
3
cos
3
isin
3
2
cos
5
6
isin
5
6
3
2
3
2
3
i
(3
i)
3
2
3
3
2
3
2
i
Therefore, the equation is z2
3
2
3
3
2
3
2
iz(33
3i) 0.
33. r52(
12)2
Arctan
5
12
2
169
or 13 5.11
5 12i13(cos 5.11 isin 5.11)
x
lb
x
lb
x
lb
x
lb
23 lb 23 lb
Prop
130˚
50˚
3. Rewrite 1 iin polar form as 2
cos
4
isin
4
.
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 a0 and b0, then abia. 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 Check for Understanding
1. Same results, 4 4i; answers may vary.
(1 i)(1 i)(1 i)(1 i)(1 i)
(1 2ii2)(1 2ii2)(1 i)
(2i)(2i)(1 i)
4(1 i)
4 4i
(1 i)5
→r2
, v
4
(2
)5
cos (5)
4
isin (5)
4
42
cos
5
4
isin
5
4
42
2
2
i
2
2
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.
3.
4. Shembala is correct. The polar form of aaiis
a2
cos
4
isin
4
. By De Moivre’s Theorem,
the polar form of (aai)2is 2a2
cos
2
isin
2
.
Since cos
2
0, this is a pure imaginary number.
5. r(3
)2
(
1)2
or 2 vArctan
1
3
or
6
23
cos (3)
6
isin (3)
6
8
cos
2
isin
2
8(0 i(1))
8i
6. r32(
5)2
or 34
vArctan
3
5
1.030376827
34
4(cos (4)() isin (4)(v))
644 960i
7. r021
2
or 1 v
2
1
1
6
cos
1
6
2
isin
1
6
2
1
cos
1
2
isin
1
2
0.97 0.26i
8. r(2)2
(1
)2
or 5
vArctan
1
2
2.677945045
5
1
3
cos
1
3
() isin (3)()
0.82 1.02i
9. x4i0 →x4i
Find the fourth roots of i.
r02(
1)2
1v
3
2
(i)
1
4
1 (cos
3
2
2n
isin
3
2
2n
4
1
cos
3
8
4n
isin
3
8
4n
x1cos
3
8
isin
3
8
0.38 0.92i
x2cos
7
8
isin
7
8
0.92 0.38i
x3cos
11
8
isin
11
8
0.38 0.92i
x4cos
15
8
isin
15
8
0.92 0.38i
10. 2x34 2i0 →x32 i
Find the third roots of 2 i.
r(2)2
(1
)2
5
vArctan
1
2
3.605240263
(2 i)
1
3
[5
(cos (v2i) isin (v2n))]
1
3
(5
)
1
3
cos
v
3
2n
isin
v
3
2n
x1(5
)
1
3
cos
3
v
isin
3
v
0.47 1.22i
x2(5
)
1
3
cos
v
3
2
isin
v
3
2
1.29 0.20i
x3(5
)
1
3
cos
v
3
4
isin
v
3
4
0.81 1.02i
Chapter 9 298
i
O
a
a
a
a
a
ai
a
ai a
ai
a
ai
i
O
1
1
0.38 0.92
i
0.92 0.38
i
0.38 0.92
i
0.92 0.38
i
1
1
i
O
1
1
1
1
1.29 0.20
i
0.82 1.02
i
0.47 1.22
i
11. For w1, the modulus (0.82
(0.7
)2
)2or 1.13.
For w2, the modulus 1.132or 1.28.
For w3, the modulus 1.282or 1.64.
This moduli will approach infinity as the number
of iterations increases. Thus, it is an escape set.
Pages 605–606 Exercises
12. 33
cos (3)
6
isin (3)
6
27
cos
2
isin
2
27(0 i(1))
27i
13. 25
cos (5)
4
isin (5)
4
32
cos
5
4
isin
5
4
32
2
2
i
2
2
162
162
i
14. r(2)2
22
22
vArctan
2
2
7
4
(22
)3
cos (3)
7
9
isin (3)
7
4
162
cos
21
4
isin
21
4
162
2
2
i
2
2
16 16i
15. r12(
3
)2
2vArctan
1
3
3
24
cos (4)
3
isin (4)
3
16
cos
4
3
isin
4
3
16
1
2
i
2
3
8 83
i
16. r32(
6)2
35
vArctan
3
6
1.107148718
(35
)4(cos (4)(v) isin (4)(v))
567 1944i
17. r223
2
13
vArctan
3
2
0.9827937232
(13
)2(cos (2)(v) isin (2)(v))
0.03 0.07i
18. r224
2
25
vArctan
4
2
1.107148718
(25
)4(cos (4)(v) isin (4)(v))
112 384i
19. 32
1
5
cos
1
5
2
3
isin
1
5
2
3
2
cos
2
1
5
isin
2
1
5
1.83 0.81i
20. r(1)2
02
1v
1
1
4
cos
1
4
() isin
1
4
()
cos
4
isin
4
0.71 0.71i
21. r(2)2
12
5
vArctan
1
2
2.677945045
(5
)
1
4
cos
1
4
(v) isin
1
4
(v)
0.96 0.76i
22. r42(
1)2
17
vArctan
4
1
0.2449786631
(17
)
1
3
cos
1
3
(v) isin
1
3
(v)
1.60 0.13i
23. r222
2
22
vArctan
2
2
4
(22
)
1
3
cos
1
3
4
isin
1
3
4
1.37 0.37i
24. r(1)2
(1
)2
2
vArctan
1
1
3
4
(2
)
1
4
cos
1
4
3
4
isin
1
4
3
4
0.91 0.61i
25. r021
2
1v
2
1
1
2
cos
1
2
2
isin
1
2
2
0.71 0.71i
26. x31 0 →x31
Find the third roots of 1.
r120
2
1v0
1
1
3
[1 (cos (0 2n) isin (0 2n))]
1
3
cos
2n
3
isin
2n
3
x1cos 0 isin 0 1
x2cos
2
3
isin
2
3
1
2
2
3
i
x3cos
4
3
isin
4
3
1
2
2
3
i
299 Chapter 9
i
O
11
1
1
i
2
123
i
2
123
27. x51 →x51
Find the fifth roots of 1.
r(1)2
02
1v
(1)
1
5
[1 (cos (2n) isin (2n))]
1
5
cos
5
2n
isin
5
2n
x1cos
5
isin
5
0.81 0.59i
x2cos
3
5
isin
3
5
0.31 0.95i
x3cos
5
5
isin
5
5
1
x4cos
7
5
isin
7
5
0.31 0.95i
x5cos
9
5
isin
9
5
0.81 0.59i
28. 2x4128 0 →x464
Find the fourth roots of 64.
r642
02
64 v0
64
1
4
[64 (cos (0 2n) isin (0 2n))]
1
4
22
cos
n
2
isin
n
2
x122
(cos 0 isin 0) 22
x222
cos
2
isin
2
22
i
x322
(cos isin ) 22
x422
cos
3
2
isin
3
2
22
i
29. 3x448 0 →x416
Find the fourth roots of 16.
r(16)
202
16 v
(16)
1
4
[16 (cos (2n) isin (2n))]
4
1
2
cos
4
2n
isin
4
2n
x12
cos
4
isin
4
2
2
i
x22
cos
3
4
isin
3
4
2
2
i
x32
cos
5
4
isin
5
4
2
2
i
x42
cos
7
4
isin
7
4
2
2
i
30. x4(1 i) 0 →x41 i
Find the fourth roots of 1 i.
r121
2
2
vArctan
1
1
4
(1 i)
1
4
2
cos
4
2n
isin
4
2n
1
4
(2
)
1
4
cos
16
8n
isin
16
8n
x1(2
)
1
4
cos
1
6
isin
1
6
1.07 0.21i
x2(2
)
1
4
cos
9
1
6
isin
9
1
6
0.21 1.07i
x3(2
)
1
4
cos
1
1
7
6
isin
1
1
7
6
1.07 0.21i
x4(2
)
1
4
cos
2
1
5
6
isin
2
1
5
6
0.21 1.07i
31. 2x42 23
i0 →x41 3
i.
Find the fourth roots of 1 3
i.
r(1)2
(
3
)2
2
vArctan
1
3
3
or
4
3
(13
i)
1
4
2
cos
4
3
2n
isin
4
3
2n
1
4
2
1
4
cos
4
12
6n
isin
4
12
6n
x12
1
4
cos
4
1
2
isin
4
1
2
0.59 1.03i
x22
1
4
cos
1
1
0
2
isin
1
1
0
2
1.03 0.59i
x32
1
4
cos
2
1
2
2
isin
2
1
2
2
0.59 1.03i
x42
1
4
cos
2
1
8
2
isin
2
1
8
2
1.03 0.59i
Chapter 9 300
i
O
1
1
1
0.31 0.95
i
0.81 0.59
i
0.81 0.59
i
0.31 0.95
i
1
i
O
21
1
2
2
3
3
31
2
3
1
2
i
2
22 2 2
2
i
2
i
O
1
1
2
i
2
2
i
2
2
i
2
2
i
2
1
1
i
O
1
1
0.21 1.07
i
1.07 0.21
i
1.07 0.21
i
0.21 1.07
i
1
1
i
O
1
1
1.03 0.59
i
0.59 1.03
i
0.59 1.03
i
1.03 0.59
i
1
1
32. Rewrite 10 9iin polar form as
181
cos
tan1
10
9
isin
tan1
10
9
.
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 4iin polar form as
25
[cos (tan1(2)) isin (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 20iin polar form as
4106
cos
tan1
5
9
isin
tan1
5
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.
35. For w1, the modulus
1
2
2
3
4
2
2or 0.81.
For w2, the modulus (0.81)2or 0.66.
For w3, the modulus (0.66)2or 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
cos
2
3
n
1
isin
2
3
n
1
, n0, 1, . . . , 30. Then
acos
2
3
n
1
. The maximum value of a cosine
expression is 1, and it is achieved in this
situation when n0.
36b. From the polar form in the solution to part a, we
get bsin
2
3
n
1
. bwill be maximized when
2
3
n
1
is
as close to
2
as possible. This occurs when n8,
so the maximum value of bis sin
1
3
6
1
, or about
0.9987.
37. x61 0 →x61
Find the sixth roots of 1.
r120
2
1v0
1
1
6
[1 (cos (0 2n) isin (0 2n))]
1
6
cos
n
3
isin
n
3
x1cos 0 isin 0 1
x2cos
3
isin
3
1
2
2
3
i
x3cos
2
3
isin
2
3
1
2
2
3
i
x4cos
3
3
isin
3
3
1
x5cos
4
3
isin
4
3
1
2
2
3
i
x6cos
5
3
isin
5
3
1
2
2
3
i
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
of 2. The dilation factor is
2
2
.
2
2
(cos 45° isin 45°)
2
2
2
2
isin
2
2
0.5 0.5i
38b.
2
2
(cos 45° isin 45°)
2
1
2
(cos 90° isin 90°)
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.
40. r2(3) or 6 v
6
5
3
6
10
6
or
11
6
6
cos
11
6
isin
11
6
6
2
3
i
1
2
33
3i
41. (2 5i) (3 6i) (6 2i)
(2 (3) (6)) (5i6i2i)
5 i
42. xt, y2t7
43. cos 22.5° cos
4
2
5°
1c
2
os 45°
1
2
2
2
301 Chapter 9
y
x
O
(2, 2)
1
1122
1
2
2
(0, 2)
22
2
4
2
2
2
2
44. Find B.
B180° 90° 81°15
8°45
Find a.
tan 81°15
2
a
8
28 tan 81°15a
181.9 a
Find c.
cos 81°15
2
c
8
c
cos 8
2
1
8
°15
c184.1
45. Let xthe number of large bears produced.
Let ythe number of small bears produced.
x300
y400
xy1200
f(x, y) 9x5y
f(300, 400) 9(300) 5(400)
4700
f(300, 900) 9(300) 5(900)
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
6
2
4
.4
3
1
.
0
6
or 36% alcohol
The correct choice is A.
Chapter 9 Study Guide and Assessment
Pages 607 Check for Understanding
1. absolute value 2. Polar
3. prisoner 4. iteration
5. pure imaginary 6. cardioid
7. rectangular 8. spiral of Archimedes
9. Argand 10. modulus
Pages 608–610 Skills and Concepts
11. 12.
13. 14.
15. Sample answer: (4, 585°), (4, 945°), (4, 45°),
(4, 405°)
(r, v360k°)
→(4, 225° 360(1)°) →(4, 585°)
→(4, 225° 360(2)°) →(4, 945°)
(r, v(2k1)180°)
→(4, 225° (1)180°) →(4, 45°)
→(4, 225° (1)180°) →(4, 405°)
16. 17.
18. 19.
20. 21.
circle Spiral of Archimedes
22. 23.
limaçon rose
24. x6 cos 45° y6 sin 45°
6
2
2
6
2
2
32
32
(32
, 32
)
25. x2 cos 330° y2 sin 330°
2
2
3
2
1
2
3
1
(3
, 1)
Chapter 9 302
y
x
O
200
200 600 1000
400
600
800
1000
1200 (300, 900)
(800, 400)
(300, 400)
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
A
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
B
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
C
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
D
12340˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
8642 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
3224168
8642 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
8642
26. x2 cos
3
4
y2 sin
3
4
2
2
2
2
2
2
2
2
(2
, 2
)
27. x1 cos
2
y1 sin
2
01
(0, 1)
28. r(3
)2(
3)2
vArctan
3
3
12
or 23
4
3
23
,
4
3
29. r525
2
vArctan
5
5
50
or 52
4
52
,
4
30. r(3)2
12
vArctan
1
3
10
3.16 2.82
(3.16, 2.82)
31. r422
2
vArctan
2
4
20
4.47 0.46
(4.47, 0.46)
32. A2
B2
221
2
5
Since Cis positive, use 5
.
2
5
x
1
5
y
3
5
0
cos f
2
5
5
, sin f
5
5
, p
3
5
5
fArctan
1
2
27°
Since cos f0 and sin f0, the normal lies in
the third quadrant.
f180° 27° or 207°
prcos (vf)
3
5
5
rcos (v207°)
33. A2
B2
321
2
10
Since Cis positive, use 10
.
3
10
x
1
10
y
4
10
0
cos f
3
10
10
, sin f
1
1
0
0
, p
2
5
10
fArctan
1
3
18°
Since cos f0 and sin f0, the normal lies in
the third quadrant.
f180° 18° or 198°
prcos (vf)
2
5
10
rcos (v198°)
34. 3 rcos
v
3
0 rcos vcos
3
rsin vsin
3
3
0
1
2
rcos v
2
3
rsin v3
0
1
2
x
2
3
y3
0 x3
y6 or
x3
y6 0
35. 4 rcos
v
2
0 rcos vcos
2
rsin vsin
2
4
0 0 rsin v4
0 y4
0 y4 or
y4 0
36. i10 i25 (i4)2i2(i4)6i
(1)2(1) (1)6i
1 i
37. (2 3i) (4 4i) (2 (4)) (3i(4i))
2 7i
38. (2 7i) (3 i) (2 (3)) (7i(i))
1 6i
39. i3(4 3i) 4i33i4
4(i) 3(1)
3 4i
40. (i7)(i7) i214i49
1 14i49
48 14i
41.
4
5
2
2
i
i
4
5
2
2
i
i
5
5
2
2
i
i
20
25
1
8i
4
i2
4i2
16
29
18i
1
2
6
9
1
2
8
9
i
42.
1
5
i
2
i
1
5
i
2
i
1
1
2
2
i
i
5 52
ii2
i2
1 2i2
303 Chapter 9
(5 2
) (52
1)i
3
5
3
2
1
3
52
i
43. r222
2
vArctan
2
2
8
or 22
4
22
cos
4
isin
4
44. r12(
3)2
vArctan
1
3
2
10
5.03
10
(cos 5.03 isin 5.03)
45. r(1)2
(3
)2
vArctan
1
3
4
or 2
2
3
2
cos
2
3
isin
2
3
46. r(6)2
(4
)2
vArctan
4
6
52
or 213
3.73
213
(cos 3.73 i sin 3.73)
47. r(4)2
(1
)2
vArctan
1
4
17
3.39
17
(cos 3.39 isin 3.39)
60. r(3
)2
(1
)2
2
v Arctan
6
1
3
Chapter 9 304
6
2,
()
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
i
3
3,
()
5
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
i
48. r420
2
v0
16
or 4
4(cos 0 isin 0)
49. r(2
2
)2
02
v
8
or 22
22
(cos isin )
50. r023
2
v
2
9
or 3
3
cos
2
isin
2
51. 52.
2
cos
6
isin
6
3
cos
5
3
isin
5
3
2
2
3
i
1
2
3
1
2
i
2
3
3
i
3
2
3
2
3
i
53. r4(3) or 12 v
3
3
or
2
3
12
cos
2
3
isin
2
3
12
1
2
i
2
3
6 63
i
54. r8(4) or 32 v
4
2
32
cos
3
4
isin
3
4
4
2
4
or
3
4
32
2
2
i
2
2
162
162
i
55. r2(5) or 10 v2 0.5 or 2.5
10 (cos 2.5 isin 2.5)
8.01 5.98i
56. r
8
2
or 4 v
7
6
5
3
4
cos
2
isin
2
7
6
10
6
or
2
4(0 i(1))
4i
57. r
6
4
or
3
2
v
2
6
3
2
cos
3
isin
3
3
6
6
or
3
3
2
1
2
i
2
3
3
4
3
4
2
i
58. r
2
4
.
.
2
4
or 0.5 v1.5 0.6 or 0.9
0.5 (cos 0.9 isin 0.9)
0.31 0.39i
59. r222
2
22
vArctan
2
2
4
(2
)8
cos (8)
4
isin (8)
4
4096 (cos 2isin 2)
4096
27
cos (7)
6
isin (7)
6
128
cos
7
6
isin
7
6
128
2
3
i
1
2
643
64i
61. r(1)2
12
2
vArctan
1
1
3
4
(2
)4
cos (4)
3
4
isin (4)
3
4
4(cos 3isin 3)
4
62. r(2)2
(2
)2
22
vArctan
2
2
5
4
(22
)3
cos (3)
5
4
isin (3)
5
4
162
cos
15
4
isin
15
4
162
2
2
i
2
2
16 16i
63. r021
2
1v
2
1
1
4
cos
1
4
2
isin
1
4
2
cos
8
isin
8
0.92 0.38i
64. r(3
)2
12
2vArctan
6
2
1
3
cos
1
3
6
isin
1
3
6
2
1
3
cos
1
8
isin
1
8
1.24 0.22i
Page 611 Applications and Problem Solving
65. lemniscate
66. r752
1252
vArctan
1
7
2
5
5
21,25
0
145.77 59.04°
(145.77, 59.04°)
67. rcos
v
2
5 0
rcos cos
2
rsin vsin
2
5 0
rsin v5 0
y5 0
y5
68. I
E
Z
50
4
1
5
8
j
0j
50
4
1
5
8
j
0j
4
4
5
5
j
j
1100
41
470j2
26.83 11.46jamps
200 470j900j2
16 25j2
1
3
Page 611 Open-Ended Assessment
1a. Sample answer: 4 6iand 3 2i
(4 6i) (3 2i) (4 3) (6i2i)
7 4i
1b. No. Sample explanation: 2 3iand 5 ialso
have this sum.
(2 3i) (5 i) (2 5) (3i(i))
7 4i
2a. Sample answer: 4 i
z421
2
17
2b. No. Sample explanation: 1 4ialso has this
absolute value.
z124
2
17
Chapter 9 SAT & ACT Preparation
Page 613 SAT and ACT Practice
1. ∠aand ∠bform a linear pair, so ∠bis
supplementary to ∠a. Since ∠band ∠dare
vertical angles, they are equal in measure. So ∠d
is also supplementary to ∠a. Since ∠dand ∠fare
alternate interior angles, they are equal. So ∠fis
supplementary to ∠a. And since ∠fand ∠hare
vertical angles, ∠his supplementary to ∠a. The
angles supplementary to ∠aare angles b, d, f, and
h. The correct choice is A.
2. Draw the given triangle and draw the height h
from point B.
The answer choices include sin x. Write an
expression for the height, using the sine of x.
sin x
h
8
A
1
2
bh
8 sin xh
1
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∠Qm∠Sb.
In SMO, cba180 or a180 (cb).
Also, xa180 or a180 xsince consecutive
interior angles are supplementary.
180 (cb) 180 x
xcb
The correct choice is E.
4. Volume wh
16,500 75 w10
16,500 750w
22 w
The correct choice is A.
5.
100
1100
10
1
99
100
1100
10
11
0
00
1
01
9
00
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.
5x26x70
5x26y10
6x6y60
xy10, so 10x10y100.
The correct choice is E.
8. Since ∠Bis a right angle, ∠Cis a right angle also,
because they are alternate interior angles.
In the triangle containing ∠C, 90 xy180 or
xy90.
The straight angle at Dis made up of 3 angles.
120 xx180
2x60 or x30
xy90
(30) y90
y60
The correct choice is B.
9. In the slope-intercept form of a line, y mx b,
mrepresents the slope of the line, and b
represents the y-intercept. Since the slope is
given as
3
2
, the slope-intercept form of the line is
y
3
2
x b.
Since (–3, 0) is on the line, it satisfies the
equation. 0
3
2
(–3) b. So b
9
2
.
The correct choice is D.
10. Note that consecutive interior angles are
supplementary.
110 2x180 yx180
2x70 y(35) 180
x35 y145
The answer is 145.
305 Chapter 9
h
10
8
B
C
A
x
˚
MQ
T
O
R
P
NS
x
˚
b
˚
c
˚
a
˚
h
10 ft
w
75 ft
Chapter 10 306
Introduction to Analytic Geometry
Pages 619–620 Check for Understanding
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.
3a. Yes; the distance from Bto Ais
a
2
5
and the
distance from Bto Cis also
a
2
5
.
3b. Yes; the distance from Bto Ais a2b
2
and
the distance from Cto Ais also a2b
2
.
3c. No; the distance from Ato Bis a2b
2
, the
distance from Ato Cis ab, and the distance
from Bto Cis 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
d021
02
d100
or 10
x1
2
x2
,
y1
2
y2
5
2
5
,
1
2
11
(5, 6)
6. d(x2
x2)2
(y2
y1)2
d(4
0)2
(3
0)2
d(4)2
(3
)2
d25
or 5
x1
2
x2
,
y1
2
y2
0
2
(4)
,
0
2
(3)
(2, 1.5)
7. d[0 (
2)]2
(4
2)2
d222
2
d8
or 22
x1
2
x2
,
y1
2
y2
2
2
0
,
2
2
4
(1, 3)
10-1 8. AB (x2
x1)2
(y2
y1)2
(6 3
)2(2
4)2
13
slope of A
B
m
y
x
2
2
y
x
1
1
2
6
4
3
or
2
3
DC (x2
x1)2
(y2
y1)2
(8 5
)2(7
9)2
13
slope of D
C
m
y
x
2
2
y
x
1
1
7
8
9
5
or
2
3
Yes; A
B
D
C
since AB 13
and DC 13
,
and A
B
D
C
since the slope of A
B
is
2
3
and the
slope of D
C
is also
2
3
.
9. XY (x2
x1)2
(y2
y1)2
[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
Y
X
Z
, since XY 217
and XZ 217
,
therefore XYZ is isosceles.
10a.
10b. BD (c0
)2(a
0)2
c2a
2
AC (c0
)2(0
a)2
c2a
2
Thus, A
C
B
D
.
10c. The midpoint of A
C
is
2
c
,
a
2
. The midpoint of
B
D
is
2
c
,
a
2
. Therefore, the diagonals intersect
at their common midpoint, E
2
c
,
a
2
. Thus,
A
E
E
C
and B
E
E
D
.
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.
x1
2
x2
,
y1
2
y2
0
2
80
,
0
2
120
(40, 60)
Chapter 10 Conics
y
x
B
(0, 0)
A
(0,
a
)
D
(
c
,
a
)
E
C
(
c
, 0)
O
11b. d(x2
x1)2
(y2
y1)2
d(40
0)2
(60
0)2
d402
602
d5200
d2013
or about 72 yards
Pages 620–622 Exercises
12. d(x2
x1)2
(y2
y1)2
d[4 (
1)]2
(13
1)2
d521
22
d169
or 13
x1
2
x2
,
y1
2
y2
1
2
4
,
1
2
13
(1.5, 7)
13. d(x2
x1)2
(y2
y1)2
d(1
1)2
(3
3)2
d(2)2
(6
)2
d40
or 210
x1
2
x2
,
y1
2
y2
1
2
(1)
,
3
2
(3)
(0, 0)
14. d(x2
x1)2
(y2
y1)2
d(0 8
)2(8
0)2
d(8)2
82
d128
or 82
x1
2
x2
,
y1
2
y2
8
2
0
,
8
2
0
(4, 4)
15. d(x2
x1)2
(y2
y1)2
d[5 (
1)]2
[3
(
6)]2
d623
2
d45
or 35
x1
2
x2
,
y1
2
y2
1
2
5
,
6
2
(3)
(2, 4.5)
16. d(x2
x1)2
(y2
y1)2
d(72
3
2
)2
[1
(5)]
2
d(42
)
242
d48
or 43
x1
2
x2
,
y1
2
y2
,
(52
, 3)
17. d(x2
x1)2
(y2
y1)2
d(aa
)2(
9 7
)2
d02(
16)2
d256
or 16
x1
2
x2
,
y1
2
y2
a
2
a
,
7
2
(9)
(a, 1)
18. d(x2
x1)2
(y2
y1)2
d[r2
(6
r)]2
(s
s)2
d(8)2
02
d64
or 8
x1
2
x2
,
y1
2
y2
6r
2
r2
,
s
2
s
2r
2
4
,
2
2
s
(r2, s)
5(1)
2
32
72
2
19. d(x2
x1)2
(y2
y1)2
d(c2
c)2
(d
1
d)2
d22(
1)2
d5
x1
2
x2
,
y1
2
y2
cc
2
2
,
dd
2
1
2c
2
2
,
2d
2
1
c1, d
1
2
20. d(x2
x1)2
(y2
y1)2
d[w(
w2
)]2(
4w
w)2
d22(
3w)2
d4 9
w2
or 9w2
4
x1
2
x2
,
y1
2
y2
w2
2
w
,
w
2
4w
w1,
5
2
w
21. d(x2
x1)2
(y2
y1)2
20 (2a
a)2
[7
(9)]
2
20 (3a)
216
2
20 9a2
256
400 9a2256
144 9a2
a216
a16
or 4
22. Let Dhave coordinates (x2, y2).
4
2
x2
,
1
2
y2
3,
5
2
4
2
x2
3
1
2
y2
5
2
4 x261 y25
x210 y26
Then Dhas coordinates (10, 6).
23. Let the vertices
of the quadrilateral
be A(2, 3), B(2, 3),
C(2, 3), and D(3, 2).
Aquadrilateral is a
parallelogram if one pair
of opposite sides are
parallel and congruent.
A
D
and B
C
are one pair of opposite sides.
slope of A
D
slope of B
C
m
y
x
2
2
y
x
1
1
m
y
x
2
2
y
x
1
1
3
2
(
3
2)
2
3
(
(
3
2
)
)
1
5
1
5
Their slopes are equal, therefore A
D
B
C
.
AD (x2
x1)2
(y2
y1)2
[3 (
2)]2
(2
3)2
52(
1)2
26
BC (x2
x1)2
(y2
y1)2
[2 (
3)]2
[3
(
2)]2
52(
1)2
26
The measures of A
D
and B
C
are equal.
Therefore A
D
B
C
. Since A
D
B
C
and A
D
B
C
,
quadrilateral ABCD is a parallelogram; yes.
307 Chapter 10
O
y
x
A
(2, 3)
B
(3, 2)
C
(2, 3)
D
(3, 2)
24. Let the vertices of the
quadrilateral be
A(4, 11), B(8, 14),
C(4, 19), and D(0, 15).
Aquadrilateral is a
parallelogram if both
pairs of opposite sides
are parallel.
A
B
and D
C
are one pair
of opposite sides.
slope of A
B
slope of D
C
m
y
x
2
2
y
x
1
1
m
y
x
2
2
y
x
1
1
1
8
4
4
11
19
4
1
0
5
3
2
4
4
or 1
Since A
B
⁄D
C
, 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)
m
y
x
2
2
y
x
1
1
m
y
x
2
2
y
x
1
1
3
8
1
1
5
k
3
(
(
8
3
)
)
1
9
8
or
1
2
k
6
8
k
6
8
1
2
⇒k5
26. d(x2
x1)2
(y2
y1)2
AB [1
(3)]
2(2
3
0)2
16
or 4
BC [1 (
1)]2
(0
23
)
2
16
or 4
CA (3
1)2
(0
0)2
16
or 4
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
(4 2
)2(4
5)2
5
HG (x2
x1)2
(y2
y1)2
(2 0
)2(0
1)2
5
slope of E
F
m
y
x
2
2
y
x
1
1
4
4
5
2
or
1
2
slope of F
G
m
y
x
2
2
y
x
1
1
0
2
4
4
or
2
1
slope of H
G
m
y
x
2
2
y
x
1
1
0
2
1
0
or
1
2
E
F
H
G
since EF 5
and HG 5
.
E
F
H
G
since the slope of E
F
is
1
2
and the slope
of H
G
is
1
2
. Thus the points form a
parallelogram. E
F
⊥F
G
since the product of the
slopes of E
F
and F
G
,
1
2
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 Dbe the midpoint of A
B
and E
be the midpoint of B
C
.
The coordinates of Dare
0
2
b
,
0
2
c
or
b
2
,
2
c
.
The coordinates of Eare
b
2
a
,
c
2
0
or
b
2
a
,
2
c
.
AC (a0
)2(0
0)2
a2
or a
DE
b
2
a
b
2
2
2
c
2
c
2
a
4
2
or
a
2
Since DE
1
2
AC, 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 Aand Bhave coordinates
(0, 0)and(b,0), respectively. To make the trapezoid
isosceles, let Chave coordinates (ba, c) and let
Dhave coordinates (a, c).
AC (a0
)2(c
0)2
a2c
2
BD (ba
b)2
(c
0)2
a2c
2
AC a2c
2
a2c
2
BD, so the
diagonals of an isosceles trapezoid are congruent.
Chapter 10 308
O
y
x
44
8
4
12
16
20
812
A
(4, 11)
D
(0, 15)
B
(8, 14)
C
(4, 19)
y
x
B
(
b
,
c
)
DE
C
(
a
, 0)
A
(0, 0)
O
y
x
A
(0, 0)
C
(
a
,
c
)
D
(
b
a
,
c
)
B
(
b
, 0)
O
30. In ABC, let the vertices be A(0, 0) and B(a, 0).
Since A
C
and B
C
are congruent sides, let the third
vertex be C
a
2
, b
. Let Dbe the midpoint of A
C
and let Ebe the midpoint of B
C
.
The coordinates of Dare:
,
b
2
0
or
a
4
,
b
2
.
The coordinates of Eare:
,
b
2
0
or
3
4
a
,
b
2
.
AE
3
4
a
0
2
b
2
0
2
1
2
9
4
a2
b2
BD
a
a
4
2
0
b
2
2
1
2
9
4
a2
b2
Since AE BD, the medians to the congruent
sides of an isosceles triangle are congruent.
31. Let Aand Bhave coordinates (0, 0) and (b, 0)
respectively. To make a parallelogram, let Chave
coordinates (ba, c) and let Dhave coordinates
(a, c).
The midpoint of B
D
is
a
2
b
,
0
2
c
or
a
2
b
,
2
c
.
The midpoint of A
C
is
a
2
b0
,
c
2
0
or
a
2
b
,
2
c
.
Since the diagonals have the same midpoint, the
diagonals bisect each other.
a
2
a
2
a
2
0
2
32. Let the vertices of quadrilateral ABCD be A(a, e),
B(b, f), C(c, g), and D(d, h). The midpoints of A
B
,
B
C
, C
D
, and D
A
, respectively,
are L
a
2
b
,
e
2
f
, M
b
2
c
,
f
2
g
,
N
c
2
d
,
g
2
h
, and P
a
2
d
,
e
2
h
.
309 Chapter 10
y
x
A
(0, 0)
C
,
b
()
a
2
ED
B
(
a
, 0)
O
y
x
A
(0, 0)
C
(
a
b
,
c
)
D
(
a
,
c
)
B
(
b
, 0)
O
The slope of L
M
is or
c
g
a
e
.
f
2
g
e
2
f
b
2
c
a
2
b
The slope of N
P
is or
a
e
g
c
.
These slopes are equal, so L
M
N
P
.
The slope of M
N
is or
b
f
h
d
.
f
2
g
g
2
h
b
2
c
c
2
d
e
2
h
g
2
h
a
2
d
c
2
d
y
x
A
(
a
,
e
)
B
(
b
,
f
)
D
(
d
,
h
)
C
(
c
,
g
)
M
P
L
N
OO
O
y
x
B
A
C
D
y
x
101020 20 30
10
10
20
O
The slope of P
L
is or
d
h
b
f
.
e
2
h
e
2
f
a
2
d
a
2
b
These slopes are equal, so M
N
P
L
. Since
L
M
N
P
, and M
N
P
L
, PLMN is a parallelogram.
33. Let the vertices of the
rectangle be A(3, 1),
B(1, 3), C(3, 1), and
D(1, 3). Since the area
of a rectangle is the length
times the width, find the
measure of two consecutive
sides, A
D
and D
C
.
AD (x2
x1)2
(y2
y1)2
[1 (
3)]2
(3
1)2
42(
4)2
32
or 42
DC (x2
x1)2
(y2
y1)2
(3 1
)2[
1 (
3)]2
222
2
8
or 22
Area w
(42
)(22
)
16
The area of the rectangle is 16 units2.
34a.
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
432
(12
)2
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(2a2c, 2b),
D(2c, 2b). The coordinates of the midpoints are:
P(a, 0), Q(2ac, b), R(a, 2b), S(c, b).
PQ (2a
ca
)2(b
0)2
(ac
)2b
2
QR (2a
ca
)2(b
2b
)2
(ac
)3b
2
RS (ac
)2(2
bb
)2
(ac
)2b
2
PS (ac
)2(0
b)2
(ac
)2b
2
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
d[1 (
3)]2
(3
2)2
d41
or 241
meters
distance from rosebushes to bench:
d(x2
x1)2
(y2
y1)2
d(3 1
)2[3
(
3)]2
d210
or 410
meters
distance from bench to fountain:
d(x2
x1)2
(y2
y1)2
d(3
3)2
(2
3)2
d37
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).
x1
2
x2
,
y1
2
y2
3
2
1
,
2
2
(3)
1,
1
2
37a. Find a representation for MA and for MB.
MA t2(
3t1
5)2
t29
t29
0t2
25
10t2
90t
225
MB (t9
)2(3
t12
)2
t21
8t8
1 9
t27
2t1
44
10t2
90t
225
By setting these representations equal to each
other, you find a value for tthat would make the
two distances equal.
MA MB
10t2
90t
225
10t2
90t
225
Since the above equation is a true statement,
tcan take on any real values.
37b. Aline; this line is the perpendicular bisector of
A
B
.
38. ra2b
2
vArctan
a
b
(5)2
122
Arctan
12
5
169
or 13 1.176005207
(5 12i)2132[cos 2vi sin 2v]
119 120i
39. If v
u
(115, 2018, 0), then
v
u
1152
2018
202
40855
49
or about 2021
The magnitude of the force is about 2021 N.
40. 2 sec2x
1
1
sin x
1
1
sin x
2 sec2x
2 sec2x
1s
2
in2x
2 sec2x
cos
2
2x
2 sec2x2 sec2x
41. srv
11.5 12v
v
1
1
1
2
.5
radians
1
1
1
2
.5
18
0°
54.9°
42. sin 390° sin (390° 360°)
sin 30° or
1
2
43. z28z14
z28z16 14 16
(z4)22
z4 2
z4 2
44. x216
x16
or 4
y24
y4
or 2
Evaluating (xy)2when x4 and y2
results in the greatest possible value, [4 (2)]2
or 36.
Circles
Page 627 Check for Understanding
1. complete the square on each variable
10-2
(1 sin x) (1 sin x)
(1 sin x)(1 sin x)
Chapter 10 310
y
x
A
(0, 0)
C
(2
a
2
c
, 2
b
)
D
(2
c
, 2
b
)
R
SQ
PB
(2
a
, 0)
O
2. Sample answer: (x4)2(y9)21,
(x4)2(y9)22, (x4)2(y9)23,
(x4)2(y9)24, (x4)2(y9)25
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
(xh)2(yk)2r2.
4. The equation x2y28x8y36 0 written
in standard form is (x4)2(y4)24.
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. (xh)2(yk)2r2
(x0)2(y0)292
x2y281
7. (xh)2(yk)2r2
[x(1)]2(y4)2[3 (1)]2
(x1)2(y4)216
8. x2y24x14y47 0
x24x4 y214y49 47 4 49
(x2)2(y7)2100
9. 2x22y220x8y34 0
2x220x2y28y34
2(x210x25) 2(y34y4) 34 2(25)
2(4)
2(x5)22(y2)224
(x5)2(y2)212
10. x2y2Dx Ey F0
0202D(0) E(0) F0
4202D(4) E(0) F0
0242D(0) E(4) F0
F0F0
4DF16 ⇒D4
4EF16 E4
x2y24x4y0
x24x4 y24y4 0 4 4
(x2)2(y2)28
center: (h, k) (2, 2)
radius: r28
r8
or 22
11. x2y2Dx Ey F0
1232D(1) E(3) F0⇒D3EF10
5252D(5) E(5) F0⇒5D5EF50
5232D(5) E(3) F0⇒5D3EF34
D3EF10
(1)(5D5EF) (1)(50)
4D2E40
5D5EF50
(1)(5D3EF) (1)(34)
2E16
E8
4D2(8) 40
4D24 (6) 3(8) F10
D6F20
x2y26x8y20 0
x26x9 y28y16 20 9 16
(x3)2(y4)25
center: (h, k) (3, 4)
radius: r25
r5
12. (xh)2(yk)2r2
[x(2)]2(y1)2r2
(x2)2(y1)2r2
(1 2)2(5 1)2r2
25 r2
(x2)2(y1)225
311 Chapter 10
y
x
(9, 0)
(0, 9)
O
y
x
(3, 4)
(1, 4)
(1, 8)
O
y
x
(2, 3)
(2, 7)(8, 7)
O
y
x
(5, 2)
(5,2 23)
(5 23, 2)
O
13. midpoint of diameter:
x1
2
x2
,
y1
2
y2
2
2
10
,
6(
2
10)
(4, 2)
radius: r(x2
x1)2
(y2
y1)2
[4 (
2)]2
[6
(2)]
2
628
2
100
or 10
(xh)2(yk)2r2
(x4)2[y(2)]2102
(x4)2(y2)2100
14. (xh)2(yk)2r2
(x0)2(y0)2(1740 185)2
x2y219252
Pages 627–630 Exercises
15. (xh)2(yk)2r2
(x0)2(y0)252
x2y225
16. (xh)2(yk)2r2
[x(4)]2(y7)2(3
)2
(x4)2(y7)23
17. (xh)2(yk)2r2
[x(1)]2[y(3)]2
2
(x1)2(y3)2
1
2
2
2
18. (xh)2(yk)2r2
[x(5)]2(y0)2
9
2
2
(x5)2y2
8
4
1
19. (xh)2(yk)2r2
(x6)2(y1)262
(x6)2(y1)236
20. (xh)2(yk)2r2
(x3)2[y(2)]2[2 (2)]2
(x3)2(y2)216
21. 36 x2y2
x2y236
22. x2y2y
3
4
x2y2y
1
4
3
4
1
4
x2
y
1
2
21
Chapter 10 312
y
x
(0, 5)
(5, 0)
O
(4 3, 7)
(4, 7
(4, 7)
3)
y
x
O
6
2
2
22
2
(1, 3)
()
()
1,
, 3
y
x
O
9
2
(5, 0)
()
5,
19
2
()
, 0
y
x
O
(0, 1) (6, 1)
(6, 7)
y
x
O
y
x
(3, 2)
(3, 2)
(7, 2)
O
y
x
(0, 6)
(6, 0)
O
y
x
O
(
1,
)
1
2
(
0,
)
1
2
(
0,
)
1
2
23. x2y24x12y30 0
x24x4 y212y36 30 4 36
(x2)2(y6)210
24. 2x22y22x4y1
2x22x2y24y1
2
x21x
1
4
2(y22y1) 12
1
4
2(1)
2
x
1
2
22(y1)2
3
2
x
1
2
2(y1)2
3
4
25. 6x212x6y236y36
6(x22x1) 6(y26y9) 36 6(1) 6(9)
6(x1)26(y3)296
(x1)2(y3)216
26. 16x216y28x32y127
16x28x16y232y127
16
x2
1
2
x
1
1
6
16(y22y1)2127 16
1
1
6
16(1)
16
x
1
4
216(y1)2144
x
1
4
2(y1)29
27. x2y214x24y157 0
x214x49 y224y144 157 49 144
(x7)2(y12)236
28. x2y2Dx Ey F0
02(1)2D(0) E(1) F0 ⇒
EF1
(3)2(2)2D(3) E(2) F0 ⇒
3D2EF13
(6)2(1)2D(6) E(1) F0 ⇒
6DEF37
EF1
(1)(3D2EF) (1)(13)
3DE12
3D2EF13
(1)(6D2EF) (1)(37)
3D2E24
3DE12 3(6) E12
3DE24 E6
6D36
D6(6) F1
F7
x2y26x6y7 0
x26x9 y26y9 7 9 9
(x3)2(y3)225
center: (h, k) (3, 3)
radius: r225
r25
or 5
29. x2y2Dx Ey F0
72(1)2D(7) E(1) F0 ⇒
7DEF50
112(5)2D(11) E(5) F0 ⇒
11D5EF146
32(5)2D(3) E(5) F0 ⇒
3D5EF34
7DEF50
(1)(11D5EF) 1(146)
4D4E96
11D5EF146
(1)(3D5EF) 1(34)
8D12
D14
4(14) 4E96
4E40 7(14) (10) F50
E10 F58
x2y214x10y58 0
x214x49 y210y25 58 49 25
(x7)2(y5)216
center: (h, k) (7, 5)
radius: r216
r16
or 4
313 Chapter 10
y
x
O
(2, 6 10)
(2, 6)
(2 10, 6)
y
x
O
(
1 3, 1
)
2
2
(
1, 1
)
2
(
1, 1
2 3
)
y
x
O
(1, 3)
(3, 3)
(1, 7)
()
,1
2
13
4
()
,
1
4
1
()
,
1
4
y
x
O
(7, 12)
(1, 12)
(7, 6)
y
x
O
30. x2y2Dx Ey F0
(2)272D(2) E(7) F0 ⇒
2D7EF53
(9)202D(9) E(0) F0 ⇒
9DF81
(10)2(5)2D(10) E(5) F10 ⇒
10D5EF125
2D7EF 53
(1)(9DF) (1)(81)
7D7E28
DE4
10D5EF125
(1)(9DF) (1)(81)
D5E44
D5E44
DE4
4E40
E10
D(10) 4
D6
9(6) F81
F135
x2y26x10y135 10
x26x9 y210y25 135 9 25
(x3)2(y5)2169
center: (h, k) (3, 5)
radius: r2169
r169
or 13
31. x2y2Dx Ey F0
(2)232D(2) E(3) F0 ⇒
2D3EF13
62(5)2D(6) E(5) F0 ⇒
6D5EF61
0272D(0) E(7) F0 ⇒
7EF49
2D3EF13
(1)(6D5EF) (1)(61)
8D8E48
DE6
6D5EF61
(1)(7EF) (1)(49)
6D12E12
D2E2
DE6
D2E2
E4
E4
D2(4) 2
D10
7(4) F49
F21
x2y210x4y21 0
x210x25 y24y4 21 25 4
(x5)2(y2)250
center: (h, k) (5, 2)
radius: r250
r50
or 52
32. x2y2Dx Ey F0
4252D(4) E(5) F0 ⇒
4D5EF41
(2)232D(2) E(3) F0 ⇒
2D3EF13
(4)2(3)2D(4) E(3) F0 ⇒
4D3EF25
4D5EF41
(1)(2D3EF) (1)(13)
6D2E28
3DE14
4D5EF41
(1)(4D3EF) (1)(25)
8D8E16
DE2
3DE14
(1)(DE)(1)(2)
2D12
D6
6E2
E4
2(6) 3(4) F13
F37
x2y26x4y37 0
x26x9 y24y4 37 9 4
(x3)2(y2)250
center: (h, k) (3, 2)
radius: r250
r50
or 52
33. x2y2Dx Ey F0
1242D(1) E(4) F0 ⇒
D4EF17
22(1)2D(2) E(1) F0 ⇒
2DEF5
(3)202D(3) E(0) F0 ⇒
3DF9
D4EF 17
(1)(2DEF) (1)(5)
D5E12
D4EF17
(1)(3DF) (1)(9)
4D4E8
DE2
D5E12
DE2
6E14
E
7
3
D
7
3
2
D
1
3
3
1
3
F9
F8
x2y2Dx Ey F0
x2y2
1
3
x
7
3
y8 0
x2
1
3
x
3
1
6
y2
7
3
y
4
3
9
6
8
3
1
6
4
3
9
6
x
1
6
2
y
7
6
2
1
1
6
8
9
Chapter 10 314
center: (h, k)
1
6
,
7
6
radius: r2
1
1
6
8
9
r
1
1
6
8
9
3
1
3
2
or
13
6
2
34. x2y2Dx Ey F0
0202D(0) E(0) F0 ⇒F0
(2.8)202D(2.8) E(0) F0 ⇒
2.8DF7.84
(5)222D(5) E(2) F0 ⇒
5D2EF29
2.8D0 7.84 5(2.8) 2E(0) 29
2.8D7.84 2E15
D2.8 E7.5
x2y22.8x7.5y0 0
x22.8x1.96 y27.5y14.0625 1.96 14.0625
(x1.4)2(y3.75)216.0225
or about (x1.4)2(y3.75)216.02
35. (xh)2(yk)2r2
[x(4)]2(y3)2r2
(x4)2(y3)2r2
(0 4)2(0 3)2r2
25 r2
(x4)2(y3)225
36. (xh)2(yk)2r2
(x2)2(y3)2r2
(5 2)2(6 3)2r2
18 r2
(x2)2(y3)218
37. midpoint of diameter:
x1
2
x2
,
y1
2
y2
2
2
(6)
,
3
2
(5)
(2, 1)
r(x2
x1)2
(y2
y1)2
(2
2)2
(1
3)2
(4)2
(4
)2
32
(xh)2yk)2r2
[x(2)]2[y(1)]2
32
2
(x2)2(y1)232
38. midpoint of diameter:
x1
2
x2
,
y1
2
y2
3
2
2
,
4
2
1
1
2
,
5
2
r(x2
x1)2
(y2
y1)2
1
2
2
2
5
2
1
2
5
2
2
3
2
2
1
2
7
(xh)2(yk)2r2
x
1
2
2
y
5
2
2
1
2
7
2
x
1
2
2
y
5
2
2
1
2
7
39. (xh)2(yk)2r2
(x5)2(y1)2r2
x3y2
x3y2 0 ⇒A1, B3, and C2
r
Ax1By1C
A2
B2
315 Chapter 10
10
10
or 10
(x5)2(y1)210
2
(x5)2(y1)210
40. center: (h, 0), radius: r1
(xh)2(yk)2r2
2
2
h
2
2
2
0
212
1
2
2
hh2
1
2
1
h22
h1 1
hh2
0
h0 or h2
(x0)2(y0)21x2
2(y0)21
x2y21orx2
2y21
41a. (xh)2(yk)2r2
(x0)2(y0)2
1
2
2
2
x2y236
41b. x2y236
y236 x2
y36
x2
dimensions of rectangle:
2xby 2y⇒2xby 236
x2
41c. A(x) 2x236
x2
4x36
x2
41d.
[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.
42a.
[15.16, 15.16] scl:1 by [5, 5] scl:1
42b. a circle centered at (2, 3) with radius 4
42c. (x2)2(y3)216
(1)(5) (3)(1) 2
123
2
42d. center: (h, k) (4, 2)
radius: r236
r36
or 6
[DRAW] 9:Circle( 4 2 6
[15.16, 15.16] scl:1 by [5, 5] scl:1
43a. (xh)2(yk)2r2
(x0)2(y0)2
2
2
4
2
x2y2144
43b. x2y26.25 ⇒r126.25
r16.25
or 2.5
If the circles are equally spaced apart then
radius r2of the middle circle is found by adding
the radius of the smallest circle to the radius of
the largest circle and dividing by two.
r2
12
2
2.5
or 7.25
area of area of area of
region Bmiddle circle smallest circle
r22r12
(r22r12)
(7.2522.52)
(46.3125) or about 145.50
The area of region Bis about 145.50 in2.
44.
x2y2Dx Ey F0
2212D(2) E(1) F0 ⇒
2DEF5
52(1)2D(5) E(1) F0 ⇒
5DEF26
12232D(12) E(3) F0 ⇒
12D3EF153
2DEF5
(1)(5DEF) (1)(26)
3D2E21
2DEF5
(1)(12D3EF) (1)(153)
10D2E148
3D2E21
10D2E148
13D169
D13
3(13) 2E21
2E18
E9
)
,,-
2nd
2(13) (9) F5
F30
x2y213x9y30 0
x213x42.25y29y20.253042.2520.25
(x6.5)2(y4.5)232.5
45a. (xh)2(yk)2r2
(xk)2(yk)222
(xk)2(yk)24
45b. (x1)2(y1)24
(x0)2(y0)24
(x1)2(y1)24
45c. All of the circles in this family have a radius of 2
and centers located on the line yx.
46a. (xh)2(yk)2r2
(x0)2(y0)2142
x2y2196
y2196 x2
y196
x2
y196
x2
46b. No, if x7, then y147
12.1 ft, so the
truck cannot pass.
47. x2y28x6y25 0
x28x16 y26y9 25 16 9
(x4)2(y3)20
radius: r20
r0
or 0
center: (h, k) (4, 3)
Graph is a point located at (4, 3).
48a. (xh)2(yk)2r2
(x0)2(y0)2r2
x2y2r2
(475)2(1140)2r2
1,525,225 r2
x2y21,525,225
48b. r21,525,225
r1,525
,225
or 1235
Ar2
(1235)2or approximately 4,792,000 ft2
48c. 0.23328
about 23%
49a. P
A
has a slope of
y
x
4
3
and P
B
has slope of
y
x
4
3
. If P
A
⊥P
B
then
y
x
4
3
y
x
4
3
1.
y
x
4
3
y
x
4
3
1
y
x
2
2
1
9
6
1
y216 x29
x2y225
49b. If P
A
⊥P
B
, then A, P, and Bare on the circle
x2y225.
50. d(x2
x1)2
(y2
y1)2
d(2
4)2
[6
(3)]2
d(6)2
92
d117
250024,792,000
25002
Chapter 10 316
O
y
x
(5, 1)
(2, 1) (12, 3)
2
x
3
y
7 0
4
x
7
y
27
x
5
y
3 0
y
x
y
x
O
51. (2 i)(3 4i)(1 2i)(6 8i3i4i2)(1 2i)
[6 8i3i4(1)](1 2i)
(10 5i)(1 2i)
10 20i5i10i2
10 20i5i10(1)
20 15i
52. xtv
u
cos vytv
u
sin v
1
2
gt2
xt(60) cos 60° yt(60) sin 60°
1
2
(32)t2
x60tcos 60° y60tsin 60° 16t2
x60(0.5) cos 60° y60(0.5) sin 60° 16(0.5)2
x15 y21.98076211
15 ft horizontally, about 22 ft vertically
53. A
5
2
or 2.5
20
2
k
and k
1
0
yAcos (kt)
y2.5 cos
1
0
t
54. s
1
2
(abc)
1
2
(15 25 35)
37.5
ks(s
a)(s
b)(s
c)
37.5(3
7.5
15)(3
7.5
25)(3
7.5
3.5)
26,36
7.187
5
162 units2
55. vv0
26
4h
95 152
64h
95215264h
h
952
6
4
152
h137.5 ft
56. y6x43x21
b6a43a21(x, y) (a, b)
x-axis: (x, y) (a, b)
b6a43a21; no
y-axis: (x, y) (a, b)
b6(a)43(a)21
b6a43a21; yes
yx: (x, y) (b, a)
a6b43b21; no
origin: f(x) f(x)
f(x)6(x)43(x)21f(x)(6x43x21)
f(x)6x43x21f(x)6x43x21
no
The graph is symmetric with respect to the y-axis.
57a. Let xnumber of cases of drug A.
Let ynumber of cases of drug B.
x10
y9
3x5y55
x5y45
(0, 0), (0, 9), (5, 8), (10, 5)
f(x, y) 320x500y
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.
xxxxxyy5x2x
The correct choice is A
Ellipses
Pages 637–638 Check For Understanding
1.
a
y2
2
b
x2
2
1
8
y2
2
5
x2
2
1
6
y
4
2
2
x
5
2
1
2. Since the foci lie on the major axis, determine
whether the major axis is horizontal or vertical. If
the a2is the denominator of the xterms, the major
axis is horizontal. If the a2is the denominator of
the yterms, the major axis is vertical.
3. When the foci and center of an ellipse coincide,
c0.
c2a2b2
0 a2b2
b2a2
ba
The figure is a circle.
4. In an ellipse, b2a2c2and
a
c
e.
a
c
e
cae
c2a2e2
5. Shanice; an equation with only one squared term
cannot be the equation of an ellipse.
6. center: (h, k) (7, 0)
a0 6or 6
b7 (4)or 3
(y
a2k)2
(x
b2h)2
1
(y
620)2
[x
3
(
27)]2
1
3
y
6
2
(x
9
7)2
1
ca2b
2
foci: (h, kc) (7, 0 33
)
c623
2
(7, 33
)
c33
10-3
317 Chapter 10
O
y
x
4
8
4
12
812162024
(0, 9)
(0, 0)
(5, 8)
(10, 5)
x
5
y
45
3
x
5
y
55
x
10
y
9
y
0
x
0
xxxxx
y
y
e
a
c
e
a
0
e0
b2a2c2
b2a2a2e2
b2a2(1 e2)
7. center: (h, k) (0, 0)
a236 b24ca2b
2
a36
or 6 b4
or 2 c36
4
or 42
foci: (hc, k) (0 42
, 0) or (42
, 0)
major axis vertices: (ha,k)(0 6, 0) or (6, 0)
minor axis vertices: (h,kb)(0, 0 2) or (0, 2)
8. center: (h, k) (0, 4)
a281 b249 ca2b
2
a81
or 9 b49
or 7 c81 4
9
or 42
foci: (hc, k) (0 42
, 4) or (42
, 4)
major axis vertices: (ha, k) (0 9, 4) or
(9, 4)
minor axis vertices: (h, kb) (0, 4 7) or
(0, 11), (0, 3)
9. 25x29y2100x18y116
25(x24x?) 9(y22y?) 116 ??
25(x24x4) 9(y22y1) 116 25(4) 9(1)
25(x2)29(y1)2225
(x
9
2)2
(y
25
1)2
1
center: (h, k) (2, 1)
a225 b29ca2b
2
a25
or 5 b9
or 3 c25
9
or 4
foci: (h, kc) (2, 1 4) or (2, 5), (2, 3)
major axis vertices: (h, ka) (2, 1 5) or
(2, 6), (2, 4)
minor axis vertices: (hb, k) (2 3, 1) or
(1, 1), (5, 1)
10. 9x24y218x16y11
9(x22x?) 4(y24y?) 11 ? ?
9(x22x1) 4(y24y4) 11 9(1) (4)
9(x1)24(y2)236
(x
4
1)2
(y
9
2)2
36
center: (h, k) (1, 2)
a29b24ca2b
2
a9
or 3 b4
or 2 c9 4
or 5
foci: (h, kc) 1, 2 5
major axis vertices: (h, ka) (1, 2 3) or
(1, 1), (1, 5)
minor axis vertices: (hb, k) (1 2, 2) or
(3, 2), (1, 2)
11. center: (h, k) (2, 3)
a
8
2
or 4
b
2
2
or 1
(y
a2
k)2
(x
b2
h)2
1
[y
4
(
2
3)]2
[x
1
(
2
2)]2
1
(y
16
3)2
(x
1
2)2
1
12. The major axis contains the foci and it is located
on the x-axis.
center: (h, k)
1
2
1
,
0
2
0
or (0, 0)
c1, a4
c2a2b2
1242b2
b215
(x
a2
h)2
(y
b2
k)2
1
(x
42
0)2
(y
15
0)2
1
1
x
6
2
1
y
5
2
1
13. center: (h, k) (1, 2)
The points at (1, 4) and
(5, 2) are vertices of the
ellipse.
a4, b2
(x
a2
h)2
(y
b2
k)2
1
(x
42
1)2
(y
22
2)2
1
(x
16
1)2
(y
4
2)2
1
Chapter 10 318
y
x
(6, 0)
(0, 2)
(6, 0)
(0, 2)
O
y
x
(9, 4)
(0, 11)
(0, 3)
(0, 4) (9, 4)
O
y
x
(2, 6)
(5, 1)
(2, 4)
(2, 1) (1, 1)
O
y
x
(1, 2) (3, 2)
(1, 5)
(1, 2)
(1, 1)
O
O
y
x
(1, 4)
(1, 2) (5, 2)
14. center: (h, k) (3, 1) foci: (h, k c)
3, 4 39
19. center: (h, k) (2, 1)
a24b21ca2b
2
a4
or 2 b1
or 1 c4 1
or 3
foci: (h, kc) (2, 1 3
)
major axis vertices: (h, ka) (2, 1 2) or
(2, 3), (2, 1)
minor axis vertices: (hb, k) (2 1, 1) or
(1, 1), (3, 1)
20. center: (h, k) (6, 7)
a2121 b2100
a121
or 11 b100
or 10
ca2b
2
c121
100
or 21
foci: (h, kc) (6, 7 21
)
major axis vertices: (h, ka) (6, 7 11) or
(6, 18), (6, 4)
minor axis vertices: (hb, k) (6 10, 7) or
(16, 7), (4, 7)
21. center: (h, k) (4, 6)
a216 b29ca2b
2
a16
or 4 b9
or 3 c16
9
or 7
foci: (hc, k) (4 7
, 6)
major axis vertices: (ha, k) (4 4, 6) or
(8, 6), (0, 6)
minor axis vertices: (h, kb) (4, 6 3) or
(4, 3), (4, 9)
319 Chapter 10
a6
e
a
c
1
3
6
c
2 c
c2a2b2
2262b2
4 36 b2
b232
(y
a2
k)2
(h
b2
h)2
1
(y
62
1)2
(x
32
3)2
1
(y
36
1)2
(x
32
3)2
1
15. The major axis contains the foci and is located on
the x-axis.
center: (h, k) (0, 0)
c0.141732
a
1
2
(3.048) or 1.524
c2a2b2
(0.141732)2(1.524)2b2
0.020 2.323 b2
b22.302
b1.517
(x
a2
h)2
(y
b2
k)2
1
(
1
x
.
52
0
4
)
2
2
(
1
y
.
51
0
7
)
2
2
1
1.5
x
2
2
42
1.5
y
1
2
72
1
Pages 638–641 Exercises
16. center: (h, k) (0, 5)
a0 (7)or 7
b5 0or 5
(x
a2
h)2
(y
b2
k)2
1
(x
72
0)2
(y
5
(
2
5)]2
1
4
x
9
2
(y
25
5)2
1
ca2b
2
c725
2
or 26
foci: (hc, k) (0 26
, 5)
(26
, 5)
17. center: (h, k) (2, 0)
a2 2or 4
b0 2or 2
(x
a2
h)2
(y
b2
k)2
1
[x
4
(
2
2)]2
(y
22
0)2
1
(x
16
2)2
y
4
2
1
ca2b
2
c422
2
or 23
foci: (hc, k) (2 23
, 0)
18. centers: (h, k) (3, 4)
a4 12or 8
b3 2or 5
(y
a2
k)2
(x
b2
h)2
1
(y
82
4)2
[x
5
(
2
3)]2
1
(y
64
4)2
(x
25
3)2
1
ca2b
2
c825
2
or 39
y
x
(2, 3)
(1, 1)
(3, 1)
(2, 1)
(2, 1)
O
y
x
(16, 7)
(6, 7)
(6, 18)
(4, 7)
(6, 4)
O
y
x
(4, 3)
(4, 6)
(4, 9)
(8, 6)
(0, 6)
O
22. (h, k) (0, 0)
a29b24ca2b
2
a9
or 3 b4
or 2 c9 4
or 5
foci: (h, kc) 0, 0 5
or 0, 5
major axis vertices: (h, ka) (0, 0 3) or
(0, 3)
minor axis vertices: (hb, k) (0 2, 0) or
(2, 0)
23. 4x2y28x6y9 0
4(x22x?) (y26y?) 9 ? ?
4(x22x1) (y26y9) 9 4(1) 9
4(x1)2(y3)24
(x
1
1)2
(y
4
3)2
1
center: (h, k) (1, 3)
a24b21ca2b
2
a4
or 2 b1
or 1 c4 1
or 3
foci: (h, kc) 1, 3 3
major axis vertices: (h, ka) (1, 3 2)
or (1, 1), (1, 5)
minor axis vertices: (hb, k) (1 1, 3)
or (2, 3), (0, 3)
24. 16x225y296x200y144
16(x26x?) 25(y28y?) 144 ??
16(x26x9) 25(y28y16)
144 16(9) 25(16)
16(x3)225(y4)2400
(x
25
3)2
(y
16
4)2
1
center: (h, k) (3, 4)
a225 b216 ca2b
2
a25
or 5 b16
or 4 c25
16
or 3
foci: (hc, k) (3 3, 4) or (6, 4), (0, 4)
major axis vertices: (ha, k) (3 5, 4) or
(8, 4), (2, 4)
major axis vertices: (h, kb) (3, 4 4) or
(3, 8), (3, 0)
25. 3x2y218x2y4 0
3(x26x?) (y22 ?) 4
3(x26x9) (y22y1) 4 3(9) 1
3(x3)2(y1)224
(x
8
3)2
(y
24
1)2
1
center: (h, k) (3, 1)
a224 b28
a24
or 26
b8
or 22
ca2b
2
c24
8
or 4
foci: (h, kc) (3, 1 4) or (3, 5), (3, 3)
major axis vertices: (h, ka) (3, 1 26
)
minor axis vertices: (hb, k) (3 22
, 1)
26. 6x212x6y36y36
6(x22x?) 6(y26y?) 36
6(x22x1) 6(y26y9) 36 6(1) 6(9)
6(x1)26(y3)296
(x
16
1)2
(y
16
3)2
1
center: (h, k) (1, 3)
a216 b216 ca2b
2
a16
or 4 b16
or 4 c16
16
or 0
foci: (hc, k) or (h, kc) (1, 3)
Since ab4, the vertices are (h4, k) and
(h, k4) or (5, 3), (3, 3), (1, 1), (1, 7)
Chapter 10 320
y
x
(0, 3)
(0, 3)
(2, 0)(2, 0)
O
y
x
(1, 1)
(1, 5)
(0, 3)
(1, 3)
(2, 3)
O
y
x
(2, 4)
(3, 0)
(3, 4)
(3, 8)
(8, 4)
O
y
x
(3 22, 1)
(3 2
(3, 1) 2, 1)
(3, 1 26)
(3, 1 26)
O
(3, 3)
(1, 7)
(1, 1)
(1, 3) (5, 3)
y
x
O
27. 18y212x2144y48x120
18(y28y?) 12(x24x?) 120 ??
18(y28y16) 12(x24x4)
120 18(16) 12(4)
18(y4)212(x2)2216
(y
12
4)2
(x
18
2)2
1
center: (h, k) (2, 4)
a218 b212
a18
or 32
b12
or 23
ca2b
2
c18
12
or 6
foci: (hc, k) (2 6
, 4)
major axis vertices: (hg, k) (2 32
, 4)
minor axis vertices: (h, kb) (2, 4 23
)
28. 4x28y9x254x49 0
4(y22y?) 9(x26x?) 49 ? ?
4(y22y1) 9(x26x9) 49 4(1) 9(9)
4(y1)29(x3)236
(y
9
1)2
(x
4
3)2
1
center: (h, k) (3, 1)
a29b24ca2b
2
a9
or 3 b4
or 2 c9 4
or 5
foci: ( j, k c) (3, 1 5)
major axis vertices: (h, ka) (3, 1 3) or (3, 4),
(3, 2)
minor axis vertices: (hb, k) (3 2, 1) or
(5, 1), (1, 1)
29. 49x216y2160y384 0
49x216(y210y?) 384 ?
49x216(y210y25) 384 16(25)
49x216(y5)2784
1
x
6
2
(y
49
5)2
1
center: (h, k) (0, 5)
a249 b216 ca2b
2
a49
or 7 b16
or 4 c49
16
or 33
foci: (h, kc) (0, 5 33
)
major axis vertices: (h, ka) (0, 5 7) or
(0, 2), (0, 12)
minor axis vertices: (hb, k) (0 4, 5) or
(4, 5)
30. 9y2108y4x256x484
9(y212y?) 4(x214x?) 484 ? ?
9(y212y36) 4(x214x49)
484 9(36) 4(49)
9(y6)24(x7)236
(y
4
6)2
(x
9
7)2
1
center: (h, k) (7, 6)
a29b24ca2b
2
a9
or 3 b4
or 2 c9 4
or 5
foci: (hc, k) (7 5
, 6)
major axis vertices: (ha, k) (7 3, 6) or
(10, 6)(4, 6)
minor axis vertices: (h, kb) (7, 6 2) or
(7, 4), (7, 8)
31. a7, b5
(x
a2
h)2
(y
b2
k)2
1
[x
7
(
2
3)]2
[y
5
(
2
1)]2
1
(x
49
3)2
(y
25
1)2
1
321 Chapter 10
y
x
(2 32, 4) (2 3
(2, 4)
2, 4)
(2, 4 23)
(2, 4 23)
O
y
x
(1, 1) (3, 1)
(3, 2)
(3, 4)
(5, 1)
O
y
x
(0, 5) (4, 5)(4, 5)
(0, 12)
(0, 2)
O
y
x
(4, 6) (7, 6)
(7, 8)
(7, 4)
(10, 6)
O
Chapter 10 322
32. The major axis contains the foci and it is located
on the x-axis.
center: (h, k)
2
0
2
,
0
2
0
or (0, 0)
c2, a7
c2a2b2
2272b2
b245
(x
a2
h)2
(y
b2
k)2
1
(x
72
0)2
(y
45
0)2
1
4
x
9
2
4
y
5
2
1
33. b
3
4
a
6
3
4
a
8 a
(x
a2
h)2
(y
b2
k)2
1
(x
82
0)2
(y
62
0)2
1
6
x
4
2
3
y
6
2
1
34. The major axis contains the foci and it is the
vertical axis of the ellipse.
center: (h, k)
1
2
(1)
,
1
2
(5)
or (1, 2)
c1 ka213
c1 (2)or 3
c2a2b2
32213
2b2
b252 9
b243
(y
a2
k)2
(x
b2
h)2
1
[y
(2
(
1
3
2
)
)
2
]2
[x
4
(
3
1)]2
1
(y
52
2)2
(x
43
1)2
1
35. The horizontal axis of
the ellipse is the major
axis.
enter: (h, k)
11
2
7
,
5
2
5
or (2, 5)
ha7kb9
2 a75 b9
a9b4
(x
a2
h)2
(y
b2
k)2
1
[x
9
(
2
2)]2
(y
42
5)2
1
(x
81
2)2
(y
16
5)2
1
(0, 7)
(4, 7)
(4, 0)
4
4812
8
12
O
y
x
36. The major axis contains the foci and it is the
vertical axis of the ellipse.
c
5
2
(1)
or 3
center: (h, k)
1
2
1
,
1
2
5
or (1, 2)
(y
a2
k)2
(x
b2
h)2
1
(2
a2
2)2
(y
b2
1)2
1
a
02
2
3
b
2
2
1
b
9
2
1
9 b2
(y
a2
k)2
(x
b2
h)2
1
(y
18
2)2
(x
9
1)2
1
37.
1
2
a
c
a
2
c
1
2
0
c
5 c
(x
a2
h)2
(y
b2
k)2
1
(x
1
02
0)2
(y
75
0)2
1
1
x
0
2
0
7
y
5
2
1
38. tangent vertices:
(4, 0), (0, 7)
a7 0or 7
b4 0or 4
(y
a2
k)2
(x
b2
h)2
1
[y
7
(
2
7)]
(x
42
4)2
1
(y
49
7)2
(x
16
4)2
1
39. b2a2(1 e2)
b222
1
3
4
2
b2
2
1
8
6
or 1.75
Case 1: Horizontal axis is major axis.
(x
a2
h)2
(y
b
2
k)
1
(x
22
0)2
(y
1
.75
0)2
1
x
4
2
1
y
.7
2
5
1
Case 2: Vertical axis is major axis.
(y
a2
k)2
(x
b2
h)2
1
(y
22
0)2
(x
1
.75
0)2
1
y
4
2
1
x
.7
2
5
1
b2a2c2
b210252
b275
c2a2b2
32a29
18 a2
O
y
x
(2, 9)
(2, 1)
(11, 5) (7, 5)
4
48812 4
8
12
40. The major axis contains the foci and it is the
horizontal axis of the ellipse.
center: (h, k)
3
2
1
,
5
2
5
or (2, 5)
foci: (3, 5) (hc, k)
3 hc
3 2 c
1 c
e
a
c
0.25
a
1
a4
(x
a2
h)2
(y
b2
k)2
1
(x
42
2)2
(y
15
5)2
1
(x
16
2)2
(y
15
15)2
1
41. a
2
2
0
or 10
b2a2(1 e2)
b2102
1
1
7
0
2
or 51
(y
a2
k)2
(x
b2
h)2
1
(y
1
02
0)2
(x
51
3)2
1
1
y
0
2
0
(x
51
3)2
1
42. focus: (1, 1 5
) (h, kc)
1 5
kc
1 5
1 c
5
c
e
a
c
3
5
a
5
a3
major axis: vertical axis
(y
a2
k)2
(x
b2
h)2
1
[y
3
(
2
1)]2
(x
4
1)2
1
(y
9
1)2
(x
4
1)2
1
43. x24y26x24y41
(x26x?) 4(y26y?) 41 ? ?
(x26x9) 4(y26y9) 41 9 4(9)
(x3)24(y3)24
4(y3)24 (x3)2
(y3)2
4(x
4
3)2
y3
4(x
4
3)
2
y
4(x
4
3)
2
3
vertices: (5, 3),
(1, 3), (3, 2),
(3, 4)
[7.28, 7.28] scl:1 by [4.8, 4.8] scl:1
b2a2(1 e2)
b232
1
3
5
2
b24
b2a2(1 e2)
b242(1 0.252)
b215
323 Chapter 10
44. 4x2y28x2y1
4(x22x?) (y22y?) 1 ? ?
4(x22x1) (y22y1) 1 4(1) 1
4(x1)2(y1)24
(y1)24 4(x1)2
y1 4
4(x
1)2
y4 4
(x1
)2
1
Vertices: (0, 1),
(2, 1), (1, 1),
(1, 3)
[4.7, 4.7] scl:1 by [3.1, 3.1] scl:1
45. 4x29y216x18y11
4(x24x?) 9(y22y?) 11 ? ?
4(x24x4) 9(y22y1) 11 4(4) 9(1)
4(x2)29(y1)236
9(y1)236 4(x2)2
y1
36
4
9
(x
2)2
y
36
4
9
(x
2)2
1
Vertices: (1, 1),
(5, 1), (2, 3),
(2, 1)
[7.28, 7.28] scl:1 by [4.8, 4.8] scl:1
46. 25y216x2150y32x159
25(y26y?) 16(x22x?) 159 ? ?
25(y26y9) 16(x22x1)
159 25(9) 16(1)
25(y3)216(x1)2400
25(y3)2 400 16(x1)2
(y3)
400
1
2
6
5
(x
1)2
y
400
1
5
6(x
1)2
3
Vertices: (4, 3),
(6, 3), (1, 7),
(1, 1)
[15.16, 15.16] scl:1 by [10, 10] scl:1
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 c2a2b2, 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. Ahorizontal line; see students’ work.
49a. a
9
2
6
or 48
b
4
2
6
or 23
(h, k) (0, 0)
(x
a2
h)2
(y
b2
k)2
1
(x
4
82
0)2
(y
2
32
0)2
1
23
x
0
2
4
5
y
2
2
9
1
49b. ca2b
2
c2304
529
c42.13
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.
2c2(42)
84
about 84 ft
50a. x2y2r2
x
r2
2
y
r2
2
1Ar2
a
x2
2
b
y2
2
1Arr
Aab
Aab
50b.
x
9
2
y
4
2
1
a29b24
a3b2
Aab
A(3)(2)
A6units2
51. If (x, y) is a point on the ellipse, then show that
(x, y) is also on the ellipse.
a
x2
2
b
y2
2
1
(
a
x
2
)2
(
b
y
2
)2
1
a
x2
2
b
y2
2
1
Thus, (x, y) is also a point on the ellipse and
the ellipse is symmetric with respect to the origin.
Chapter 10 324
O
y
x
ca
x
52a. a
8
2
or 4
b3
ca2b
2
c423
2
c7
foci: (hc, 0) (0 7
, 0) or (7
, 0)
The thumbtacks should be placed (7
, 0) from
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.
53b.
a
c
e
69
c
55
0.052
c361.66
[figure not drawn to scale]
xacEarth’s radius
x6955 361.66 6357
x959.66
x960 km
54. x2y2Dx Ey F0
02(9)2D(0) E(9) F0 ⇒
9EF81
72(2)2D(7) E(2) F0 ⇒
7D2EF53
(5)2(10)2D(5) E(10) F0 ⇒
5D10EF125
9EF81
(1)(7D2EF) (1)(53)
7D7E28
DE4
7D2EF53
(1)(5D10EF) (1)(125)
12D8E72
(8)(DE) (8)(4)
12D8E72
4D40
D10
DE49EF81
10 E49(6) F81
E6F135
x2y2Dx Ey F0
x2y210x6y135 0
(x210x25) (y26y9) 135 25 9
(x5)2(y3)2169
60. Let h0.1.
xhx0.1
1 0.1 or 1.1
f(x0.1) f(1.1)
(1.1)24(1.1) 12
15.19
xhx0.1
1 0.1 or 0.9
f(x0.1) f(0.9)
(0.9)24(0.9) 12
14.79
f(x) 16
f(x) f(x0.1) and f(x) f(x0.1), so the point
is a location of a minimum.
61. The graph of the parent function g(x) xis
translated 2 units right.
62. Initial location: (2, 0)
Rot90
or (0, 2)
0
2
2
0
1
0
0
1
325 Chapter 10
(5, 16)
(8, 3)(5, 3)
y
x
O
O
y
x
D
(5, 4)
A
(1, 2)
B
(5, 4)
C
(4, 1)
center: (h, k) (5, 3)
radius: r2169
r13
55. Graph the quadrilateral with vertices A(1, 2),
B(5, 4), C(4, 1), and D(5, 4).
Aquadrilateral is a
parallelogram if one
pair of opposite sides
are parallel and
congruent.
slope of D
A
slope of C
B
m
y
x
2
2
y
x
1
1
m
y
x
2
2
y
x
1
1
1
2
(
4
5)
5
4
4
1
4
6
or
3
2
5
1
The slopes are not equal, so D
A
C
B
.The
quadrilateral is not a parallelogram; no.
56. cos 2v1 2 sin2v
cos 2v1 2
7
8
2
cos 2v
6
3
4
4
or
1
3
7
2
57. A4
2
k
180
A4k2
k
c
20° h0
2
c
20°
c40°
yAcos(kx c) h
y4 cos[2x(40°)] 0
y4 cos(2x40°)
58. A180° (121° 3242° 5) or 16° 23
sin
a
A
sin
b
B
sin
a
A
sin
c
C
sin 1
4
6
.1
°23
sin 4
b
2° 5
sin 1
4
6
.1
°23
sin 12
c
1° 32
4.
s
1
in
si
1
n
6°
42
2
°
3
5
b
4.1
si
s
n
in
16
1
°
21
2
°
3
32
c
9.7 b12.4 c
59. P(x) x44x32x21
P(5) 544(5)32(5)21
P(5) 74
P(5) 0; no, the binomial is not a factor of the
polynomial.
g
(
x
)
x
O
Rot80
or (2, 0)
2
0
2
0
0
1
1
0
Rot270
or (0, 2)
63. mQTS mTSR 180
abcd180 pbc180
bbcc180 p90 180
2b2c180 p90
bc90
The correct choice is C.
Page 641 Graphing Calculator Exploration
1. Sample answer: The graph will shift 4 units to the
right.
[15.16, 15.16] scl:2 by [10, 10] scl:2
0
2
2
0
1
0
0
1
2. Sample answer: The graph will shift 4 units to the
left.
[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.
[18.19, 18.19] scl:2 by [12, 12] scl:2
5. Sample answer: The graph will rotate 90°.
[15.16, 15.16] scl:2 by [10, 10] scl:2
6. For (xc), the graph will shift cunits to the left.
For (xc), the graph will shift cunits to the
right.
7. For (yc), the graph will shift cunits down. For
(yc), the graph will shift cunits up.
8. The graph will rotate 90°.
Hyperbolas
Pages 649–650 Check For Understanding
1. The equations of both hyperbolas and ellipses
have x2terms and y2terms. In an ellipse, the
terms are added and in a hyperbola these terms
are subtracted.
10-4
2. transverse axis: vertical
2a4
a2
An equation in standard form of the hyperbola
must have
2
y2
2
or
y
4
2
as the first term; b.
3. e
a
c
, so ae cand a2e2c2.
Since c2a2b2we have
a2e2a2b2
a2e2a2b2
a2(e21) 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)
a225 b24ca2b
2
a5b2c25
4
or 29
transverse axis: horizontal
foci: (hc, k) (0 29
, 0) or (29
, 0)
vertices: (ha, k) (0 5, 0) or (5, 0)
asymptotes: yk
a
b
(xh)
y0
2
5
(x0)
y
2
5
x
6. center: (h, k) (2, 3)
a216 b24ca2b
2
a16
or 4 b4
or 2 c16
4
or 25
transverse axis: vertical
foci: (h, kc)
2, 3 25
vertices: (h, ka) (2, 3 4) or (2, 7), (2, 1)
asymptotes: yk
a
b
(xh)
y3
4
2
(x2)
y3 2(x2)
Chapter 10 326
y
x
(5, 0) (5, 0)
4
8
8
4
4884
O
y
x
(2, 1)
(2, 7)
(2, 3)
O
7. y25x220x50
y25(x24x?) 50 ?
y25(x24x4) 50 (5)(4)
y25(x2)230
3
y
0
2
(x
6
2)2
1
xhx2yky
h2k0
center: (h, k) (2, 0)
a230 b26ca2b
2
a30
b6
c30
6
or 6
transverse axis: vertical
foci: (h, kc) (2, 0 6) or (2, 6)
vertices: (h, ka)
2, 0 30
or
2, 30
asymptotes: yk
a
b
(xh)
y0
3
6
0
(x2)
y5
(x2)
8. center: (h, k) (0, 5)
transverse axis: horizontal
a5, b3
(x
a2
h)2
(y
b2
k)2
1
(x
52
0)2
(y
32
5)2
1
2
x
5
2
(y
9
5)2
1
9. c9
quadrants: II and IV
transverse axis: yx
vertices: xy 9xy 9
3(3) 93(3) 9
(3, 3) (3, 3)
10. center: (h, k) (1, 4)
(x
a2
h)2
(y
b2
k)2
1
(x
52
1)2
[y
2
(
2
4)]2
1
(x
25
1)2
(y
4
4)2
1
11. 2b6
b3
center:
x1
2
x2
,
y1
2
y2
3
2
3
,
4
2
0
(3, 2)
transverse axis: vertical
a4 2or 2
(y
a2
k)2
(x
b2
h)2
1
(y
22
2)2
(x
32
3)2
1
(y
4
2)2
(x
9
3)2
1
12. center:
x1
2
x2
,
y1
2
y2
0
2
0
,
6
2
(6)
(0, 0)
transverse axis: vertical
cdistance from center to a focus
0 6or 6
b2c2a2b2a2
a2c2a2b218
2a2c2
a2
c
2
2
a2
6
2
2
or 18
(y
a2
k)2
(x
b2
h)2
1
(y
18
0)2
(x
18
0)2
1
1
y
8
2
1
x
8
2
1
13. center:
x1
2
x2
,
y1
2
y2
10
2
(10)
,
0
2
0
(0, 0)
transverse axis: horizontal
cdistance from center to a focus
10 0or 10
e
a
c
b2c2a2
5
3
1
a
0
b210262
b264
a6
(x
a2
h)2
(y
b2
k)2
1
(x
62
0)2
(y
64
0)2
1
3
x
6
2
6
y
4
2
1
14a. The origin is located midway between stations A
and B; (h, k) (0, 0). The stations are located at
the foci, so 2c130 or c65.
The difference of the distances from the plane to
each station is 50 miles.
50 2a(Definition of hyperbola)
25 a
b2c2a2
b2652252
b23600
327 Chapter 10
y
x
(2,
(2, 0)
30)
(2,
484
4
4
8
8
30)
O
y
x
(3, 3)
(3, 3)
O
O
y
x
A
(65, 0)
B
(65, 0)
transverse axis: horizontal
(x
a2
h)2
(y
b2
k)2
1
(x
2
52
0)2
(y
3
60
0
0
)2
1
6
x
2
2
5
36
y
0
2
0
1
14b. Vertices: (ha, k) (0 25, 0) or (25, 0)
asymptotes: yk
a
b
(xh)
y0
3
2
6
5
00
(x0)
y
1
5
2
x
14c. Let y6.
6
x
2
2
5
36
y
0
2
0
1
6
x
2
2
5
36
6
0
2
0
1
6
x
2
2
5
3
3
6
6
00
1
6
x
2
2
5
1
3
3
6
6
00
6
x
2
2
5
1.01
x2625(1.01)
x2631.25
x631.2
5
x25.1
Since the phase is closer to station Athan
station B, use the negative value of xto locate
the ship at (25.1, 6).
Pages 650–652 Exercises
15. center: (h, k) (0, 0)
a2100 b216
a100
or 10 b16
or 4
ca2b
2
c100
16
or 229
transverse axis: horizontal
foci: (hc, k) (0 229
, 0) or (229
, 0)
vertices: (ha, k) (0 10, 0) or (10, 0)
asymptotes: yk
a
b
(xh)
y0
1
4
0
(x0)
y
2
5
x
Chapter 10 328
y
x
20204060 40 60
20
20
40
60
40
60
Station
B
(65, 0)
Station
A
(65, 0)
Plane
located
on this
branch
O
y
x
(3, 5) (3, 5)
(0, 5)
O
y
x
(2, 0) (2, 0)
2442
4
8
8
4
O
y
x
(1, 15)
(1, 7)
(1, 1)
O
16. center: (h, k) (0, 5)
a29b281
a9
or 3 b81
or 9
ca2b
2
c9 8
1
or 310
transverse axis: horizontal
foci: (hc, k) (0 310
, 5) or (310
, 5)
vertices: (ha, k) (0 3, 5) or (3, 5)
asymptotes: yk
a
b
(xh)
y5
9
3
(x0)
y5 3x
17. center: (h, k) (0, 0)
a24b249
a4
or 2 b49
or 7
ca2b
2
c4 4
9
or 53
transverse axis: horizontal
foci: (hc, k) (0 53
, 0) or (53
, 0)
vertices: (ha, k) (0 2, 0) or (2, 0)
asymptotes: yk
a
b
(xh)
y0
7
2
(x0)
y
7
2
x
18. center: (h, k) (1, 7)
a264 b24
a64
or 8 b4
or 2
ca2b
2
c64
4
or 217
transverse axis: vertical
foci: (h, kc) (1, 7 217
)
vertices: (h, ka)
(1, 7 8) or (1, 15), (1, 1)
asymptotes: yk
a
b
(xh)
y7
8
2
[x(1)]
y7 4(x1)
y
x
(10, 0)
8816
4
8
16
4
8
(10, 0)
O
19. x24y26x8y11
(x26x?) 4(y22y?) 11 ? ?
(x26x9) 4(y22y 1) 11 9 (4)(1)
(x3)24(y1)216
(x
16
3)2
(y
4
1)2
1
center: (h, k) (3, 1)
a216 b24ca2b
2
a16
or 4 b24
or 2 c16
4
or 25
transverse axis: horizontal
foci: (hc, k) (3 25
, 1)
vertices: (ha, k) (3 4, 1) or (1, 1),
(7, 1)
asymptotes: yk
a
b
(xh)
y(1)
2
4
[x(3)]
y1
1
2
(x3)
20. 4x9y224x90y153 0
9(y210y?) 4(x26x?) 153
9(y210y25) 4(x26x9) 153
9(25) 4(9)
9(y5)24(x3)236
(y
4
5)2
(x
9
3)2
1
center: (h, k) (3, 5)
a24b29ca2b
2
a4
or 2 b9
or 3 c4 9
or 13
transverse axis: vertical
foci: (h, kc) (3, 5 13
)
vertices: (h, ka) (3, 5 2) or (3, 7), (3, 3)
asymptotes: yk
a
b
(xh)
y5
2
3
[x(3)]
y5
2
3
(x3)
21. 16y225x296y100x3560
16(y26y?)25(x24x?)356
16(y26y9)25(x24x4)35616(9)25(4)
16(y3)225(x2)2400
(y
25
3)2
(x
16
2)2
1
center: (h, k) (2, 3)
a225 b216 ca2b
2
a25
b16
c25
16
or 5 or 4 or 41
transverse axis: vertical
foci: (h, kc) (2, 3 41
)
vertices: (h, ka) (2, 3 5) or (2, 8), (2, 2)
asymptotes: yk
a
b
(xh)
y3
5
4
(x2)
22. 36x249y272x294y2169
36(x22x?)49(y26y?)2169??
36(x2x1)49(y26y9)216936(1)49(9)
36(x1)249(y3)21764
(x
49
1)2
(y
36
3)2
1
center: (h, k) (1, 3)
a249 b236 ca2b
2
a49
b36
c49
36
or 7 or 6 or 85
transverse axis: horizontal
foci: (hc, k) (1 85
, 3)
vertices: (ha, k) (1 7, 3) or (8, 3),
(6, 3)
asymptotes: yk
a
b
(xh)
y(3)
6
7
(x1)
y3
6
7
(x1)
329 Chapter 10
y
x
(7, 1) (3, 1) (1, 1)
O
y
x
(3, 7)
(3, 5)
(3, 3)
O
y
x
(2, 8)
(2, 3)
(2, 2)
O
y
x
(6, 3) (8, 3)
(1, 3)
O
23. 25y29x2100y72x2690
25(y24y?)9(x28x?)269??
25(y24y4)9(x28x16)26925(4)9(16)
25(y2)29(x4)2225
(y
9
2)2
(x
25
4)2
1
center: (h, k) (4, 2)
a29b225 ca2b
2
a9
b25
c9 2
5
or 3 or 5 or 34
transverse axis: vertical
foci: (h, kc)
4, 2 34
vertices: (h, ka)
(4, 2 3) or (4, 5), (4, 1)
asymptotes: yk
a
b
(xh)
y2
3
5
[x(4)]
y2
3
5
(x4)
24. center: (h, k) (4, 3)
transverse axis: vertical
a4, b3
(y
a2k)2
(x
b2h)2
1
(y
423)2
(x
324)2
1
(y
16
3)2
(x
9
4)2
1
25. center: (h, k) (0, 0)
transverse axis: horizontal
a3, b3
(x
a2h)2
(y
b2k)2
1
(x
320)2
(y
320)2
1
x
9
2
y
9
2
1
26. center: (h, k) (4, 0)
transverse axis: vertical
a2, b1
(y
a2k)2
(x
b2b)2
1
(y
220)2
[x
1
(
24)]2
1
y
4
2
(x
1
4)2
1
27. c49
quadrants: I and III
transverse axis: yx
vertices: xy 49 xy 49
7(7) 49 7(7) 49
(7, 7) (7, 7)
28. c36
quadrants: II and IV
transverse axis: yx
vertices: xy 36 xy 36
6(6) 36 6(6) 36
(6, 6) (6, 6)
29. 4xy 25
xy
2
4
5
c
2
4
5
quadrants: II and IV
transverse axis: yx
vertices: xy
2
4
5
xy
2
4
5
5
2
5
2
2
4
5
5
2
5
2
2
4
5
5
2
,
5
2
5
2
,
5
2
Chapter 10 330
y
x
(4, 1)
(4, 5)
y
2 (
x
4)
3
5(4, 2)
(4, 2 34)
(4, 2 34)
O
y
2 (
x
4)
3
5
y
x
(7, 7)
(7, 7)
O
y
x
(6, 6)
(6, 6)
O
y
x
()
5
2,
O
5
2
()
5
2,5
2
30. 9xy 16
xy
1
9
6
c
1
9
6
quadrants: I and III
transverse axis: yx
vertices: xy
1
9
6
xy
1
9
6
4
3
4
3
1
9
6
4
3
4
3
1
9
6
4
3
,
4
3
4
3
,
4
3
31. center: (h, k) (4, 2)
(y
a2k)2
(x
b2h)2
1
[y
2
(
22)]2
(x
324)2
1
(y
4
2)2
(x
9
4)2
1
32. center:
x1
2
x2
,
y1
2
y2
0
2
0
,
3
2
(3)
(0, 0)
transverse axis: vertical
adistance from center to a vertex
0 3] or 3
cdistance from center to a focus
0 (9)or 9
b2c2a2
b29232or 72
(y
a2k)2
(x
b2h)2
1
(y
320)2
(x
72
0)2
1
y
9
2
7
x
2
2
1
33. 2a6
a3
center:
x1
2
x2
,
y1
2
y2
5
2
(5)
,
2
2
2
(0, 2)
transverse axis: horizontal
cdistance from center to a focus
0 5or 5
b2c2a2
b25232or 16
(x
a2b)2
(y
b2k)2
1
(x
320)2
(y
16
2)2
1
x
9
2
(y
16
2)2
1
34. 2b8
b4
center:
x1
2
x2
,
y1
2
y2
3
2
(3)
,
9
2
(5)
(3, 2)
transverse axis: vertical
adistance from center to a vertex
2 9or 7
(y
a2k)2
(x
b2h)2
1
(y
722)2
[x
4
(
23)]2
1
(y
49
2)2
(x
16
3)2
1
35. center:
x1
2
x2
,
y1
2
y2
8
2
(8)
,
0
2
0
(0, 0)
transverse axis: horizontal
cdistance from center to a focus
0 8or 8
b2c2a2b2a2
a2c2a2b232
2a2c2
a2
c
2
2
a2
8
2
2
or 32
(x
a2h)2
(y
b2k)2
1
(x
32
0)2
(y
32
0)2
1
3
x
2
2
3
y
2
2
1
36. centers: (h, k) (3, 1)
cdistance from center to a focus
3 2or 5
e
a
c
b2c2a2
5
4
a
5
b25242
a4b29
transverse axis: horizontal
(x
a2h)2
(y
b2k)2
1
[x
4
(
23)]2
(y
9
1)2
1
(x
16
3)2
(y
9
1)2
1
37. center: (h, k) (4, 2)
adistance from center to a vertex
2 5or 3
transverse axis: vertical
4y4 3x
4y4 12 3x12
4y8 3x12
4(y2) 3(x4)
y2
3
4
(x4)
a
b
3
4
3
b
3
4
b4
(y
a2k)2
(x
b2h)2
1
(y
322)2
(x
424)2
1
(y
9
2)2
(x
16
4)2
1
331 Chapter 10
y
x
()
4
3,
O
4
3
()
4
3,4
3
38. center: (h, k) (3, 1)
adistance from center to a vertex
3 5or 2
transverse axis: horizontal
3x11 2y
3x11 4 2y4
3x15 2y4
3(x5) 2(y2)
3
2
(x5) y2
y2
3
2
(x5)
a
b
3
2
b
2
3
2
b3
(x
a2
b)2
(y
b2
k)2
1
(x
22
3)2
[y
3
(
2
1)]2
1
(x
4
3)2
(y
9
1)2
1
39. center:
x1
2
x2
,
y1
2
y2
0
2
0
,
8
2
(8)
(0, 0)
cdistance from center to a focus
0 8or 8
e
a
c
b2c2a2
4
3
a
8
b28262
a6b228
transverse axis: vertical
(y
a2k)2
(x
b2h)2
1
(y
620)2
(x
28
0)2
1
3
y
6
2
2
x
8
2
1
40. centers:
x1
2
x2
,
y1
2
y2
10
2
(2)
,
3
2
(3)
(4, 3)
cdistance from center to a focus
4 10or 6
e
a
c
b2c2a2
6
5
a
6
b26252
a5b211
transverse axis: horizontal
(x
a2h)2
(y
b2k)2
1
(x
524)2
[y
1
(
1
3)]
1
(x
25
4)2
(y
11
3)2
1
41. center:
x1
2
x2
,
y1
2
y2
9
0
(9)
,
0
2
0
(0, 0)
cdistance from center to a focus
0 9or 9
b2c2a2b2a2
a292a2b2
8
2
1
2a281
a2
8
2
1
transverse axis: horizontal
(x
a2h)2
(y
b2k)2
1
1
2
8
x
1
2
2
8
y
1
2
1
42. center:
x1
2
x2
,
y1
2
y2
1
2
1
,
5
2
(3)
(1, 1)
cdistance from center to a focus
1 5or 4
transverse axis: vertical
a
b
2
a2b
c2a2b2
42(2b)2b2
16 5b2
1
5
6
b2
(y
a2
k)2
(x
b2h)2
1
1
5(y
6
4
1)2
5(x
1
6
1)2
1
43a. quadrants: I and II
transverse axis: yx
43b. PV 505
(101)V505
V5.0 dm3
43c. PV 505
(50.5)V505
V10.0 dm3
43d. If the pressure is halved, then the volume is
doubled, or V2(original V).
44. In an equilateral hyperbola, aband
c2a2b2.
c2a2a2ab
c22a2
ca2
Since e
a
c
, we have
e
a
c
e
e2
Thus, the eccentricity of any equilateral hyperbola
is 2
.
a2
a
(x1)2
1
5
6
(y1)2
6
5
4
a2(2b)2
a24
1
5
6
a2
6
5
4
(y0)2
__
8
2
1
(x0)2
__
8
2
1
Chapter 10 332
V
P
2
250
200
150
100
50
46810
O
45a.
2a150
a75
e
a
c
5
3
7
c
5
125 c
transverse axis: horizontal
center: (h, k) (0, 0)
(x
a2h)2
(y
b2k)2
1
(x
7
52
0)2
(y
1
00
0
2)2
1
7
x
5
2
2
10
y2
02
1
45b.
top:
7
x
5
2
2
10
y2
02
1
7
x
5
2
2
1
1
0
0
0
0
2
2
1(x, y) (x, 100)
7
x
5
2
2
1 1
7
x
5
2
2
2
x211,250
x106.07 ft
base:
7
x
5
2
2
10
y2
02
1
7
x
5
2
2
(
1
3
0
5
0
0
2)2
1(x, y) (x, 350)
7
x
5
2
2
12.25 1
7
x
5
2
2
13.25
x274,531.25
x273.00 ft
46. The origin is located midway between stations A
and B. The stations are located at the foci, so
2c4 or c2 miles.
c2 mi
c2 mi
52
1
8
m
0
i
ft
c10,560 ft
drt
d1100(2) or 2200 ft
b2c2a2
b21252752
b210,000
b100
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
b2c2a2
bc2a
2
b10,56
021
1002
b10,503
center: (h, k) (0, 0)
transverse axis: horizontal
(x
a2h)2
(y
b2k)2
1
(x
11
00
02
)2
(
1
y
0
,50
0
3
)2
2
1
11
x
0
2
02
10,
y
5
2
032
1
47. center: (h, k) (0, 0)
c0 6or 6
PF1PF210
2a10 PF1PF22a
a5
b2c2a2
b26252
b211
transverse axis: horizontal
(x
a2h)2
(y
b2k)2
1
(x
520)2
(y
11
0)2
1
2
x
5
2
1
y
1
2
1
48a.
1
x
6
2
y
9
2
1
center: (h, k) (0, 0)
a216 b29
a16
or 4 b9
or 3
transverse axis: horizontal
vertices: (ha, k) (0 4, 0) or (4, 0)
asymptotes: yk
a
b
(xh)
y0
3
4
(xh)
y
3
4
x
y
9
2
1
x
6
2
1
center: (h, k) (0, 0)
a29b216
a9
or 3 b16
or 4
transverse axis: vertical
vertices: (h, ka) (0, 0 3) or (0, 3)
asymptotes: yk
a
b
(xh)
y0
3
4
(x0)
y
3
4
x
333 Chapter 10
y
x
150 ft
O
y
x
(0, 100)
(
x
, 100)
(
x
, 350)
(0, 350)
O
O
y
x
(10,560, 0)
A
(10,560, 0)
B
y
x
O
48b. They are the same lines.
48c.
(y
25
2)2
(x
16
3)2
1
48d.
(x
16
3)2
(y
25
2)2
1
center: (h, k) (3, 2)
a216 b225
a16
or 4 b25
or 5
transverse axis: horizontal
vertices: (ha, k) (3 4, 2) or (7, 2), (1, 2)
asymptotes: yk
a
b
(xh)
y2
5
4
(x3)
(y
25
2)2
(x
16
3)2
1
center: (h, k) (3, 2)
a225 b216
a25
or 5 b16
or 4
transverse axis: vertical
vertices: (h, ka) (3, 2 5) or (3, 7), (3, 3)
asymptotes: yk
a
b
(xh)
y2
5
4
(x3)
49. center:
x1
2
x2
,
y1
2
y2
2
2
2
,
3
2
(3)
(2, 0)
a4
cdistance from center to a focus
0 3or 3
b2a2c2
b24232or 7
major axis: vertical
(y
a2
k)2
(x
b2
h)2
1
(y
42
0)2
(x
7
2)2
1
1
y
6
2
(x
7
2)2
1
50. x2y24x14y28 0
(x24x?) (y214y?) 28 ? ?
(x24x4) (y214y49) 28 4 49
(x2)2(y7)281
51.
AB (2 1
)2(
1 3
)2
5
BC (2 6
)2(
1 2
)2
5
CD (6 3
)2(2
6)2
5
AD (3 1
)2(6
3)2
5
Thus, ABCD is a rhombus. The slope of A
D
6
3
3
1
or
3
4
and the slope of A
B
3
1
1
2
or
4
3
.
Thus, A
D
is perpendicular to A
B
and ABCD is a
square.
52. (r, v) (90, 208°)
(r, v360 k°) (90, 208° 360(1)°)
(90, 152°)
(r, v(2k1)(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. xcos fysin fp0
xcos 60 ysin 60 3 0
1
2
x
2
3
y3 0
x3
y6 0
55.
tan 30°
90
x
00
9000 tan 30° x
5196 x
drt
d
t
r
51
1
9
5
6
r
346.4 r
about 346 m/s
56.
Since 0.2506 is closer to zero than 0.6864, the
zero is about 1.3.
Chapter 10 334
y
x
O
44
4
4
8
8
(2, 7)
(11, 7)
(2, 16)
y
x
O
y
x
O
A
(1, 3)
B
(2, 1)
C
(6, 2)
D
(3, 6)
x
9000 m
60˚
30˚
Since 0.0784 is closer to zero than 0.2446, the
zero is about 0.6.
57. Case 1: ris positive and sis negative.
Case 2: ris negative and sis positive.
I. r3s3is false if ris negative.
II. r3s2is false for each case.
III. r4s4is true for each case.
The correct choice is C.
Parabolas
Pages 658–659 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)
p4
(xh)24p(yk)
(x2)24(4)(y1)
(x2)216(y1)
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)
p5
(yk)24p(xh)
(y5)24(5)[x(4)]
(y5)220(x4)
5a. ellipse 5b. parabola
5c. hyperbola 5d. circle
6. vertex: (h, k) (0, 1)
4p12
p3
focus: (h, kp) (0, 1 3) or (0, 4)
directrix: ykp
y1 3
y2
axis of symmetry: xh
x0
10-5
7. y24x2y5 0
y22y4x5
y22y? 4x5 ?
y22y1 4x5 1
(y1) 4(x1)
vertex: (h, k) (1, 1)
4p4
p1
focus: (hp, k) (1 1, 1) or (2, 1)
directrix: xhp
x1 1
x0
axis of symmetry: yk
y1
8. x28x4y8 0
x28x4y8
x28x? 4y8 ?
x28x16 4y8 16
(x4)24(y2)
vertex: (h, k) (4, 2)
4p4
p1
focus: (h, kp) (4, 2 (1))or (4, 1)
directrix: ykp
y2 (1)
y3
axis of symmetry: xh
x4
9. vertex: (h, k) (0, 0)
opening: downward
p 4
(xh)24p(yk)
(x0)24(4)(y0)
x216y
335 Chapter 10
y
x
(0, 4)
(0, 1)
y
2
O
y
x
(1, 1)
(2, 1)
x
0
O
y
x
(4, 1)
(4, 2)
y
3
O
y
x
O
10. (yk)24p(xh)
(1 5)24p[2 (7)] (h, k) (7, 5);
(6)236p (x, y) (2, 1)
1 p
(yk)24p(xh)
(y5)24(1)[x(7)]
(y5)24(x7)
11. vertex: (h, k) (4, 3)
opening: upward
(xh)24p(yk)
(5 4)24p[2 (3)] (h, k) (4, 3);
1220p (x, y) (5, 2)
2
1
0
p
(xh)24p(yk)
(x4)24
2
1
0
(y3)
(x4)2
1
5
(y3)
12a. sv0t16t23
s56t16t23
16t256ts3
16
t2
7
5
t?
s3 ?
16
t2
7
2
t
4
1
9
6
s3 (16)
4
1
9
6
16
t
7
4
2s52
t
7
4
2
1
1
6
(s52)
t
7
4
24
6
1
4
(s52)
vertex: (h, k)
7
4
, 52
opening: downward
12b. The maximum height is s52 ft.
12c. Let s0.
s56t16t23
0 16t256t3
t
t
t3.6 or 0.05
3.6 s
Pages 659–661 Exercises
13. vertex: (h, k) (0, 0)
4p8
p2
focus: (hp, k) (0 2, 0) or (2, 0)
directrix: xhp
x0 2
x2
axis of symmetry: yk
y0
14. vertex: (h, k) (0, 3)
4p4
p1
focus: (h, kp) (0, 3 (1)) or (0, 2)
directrix: ykp
y3 (1)
y4
axis of symmetry: xh
x0
15. vertex: (h, k) (0, 6)
4p4
p1
focus: (hp, k) (0 1, 6) or (1, 6)
directrix: xhp
x0 1
x1
axis of symmetry: yk
y6
56 562
4(1
6)(3)
2(16)
bb24
ac
2a
Chapter 10 336
y
x
(7, 5)
O
y
x
(4, 3)
O
s
x
1
5
10
15
20
25
30
35
40
45
55
50
234
O
y
x
(2, 0)
x
2
O
y
x
(0, 2)
(0, 3)
y
4
O
y
x
(1, 6)
(0, 6)
x
1
O
16. y212x2y13
y22y12x13
y22y? 12x13 ?
y22y1 12x13 1
(y1)212(x1)
vertex: (h, k) (1, 1)
4p12
p3
focus: (hp, k) (1 (3), 1) or (4, 1)
directrix: xhp
x1 (3)
x2
axis of symmetry: yk
y1
17. y2 x24x
x24xy2
x24x? y2 ?
x24x4 y2 4
(x2)2y2
vertex: (h, k) (2, 2)
4p1
p
1
4
focus: (h, kp)
2, 2
1
4
or
2,
7
4
directrix: ykp
y2
1
4
y
9
4
axis of symmetry: xh
x2
18. x210x25 8y24
(x5)28(y3)
vertex: (h, k) (5, 3)
4p8
p2
focus: (h, kp) (5, 3 (2)) or (5, 1)
directrix: ykp
y3 (2)
y5
axis of symmetry: xh
x5
19. y22x14y41
y214y2x41
y214y? 2x41 ?
y214y49 2x41 49
(y7)2(2x4)
vertex: (h, k) (4, 7)
4p2
p
2
4
or
1
2
focus: (hp, k)
4
1
2
, 7
or
7
2
, 7
directrix: xhp
x4
1
2
x
9
2
axis of symmetry: yk
y7
337 Chapter 10
y
x
(4, 1) (1, 1)
x
2
O
y
x
1
123
(2, 2)
y
4
9
O
2,
)(
4
7
y
x
(5, 3)
(5, 1)
y
5
O
y
x
(4, 7)
O
x
2
9
, 7
)(
2
7
20. y22y12x13 0
y22y12x13
y22y? 12x13 ?
y22y1 12x13 1
(y1)212(x1)
vertex: (h, k) (1, 1)
4p12
p3
focus: (hp, k) (1 3, 1) or (4, 1)
directrix: xhp
x1 3
x2
axis of symmetry: yk
y1
21. 2x212y16x20 0
2x216x12y20
2(x28x?) 12y20 ?
2(x28x16) 12y20 2(16)
2(x4)212y12
(x4)26(y1)
vertex: (h, k) (4, 1)
4p6
p
6
4
or
3
2
focus: (h, kp)
4, 1
3
2
or
4,
1
2
directrix: ykp
y1
3
2
y
5
2
axis of symmetry: xh
x4
22. 3x230x18x87 0
3x218x30y87
3(x26x?) 30y87 ?
3(x26x9) 30y87 3(9)
3(x3)230y60
(x3)210y20
(x3)210(y2)
vertex: (h, k) (3, 2)
4p10
p
1
4
0
or
5
2
focus: (h, kp)
3, 2
5
2
or
3,
9
2
directrix: ykp
y2
5
2
y
1
2
axis of symmetry: xh
x3
23. 2y216y16x64 0
2y216y16x64
2(y28y?) 16x64 ?
2(y28y16) 16x64 2(16)
2(y4)216x32
(y4)28x16
(y4)28(x2)
vertex: (h, k) (2, 4)
4p8
p2
focus: (hp, k) (2 (2), 4) or (4, 4)
directrix: xhp
x2 (2)
x0
axis of symmetry: yk
y4
24. vertex: (h, k) (5, 1)
opening: right
hp2
5 p2
p7
(yk)24p(xh)
(y1)24(7)[x(5)]
(y1)228(x5)
Chapter 10 338
y
x
(4, 1)
(1, 1)
x
2
O
y
x
(4, 1)
y
2
5
(
4,
)
2
1
O
y
x
(3, 2)
y
2
1
O
3,
)(
2
9
y
x
(2, 4)
x
0
(4, 4)
O
y
x
(5, 1)
4
4812
8
4
8
O
25. opening: left
focus: (hp, k) (0, 6)
hp0k6
h(3) 0
h3
(yk)24p(xh)
(y6)24(3)(x3)
(y6)212(x3)
26. opening: upward
vertex: (h, k)
4,
5
2
1
or (4, 3)
focus: (h, kp) (4, 1)
kp1
3 p1
p2
(xb)24p(yk)
(x4)24(2)[y(3)]
(x4)28(y3)
27. opening: downward
vertex: (h, k) (4, 3)
(xh)24p(yk)
(5 4)24p(2 3) (h, k) (4, 3);
124p (x, y) (5, 2)
1
4
p
(xh)24p(yk)
(x4)24
1
4
(y3)
(x4)2(y3)
28. vertex: (h, k) (2, 3)
(yk)24p(xh)
[1 (3)]24p[3 (2)] (h, k) (2, 3);
424p(x, y) (3, 1)
4 p
(yk)24p(xh)
(y3)24(4)(x2)
(y3)216(x2)
29. opening: upward
p2
focus: (h, kp) (1, 7)
h1kp7
k2 7
k5
(xh)24p(yk)
[x(1)]24(2)(y5)
(x1)28(y5)
30. opening: downward
vertex: (h, k) (5, 3) (maximum)
(xh)24p(yk)
(1 5)24p[7 (3)] (h, k) (5, 3);
(4)216p (x, y) (1, 7)
1 p
(xh)24p(yk)
(x5)24(1)(y3)
(x5)24(y3)
31. opening: right
vertex: (h, k) (1, 2)
(yk)24p(xh)
(0 2)24p[0 (1)] (h, k) (1, 2);
(2)24p (x, y) (0, 0)
1 p
(yk)24p(xh)
(y2)24(1)[x(1)]
(y2)24(x1)
339 Chapter 10
y
x
(3, 6)
4
4
88
4
8
12
16
4
O
y
x
(4, 3)
O
y
x
(4, 3)
O
y
x
O
(2, 3)
y
x
O
(1, 5)
y
x
O
(5, 3)
y
x
O
(1, 2)
32. opening: upward
h
x1
2
x2
1
2
2
or
3
2
vertex: (h, k)
3
2
, 0
(xh)24p(yk)
1
3
2
24p(1 0) (h, k)
3
2
, 0
;
1
4
4p (x, y) (1, 1)
1
1
6
p
(xh)24p(yk)
x
3
2
24
1
1
6
(y0)
y
3
2
1
4
y
33a. vertex: (h, k) (0, 0)
depth: x4
p2
(yk)24p(xh)
(y0)24(2)(x0) (h, k) (0, 0);
y28xp 2
y8x
y8(4)
y42
y142
, y242
diameter y1y2
42
(42
)
82
in.
33b. depth: x1.25(4)
x5
(yk)24p(xh)
(y0)24(2)(x0) (h, k) (0, 0);
y28xp 2
y8x
y8(5)
y210
y1210
, y2210
diameter y1y2
210
(210
)
410
in.
34a. Let yincome per flight.
Let xthe number of $10 price decreases.
Income number of passengers cost of a ticket
y(110 20x) (140 10x)
y
15,400 1100x2800x200x2
y200
x2
1
2
7
x
15,400
y15,400 200
x2
1
2
7
x
y15,400 200
1
4
7
2200
x2
1
2
7
x
1
9
7
2
2
1
00
(y19,012.5)
x
1
4
7
2
The vertex of the parabola is at
1
4
7
, 19,012.5
,
and because pis negative it opens downward. So
the vertex is a maximum and the number of $10
price decreases is
1
4
7
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 yincome per flight.
Let xthe number of $10 price decreases.
Income number of passengers cost of a ticket
y(110 10x)(140 10x)
y15,400 300x100x2
y100(x23x) 15,400
y15,400 100(x23x)
y15,400 100
3
2
2100
x23x
3
2
2
y15,625 100
x
3
2
2
1
1
00
(y15,625)
x
3
2
2
The vertex of the parabola is at
3
2
, 15,625
, and
because pis negative, it opens downward. So the
vertex is a maximum and the number of $10
price decreases is
3
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
Chapter 10 340
y
x
1
1
2
O
, 0
)(
2
3
y
x
(4,
y
1)
(4,
y
2)
(2, 0) (4, 0)
O
35a. Let (h, k) (0, 0).
x24py x24py
x24
1
8
yp
1
8
x24
1
4
yp
1
4
2x2yx
2y
x24py
x21(1)yp 1
1
4
xy
The opening becomes
narrower.
35b. The opening becomes wider.
36a. Sample answer:
opening: upward
vertex: (h, k) (0, 0)
(xh)24p(yk)
(2100 0)24p(490 0) (h, k) (0, 0)
2250 p (x, y) (2100, 490)
(xh)24p(yk)
(x0)24(2250)(y0)
x29000y
36b. x29000y
(720)29000y
57.6 y
57.6 10 67.6 ft
37. (yk)24p(xh)
y22ky k24px 4ph
y24px 2ky k24ph 0
y2Dx Ey F0
(xh)24p(yk)
x22hx h24py 4pk
x24py 2hx h24pk 0
x2Dx Ey F0
38a. 4p8
p2 or p2
opening: right
(yk)24p(xh)
(y2)24(2)[x(3)]
(y2)28(x3)
opening: left
(yk)24p(xh)
(y2)24(2)[x(3)]
(y2)28(x3)
38b. 4p16
p4
focus of parabola center of circle
vertex: (h, k) (1, 4)
focus: (h, kp) (1, 4 (4)) or (1, 0)
diameter latus rectum
16
radius
1
2
(16)
8
(xh)2(yk)2r2
(x1)2(y0)282
(x1)2y264
39. center: (h, k) (2, 3)
a225 b216
a25
or 5 b16
or 4
ca2b
2
c25
16
or 41
transverse axis: vertical
foci: (h, kc) (2, 3 41
)
vertices: (h, ka) (2, 3 5) or (2, 8), (2, 2)
asymptotes: yk
a
b
(xh)
y3
5
4
(x2)
40. 4x225y2250y525 0
4x225(y210y?) 525 ?
4x225(y210y25) 525 25(25)
4x225(y5)2100
2
x
5
2
(y
4
5)2
1
center: (h, k) (0, 5)
a225 b24
a25 or 5 b4
or 2
ca2b
2
c25
4
or 21
foci: (hc, k) (0 21
, 5) or (21
, 5)
major axis vertices: (ha, k) (0 5, 5) or
(5, 5)
minor axis vertices: (h, kb) (0, 5 2) or
(0, 3), (0, 7)
341 Chapter 10
O
y
x
2
x
2
yx
2
y
x
2
y
1
4
O
y
x
(2100, 490)
(0, 0)
10 ft
500 ft
roadway
2100 2100
y
x
(2, 8)
(2, 3)
(2, 2)
O
y
x
(5, 5) (0, 5)
(5, 5)
(0, 3)
(0, 7)
O
45. 19 t14 19
19 t33
t33or 27
perimeter 14 19 t
14 19 27
60
The correct choice is C.
Page 661 Mid-Chapter Quiz
1a. AB (x2
x1)2
(y2
y1)2
(6 3
)2(9
3)2
326
2
45
BC (x2
x1)2
(y2
y1)2
(9 6
)2(3
9)2
32(
6)2
45
Since AB BC, triangle ABC is isosceles.
1b. AC (x2
x1)2
(y2
y1)2
(9 3
)2(3
3)2
620
2
6
perimeter AB BC AC
45
45
6
19.42 units
2. Diagonals of a rectangle intersect at their
midpoint.
midpoint of A
C
x1
2
x2
,
y1
2
y2
4
2
5
,
9
2
5
(0.5, 7)
3. x2y26y8x16
(x28x?) (y26y?) 16 ? ?
(x28x16) (y26y9) 16 16 9
(x4)2(y3)29
center: (4, 3): radius: 9
or 3
4. (xh)2(yk)2r2
[x(5)]2(y2)2
7
2
(x5)2(y2)27
5a.
Let d1be the greatest
distance from the
satellite to Earth. Let
d2be the least
distance from the
satellite to Earth.
Chapter 10 342
12 cos 2(r, )
012(12, 0)
6
6
6,
6
3
6
6,
3
2
12
12,
2
2
3
6
6,
2
3
5
6
6
6,
5
6
12 (12, )
7
6
6
6,
7
6
4
3
6
6,
4
3
3
2
12
12,
3
2
5
3
6
6,
5
3
11
6
6
6,
11
6
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
369
y
x
O
(4, 6)
(4, 3)
(7, 3)
O
y
x
ca
d
1
d
2
6.4 cm
a
41.
42. 2n, where nis any integer
43. The measure of ais
360° 12 or 30°.
cos 30°
6
a
.4
6.4 cos 30° a
5.5 a; 5.5 cm
44.
y→0 as x→, y→0 as x→
g(x)
x2
4
1
xyg(x)
10,000 4 108
1000 4 106
100 4 104
10 0.04
04
10 0.04
100 4 104
1000 4 106
10,000 4 108
a
1
2
(10,440)
a5220
a
c
e
52
c
20
0.16
c835.20
radius of Earth
1
2
(7920)
3960
d1acEarth radius
d15220 835.20 3960
d12095.2 miles
d2major axis d1Earth diameter
d210,440 2095.2 7920
d2424.8 miles
5b. (h, k) (0,0)
a5220
b2a2(1 e2)
b2(5220)2(1 0.162)
b226,550,840.96
(x
a2
b)2
(y
b2
k)2
1
(x
5
22
0
0
)2
26,5
(y
50
,8
0
4
)
0
2
.96
1
27,24
x
8
2
,400
26,550
y
,
2
840.96
1
6. 9x225y272x250y554 0
9(x28x?) 25(y210y?)
544 ? ?
9(x28x16) 25(y210y25)
544 9(16) 25(25)
9(x4)225(y5)2225
(x
25
4)2
(y
9
5)2
1
center: (h, k) (4, 5)
a225 b29ca2b
2
a5b3c25
9
or 4
major axis vertices: (ha, k) (4 5, 5) or
(9, 5), (1, 5)
minor axis vertices: (h, kb) (4, 5 3) or
(4, 2), (4, 8)
foci: (hc, k) (4 4, 5) or (8, 5), (0, 5)
7. 3y224yx22x41 0
3(y28y?) (x22x?) 41 ??
3(y28y16) (x22x1) 41 3(16) 1
3(y4)2(x1)26
(y
2
4)2
(x
6
1)2
1
center: (h, k) (1, 4)
a22b26ca2b
2
a2
b6
c2 6
or 22
transverse axis: vertical
vertices: (h, ka)
1, 4 2
foci: (h, kc)
1, 4 22
asymptotes: yk
a
b
(xh)
y(4)
2
6
[x(1)]
y4
3
3
(x1)
8. To find the center, find the intersection of the
asymptotes.
y2x4
2x2x4
4x4
x1
y2 1
y2
The center is at (1, 2).
Notice that (4, 2) must be a vertex and aequals
4 1 or 3.
Point Ahas an x-coordinate of 4.
Since y2x, the y-coordinate is 2 4 or 8.
The value of bis 8 2 or 6.
The equation is
(x
9
1)2
(y
36
2)2
1.
9. y24x2y5 0
y22y4x5
y22y? 4x5 ?
y22y1 4x5 1
(y1)24(x1)
vertex: (h, k) (1, 1)
4p4
p1
focus: (hp, k) (1 1, 1) or (2, 1)
axis of symmetry: yk
y1
directrix: xhp
x1 1
x0
343 Chapter 10
y
x
O
(4, 2)
(4, 5)
(9, 5)
(4, 8)
(1, 5)
yx
O
(1, 4)
(1, 4 2)
(1, 4 2)
y
x
O
(1, 1)
(2, 1)
x
0
Chapter 10 344
10. vertex: (h, k) (5, 1)
(xh)24p(yk)(h, k) (5, 1)
(9 5)24p[2 (1)] (x, y) (9, 2)
424p
4 p
(xh)24p(yk)
(x5)24(4)[y(1)]
(x5)216(y1)
Rectangular and Parametric
Forms of Conic Sections
Page 665 Graphing Calculator Exploration
1.
Tmin: [0, 6.28] step: 0.1
[7.58, 7.58] scl1 by [5, 5] scl1
1a. (1, 0) 1b. clockwise
2.
Tmin: [0, 6.28] step: 0.1
[7.58, 7.58] scl:1 by [5, 5] scl1
2a. (0, 1) 2b. clockwise
3.
Tmin: [0, 6.28] step: 0.1
[7.58, 7.58] scl1 by [5, 5] scl1
an ellipse
4. The value of adetermines the length of the radius
of the circle.
5. Each graph is traced out twice.
Page 667 Check for Understanding
1. For the general equation of a conic, Aand Chave
the same sign and ACfor an ellipse. Aand C
have opposite signs for a hyperbola. ACfor a
circle. Either A0 or C0 for a parabola.
2. t
10-6
y
x
O
(1, 2)
(2, 1)
(4, 1)
(1, 1)
(1, 0)
y
x
O
(1, 0)
3. Sample answer:
rectangular equation: y2x
parametric equations: yty
2x
t2x
t2x
yt, xt2, t
4. A1, c9; since Aand Chave the same sign
and are not equal, the conic is an ellipse.
x29y22x18y1 0
(x22x?) 9(y22y?) 1 ? ?
(x22x1) 9(y22y1) 1 1 9(1)
(x1)29(y1)29
(x
9
1)2
(y
1
1)2
1
center: (h, k) (1, 1)
a29b21
a3b1
vertices: (ha, k) (1 3, 1) or (2, 1), (4, 1)
(h, kb) (1, 1 1) or (1, 2), (1, 0)
5. A1, C0; since C0, the conic is a parabola.
y28x8
y28x8
y28(x1)
vertex: (h, k) (1, 0)
opening: right
6. A1, C1; since Aand Chave different signs,
the conic is a hyperbola.
x24xy25 4y0
(x24x?) (y24y?) 5 ? ?
(x24x4) (y24y4) 5 4 4
(x2)2(y2)25
(x
5
2)2
(y
5
2)2
1
center: (h, k) (2, 2)
a25
a5
vertices: (ha, k) (2 5
, 2)
asymptotes: yk
a
b
(xh)
y(2)
5
5
(x2)
y2 (x2)
y
x
(2, 2)
(2 5, 2) (2 5, 2)
O
7. A1, C1; since AC, the conic is a circle.
x26xy212y41 0
(x26x?) (y212y?) 41 ? ?
(x26x9) (y212y36) 41 9 36
(x3)2(y6)24
center: (h, k) (3, 6)
radius: r24
r2
8. yt26t2
yx26x2
9. x2 cos ty3 sin t
2
x
cos t
3
y
sin t
cos2tsin2t1
2
x
2
3
y
21
x
4
2
y
9
2
1
10. Sample answer:
Let xt.
y2x25x
y2t25t
xt, y2t25t, t
11. Sample answer:
x2y236
3
x
6
2
3
y
6
2
1
6
x
2
6
y
21
cos2tsin2t1
6
x
2cos2t
6
y
2sin2t
6
x
cos t
6
y
sin t
x6 cos ty6 sin t
x6 cos t, y6 sin t, 0 t2
12. x
8
t2
0
x
8
y
0
2
80xy2
y280x
Pages 667–669 Exercises
13. A1, C0; since C0, the conic is a parabola.
x24y6x9 0
x26x? 4y9 ?
x26x9 4y9 9
(x3)24y
vertex: (h, k) (3, 0)
opening: upward
14. A1, C1; since AC, the conic is a circle.
x28xy26y24 0
(x28x?) (y26y?) 24 ? ?
(x28x16) (y26y9) 24 16 9
(x4)2(y3)21
center: (h, k) (4, 3)
radians: r21
r1
345 Chapter 10
y
x
O
(3, 8)
(5, 6)
(3, 6)
txy(x, y)
662(6, 2)
557(5, 7)
4410(4, 10)
3311(3, 11)
2210(2, 10)
117(1, 7)
00 2 (0, 2)
y
x
O
y
x
3
2
t
2
t
t
0
t
O
txy(x, y)
020(2, 0)
2
03(0, 3)
20(2, 0)
3
2
03(0, 3)
y
x
(3, 0)
O
(3, 3) (4, 3)
(4, 4)
y
x
O
15. A1, C3; since Aand Chave different signs,
the conic is a hyperbola.
x23y22x24y41 0
x23y22x24y41 0
3(y28y?) (x22x?) 41 ? ?
3(y28y16) (x22x1) 41 3(16) 1
3(y4)2(x1)26
(y
2
4)2
(x
6
1)2
1
center: (h, k) (1, 4)
a22
a2
vertices: (h, ka) (1, 4 2
)
asymptotes: yk
a
b
(xh)
y(4)
2
6
[x(1)]
y4
3
3
(x1)
16. A9, C25; since Aand Chave the same sign
and are not equal, the conic is an ellipse.
9x225y254x50y119 0
9(x26x?) 25(y22y?) 119 ? ?
9(x26x9) 25(y22y1)
119 9(9) 25(1)
9(x3)225(y1)2225
(x
25
3)2
(y
9
1)2
1
center: (h, k) (3, 1)
a225 b29
a5b3
vertices: (ha, k) (3 5, 1) or (8,1), (2, 1)
(h, kb) (3, 1 3) or (3, 4), (3, 2)
17. A1, C0; since C0, the conic is a parabola.
x2y8x16
x28x? y16 ?
x28x16 y16 16
(x4)2y
vertex: (h, k) (4, 0)
opening: upward
18. ACDE0; the conic is a hyperbola.
2xy 3
xy
3
2
quadrants: I and III
transverse axis: yx
vertices:
3
2
,
3
2
or
6
2
,
6
2
,
3
2
,
3
2
or
2
6
,
2
6
19. A5, C2; since Aand Chave the same sign
and are not equal, the conic is an ellipse.
5x22y240x20y110 0
5(x28x?) 2(y210y?) 110 ? ?
5(x28x16) 2(y210y25)
110 5(16) 2(25)
5(x4)22(y5)220
(x
4
4)2
(y
10
5)2
1
(y
10
5)2
(x
4
4)2
1
center: (h, k) (4, 5)
a210 b24
a10
b2
vertices: (h, ka) (4, 5 10
)
(hb, k) (4 2, 5) or (6, 5), (2, 5)
Chapter 10 346
y
x
(1, 4)
(1, 4 2)
(1, 4 2)
O
y
x
(2, 1)
(3, 4)
(3, 1)
(3, 2)
(8, 1)
O
y
x
(4, 0)
O
y
x
O
()
,6
2
6
2
()
,6
2
6
2
y
x
(2, 5)
(4, 5) (6, 5)
(4, 5 10)
(4, 5 10)
O
20. A1, C1; since AC, the conic is a circle.
x28x11 y2
(x28x?) y211 ?
(x28x16) y211 16
(x4)2y25
center: (h, k) (4, 0)
radius: r25
r5
21. A9, C8; since Aand Chave different signs,
the conic is a hyperbola.
8y29x216y36x100 0
8(y22y?) 9(x24x?) 100 ? ?
8(y22y1) 9(x24x4) 100 8(1) 9(4)
8(y1)29(x2)272
(y
9
1)2
(x
8
2)2
1
center: (h, k) (2, 1)
a29
a3
vertices: (h, ka) (2, 1 3) or (2, 4), (2, 2)
asymptotes: yk
a
b
(xh)
y1
2
3
2
(x2)
y1
3
4
2
(x2)
22. A0, C4; since A0, the conic is a parabola.
4y24y8x15
4(y2y?) 8x15 ?
4
y2y
1
4
8x15 4
1
4
4
y
1
2
28x16
y
1
2
22x4
y
1
2
22(x2)
vertex: (h, k)
2,
1
2
opening: left
347 Chapter 10
y
x
(4
(4, 0)
5, 0)
(4, 5)
O
y
x
(2, 4)
(2, 1)
(2, 2)
O
()
2 , 1
2
y
x
O
y
x
(7, 2)
(5, 2)
(3, 2)
O
y
x
(2, 3
(2, 3)
10)
(2 10, 3)
O
txy(x, y)
117(1, 7)
001(0, 1)
111(1, 1)
227(2, 7)
y
x
O
23. 4y210x16yx25
x24y210x16y5 0
A1, C4; since Aand Chave different signs,
the conic is a hyperbola.
x24y210x16y5 0
(x210x?) 4(y24y?) 5 ? ?
(x210x25) 4(y24y4) 525 4(4)
(x5)24(y2)24
(x
4
5)2
(y
1
2)2
1
center: (h, k) (5, 2)
a24
a2
vertices: (ha, k) (5 2, 2) or (3, 2),
(7, 2)
asymptotes: yk
a
b
(xh)
y(2)
1
2
[x(5)]
y2
1
2
(x5)
24. Ax2Bxy Cy2Dx Ey F0
2x20 2y2(8)x12y6 0
2x22y28x12y6
AC; circle
2(x24x?) 2(y26y?) 6 ? ?
2(x24x4) 2(y26y6y9)
62(4) 2(9)
2(x2)22(y3)220
(x2)2(y3)210
center: (h, k) (2, 3)
radius: r210
r10
25. y2t24t1
y2x24x1
26. xcos 2tysin 2t
cos22tsin22t1
x2y21
27. xcos tysin t
xcos t
cos2tsin2t1
(x)2y21
x2y21
28. x3 sin ty2 cos t
3
x
sin t
2
y
cos t
cos2tsin2t1
2
y
2
3
x
21
y
4
2
x
9
2
1
x
9
2
y
4
2
1
29. xsin 2ty2 cos 2t
xsin 2t
2
y
cos 2t
cos22tsin22t1
2
y
2(x)21
y
4
2
x21
x2
y
4
2
1
30. x2t1
x1 2t
x
2
1
t
yt
y
x
2
1
31. x3 cos 2ty3 sin 2t
3
x
cos 2t
3
y
sin 2t
cos22tsin22t1
3
x
2
3
y
21
x
9
2
y
9
2
1
x2y29
32. Sample answer:
x2y225
2
x
5
2
2
y
5
2
1
5
x
2
5
y
21
cos2tsin2t1
5
x
2cos2t
5
y
2sin2t
5
x
cos t
5
y
sin t
x5 cos ty5 sin t
x5 cos t, y5 sin t,0 t2
Chapter 10 348
txy(x, y)
010(1, 0)
2
01(0, 1)
10(1, 0)
3
2
01(0, 1)
y
x
t
0
O
txy(x, y)
010(1, 0)
2
01(0, 1)
10(1, 0)
3
2
01(0, 1)
y
x
t
0
O
t
0
y
x
O
txy(x, y)
020(0, 2)
2
30(3, 0)
02(0, 2)
3
2
30(3, 0)
txy(x, y)
002(0, 2)
4
10(1, 0)
2
02(0, 2)
3
4
10(1, 0)
t
0
y
x
O
y
x
O
txy(x, y)
010(1, 0)
111(1, 1)
232
(3, 2
)
353
(5, 3
)
472(7, 2)
33. Sample answer:
x2y216 0
x2y216
1
x
6
2
1
y
6
2
1
4
x
2
4
y
21
cos2tsin2t1
4
x
2cos2t
4
y
2sin2t
4
x
cos t
4
y
sin t
x4 cos ty4 sin t
x4 cos t, y4 sin t, 0 t2
34. Sample answer:
x
4
2
2
y
5
2
1
2
x
2
5
y
21
cos2tsin2t1
2
x
2cos2t
5
y
2sin2t
2
x
cos t
5
y
sin t
x2 cos ty5 sin t
x2 cos t, y5 sin t, 0 t2
35. Sample answer:
1
y
6
2
x21
x2
4
y
21
cos2tsin2t1
x2cos2t
4
y
2sin2t
xcos t
4
y
sin t
y4 sin t
xcos t, y4 sin t, 0 t2
36. Sample answer:
Let xt.
yx24x7
yt24t7
xt, yt24t7, t
37. Sample answer:
Let yt.
xy22y1
xt22t1
xt22t1, yt, t
38. Sample answer:
Let yt.
(y3)24(x2)
(t3)24(x2)
0.25(t3)2x2
0.25(t3)22 x
x0.25(t3)22, yt, t
39a. Answers will vary. Sample answers:
Let xt.
xy
ty
t2y
xt, yt2, t0
Let yt.
xy
xt
xt
, yt, t0
39b.
Tmin: [0, 5] step: 0.1
[7.58, 7.58] scl1 by [5, 5] scl1
39c. yes
39d. There is usually more than one parametric
representation for the graph of a rectangular
equation.
40a. a circle with center (0, 0) and radius 6 feet
(xh)2(yk)2r2
(x0)2(y0)262
x2y236
40b. Sample answer:
x2y236
3
x
6
2
3
y
6
2
1
6
x
2
6
y
21
sin2(qt) cos2(qt) 1
Since the paddlewheel completes a revolution in
2 seconds, the period is
2
q
2, so q.
sin2(t) cos2(t) 1
6
x
2sin2(t)
6
y
2cos2(t)
6
x
sin(t)
6
y
cos(t)
x6 sin(t)y6 cos (t)
x6 sin (t), y6 cos (t), 0t2
40c. C2r
C26
C37.7 ft
The paddlewheel makes 1 revolution, or moves
37.7 ft in 2 seconds.
37
2
.7
s
ft
60 s 1131 ft
The paddlewheel moves about 1131 ft in
1 minute.
349 Chapter 10
txy(x, y)
000(0, 0)
111(1, 1)
224(2, 4)
339(3, 9)
y
x
O
41a. A2, C5; since Aand Chave the same sign
and AC, the graph is an ellipse.
2x25y20
5y22x
y2
2
5
x
y
2
5
x
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. A1, C1; since AC, the graph is a circle.
x2y24x6y13 0
(x24x?) (y26y?) 13
(x24x4) (y26y9) 13 4 9
(x2)2(y3)20
center: (h, k) (2, 3)
radius: 0
The graph is a point at (2, 3); the equation is
that of a degenerate circle.
41c. A9, C1; since Aand Chave different
signs, the graph is a hyperbola.
y29x20
y29x2
y3x
The graph is two intersecting lines y3x; the
equation is that of a degenerate hyperbola.
42. The substitution for xmust be a function that
allows xto take on all of the values stipulated by
the domain of the rectangular equation. The
domain of yx25 is all real numbers, but
using a substitution of xt2would only allow for
values of xsuch that x0.
43a. center: (h, k) (0, 0)
(xh)2(yk)2r2
(x0)2(y0)26
x2y236
43b. x2y236
3
x
6
2
3
y
6
2
1
6
x
2
6
y
21
sin2tcos2t1
6
x
2sin2t
6
y
2cos2t
6
x
sin t
6
y
cos t
x6 sin ty6 cos t
x6 sin t, y6 cos t
Since the second hand makes 2 revolutions,
0 t4.
43c.
Tmin: [0, 4] step: 0.1
[9.10, 9.10] scl1 by [6, 6] scl1
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,
cos t
a
x
or xacos t. Using the smaller triangle,
sin t
b
y
or ybsin t.
45. x212y10x25
x210x? 12y25 ?
x210x25 12y25 25
(x5)212y
vertex: (h, k) (5, 0)
4p12
p3
focus: (h, kp) (5, 0 3) or (5, 3)
axis of symmetry: xh
x5
directrix: ykp
y0 3
y3
46. c25
quadrants: II and IV
transverse axis: yx
vertices: xy 25 xy 25
5(5) 25 5(5) 25
(5, 5) (5, 5)
47. 3x23y218x12y9
3(x26x?) 3(y24y?) 9 ? ?
3(x26x9) 3(y24y4) 9 3(9) 3(4)
3(x3)23(y2)248
(x3)2(y2)216
center: (h, k) (3, 2)
radians: r216
r4
Chapter 10 350
(5, 0)
y
x
O
(5, 5)
(5, 5)
y
x
O
y
x
(3, 2)
(3, 2)
(7, 2)
O
48.
cos 60°
3
x
0
sin 60°
3
y
0
x30 cos 60° y30 sin 60°
x15 lb y153
lb
49. y0.13x37.8
0.13xy37.8 0
A0.13, B1, C37.8
Car 1: (x1, y1) (135, 19)
d1Ax1By1C
A2
B2
53. ykxz y kxz
16 k(5)(2) y1.6(8)(3)
1.6 ky38.4
54. 595(3) 7(9)
73
78
Yes, an inverse exists since the determinant of the
matrix 0.
55. m
y
x
2
2
y
x
1
1
3
7
(
4
6)
1
3
yy1m(xx1)
y4
1
3
(x6) or y7
1
3
(x3)
ymx bymx b
4
1
3
(6) by
1
3
x6
6 b
56. (1 # 4) @ (2 # 3) 1 @ 2
2
The correct choice is B.
Transformation of Conics
Pages 674–675 Check for Understanding
1. Sample answers:
(h, k) (0, 0)
xy2
(h, k) (3, 3)
(xh) (yk)2
(x3) (y3)2
2. Replace xwith xcos 30° ysin 30° or
x
1
2
y.
Replace ywith xsin 30° ycos 30° or
1
2
x, y.
2
2
3
2
10-7
351 Chapter 10
x
y
30
60˚
d1
d11.24
The point (135, 19) is about 1 unit from the line
y0.13x37.8.
Car 2: (x2, y2) (245, 16)
d2Ax2By2C
A2
B2
0.13(135) 1(19) (37.8)
(0.13)
212
d2
d29.97
The point (245, 16) is about 10 units from the line
y0.13x37.8.
Car 1: the point (135, 19) is about 9 units closer
to the line y0.13x37.8 than the point
(245, 16).
50. Let vSin1
1
2
.
Sin v
1
2
v30
sin
2 Sin1
1
2
sin (2v)
sin (2 30)
sin 60
51. s
1
2
(abc)
s
1
2
(48 32 44)
s62
Ks(s
a)(s
b)(s
c)
K62(62
48
)(62
32)(6
2 4
4)
K46872
0
K685 units2
52. 2y3
2y3
1
2y3
2y3
1
2y3 2y3 22y3
1
7 22y3
7
2
2y3
4
4
9
2y3
3
4
7
2y
3
8
7
y
3
2
0.13(245) 1(16) (37.8)
(0.13)
212
x
y
2
y
x
O
y
x
x
3 (
y
3)2
O
3. 90° or 270°
4. Ebony; B24AC
63
24(7)(13) 0
and AC
5. B24AC 0 4(1)(1)
4
AC1; circle
(xh)2(yk)27
(x3)2(y2)27(h, k) (3, 2)
x26x9 y24y4 7
x2y26x4y6 0
6. B24AC 0 4(2)(0)
0
parabola
y2x27x5
y5 2x27x
y5 2
x2
7
2
x
y5 2
7
4
22
x2
7
2
x
7
4
2
y5
4
8
9
2
x
7
4
2
y
9
8
2
x
7
4
2
y
9
8
k2
x
7
4
h
2
y
9
8
5 2
x
7
4
4
2(h, k) (4, 5)
y
3
8
1
2
x
9
4
2
y
3
8
1
2
x2
1
4
8
x
8
1
1
6
y
3
8
1
2x29x
8
8
1
0 2x29xy 14
2x29xy14 0
7. B24AC 0 4(1)(1)
4
hyperbola
x2y29
(xcos 60° ysub 60°)2
(xsin 60° ycos 60°)29
1
2
x
2
3
y
2
2
3
x
1
2
y
29
1
4
(x)2
2
3
xy
3
4
(y)2
3
4
(x)2
2
3
xy
1
4
(y)2
9
1
2
(x)23
xy
1
2
(y)29
(x)223
xy(y)218
(x)223
xy(y)218 0
8. B24AC 0 4(1)(1)
4
AC1; circle
x25xy23
xcos
4
ysin
4
25
xcos
4
ysin
4
xsin
4
ycos
4
23
2
2
x
2
2
y
25
2
2
x
2
2
y
2
2
x
2
2
y
23
1
2
(x)2xy
1
2
(y)2
5
2
2
x
5
2
2
y
1
2
(x)2xy
1
2
(y)23
(x)2(y)2
5
2
2
x
5
2
2
y3
2(x)22(y)252
x52
y6
2(x)22(y)252
x52
y60
9. B24AC 424(9)(4)
128
AC; ellipse
tan 2v
A
B
C
tan 2v
9
4
4
tan 2v0.8
2v38.65980825°
v19°
10. B24AC 524(8)(4)
153
hyperbola
tan 2v
A
B
C
tan 2v
8
5
(4)
tan 2v0.416666667
2v22.61986495°
v11°
11. 3(x1)24(y4)20
3(x22x1) 4(y28y16) 0
3x26x3 4y232y64 0
4y232y(3x26x67) 0
↑↑ ↑
ab c
ybb24
ac
2a
Chapter 10 352
O
y
x
1
90˚
270˚
x
2
100
y
2
25
1
x
2
25
y
2
100
y32 322
4(4)(3
x26
x67
)
2(4)
y32 48x2
96
x48
8
y32 48(x
1)2
8
x1, y4; point
y
x
(1, 4)
O
12a. y
1
6
x2
xsin 30° ycos 30°
1
6
(xcos 30° ysin 30°)2
1
2
x
2
3
y
1
6
2
3
x
1
2
y
2
1
2
x
2
3
y
1
6
3
4
(x)2
2
4
3
xy
1
4
(y)2
1
2
x
2
3
y
1
8
(x)2
12
3
xy
2
1
4
(y)2
12x123
y3(x)223
xy(y)2
0 3(x)223
xy
(y)212x123
y
3(x) 23
xy(y)212x123
y0
12b. 3x223
xy y212x123
y0
1y2(23
x123
)y(3x212x)0
↑↑ ↑
ab c
y
y
y3
x63
y3
x63
and y
1
6
x2
Pages 675–677 Exercises
13. B24AC 0 4(3)(0)
0
parabola
y3x22x5
y5 3x22x
y5 3
x2
2
3
x
y5 3
1
3
23
x2
2
3
x
1
3
2
y
1
3
4
3
x
1
3
2
y
1
3
4
k3
x
1
3
h
2
y
1
3
4
3 3
x
1
3
2
2(h, k) (2, 3)
y
5
3
3
x
7
3
2
y
5
3
3
x2
1
3
4
x
4
9
9
y
5
3
3x214x
4
3
9
0 3x214xy18
3x214xy18 0
192
x43
2
2
12x2
144x
432
12
x24
8x
2
(23
x123
) (23
x
12
3
)2
4(1
)(3x2
12x)
2(1)
bb24
ac
2a
14. B24AC 0 4(4)(5)
80
AC; ellipse
4x25y220
4(xh)25(yk)220
4(x5)25(y6)220
(h, k) (5, 6)
4(x210x25) 5(y212y36) 20
4x240x100 5y260y180 20
4x25y240x60y260 0
15. B24AC 0 4(3)(1)
12
AC; ellipse
3x2y29
3(xh)2(yk)29
3(x1)2(y3)29(h, k) (1, 3)
3(x22x1) y26y9 9
3x26x3 y26y9 9
3x2y26x6y3 0
16. B24AC 0 4(12)(4)
192
AC; ellipse
4y212x224
4(yk)212(xh)224
4(y4)212(x1)224
(h, k) (1, 4)
4(y28y16) 12(x22x1) 24
4y232y64 12x224x12 24
y28y16 3x26x3 6
3x2y26x8y13 0
17. B24AC 0 4(9)(25)
900
hyperbola
9x225y2225
9(xh)225(yk)2225
9(x0)225(y5)2225 (h, k) (0, 5)
9x225(y210y25) 225
9x225y2250y850 0
18. (x3)24y
x26x9 4y0
B24AC 0 4(1)(0)
0
parabola
(x3)24y
(x3 h)24(yk)
(x3 7)24(y2) (h, k) (7, 2)
(x10)24y8
x220x100 4y8
x220x4y108 0
19. B24AC 0 4(1)(0)
0
parabola
x28y0
(xcos 90° ysin 90°)28(xsin 90° ycos 90°) 0
(y)28(x)0
(y)28x0
353 Chapter 10
20. B24AC 0 4(2)(2)
16
AC; circle
2x22y28
2(xcos30°ysin30°)2
2(xsin30°ycos30°)28
2
2
3
x
1
2
y
22
1
2
x
2
3
y
28
2
3
4
(x)2
2
3
xy
1
4
(y)2
2
1
4
(x)2
2
3
xy
3
4
(y)2
8
3
2
(x)23
xy
1
2
(y)2
1
2
(x)2
3
xy
3
2
(y)28
2(x)22(y)28
(x)2(y)240
21. B24AC 0 4(1)(0)
0
AC; parabola
y28x0
xsin
6
ycos
6
28
xcos
6
ysin
6
0
1
2
x
2
3
y
28
2
3
x
1
2
y
0
1
4
(x)2
2
3
xy
3
4
(y)243
x4y0
(x)223
xy3(y)2163
x16y0
22. B24AC 124(0)(0)
1
hyperbola
xy 8
xcos
4
ysin
4
xsin
4
ycos
4
8
2
2
x
2
2
y
2
2
x
2
2
y
8
1
2
(x)2
1
2
xy
1
2
xy
1
2
(y)28
(x)2(y)216
(x)2(y)216 0
23. B24AC 0 4(1)(1)
4
AC; circle
x25xy23
xcos
3
ysin
3
25
xcos
3
ysin
3
xsin
3
ycos
3
23
1
2
x
2
3
y
25
1
2
x
2
3
y
2
3
x
1
2
y
23
1
4
(x)2
2
3
xy
3
4
(y)2
5
2
x
5
2
3
y
3
4
(x)2
2
3
xy
1
4
(y)23
(x)2(y)2
5
2
x
5
2
3
y3
2(x)22(y)25x53
y6 0
24. B24AC 0 4(16)(4)
256
hyperbola 16x24y264
16(xcos 60° ysin 60°)2
4(x sin 60° ycos 60°) 64
16
1
2
x
2
3
y
24
2
3
x
1
2
y
264
16
1
4
(x)2
2
3
xy
3
4
(y)2
4
3
4
(x)2
2
3
xy
1
4
(y)2
64
4(x)283
xy12(y)2
3(x)223
xy(y)264
(x)2103
xy11y)264 0
25. 6x25y230
6(xcos 30° ysin 30°)2
5(xsin 30° ycos 30°)230
6
2
3
x
1
2
y
25
1
2
x
2
3
y
230
6
3
4
(x)2
2
3
xy
1
4
(y)2
5
1
4
(x)2
2
3
xy
3
4
(y)2
30
1
4
8
(x)2
6
2
3
xy
6
4
(y)2
5
4
(x)2
5
2
3
xy
1
4
5
(y)230
2
4
3
(x)2
2
3
xy
2
4
1
(y)230 0
23(x)223
xy21(y)2120 0
26. 324AC 424(9)(5)
164
AC; ellipse
tan 2v
A
B
C
tan 2v
9
4
5
tan 2v1
2v45°
v23°
27. B24AC (1)24(1)(4)
17
hyperbola
tan 2v
A
B
C
tan 2v
1
(
1
4)
tan 2v
1
5
2v11.30993247°
v6°
28. B24AC 824(8)(2)
0
parabola
tan 2v
A
B
C
tan 2v
8
8
2
tan 2v
4
3
2v53.13010235°
v27°
29. B24AC 924(2)(14)
31
AC; ellipse
tan 2v
A
B
C
tan 2v
2
9
14
tan 2v
3
4
2v36.86989765°
v18°
Chapter 10 354
30. B24AC 424(2)(5)
24
AC; ellipse
tan 2v
A
B
C
tan 2v
2
4
5
tan 2v
4
3
2v53.13010235°
v27°
31. B24AC
43
24(2)(6)
0
parabola
tan 2v
A
B
C
tan 2v
tan 2v3
2v60°
v30°
32. B24AC 424(2)(2)
0
parabola
AC; v
4
or 45°
33. (x2)2(x3)25(y2)
x24x4 x26x9 5(y2)
10x5 5(y2)
2x1 y2
2x3 y
y2x3line
34. 2x26y28x12y14 0
x23y24x6y7 0
3y2(6)y(x24x7) 0
↑↑ ↑
ab c
ybb24
ac
2a
43
2 6
36. (x2)2(y2)24(xy) 8
x24x4 y24y4 4x4y8
(1)y20yx20
↑↑↑
abc
ybb24
ac
2a
355 Chapter 10
y
x
y
2
x
3
O
y
x2, y1; point
35. y29x20
y29x2
y9x2
y3x
intersecting lines
6 12(x
2)2
6
y
x
(2, 1)
O
y
x
y
3
x
y
3
x
O
y0 0 4
(1)(x2)
2(1)
y
x0, y0; point
37. x22xy y25x5y0
(1)y2(2x5)y(x25x) 0
↑↑ ↑
ab c
ybb24
ac
2a
4x2
2
y(2x5) (2x
5)2
4(1)(x
25x
)
2(1)
y2x5 4x2
20x
25
4x2
20x
2
y
x
(0, 0)
O
y
[6.61, 14.6] scl1 by [2, 12] scl1
38. 2x29xy 14y25
14y2(9x)y(2x25) 0
↑↑ ↑
ab c
ybb24
ac
2a
2x5 40x
2
5
2
y9x(9x)2
4(14)
(2x2
5)
2(14)
y
[7.58, 7.58] scl1 by [5, 5] scl1
9x21x2
28
0
28
y(6) (6)2
4(3
)(x2
4x
7)
2(3)
y6 12x2
48
x48
6
y
[7.58, 7.58] scl1 by [5, 5] scl1
40. 2x243
xy 6y23xy
6y2(43
x1)y(2x23x) 0
↑↑ ↑
ab c
ybb24
ac
2a
5x153x2
32
39. 8x25xy 4y22
(4)y2(5x)y(8x22) 0
↑↑ ↑
ab c
ybb24
ac
2a
42. 9x24xy 6y220
6y2(4x)y(9x220) 0
↑↑ ↑
ab c
ybb24
ac
2a
Chapter 10 356
y5x(5x)2
4(4
)(8x2
2)
y(43
x1) (43
x
1)2
4(6
)(2x2
3x)
2(6)
y43
x1 48x2
83
x
1
48x2
72x
12
y
[8.31, 2.31] scl1 by [2, 5] scl1
41. 2x24xy 2y222
x22
y12
2y2(4x22
)y(2x222
x12) 0
↑↑ ↑
ab c
ybb24ac
2a
43
x1 83
x72
x1
12
y(4x22
)(4x
22
)2
4(2
)(2x2
22
x
12
)
2(2)
y4x22
16x2
162
x8
16x
216
2
x
96
4
y
[10.58, 4.58] scl1 by [2, 8] scl1
4x22
32
2
x8
8
4
y4x(4x)2
4(6)(9
x22
0)
2(6)
y
[7.85, 7.85] scl1 by [5, 5] scl1
43a.
T(1320, 1320)
43b. circle
center: (h, k) (1320, 1320)
radius: r1320
(xh)2(yk)2r2
(x1320)2(y1320)213202
(x1320)2(y1320)21,742,400
44a. B24AC 0 4(1)(0)
0
parabola; 360°
44b. B24AC 0 4(8)(6)
192
AC; ellipse; 180°
44c. B24AC 424(0)(0)
16
hyperbola; 180°
44d. 324AC 0 4(15)(15)
900
AC; 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 xxcos vysin vand
yxsin vycos v.
x2y2r2
(xcos vysin v)2(xsin vycos v)2r2
(x)2cos2vxycos vsin v(y)2sin2v
(x)2sin2vxycos vsin v(y)2cos vr2
[(x)2(y)2] cos2v[(x)2(y)2] sin2vr2
[(x)2(y)2](cos2vsin2v) r2
[(x)2(y)2](1) r2
(x)2(y)2r2
46a. B24AC
103
24(31)(21)
2304
AC; elliptical
4x212
x24
80
12
O
y
x
(0, 5280)
(5280, 0)
(1320, 1320)
(y3)225(x2)21
51.
46b. 31x2103
xy 21y2144
21y2(103
x)y(31x2144) 0
↑↑ ↑
ab c
ybb24
ac
2a
48a. center: (h, k) (0, 0)
major axis: horizontal
a81
or 9
b36
or 6
ca2b
2
c81
36
c45
or 35
48b. T(35
, 0)
8
x
1
2
3
y
6
2
1
(x
81
h)2
(y
36
k)2
1
48c.
(y
8
1
)2
1
49. A3, C5; since Aand Chave different signs,
the conic is a hyperbola.
50. (h, k) (2, 3)
e
a
c
2
5
6
1
c
2
5
6
c
b2a2c2
b212
2
5
6
2
b2
2
1
5
major axis: horizontal
(x
a2h)2
(y
b2k)2
1
(x
1
2)2
1
(x2)225(y3)21
major axis : vertical
(y
a2k)2
(x
b2h)2
1
[y
1
(
23)]2
1
(x2)2
2
1
5
[y(3)]2
2
1
5
(x35
)2
36
357 Chapter 10
y(103
x) (10
3
x)2
4(21)
(31x2
144
)
2(21)
y
46c. tan 2v
A
B
C
tan 2v
tan 2v3
2v60°
v30°
47a. tan 2v
A
B
C
tan 2v
9
2
1
3
1
tan 2v3
2v60°
v30°
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°.
47b. B24AC
23
24(9)(11)
384
AC; the graph is an ellipse.
9x223
xy 11y224 0
9[x cos (30°) y sin (30°)]223
[(x cos (30°) y sin (30°)]
[x sin (30°) y cos (30°)] 11
[x sin (30°) y cos (30°)2] 24 0
9
2
3
x
1
2
y
223
2
3
x
1
2
y
1
2
x
2
3
y
11
1
2
x
2
3
y
224 0
9
3
4
(x)2
2
3
xy
1
4
(y)2
23
4
3
(x)2
3
4
xy
1
4
xy
4
3
(y)2
11
1
4
(x)2
2
3
xy
3
4
(y)2
24 0
2
4
7
(x)2
9
2
3
xy
9
4
(y)2
6
4
(x)2
6
4
3
xy
2
4
3
xy
6
4
(y)2
1
4
1
(x)2
11
2
3
xy
3
4
3
(y)224 0
8(x)212(y)224 0
(x
3
)2
(y
2
)2
1
103
31 21
103
x230
4x2
12,09
6
42
y
x
O
y
x
O
4
4848
4
8
8
v030°60° 90° 120° 150° 180°
r11.43.9 3.9 1.4 11
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
58.
59. The expression
15
8a
0a
8b
6
5
b2
simplifies to
a3
2
6
b3
. Since
1 band 2 a, the expression is always larger
than
23
21
6
3
1
9
. Since b2 and a3, the
expression is always less than
33
22
6
3
7
32
6
or 2.
The correct choice is B.
Systems of Second-Degree
Equations and Inequalities
Page 682 Check for Understanding
1. Possible number of solutions: 0, 1, 2, 3, or 4
10-8
Chapter 10 358
N
WE
S
8
5
c
52.
c2a2b2
c28252
c89
c9.4
about 9.4 m/s
53. cos 70° 0.34
cos 170° 0.98
cos 70°
54.
5
1
6
5
1
6
18
0°
56.25
56
1
6
5
0
56° 15
55.
y22
y
3
y
5
2
(y
2y
2)
(y
5
1)
y22
y
3
y
5
2
y
A
2
y
B
1
2y5 A(y1) B(y2)
2(2) 5 A(2 1) B(2 2)
1 A
1 A
2y5 A(y1) B(y2)
2(1) 5 A(1 1) B(1 2)
3 B
y
A
2
y
B
1
y
1
2
y
3
1
56.
x
y
1
2
x
y
2
1
1
y
2
2
5
4
y29.6
57. 8m3n4p6
8m6 3n4p
m
3
4
3
8
n
1
2
p
4m9n2p4
4
3
4
3
8
n
1
2
p
9n2p4
3
3
2
n2p9n2p4
2
2
1
n7
n
2
3
8m3n4p66m12n5p1
8m3
2
3
4p66m12
2
3
5p1
8m2 4p66m8 5p1
8m4p46m5p7
2mp1
2mp1→10m5p5
6m5p7→()6m5p7
16m12
m
3
4
8m3n4p6
8
3
4
3
2
3
4p6
6 2 4p6
p
1
2
3
4
,
2
3
,
1
2
xh(x) [[x]] 3
3 x26
2 x15
1 x04
0 x13
1 x22
2 x31
3 x40
4 x51
5 x62
6 x73
h
(
x
)
x
359 Chapter 10
O
y
x
x
2 (
y
3)2 9
y
x
2
(1, 1)
(3, 3)
y
x
O
y(40) (40)
24(
5)(84)
2(5)
y
no solution
40 80
10
y
x
O
y
x
O
(2, 0) (2, 0)
y
x
O
(1, 1)
y
x
O
4812
4
4
8
12
y
x
O
y
x
O
4884
4
8
4
8
2. Sample answer: yx2, x2(y3)29
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. xy0
xy
(x
20
1)2
(y
5
1)2
1
(x
20
1)2
(x
5
1)2
1
(x1)24(x1)220
x22x1 4(x22x1) 20
5x210x5 20 0
5(x22x3) 0
5(x3)(x1) 0
x3 0x1 0
x3x1
(3, 3), (1, 1)
6. x2y10
x10 2y
x2y216
(10 2y)2y216
100 40y5y216
5y240y84 0
ybb24
ac
2a
7. x2y24
y24 x2
9x24y236 x2y24
9x24(4 x2) 36 22y24
13x216 36 y20
x24y0
x2
(2, 0)
8. xy 1x2y
x(x2) 11
2y
x311 y
x1
(1, 1)
9. 10.
11.
(2, 1), (2, 1)
15. 1 2xy
1 2xy
4x2y225
4x2(1 2x)225
4x21 4x4x225
8x24x24 0
4(2x2x6) 0
4(2x3)(x2) 0
2x3 0x2 0
x1.5 x2
1 2xy1 2xy
1 2(1.5) y1 2(2) y
4 y3 y
(1.5, 4), (2, 3)
16. xy2
x2 y
x2100 y2
(2 y)2100 y2
4 4yy2100 y2
2y24y96 0
2(y22y48) 0
2(y6)(y8) 0
y6 0y8 0
y6y8
xy2xy2
x6 2x(8) 2
x8x6
(8, 6), (6, 8)
Chapter 10 360
y
x
(22, 14)
20
20
20
20
40
4040
40
O
12a. Let xside length of flowerbed 1.
Let yside length of flowerbed 2.
A1xxor x2
A2yyor y2
Total Area x2y2
680 x2y2
x2y2680
Difference of Areas x2y2
288 x2y2
x2y2288
12b. Since side length
cannot be negative, an
estimated solution is
(22, 14).
12c. x2y2680
y2680 x2
x2y2288 x2y2680
x2(680 x2) 288 222y2680
2x2968 y2196
x2484 y14
x22
22 ft and 14 ft
Pages 682–684 Exercises
13. x1 0
x1
y249 x2
y249 (1)2
y248
y6.9
(1, 6.9)
14. xy 2
y
2
x
x23 y2
x23
2
x
2
x23
x
4
2
x43x24
x43x24 0
(x24)(x21) 0
x24 0x21 0
x24x21
x2
xy 2xy 2
2(y) 2(2)y2
y1y1
(1, 6.9)
(1, 6.9)
y
x
O
y
x
(2, 1)
(2, 1)
O
y
x
(2, 3)
(1.5, 4)
O
(6, 8)
(8, 6)
y
x
O
y
x
(3, 2)
(2, 3)
(3, 2)
(2, 3)
O
y
x
(3, 2)
(0, 1)
O
17. xy0
xy
(x
9
1)2
y21
(y
9
1)2
y21
(y1)29y29
y22y1 9y29
8y22y8 0
4y2y4 0
ybb24
ac
2a
19. xy1
y1 x
(y1)24 x
(1 x1)24 x
(2 x)24 x
4 4xx24 x
x23x0
x(x3) 0
x0x3 0
x3
xy1xy1
0 y13 y1
y1y2
(0, 1), (3, 2)
20. xy 6 0
y
6
x
x2y213
x2
6
x
213
x2
3
x
6
2
13
x436 13x2
x413x236 0
(x29)(x24) 0
x29 0x24 0
x29x24
x3x2
xy 6 0xy 6 0
3y6 02y6 0
y2y3
xy 6 0xy 6 0
3y6 02y6 0
y2y3
(3, 2), (3, 2) (2, 3), (2, 3)
361 Chapter 10
y1 124
(4)(4)
2(1)
y
no solution
18. x22y210
x210 2y2
3x29 y2
3(10 2y2) 9 y2
30 6y29 y2
5y221
y24.2
y2.0
3x29 y23x29 y2
3x29 (2.0)23x29 (2.0)2
x21.6 x21.6
x1.3 x1.3
(1.3, 2.0), (1.3, 2.0)
1 63
2
y
x
O
y
x
(1.3, 2.0) (1.3, 2.0)
(1.3, 2.0)
(1.3, 2.0)
O
21. x2y3 0
x2y3
x24y236
y3 4y236
4y2y33 0
(y3)(4y11) 0
y3 04y11 0
y3y2.8
x2y3 0x2y3 0
x23 3 0x2(2.8) 3 0
x20x25.8
x0x2.4
(0, 3), (2.4, 2.8)
22. 2yx3 0
2y3 x
x216 y2
(2y3)216 y2
4y212y9 16 y2
5y212y7 0
ybb24
ac
2a
23. xy 4
y
4
x
x225 9y2
x225 9
4
x
2
x225
1
x
4
2
4
x425x2144
x425x2144 0
(x29)(x216) 0
x29 0x216 0
x29x216
x3x4
xy 4xy 4
3y44y4
y1.3 y1
xy 4xy 4
3y44y4
y1.3 y1
(3, 1.3), (3, 1.3) (4, 1), (4, 1)
24. 25.
26.
27.
Chapter 10 362
y
x
(2.4, 2.8)
(0, 3)
(2.4, 2.8)
O
y
y
y0.5 or y2.9
2yx3 02yx3 0
2(0.5) x3 02(2.9) x3 0
x4.0 x2.8
(4.0, 0.5), (2.8, 2.9)
12 284
10
12 122
4(5)(
7)
2(5)
(4, 0.5)
(2.8, 2.9)
y
x
O
4
4488
8
4
8
y
x
O
y
x
O
y
x
O
y
x
O
4884
4
4
8
8
28. 29.
30. 31.
32. 33.
34. parabola:
vertex: (1, 3)
(yk)24p(xh)
(5 3)24p(1 1)
1
2
p
(y3)24
1
2
(x1)
(y3)22(x1)
line:
m2, b7
ymx b
y2x7
35. circle:
center: (0, 0), radius: 22
x2y2r2
x2y28
hyperbola:
(2)(2) 4
xy 4
36. large ellipse:
a5, b4, center (0, 0)
a
y2
2
b
x2
2
1
2
y
5
2
1
x
6
2
1(interior is shaded)
small ellipse:
a3, b2, center (0, 1)
a
x2
2
(y
b2k)2
1
x
9
2
(y
4
1)2
1(exterior is shaded)
37a. 2x2yPxyA
2x2y150 xy 800
37b. Asystem of a line and a hyperbola may have 0,
1, or 2 solutions.
37c.
37d. xy 800
y
80
x
0
2x2y150
2x2
80
x
0
150
2x
16
x
00
150
2x21600 150x
x275x800 0
xbb24
ac
2a
363 Chapter 10
y
x
O
4488
4
8
8
4
y
x
O
y
x
O
8
4
8
4
84
4
8
y
x
O
4488
4
4
8
8
y
x
O
448
4
4
8
y
x
O
y
x
O
40
40 804080
40
80
80
x
x
x12.88 or x62.12
xy 800 xy 800
12.88y800 62.12y800
y62.11 y12.88
12.9 m by 62.1 m or 62.1 m by 12.9 m
38a. (h, k) (0, 4)
(x, y) (6, 0)
(xh)24p(yk)
(6 0)24p(0, 4)
36 16p
2.25 p
(xh)24p(yk)
(x0)24(2.25)(y4)
x29(y4)
x29(y4), y0
38b.
75 2425
2
(75) (75)
24(
1)(800
)
2(1)
y
x
O
38c. (h, k) (0, 3)
(x, y) (6, 0)
(xh)24p(yk)
(6 0)24p(0 3)
36 12p
3 p
(xh)24p(yk)
(x0)24(3)(y3)
x212(y3)
x212(y3), y 0
39. xy 12
y
1
x
2
xy1
x
1
x
2
1
x212 x
x2x12 0
(x4)(x3) 0
x4 0x3 0
x4x3
xy 12 xy 12
4y12 3y12
y3y4
(4, 3) (3, 4)
Check that (4, 3) and (3, 4) are also solutions
of y225 x2.
y225 x2y225 x2
(3)225 (4)2(4)225 (3)2
9 916 16
(4, 3), (3, 4)
40a. first station: (h, k(0, 0)
x2y2r2
x2y2502
x2y22500
second station: (h, k) (0, 30)
(xh)2(yk)2r2
x2(y30)2402
x2(y30)21600
third station: (h, k) (35, 18)
(xh)2(yk)2r2
(x35)2(y18)2132
(x35)2(y18)2169
40b. estimate: (40, 30)
40c. x2y22500
x22500 y2
x2(y30)21600
2500 y2y260y900 1600
60y1800 0
y30
(x35)2(y18)2169
x270x1225 (30 18)2169
x270x1200 0
(x30)(x40) 0
x30 0orx40 0
x30 x40
Check (30, 30) and (40, 30):
x2y22500 x2y22500
3023022500 4023022500
1800 2500 2500 2500
(40, 30)
41. x3yk
2y23yk
2y23yk0
y2
3
2
y
1
2
k0
y2
3
2
y
3
4
20Complete the square.
y
3
4
20
1
2
k
3
4
2
1
2
k
1
9
6
k
9
8
42a.
42b. yes; (6, 2) or (6, 2)
42c. Earth’s surface:
x12y1240
x
4
1
0
2
y
4
1
0
2
1
x
4
1
0
2
y
4
1
0
21
cos2tsin2t1
x
4
1
0
2cos2t
y
4
1
0
2sin2t
2
x1
10
cos t
2
y1
10
sin t
x1210
cos ty
1210
sin t
Chapter 10 364
(3, 4)
(4, 3)
y
x
O
4080 40 80
40
40
80
y
x
O
y
x
4
4
8
4
848
8
O
asteroid:
Let y2t.
x20.25y25
x20.25 t25
42d.
Tmin 8, Tmax 8, Tstep 0.13
[15.16, 15.16] scl1 by [10, 10] scl 1
43.
x
9
2
y21
(xsin 30° ycos 30°)21
(xcos 30° ysin 30°)2
9
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.
Graphing Calculator Exploration:
Shading Areas on a Graph
Page 686
1.
[9.1, 9.1] scl1 by [6, 6] scl1
2.
[9.1, 9.1] scl1 by [6, 6] scl1
3.
[9.1, 9.1] scl1 by [6, 6] scl1
4.
[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.
10-8B
365 Chapter 10
1
2
x
2
3
y
21
1
4
(x)2
2
3
xy
3
4
(y)21
3
4
(x)2
2
3
xy
1
4
(y)2
9
4
(x)2
9
2
3
xy
2
4
7
(y)29
3(x)243
xy7(y)290
44. x4t1y5t7
x
4
1
t
y
5
7
t
x
4
1
y
5
7
(x1)(5) (4)(y7)
5x5 4y28
5x4y33 0
45. 4 csc vcos vtan v4
si
1
nv
(cos v)
c
s
o
in
s
v
v
4
46. r10 cm or 0.10 m
vrq
v(0.10)
5
1
2p
v3.14 m/s
47.
between 1 and 2
48. y(x2)23
x(y2)23
x3 (y2)2
x3
y2
x3
2 y
3
4
(x)2
2
3
xy
1
4
(y)2
9
2
3
x
1
2
y
2
9
r1004
01004
11113
2124 4
y
x
y
y
(
x
2)2 3
x
3 2
O
5c. SHADE(
((36X2)/9),
((36X2)/9),6,2,3,4);
SHADE(X22,
((36X2)/9),2,2,3,4,);
SHADE(
((36X2)/9),
((36X2)/9),2,6,3,4)
[9.1, 9.1] scl1 by [6, 6] scl1
6. See students’ work.
Chapter 10 Study Guide and Assessment
Page 687 Understanding and Using the
Vocabulary
1. true 2. false; center
3. false; transverse 4. true
5. false; hyperbola 6. false; axis or axis of
symmetry
7. true 8. false; parabola
9. true 10. false; ellipse
Pages 688–690 Skills and Concepts
11. d(x2
x1)2
(y2
y1)2
d(3
1)2
[4
(6)]2
d20
or 25
x1
2
x2
,
y1
2
y2
1
2
(3)
,
6
2
(4)
(1, 5)
12. dx2x
1)2
(y2
y1)2
d(a3
a)2
(b
4
b)2
d25
or 5
x1
2
x2
,
y1
2
y2
aa
2
3
,
bb
2
4
(a1.5, b2)
13. AB (x2
x1)2
(y2
y1)2
[3 (
5)]2
[4
(2)]
2
100
or 10
DC (x2
x1)2
(y2
y1)2
(10
2)2
[3
(3)]2
100
or 10
BC (x2
x1)2
(y2
y1)2
(10
3)2
(3
4)2
50
or 52
AD (x2
x1)2
(y2
y1)2
[2 (
5)]2
[3
(2
)]2
50
or 52
Yes; AB DC 10 and BC AD 52
. Since
opposite sides of quadrilateral ABCD are
congruent, ABCD is a parallelogram.
14. (xh)2(yk)2r2
(x0)2(y0)2
33
2
x2y227
15. (xh)2(yk)2r2
(x2)2(y1)222
(x2)2(y1)24
16. x2y26y
x2(y26y?) 0 ?
x2(y26y9) 0 9
x2(y3)29
17. x214xy26y23
(x214x?) (y26y?) 23 ? ?
(x214x49) (y26y9) 23 49 9
(x7)2(y3)281
18. 3x23y26x12y60 0
x2y22x4y20 0
(x22x?) (y24y?) 20 ? ?
(x22x1) (y24y4) 20 1 4
(x1)2(y2)225
Chapter 10 366
(33, 0)
(0, 33)
y
x
O
(2, 1) (4, 1)
(2, 1)
y
x
O
(0, 3) (3, 3)
(0, 6)
y
x
O
(7, 6)
(16, 3)
(7, 3)
y
x
O
(1, 3)
(1, 2) (4, 2)
y
x
O
19. x2y2Dx Ey F0
1212D(1) E(1) F0⇒
DEF2
(2)222D(2) E(2) F0⇒
2D2EF 8
(5)212D(5) E(1) F0⇒
5DEF 26
DEF2
(1)(5DEF) (1)(26)
6D24
D4
2D2EF8
(1)(DEF) (1)(2)
3DE6
3(4) E6
E6
DEF2
4 (6) F2
F12
x2y2Dx Ey F0
x2y24x6y12 0
(x24x?) (y26y?) 12 ? ?
(x24x4) (y26y9) 12 4 9
(x2)2(y3)225
center: (h, k) (2 3)
r25
or 5
20. center: (h, k) (5, 2)
a236 b216
a36
or 6 b16
or 4
ca2b
2
c6 4
or 2
foci: (h, kc) (5, 2 2
)
major axis vertices: (h, ka) (5, 2 6) or
(5, 8), (5, 4)
minor axis vertices: (hb, k) (5 4, 2) or
(9, 2), (1, 2)
21. 4x225y224x50y39
4(x26x?) 25(y22y?) 39 ??
4(x26x9) 25(y22y1) 39 4(9) 25(1)
4(x3)225(y1)2100
(x
25
3)2
(y
4
1)2
1
center: (h, k) (3, 1)
a225 b24
a25
or 5 b4
or 2
ca2b
2
c25
4
or 21
foci: (hc, k) (3 21
, 1)
major axis vertices: (ha, k) (3 5, 1) or
(8, 1), (2, 1)
minor axis vertices: (h, kb) (3, 1 2) or
(3, 1), (3, 3)
22. 6x24y224x32y64 0
6(x24x?) 4(y28y?) 64 ??
6(x24x4) 4(y28y16) 64 6(4) 4(16)
6(x2)24(y4)224
(x
4
2)2
(y
6
4)2
24
center: (h, k) (2, 4)
a26b24
a6
b4
or 2
ca2b
2
c6 4
or 2
foci: (h, kc) (2, 4 2
)
major axis vertices: (h, ka)
2, 4 6
minor axis vertices: (hb, k) (2 2, 4) or
(0, 4), (4, 4)
23. x24y21248x48y
(x28x?)4(y212y?)124??
(x28x16)4(y212y36)12416 4(36)
(x4)24(y6)236
(x
36
4)2
(y
9
6)2
36
center: (h, k) (4, 6)
a236 b29
a36
or 6 b9
or 3
ca2b
2
c36
9 or 33
foci: (hc, k) (4 33
, 6)
major axis vertices: (ha, k) (4 6, 6) or
(10, 6), (2, 6)
minor axis vertices: (h, kb) (4, 6 3) or
(4, 9), (4, 3)
367 Chapter 10
(5, 8)
(5, 2) (9, 2)
(1, 2)
(5, 4)
y
x
O
y
x
(2, 1)
(8, 1)
(3, 1)
(3, 3)
(3, 1)
O
y
x
(2, 4 6)
(2, 4 6)
(4, 4)
(2, 4)
(0, 4)
O
y
x
(2, 6)
(4, 3)
(4, 6)
(10, 6)
(4, 9)
O
24. (h, k) (4, 1)
a9
b6
(y
a2
k)2
(x
b2
h)2
1
(y
92
1)2
[x
6
(
2
4)]2
1
(y
81
1)2
(x
36
4)2
1
25. center: (h, k) (0, 0)
a225 b216
a25
or 5 b16
or 4
ca2b
2
c25
16
or 41
transverse axis: horizontal
foci: (hc, k) (0 41
, 0) or (41
, 0)
vertices: (ha, k) (0 5, 0) or (5, 0), (5, 0)
asymptotes: yk
a
b
(xh)
y0
4
5
(x0)
y
4
5
x
26. center: (h, k) (1, 5)
a236 b29
a36
or 6 b9
or 3
ca2b
2
c36
9
or 35
transverse axis: vertical
foci: (h, kc)
1, 5 35
vertices: (h, ka) (1, 5 6) or (1, 1), (1, 11)
asymptotes: yk
a
b
(xh)
y(5)
6
3
(x1)
y5 2(x1)
27. x24y216y20
x24(y24y?) 20 ?
x24(y24y4) 20 4(4)
x24(y2)24
x
4
2
(y
1
2)2
1
center: (h, k) (0, 2)
a24b21
a4
or 2 b1
or 1
ca2b
2
c4 1
or 5
transverse axis: horizontal
foci: (hc, k) (0 5
, 2) or (5
, 2)
vertices: (ha, k) (0 2, 2) or (2, 2),
(2, 2)
asymptotes: yk
a
b
(xh)
y(2)
1
2
(x0)
y2
1
2
x
28. 9x216y236x96y36 0
9(x24x?) 16(y26y?) 36 ??
9(x24x4) 16(y26y9) 36 9(4) 16(9)
9(x2)216(y3)2144
(y
9
3)2
(x
16
2)2
1
center: (h, k) (2, 3)
a29b216
a9
or 3 b16
or 4
ca2b
2
c9 1
6
or 5
transverse axis: vertical
foci: (h, kc) (2, 3 5) or (2, 2), (2, 8)
vertices: (h, ka) (2, 3 3) or (2, 0), (2, 6)
asymptotes: yk
a
b
(xh)
y(3)
3
4
(x2)
y3
3
4
(x2)
Chapter 10 368
y
x
(5, 0) (5, 0)
O
y
x
(1, 11)
(1, 5)
(1, 1)
O
y
x
(2, 2)
(0, 2)
(2, 2)
O
(2, 0)
(2, 3)
(2, 6)
y
x
O
29. c9
quadrants: I and III
transverse axis: yx
vertices: xy 9xy 9
3(3) 9(3)(3) 9
(3, 3) (3, 3)
30. 2b10
b5
center:
x1
2
x2
,
y1
2
y2
1
2
1
,
1
2
5
(1, 2)
transverse axis: vertical
adistance from center to a vertex
2 (1)or 3
(y
a2
k)2
(x
b2
h)2
1
(y
32
2)2
(x
52
1)2
1
(y
9
2)2
(x
25
1)2
1
31. center:
x1
2
x2
,
y1
2
y2
2
2
6
,
3
2
(3)
(2, 3)
adistance from center to a vertex
2 (2)or 4
cdistance from center to a focus
2 (4)or 6
b2c2a2
b26242
b220
transverse axis: horizontal
(x
a2
h)2
(y
b2
k)2
1
(x
42
2)2
[y
2
(
0
3)]2
1
(x
16
2)2
(y
20
3)2
1
32. vertex: (h, k) (5, 3)
4p8
p2
focus (h, kp) (5, 3 2) or (5, 5)
directrix: ykp
y3 2
y1
axis of symmetry: xh
x5
33. vertex: (h, k) (1, 2)
4p16
p4
focus: (hp, k) (1 (4), 2) or (3, 2)
directrix: xhp
x1 (4)
x5
axis of symmetry: yk
y2
34. y26y4x25
y26y? 4x25 ?
y26y9 4x25 9
(y3)24(x4)
vertex: (h, k) (4, 3)
4p4
p1
focus: (hp, k) (4 1, 3) or (5, 3)
directrix: xhp
x4 1
x3
axis of symmetry: yk
y3
35. x24xy8
x24x4 y8 4
(x2)2y4
vertex: (h, k) (2, 4)
4p1
p
1
4
focus: (h, kp)
2, 4
1
4
or (2, 4.25)
directrix: ykp
y4
1
4
y3.75
axis of symmetry: xh
x2
369 Chapter 10
(3, 3)
(3, 3)
y
x
O
y
x
y
1
(5, 3)
(5, 5)
O
y
x
x
5
(1, 2)
(3, 2)
O
y
x
(4, 3) (5, 3)
O
x
3
y
x
(2, 4)
(2, 4.25)
O
y
3.75
36. vertex: (h, k) (1, 3)
(yk)24p(xh)
(7 3)24p[3 (1)]
16 8p
2 p
Since parabola opens left, p2.
(yk)24p(xh)
(y3)24(2)[x(1)]
(y3)28(x1)
37. vertex: (h, k)
5,
2
2
4
or (5, 1)
focus: (h, kp) (5, 2)
kp2
1 p2
p3
(xh)24p(yk)
(x5)24(3)[y(1)]
(x5)212(y1)
38. A5, c2; ellipse
39. AC0; equilateral hyperbola
40. AC5; circle
41. C0; parabola
42. yt23
yx23
43. xcos 4tysin 4t
cos24tsin24t1
x2y21
44. x2 sin ty3 cos t
2
x
sin t
3
y
cos t
sin2tcos2t1
2
x
2
3
y
21
x
4
2
y
9
2
1
45. xt
x2t
y
2
t
1
y
x
2
2
1
46. Sample answer:
Let xt.
y2x24
y2t24, t
47. Sample answer:
x2y249
4
x
9
2
4
y
9
2
1
7
x
2
7
y
21
sin2tcos2t1
7
x
2sin2t
7
y
2cos2t
7
x
sin t
7
y
cos t
x7 sin ty7 cos t, 0 t2p
Chapter 10 370
txy (x, y)
221(2, 1)
112(1, 2)
003(0, 3)
112(1, 2)
221(2, 1)
y
x
O
txy(x, y)
010(1, 0)
8
01(0, 1)
4
10(1, 0)
3
8
01(0, 1)
t
0
t
4
t
8
t
8
3
y
x
O
txy(x, y)
003(0, 3)
2
20(2, 0)
03(0, 3)
3
2
20(2, 0)
t
2
t
t
0
t
3
2
y
x
O
txy(x, y)
001(0, 1)
421(2, 1)
933.5(3, 3.5)
t
0
t
9
y
x
O
48. Sample answer:
3
x
6
2
8
y
1
2
1
6
x
2
9
y
21
cos2tsin2t1
6
x
2cos2t
9
y
2sin2t
6
x
cos t
9
y
sin t
x6 cos ty9 sin t, 0 t2p
49. Sample answer:
Let yt.
xy2
xt2, t
50. B24AC 0 4(4)(9)
144
AC; ellipse
4x29y236
4
xcos
p
6
ysin
p
6
29
xsin
p
6
ycos
p
6
236
4
2
3
x
1
2
y
29
1
2
x
2
3
y
236
4
3
4
(x)2
2
3
xy
1
4
(y)2
9
1
4
(x)3
2
3
xy
3
4
(y)2
36
3(x)223
xy(y)2
9
4
(x)2
9
2
3
xy
2
4
7
(y)236
2
4
1
(x)2
5
2
3
xy
3
4
1
(y)236
21(x)2103
xy31(y)2144 0
51. B24AC 0 4(0)(1)
0
parabola
y24x0
(xsin 45° ycos 45°)24(xcos 45° ysin 45°) 0
2
2
x
2
2
y
24
2
2
x
2
2
y
0
1
2
(x)2xy
1
2
(y)222
x22
y0
(x)22xy(y)242
x42
y0
52. B24AC 0 4(4)(16)
256
hyperbola
4x216(y1)264
4(xh)216(yk1)264
4(x1)216(y2 1)264
4(x1)216(y1)264
4(x22x1) 16(y22y1)64
4x28x4 16y232y16 64 0
x24y22x8y19 0
53. B24AC
23
24(6)(8)
180
AC; ellipse
tan 2v
A
B
C
tan 2v
6
2
3
8
tan 2v3
2v60°
v30°
54. B24AC (6)24(1)(9)
0
parabola
tan 2v
A
B
C
tan 2v
1
6
9
tan 2v
3
4
2v36.86989765°
v18°
55. (x1)24(y1)220
(y1)24(y1)220
5(y1)220
(y1)24
y1 2
y3 or 1
y3xy
x3
y1xy
x1
(3, 3), (1, 1)
56. 2xy0
2xy
y249 x2
(2x)249 x2
4x249 x2
3x249
x4.04
2xy02xy0
2(4.04) y02(4.04) y0
y8.08 y8.08
(4.0, 8.1), (4.0, 8.1)
371 Chapter 10
y
x
(1, 1)
(3, 3)
O
y
x
(4.0, 8.1)
(4.0, 8.1)
O
57. x24x4y4
x24x4 4y
(x2)24y0
(x2)2x24x4 0
x24x4 x24x4 0
2x28x0
2x(x4) 0
2x0x4 0
x0x4
(x2)24y0(x2)24y0
(0 2)24y0(4 2)24y0
y1y1
(0, 1), (4, 1)
58. xy 4
y
4
x
x2y212
x2
4
x
212
x2
1
x
6
2
12
x416 12x20
(x2)212(x2) 16 0
x2bb24
ac
2a
61. 62.
Page 691 Applications and Problem Solving
63a. r(x2
x1)2
(y2
y1)2
r(12
0)2
(16
0)2
r20
x2y2r2
x2y2202
x2y2400
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
percent not watered
7
2
4
0
3
0
.
0
4
0.37
about 37%
64. 2a2,000 e
a
c
a6000 0.2
60
c
00
1200 c
b2a2c2
b26000212002
b234,560,000
(x
a2
h)2
(y
b2
k)2
1
(x
60
00
0
2
)2
34
(y
,5
60
0
,0
)2
00
1
36,00
x
0
2
,000
34,56
y
0
2
,000
1
65. a3.5
b3
ca2b
2
c3.52
32
c1.8
about 1.8 feet from the center
Chapter 10 372
y
x
(0, 1)
(4, 1)
O
x2
x210.472 or x21.528
x3.236 x1.236
xy 4xy 4
3.236y41.236y4
y1.236 y3.236
xy 4xy 4
3.236y41.236y4
y1.236 y3.236
(3.2, 1.2), (3.2, 1.2), (1.2, 3.2), (1.2, 3.2)
59. 60.
12 122
4(1)(1
6)
2(1)
y
x
(1.2, 3.2)
(3.2, 1.2)
(1.2, 3.2)
(3.2, 1.2)
O
y
x
O
y
x
O
4
4
4
4
y
x
O
2
2
y
x
O
Page 691 Open-Ended Assessment
1. Sample answer:
e
a
c
1
9
a
c
Let a9.
1
9
9
c
c1
a
x2
2
b
y2
2
1
9
x2
2
8
y
0
2
1
8
x
1
2
8
y
0
2
1
2. Sample answer:
axis of symmetry: xh
x2, so h2
focus: (h, kp) (2, 5)
kp5
Let k2, p3.
(xh)24p(yk)
(x2)24(3)(y2)
(x2)212(y2)
SAT & ACT Preparation
Page 693 SAT & ACT Practice
1. Add the two numbers of parts to get the whole, 8.
The fraction of red jelly beans to the whole is
3
8
.
The total number of jelly beans is 160. The
number of red jelly beans is
3
8
(160) or 60. The
correct choice is C.
Or you can use a ratio box. Multiply by 20.
2. Notice the capitalized word EXCEPT. You might
want to try the plug-in method on this problem.
Choose a value for bthat is an odd integer, say 1.
Then substitute that value for bin the equation.
a2b122
a2(1) 122
a2122
a12
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.
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.
A
1
2
bh
1
2
(6)(7) or 21
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
fraction of girls in the whole class is
4
7
. Find the
number of girls in the class by multiplying this
fraction by 35.
4
7
(35) 20 There are 20 girls in the class.
Using the same process, the fraction of boys is
3
7
.
3
7
(35) 15 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.
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.
373 Chapter 10
Green Red Whole
53 8
60 160
Girls Boys Total
437
555
20 15 35
b2a2c2
b29212
b280
6. For a quadratic equation in the form
ya(x – h)2k, 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
of y
1
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
8
2
8
2
2
2
8
2
2
or 42
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 xas the unknown. Multiply the cross-
products. Solve for x.
10
x
0
1
7
x
1
5
(100)
1
7
x20
x140
The correct choice is E.
1
7
1
5
9. Let drepresent the number of dimes in the jar.
Since there are 4 more nickels than dimes, there
are d4 nickels in the jar. So, the ratio of dimes
to nickles in the jar is
d
d4
. This ratio is less
than 1. The only answer choice that is less than
1 is choice A,
1
8
0
. If
d
d4
1
8
0
, then d16. So,
there are 16 dimes and 16 4 or 20 nickels in the
jar, and
1
26
0
1
8
0
.
The correct choice is A.
10. Set up a proportion.
t
t
o
o
t
t
a
a
l
l
b
l
o
it
t
e
tl
r
e
s
s
1
x
b
li
o
t
t
e
t
r
l
s
e
2
8
0
1
x
20x8
x0.4 or
2
5
The correct answer is .4 or 2/5.
Chapter 10 374
Real Exponents
Page 695 Graphing Calculator Exploration
1.
2.
3. amanamn
4. (am)namn
5.
a
b
m
a
bm
m
, when b0
Page 700 Check for Understanding
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.
4. 54
5
1
4
6
1
25
11-1 8. 32
3
5
(25)
3
5
2
1
5
5
23
8
9. (3a2)33a533a63a5
34a1
81a1or
8
a
1
10. m3n2
m4n5
(m3n2)
1
2
(m4n5)
1
2
m
3
2
n
2
2
m
4
2
n
5
2
m
7
2
n
7
2
or m3n3mn
11.
8n
4
n
2
7
8n
4
n
27
1
2
8
n
2
2
7
2
4
n
2
375 Chapter 11
Chapter 11 Exponential and Logarithmic Function
(23)
n
2
2
7
2
(22)
n
2
2
3
2
n
2
7
2
2
2
2
n
2
3
2
n
2
7
2
2
2
2
n
2
5n
2
7
22n32
n
2
1
22n32n1
12. (2x4y8)
1
2
2
1
2
(x4)
1
2
(y8)
1
2
2
1
2
x
4
2
y
8
2
2
1
2
x2y4or x2242
13. 169x5
(169x5)
1
2
169
1
2
(x5)
1
2
13x
5
2
14.
4a2b3c4
d5
(a2b3c4d5)
1
4
(a2)
1
4
(b3)
1
4
(c4)
1
4
(d5)
1
4
a
1
2
b
3
4
cd
5
4
15. 6
1
4
b
3
4
c
1
4
(6b3c)
1
4
46b3c
5.
1
9
6
2
1
9
6
2
1
9
6
2
2
2
8
5
1
6
1
1
9
6
2
6. 216
1
3
(63)
1
3
6
3
3
6
7. 27
3
27
1
2
3
1
2
(33)
1
2
3
1
2
3
3
2
3
1
2
3
4
2
32
9
16. 15x
1
3
y
1
5
15x
1
5
5
y
1
3
5
15 (x5y3)
1
1
5
15
15 x5y3
17.
3p4q6r5
(p4q6r5)
1
3
(p4)
1
3
(q6)
1
3
(r5)
1
3
p
4
3
q2r
5
3
pq2r
3pr2
18. y
4
5
34
y
4
5
5
4
34
5
4
y(345)
1
4
y82.1
19. Ar2r3.875 107m
A(3.875 107 m)2
(3.875)2(107)2m2
(15.015625 1014) m2
4.717 1013 m2
Pages 700–703 Exercises
33.
3216
2(216
1
3
)2
216
2
3
(63)
2
3
62
36
34. 81
1
2
81
1
2
(92)
1
2
9
1
9
8
8
9
35.
7(
1
128
)4
1
(128)
4
7
1
(92)
1
2
Chapter 11 376
24.
7
8
31
7
8
3
8
7
3
8
7
3
3
5
3
1
4
2
3
25. (3133)1
31
1
31
1
1
3
1
9
1
4
9
3
3
3
2
3
29. 2
1
2
12
1
2
2
1
2
(2 6)
1
2
2
1
2
2
1
2
6
1
2
2 6
1
2
26
30. 64
1
2
(26)
1
1
2
2
1
2
or 2
31. 16
1
4
1
16
1
4
32.
(
37)
2
(9
7
4
6
)
37(32)4
(33)
6
2
1
[(2)7]
4
7
(
1
2)4
1
1
6
36. (3n2)333(n2)3
27n6
37. (y2)4y8
(y
1
2)4
y8
y
1
8
y8
1
38. (4y4)
3
2
4
3
2
(y4)
3
2
(22)
3
2
(y4)
3
2
2
6
2
y
1
2
2
8y6
39. (27p3q6r1)
1
3
27
1
3
(p3)
1
3
(q6)
1
3
1
r
1
3
(33)
1
3
(p3)
1
3
(q6)
1
3
1
r
1
3
3
3
3
p
3
3
q
6
3
1
r
1
3
3pq2r
1
3
40. [(2x4]2
[(2x
1
)4]2
28
1
x8
28x8or
25
1
6x8
43. 2nn
1
2
4
2n
4n
1
2
20. (6)4
(
1
6)4
12
1
96
21. 64
6
1
4
12
1
96
22. (5 3)2152
225
23.
2
2
4
1
24(1)
25
32
9
4
26. 81
1
2
(92)
1
2
9
27. 729
1
3
(93)
1
3
9
28. 33
(33)
2
3
27
27
2
3
1
(24)
1
4
1
2
3
3
1
9
5
36
729
41. (36x6)
1
2
36
1
2
(x6)
1
2
6x3
42.
b
b
2
2
n
n
1
2
(b2nb2n)
1
2
(b4n)
1
2
b2n
2n
1
2
4
n
2
44.
3m
1
2
27n
1
4
434
m
1
2
4274
n1
4
4
34m2(33)4n
316m2n
45.
256
f
g4
1
h
6
4
1
4
(f16 2561g4h4)
1
4
(f16)
1
4
(2561)
1
4
(g4)
1
4
(h4)
1
4
f4256
1
4
gh1
4 f4gh1or
4
f4
h
g
46.
6x2
x
3
4
x
3
4
x2
x
3
4
x
3
4
1
6
x2x
3
4
x
6
4
1
1
6
x
5
4
x
3
2
1
1
6
47.
2x
1
4
y
1
3
3x
1
4
y
2
3
6x
2
4
y
3
3
6x
1
2
y
61. x
3(245
)
1
5
30.33
0.69
62. d3e2f2(d3)
1
2
(e2)
1
2
(f2)
1
2
defd
63.
3a5b7c
(a5)
1
3
(b7)
1
3
(c)
1
3
ab2
3a2bc
64. 20x3y
6
(20)
1
2
(x3)
1
2
(y6)
1
2
2xy35x
377 Chapter 11
x3xy
83
8
65
9
61
63
6
7
1
29
53
5
2
1
43
1
3
0
3
3
1
3
0
3
1.395
1
2
3
1
2
1.732
2
3
3
2
3
2.080
1
9
0
3
1
9
0
3.389
5
3
3
5
3
14.620
7
2
3
7
2
46.765
50. xy3
(x)
1
2
(y3)
1
2
x
1
2
y
3
2
51.
38x3y6
8
1
3
(x3)
1
3
(y6)
1
3
2xy2
52. 17
7x14y7z
12
17(x14)
1
2
(y7)
1
7
(z12)
1
7
17x2yz
1
7
2
53.
5a10b2
4c2
(a10)
1
5
(b2)
1
5
(c2)
1
4
a2b
2
5
c
1
2
54. 60
8r80s56
t27
60(r80)
1
8
(s56)
1
8
(t27)
1
8
60r10st
2
8
7
48.
m
n
a
a
n
1
m
1
a
n
1
m
1
a
n
1
m
a
m
1
n
mn
a
49. m6n
(m6)
1
2
n
1
2
m3n
1
2
55. 16
1
5
516
56. (7a)
5
8
b
3
8
875a5b
3
57. p
2
3
q
1
2
r
1
3
p
4
6
q
3
6
r
2
6
6p4q3r2
58. 2
1
3
2
2
3
2
1
3
32
59. 13a
1
7
b
1
3
13a
2
3
1
b
2
7
1
13
21 a3b7
60. (n3m9)
1
2
(n3)
1
2
(m9)
1
2
n
3
2
m
9
2
nm4mn
65. 14.2 x
3
2
(14.2)
2
3
(x
3
2
)
2
3
0.17 x
66. 724 15a
5
2
12
712 15a
5
2
7
1
1
5
2
a
5
2
7
1
1
5
2
2
5
(a
5
2
)
2
5
4.68 a
67.
1
8
x5
3.5
x
5
2
28
(x
5
2
)
2
5
(28)
2
5
x3.79
68. d6.794 103km so r3.397 103km
V
4
3
r3
4
3
(3.397 103km)3
1.64 1011 km3
69. y3x; x8, 6, 5,
1
3
0
3
,
1
2
,
2
3
,
1
9
0
,
5
3
,
7
2
69a. If x0then y0. If x0then y1. Since
x0, y0 and y1. So, 0 y1.
69b. If x0 then y1. If x1 then y3. So,
1 y3.
69c. If x1 then y3. So, y3.
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.
70. r(1.2 1015)A
1
3
If r2.75 1015 then
2.75 105(1.2 1015)A
1
3
2
1
.
.
7
2
5
1
1
0
0
1
1
5
5
A
1
3
2.29 A
1
3
12.04 A
Since 12.04 12, which is the mass number of
carbon, the atom is carbon.
71. 32(x24x)16(x24x3)
(25)(x24x)(24)(x24x3)
2(5x220x)2(4x216x12)
5x220x4x216x12
x24x12 0
(x6)(x2) 0
x6 0x2 0
x6x2
72a.
72b. A5-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
G
4
M
e
2
t2
G6.67 1011
Mt5.98 1024
t1 day 86,4000 seconds
r3
r42,250,474.31 m
73b. 42,250,474.31 m 42,250.47431 km
42.250.47431 6380 35870.47431
35,870 km
mfactors nfactors mnfactors
74a. amanaa... aaa... aaa... aamn
(6.67 1011)(5.98 1024)(86,400)2
42
nfactors
mfactors mfactors mfactors
74b. (am)naa... aaa... a... aa... a
mnfactors
aa... aamn
mfactors mfactors mfactors
74c. (ab)mab ab ... ab aa... abb... bambm
mfactors
74d.
a
b
m
a
b
a
b
...
a
b
a
bm
m
mfactors
74e.
a
a
m
n
a
n
a
fa
ct
..
o
.
r
s
a
amn
aa... a
75.
76. y212x
(y0)24(3)(x0)
Vertex is at (0, 0) and p3. 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 x3.
77. (23
2i)
1
5
Convert to polar form r(cos visin v).
r
23
22
2
vArctan
2
2
3
16
Arctan
3
3
4
6
So, (23
2i) 4
cos
6
isin
6
.
Use De Moivre’s Theorem.
4
cos
6
isin
6
1
5
4
1
5
cos
1
5
6
sin
1
5
6
4
1
5
cos
3
0
4
1
5
isin
3
0
1.31 0.14i
78.
Lemniscate
Chapter 11 378
Wind Speed Wind Chill
50.8
10 12.7
15 22.6
20 29.9
25 35.3
30 39.2
y
x
O
y
x
O
246
4
2
2
4
6
8
8
6
0
6
11
2
6
3
3
3
4
6
5
3
2
6
7
3
2
5
1234
79. Use the equation ytv
u
sin v
1
2
qt2h.
v
u
105 g32 h3v42
yt(105)(sin 42°)
1
2
(32)t23
16t2(105 sin 42°)t3
Find twhen y0 (i.e., the ball is on the ground).
t
t0.04, 4.43
So, the ball hits the ground after about 4.43 s.
80. TC
u
(2 3), (6 (4)), (5 6)
1, 10, 11
TC
u
(2 3
)2(6
(
4))2
(5
6)2
222
81. Sample answer:
tan Scos S
1
2
c
s
o
in
s
S
S
co
1
sS
1
2
sin S
1
2
82. cot v0
v
2
Period of cot vis so
cot v0
v
2
n, where nis an integer.
83. r6h150m
r
15
6
0
h
m
r25m/h
84. 90°, 270°
85. x325x0
Degree of 3 so there are 3 complex roots.
x325x0
x(x5)(x5) 0
x0x5 0x5 0
x5x5
3; 5, 0, 5
86.
[5, 5] scl:0.5 by [5, 5] scl:0.5
abs. min; (0.75, 1.88)
105 sin 42° (105 s
in 42°
) 4(
16)(
3)
2(16)
87. The time it takes to paint a building is inversely
proportional to the number of painters.
48
k
8
k384
So t
3
1
8
6
4
t24
The correct choice is E.
Exponential Functions
Page 705 Graphing Calculator Exploration
1. positive reals
2. (0, 1)
3. For a0.5 and 0.75, y→as x→, y→0 as
x→. For a2 and 5, y→0 as x→, y→as
x→.
4. horizontal asymptote at y0, 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 ybxis decreasing when 0 b1
because multiplying by number between 0 and 1
results in a product less than the original number.
The graph of ybxis increasing when b1
because multiplying by a number greater than 1
results in a product greater than the original
number.
8. There is a horizontal asymptote at y0 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.
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 y0, a y-intercept of (0, 1), and no
vertical asymptote. The graph of ybxis
decreasing when 0 b1 and increasing when
b1.
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 y4xand y4x3 are the same
except the graph of y4x3 is shifted down
three units from the graph of y4x.
11-2
379 Chapter 11
5. xy
1
1
9
01
13
29
6. xy
29
13
01
1
1
9
7. xy
23
3
4
Use (0, 0) as a test point.
13
1
2
0?
204
040
?
1 4
1203
20 The statement is true so shade
the region containing (0, 0).
8. Use NN0(1 r)twhere N03750, r0.25,
and t2.
N3750 (1 0.25)2
3750 (0.75)2
2109.38
9a. 9,145,219 8,863,052 282,167
282
7
,167
40,309.57
4
8
0
,8
,3
6
0
3
9
,0
.5
5
7
2
0.0045
0.45%
9b. Use NN0(1 r)t.
N8,863,052(1 0.0045)20
N9,695,766
Pages 708–711 Exercises
10. xy
1
1
2
01
12
24
11. xy
1
1
2
01
12
24
12. xy
24
12
01
1
1
2
13. xy
31
22
14
08
14. xy
31
22
14
08
15. xy(0, 0)
11
3
4
0 ?
422
010
?
1 2
120 1; no
214
16. xy
15
01
1
1
5
17. xy
24
12
01
1
1
2
Chapter 11 380
y
x
y
3
x
O
y
x
y
3
x
O
y
x
y
2
x
4
O
y
x
y
2
x
O
y
x
y
2
x
O
y
x
y
2
x
O
y
x
y
2
x
3
O
yx
y
2
x
3
O
y
x
y
4
x
2
O
y
x
y
x
()
1
5
O
y
x
y
x
()
1
2
O
(0, 0)
0 ?
1
2
0
0 ?
1
18. xy(0, 0)
3
1
2
0 ?
204
410
?
24
520
1
1
6
64
19. xy
1 100
01
1
1
1
00
B
20. xy
15
01
1
1
5
C
21. xy
149
07
11
A
22.
[5, 5] scl:1 by [5, 5] scl:1
22a. The graph of y5xis a reflection of y5x
across the x-axis. The graph of y5xis a
reflection of y5xacross the y-axis.
22b. The graph of y5x2 is shifted up two units,
while the graph of y5x2 is shifted down
two units.
22c. The graph of y10xincreases more quickly
than the graph of y5x. The graphs are not the
same because 52x10x.
23a. The graph of y6x4 is shifted up four units
from the graph of y6x.
[10, 10] scl:1 by [1, 9] scl:1
23b. The graph of y3xis a reflection of the graph
of y3xacross the x-axis.
[10, 10] scl:1 by [10, 10] scl:1
23c. The graph of y7xis a reflection of the graph
of y7xacross the y-axis.
[10, 10] scl:1 by [1, 9] scl:1
23d. The graph of y
1
2
xis a reflection of the graph
of y2xacross the y-axis.
[10, 10] scl:1 by [1, 9] scl:1
24a.
24b. y9.25(1.06)50
y170.386427 thousand
y$170,400
25a. y(0.85)x
381 Chapter 11
O
y
x
(0, 1)
()
1, 1
100
y
0.01
x
y
x
y
2
x
4
O
O
y
x
(1, 5)
(0, 1)
()
1, 1
5
y
5
x
O
y
x
(0, 7)
(1, 1)
y
71
x
2
24681012 141618 2022
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
y
x
O
25b.
25c. y(0.85)12
0.14 or 14%
25d. No; the graph has an asymptote at y0, so the
percent of impurities, y, will never reach 0.
26a. N876,156(1 0.0074)15
978,612.2261 or 978,612
26b. N2,465,326(1 0.0053)15
2,668,760.458 or 2,668,760
26c. 152,307 139,510(1 r)10
(1 r)10
1r
N139,510(1 r)25
N 173,736.7334 or 173,737
26d. 191,701 168,767(1 r)10
(1 r)10
1r
N168,767(1 r)25
N232,075.6889 or 232,076
27a. O100(3
3
5
5
)
100(33)
2700 units
27b. s
4
1
.2
h
m
r
i
52
1
8
m
0
i
ft
3
1
60
h
0
r
s
6.16 ft/s
O100(3
36
5
.16
)
5800.16
5800 units
28a. Pn121,000, n30 12 or 360,
i0.075 12 or 0.00625
121,000 P
P
846.04955; $846.05
28b. Pn121,000, n20 12 or 240,
i0.0725 12 or 0.00604
121,000 P
P956.35494; $956.35
28c. 30 year: I360(846.05) 121,000
$183.578
20 year: I240(956.35) 121,000
$108,524
1
1
0.0
1
7
2
25
240
———
0.0
1
7
2
25
1 (1 0.00625)360
0.00625
191,701
168,767
191,701
168,767
152,307
139,510
152,307
139,510
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. P4000, n43, i0.0475
Fn4000
535,215.918; $535,215.92
29b. P4000, n43, i0.0525
Fn4000
611,592.1194 or $611,592.12
$611,592.12 535,215.92 $76,376.20
30. The function yaxis undefined when a0 and
the exponent xis a fraction with an even
denominator.
31a. Compounded once:
I1000[(1 0.05)11]
50; $50
Compounded twice:
I1000[
1
0.
2
05
21]
50.625; $50.63
Compounded four times:
I1000[
1
0.
4
05
41]
50.9453; $50.94
Compounded twelve times:
I1000
1
0
1
.0
2
5
12 1
51.1619; $51.16
Compounded 365 times:
I1000
1
0
3
.
6
0
5
5
365 1
51.2675; $51.26
31b. Let xrepresent the investment.
Statement savings: Ix[(1 0.051)11]
0.051x
The return is 5.1%
Money Market Savings: Ix
1
0.0
1
5
2
05
12 1
0.517x
The return is 5.17%
Super Saver: Ix
1
0
3
.
6
0
5
5
365 1
0.513x
The return is 5.13%
Money Market Savings
31c. (1 0.05)
1
36
x
5
365
(1.05)
3
1
65
1
36
x
5
365[(1.05)
3
1
65
1] x
0.04879 x; 4.88%
32. 4x2(4x)24x2(4)2(x)2
1
4
6
x
x
2
2
1
4
33. y15
Use yrsin v.
y15
So, 15 rsin v.
34. 3, 92, 1(3)(2) (9)(1)
3
3; no because the inner product does not equal 0.
(1 0.0525)43 1
0.0525
(1 0.0475)43 1
0.0475
Chapter 11 382
0.2
0.1
12345678
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
y
x
O
1
1
0
1
1
0
35.
1
3
cos
7
8
isin
7
8
33
cos
4
isin
4
1
3
33
cos
7
8
4
isin
7
8
4
3
cos
5
8
isin
5
8
0.66 1.60i
36. s72t16t24
s4 16t272t
s4 (16)(5.0625) 16(t24.5 5.0625)
(s85) 16(t2.25)2
Vertex: (2.25, 85)
Maximum height: 85 feet.
37a.
37b.
(12
2
(7))
,
(8
2
6)
(2.5, 7)
38. sin4Acos2Acos4Asin2A
(sin2A)2cos2Acos4Asin2A
(1 cos2A)2cos2Acos4Asin2A
(1 2cos2Acos4A) cos2Acos4Asin2A
cos4A1 cos2Acos4Asin2A
cos4Asin2Acos4Asin2A
39.
2900
1
rev
1
2
r
ev
4800
V9.2
480
1
0
138,732.73 or about 139,000 cm/s
40.
tan 69°
20
x
0
200 tan 69° x
521.02 x
about 521 feet
41.
Sample answer: y948.4x4960.6
42.
The parent graph is translated 3 units left. The
vertical asymptote is now x3. The horizontal
asymptote, y0, is unchanged.
43. CAC 32CAB 16
100.53 50.27
100.53 50.27 50.26 or about 50.
The correct choice is E.
The Number e
Page 714 Check for Understanding
1. C
2. If kis positive, the equation models growth.
If kis 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. y33,430e0.0397(60)
361,931.0414 or 361,931
7. A12,000e0.064(12)
25,865.412 or $25,865.41
Pages 714–717 Exercises
8. p(100 18)e06(2) 18
42.6 or 43%
9a. y84e0.23(15) 76
78.66 or 78.7°F
9b. too cold; After 5 minutes, his coffee will be about
90°F.
10a.
[4, 4] scl:1 by [1, 10] scl:1
10b. symmetric about y-axis
11-3
383 Chapter 11
y
x
(7, 6)
(12, 8)
O
200 ft
21˚
x
O
y
x
1
x
y
1
x
3
y
11a. Annually: I100[(1 0.08)11]
80 $80.00; 8%
Semi-annually: I1000
1
0.
2
08
21
81.6 $81.60; 8.16%
Quarterly: I1000
1
0.
4
08
41
82.4316 $82.43; 8.243%
Monthly: I1000
1
0
1
.0
2
8
12 1
82.9995 $83.00; 8.3%
Daily: I1000
1
0
3
.
6
0
5
8
365 1
83.2776 $83.28; 8.328%
Continuously: I1000(e0.08(1) 1)
83.2871 $83.98; 8.329%
11b. continuously 11c. E
1
n
r
n1
11d. Eer1
12a. y525(1 e0.038(24))
314.097 314 people
12b. after about 61h
[0, 100] scl:10 by [0, 550] scl:50
13a. P1 e6(0.5)
0.95021 95%
13b.
x0.02; about 0.02 h
0.0
1
2h
60
1
m
h
in
1.2
[0, 1] scl:0.1 by [0, 1] scl:0.1 about 1.2 min
14a. For x10:
2
2
(
(
1
1
0
0
)
)
1
1
10
2
1
1
9
10
2.720551414
For x100:
2
2
(
(
1
1
0
0
0
0
)
)
1
1
100
2
1
0
9
1
9
100
2.718304481
For x1000:
2
2
(
(
1
1
0
0
0
0
0
0
)
)
1
1
1000
2
1
0
9
0
9
1
9
1000
2.718282055
2.720551414; 2.718304481; 2.718282055
14b. 2 decimal places; 4 decimal places; 6 decimal
places
14c. always greater
15a. 5 days: P1 e0.047(5)
0.20943 20.9%
20 days: P1 e0.047(26)
0.60937 60.9%
90 days: P1 e0.047(90)
0.98545 98.5%
20.9%; 60.9% 98.5%
15b.
about 29 days
15c. Sample answer: The probability that a person
who is going to respond has responded
approaches 100% as tapproaches 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. cshifts the graph to the right or left
17. 120,000 P
(1 0
0
.
.
0
0
3
3
5
5
)81
120,000 P(9.051687)
P13,257.19725
$13,257.20
18. x
8
5
y
3
5
z
1
5
x
5x3y3z
19. y6x2v45°
xsin 45° ycos 45° 6(xcos 45° ysin 45°)2
xy6
xy
2
xy6
1
2
x2xy
1
2
y2
xy3x26xy 3y2
2
x2
y6x212xy 6y2
6x212xy 6y22
x2
y0
20. r(5)2
(1
)2
vArctan
1
5
26
3.34
26
(cos 3.34 i sin 3.34)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
Chapter 11 384
Effective
Interest Annual
Compounded Interest Yield
Annually $80.00 8%
Semi-annually $81.60 8.16%
Quarterly $82.43 8.243%
Monthly $83.00 8.3%
Daily $83.28 8.328%
Continuously $83.29 8.329%
21.
d
u
x, yF
u
0, 150
cos 28°
1
x
0
sin 28°
1
x
0
x10 cos 28° x10 sin 28°
x8.8295 x4.6947
d8.8295, 4.6947
WF
u
d
u
W0, 1508.8295, 4.6947
0 704.205
704.2 ft-lb
22. y1.5(10)213.3(14) 19.4
2.4
23. 2x3
4
2x3 16
2x13
x
1
2
3
24. 3x26
3x2 63x2 6
3x43x8
x
4
3
x
8
3
x
8
3
x
4
3
25. 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.
26.
4xy64(2y12) y6x2(6) 12
x2y12 9y54 x0
y6
(0, 6)
27. {4, 2, 5}; {5, 7}; yes
28. The correct choice is D.
6
2y12
4xy
x
9
3
6
15
6
18
9
6
3
1
2
5
2
6
3
2
Page 717 Mid-Chapter Quiz
1. 6428
2. (
3343
)2(343
1
3
)2
((73)
1
3
)2
72
4
1
9
3.
2
8
7
x
w
3y
6z
6
9
1
3
2
8
7
x
w
3
6
z
y
9
6
1
3
(23)
1
3
(x3)
1
3
(z9)
1
3
(33)
1
3
(w6)
1
3
(y6)
1
3
385 Chapter 11
y
x
28˚
O
(
x,
y
)
(0, 150)
y
x
d
4
3
2
w
x
2
z
y
3
2
or
2
3
w2xy2z3
4. a6b3
(a4b3)
1
2
(a6)
1
2
(b3)
1
2
a3b
3
2
5. (125a2b3)
1
3
3125a2
b3
353a2b
3
5b
3a2
6. 1.75 1020.094 A3
1.75 1020.94 A
3
2
1.75
0.
94
102
A
3
2
1.75
0.
94
102
2
3
(A
3
2
)
2
3
151.34 A
1.51 102mm2
7. 1,786,691 1,637,859 (1 r)8
1
1
,
,
7
6
8
3
6
7
,
,
6
8
9
5
1
9
(1 r)8
1
1
,
,
7
6
8
3
6
7
,
,
6
8
9
5
1
9
1
8
[(1 r)8]
1
8
1
1
,
,
7
6
8
3
6
7
,
,
6
8
9
5
1
9
1
8
1 r
r0.011 Store the exact value in
your calculator’s
memory.
N1,637,859 (1 0.011)24 Use the stored
value for r.
2,216,156.979
2,126,157
8. A3500
1
0.0
4
52
(4)(3.5)
3500(1.013)14
4193.728
$4193.73
9a. y6.7e
4
1
8
5
.1
0.271292
271,292 ft3
9b. y6.7e
4
5
8
0
.1
2.560257
2,560,257 ft3
10. 2 years: n
12
2
0
0
e
0
0.35(2)
18.3
15 years: n
120
2
e
0
0
0.35(15)
181
60 years: n
120
2
e
0
0
0.35(60)
200
18.3; 181; 200
Logarithmic Functions
Pages 722–723 Check for Understanding
1. y3xand log3xare
similar in that they are
both continuous, one-to-
one, increasing and
inverses.
y3xand log3xare not
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. y3xhas a
y-intercept and a
horizontal asymptote
whereas ylog3xhas a
x-intercept and a
vertical asymptote.
2. Let bxm, then logbmx.
(bx)pmp
bxp mp
logbbxp logbmp
xp logbmp
plogbmlogbmp
3.
Log5xis an increasing function and log
1
5
xis a
decreasing function.
4. Sean is correct. The product property states that
logbmn logbmlogbn.
5. In half-life applications r
1
2
. So, (1 r)
becomes
1
1
2
or
1
2
. Thus, the formula
NN0(1 r)tbecomes NN0
1
2
t.
6. 9
3
2
27 7.
2
1
5
1
2
5
8. log7y69. log8
1
4
2
3
11-4 14. log7n
2
3
log78
log7nlog78
2
3
n8
2
3
n4
15. log6(4x4) log664
4x4 64
x15
16. 2 log64
1
4
log616 log6x
log642log616
1
2
log6x
log6log6x
42
16
1
4
Chapter 11 386
O
y
x
y
3
x
y
log3
x
O
y
x
y
log
x
y
log5
x
1
5
x
x8
42
16
1
4
17. xy
10
21
42
18. xy
10
61
19. 16
3.3 log
t
41024
t16(33 log41024) log41024 x
t16(33 5) 4x1024
t264 h 22x210
2x10
x5
Pages 723–725 Exercises
20. 27
1
3
321. 16
1
2
4
22. 74
24
1
01
23. 4
5
2
32
24. ex65.98 25.
6
436
26. log81 9
1
2
27. log36 216
3
2
28. log
1
8
512 329. log6
3
1
6
2
y
x
y
log
x
2
1
O
y
x
y
log6
x
O
10. log2
1
1
6
x
2x
1
1
6
2x24
x4
11. log10 0.01 x
10x0.01
10x102
x2
12. log7
3
1
43
x
7x
3
1
43
7x73
x3
13. log2x5
25x
32 x
30. log16 1 031. logx14.36 1.238
32. log864 x
x2
47. log10
310
x
10x
310
10x10
1
3
x
1
3
48. log12 x
1
2
log12 9
1
3
log12 27
log12 xlog12 9
1
2
log12 27
1
3
log12 xlog12 9
1
2
27
1
3
x9
1
2
27
1
3
x3 3
x9
49. log5(x4) log58 log564
log5(x4)(8) log564
(x4)(8) 64
x4 8
x4
50. log4(x3) log4(x3) 2
log4(x3)(x3) 2
42(x3)(x3)
16 x29
25 x2
5 x
51.
1
2
(log7xlog78) log716
1
2
(log78x) log716
log7(8x)
1
2
log716
(8x)
1
2
16
8x256
x32
52. 2 log5(x2) log536
log5(x2)2log536
(x2)236
x2 6
x8
53. xy
10
2
1
2
41
387 Chapter 11
y
x
y
log4
x
O
8x64
8x82
x2
33. log125 5 x
125x5
(53)x51
3x1
x
1
3
34. log232 x
2x32
2x25
x5
35. log4128 x
4x128
22x27
2x7
x
7
2
or 3.5
36. log996x
9x96
x6
37. log49 343 x
49x343
72x73
2x3
x
3
2
or 1.5
38. log816 x
8x16
23x24
3x4
x
4
3
39. log8
4096 x
8
x4096
8
2
x
84
2
x
4
x8
40. 104 log102 x
10log1024 x
24x
16 x
41. logx49 2
x249
x7
42. log33xlog336
3x36
x12
43. log6xlog69 log654
log69xlog654
9x54
x6
44. log848 log8wlog86
log8
4
w
8
log86
4
w
8
6
6w48
w8
45. log6216 x
6x216
6x63
x3
46. log50.04 x
5x0.04
5x52
54. xy
10
23
46
55. xy
20
61
56. xy
10
21
42
57. xy
10
22
44
58. xy
00
91
59. Use NN0(1 r)t; r1 since the rate of growth
is 100% every ttime periods.
64,000 1000 (1 1)t
64 2t
log226log22t
6 t
t15 90 min
60. All powers of 1 are 1, so the inverse of y1xis
not a function.
61. Let logbmxand logbny.
So, bxmand byn.
m
n
b
b
x
y
bxy
m
n
bxy
logb
m
n
xy
logb
m
n
logbmlogbn
62a. 5000 2500
1
4
r
410
27
1
4
r
40
62b. 2
1
4
r
40
2
4
1
0
1
4
r
40
1.0175 1
4
r
0.0699 r
6.99%
62c. 2
1
4
r
28
2
2
1
8
1
4
r
1.0251 1
4
r
0.1004 r
10.04%
1
40
Chapter 11 388
y
x
O
20
15
10
5510152025303540
5
10
15
20
y
3 log2
x
y
x
y
log5 (
x
1)
y
x
y
log2
x
y
x
y
2 log2
x
y
x
y
log10 (
x
1)
63a. nlog2
nlog24
2n4
n2
1
—
1
4
63b. 3 log2
p
1
23
p
1
8 p1
1
8
p
less light;
1
8
64. Let ylogax, so xay.
xay
logbxlogbay
logbxylogba
y
l
l
o
o
g
g
b
b
a
x
logax
l
l
o
o
g
g
b
b
a
x
65a.
65b. log2.72
14
P
.7
0.02(1)
14
P
.7
2.720.02
P14.7 (2720.02)
14.4 psi
65c. log2.72
14
P
.7
0.02(6.8)
P14.7(2.720.136)
16.84 psi
66. 6.8 38
1
2
t
6
3
.
8
8
1
2
t
log
6
3
.
8
8
log
1
2
t
log
6
3
.
8
8
tlog
1
2
t
t2.5
2.5 3.82 9.55
about 9 days
log
6
3
.
8
8
__
log
1
2
P
h
log2.72
42
13
12.5
13.5
14
14.5
15
15.5
16
24
0.02
h
()
P
14.7
O
67. 69.6164
68. 90,000 P
P891.262
$891.26
69. ellipse
9x218x4y216y11 0
9x218x4y216y11
9(x22x1) 4(y24y4) 11 9 16
9(x1)24(y2)236
(x
4
1)2
(y
9
2)2
1
70. r3, v2
(3 cos 2t, 3 sin 2t)
71. AB (1
(1))2
(3
3)2
02(
6)2
6
BC (3 (
1))2
(0 (
3))2
423
2
5
AC (3 (
1))2
(0 3)
2
423
2
5
72. 5
cos
3
4
isin
3
4
2
cos
2
3
isin
2
3
5 2
cos
3
4
2
3
isin
3
4
2
3
10
cos
1
1
7
2
isin
1
1
7
2
10 cos
1
1
7
2
10isin
1
1
7
2
2.59 9.66i
10
cos
1
1
7
2
isin
1
1
7
2
, 2.59 9.66i
73. (3 4j)(12 7j) 36 21j48j28j2
64 27j
64 27jvolts
74.
Both vectors have the same direction. 50° south of
east. Therefore, AB
u
and CD
u
are parallel.
1
1
0.
1
1
2
15
12.30
______
0.
1
1
2
15
389 Chapter 11
y
x
O
y
x
50˚
50˚
O
A
D
C
B
c
˚
c
˚
b
˚
x
˚
S
N
Q
M
P
75. cos (AB) cos Acos Bsin Asin B
cos A
1
5
3
cos B
3
3
5
7
x252132x2352372
x2144 x2144
x12 x12
So, sin A
1
1
2
3
So, sin B
1
3
2
7
cos (AB)
1
5
3
3
3
5
7
1
1
2
3
1
3
2
7
1
4
7
8
5
1
1
4
4
8
4
1
4
3
8
1
1
76. yAsin (kt c) h
A
90
2
64
h
90
2
64
2
k
4
13 77 k
2
y13 sin
2
tc
77
64 13 sin
2
(1) c
77
13 13 sin
2
c
1 sin
2
c
sin1(1)
2
c
3.14 c
So, y13 sin
2
k3.14
77
77. c2(6.11)2(5.84)22(6.11)(5.84) cos 105.3
c237.3321 34.1056 71.3648 cos 105.3
c290.2689
c9.5
(6.11)2(5.84)2(9.5)22(5.84)(9.5) cos A
37.3321 34.1056 90.25 110.96 cos A
cos A0.7843
A38.34 or 38° 20
B 180 (105° 1838° 20)
36° 22
c9.5, A38° 20, B36° 22
78.
mSMN mQNM alternate interior angles
x°b°c°Exterior Angle Theorem
The correct choice is E.
Common Logarithms
Page 730 Check for Understanding
1. log 1 0 means log10 1 0. So, 10° 1.
log 10 1 means log10 10 1. So, 10110.
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 104log 8
4 0.9031
4.9031
6. log 0.003 log (0.001 3)
log 103log 3
3 0.4771
2.5229
7. log 0.0081 log (0.0001 34)
log 1044 log 3
4 4(0.4771)
2.0915
8. 2.6274 9. 74,816.95
10.
11. log12 18
l
l
o
o
g
g
1
1
8
2
1.1632
12. log815
l
l
o
o
g
g
1
8
5
1.3023
13. 2.2x59.32
(x5) log 2.2 log 9.32
(x5)
l
l
o
o
g
g
9
2
.
.
3
2
2
x7.83
14. 6x24x
(x2) log 6 xlog 4
xlog 6 2 log 6 xlog 4
2 log 6 xlog 4 x log 6
2 log 6 x(log 4 log 6)
log
4
2
log
lo
6
g6
x
8.84 x
15. 4.3x76.2
xlog 4.3 log 76.2
x
l
l
o
o
g
g
7
4
6
.
.
3
2
x2.97
11-5 16. 3x32
44x1
3x32
4
x
4
1
(x3) log 3 log 2
x
4
1
log 4
(4x12)log 3 4 log 2 (x1) log 4
4xlog 3 12 log 3 4 log 2 xlog 4 log 4
4xlog 3 xlog 4 4 log 2 log 4 12 log 3
x(4 log 3 log 4) 4 log 2 log 4 12 log 3
x
x4.84
17.
[10, 10] scl:1 by [20, 100] scl:10
5.5850
18a. Rlog
2
1
0
.6
0
4.2
6.3
18b. 10 times; According to the definition of
logarithms, Rin the equation
Rlog
T
a
Bis an exponent of the base of the
logarithm, 10. 105is ten times greater than 104.
Pages 730–732 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
22. log 0.06 log
0.01
1
2
2
log 0.01 log
log 0.01 log 12
1
2
log 4
2 1.0792
1
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
12
4
1
2
4 log 2 log 4 12 log 3
4 log 3 log 4
Chapter 11 390
y
x
y
log (
x
3)
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
27. log 1800 log
100 9 4
1
2
log 100 log 9
1
2
log 4
2 0.9542
1
2
(0.6021)
3.2553
28. 1.9921 29. 2.9515
30. 0.871 31. 2.001
32. 3.2769 33. 2.1745
34. log28
l
l
o
o
g
g
8
2
35. log5625
lo
l
g
og
62
5
5
34
36. log624
l
l
o
o
g
g
2
6
4
37. log74
l
l
o
o
g
g
4
7
1.7737 0.7124
38. log6.5 0.0675
log
lo
0
g
.
0
0
.
6
5
75
3.8890
39. log
1
2
15
41. 5x4x3
x log 5 (x3) log 4
xlog 5 xlog 4 3 log 4
x(log 5 log 4) 3 log 4
x
log
3
5
l
og
l
4
og 4
x18.6377
42.
1
3
log xlog 8
x
1
3
8
x512
43. 0.1643x0.38x
(43x) log 0.16 (8 x) log 0.3
4 log 0.16 3xlog 0.16 8 log 0.3 xlog 0.3
3xlog 0.16 xlog 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
x0.3434
44. 4 log (x3) 9
log (x3)
9
4
(x3) antilog
9
4
xantilog
9
4
3
x174.8297
45. 0.25 log 16x
0.25 xlog 16
x
lo
0
g
.2
1
5
6
x0.2076
8 log 0.3 4 log 0.16
3 log 0.16 log 0.3
log 15
log
1
2
46. 3x12x7
(x1) log 3 (x7) log 2
xlog 3 log 3 xlog 2 7 log 2
xlog 3 xlog 2 log 3 7 log 2
x(log 3 log 2) log 3 7 log 2
x
lo
lo
g
g
3
3
7
lo
lo
g
g
2
2
x9.2571
47. logx6 1
l
l
o
o
g
g
6
x
1
log 6 log x
6 x
When x1, log 1 0, which means
l
l
o
o
g
g
6
x
is
undefined. When x1,
l
l
o
o
g
g
6
x
is negative, which is
not greater than 1. So, xmust also be greater than
1. Therefore, 1 x6.
48. 42x53x3
(2x5) log 4 (x3) log 3
2xlog 4 5 log 4 xlog 3 3 log 3
2xlog 4 xlog 3 5 log 4 3 log 3
x(2 log 4 log 3) 5 log 4 3 log 3
x
5
2
lo
lo
g
g
4
4
3
lo
lo
g
g
3
3
x2.1719
49. 0.52x40.15x
(2x4) log 0.5 (5 x) log 0.1
2x(log 0.5) 4 log 0.5 5 log 0.1 xlog 0.1
2xlog 0.5 xlog 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.
x3.8725
50. log2x3
x23
x0.1250
52.
[10, 10] scl:1 by [3, 3] scl:1
53.
[1, 10] scl:1 by [1, 3] scl:1
5 log 0.1 4log 0.5
2 log 0.5 log 0.1
391 Chapter 11
40. 2x95
xlog 2 log 95
x
l
l
o
o
g
g
9
2
5
x6.5699
51. x
lo
l
g
og
52
3
.7
x3.6087
54.
[10, 1] scl:1 by [2, 10] scl:1
55. x0.3210
[5, 5] scl:1 by [10, 50] scl:10
56. x0.1975
[5, 5] scl:1 by [3, 10] scl:1
57. x2
[5, 5] scl:1 by [5, 10] scl:1
58a. h
10
9
0
log
1
1
0
4
.
.
3
7
1.7 mi
58b. 4.3
10
9
0
log
14
P
.7
0.3870 log Plog 14.7
0.3870 log 14.7 log P
0.7803 log P
6.03 P; 6 psi
59a. M5.3 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
60a. q
1
2
0.89
1
2
0.1342
0.9112
$91,116
60b. 0.9535
1
2
0.8t
log 0.9535 0.8tlog
1
2
0.8t
log 0.9535
log
1
2
Chapter 11 392
log
tlog 0.8
12.0016 t
12 years
61. Sample answer: xis between 2 and 3 because 372
is between 100 and 1000, and log 100 2 and log
1000 3.
62a. L10 log
1.0
1
1012
10(log 1 log (1.0 1012))
120 dB
62b. 20 log
1.0
I
1012
2 log Ilog (1.0 1012)
2 log I12
10 log I
1 1010 I; 1 1010 W/m2
63. Use NN0
1
2
t.
N630 micrograms 63 104gram
N01 milligram 1.0 103gram
6.3 104(1.0 103)
1
2
t
log
6
1
.
.
3
0
1
1
0
0
4
3
tlog
1
2
0.6666 t
0.6666 5730 3819 yr
64. logaylogaPlogaqlogar
logayloga
p
q
logar
logayloga
p
q
r
y
p
q
r
65. logx243 5
x5243
x3
66.
increasing from to
67. (a4b2)
1
3
c
2
3
(a4)
1
3
(b2)
1
3
(c2)
1
3
a
3ab2c2
68. (5)2(0)2D(5) E(0) F0
5DF25 0
(1)2(2)2D(1) E(2) F0
D2EF5 0
(4)2(3)2D(4) E(3) F0
4D3EF25 0
log 0.9535
log
1
2
5D0EF25 0
()D2EF5 0
4D2E20 0
4D3EF25 0
()D2EF5 0
3DE20 0
4D2E20 0
2(3DE20 0)
10D 60 0
D6
4(6) 2E20 0
2E4 0
E2
5(6) 0(2) F25 0
F5 0
F5
x2y26x2y5 0
(x26x9) (y22y1) 5 9 1
(x3)2(y1)25
69.
65
2
25
,
18
2
4
(25
, 11)
70. r6
r236
x2y236
71.
72. AB
u
(6 5), (5 6)
1, 1
AB
u
(6 5
)2(
5 6
)2
2
1.414
73.
v36 10 360°
cos 36°
3.
a
65
sin 36°
3.
b
65
a3.65 cos 36° b3.65 sin 36°
2.9529 2.1454
Use A
1
2
aP, where P10(2.1454) 21.454.
A
1
2
(2.9529)(21.454)
31.6758 or 31.68 cm2
74. f(x) x32x211x12
f(1) 1 2(1) 11(1) 12 Test f(1).
f(1) 0(x1) is a factor.
11211 12
1112
1112 0
x2x12 0
(x4)(x3) 0
So, the factors are (x4)(x3)(x1).
75.
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.
Natural Logarithms
Page 735 Check for Understanding
1. ln e1 is the same as logee 1. And e1e. So,
ln e1.
2. The two logarithms have different bases.
log 17 ⇒10x17 or x1.23
ln 17 ⇒ex17 or x2.83
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,
APert is a special case of the equation
NN0ekt and is used primarily for computations
involving money.
5. 4.7217 6. 1.1394
7. 15.606 8. 0.4570
9. log5132
ln
ln
13
5
2
3.0339
10. log364
l
l
n
n
6
3
4
3.7856
11. 18 e3x
ln 18 3xln e
ln
3
18
x
0.9635 x
11-6
393 Chapter 11
246 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
b
a
3.65 cm
O
y
x
y
5
x
3
2
x
5
12. 10 5e5k
2 e5k
ln 2 5kln e
ln
5
2
k
0.1386 k
13. 25ex100
ex4
xln eln 4
x1.3863
14. 4.5 e0.031t
ln 4.5 0.031tln e
l
0
n
.0
4
3
.
1
5
t
48.5186 t
15. x13.57
[20, 20] scl:2 by [4, 20] scl:2
16. x26.90
[15, 30] scl:5 by [50, 150] scl:10
17a. p760e0.125(3.3)
760e0.4125
503.1 torrs
17b. 450 760e0.125a
4
7
5
6
0
0
e0.125a
ln
4
7
5
6
0
0
0.125aln e
a
4.1926 a; 4.2 km
Pages 736–737 Exercise
18. 5.4931 19. 0.2705
20. 6.8876 21. 0.9657
22. 023. 2.2322
24. 10.4395 25. 1.2134
26. 0.0233 27. 0.9966
28. 146.4963 29. 0.2417
30. log12 56
l
l
n
n
5
1
6
2
31. log536
l
l
n
n
3
5
6
1.6199 2.2266
32. log483
l
l
n
n
8
4
3
3.1875
ln
4
7
5
6
0
0
0.125
33. log80.512
ln
l
0
n
.5
8
12
0.3219
34. log6323
ln
1
3
.6
03
3.2246
35. log5288
ln
l
n
2
5
88
1.7593
36. 6x72 37. 2x27
x ln 6 ln 72 x ln 2 ln 27
x
l
l
n
n
7
6
2
x
l
l
n
n
2
2
7
2.3869 4.7549
38. 9x47.13
(x 4) ln 9 ln 7.13
x ln 9 4 ln 9 ln 7.13
xln 9 ln 7.13 4 ln 9
x
ln 7.13
ln
9
4ln9
x4.8940
39. 3x32
x ln 3 ln 3 ln 2
x
ln 3
ln
l
3
n2
x1.3155
41. 60.3 e0.1t
ln 60.3 0.1tln e
ln
0
6
.1
0.3
t
40.9933 t
42. 6.2e0.64t3et1
ln 6.2 0.64tln eln 3 (t1) ln e
ln 6.2 ln 3 1 0.36t
ln 6.2
0
.3
l
6
n31
t
0.7613 t
43. 22 44 (1 e2x)
1
2
1 e2x
1
2
e2x
ln
1
2
2xln e
x
0.3466 x
44. 25 e0.075y
ln 25 0.075yln e
0
ln
.0
2
7
5
5
y
y42.9183
ln
1
2
2
Chapter 11 394
40. 25ex1000
ex40
xln eln 40
x3.6889
45. 5x76
xln 5 ln 7 ln 6
x
ln 7
ln
l
5
n6
x 1.7657
46. 12x44x
(x4) ln 12 xln 4
xln 12 4 ln 12 xln 4
xln 12 xln 4 4 ln 12
x(ln 12 ln 4) 4 ln 12
x
ln
4
12
ln
1
l
2
n4
x9.0474
47. x
2
3
27.6
x(27.6)
3
2
x144.9985
48. x3.76
[5, 5] scl:1 by [50, 700] scl:50
49. x7.64
[70, 10] scl:1 by [3, 10] scl:1
50. t133.14
[10, 150] scl:10 by [100, 2000] scl:100
51. x2.14
[6, 6] scl:1 by [4, 24] scl:2
52. x4.72
[10, 10] scl:1 by [10, 75] scl:5
53. x0.37
[5, 5] scl:1 by [2, 5] scl: 0.5
54. 0.6 1e
ln 0.6 ln e
t
20,000(4 1011)
t
20,000(41011)
395 Chapter 11
20,000(4 1011) ln 0.6 t
4.09 107t
4.09 107s
55. 2.8 9
1
2
t
ln
2
9
.8
tln
1
2
t
1.6845 t
1.6845 8 24 323.4236
324 h
56a. ln 180 72k(0) c
4.6821 c
56b. ln 150 72k(2) 4.6821
ln 78
2
4.6821
k
0.1627 k
56c. ln 100 72(0.1622)t4.6821
ln 2
8
0
.16
4
2
.6
7
821
t
8.3 t8.3 2 6.3
about 6.3 min
57. e2x4ex3 0
(ex3)(ex1) 0
ex3 0ex1 0
ex3ex1
xln eln 3 xln eln 1
x1.0986 x0
0 or 1.0986
58a. 2 e0.063t
ln 2 0.063tln e
0
l
.
n
06
2
3
t
11.0023 t
about 11 years
58b. See students’ work.
ln
2
9
.8
__
ln
1
2
59. 1800 5000 ln r
1
5
8
0
0
0
0
0
ln r
antiln
5
1
0
8
0
0
0
0
r
0.6977 r; about 70%
60a.
1
2
1ek(1622)
ln
1
2
1622 kln e
k
0.000427 k
60b. 1.7 23e(0.000427)(t)
ln
1
2
.
.
7
3
0.000427tln e
t
707.9177 t
about 708 yr
61. yis a logarithmic function of x. The pattern in the
table can be determined by 3yxwhich can be
expressed as log3xy.
62. 1.2844
63. 16
3
4
8
64. x2y4
x24y28
(y4) 4y28
4y2y4 0
y1 1 4
(4)(4
)
8
ln
1
2
.
.
7
3
0.000427
ln
1
2
1622
68. 2x5y3 0
A2
B2
22(
5)2
29
2
2
x
9
5
2
y
9
2
3
9
0
2
29
29
x
5
29
29
y
3
29
29
0
p
3
29
29
0.56 units
sin
5
29
cos
2
29
tan
tan
5
2
112°
2
29
29
x
5
29
29
y
3
29
29
0;
3
29
29
0.56; 112°
69. y70 cos 4v
70. d800 (10 55)
250
The correct answer is 250.
Graphing Calculator Exploration:
Natural Logarithms and Area
Pages 738–739
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 k1, then the area of the region is equal to
ln k. If 0 k1, then the opposite of the area is
equal to ln k.
7. The value of ashould be equal to or very close to
1, and the value of bshould be very close to e. This
prediction is confirmed when you display the
actual regression equation.
8. Sample answer: Define ln kfor k0 to be
the area between the graph of y
1
x
, the x-axis,
and the vertical lines x1 and xkif k1 and
to be the opposite of this area if 0 k1. Define
eto be the value of kfor which the area of the
region is equal to 1.
11-6B
5
29
29
___
2
29
29
Chapter 11 396
y
x
O
y
1
8
65
y0.9, 1.1
x20.9 4x211 4
x4.9
x2.9
x2.2, 2.2 x1.7, 1.7
65.
52
m
.4
2N
146 cm3
100
m
3
3
cm3
c
0.00765 c; 0.00765 N m
66. x0.25 cos y0.25 sin
0.25 0
(0.25, 0)
67. a
u
1, 234, 3
1, 212, 9
13, 7
Modeling Real-World Data with
Exponential and Logarithmic
Functions
Page 744 Check for Understanding
1. Replace Nby 4N0in the equation NN0ekt,
where N0is the amount invested and kis 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. y2e(ln 4)xor y2e1.3863x; ln yln 2 (ln 4)xor
ln y0.6931 1.3863x
4. t
0.
l
0
n
1
2
75
5. t
0
ln
.0
2
8
39.61 yr 8.66 yr
6a. y10.0170(0.9703)x
6b. y10.0170(0.9703)x
y10.0170(eln 0.9703)x
y10.0170e(ln 0.9703)x
y10.0170e0.0301x
6c. 5 10.0170e0.0301x
ln
10.0
5
170
0.0301x
x
23.08 x; 23.08 min
ln
10.0
5
170
0.0301
11-7 15c. 0.415 1.0091e0.0197x
ln
1
0
.
.
0
4
0
1
9
5
1
0.0197x
x
ln
1
0
.
.
0
4
0
1
9
5
1
0.0197
397 Chapter 11
Pages 745–748 Exercises
7. t
0.
l
0
n
2
2
25
8. t
0
ln
.0
2
5
30.81 yr 13.86 yr
9. t
0.0
ln
71
2
25
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. y4.7818(1.7687)x
14b. y4.7818(1.7687)x
y4.7818(eln 1.7687)x
y4.7818e(ln 1.7687)x
y4.7818e0.5702x
14c. Use t
ln
k
2
; k0.5702.
t
0.
l
5
n
7
2
02
1.215 hr
15a. y1.0091(0.9805)x
15b. y1.0091(0.9805)x
y1.0091(eln 0.9805)x
y1.0091e(ln 0.9805)x
y1.0091e0.0197x
45.10 x
45.10 10 35.10 min
16a. y2137.5192(1.0534)x
16b. y2137.5192(1.0534)x
y2137.5192(eln 1.0534)x
y2137.5192e(ln 1.0534)x
y2137.5192e0.0520x
16c. 2631.74 2137.52e4r
ln
2
2
6
1
3
3
1
7
.
.
7
5
4
2
4r
ln r
2
2
6
1
3
3
1
7
.
.
7
5
4
2
4
0.0520 r; 5.2%
17. y40 14.4270 ln x
18a. y–826.4217 520.4168 ln x
18b. The year 1960 would correspond to x0 and
ln 0 is undefined.
19. Take the square root of each side.
ycx2
y
cx2
y
c
x
20a. 1034.34 1000(1 r)1
1.03034 1 r
0.03034 r; 3.034%
20b. y1000.0006(1.0303)x
20c. y1000.0006(1.0303)x
y1000.0006(eln 1.0303)x
y1000.0006e(ln 1.0303)x
y1000.0006e0.0299x
20d. 1030.34 1000er
ln
10
1
3
0
0
0
.
0
34
r
0.0299 r; 2.99%
21a.
21b. ln y0.0136x1.6889
21c. ln y0.0136x1.6889
ye0.0136x1.6889
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 y2, so you must subtract 2 from
each y-value before a calculator can perform
exponential regression.
22b. y2 1.0003(2.5710)x
23a. ln yis a linear function of ln x.
ycxa
ln yln(cxa)
ln yln cln xa
ln yln caln x
x050100 150 190 200
ln y1.81 2.07 3.24 3.75 4.25 4.38
23b. The result of part a indicates that we should
take the natural logarithms of both the x- and
y-values.
23c. ln y0.4994 ln x1.3901
23d. ln y0.4994 ln x1.3901
eln ye0.4994ln x1.3901
ye0.4994ln xe1.3901
y(eln x)0.4994 4.0153
y4.0153x0.4994
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 (5x16)
7x5x16
2x16
x8
27a. yx(400 20(x3))
yx(460 20x)
y20x2460x
y2645 20(x223x132.25)
y2645 20(x11.5)
vertex at (11.5, 2645), maximum at x11.5
$11.50
27b. At maximum, y2645.
$2645
28. 5
cos
6
isin
6
5
2
3
1
2
i
5
2
3
5
2
i
29.
sin
60
24°
sin
b
48°
60 sin 48° bsin 24°
b109.625
about 109.6 ft
30. 5x28x12 0
Discriminant: (8)24(5)(12) 176
The discriminant is negative, so there are
2 imaginary roots.
x
8
10
176
8
1
4
0
i11
or
42
5
i11
31. 4 units left and 8 units down
32.
(2, 1): f(x) 2(2) 8(1) 10
22
(4, 1): f(x) 2(4) 8(1) 10
26
(2, 8): f(x) 2(2) 8(3) 10
38
(4, 4): f(x) 2(4) 8(4) 10
50
50; 22
33. Circle Xcontains the regions a, b, d, and e.
Circle Zcontains 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 Understanding the Vocabulary
1. common logarithm 2. exponential growth
3. logarithmic function 4. scientific notation
5. mantissa 6. natural logarithm
7. linearizing data 8. exponential function
9. nonlinear regression 10. exponential equation
Pages 750–752 Skills and Concepts
11.
1
4
212. (64)
1
2
8
16
13. (27)
4
3
(33)
4
3
34
81
15. 3x2(3x)2
(
3
3
x
x
2
)2
3
9
x
x
2
2
1
3
17.
1
2
x4
3
1
2
3(x4)3
1
8
x12
18. (w3)4(4w2)2w12 42w4
16w16
19.
(2a)
1
3
(a2b)
1
3
3
(2a)
1
3
3
(a2b)
1
3
2
(2a)(a2b)
2a3b
20.
3x
1
2
y
1
4
(4x2y2) 12x
5
2
y
9
4
1
1
4
2
Chapter 11 398
ln x6.21 6.91 8.52 9.21 9.62
ln y4.49 4.84 5.65 5.99 6.19
b
60 ft
18˚
42˚
14. (
4256
)3(256)
3
4
(44)
3
4
43
64
16.
6a
1
3
363
a
1
3
3
216a
O
y
x
(4, 4)
(2, 1)
(2, 3)
(4, 1)
x
2
y
4
x
4
y
1
x
2
21. 22.
23. 24.
25. 26.
27. A2500e0.065(10)
4788.8520; $4788.85
28. A6000e0.0725(10)
12,388.3866; $12,388.39
29. A12,000e0.059(10)
21,647.8610, $21,647.86
30. 8
2
3
431. 34
8
1
1
32. log216 433. log5
2
1
5
2
34. 2x32 35. 10x0.001
2x2510x103
x5x3
36. 4x
1
1
6
37. 2x0.5
4x422x21
x2x1
38. 6x216 39. 9x
1
9
6x639x91
x3x1
40. 4x1024 41. 8x512
4x458x83
x5x3
42. x481 43.
1
2
4x
x(81)
1
4
16 x
x 3
44. log33 log3 xlog345
log33xlog345
3x45
x15
45. 2 log64
1
3
log68 log6x
log642log68
1
3
log6x
399 Chapter 11
y
x
y
3
x
O
y
x
y
x
()
1
2
O
y
x
y
2
x
1
O
y
x
y
2
x
2
O
y
x
y
2
x
1
O
y
x
y
2
x
2
O
log6log6x
42
8
1
3
x
42
8
1
3
8 x
46. log2x
1
3
log627
log2xlog227
1
3
x27
1
3
x3
47.
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. 4x6x2
xlog 4 (x2) log 6
xlog4xlog 6 2 log 6
xlog 4 xlog 6 2 log 6
x(log 4 log 6) 2 log 6
x
log
2
4
l
og
l
6
og 6
x8.84
53. 120.5x80.1x4
0.5xlog 12 (0.1x4) log 8
0.5xlog 12 0.1xlog 8 4 log 8
0.5xlog 12 0.1xlog 8 4 log 8
x(0.5 log 12 0.1 log 8) 4 log 8
x4 log 8
0.5 log 12 0.1 log 8
y
x
y
log10
x
O
x8.04
54.
1
4
3x6x2
3xlog
1
4
(x2) log 6
3xlog
1
4
xlog 6 2 log 6
3xlog
1
4
xlog 6 2 log 6
x(3 log
1
4
log 6) 2 log 6
x2 log 6
3 log
1
4
log 6
62. log4100
lo
l
g
og
10
9
0
63. log15 125
l
l
o
o
g
g
1
1
2
5
5
2.0959 1.7829
64. 4x100
x ln 4 ln 100
x
ln
ln
10
4
0
x3.3219
65. 6x230
(x2) ln 6 ln 30
x2
l
l
n
n
3
6
0
x
l
l
n
n
3
6
0
2
x3.8982
66. 3x142x
(x1) ln 3 2xln 4
x ln 3 ln 3 2xln 4
x ln 3 2x ln 4 ln 3
x(ln 3 2 ln 4) ln 3
x
ln 3
ln
2
3
ln 4
x0.6563
67. 94x5x4
4x ln 9 (x4) ln 5
4xln 9 x ln 5 4 ln 5
4x ln 9 xln 5 4 ln 5
x(4 ln 9 ln 5) 4 ln 5
x
4ln
4
9
l
n
l
5
n5
x0.8967
68. 24 e2x69. 15ex200
ln 24 2xe
x
2
1
0
5
0
x
ln
2
24
xln
2
1
0
5
0
x1.5890 x2.5903
70. x3.333
[5, 5] scl:1 by [10, 60] scl:10
71. x2.20
[1, 5] scl:1 by [1, 10] scl:1
Chapter 11 400
y
x
y
3 log (
x
2)
O
y
x
y
7
x
2
O
Change the inequality because 3 log
1
4
log 6 is
negative.
x0.6
55. 0 12x87x4
(2x8) log 0.1 (x4) log 7
2xlog 0.1 8 log 0.1 xlog 7 4 log 7
2xlog 0.1 xlog 7 4 log 7 8 log 0.1
x(2 log 0.1 log 7) 4 log 7 8 log 0.1
x
x4
56. log (2x3) log (3 x)
log (2x3) log (3 x)1
(2x3) (3 x)1
(2x3)(3 x) 1
2x23x8 0
x
3
4
73
1.39, 2.89
57.
58.
59. x3.42
[5, 5] scl:1 by [5, 5] scl:1
60. log415
l
l
o
o
g
g
1
4
5
61. log824
l
l
o
o
g
g
2
8
4
1.9534 1.5283
4 log 7 7 log 0.1
2 log 0.1 log 7
72. t
0
l
.
n
02
2
8
73. t
0.0
ln
51
2
25
24.76 13.52
74. 18
ln
k
2
k
ln
18
2
0.0385; 3.85%
Page 753 Applications and Problem solving
75. 0.065
1
2
t
log 0.65 tlog
1
2
t
0.6215 t
0.6215 5730 3561.13 or 3561 yr.
76a. 10 log
1.15
10
1
12
010
20.6 20.6 dB
76b. 10 log
9
10
1
0
2
9
39.5 39.5 dB
76c. 10 log
8.95
10
1
1
2
03
99.5 99.5 dB
77. 200,000 142,000e0.014t
1
7
0
1
0
e0.014t
ln
1
7
0
1
0
0.014t
t
ln
1
7
0
1
0
0.014
log 0.65
log
1
2
Chapter SAT & ACT Preparation
Page 755 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.
2. Since one root is
1
2
, x
1
2
, 2x1, and 2x0.
Similarly for the root that is
1
3
,
x
1
3
, 3x1, and 3x1 0.
To find the quadratic equation, multiply these two
factors and let the product equal zero.
(2x1)(3x1) 0
6x25x1 0
The correct choice is E.
3. The result of dividing Tby 6 is 14 less than the
correct average.
T
6
correct answer 14
T
6
14 correct average
The correct average is the total divided by the
number of scores, 5.
correct average
T
5
T
6
14
T
5
The correct choice is E.
4.
tan A
2
x
y
Find the area of ABC.
A
1
2
bh
1
2
xy
x
2
y
Simplify the ratio.
area
ta
of
n
A
ABC
2
x
y
x
2
y
x
4
2
The correct choice is E.
5.
x
y
1
2
0
y
2yx 10y
2x10
x5
The correct choice is B.
x
2
y
2
x
y
y
2
x
401 Chapter 11
24.4 t
1990 24 2014
78a. N65 30e0.20(2)
44.89; 45 words per minute
78b. N65 30e0.20(15)
63.50; 64 words per minute
78c. 50 65 30e0.20t
1
2
e0.20t
ln
1
2
0.20t
t
ln
1
2
0.20
3.47 t; 3.5 weeks
Page 753 Open-Ended Assessment
1. Sample answer: (n4)
1
4
(4m)1
2. Sample answer:
log 2 log(x2)
1
2
log 36
y
x
B
(
x
,
y
)
2
C
(
x
, 0)
y
A
x
2
6. CD
2
3
D
D
2
3
and, therefore, r
1
3
.
Now use this value for the radius to calculate half
of the area.
1
2
A
1
2
r2
1
2
1
3
2
1
2
1
9
1
8
The correct choice is A.
7. The average of 8 numbers is 20.
20
sum of eight numbers 160
The average of 5 of the numbers is 14.
14
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
9
3
0
30
The correct choice is D.
8. The sum of the angles in a triangle is 180°. Since
∠Bis 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.
2x3x90
5x90
x18
The question asks for the measure of ∠A.
∠A2x2(18) 36
The correct choice is C.
sum of three numbers
3
sum of five numbers
5
sum of eight numbers
8
9. Ais the arithmetic mean of three consecutive
positive even integers, so A
3x3
6
x2, where xis a positive even integer.
Then Ais also a positive even integer. Since Ais
even, when Ais 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 bmust be a prime integer. Next
notice that 3bis greater than 10. So bcould be 5,
since 3(5) 15. (bcannot be 3.) Check to be sure
that 5 fits the rest of the inequality.
3(5) 15
5
6
(5)
2
6
5
4
1
6
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.
x(x2) (x4)
3
Chapter 11 402
Arithmetic Sequences and Series
Pages 762–763 Check for Understanding
1. a16 4(1) or 2
a26 4(2) or 2
a36 4(3) or 6
a46 4(4) or 10
a56 4(5) or 14
2, 2, 6, 10, 14; yes, there is a common
difference of 4.
2a.
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.
3b. Sn
n
2
[2a, (n 1)d]
S10
1
2
0
[2 3750 (10 1)500]
$60,000
4. Negative; let n and n 1 be two consecutive
numbers in the sequence.
d(n1) nor 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. d11 6 or 5
16 5 21, 21 5 26, 26 5 31, 31 5 36
21, 26, 31, 36
7. d7 (15) or 8
1 8 9, 9 8 17, 17 8 25, 25 8 33
9, 17, 25, 33
8. d(a2) (a6)
aa2 6 or 4
a2 4 a6, a6 4 a10,
a10 4 a14, a14 4 4 a18
a6, a10, a14, a18
9. ana1(n1)d
a17 10 (17 1)(3)
38
10. ana1(n1)d
37 13 (n1)5
50 5(n1)
10 n1
11 n
12-1 11. ana1(n1)d
3 a1(7 1)(2)
3 a112
15 a1
12. ana1(n1)d
34 100 (12 1)d
66 11d
6 d
13. ana1(n1)d
24 9 (4 1)d
15 3d
5 d
9 5 14, 14 5 19
9, 14, 19, 24
14. Sn
n
2
[2a1(n1)d]
S35
3
2
5
[2.7 (35 1) 2]
1435
15. Sn
n
2
[2a1(n1)d]
210
n
2
[2 30 (n1)(4)]
420 60n4n24n
4n264n420 0
4(n21)(n5) 0
n21 or n5
Since ncannot be negative, n21.
16. n19, a19 27, d1
ana1(n1)d
27 a1(19 1)1
9 a1
S19
n
2
(a1a19)
1
2
9
(9 27)
342 seats
Pages 763–765 Exercises
17. d1 5 or 6
7 (6) 13, 13 (6) 19,
19 (6) 25, 25 (6) 31
13, 19, 25, 31
18. d7 (18) 11
4 11 15, 15 11 26, 26 11 37, 37 11 48
15, 26, 37, 48
19. d4.5 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
20. d3.8 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
21. db4 bor 4
b8 4 b12, b12 4 b16,
b16 4 b20, b20 4 b24
b12, b16, b20, b24
22. d0 (x) or x
xx2x, 2xx3x, 3xx4x, 4xx5x
2x, 3x, 4x, 5x
403 Chapter 12
Chapter 12 Sequences and Series
an
n
O
1
1
123456
2
3
23. dn5nor 6x
7n(6n) 13n13n(6n) 19n,
19n(6n) 25n, 25n(6n) 31n
13n, 19n, 25n, 31n
24. d5 (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(2a2) (2a5) or 7
2a9 7 2a16, 2a16 7 2a23,
2a23 7 2a30, 2a30 7 2a37
2a16, 2a23, 2a30, 2a37
26. d5 (3 7
) or 2 7
77
27
927
,927
27
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 (n1)(5)
60 5(n1)
12 n1
13 n
30. 138 2 (n1)7
140 7(n1)
20 n1
21 n
31. 38 a1(15 1)(3)
38 a142
80 a1
32. 10
2
3
a1(7 1)
1
3
10
2
3
a12
8
2
3
a1
33. 58 6 (14 1)d
52 13d
4 d
34. 26 8 (11 1)d
18 10d
1
4
5
d
35. d (1 5
) (4 5
) or 3
a84 5
(8 1)3
17 5
36. d6 (5 i) 1 i
a12 5 i(12 1)(1 i)
5 i11 11i
16 10i
37. d10.5 12.2 or 1.7
a33 12.2 (33 1)(1.7)
42.2
38. d4 (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
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
d
3
12
3
3
12
3
2
3
,
12
3
2
3
12
3
3
24
3
3
3
,
12
3
2
3
,
24
3
3
, 12
42. 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
43. d1
3
2
or
1
2
a11
3
2
(11 1)
1
2
3.5
S11
1
2
1
3
2
3
1
2
11
44. d4.8 (5) or 0.2
a100 5 (100 1)0.2
14.8
S100
10
2
0
(5 14.8)
490
45. d13 (19) or 6
a26 19 (26 1)6
131
S26
2
2
6
(19 131)
1456
46. 14
n
2
[2(7) (n1)1.5]
28 14n1.5n21.5n
0 1.5n215.5n28
n
n8 or n2
1
3
Since there cannot be a fractional number of
terms, n8.
47. 31.5
n
2
[2(3) (n1)2.5]
63 6n2.5n225n
0 2.5n28.5n63
n
n7 or n3.6
Since ncannot be negative, n7.
48. d7 5 or 2
an5 (n1)2
2n3
49. d2 6 or 8
an6 (n1)(8)
8n14
8.5 (8.5
)24(
2.5)(
63)
2(2.5)
15.5 (15.
5)2
4(1.5)
(28)
2(1.5)
12 3
3
Chapter 12 404
50. 9:00, 9:30, 10:00, 10:30, 11:00, 11:30, 12:00
n7, d2, a13
a73 (7 1)2
15 data items per minute
51. Let dbe the common difference. Then, yxd,
zx2d, and wx3d. Substitute these
values into the expression xwyand simplify.
x(x3d) (xd) x2dor z.
52. a15, d4, n25
a25 5 (25 1)4
101
S25
2
2
5
(5 101)
1325 bricks
53. Sn
n
2
(128° 172°); Sn(n2)180°
150n180n360
30n360
n12
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. a11, d2
S5
5
2
[2(1) (5 1)2]
25
55b. S10
1
2
0
[2(1) (10 1)2]
100
55c. Conjecture: The sum of the first nterms of the
sequence of natural numbers is n2.
Proof:
Let an2n1. The first term of the sequence
of natural numbers is 1, so a11.
Then, using the formula for the sum of an
arithmetic series,
Sn
n
2
(a1an)
Sn
n
2
[1 (2n1)]
n
2
(2n) or n2
56. a15, d7, n15
S15
1
2
5
[2(5) (15 1)7]
810 feet
57. n10, S10 5510, d100
5510
1
2
0
[2a1(10 1)100]
5510 10a14500
1010 10a1
101 a1
a10 101 (10 1)100
1001
least: $101, greatest: $1001
58. Sna1a2(a31a32)
(a41a42) (a51a52)
a1a2a2a1a3a2a4a3
a1a2a2a1(a2a1) a2a3a2
a3
0
59. APert
100e0.07(15)
$285.77
60. 4x225y2250y525 0
4x225(y210y)525
4x225(y5)2100
2
x
5
2
(y
4
5)2
1
h0, k5, a5, b2, c21
center: (0, 5)
foci: (21
, 5)
vertices: major →(5, 5)
minor →(0, 3) and (0, 7)
61. r
1
6
2
or 0.5
v
5
8
2
or
8
0.5cos
8
i sin
8
0.46 0.19i
62. 2, 1, 35, 3, 02(5) (1)(3) (3)(0)
7
63. x cos 30° y sin 30° 5 0
2
3
x
1
2
y5 0
3
xy10 0
64.
65. Find A.
A90° 19° 32
70° 28
Find a.
cos 19° 32
4
a
.5
4.2 a
Find b.
sin 19° 32
4
b
.5
1.5 b
66. discriminant (3)24(4)(2)
23
Since the discriminant is negative, this indicates
two imaginary roots.
405 Chapter 12
y
x
(0, 3)
(5, 5)
(5, 5)
(0, 7)
(0, 5)
O
y
x
O
6
4
2
2
67. x1
x3
x
2
4
x
2
(x2 3x)
x2
(x3)
1
yx1
68.
A(1, 2), B(3, 0), C(4, 1)
69. a4b15 →a15 4b
4ab15
4(15 4b) b15
60 16bb15
15b45
b3
4a(3) 15
4a12
a3
ab3 (3) or 6
The correct choice is C.
Geometric Sequences and Series
Page 771 Check for Understanding
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)11or 9
a2(3)21or 27
a3(3)31or 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.
12-2
4
1
3
0
1
2
1
4
0
3
2
1
1
0
0
1
64b.
6c. an exponential function
7. ror 6
24(6) 144, 144(6) 864, 864(6) 5184
144, 864, 5184
8. r
3
2
9
2
3
2
2
4
7
,
2
4
7
3
2
8
8
1
,
8
8
1
3
2
2
1
4
6
3
2
4
7
,
8
8
1
,
2
1
4
6
3
9. r
1
7
.8
.2
or 4
28.8(4) 115.2, 115.2(4) 460.8,
460.8(4) 1843.2
115.2, 460.8, 1843.2
10. r
2
7
.1
or 0.3
ana1rn1
a77(0.3)71
0.005103
11. ana1rn1
24 a1(2)51
24 16a1
3
2
a1
12. a3
2
2
.5
or 1.25
a2
1.
2
25
or 0.625
a1
0.8
2
25
or 0.3125
0.3125, 0.625, 1.25
13. ana1rn1
27 1(r)41
27 r3
3 r
1(3) 3, 3(3) 9
1, 3, 9, 27
14. r
0
.
1
5
or 2
Sn
a1
1
a
r
1rn
S9
0.5
1
0
(
.
5(
2
)
2)9
85.5
4
—
2
3
Chapter 12 406
Beginning of Value of
Year Computers
127,500.00
215,125.00
38318.75
44575.31
52516.42
61384.03
1023
Years 456
2500
5000
0
7500
10,000
12,500
15,000
17,500
25,000
27,500
Value
20,000
22,500
15. r1.035
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
Pages 771–773 Exercises
16. r
1
2
0
or 0.2
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
8
20
or 2.5
50(2.5) 125, 125(2.5) 312.5,
312.5(2.5) 781.25
125, 312.5, 781.25
18. r
2
3
or 3
2
9
2(3) 6, 6(3) 18, 18(3) 54
6, 18, 54
26. a58
3
2
51
8
2
1
27. ror
3
4
3
8
—
1
2
407 Chapter 12
19. r or
2
5
—
1
3
0
—
—
—
3
4
—
2
3
5
2
5
1
6
25
,
1
6
25
2
5
6
1
2
2
5
,
6
1
2
2
5
2
5
3
2
1
4
25
1
6
25
,
6
1
2
2
5
,
3
2
1
4
25
20. r
3
.
7
5
or 0.5
1.75(0.5) 0.875, 0.875(0.5) 0.4375,
0.4375(0.5) 0.21875
0.875, 0.4375, 0.21875
21. r
—
3
6
2
—
or 2
62
2
12, 122
122
, 122
2
24
12, 122
, 24
22. r
3
9
3
or
3
3
3
3
3
3
, 3
3
3
1, 1
3
3
3
3
3
, 1,
3
3
23. r
i
1
1
1
or i
i(i) 1, 1(i) i,i(i) 1
1, i, 1
24. r
t
t
5
8
or t3
t2(t3) t1, t1(t3) t4, t4(t3) t7
t1, t4, t7
25.
b
a
2
a
b
2
a
1
b
,
a
1
b
a
b
2
a
1
3
,
a
1
3
a
b
2
a
b
5
,
a
b
5
a
b
2
a
b2
7
,
a
b2
7
a
b
2
a
b3
9
a
1
b
,
a
1
3
,
a
b
5
,
a
b2
7
,
a
b3
9
a6
1
2
3
4
61
2
2
0
4
4
3
8
28. r
0
4
.
.
4
0
or 0.01
a740(0.01)71
4 1011
29. r
1
5
0
or 2
a95
2
91
165
30. 192 a1(4)61
192 1024a1
0.1875 a1
31. 322
a1(2
)51
322
4a1
82
a1
32. 6 a1
1
3
51
6
8
1
1
a1
486 a1
a2486
1
3
or 162
a3162
1
3
or 54
486, 162, 54
33. 0.32 a1(0.2)51
0.32 0.0016a1
200 a1
a2200(0.2) or 40
a340(0.2) or 8
200, 40, 8
34. 81 256r51
2
8
5
1
6
r4
3
4
r
256
3
4
192, 192
3
4
144,
144
3
4
108
256, 192, 144, 108, 81
35. 54 2r41
27 r3
3 r
2(3) 6, 6(3) 18
2, 6, 18, 54
36. 7
4
7
r31
4
4
9
r2
7
2
r
4
7
7
2
2
4
7
, 2, 7
37. ror 3
5
—
5
3
43a. r
5.
5
50
or 1.1
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
43b. S52
5
1
5(1
1
.
.
1
1
)52
$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.
44a.
1
x
1
8
3
0
1
x
8
1
0
3
x
44b.
1
8
1
4
2
1
0
1
z
1
5
1
z
5 z
45. a23(a1)by definition
3(2)
6
r
a
a
2
1
3
Then ana1rn1
So, an(2)(3)n1
46. a11, r2.5, n15
a15 1(2.5)151
372,529
47a. 251
0.
1
0
2
24
$25.05
47b. No; at the end of two years, she will have only
$615.23 in her account.
S24
25.05 25.05
1
0.
1
0
2
24
24
1
1
0.
1
0
2
24
2
1
0
1
z
2
1
5
1
8
2
Chapter 12 408
S5
60
3
5
38. r
1
6
3
5
or 0.2
S665
6
1
5
(0.
0
2
.
)
2
6
81.2448
39. ror
3
2
3
2
1
5
3
5
3
(3)5
1 3
S10
1 1
3
2
10
——
1
3
2
1
5
1
1
,6
2
05
40. r
2
2
3
or 3
S8
1
1
60
3
1
1
60
3
1
1
3
3
160(1
2
3
)
80(1 3
)
41a. The population doubles every half-hour, so r2.
After 1 hour, the number of bacteria is the third
term in the sequence and n12. After 2 hours,
it is the fifth term and n1 4. After 3 hours,
it is the seventh term and n1 6. After t
hours it is the 2t1 term and n1 2t.
btb022t
41b. bt30 22(5)
30,720
41c. 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.
42a. a7a4r3
12 4r3
3r3
33
r
42b. a4a1r41
4 a1
33
3
4 3a1
4
3
a1
a28
4
3
33
281
26,244
2 2(3
)8
1 3
47c. 750
1.5 a11 1
0.
1
0
2
24
24
$30.54 a1
a01
0.
1
0
2
24
$30.54
a0$30.48
The least monthly deposit is $30.48.
a1a1
1
0.
1
0
2
24
24
1
1
0
1
.2
2
4
$615.23
48. r
a
a
2
1
r3
ana1rn1
6561
8
1
1
3n1
(6561)(81) 3n1
(38)(34)3n1
312 3n1
12 n1
13 n
6561 is the 13th term of the sequence.
49. Sn
n
2
[2a1(n1)d]
650
n
2
[2(20) (n1)5]
1300 40n5n25n
05n235n1300
05(n13)(n20)
n13 or n20
Since ncannot be negative, n13 weeks.
50. log11 265
lo
l
g
og
2
1
6
1
5
2.3269
51.
52. A2
B2
32(
52)
34
Since Cis positive, use 34
.
3
34
x
5
34
y
5
34
0
p
5
34
or
5
34
34
, cos f
3
34
, sin f
5
34
tan f
tan f
5
3
f59°
Since cosine is negative and sine is positive,
59° 180° or 121°.
prcos(vf)
5
34
34
rcos(v121°)
53. 3x4y5
y
3
4
x
5
4
xt, y
3
4
t
5
4
54. csc v3
si
1
nv
3
1
3
sin v
5
34
3
34
2
1
7
—
8
1
1
55. Find the amplitude.
A
86
2
36
or 25
Find h.
h
86
2
36
or 61
Find k.
2
k
4
k
2
y25 sin
2
tc61
36 25 sin
2
1 c61
1sin
2
c
2
2
c
c
y25 sin
2
t3.1461
56. Since 43° 90°, consider the following.
bsin A20 sin 43°
13.64
Since 11 13.64, no triangle exists.
57. (n2) 49
n27
Solution set: {nn3 or n3}
The answer is 6.
Infinite Sequences and Series
Pages 780–781 Check for Understanding
1a.
1b. The value of anapproaches 1 as the value of n
increases.
1c. lim
n→
n
n
1
1
12-3
409 Chapter 12
nmn1nm
55 1 or 4(5)(4) or 20
44 1 or 3(4)(3) or 12
33 1 or 2(3)(2) or 6
33 1 or 4 3(4) or 12
44 1 or 5 4(5) or 20
55 1 or 6 5(6) or 30
←least
possible
value
y
x
O
an
n
O
1.0
0.5
12345678910
1d. lim
n→
n
n
1
lim
n→
n
n
lim
n→
n
1
1 0
1
The limits are equal.
2a. See students’ work. Student’s should draw the
following conclusions:
lim
n→
1
2
n0
lim
n→
1
4
n0
lim
n→(1)n1
lim
n→(2)nno limit
lim
n→(5)nno limit
2b. If r1, then lim
n→rn0. If r1, then
lim
n→rn1. If r1 then lim
n→rndoes not exist.
3. Sample answer: 2 4 8
4. Zonta is correct. As napproaches infinity the
expression 2n3 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 2n3 is 1.
5. 0; as n→, 5nbecomes increasingly large and
thus the value
5
1
n
becomes smaller and smaller,
approaching zero. So the sequence has a limit of
zero.
6. lim
n→
5
2n
n2
lim
n→
2
5
n
1
2
n
lim
n→
2
5
n
lim
n→
5
2
n
1
5
2
0 0
As napproaches infinity,
1
2
nbecomes increasingly
large, so the sequence has no limit.
7.
3
7
, lim
n→
3n
7
n
6
lim
n→
3
7
6
7
n
1
lim
n→
3
7
lim
n→
6
7
lim
n→
n
1
3
7
6
7
0 or
3
7
8. 0.7
1
7
0
1
7
00
a1
1
7
0
, r
1
1
0
Sn
7
9
9. 5.1
2
6
5
1
1
0
2
0
6
0
1,0
1
0
2
0
6
,000
a1
1
1
0
2
0
6
0
, r
10
1
00
Sn5
5
1
9
2
9
6
9
5
1
1
4
1
1
1
0
2
0
6
0
——
1
10
1
00
1
7
0
—
1
1
1
0
10. r
3
6
or
1
2
Sn6
——
1
1
2
Chapter 12 410
4
11. ror
1
3
1
4
—
3
4
Sn
3
4
—
1
1
3
1
1
8
12. r
3
3
or 3
The sum does not exist since r3
1.
13. a175, r
2
5
Sn
125 m
Pages 781–783 Exercises
14. lim
n→
7
5n
2n
lim
n→
7
5
n
1
2
5
lim
n→
7
5
lim
n→
n
1
lim
n→
2
5
7
5
0
2
5
or
2
5
15. lim
n→
n3
n
22
lim
n→n
n
2
.
lim
n→
n
2
lim
n→2
n
1
2 0 or 0, but as napproaches
infinity, nbecomes increasingly large, so the
sequence has no limit.
16. lim
n→
6n
3
2
n
25
lim
n→
6
3
3
5
n2
lim
n→2 lim
n→
5
3
lim
n→
n
1
2
2
5
3
0 or 2
17. lim
n→
9n3
2n
53
n2
lim
n→
9
2
2
5
n2
n
1
3
lim
n→
9
2
lim
n→
5
2
lim
n→
n
1
2
lim
n→
n
1
3
9
2
5
2
0 0 or
9
2
18. lim
n→
(3n4
n
)
2
(1 n)
lim
n→
3n2
n
2n4
lim
n→3
n
1
n
4
2
lim
n→(3) lim
n→
n
1
lim
n→4 lim
n→
n
1
2
3 0 4 0 or 3
19. Dividing by the highest powered term, n2, we find
lim
n→which as napproaches infinity
simplifies to
8
0
0
0
0
8
0
. Since this fraction is
undefined, the limit does not exist.
8
n
5
n
2
2
——
n
3
2
n
2
75
1
2
5
20. lim
n→
2
4
n3
3
3
n
n
2
n2
5
lim
n→
lim
n→
4
2
0
3
3
0
0
5
0
0
or 0
21. As n→, 3nbecomes increasingly large and thus
the value
3
1
n
becomes smaller and smaller,
approaching zero. So the sequence has a limit of
zero.
22. Dividing by the highest powered term, n, we find
lim
n→ lim
n→
(
n
2)n
which as napproaches infinity
lim
n→ 1, but lim
n→
(
n
2)n
has no limit since
o2o 1.
23. lim
n→
5n
n
(
2
1)n
lim
n→
5
n
n
2
lim
n→
(
n
1
2
)n
lim
n→
n
5
lim
n→
(
n
1
2
)n
lim
n→
(
n
1
2
)n
As nincreases, the value of the numerator
alternates between 1 and 1. As napproaches
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.
24. 0.4
1
4
0
1
4
00
…
a1
1
4
0
, r
1
1
0
Sn
1
4
0
—
1
1
1
0
1
n
4
1
1
n
4
1
(
n
2)n
—
n
4
1
lim
n→4lim
n→
n
1
3
lim
n→3 lim
n→
n
1
2
lim
n→
n
1
lim
n→2 lim
n→3 lim
n→
n
1
lim
n→5 lim
n→
n
1
3
n
4
3
n
3
2
n
1
2
n
3
n
5
3
n
4
3
3
n
n
3
n
n
2
3
2
n
n
3
3
3
n
n
3
2
n
5
3
27. 6.2
5
9
6
1
2
0
5
0
9
0
1,0
2
0
5
0
9
,000
…
a1
1
2
0
5
0
9
0
, r
10
1
00
Sn6
1
2
0
5
0
9
0
——
1
10
1
00
411 Chapter 12
6
2
9
5
9
9
9
6
2
7
7
28. 0.1
5
1
1
0
5
0
10
1
,0
5
00
…
a1
1
1
0
5
0
, r
1
1
00
Sn
1
1
0
5
0
—
1
1
1
00
1
9
5
9
or
3
5
3
29. 0.26
3
1
2
0
1
6
0
3
00
100
6
,
3
000
…
a1
1
6
0
3
00
, r
1
1
00
Sn
1
5
1
6
0
3
00
—
1
1
1
00
1
5
9
6
9
3
0
1
2
1
9
0
30. The series is geometric, having a common ratio of
0.1. Since this ratio is less than 1, the sum of the
series exists and is
2
9
.
31. r
1
1
2
6
or
3
4
Sn16
1
3
4
64
32. r
7
5
.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.
33. r
1
5
0
or
1
2
Sn10
1
1
2
20
34. The series is arithmetic, having a general term of
7 n. Since lim
n→7 ndoes not equal zero, this
series has no sum.
35. ror 2
1
4
—
1
8
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.
4
9
25. 0.5
1
1
5
0
1
0
10
5
,0
1
00
…
a1
1
5
0
1
0
, r
1
1
00
Sn
5
9
1
9
or
1
3
7
3
26. 0.3
7
0
1
3
0
7
0
0
0
1,0
3
0
7
0
0
,000
…
a1
1
3
0
7
0
0
0
, r
10
1
00
Sn
3
9
7
9
0
9
or
1
2
0
7
1
3
0
7
0
0
0
——
1
10
1
00
1
5
0
1
0
—
1
1
1
00
36. ror
1
6
1
9
—
2
3
41b. lim
n→
2n
n
2
1
2n
n
2
1
lim
n→
n2(2n1) n2(2n1)
(2n1)(2n1)
Chapter 12 412
Sn
2
3
——
1
1
6
4
7
37. ror
2
3
4
5
—
6
5
Sn
6
5
—
1
2
3
3
3
5
38. r
1
5
or
5
5
Sn
5
1
5
5
1
5
5
—
1
5
5
5
1
5
5
5
1
1
2
5
5
4
5
(5
1)
39. r
4
8
3
or
2
3
Sn8
1
2
3
1
2
3
——
1
2
3
8
1
2
3
81
2
3
——
1
3
4
321
2
3
32 163
40a. a135 r
2
5
a235
2
5
or 14
a314
a414
2
5
or 5.6
a55.6
35, 14, 14, 5.6, 5.6
40b. Sn35
14
1
2
5
14
1
2
5
35
2
3
0
7
3
0
81
2
3
m or about 82 m
41a. The limit of a difference equals the difference of
the limits only if the two limits exist. Since
neither lim
n→
2n
n
2
1
nor lim
n→
2n
n
2
1
exists, this
property of limits does not apply.
lim
n→
lim
n→
4n
2
2
n2
1
lim
n→
n
2
2
n2
—
4
n
n
2
2
n
1
2
2n3n22n3n2
4n21
lim
n→
1
2
42a. 12 4 3
r
33
a4a1r41
4 a1
33
3
4 a1(3)
4
3
a1
42b. a28 a1r281
4
3
33
27
4
3
(39)
4(38)
26,244
43. No; if nis even, lim
n→cos
n
2
1
2
, but if nis odd,
lim
n→cos
n
2
1
2
.
44a. After 2 hours,
1
2
Dexists. After 4 hours,
1
2
1
2
Dor
1
4
Dexists. After 6 hours and before the second
dose,
1
2
1
2
1
2
Dor
1
8
Dexists.
44b. a1D, r
1
8
SnD
1
1
8
n
——
1
1
8
2
4
n
1
2
8
7
D1
1
8
n
44c. lim
n→SnS
S
1
a
1
r
D
1
1
8
8
7
D
44d. 350
8
7
D
306.25 D
The largest possible dose is 306.25 mg.
45a. Aside of the original square measures
2
4
0
or
5feet. Half of 5 feet is
5
2
feet.
5
2
2
5
2
2s2
5
4
0
s2
5
2
2
s
Perimeter 4
5
2
2
or 102
feet.
45b. a120, r
10
2
0
2
or
2
2
S20
1
2
2
51. vy125 sin 20°
42.75 miles
vx125 cos 20°
117.46 miles
52. cos 112.5° cos
22
2
5°
1c
o
2
s225
°
1
2
2
2
413 Chapter 12
1
2
2
——
1
2
2
20
1
2
2
40 202
ft or about 68 ft
46a.
15.0, 12.9, 11.2, 9.7, 8.4, 7.2, 6.2, 5.4, 4.7, 4.0
46b. The 2000–2001 school year corresponds to the
9th term of the sequence, 4.7. The model is 0.3
below the actual statistic.
46c. The 2006–2007 school year would correspond to
the 15th term of the sequence.
17.3032181(0.864605)15 2.0
46d. Yes; as n→, 17.3032181(0.864605)n→0.
46e. No, the number of students per computer must
be greater than zero.
47. 3
2
3
2, 2
2
3
1
1
3
, 1
1
3
2
3
8
9
,
8
9
2
3
1
2
6
7
2, 1
1
3
,
8
9
,
1
2
6
7
48. a16 1.5 (16 1)0.5
9
49. x24y212x16y16
(x6)24(y2)24
(x
4
6)2
(y
1
2)2
1
h6, k2, a2, b1, c5
center: (6, 2)
foci: (6 5
, 2)
vertices: (8, 2) and (4, 2)
50.
20 102
1
2
4
n17.3032181(0.864605)n
1 15.0
2 12.9
311.2
49.7
58.4
67.2
76.2
85.4
94.7
10 4.0
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
2468
125 miles
20˚
2
4
2
53.
54. possible values of p: 1, 2
possible values of q: 1, 2, 4, 8
possible rational zeros,
p
q
: 1, 2,
1
2
,
1
4
,
1
8
55. If b1, then 4b26 30 which is divisible by 2,
5, and 6.
If b11, then 4b26 70, which is divisible
by 7.
4b26 is not divisible by 4 since 4 divides 4b
evenly, but does not divide 26 evenly. The correct
choice is B.
Graphing Calculator Exploration:
Continued Fractions
Page 784
1. 1.618181818
2. N24
3. x1
1
x
x2x1
x2x1 0
x(1) (1)2
4(1
)(1)
2(1)
12-3B
2
2
2
y
1
180˚360˚
180˚
1
O
x
1
2
5
Since the sum 1 cannot be
1
1
1
1
negative, the value of 1 is
1
2
5
.
1
1
1
1
4. Let x3 , then x3
1
x
.
1
3
1
3
11. Answers will vary. Sample answers: A1, B14;
A4, B1.
A2B: If A1 and B14, 121
4
15
.
If A4 and B1, 42(
1)
15.
Convergent and Divergent Series
Pages 790–791 Check for Understanding
1a. See students’ work.
1b. See students’ work.
1c. See students’ work.
1d. In a given trigonometric series where r1,
each succeeding term is larger than the one
preceding it. Therefore, the series approaches
and thus does not converge.
2. As n→, S→6.
3a.
3b. convergent
3c.
n
3n
2
3d. We can use the ratio test to determine whether
the series is indeed convergent.
an
n
3n
2
and an1
(n
3
n
1
1)2
rlim
n→
rlim
n→
(n
3
n
1
1)2
n
3n
2
rlim
n→
n2
3
2
n
n
21
rlim
n→
r
1
3
Since r1, the series is convergent.
4. Consider the infinite series an
1. If the lim
n→an0, the sum of the series does not
exist, and thus the series is divergent. If the
lim
n→an0, the sum may or may not exit and
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 r1 and diverges for r1.
4. Ratio test: the series converges for r1 and
diverges for r1. If r1, the test fails. This
test can only be used if all the terms of the
1
n
2
n
1
3
(n
3
n
1
1)2
—
n
3n
2
12-4
Chapter 12 414
Solve for x.
x3
1
x
x23x1
x23x1 0
x3 (3)2
4(1
)(1)
2(1)
Solve for x.
xA
1
x
x2Ax 1
x2Ax 1 0
xA(A)2
4(1
)(1)
2(1)
Since the sum of positive numbers must remain
positive, x
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.
9. xA2AB
2A
2A
B
AA2
4
2
AA2
4
2
xA2A
x
B
A
(xA)22A(xA) B
x22Ax A22Ax 2A2B
x2A2B
x
A2
B
Since the sum Acannot be
B
2A
2A
B
negative, the value of Ais
A2
B
.
10. They will be opposites.
B
2A
2A
B
sn
n
O
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
12345678
3
2
13
Since the sum of positive numbers must remain
positive, x
3
2
13
.
5. The output and the decimal approximation are
equal.
6. Let xA, then xA
1
x
.
1
A
A
1
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.
5. an
2
n
n
, an1
n
2n
1
1
rlim
n→
a
a
n
n1
lim
n→
n
2n
1
1
2
n
n
12a. The series is geometric where r
1
9
5
0
0
0
0
or
3
5
.
S10
1500 1500
3
5
10
———
1
3
5
415 Chapter 12
lim
n→
(n
n
2n1
)
1
2n
lim
n→
(n
2
n
1)
lim
n→
1
2
2
1
n
1
2
convergent
6. an
4n
4
n
1
, an1
4(n
4(
n
1)
1
)
1
rlim
n→
4(n
4(
n
1)
1
)
1
——
4n
4
n
1
lim
n→
(4n
(4
n
4)(
3
4
)
n
(4
n)
1)
lim
n→
16
1
n
6
2n
2
12
1
n
2
n
4
lim
n→
16
n
n
2
2
1
n
2
2
n
——
16
n
n
2
2
1
n
2
2
n
n
4
2
1
1
6
6
or 1
test provides no information
7. The general term is
(n
n
1)
.
n
n
1
n
1
for all n, so divergent
8. The series is arithmetic, so it is divergent.
9. The general term is
2
1
n2
.
2
1
n2
n
1
for all n, so convergent
10. an
n
1
2n
, an1
(n1
1
)2n1
r
(n1
1
)2n1
——
n
1
2n
lim
n→
(n
n
1
2
)
n
2n1
lim
n→
2(n
n
1)
lim
n→
n
n
——
2
n
n
n
1
2(1
1
0)
1
2
convergent
11. The series is geometric where r
3
4
.
Since
3
4
1, it is convergent.
3727 m
12b. S1500
1
3
5
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 Exercises
13. an
3
4
n
, an 1
3n
4
1
rlim
n→
3n
4
1
—
3
4
n
lim
n→
4
4
3
n
3n
3
1
3
convergent
14. an
5
2
n
n
, an1
5(
2
n
n
1
1)
rlim
n→
5(
2
n
n
1
1)
—
5
2
n
n
lim
n→
2n
2n
(5
2
n
5n
5)
lim
n→
5n
10
n
5
lim
n→
10
n
n
—
5
n
n
n
5
1
5
0
or 2
divergent
15. an
n
2n
2
, an1
(n
2
n
1
1
)2
rlim
n→
(n
2
n
1
1
)2
——
n
2n
2
lim
n→
2n(n
2n
2
2
2
n
n
2
1)
lim
n→
n2
2
2
n
n
2
1
lim
n→
2
n
n
2
2
——
n
n
2
2
2
n
n
2
n
1
2
2
divergent
16. an
(n1)
2
(n2)
, an1
(n2)
2
(n3)
rlim
n→
(n2)
2
(n3)
——
(n1)
2
(n2)
24. The general term is
2n
1
1
.
2n
1
1
n
1
for all n, so convergent
25. The series is geometric where r
3
4
.
Since
3
4
1, it is convergent.
26. The general term is
2
2
n
n
1
1
.
2
2
n
n
1
1
n
1
for all n, so divergent
27. The general term is
5
1
n2
.
5
1
n2
n
1
for all n, so convergent
28. The general term is
1
n
.
1
n
n
1
for all n, so divergent
29. The series is arithmetic, so it is divergent
30. an
2
2
nn
11
, an1
2(n
2
n
1)21
rlim
n→
2(n
2n
1)
21
——
2
2
nn
11
Chapter 12 416
lim
n→
2
2
(
(
n
n
2
2
3
5
n
n
2
6
)
)
lim
n→
n
n
2
2
3
n
n
2
n
2
2
——
n
n
2
2
5
n
n
2
n
6
2
1
test provides no information
17. an, an12(n1) 1
1 2 ... [2(n1) 1]
2n1
1 2 ... (2n1)
rlim
n→2n1
1 2 ... (2n1)
2(n1) 1
1 2 ... [2(n1) 1]
lim
n→
lim
n→
(2n
1
1)(2n)
lim
n→
4n21
2n
0
convergent
18. an
12
5
n
... n
, an15n1
1 2 ... (n1)
(2n1)(1 2 ... (2n1))
(2n1)(1 2 ... (2n1))
rlim
n→
12
5
n
... n
lim
n→
lim
n→
n
5
1
0
convergent
19. an
2n2
2
(
n
n1)
, an12(n1) 2(n2)
2n1
5n5(1 2 ... n)
5n(1 2 ... (n1))
5n1
1 2 ... (n1)
rlim
n→
2n2
2
(n
n
1)
lim
n→
2(n1) 2(n2) 2n
2n2(n1) 2n2
2(n1) 2(n2)
2n1
lim
n→
n
2
n
2
lim
n→
2
n
n
2
2
n
1
2
convergent
20. The general term is
(2
1
n)2
or
4
1
n2
.
4
1
n2
n
1
for all n, so convergent
21. The general term is
n31
1
.
n31
1
n
1
for all n, so convergent
22. The general term is
n
n
1
.
n
n
1
n
1
for all n, so divergent
23. The general term is
n
5
2
.
n
5
2
n
1
for all n, so divergent
lim
n→
lim
n→
2
4
n
n
1
2
lim
n→
2
n
n
n
1
—
4
n
n
n
2
(2n1) 2n2
(2n1) 2n2 2
2
4
or
1
2
convergent
31a. No, MagicSoft let a11,000,000 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 a1700,000 and
r0.70.
S
1
70
0,
0
0
.
0
7
0
0
$2.3 million
32. the harmonic series: 1
1
2
1
3
1
4
33a. Culture A: 1400 cells, Culture B; 713 cells
Culture A generates an arithmetic sequence
where a11000, d200 and n8
a81000 (8 1)200
2400
Only considering cell growth, there are
2400 1000 or 1400 new cells.
Culture B generates a geometric sequence where
a11000 and r1.08
a81000(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.
34a.
2
2
n
n
1
1
34b. Slim
n→Snlim
n→
2
2
n
n
1
1
lim
n→
2(2n
2n
1)
lim
n→2
2
2
n
2 seconds
35a. When the minute hand is at 4, 20 minutes have
passed. This is 20
6
1
0
or
1
3
hour.
35b.
1
3
the distance between 4 and 5 is
1
3
(5) or
5
3
minutes. This is
5
3
6
1
0
or
3
1
6
hour.
35c.
1
3
3
1
6
(5)
6
3
5
6
minutes
6
3
5
6
5
3
3
5
6
minute
3
5
6
6
1
0
4
1
32
hour
1
3
3
1
6
4
1
32
(5)
7
4
8
3
5
2
minutes
7
4
8
3
5
2
6
3
5
6
4
5
32
minute
4
5
32
6
1
0
51
1
84
hour
4
1
32
,
51
1
84
35d. The sequence
1
3
,
3
1
6
,
4
1
32
,
51
1
84
is geometric
where r
1
1
2
.
lim
n→SnS
1
a
1
r
1
3
—
1
1
1
2
41. AB
u
5 8, 1 (3)
3, 2
42. List all cubes from 1 to 200. There
are five. The correct choice is E.
Page 793 Mid-Chapter Quiz
1. a12 11 (19 1)(2)
25
2. S20
2
2
0
[2(14) (20 1)6]
860
3. 189 56r41
2
8
7
r3
3
2
r
56
3
2
84, 84
3
2
126
56, 84, 126, 189
4. r
3
6
or 2
S8
3
1
3
(
(
2
2
)
)8
255
5. lim
n→
n2
n
22
n
1
5
lim
n→
1
6. a1250, r0.55
a2250(0.55) or 137.5
a3137.5
a4137.5(0.55) or 75.625
a575.625
Sn250
1
1
37
0
.
.
5
55
1
1
37
0
.
.
5
55
861 ft
7. a1
2
1
5
, r
1
1
0
S
2
1
5
—
1
1
1
0
n
n
2
2
2
n
n
2
n
5
2
——
n
n
2
2
n
1
2
417 Chapter 12
nn
3
11
28
327
464
5 125
1
4
1
hour
The hands will coincide at 4
1
4
1
o’clock,
approximately 21 min 49 s after 4:00.
36. lim
n→
3
4
n
n
2
2
2
5
n
lim
n→
4
n
n
2
2
n
5
2
—
3
n
n
2
2
2
n
n
2
4
3
37. ror 2
2
2
4
2
5
8. an
12
10
n
n
, an1
12…
10n
(n
11)
rlim
n→
12
1
0
…
nn
lim
n→
lim
n→
n
1
0
1
As n→,
n
1
0
1
→. Since r1, the series is
divergent.
9. The series is geometric where r
1
3
.
Since r1, it is convergent.
10n[1 2 … (n1)]
10n10(1 2 … n)
1 2 … (n1)
10n1
a92
2
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 (v30°)
1
2
r(cos vcos 30° sin v sin 30°)
1
2
2
3
r cos v
1
2
rsin v
0
2
3
x
1
2
y
1
2
0 3
xy1
10. 5001
0.
4
12
515
a1515, r1.03 n4
S4
$2154.57
Sigma Notation and the nth term.
Pages 797–798 Check for Understanding
1. The series 4 6 8 10 12 can be
represented by
4
n0
2n4 or by
5
n1
2n2.
2a. Sample answer: (1)n1
2b. Sample answer: (1)n
3a. 9; 2, 3, 4, 5, 6, 7, 8, 9, 10
3b. tba1
3c. tba1
3 (2) 1
6
3d.
3
k2
k
1
3
2
1
3
1
1
3
0
1
5
1
1
3
2
1
3
3
1
3
1
1
1
2
1
3
1
4
1
5
1
6
1
1
2
1
3
1
4
1
5
1
5
There are 6 terms.
4.
6
n1
(n3) (1 3) (2 3) (3 3)
(4 3) (5 3) (6 3)
(2) (1) 0 1 2 3
3
5.
5
k2
4k4 2 4 3 4 4 4 5
8 12 16 20
56
6.
4
a0
2
1
a
2
1
0
2
1
1
2
1
2
2
1
3
2
1
4
1
1
2
1
4
1
8
1
1
6
1
1
1
5
6
7.
p0
5
3
4
p5
3
4
05
3
4
15
3
4
25
3
4
3
5
3
4
10
5
1
4
5
4
1
5
6
1
6
3
4
5
5
3
4
Sn
20
8.
5
n1
5n9.
3
k0
(3k1)
10.
4
n1
(8 6n)11.
n2
3
1
2
n
12.
n1
(3)n
5
1
3
4
12-5
515 515(1.03)4
1 1.03
13a.
60
n1
389(0.63)n1
S60
1051 ft
13b. S
1
38
0
9
.63
1051 ft
Pages 798–800 Exercises
14.
4
n1
(2n7) (2 1 7) (2 2 7) (2 3 7)
(2 4 7)
(5) (3) (1) 1
8
15.
5
a2
5a5(2) 5(3) 5(4) 5(5)
10 15 20 25
70
16.
8
b3
(6 4b) (6 4 3) (6 4 4) (6 4 5)
(6 4 6) (6 4 7) (6 4 8)
(6) (10) (14) (18)
(22) (26)
96
17.
6
k2
(kk2) (2 22) (3 32) (4 42)
(5 52) (6 62)
6 12 20 30 42
110
18.
8
n5
n
n
4
5
5
4
6
6
4
7
7
4
8
8
4
5 3
7
3
2
12
1
3
19.
8
j4
2j2425262728
16 32 64 128 256
496
20.
3
m0
3m1301311321331
1
3
1 3 9
13
1
3
21.
3
r1
1
2
4r
1
2
41
1
2
42
1
2
43
4
1
2
16
1
2
64
1
2
85
1
2
22.
5
i3
(0.5)i(0.5)3(0.5)4(0.5)5
8 16 32
56
23.
7
k3
k! 3! 4! 5! 6! 7!
6 24 120 720 5040
5910
24.
10
p0
4(0.75)p4(0.75)04(0.75)14(0.75)2
4(0.75)3 4(0.75)
4 3 2.25 1.6875 4(0.75)
S
1
4
0.75
16
389 389(0.63)60
1 0.63
Chapter 12 418
25.
n1
4
2
5
n4
2
5
14
2
5
24
2
5
3 4
2
5
8
5
1
2
6
5
1
3
2
2
5
4
2
5
S
8
5
—
1
2
5
47b. 0 1(1 x) 2(2 x) 3(3 x) 4(4 x)
5(5 x) 25
1 x4 2x9 3x16 4x25 5x25
55 15x25
15x30
x2
48a. false;
7
k3
3k3334 37
9
b7
3b373839
Since there are two 37terms,
7
k3
3k
9
b7
3b
9
a3
3a.
48b. true;
8
n2
(2n3) 1 3 13 49
9
m3
(2m5) 1 3 13 49
Since 49 49,
8
n2
(2n3)
9
n3
(2m5).
48c. true; 2
7
n3
n218 32 98 270
7
n3
2n218 32 98 270
Since 270 270, 2
7
n3
n2
7
n3
2n2.
48d. false;
10
k1
(5 n) 6 7 15 105
9
p0
(4 p) 4 5 13 85
Since 85 105,
10
k1
(5 n)
9
p0
(4 p).
49a. 6!
49b. 5! or 120
49c. 4! or 24, “LISTEN”
50a. 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
50c.
8
n1
n21222324252627282
1 4 9 16 25 36 49 64
204
51. The general term is
n
3
2
.
n
3
2
n
1
for all n, so divergent
52. 0.42(21) 8.82 liters
If with each stroke 20% is removed, then 80%
remains.
a121(0.80) 16.8
a216.8(0.80) 13.44
a313.44(0.80) 10.752
a410.752(0.80) 8.6016
It will take 4 strokes for 42% of the air to remain.
419 Chapter 12
8
3
or 2
2
3
26.
5
n2
nin(2 i2) (3 i3) (4 i4) (5 i5)
(1) (3 i) (5) (5 i)
14
27.
4
k1
(3k3) 28.
4
k0
4k
29.
12
k4
2k30.
3
k0
(2)3k
31.
4
k1
2 5k32.
3
k0
(13 4k)
33.
10
k2
5k
1
1
34.
k1
2k
2
k
1
35.
k2
(1)kk236.
k0
k
5
!
37.
n0
(1)n1
3
2
2
n
38.
k1
(k
k
1)!
39.
k1
2kk
3
40.
k2
k
3
2
k
k!
1
41.
k11
42.
(a
a!
2)!
a(a
(a
1
)(a
2)
!
2)!
a(a
1
1)
43.
(
(
a
a
1
2
)
)
!
!
a(a1)(a1)
44.
(a
(
a
b
b)!
1)!
a
45. 43.64
46a.
n1
500,000(0.35)n
46b. S
1
17
5,
0
0
.
0
3
0
5
269,239 people
46c.
2
5
6
0
9
0
,
,
2
0
3
0
0
0
53.8%
46d. The ad agency assumes that the people who buy
the tennis shoes will be satisfied with their
purchase.
47a. (x3) (x6) (x9) (x12) (x15)
(x18) 3
6x63 3
6x60
x60
(ab)(ab1)!
(ab1)!
(a1)(a)(a1)(a2)!
(a2)!
2
k
2
3k!
53. 322
a1
2
51
322
4a1
82
a1
82
2
16, 16
2
162
,
162
2
32
82
, 16, 162
, 32
54. log10 0.001 3
55. x2y2Dx Ey F0
(0, 9) →81 9EF0
(7, 2) →49 4 7D2EF0
(0, 5) →25 5EF0
9EF81 9(4) F81
5EF25 F45
14E56 53 7D2(4) 45 0
E 47D0
D0
x2y24y45 0
x2(y2)249
56. (2
i)(42
i)8 52
ii2
7 5i2
57. vy59 sin 63°
52.57 ft/s
vx59 cos 63°
26.79 ft/s
58. d1
2x1
3y
1
1
3
9
, d2
x1
4y
1
1
7
4
2x1
3y
1
1
3
9
x1
4y
1
1
7
4
217
x1317
y1917
13
x1413
y1413
217
13
x
413
317
y917
413
0
59.
5
9
m
m
2
3
3(5 m) 2(9 m)
15 3m18 2m
m3The correct choice is D.
The Binomial Theorem
Pages 803–804 Check for Understanding
1a. n0: 1
n1: 1 1 or 2
n2: 1 2 1 or 4
n3: 1 3 3 1 or 8
n4: 1 4 6 4 1 or 16
n5: 1 5 10 10 5 1 or 32
1, 2, 4, 8, 16, 32
1b. 2n
2a. The second term of (xy)3is 3x2y.
It is negative.
The second term of (xy)4is 4x3y.
It is negative.
The second term of (xy)5is 5x4y.
It is negative.
2b. The third term of (xy)3is 3xy2. It is positive.
The third term of (xy)4is 6x2y2. It is positive.
The third term of (yy)5is 10x3y2. It is positive.
12-6
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
yis 12 7 or 5.
To find the coefficient of the term use the formula
(xy)n
n
r0
r!(n
n
!
r)!
xnryr.
Evaluate the general term for n12 and r5.
5
1
!
2
7
!
!
x7y5792x7y5.
5. c55c4d10c3d210c2d35cd4d5
6. (a3)6a63 6a5
32
1
6
2
5
a4
33
1
6
2
5
3
4
a3
34
1
6
2
5
3
4
4
3
a2
a
a618a5135a4540a31215a2
1458a729
7. (5 y)353(y)03 52(y)
32
2
5
1
(y)2
125 75y15y2y3
8. (3p2q)4(3p)4(2q)04(3p)3(2q)1
43(3
2
p)
2
1
(2q)2
81p4216p3q216p2q296pq3
16q4
9.
r!(7
7
!
r)!
(a)7r(b)r
5!(7
7
!
5)!
(a)75(b)5
a2b5
21a2b5
10.
r!(9
9
!
r)!
(x)9r
3
r
3!(9
9
!
3)!
(x)93
3
3
x633
2523
x6
11. (H T)5H55H4T 10H3T210H2T35HT2T5
11a. 111b. 10
11c. 1 5 or 6 11d. 10 10 5 1 or 26
Pages 804–805 Exercises
12. (ab)8a88a7b28a6b256a5b370a4b4
56a3b528a2b68ab7b8
13. (n4)6n624n5240n41280n33840n2
6144n4096
14. (3cd)481c4108c3d54c2d212cd3d4
15. (2 a)9512 2304a4608a25376a3
4032a42016a5672a6144a7
18a8a9
9 8 7 6 5 4 3 2 1
3 2 1 6 5 4 3 2 1
7 6 5 4 3 2 1
5 4 3 2 1 2 1
4 3 2 1(3p)0(2q)4
4 3 2 1
4 3 2(3p)(2q)3
3 2 1
3 2 1 50(y)3
3 2 1
366 5 4 3 2 1
1 2 3 4 5 6
356 5 4 3 2
1 2 3 4 5
Chapter 12 420
59 ft/s
63˚
421 Chapter 12
16. (d2)7d7207 d621
76
2
d
1
522
76
3
5
2
d
1
423
d714d684d5280d4560d3
672d2448d128
17. (3 x)535(x)05 34(x1)1
54
2
3
3
1
(x)2
243 405x270x290x315x4x5
18. (4ab)4(4a)4(b)04(4a)3(b)1
43
2
(4
a
1
)2(b)2
43
3
2
2
(4
a
1
)1(b)3
256a4256a3b96a2b216ab3b4
19. (2x3y)3(2x)3(3y)03(2x)2(3y)1
32(2
2
x
)(
1
3y)2
8x336x2y54xy227y3
20.
3m2
4(3m)4
2
04(3m)3
2
1
43(3
2
m
)2
1
(
2
)2
4 3 2(3m)1(2
)3
3 2 1
3 2 1(2x)0(3y)3
3 2 1
4 3 2 1(4a)0(b)4
4 3 2 1
5 4 3 2 1 30(x)5
5 4 3 2 1
5 4 3 2 31(x)4
4 3 2 1
5 4 3 32(x)3
3 2 1
7 6 5 4 3 2 1 d027
7 6 5 4 3 2 1
7 6 5 4 3 2 d126
6 5 4 3 2 1
7 6 5 4 3 d225
5 4 3 2 1
7 6 5 4 d324
4 3 2 1
24. (p2q)8(p2)8(q)08(p2)7(q)1
87
2
(p
2
1
)6(q)2
87
3
6
2
(p
2
1
)5(q)3
p16 8p14q28p12q256p10q3
70p8q4 56p6q528p4q68p2q7
q8
25. (xy 2z3)6(xy)6(2x3)06(xy)5(2z3)1
65(x
2
y)
4(
1
2z3)2
x6y612x5y5z360x4y4z6
160x3y3z9 240x2y2z12
192xyz15 64z18
26.
4!(9
9
!
4)!
(x)94(y)4x5y4
126x5y4
27.
3!(8
8
!
3)!
(a)83
2
3
a522
1122
a5
28.
3!(7
7
!
3)!
(2a)73(b)335 16a4(b3)
560a4b3
29.
6!(9
9
!
6)!
(3c)96(2d)684 27c364d6
145,152c3d6
30.
7!(1
1
0
0
!
7)!
(
1
2
x)107(y)7120
1
8
x3(y7)
15x3y7
31.
5!(1
1
1
1
!
5)!
(2p)115(3q)5462 64p6(243q5)
7,185,024p6q5
32. The middle term is the fifth term.
4!(8
8
!
4)!
x
84
y
470x2y2
33. (MW)8M88M7W28M6W256M5W3
70M4W456M3W528M2W6
8MW 7W8
70 56 28 8 1 or 163
34. Sample answer: Treat abas a single term and
expand [ab) c]12 using the Binomial
Theorem. Then evaluate each (ab)nterm in the
expansion using the Binomial Theorem.
35. (T F)12 T12 12T11F 6610F2220T9F3
495T8F4792T7F5924T6F6
792T5F7495T4F8220T3F9
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 n6. Use the
8 7 6 5 4 3 2 1
3 2 1 5 4 3 2 1
9 8 7 6 5 4 3 2 1
4 3 2 1 5 4 3 2 1
6 5 4 3 2 1(xy)0(2z3)6
6 5 4 3 2 1
6 5 4 3 2(xy)1(2z3)5
5 4 3 2 1
6 5 4 3(xy)2(2z3)4
4 3 2 1
6 5 4(xy)3(2z3)3
3 2 1
8 7 6 5 4 3 2 1(p2)0(q)8
8 7 6 5 4 3 2 1
8 7 6 5 4 3 2(p2)1(q)7
7 6 5 4 3 2 1
8 7 6 5 4 3(p2)2(q)6
6 5 4 3 2 1
8 7 6 5 4(p2)3(q)5
5 4 3 2 1
8 7 6 5(p2)4(q)4
4 3 2 1
421 Chapter 12
81m41082
m3108m2
242
m4
21.
c
1
6
c
6(1)06
c
5(1)1
65(
2
c
)
1
4(1)2
6 5 4(c
)3(1)3
321
4 3 2 1(3m)0(2
)4
4 3 2 1
6 5 4 3(c
)2(1)4
4321
6 5 4 3 2(c
)1(1)5
54321
c36c2c
15c220cc
15c
6c
1
22.
1
2
n2
1
2
n
5(2)05
1
2
n
4(2)1
5 4
1
2
n
3(2)2
2 1
6 5 4 3 2 1(c
)0(1)6
6 5 4 3 2 1
5 4 3 2
1
2
n
1(2)4
4 3 2 1
5 4 3
1
2
n
2(2)3
3 2 1
5 4 3 2 1
1
2
n
0(2)5
5 4 3 2 1
3
1
2
n5
5
8
n45n320n240n32
23.
3a
2
3
b
4(3a)4
2
3
b
04(3a)3
2
3
b
1
4 3 2(3a)1
2
3
b
3
3 2 1
4 3(3a)2
2
3
b
2
2 1
4 3 2 1(3a)0
2
3
b
4
4 3 2 1
81a472a3b24a2b2
3
9
2
ab3
1
8
6
1
b4
Binomial Theorem to find the 4th term for the
expansion of 3x
4
1
x
6.
3
6
!3
!
!
(3x)3
4
1
x
3
1
1
3
6
5
37a. Sample answer: 1 0.01
37b. Sample answer: 1.04060401
(1 0.01)4144 130.0116 120.012
4 110.0130.014
1.04060401
37c. 1.04060401; the two values are equal.
38.
7
x2
5 2k(5 2 2) (5 2 3) (5 2 4)
(5 2 5) (5 2 6) (5 2 7)
1 (1) (3) (5) (7) (9)
24
39. an
2
n
n
!
an1
(n
2
n
1
1
)!
rlim
n→
lim
n→
2n
2n
(n
2
1
n
)
!
n!
lim
n→
n
2
1
0
convergent
40. This is a geometric series where r
1
2
.
S
1
1
3
41. PnP
1(1
i
i)n
150,000 P
150,000 P(136.283491)
$1100.65 P
42. MK
[4 (2)]i
u
(8 6)j
u
(2 3)k
u
6i
u
2j
u
5k
u
43. srv
0.25 r
4
r0.3183098862 mi
r0.3183098862(5280)
1681 feet
44. Test all answer choices that are prime integers.
You can eliminate answer choice C.
A: 3(2)
?10
?
5
6
(2)
6
?10
?
5
3
; false
B: 3(3)
?10
?
5
6
(3)
9
?10
?
5
2
; false
D: 3(11)
?10
?
5
6
(11)
33
?10
?
5
6
5
; true The correct choice is D.
1
1
0
1
.0
2
8
12(30)
———
0
1
.0
2
8
2
3
—
1
1
2
(n
2
n
1
1
)!
——
2
n
n
!
Special Sequences and Series
Page 809 Graphing Calculator Exploration
1. Sample answer (without zooming): 2.4 x2.4
2. Sample answer for greatest difference: about 0.08;
The least difference is 0.
3. Sample answer: 3.4 x3.4; sample answer:
about 0.15; 0
4. Sample answer: 3.8 x3.8; sample answer:
about 0.05; 0
5. larger
6.
1
x1
1
1
!
Pages 811–812 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 x1.5
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. an1anan1for n2
5. ln (7) ln (1) ln (7)
i1.9459
6. ln (0.379) ln (1) ln (0.379)
i0.9702
7. e0.8 1 0.8
(0
2
.8
!
)2
(0
3
.8
!
)3
(0
4
.8
!
)4
1 0.8 0.32 0.08 0.017
2.22
8. e1.36 1 1.36
(1.
2
3
!
6)2
(1.
3
3
!
6)3
(1.
4
3
!
6)4
1 1.36 0.925 0.419 0.143
3.85
9. sin xx
x
3
3
!
x
5
5
!
x
7
7
!
x
9
9
!
sin sin 3.1416
3.1416
(3.1
3
4
!
16)3
(3.1
5
4
!
16)5
(3.1
7
4
!
16)7
(3.1
9
4
!
16)9
3.1416 5.1677 2.5502 0.5993 0.0821
0.0069
10. 2
cos
3
4
isin
3
4
2
ei
3
4
11. 1 3
i2
1
2
2
3
i
2
cos
2
3
isin
2
3
2ei
2
3
12-7
Chapter 12 422
12a. APert
2PPe(0.06)5
Pe0.3
1 0.3
(0
2
.3
!
)2
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.
Pages 812–814 Exercises
13. ln (4) ln (1) ln 4 i1.3863
14. ln (3.1) ln (1) ln (3.1) i1.1314
15. ln (0.25) ln (1) ln (0.25) i1.3863
16. ln (0.033) ln (1) ln (0.033) i3.4112
17. ln (238) ln (1) ln (238) i5.4723
18. ln (1207) ln (1) ln (1207) i7.0959
19. e1.1 1 1.1
(1
2
.1
!
)2
(1
3
.1
!
)3
(1
4
.1
!
)4
1 1.1 0.605 0.222 0.06
2.99
20. e0.2 1 (0.2)
(0
2
.
!
2)2
(0
3
.
!
2)3
(0
4
.
!
2)4
1 0.2 0.02 0.0013 0.00007
0.82
21. e4.2 1 4.2
(4
2
.2
!
)2
(4
3
.2
!
)3
(4
4
.2
!
)4
1 4.2 8.82 12.348 12.965
39.33
22. e0.55 1 0.55
(0.
2
5
!
5)2
(0.
3
5
!
5)3
(0.
4
5
!
5)4
1 0.55 0.151 0.028 0.004
1.73
23. e3.5 1 3.5
(3
2
.5
!
)2
(3
3
.5
!
)3
(3
4
.5
!
)4
1 3.5 6.125 7.146 6.253
24.02
24. e2.73 1 2.73
(2.
2
7
!
3)2
(2.
3
7
!
3)3
(2.
4
7
!
3)4
1 2.73 3.726 3.391 2.314
13.16
25. cos x1
x
2
2
!
x
4
4
!
x
6
6
!
x
8
8
!
cos cos 3.1416
1
(3.1
2
4
!
16)2
(3.1
4
4
!
16)4
(3.1
6
4
!
16)6
(3.1
8
4
!
16)8
1 4.9348 4.0588 1.3353 0.2353
0.9760
actual value: cos 1
26. sin xx
x
3
3
!
x
5
5
!
x
7
7
!
x
9
9
!
sin
4
sin 0.7854
0.7854
(0.7
3
8
!
54)3
(0.7
5
8
!
54)5
(0.7
7
8
!
54)7
(0.7
9
8
!
54)9
0.7854
0.4
6
845
0.
1
2
2
9
0
89
0
5
.1
0
8
4
4
0
3
3
0
6
.
2
1
,
1
8
3
8
7
0
0.7071
actual value: sin
4
2
2
0.7071
27. cos x1
x
2
2
!
x
4
4
!
x
6
6
!
x
8
8
!
cos
6
cos 0.5236
1
(0.5
2
2
!
36)2
(0.5
4
2
!
36)4
(0.5
6
2
!
36)6
(0.5
8
2
!
36)8
1
0.2
2
742
0.7
2
5
4
16
0.
7
0
2
2
0
06
0
4
.
0
0
,
0
3
5
2
6
0
0.8660
actual value: cos
6
2
3
0.8660
28. sin xx
x
3
3
!
x
5
5
!
x
7
7
!
x
9
9
!
sin
2
sin 1.5708
1.5708 0.6460 0.0797 0.0047 0.0002
1.0000
actual value: sin
2
1
29. 5
cos
5
3
isin
5
3
5ei
5
3
30. icos
2
isin
2
ei
2
31. r
121
2
or 2
Q Arctan
1
1
or
4
1 i2
cos
4
i sin
4
2
ei
4
32. r
(3
)2
12
or 2
vArctan
1
3
or
6
3
i 2
cos
6
i sin
6
2ei
6
33. r
2
2
2
2
or 2
vArctan
2
2
or 3
4
2
2
i2
cos
3
4
i sin
3
4
2ei
3
4
34. r
4
3
2
(4)2
or 8
vArctan
4
4
3
or
7
6
43
4i8
cos
7
6
i sin
7
6
8ei
7
6
35. r
323
2
or 32
vArctan
3
3
or
4
3 3i32
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.
37.
eix
2i
eix
2i
2
si
i
nx
sin x
eix
2
eix
cos xisin xcos xisin x
2
cos xisin x(cos xisin x)
2i
423 Chapter 12
2c
2
os x
cos x
38. See students’ work.
39. If you add the numbers on the diagonal lines as
shown, the sums are the terms of the Fibonacci
sequence.
40a.
1
1
,
2
1
,
3
2
,
5
3
,
8
5
,
1
8
3
,
2
1
1
3
,
3
2
4
1
,
5
3
5
4
,
8
5
9
5
40b. neither
40c.
40d. yes; 1.618
40e. The two ratios are equivalent to three decimal
places.
40f. See students’ work.
41a. APert
5000e0.05(13)
5000e0.65
50001 0.65
(0.
2
6
!
5)2
(0.
3
6
!
5)3
(0.
4
6
!
5)4
$9572.29
41b. No, the account will be short by more than
$30,000.
41c. about 42 years; 47 years old
41d. 40,000 Pe0.65
$20,882 P
42a. Every third Fibonacci number is an even
number.
42b. Every fourth Fibonacci number is a multiple of 3.
43. (2xy)6(2x)6(y)06(2x)5(y)1
65
2
(2
x
1
)4(y)2
65
3
4
2
(2
x
1
)3(y)3
64x6192x5y240x4y2160x3y3
60x2y412xy5y6
44.
6
k1
2k
6 5 4 3 2 1(2x)0(y)6
6 5 4 3 2 1
6 5 4 3 2(2x)1(y)5
5 4 3 2 1
6 5 4 3(2x)2(y)4
4 3 2 1
45a. a10.005, r
0
0
.
.
0
0
0
1
5
or 2
a30.005(2)31
0.020 cm
a40.020(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
46. 8
1
3
or 2
47. y2Dx Ey F0
(0, 0): F0
(2, 1): 1 2DEF0
(4, 4): 16 4D4EF0
2DE1 →4D2E2
4D4E16 4D4E16
2E14
2DE1E7
2D7 1y23x7y0
2D6
D3
48. r
1
6
4
or 4
v
8
4
8
or
15
8
4cos
15
8
i sin
15
8
49. r
u
23025022(30)(50) cos 140°
r
u
75.5 N
sin
75
1
.
4
5
0°
si
3
n
0
v
sin v
30 s
7
i
5
n
.5
140°
vArcsin
30 s
7
i
5
n
.5
140°
14°48
50a.
v
t
5(2
1
s
)
e
r
c
a
o
d
n
i
d
ans
10radians per second
50b. vr
v
t
1
2
ft
(10radians/s)
15.7 ft/s
51. Let mmultiple choice. Let eessay.
me30
1m12e96
f(m, e) 5m20e
f(24, 6) 5(24) 20(6)
240
f(0, 8) 5(0) 20(8)
160
f(30, 0) 5(30) 20(0)
150
f(0, 0) 0
To receive the highest score, answer 24 multiple
choice and 6 essay.
8
2
3
8
1
3
Chapter 12 424
161520 15 6 1
151010 5 1
14641
1331
13
8
5
3
2
1
1
121
11
1
.
.
.
n
O
0.25
0.5
0.75
1
1.25
1.5
1.75
2
12345678910
Fn
Fn
1
e
m
m
e
30
m
12
e
96
(24, 6)
(0, 8)
(30, 0)
(0, 0)
m
0
e
0
10
10 20
20
30
52. (DC)222
18
2
(DC)214
DC 14
14
252(BC)2
39 (BC)2
39
BC The correct choice is C.
Sequences and Iteration
Page 819 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)21
f(1) 121
f(1) 121
f(1) 121
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. z06i
z10.6(6i) 2i5.6i
z20.6(5.6i) 2i5.36i
z30.6(5.36i) 2i5.216i
7. z025 40i
z10.6(25 40i) 2i 15 26i
z20.6(15 26i) 2i9 17.6i
z30.6(9 17.6i) 2i5.4 12.56i
8. z00, f(z) z2(1 2i)
z102(1 2i) 1 2i
z2(1 2i)2(1 2i)
1 4i4i21 2i2 6i
z3(2 6i)2(1 2i)
4 24i35i21 2i31 22i
9. z01 2i, f(z) z2(2 3i)
z1(1 2i)2(2 3i)
1 4i4i22 3i1 i
z2(1 i)2(2 3i)
1 2ii22 3i2 5i
z3(2 5i)2(2 3i)
4 20i25i22 3i19 23i
12-8
10. pn
2
4
4
0
0.60
p10.60 1.75(0.60)(1 0.60)
1.02
(1.02)(40) 41
p21.02 1.75(1.02)(1 1.02)
0.9843
(0.9843)(40) 39
p30.9843 1.75(0.9843)(1 0.9843)
1.0113
(1.0113)(40) 40
p41.0113 1.75(1.0113)(1 1.0113)
0.9913
(0.9913)(40) 40
p50.9913 1.75(0.9913)(1 0.9913)
1.0064
(1.0064)(40) 40
p61.0064 1.75(1.0064)(1 1.0064)
0.9951
(0.9951)(40) 40
p70.9951 1.75(0.9951)(1 0.9951)
1.0036
(1.0036)(40) 40
p81.0036 1.75(1.0036)(1 1.0036)
0.9973
(0.9973)(40) 40
p90.9973 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
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)24
f(4) 4216
f(16) 162256
f(256) 256265.536
13. f(4) (4 5)21
f(1) (1 5)216
f(16) (16 5)2121
f(121) (121 5)213,456
14. f(1) (1)21 0
f(0) 021 1
f(1) (1)21 0
f(0) 021 1
425 Chapter 12
15. f(0.1) 2(0.1)20.1
0.08
f(0.08) 2(0.08)2(0.08)
0.09
f(0.09) 2(0.09)20.09
0.07
f(0.07) 2(0.07)2(0.07)
0.08
16a. t1
2
1
2
t2
2
2
1
t3
2
1
2
t4
2
2
1
t10
2
2
1
16b. t1
2
4
1
2
t24
t3
2
4
1
2
t44
t10 4
16c. t1
2
7
t27
2
2
7
2
1
2
2
1
2
2
1
2
19. z01 2i
z12(1 2i) (3 2i)
2 4i3 2i
5 2i
z22(5 2i) (3 2i)
10 4i3 2i
13 2i
z32(13 2i) (3 2i)
26 4i3 2i
29 2i
20. z01 2i
z12(1 2i) (3 2i)
2 4i3 2i
1 6i
z22(1 6i) (3 2i)
2 12i3 2i
5 14i
z32(5 14i) (3 2i)
10 28i3 2i
13 30i
21. z06 2i
z12(6 2i) (3 2i)
12 4i3 2i
15 2i
z22(15 2i) (3 2i)
30 4i3 2i
33 2i
z32(33 2i) (3 2i)
66 4i3 2i
69 2i
22. z00.3 i
z12(0.3 i) (3 2i)
0.6 2i3 2i
3.6 4i
z22(3.6 4i) (3 2i)
7.2 8i3 2i
10.2 10i
z32(10.2 10i) (3 2i)
20.4 20i3 2i
23.4 22i
23. z0
1
3
2
3
i
z13
1
3
2
3
i
2i
1 2i2i
1
z23(1) 2i
3 2i
z33(3 2i) 2i
9 6i2i
9 8i
24. z00 i, f(z) z21
z1(i)21
2
z2(2)21
3
z3321
8
Chapter 12 426
t3
2
7
t47
2
2
7
t10 7
2
2
7
16d. The values of the iterates alternate between
x
2
0
and x0.
17. z05i
z12(5i) (3 2i)
3 8i
z22(3 8i) (3 2i)
6 16i3 2i
9 14i
z32(9 14i) (3 2i)
18 28i3 2i
21 26i
18. z04
z12(4) (3 2i)
11 2i
z22(11 2i) (3 2i)
22 4i3 2i
25 6i
z32(25 6i) (3 2i)
50 12i3 2i
53 14i
25. z0i, f(z) z21 3i
z1i21 3i
3i
z2(3i)21 3i
8 3i
z3(8 3i)21 3i
64 48i9 1 3i
56 45i
26. z01, f(z) z23 2i
z1123 2i
4 2i
z2(4 2i)23 2i
16 16i4 3 2i
15 18i
z3(15 18i)23 2i
225 540i324 3 2i
96 542i
27. z01 i, f(z) z24i
z1(1 i)24i
1 2i1 4i
2i
z2(2i)24i
4 4i
z3(4 4i)24i
16 32i16 4i
28i
28. z0
2
2
2
2
i, f(z) z2
z1
2
2
2
2
i2
1
2
i
1
2
i21
z2(i)21
z3121
29. z01 i, f(z) z22 3i
z1(1 i)22 3i
1 2i1 2 3i
2 i
z2(2 i)22 3i
4 4i1 2 3i
5 7i
z3(5 7i)22 3i
25 70i49 2 3i
22 73i
30. p1p0rp0
p12000 (0.052)(2000)
$2104
p22104 (0.052)(2104)
$2213.41
p32213.41 (0.052)(2213.14)
$2328.51
p42328.51 (0.052)(2328.51)
$2449.59
p52449.59 (0.052)(2449.59)
$2576.97
31.
2002 1984 18; After 18 years, about 54% of
the maximum sustainable population is present.
32. f(z) z2c
1 15i(2 3i)2c
1 15i4 12i9i2c
4 3ic
33. 2
x1
2
2
2
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
, z02
35c. 1
35d. 1
36. 2
cos
3
i sin
3
2ei
3
37.
4!(8
8
!
4)!
(2a)84(3b)470 16a481b4
90,720a4b4
38. Convergent; the series is geometric with
r
1
4
1.
39. The distance between the vertices is 130 ft.
2a130, so a75.
e
a
c
7
5
9
6
1
5
b2c2a2
b2912652
4056
a
x2
2
b
y2
2
1 ⇒
42
x
2
2
5
40
y
5
2
6
1
427 Chapter 12
Iteration xpercent x2.5x(1 x)
10.10 0.325
20.325 0.8734
30.8734 1.1498
41.1498 0.7192
50.7192 1.2241
61.2241 0.5383
70.5383 1.1596
81.1596 0.6969
90.6969 1.225
10 1.225 0.5359
11 0.5359 1.1577
12 1.1577 0.7013
13 0.7013 1.225
14 1.225 0.5359
15 0.5359 1.1577
16 1.1577 0.7013
17 0.7013 1.225
18 1.225 0.5359
40. n1sin In2sin r
1.00 sin 42° 2.42 sin r
rArcsin
1.00
2
s
.4
in
2
42°
r16°
41a.
41b. Let hheight of the building.
Let xdistance from the point of elevation to
the center of the base of the building.
tan 42°
h
x
tan 56°
h
x
40
x
tan
h
42°
tan 56°
h
h
40
tan 56°
(h40
h
)tan42°
1.6466 1
4
h
0
0.6466
4
h
0
h62 feet
No, the height of the building is about 62 feet for
a total of about 102 feet with the tower.
42. x2 for all x, so infinite discontinuity
43.
f(x, y) 2x8y10
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
2
L
2H2LHL
H3L
H
L
3The correct choice is D.
Mathematical Induction
Page 826 Check for Understanding
1. The n1 case shows that the premise is true for
an infinite number of cases.
2. Provide a counterexample.
3a. n(n2)
3b. Since 3 is the first term in the sequence of
partial sums and 1(1 2) 3, the formula is
valid for n1.
Since 8 is the second term in the sequence of
partial sums and 2(2 2) 8, the formula is
valid for n2.
Since 15 is the third term in the sequence of
partial sums and 3(3 2) 15, the formula is
valid for n3.
3c. Sk⇒k(k2); Sk1⇒(k1)(k3)
4. 8n1 7rfor 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 k1.
6. Step 1: Verify that the formula is valid for n1.
Since 3 is the first term in the sequence and
1(1 2) 3, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒3 5 7 …(2k1) k(k2)
Sk1⇒3 5 7 …
(2k1) (2k3) k(k2) (2k3)
k24k3
(k1)(k3)
Apply the original formula for nk1.
(k1)[(k1) 2] (k1)(k3)
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, 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 n1.
Since 2 is the first term in the sequence and
2(211) 2, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒2 2223…2k2(2k1)
Sk1⇒2 2223… 2k2k1
2(2k1) 2k1
2 2k12
2(2k11)
When the original formula is applied for nk1,
the same result is obtained. Thus if the formula is
valid for nk, it is also valid for nk1. Since
the formula is valid for n1, it is also valid for
n2, n3, and so on indefinitely. Thus, the
formula is valid for all positive integral values
of n.
12-9
Chapter 12 428
40 ft
56˚
42˚
y
x
x
y
14
(8, 6)
(5, 9)(3, 9)
(8, 5)(3, 5)
y
9
y
5
x
3
x
8
O
8. Step 1: Verify that the formula is valid for n1.
Since
1
2
is the first term in the sequence and
1
2
1
1
1
2
, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒
1
2
2
1
2
2
1
3
…
2
1
k
1
2
1
k
Sk1⇒
1
2
2
1
2
2
1
3
…
2
1
k
2k
1
1
1
2
1
k
2k
1
1
1
2
2
2k
2k
1
1
1
2k
2
1
2k
1
1
1
2k
1
1
When the original formula is applied for nk1,
the same result is obtained. Thus if the formula is
valid for nk, it is also valid for nk1. Since
the formula is valid for n1, it is also valid for
n2, n3, and so on indefinitely. Thus, the
formula is valid for all positive integral values
of n.
9. Sn: 3n1 2rfor some integer r
Step 1: Verify that Snis valid for n1.
S1⇒311 or 2. Since 2 2 1, Snis valid for
n1.
Step 2: Assume that Snis valid for nkand show
that it is also valid for nk1.
Sk⇒3k1 2r for some integer r
Sk1⇒3k11 2t for some integer t
3k1 2r
3(3k1) 3 2r
3k13 6r
3k11 6r2
3k11 2(3r1)
Thus, 3k11 2t, where t3r1 is an
integer, and we have shown that if Snis valid,
then Sk1is also valid. Since Snis valid for n1,
it is also valid for n2, n3, and so on
indefinitely. Hence, 3n1 is divisible by 2 for all
integral values of n.
10a. 6 4 10 10b. an
n(n
2
1)
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 n1.
Since 1 is the first term in the sequence and
1(1 1
6
)(1 2)
1, the formula is valid for n1.
Step 2: Assume that the formula is valid for
nkand derive a formula for nk1.
Sk⇒1 3 6 …
k(k
2
1)
(k1)
6
(k2)
Sk⇒1 3 6 …
k(k
2
1)
(k1)
2
(k2)
k(k1
6
)(k2)
(k1)
2
(k2)
Apply the original formula for nk1.
(k1)(k2)(k3)
6
(k1)[(k1) 1][(k1) 2]
6
(k1)(k2)(k3)
6
k(k1)(k2) 3(k1)(k2)
6
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, and so on indefinitely. Thus, the formula
is valid for all positive integral values of n.
Pages 826–828 Exercises
11. Step 1: Verify that the formula is valid for n1.
Since 1 is the first term in the sequence and
(1)[2(1) 1] 1, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒1 5 9 …(4k3) k(2k1)
Sk1⇒1 5 9 …(4k3) (4k1)
k(2k1) (4k1)
2k23k7
(k1)(2k1)
Apply the original formula for nk1.
(k1)[2(k1) 1] (k1)(2k1)
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2.
Since it is valid for n2, it is also valid for n3,
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 n1.
Since 1 is the first term in the sequence and
(1)[3(1
2
)1]
1, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒1 4 7 …(3k2)
k(3k
2
1)
Sk1⇒1 4 7 …(3k2) (3k1)
k(3k
2
1)
(3k1)
k(3k
2
1)
2(3k
2
1)
3k2
2
5k2
(k1)(
2
3k2)
Apply the original formula for nk1.
(k1)(
2
3k2)
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
(k1)[3(k1) 1]
2
429 Chapter 12
13. Step 1: Verify that the formula is valid for n1.
Since
1
2
is the first term in the sequence and
2
1
1
1
1
2
, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒
1
2
1
4
1
8
…
2
1
k
2
1
k
1
Sk1⇒
1
2
1
4
1
8
…
2
1
k
2k
1
1
2
1
k
1
2k
1
1
2
2
2k
1
2k
1
1
2k
2
1
1
2k
1
1
2k
1
1
1
When the original formula is applied for nk1,
the same result is obtained. Thus if the formula is
valid for nk, it is also valid for nk1. Since
the formula is valid for n1, it is also valid for
n2, n3, 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 n1.
Since 1 is the first term in the sequence and
12(1
4
1)2
1, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒1 8 27 …k3
k2(k
4
1)2
Sk1⇒1 8 27 …k3(k1)3
k2(k
4
1)2
(k1)3
k2(k1)24(k1)3
4
15. Step 1: Verify that the formula is valid for n1.
Since 1 is the first term in the sequence and
1, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒123252…(2k1)2
Sk1⇒123252…(2k1)2(2k1)2
(2k1)2
Apply the original formula for nk1.
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, 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 n1.
Since S1⇒1 and 211 1, the formula is valid
for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒1 2 4 …2k12k1
Sk1⇒1 2 4 …2k12k2k1 2k
2(2k) 1
2k11
When the original formula is applied for nk1,
the same result is obtained. Thus if the formula is
valid for nk, it is also valid for nk1. Since
the formula is valid for n1, it is also valid for
n2, n3, and so indefinitely. Thus, the
formula is valid for all positive integral values of n.
17. Sn⇒7n5 6rfor some integer r
Step 1: Verify that Snis valid for n1.
S1⇒715 or 12. Since 12 6 2, Snis valid for
n1.
Step 2: Assume that Snis valid for nkand show
that it is also valid for nk1.
Sk⇒7k5 6r for some integer r
Sk1⇒7k15 6t for some integer t
7k5 6r
7(7k5) 7 6r
7k135 42r
7k15 42r30
7k15 6(7r5)
Thus, 7k15 6t, where t7r5 is an
integer, and we have shown that if Snis valid,
then Sk1is also valid. Since Snis valid for n1,
it is also valid for n2, n3, and so on
indefinitely. Hence, 7n5 is divisible by 6 for all
integral values of n.
(k1)(2k1)(2k3)
3
(k1)[2(k1) 1][2(k1) 1]
3
(2k3)(k1)(2k1)
3
(2k25k3)(2k1)
3
[k(2k1) 3(2k1)](2k1)
3
k(2k1)(2k1) 3(2k1)2
3
k(2k1)(2k1)
3
k(2k1)(2k1)
3
1[2(1) 1][2(1) 1]
3
Chapter 12 430
(k1)2[k24(k1)]
4
(k1)2
4
(k2)2
Apply the original formula for nk1.
(k1)2
4
(k2)2
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
(k1)2[(k1) 1]2
4
(k1)2(k24k4)
4
18. Sn⇒8n1 7rfor some integer r
Step 1: Verify that Snis valid for n1.
S1⇒811 or 7. Since 7 7 1, Snis valid for
n1.
Step 2: Assume that Snis valid for nkand show
that it is also valid for nk1.
Sk⇒8k1 7r for some integer r
Sk1⇒8k11 7t for some integer t
8k1 7r
8(8k1) 8 7r
8k18 56r
8k11 56r7
8k11 7(8r1)
Thus, 8k11 7t, where t8r1 is an
integer, and we have shown that if Snis valid,
then Sk1is also valid. Since Snis valid for n1,
it is also valid for n2, n3, and so on
indefinitely. hence, 8n1 is divisible by 7 for all
integral values of n.
19. Sn⇒5n2n3r for some integer r
Step 1: Verify that Snis valid for n1.
S1⇒5121or 3. Since 3 3 1, Snis valid for
n1.
Step 2: Assume that Snis valid for nkand show
that it is also valid for nk1.
Sk⇒5k2k3r for some integer r
Sk1⇒5k12k13t for some integer t
5k2k3r
5k2k3r
5k5 (2k3r)(2 3)
5k12k13(2k) 6r9r
5k12k12k13(2k) 6r9r
2k1
3(2k) 15r
3(2k5r)
Thus, 5k12k13t, where t2k5ris an
integer, and we have shown that if Snis valid,
then Sk1is also valid. Since Snis valid for n1,
it is also valid for n2, n3, and so on
indefinitely. Hence, 5n2nis divisible by 3 for all
integral values of n.
20. Step 1: Verify that the formula is valid for n1.
Since ais the first term in the sequence and
1
2
[2a(1 1)d] a, the formula is valid for
n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒a(ad) (a2d) …[a(k1)d]
k
2
[2a(k1)d]
Sk1⇒a(ad) (a2d) …[a(k1)d]
(akd)
k
2
[2a(k1)d] (akd)
k[2a(k1)d] 2(akd)
2
Apply the original formula for nk1.
(k
2
1)
{2a[(k1) 1]d}
(k
2
1)
(2akd)
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, 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 n1.
Since
1
2
is the first term in the sequence and
1
1
1
1
2
, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒
2
1
3
3
1
4
…
k(k
1
1)
k
k
1
Sk1⇒
1
1
2
2
1
3
3
1
4
…
k(k
1
1)
(k1)
1
(k2)
k
k
1
(k1)
1
(k2)
(
k
k
(
k
1)
2
(k
)
1
2)
(k
k2
1
2
)(
k
k
1
2)
(k
(k
1
)(k
1)
2
2)
k
k
1
2
Apply the original formula for nk1.
(k
(
k
1)
1
)
1
k
k
1
2
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
22. Sn⇒22n132n15rfor some integer r
Step 1: Verify that Snis valid for n1.
S1⇒22(1)132(1)1or 35. Since 35 5 7, Snis
valid for n1.
Step 2: Assume that Snis valid for nkand show
that it is also valid for nk1.
Sk⇒22k13k15r for some integer r
Sk1⇒2k332k35t for some integer t
22k132k15r
22k15r32k1
22k122(5r32k1)(325)
22k345r25r32k35(32k1)
22k332k345r25r32k35(32k1)
32k3
20r5(32k1)
5(4 32k1)
Thus, 22k332k35t, where t4 32k1is
an integer; and we have shown that if Snis valid,
then Sk1is also valid. Since Snis valid for n1,
it is also valid for n2, n3, and so on
indefinitely. Hence, 22n132k1is divisible by 5
for all integral values of n.
431 Chapter 12
2ak k(k1)d2a2kd
2
(k1)2a[k(k1) 2k]d
2
(k1)2a(k2k)d
2
(k
2
1)
(2akd)
(k1)2ak(k1)d
2
23. Step 1: Verify that the formula is valid for n1.
Since S1⇒[r(cos visin v]1or r(cos visin v)
and r1[cos (1)visin (1)v] r(cos visin v), the
formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
That is, assume that [r(cos visin v)]k
rk(cos kvisin kv).
Multiply each side of the equation by
r(cos visin v).
[r(cos visin v)]k1
[rk(cos kvisin kv] [r(cos visin v)]
rk1[cos kvcos v(cos kv(isin v)
isin kvcos vi2sin kvsin v]
rk1[(cos kvcos vsin kvsin v)
i(sin kvcos vcos kvsin v)]
rk1[cos (k1)visin (k1)v]
When the original formula is applied for
nk1, the same result is obtained. Thus if
the formula is valid for nk, it is also valid for
nk1. Since the formula is valid for n1, it
is also valid for n2, n3, and so on
indefinitely. Thus, the formula is valid for all
positive integral values of n.
24a.
24b. 4 1 3
9 4 5
16 9 7
1, 3, 5, 7, …
24c. 2n1
24d. n2
24e. Step 1: Verify that the formula is valid for n1.
Since 1 is the first term in the sequence and
121, the formula is valid for n1.
Step 2: Assume that the formula is valid for
nkand derive a formula for nk1.
Sk⇒1 3 5 7 …(2k1) k2
Sk1⇒1 3 5 7 …(2k1) (2k1)
k2(2k1)
k22k1
(k1)2
When the original formula is applied for
nk1, the same result is obtained. Thus if
the formula is valid for nk, it is also valid for
nk1. Since the formula is valid for n1,
it is also valid for n2, n3, and so on
indefinitely. Thus, the formula is valid for all
positive integral values of n.
25. S1⇒n25n2rfor some positive integer r
Step 1: Verify that S1is valid for n1.
S1⇒125 1 or 6. Since 6 2 3, S1is valid for
n1.
Step 2: Assume that Snis valid for nkand show
that it is valid for nk1.
Sk⇒k25k2r for some positive integer r
Sk1⇒(k1)25(k1) 2t for some positive
integer t
Chapter 12 432
(k1)25(k1) k22k1 5k5
(k25k) (2k6)
2r2(k3)
2(rk3)
Thus, k25k2t, where trk3 is an
integer, and we have shown that if Snis valid,
then Sk1is also valid. Since Snis valid for n1,
it is also valid for n2, n3, and so on
indefinitely. Hence, n25nis divisible by 2 for all
positive integral values of n.
26a. Number of people Number of Interactions
212 1
312 3 3
4123 4 6
n
n(n
2
1)
26b. Step 1: Verify that Sn⇒0 1 2 3 …
(n1)
n(n
2
1)
is valid for n1.
Since 0 is the first term in the sequence and
1(1
2
1)
0, the formula is valid for n1.
Step 2: Assume that the formula is valid for
nkand derive a formula for nk1.
Sk⇒0 1 2 …(k1)
k(k
2
1)
Sk1⇒012…(k1) k
k(k
2
1)
k
k(k1
2
)2k
k2k
2
2k
k(k
2
1)
Apply the original formula for nk1.
k(k
2
1)
The formula gives the same result as adding the
(k1) term directly. Thus if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for
n2, n3, and so on indefinitely. Thus, the
formula is valid for all positive integral values
of n.
26c. Yes; 15 people will require
15(
2
14)
or 105
interactions and last approximately 105(0.5) or
52.5 minutes.
(k1)[(k1) 1]
2
27. Step 1: Verify that Sn⇒(xy)nxnnxn1y
n(n
2
!
1)
xn2y2
n(n1
3
)
!
(n2)
xn3y3…ynis
valid for n1. Since S1⇒(xy)1x1y1or
xy, Snis valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒(xy)kxkkxk1y
k(k
2
!
1)
xk2y2
k(k1
3
)
!
(k2)
xk3y3…yk
Sk1⇒(xy)k(xy) (xy)(xkkxk1y
k(k
2
!
1)
xk2y2
k(k1
3
)
!
(k2)
xk3y3…yk)
(xy)k1x(xkkxk1y
k(k
2
!
1)
xk2y2
k(k1
3
)
!
(k2)
xk3y3…yk) y(xkkxk1y
k(k
2
!
1)
xk2y2
k(k1
3
)
!
(k2)
xk3y3…yk)
xk1kxky
k(k
2
!
1)
xk1y2…
xykxkykxk1y2
k(k
2
!
1)
xk2y3
…yk1
xk1(k1)xkykxk1y2
k(k
2
!
1)
xk1y2…yk1
xk1(k1)xky
k(k
2
!
1)
xk1y2
…yk1
When the original formula is applied for nk1,
the same result is obtained. Thus if the formula is
valid for nk, it is also valid for nk1. Since
the formula is valid for n1, it is also valid for
n2, n3, and so on indefinitely. Thus, the
formula is valid for all positive integral values of n.
28a. 0.9 0.09 0.009 …
28b.
1
9
0n
28c. S
a1
1
a1
r
rn
1
9
0
1
9
0
1
1
0
n
1
1
1
0
When the original formula is applied for
nk1, the same result is obtained. Thus if
the formula is valid for nk, it is also valid for
nk1. Since the formula is valid for n1,
it is also valid for n2, n3, and so on in
indefinitely. Thus, the formula is valid for all
positive integral values of n.
28e. lim
n→
10
1
k
0
n1
lim
n→1
1
1
0n
1 0 or 1
Thus, 0.999… 1.
29. z12(4 i) i
8 i
z22(8 i) i
16 i
z32(16 i) i
32 i
30. 64 6 (29 1)d
70 28d
5
2
d
31. 25x24y2100x40y100 0
25(x24x) 4(y210y) 100
25(x2)24(y5)2100 100 100
25(x2)24(y5)2100
(x
4
2)2
(y
25
5)2
1
ellipse
32. 250, k4
A
(25
4
0
(4
)2
)
12,271.85 m2
33.
2
2; y
3
4
sin 2x
34. cos1
2
3
30° Since
2
3
is
negative, xis
in the second
or third quadrants.
x180° 30° or 150°
x180° 30° or 210°
35. Since the area
of the square
is 25, each side
measures 5.
Since A
E
and B
E
measure 1 unit, A
H
and B
F
each measure 4 units. A
B
is the hypotenuse of
an isosceles right triangle with legs that measure
1 unit. So A
B
measures 2
units. C
A
is the
hypotenuse of an isosceles right triangle with
legs that measure 4 units. So C
A
measures
42
units. Since ABCD is a rectangle, the
perimeter is 2(2
42
) 2 (52
)
102
units.
The correct choice is B.
433 Chapter 12
1
1
1
01
10
1
n
0
n1
28d. Step 1: Verify that Sn: 0.9 0.9 0.009 …
1
9
0n
10
1
n
0
n1
is valid for n1.
Since 0.9 is the first term in the sequence and
10
1
1
0
11
0.9, the formula is valid for n1.
Step 2: Assume that the formula is valid for
nkand derive a formula for nk1.
Sk⇒0.9 0.09 0.009 …
1
9
0k
10
1
k
0
k1
Sk1⇒0.9 0.09 0.009 …
1
9
0k
10
9
k1
10
1
k
0
k1
10
9
k1
10(10
1
k
0
k
1
1)9
10k1
10
k
1
1
09
10
1
k
0
k
1
11
1
9
0
1
1
1
0
n
1
9
0
1
2
60˚
30˚3
A
HBG
F
E
D
C
Chapter 12 Study Guide and Assessment
Page 829 Understanding the Vocabulary
1. d2. i3. m4. j
5. k6. f7. c8. e
9. b10. h
Pages 830–831 Skills and Concepts
11. d4.3 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. d23 (30) or 7
a14 30 (14 1) 7
61
S14
1
2
4
(30 61)
217
15. Sn
n
2
[2a1(n1)d]
250.2
n
2
[2(2) (n1)(1.4)]
250.2 2n0.7n(n1)
0 0.7n21.3n250.2
n
1.3
1
.4
26.5
n18 or n19.86
Since nis a positive whole number, n18.
16. r
4
7
9
or
1
7
1
1
7
1
7
,
1
7
1
7
4
1
9
,
4
1
9
1
7
3
1
43
1
7
,
4
1
9
,
3
1
43
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) 5, 5(5) 25
0.2, 1, 5, 25, 125
20. r
1
2
.2
.4
or 2
S9
1.2
1
1
(
.
2(
2
)
2)9
1.2
3
614.4
205.2
1.3 (1.3)3
4(0
.7)(2
50.2)
2(0.7)
21. r
4
4
2
or 2
S8
4
1
4(
2
2
)8
1
6
0
2
1
1
2
2
60(
1
1
2
2
)
60
1 2
Chapter 12 434
3
4
23. lim
n→
6n
n
3
lim
n→
6
n
n
lim
n→
n
3
6 0
6
24. Does not exist; lim
n→
2
3
n
n
n
3
3
lim
n→
2
3
n
; since lim
n→
2
3
n
becomes increasingly large as napproaches
infinity, the sequence has no limit.
25. lim
n→
4
n
n
4
3
4
3
n
n
3
lim
n→
4
n
n
4
3
3
n
n
4
n
n
4
4
4
n
n
4
3
0
1
0
26. 5.1
2
3
5
1
1
0
2
0
3
0
1,0
1
0
2
0
3
,000
…
a1
1
1
0
2
0
3
0
, r
10
1
00
22. lim
n→
4n
3
n
1
lim
n→
3
n
n
4
n
n
n
1
Sn5
1
1
0
2
0
3
0
1
10
1
00
5
1
9
2
9
3
9
5
3
4
3
1
3
27. r
1
5
2
0
6
4
0
or 0.4
Sn
1
1
26
0
0
.4
2100
28. an
n
5n
2
, an1
(n
5
n
1
1)2
rlim
n→
(n
5
n
1
1)2
5
n
n
2
lim
n→
5n(n
5n
2
5
2
n
n
21)
lim
n→
5
n
n
2
2
lim
n→
2
n
n
2
lim
n→
n
1
2
1
5
0 0
1
5
convergent
29. The general term is
n
n
5
or 1
n
5
.
1
n
5
n
1
for all n, so divergent
30. The general term is
n
2
.
n
2
n
1
for all n, so divergent
31.
9
a5
(3a3) (3 5 3) (3 6 3) (3 7 3)
(3.8 3) (3.9 3)
12 15 18 21 24
90
32.
k1
(0.4)k(0.4)1(0.4)2(0.4)3…(0.4)
S
1
0.4
0.4
2
3
33.
a0
(2n1) 34.
9
a1
1(n21)
35. (a4)6a6
1!(6
6
!
1)!
a5(4)1
2!(6
6
!
2)!
a4
(4)2
3!(6
6
!
3)!
a3(4)3
4!(6
6
!
4)!
a2(4)4
5!(6
6
!
5)!
a1(4)5
6!(6
6
!
6)!
a0(4)6
a624a5240a41280a33840a2
6144a4096
36. (2r3s)4 (2r)4
1!(4
4
!
1)!
(2r)3(3s)1
2!(4
4
!
2)!
(2r)2(3s)2
3!(4
4
!
3)!
(2r)(3s)3
4
4
!0
!
!
(2r)0(3s)4
16r496r3s216r2s2216rs3
81s4
37.
4!(1
1
0
0
!
4)!
x104(2)4210 x616
3360x6
38.
2!(8
8
!
2)!
4m821228 4096m61
114,688m6
39.
7!(1
1
0
0
!
7)!
x107(3y)7120 x32187y7
262,440x3y7
40.
5!(1
1
2
2
!
5)!
(2c)125(d)4792 128c7(d)5
101,376c7d5
41. 2cos
3
4
i sin
3
4
2ei
3
4
42. 4i4cos
2
i sin
2
4ei
2
43. r22(
2)2
or 22
vArctan
2
2
or
7
4
2 2i22
cos
7
4
i sin
7
4
22
ei
7
4
44. r(33
)
232
or 6
vArctan
3
3
3
or
6
33
3i6cos
6
i sin
6
6ei
6
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
46. f(3) (3)24 13
f(13) 1324 173
f(173) 17324 29,933
f(29,933) 29,93324 895,984,493
13; 173; 29.933; 895, 984, 493
47. z04i
z10.5(4i) (4 2i) 2i4 2i4
z20.5(4) (4 2i) 2 4 2i6 2i
z30.5(6 2i) (4 2i)
3 i4 2i7 3i
48. z08
z10.5(8) (4 2i) 4 4 2i2i
z20.5(2i) (4 2i) i4 2i4 3i
z30.5(4 3i) (4 2i)
2 1.5i4 2i6 3.5i
49. z04 6i
z10.5(4 6i) (4 2i)
2 3i4 2i2 i
z20.5(2 i) (4 2i)
1 0.5i4 2i5 1.5i
z30.5(5 1.5i) (4 2i)
2.5 0.5i4 2i6.5 2.75i
50. z012 8i
z0.5(12 8i) (4 2i)
6 4i4 2i10 6i
z20.5(10 6i) (4 2i)
5 3i4 2i9 5i
z30.5(9 5i) (4 2i)
4.5 2.5i4 2i8.5 4.5i
51. Step 1: Verify that the formula is valid for n1.
Since the first term in the sequence is 1 and
1(1
2
1)
1, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive a formula for nk1.
Sk⇒1 2 3 …k
k(k
2
1)
Sk1⇒123…k(k1)
k(k
2
1)
2(k
2
1)
k2
2
k
2k
2
2
k2k
2
2k2
k23
2
k2
(k1)
2
(k2)
Apply the original formula for nk1.
(k1)
2
(k2)
The formula gives the same result as adding the
(k1) term directly. Thus, if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
(k1)[(k1) 1]
2
435 Chapter 12
52. Step 1: Verify that the formula is valid for n1.
Since the first term in the sequence is 3 and
3, the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and derive formula for nk1.
Sk⇒3 8 15 …k(k2)
k(k1)
6
(2k7)
Sk1⇒3 8 15 …k(k2) (k1)(k3)
k(k1)
6
(2k7)
(k1)(k3)
k(k1)
6
(2k7)
6(k1
6
)(k3)
1(1 1)(2 1 7)
6
54b. S12
1
2
2
(16 368)
2304 ft
55. If the budget is cut 3% each year, 97% remains
after each year.
a1160,000,000, r0.97
a11 160,000,000(0.97)111
$117,987,860.30
56a. One side of the original triangle measures
6
3
or
2units. 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.
Chapter 12 436
k(k1)(2k7) 6(k1)(k3)
6
(k1)[k(2k7) 6(k3)]
6
(k1)(2k27k6k18)
6
Apply the original formula for nk1.
The formula gives the same result as adding the
(k1) term directly. Thus, if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, and so on indefinitely. Thus, the formula is
valid for all positive integral values of n.
53. Sn⇒9n4n5rfor some integer r
Step 1: Verify that Snis valid for n1.
S1⇒9141or 5. Since 5 5 1, Snis valid for
n1.
Step 2: Assume that Snis valid for nkand show
that it is also valid for nk1.
Sk⇒9k4k5rfor some integer r
Sk1⇒9k14k15tfor some integer t
9k4k5r
9k4k5r
9(9k)(4k5r)(4 5)
9k14k15(4k) 20r25r
9k14k14k15(4k) 20r25r
4k1
5(4k) 45r
5(4k9r)
Thus, 9k14k15t, where t4k9ris an
integer, and we have shown that if Snis valid,
then Sk1is also valid. Since Snis valid for n1,
it is also valid for n2, n3, and so on
indefinitely. Hence, 9n4nis divisible by 5 for all
integral values of n.
Page 833 Applications and Problem Solving
54a. a116, d48 16 or 32
A12 16 (12 1)32
368 ft
(k1)(k2)(2k9)
6
(k1)[(k1) 1][(2(k1) 7]
6
(k1)(k2)(2k9)
6
(k1)(2k213k18)
6
56b. a16, r
3
6
or
1
2
S
1
Page 833 Open-Ended Assessment
1a. Arithmetic; arithmetic sequences have common
differences, while geometric sequences have
common ratios.
1b. Sample answer: 1, 4, 7, 10, …; an1 3(n1)
2. Sample answer:
6
3n
5n2
; lim
n→
6
3n
5n2
lim
n→
n
2
5
3
n
lim
n→
n
2
0, but since lim
n→
5
3
n
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, x20 90 or x70.
The correct choice is C.
3. fraction pumped
d
kg
g
a
a
l
l
l
l
o
o
n
n
s
s
d
k
Change this fraction into a percent by multiplying
by 100.
percent pumped
d
k
100 or
10
k
0d
%
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.
6
1
1
2
Percent increase
54
4
8
48
100
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∠Bm∠ACB.
m∠Bm∠ACB 80 180
2 m∠ACB 100
m∠ACB 50
Since AD is a line segment, x70 50 180.
So, x60.
Consider right triangle CDE.
xy90
60 y90
y30
xy60 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)2or 800,000,000. Choice C
is incorrect since f(8) . 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.
Slope of AB
1
6
0
5
4
6
1
Slope of AC
1
6
Since the slopes are negative reciprocals, the line
segments are perpendicular, so m∠A90.
Therefore, ABC is a 30°-60°-90° right triangle.
The hypotenuse, AC, is twice the length of the leg
opposite the 30° angle, AB.
AB (6 5
)2(1
0 4
)2
1 3
6
or 37
AC 237
The correct choice is D.
8. Method 1: Substitute each answer choice for xto
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.
x6 0and1 2x1
x62x2
x1
The solution is 6 x1. All of the answer
choices except A are in this range.
The correct choice is A.
20,000
64
9. The increase from 99 to 100 is 1. So the percent
increase from 99 to 100 is
9
1
9
.
9
1
9
is greater than
11
00
, or 1%.
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
40 8 blue cars
1
2
8 blue cars 4 blue convertibles
1
4
the number of convertibles 4
the number of convertibles 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
2
4
0
0
100
50%
The answer is 50.
437 Chapter 12
Not Blue and Not Convertible
4
8 Blue Convertible
16
Chapter 13 438
Permutations and Combinations
Pages 842–843 Check for Understanding
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
Sgreen ———— S-green
gray————— S-gray
blue ———— M-blue
Mgreen———— M-green
gray ———— M-gray
blue ————— L-blue
Lgreen ———— 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
7. P(6, 6)
(6
6!
6)!
720
8. P(5, 3)
(5
5!
3)!
54
2
3
1
21
60
9.
P
P
(
(
1
6
2
,
,
4
8
)
)
(12
1
2!
8)!
——
(6
6!
4)!
6 5 4 3 2 1
1
13-1 12. C(4, 3) C(5, 2)
(4
4
3
!
)!3!
(5
5
2
!
)! 2!
4
1
3
3
2
2
1
1
5
3
4
2
3
1
2
2
1
1
4 10 or 40
13. Using the Basic Counting Principle,
10 9 8 7 6 5 4 3 2 1 3,628,800.
14. C(15, 9)
(15
15
9
!
)! 9!
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
Pages 543–545 Exercises
16. Using the Basic Counting Principle, 2 6 4 48.
17. P(7, 7)
(7
7!
7)!
5040
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
22. P(8, 8)
(8
8!
8)!
40,320
23. P(6, 4)
(6
6!
4)!
360
24. P(5, 3)
(5
5!
3)!
54
2
3
1
21
60
25. P(7, 4)
(7
7!
4)!
840
7 6 5 4 3 2 1
3 2 1
6 5 4 3 2 1
2 1
8 7 6 5 4 3 2 1
1
7 6 5 4 3 2 1
1
15 14 13 12 11 10 987654321
654321987654321
Chapter 13 Combinatorics and Probability
1
6
2
!
!
4
2
!
!
55,440
10. C(7, 4)
(7
7
4
!
)! 4!
35
11. C(20, 15)
(20
2
1
0
5
!
!) 15!
15,504
2019 1817 1615 1413 1211 10 987654321
543211514 1312 11 10987654321
7 6 5 4 3 2 1
3 2 1 4 3 2 1
12 11 10 9 8 7 6 5 4 3 2 1 2 1
6 5 4 3 2 1 4 3 21
26. P(9, 5)
(9
9!
5)!
15,120
27. P(10, 7)
(10
1
0!
7)!
604,800
28.
P
P
(
(
6
4
,
,
3
2
)
)
6
4
!
!
2
3
!
!
10
29.
P
P
(
(
6
5
,
,
4
3
)
)
6
5
!
!
2
2
!
!
6
30.
P(6,
P
3
(
)
9
,
P
6
(
)
7, 5)
6
9
!
!
7
3
!
!
3
2
!
!
5
31. C(5, 3)
(5
5
3
!
)! 3!
5
2
4
1
3
3
2
2
1
1
10
32. C(10, 5)
(10
10
5
!
)! 5!
252
33. C(4, 2)
(4
4
2
!
)! 2!
4
2
3
1
2
2
1
1
6
34. C(12, 4)
(12
12
4
!
)! 4!
495
35. C(9, 9)
(9
9
9
!
)! 9!
1
36. C(14, 7)
(14
14
7
!
)! 7!
3432
37. C(3, 2) C(8, 3)
(3
3
2
!
)! 2!
(8
8
3
!
)! 3!
3
1
2
2
1
1
3 56 or 168
8 7 6 5 4 3 2 1
5 4 3 2 1 3 2 1
14 13 12 11 10 9 8 7 6 5 4 3 2 1
7 6 5 4 3 2 1 7 6 5 4 3 2 1
12 11 10 9 8 7 6 5 4 3 2 1
8 7 6 5 4 3 2 1 4 3 2 1
10 9 8 7 6 5 4 3 2 1
5 4 3 2 1 5 4 3 2 1
6 5 4 3 2 1 7 6 5 4 3 2 1
9 8 7 6 5 4 3 2 1 2 1
(6
6!
3)!
(7
7!
5)!
———
(9
9!
6)!
6 5 4 3 2 1
5 4 3 2 1
(6
6!
4)!
—
(5
5!
3)!
6 5 4 3 2 1 2 1
4 3 2 1 3 2 1
(6
6!
3)!
—
(4
4!
2)!
10 9 8 7 6 5 4 3 2 1
3 2 1
9 8 7 6 5 4 3 2 1
4 3 2 1
38. C(7, 3) C(8, 5)
(7
7
3
!
)! 3!
(8
8
5
!
)! 5!
35 56 or 1960
39. C(5, 1) C(4, 2) C(8, 2)
(5
5
1
!
)! 1!
(4
4
2
!
)! 2!
(8
8
2
!
)! 2!
5
4
4
3
3
2
2
1
1
1
4
2
3
1
2
2
1
1
5 6 28 or 840
40. C(14, 4)
(14
14
4
!
)! 4!
1001
41. C(14, 5)
(14
14
5
!
)! 5!
2002
42. C(18, 12)
(18
1
1
8
2
!
)! 12!
18,564
43. C(3, 2) C(5, 1) C(8, 2)
(3
3
2
!
)! 2!
(5
5
1
!
)! 1!
(8
8
2
!
)! 2!
3
1
2
2
1
1
5
4
4
3
3
2
2
1
1
1
3 5 28 or 420
44. P(11, 11)
(11
1
1!
1)!
39,916,800
45a. C(13, 3) C(13, 2)
(13
13
3
!
)! 3!
(13
13
2
!
)! 2!
13
3
1
2
2
1
11
13
2
1
1
2
286 78 or 22,308
45b. C(4, 1) C(4, 2) C(4, 2)
(4
4
1
!
)! 1!
(4
4
2
!
)! 2!
(4
4
2
!
)! 2!
4
3
3
2
2
1
1
1
4
2
1
3
2
2
1
1
4
2
3
1
2
2
1
1
4 6 6 or 144
45c. C(12, 5)
(12
12
5
!
)! 5!
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.
P(5, 2) P(4, 3)
(5
5!
2)!
(4
4!
3)!
54
3
3
2
2
1
1
43
1
21
20 24 or 480
12 11 10 9 8 7 6 5 4 3 2 1
7 6 5 4 3 2 1 5 4 3 2 1
11 10 9 8 7 6 5 4 3 2 1
1
8 7 6 5 4 3 2 1
6 5 4 3 2 1 2 1
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
14 13 12 11 10 9 8 7 6 5 4 3 2 1
9 8 7 6 5 4 3 2 1 5 4 3 2 1
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
8 7 6 5 4 3 2 1
6 5 4 3 2 1 2 1
8 7 6 5 4 3 2 1
3 2 1 5 4 3 2 1
7 6 5 4 3 2 1
4 3 2 1 3 2 1
439 Chapter 13
47. C(3, 1) C(4, 1) C(6, 1) C(14, 6)
(3
3
1
!
)! 1!
(4
4
1
!
)! 1!
(6
6
1
!
)! 1!
(14
14
6
!
)! 6!
3
2
2
1
1
1
4
3
3
2
2
1
1
1
6 5 4 3 2 1
5 4 3 2 11
53b. Yes; let h, t, and ube the digits.
100h10tu
100h10ut
100t10hu
100t10uh
100u10th
100u10ht
200(htu) 20(htu) 2(htu)
222(htu)
222(h
6
tu)
37(htu)
54. 2140 (1.058) $2264.12
2264.12(1.058) $2395.44
2395.44(1.058) $2534.38
55.
10
n1
n3132 3... 1033025
56. 7.1x83.1
x ln 7.1 ln 83.1
x
l
l
n
n
8
7
3
.
.
1
1
x2.26
57. xe0.346 Use a calculator.
1.4
58. y4x2
xsin 45° ycos 45° 4(xcos 45°
ysin 45°)2
2
2
x
2
2
y4
2
2
x
2
2
y
2
2
x2
y8
1
2
x2xy
1
2
y2
0 4(x)28xy4(y)2
2
x2
y
59.
r2, r2 cos 2v
2 2 cos 2v
1 cos 2v
2v0° or 2v360°
v0° v180°
(2, 180°), (2, 0°)
60. vx28 cos 45°, vy28 sin 45°
28
2
2
28
2
2
19.80 ft/s 19.80 ft/s
61. sin 2x2 sin x0
2 sin xcos x2 sin x0
2 sin x(cos x1) 0
2 sin x0 or cos x1 0
sin x0cos x1
x0°, x180°, x360°, or x180°
So, x0°, 180°, 360°
62. y8 cos (v30°)
amplitude 8
period
36
1
0°
or 360°
phase shift 30°
Chapter 13 440
3 4 6 3003 or 216,216
48a. P(42, 42)
(42
4
2!
42)!
42!
1.4 1051
48b. C(42, 30)
(42
4
3
2
0
!
)! 30!
12
4
!
2
3
!
0!
1.1 1010
48c. C(5, 3) C(12, 6) C(10, 6) C(15, 5)
(5
5
3
!
)! 3!
(12
12
6
!
)! 6!
(10
10
6
!
)! 6!
(15
15
5
!
)! 5!
10 924 210 3003
5.8 109
49. P(n, n1) P(n, n)
[n(
n
n
!
1)]!
(n
n!
n)!
n
1!
!
n
0!
!
0! 1
n! n!
50a. P(6, 6)
(6
6!
6)!
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
3!
3)!
(2
2!
2)!
(2
2!
2)!
(2
2!
2)!
32
1
1
2
1
1
2
1
1
2
1
1
6 2 2 2 or 48
51a. C(11, 4)
(11
11
4
!
)! 4!
330
51b. C(6, 2) C(5, 2)
(6
6
2
!
)! 2!
(5
5
2
!
)! 2!
5
3
4
2
3
1
2
2
1
1
15 10 or 150
52. C(10, 2)
(10
10
2
!
)! 2!
45
53a.
35
6
52
or 37(2 5 9) 592
10 9 8 7 6 5 4 3 2 1
8 7 6 5 4 3 2 1 2 1
6 5 4 3 2 1
4 3 2 1 2 1
11 10 9 8 7 6 5 4 3 2 1
7 6 5 4 3 2 1 4 3 2 1
6 5 4 3 2 1
1
14 13 12 11 10 9 8 7 6 5 4 3 2 1
8 7 6 5 4 3 2 1 6 5 4 3 2 1
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
63. Find B.
B180° 90° 27° or 63°
Find a.
tan 27°
15
a
.2
15.2 tan 27° a
7.7 a
Find c.
cos 27°
15
c
.2
c
co
1
s
5
2
.2
7°
c17.1
64. Each hour, an hour hand moves through
3
1
6
2
0
or
30°. Since 12 minutes is
1
5
of an hour, the hour
hand moves through an additional
1
5
(30) or 6°.
2(30°) 6° 66°
The correct choice is A.
Permutations with Repetitions
and Circular Permutations
Pages 848–849 Check for Understanding
1. The circular permutation has no beginning or end.
Therefore, the number of different arrangements
is always
n
1
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
charms on a bracelet with a clasp is
n
2!
.
4.
2!
8!
2!
10,080
5.
2! 2!
9!
2! 2!
22,680
6.
2! 4
1
!
4
5
!
!2!
7,567,560
7. circular; (11 1)! or 3,628,800
8. circular; (8 1)! or 5040
9. circular; (12 1)! or 39,916,800
10. linear;
5
2!
or 60
11.
5!
9
2
!
!2!
756
Pages 849–851 Exercises
12.
2!
8!
2!
10,080
13.
2
1
!
0
2
!
!
907,200
14.
8
2
!
!
20,160
15.
3
1
!
0
2
!
!
302,400
16.
2!
1
2
2
!
!
2!
59,875,200
17.
4!
9
2
!
!2!
3780
18.
2!
7
2
!
!2!
630
19.
5!
9!
4!
126
20. 2.35 1048
21. circular; (12 1)! 39,916,800
50!
4! 3! 4! 2! 8! 8! 3! 2! 2! 3!
13-2
22. linear;
6! 3!
26
7
!
!10!
5.1 1012
23. circular; (9 1)! 40,320
24. circular; (5 1)! 24
25. circular; (8 1)! 5040
26. linear; 6! 720
27. linear; 10! 3,628,800
28. circular; (9 1)! 40,320
29. circular;
(14
2
1)!
3,113,510,400
30. circular; (20 1)! 1.22 1017
31. circular; (32 1)! 8.22 1033
32. linear; 25! 1.55 1025
33.
2!2
8
!2
!
!2!
2520
34a. (7 1)! 720
34b. 7! 5040
35.
3!
1
4
1
!
!
3!
46,200
36a.
2!
1
2
1
!
!
2!
4,989,600
36b. integral calculus
37a.
20!
4
1
3
4
!
!9!
7.85 1017
37b. (43 1)! 1.41 1051
37c. 43! 6.04 1052
38. Let ntotal number of symbols.
Let n3 number of dashes.
3! (n
n
!
3)!
35
210
n33n22n210 0
Use a graphing calculator to find the solution at
n7.
39. C(6, 3)
(6
6
3
!
)! 3!
20
40. (5x1)3C(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
125x375x215x1
41. xlog2413
x
lo
l
g
o
1
g
0
10
41
2
3
x8.69
42. h6, k1
3 hp
3 6 p
3 p
(yk)24p(xh)
(y1)212(x6)
43. 2(4 3i)(7 2i) 2(28 29i6i2)
56 58i12(1)
44 58i
6 5 4 3 2 1
3 2 1 3 2 1
n(n1)(n2)(n3)!
(n3)!
441 Chapter 13
2! 4!
50!
i
u
j
u
k
u
44. v
u
w
u
3
0
0
5
2
2
10. P(s)
C(4,
C
1
(
)
7
,
C
3
(
)
3, 2)
P(f) 1 P(s)
4
3
5
3
1
1
3
2
5
1
3
2
5
2
3
3
5
odds or
1
2
2
3
11. P(s)
C(3,
C
1
(
)
7
,
C
3
(
)
4, 2)
P(f) 1 P(s)
3
3
5
6
1
1
3
8
5
1
3
8
5
1
3
7
5
odds or
1
1
8
7
12. P(rain)
1
8
0
0
0
or
4
5
P(not rain) 1
4
5
1
5
odds or
1
4
Pages 856–858 Exercises
13. P(face card)
4
5
4
2
4
1
5
2
2
or
1
3
3
14. P(a card of 6 or less)
66
5
2
66
2
5
4
2
or
1
6
3
15. P(a black, non-face card)
10
5
2
10
2
5
0
2
or
1
5
3
16. P(not a face card)
4
5
0
2
or
1
1
0
3
17. P(red)
5
5
23
1
5
0
or
1
2
18. P(white)
5
2
23
1
2
0
or
1
5
19. P(not pink)
5
5
2
2
3
or
1
7
0
20. P(red or pink)
5
5
2
3
3
1
8
0
or
4
5
21. P(2 pop)
C
C
(
(
4
4
0
,
,
2
2
)
)
7
6
80
or
1
1
30
22. P(2 country)
C
C
(
(
4
8
0
,
,
2
2
)
)
7
2
8
8
0
or
1
7
95
23. P(1 rap and 1 rock)
10
73
5
18
1
7
8
8
0
0
or
1
3
3
24. P(not rock)
C
C
(
(
2
4
2
0
,
,
2
2
)
)
2
7
3
8
1
0
or
2
7
6
7
0
25. Using the Basic Counting Principle, there are 1 1
or 1 way to roll both fives. Using the Basic
Counting Principle, P(both fives)
3
1
6
.
C(10, 1) C(18, 1)
C(40, 2)
10 10 10 10
52
1
5
—
4
5
1
3
8
5
—
1
3
7
5
1
3
2
5
—
2
3
3
5
Chapter 13 442
i
u
j
u
k
u
0
5
2
2
3
0
2
2
3
0
0
5
15i
u
6j
u
10k
u
15, 6, 10
since 2, 0, 315, 6, 1030 0 30 or
0 and 2, 5, 0 15, 6, 1030 30 0 or 0,
then the resulting vector is perpendicular to v
u
and w
u
.
45. xcos 45° ysin 45° 8 0
2
2
x
2
2
y8 0
2
x2
y16 0
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 x236, then x6 or x6.
2x1261or 32
2x1261or
1
1
28
The correct choice is E.
Probability and Odds
855–856 Check for Understanding
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.
4. Geraldo is correct. P(win)
2
3
3
3
5
or 60%.
5. P(softball)
37
7
11
2
7
1
or
1
3
6. P(not a baseball)
3
3
7
7
11
or
1
2
0
1
7. P(golf ball)
37
0
11
or 0
8. P(woman)
7
7
4
or
1
7
1
9. P(s)
C
C
(
(
4
7
,
,
3
3
)
)
P(f) 1 P(s)
3
4
5
1
3
4
5
or
3
3
1
5
odds or
3
4
1
3
4
5
—
3
3
1
5
13-3
4
4
45˚
4
4
8
8
8
8
y
x
O
26. P(s)
C
C
(
(
3
6
,
,
2
2
)
)
P(f) 1 P(s)
1
3
5
1
1
5
1
5
4
5
odds or
1
4
27. P(s)
C
C
(
(
4
6
,
,
2
2
)
)
P(f) 1 P(s)
1
6
5
1
2
5
2
5
3
5
odds or
2
3
28. P(s)
C(1,
C
1
(
)
6
,
C
2
(
)
3, 1)
P(f) 1 P(s)
1
1
5
3
1
1
5
1
3
5
or
1
5
4
5
odds or
1
4
29. P(s)
C(1,
C
1
(
)
6
,
C
2
(
)
2, 1)
C(1,
C
1
(
)
6
,
C
2
(
)
3, 1)
C(2,
C
1
(
)
6
,
C
2
(
)
3, 1)
1
2
5
1
3
5
1
6
5
or
1
1
5
1
P(f) 1 P(s)
1
1
1
5
1
or
1
4
5
odds or
1
4
1
30. P(s)
C
C
(
(
2
1
7
1,
,
3
3
)
)
P(f) 1 P(s)
2
1
9
6
2
5
5
1
1
1
9
1
5
1
1
9
1
5
1
1
8
9
4
5
odds or
1
1
8
1
4
1
1
9
1
5
—
1
1
8
9
4
5
1
1
5
1
—
1
4
5
1
5
—
4
5
2
5
—
3
5
1
5
—
4
5
33. P(s)
C(3,
C
1
(
)
2
7
C
,
(
3
2
)
4, 2)
P(f) 1 P(s)
3
2
9
2
2
7
5
6
1
3
9
2
2
5
2
8
9
2
2
8
5
or
3
9
2
2
5
2
3
3
2
3
5
odds or
2
9
3
2
3
34. P(s)
1
1
249
or
2
1
50
35. P(s)
4
5
P(f) 1 P(s)
1
4
5
or
1
5
odds or
4
1
4
5
—
1
5
3
9
2
2
5
—
2
3
3
2
3
5
443 Chapter 13
31. P(s) P(f) 1 P(s)
7
2
8
9
2
1
5
1
1
2
7
2
5
2
8
9
5
2
8
5
or
2
7
2
5
5
7
3
5
odds or
2
5
2
3
32. P(s)
C
C
(
(
1
2
4
7
,
,
3
3
)
)
P(f) 1 P(s)
2
3
9
6
2
4
5
1
2
2
2
8
5
2
2
2
8
5
1
2
9
2
7
5
odds or
1
2
9
8
7
2
2
2
8
5
—
1
2
9
2
7
5
2
7
2
5
—
5
7
3
5
C(13, 2) C(11, 1)
C(27, 3)
36. P(s) P(4, 2)
2
2
,5
8
9
6
8
,9
7
6
8
0
12
2
2
,5
6
9
7
8
,6
,9
9
6
6
0
4
4
1
2
6
9
5
P(f) 1 P(s)
1
4
4
1
2
6
9
5
or
3
4
7
1
3
6
6
5
odds or
3
4
7
2
3
9
6
37. P(s)
1
1
4
or
1
5
38. P(s) 0.325 P(f) 1 P(s)
1 0.325 or 0.675
odds
0
0
.
.
3
6
2
7
5
5
or
1
2
3
7
39a. P(s)
1
1
0
1
9
1
8
or
7
1
20
39b. P(f)
1
1
0
1
1
0
1
1
0
or
10
1
00
P(s) 1 P(f)
1
10
1
00
or
1
9
0
9
0
9
0
odds or
99
1
9
40a. P(both males)
1
2
1
2
or
1
4
40b. P(s) 1
1
2
P(f) 1 P(s)
1
2
1
1
2
or
1
2
odds or
1
1
41a. P(s)
3
1
0
8
0
4
3
,7
56
1
or
1
2
2
1
92
41b. P(s)
C(15
C
,
(
8
2
)
0,
1
C
0
(
)
5, 2)
P(f) 1 P(s)
6
1
4
8
3
4
5
,7
5
1
6
0
1
2
6
2
4
5
6
1
6
8
4
4
,
,
3
7
5
5
0
6
4
6
2
4
1
6
2
6
2
4
5
6
odds or
2
4
2
2
5
1
2
6
2
4
5
6
—
4
6
2
4
1
6
C(15, 10) C(5, 0)
C(20, 10)
1
2
_
1
2
1
9
0
9
0
9
0
—
10
1
00
4
4
1
2
6
9
5
—
3
4
7
1
3
6
6
5
C(13, 3) C(13, 2)
C(52, 5)
42a. P
3
4
3
1
1
8
,
,
1
8
4
9
2
0
0.791
42b. P(f)
1
4
5
1
1
8
,
,
3
8
2
9
2
0
P(s) 1 P(f)
1
1
4
5
1
1
8
,
,
3
8
2
9
2
0
2
4
6
1
7
8
,
,
5
8
6
9
8
0
odds
2
1
6
5
7
1
,
,
5
3
6
2
8
2
or
1
7
3
5
3
,
,
6
7
6
8
1
4
43.
Given a pipe PQ
u
and a random cut point, A,
APAQ 18. If AP is xinches long, then AQ is 8x
inches long. Now, the cut must be made along AP
u
so that the longer piece will be 8 or more times as
long as the shorter piece. Thus, the probability
that the cut is on AP
u
is
x
x
8x
1
9
. Since the cut
can be made on either end of the pipe, the actual
probability is
2
9
.
44. This is a circular permutation.
(6 1)! 5! or 120
45. C(10, 4)
(10
10
4
!
)! 4!
210
46. Sn
n
2
[2a1(n1)d]
S14
1
2
4
[2(3.2) (14 1)1.5]
7(25.9)
181.3
47. Let y7log72x
log7ylog72x
y2x
So, 7log72x2x.
48. Center: (7, 2)
r2(10 7)2(8 2)2
32(10)2or 109
(x7)2(y2)2109
49. r3 2 or 6
v(
4
) or
5
4
6(cos
5
4
isin
5
4
) 6
2
2
2
2
i
32
32
i
50. u
u
3, 54, 2
3 (4), 5 2
1, 3
10 9 8 7 6 5 4 3 2 1
6 5 4 3 2 1 4 3 2 1
2
4
6
1
7
8
,
,
5
8
6
9
8
0
——
1
4
5
1
1
8
,
,
3
8
2
9
2
0
179,820 151,322
84,475 3273 179,820 151,322
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 s3
. This is the height of the equilateral
triangle. The base measures 2s. So the area of the
equilateral triangle is
1
2
2ss3
3s2
.
The correct choice is B.
Page 858 Mid-Chapter Quiz
1. P(15, 5)
(15
1
5!
5)!
360,360
2. C(20, 9)
(20
20
9
!
)! 9!
167,960
3. Using the Basic Counting Principle,
26 26 26 10 10 10 10 175,760,000.
4. P(12, 5)
(12
1
2!
5)!
95,040
5. Using the Basic Counting Principle,
18 18 3 6 5832.
6.
9
2
!
!
181,440
7.
3!
1
4
0
!
!
3!
4200
8. This is a circular permutation.
(8 1)! 5040
9. P(both hearts)
C
C
(
(
1
5
3
2
,
,
2
2
)
)
1
7
3
8
26
or
1
1
7
10. P(s)
C(3,
C
1
(
)
12
,
C
2
(
)
3, 1)
P(f) 1 P(s)
3
6
6
3
1
2
3
2
6
9
6
or
2
3
2
1
2
9
2
odds or
1
3
9
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.
13-4
2
3
2
—
1
2
9
2
12 11 10 9 8 7 6 5 4 3 2 1
7 6 5 4 3 2 1
15 14 13 12 11 10 987654321
10 987654321
Chapter 13 444
8
xx
PA Q
2
s
2
s
3
ss
s
2a.
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,
3
6
6
3
3
6
7
1
2
5. dependent,
1
4
0
3
9
1
2
5
6. dependent,
4
6
3
5
2
5
7. exclusive,
1
4
3
1
6
3
1
1
0
3
8. inclusive,
1
2
5
7
2
1
7
1
2
6
7
2
2
0
7
9. exclusive,
5
4
2
5
4
2
5
8
2
or
1
2
3
10. P(selecting 5 even numbers)
3
7
7
5
3
7
6
4
3
7
5
3
3
7
4
2
3
7
3
1
0.025
11. P(selecting 5 two digit numbers)
6
7
6
5
6
7
5
4
6
7
4
3
6
7
3
2
6
7
2
1
0.518
12. P(5 odd numbers or 5 multiples of 4)
P(5 odd numbers) P(5 multiples of 4)
3
7
8
5
3
7
7
4
3
7
6
3
3
7
5
2
3
7
4
1
1
7
7
5
1
7
6
4
1
7
5
3
1
7
4
2
1
7
3
1
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)
3
7
7
5
3
7
6
4
3
7
5
3
3
7
4
2
3
7
3
1
2
7
9
5
2
7
8
4
2
7
7
3
2
7
6
2
2
7
5
1
1
7
4
5
1
7
3
4
1
7
2
3
7
1
2
1
1
7
0
1
0.025 0.007 0.001
0.032
14. P(none if 6 clocks are damaged)
1
9
0
6
0
9
9
5
9
9
9
4
8
9
9
3
7
9
9
2
6
9
9
1
5
4
5
3
6
5
0
,
,
6
1
4
7
3
5
15. P(at least 1 right handed pitcher
P(1 right-handed pitcher) P(2 right-handed
pitchers)
C(8,
C
1
(
)
13
,
C
2
(
)
5, 1)
C(8,
C
2
(
)
13
,
C
2
(
)
5, 0)
4
7
0
8
2
7
8
8
6
7
8
8
or
3
3
4
9
Pages 864–867 Exercises
16. dependent,
5
9
4
8
1
5
8
17. independent,
5
9
5
9
2
8
5
1
18. independent,
1
6
1
6
3
1
6
19. dependent,
4
7
3
6
2
7
20. dependent,
1
4
5
1
4
4
1
7
3
1
8
95
21. dependent,
2
5
6
2
2
5
5
1
2
5
4
0
2
4
3
9
2
4
2
8
2
4
1
7
2
4
0
6
1
4
9
5
1
4
8
4
1
4
7
3
1
4
6
2
1
4
5
1
1
4
4
0
1,16
1
0
9
,054
22. dependent,
1
2
2
8
2
8
7
2
8
6
8
3
1
2
9
23. independent,
1
5
6
1
4
6
1
7
6
1
3
0
5
24
24. independent
P(winning)
4
7
P(winning next four games)
4
7
4
7
4
7
4
7
2
2
4
5
0
6
1
25. inclusive,
1
6
1
6
3
1
6
3
1
6
1
26. inclusive,
5
4
2
2
5
6
2
5
2
2
1
7
3
27. inclusive,
2
5
6
2
1
5
2
2
5
6
2
1
8
3
28. exclusive
P(at least 3 males)
P(3 males) P(4 males) P(5 males)
C(5,
C
3
(
)
9
,
C
5
(
)
4, 2)
C(5,
C
4
(
)
9
,
C
5
(
)
4, 1)
C(5,
C
5
(
)
9
,
C
5
(
)
5, 0)
1
6
2
0
6
1
2
2
0
6
1
1
26
1
8
2
1
6
or
1
9
4
29. exclusive
P(sum of 6 or sum of 9) P(sum of 6) P(sum of 9)
3
5
6
3
4
6
3
9
6
or
1
4
30. exclusive
P(at least three women)
P(3 women) P(4 women) P(5 women)
P(6 women)
C(7,
C
3
(
)
14
,
C
6
(
)
7, 3)
C(7,
C
4
(
)
14
,
C
6
(
)
7, 2)
C(7,
C
5
(
)
14
,
C
6
(
)
7, 1)
C(7,
C
6
(
)
14
,
C
6
(
)
7, 0)
1
3
2
0
2
0
5
3
3
7
0
3
0
5
3
3
1
0
4
0
7
3
30
7
03
3
2
0
1
0
14
3
or
3
4
0
2
2
9
31. exclusive
P(at least 4 tails)
P(4 tails) P(5 tails) P(6 tails)
C(6, 4)
1
2
6C(6, 5)
1
2
6C(6, 6)
1
2
6
1
6
5
4
6
6
4
6
1
4
2
6
2
4
or
3
1
2
1
32. inclusive
5
4
2
5
3
1
2
5
6
2
2
5
5
1
5
2
2
5
1
1
2
1
6
2
52
2
6
6
5
5
0
2
26
2
52
2
6
6
6
5
0
2
or
2
5
2
5
1
33. exclusive
P(at least 2 rock) P(2 rock) P(3 rock)
C(6,
C
2
(
)
11
,
C
3
(
)
5, 1)
C(6,
C
3
(
)
11
,
C
3
(
)
5, 0)
1
7
6
5
5
1
2
6
0
5
1
9
6
5
5
or
1
3
9
3
445 Chapter 13
diamonds spades,
clubs,
hearts
aces
34. P(all red cards)
2
5
6
2
2
5
5
1
2
5
4
0
2
4
3
9
2
4
2
8
9
2
9
5
9
3
6
35. P(both kings or both aces)
P(both kings) P(both aces)
5
4
2
5
3
1
5
4
2
5
3
1
2
1
6
2
52
2
1
6
2
52
2
2
6
4
52
or
2
2
21
36. P(all diamonds)
1
5
3
2
10
37. P(both red or both queens)
P(both red) P(both queens) P(both red queens)
2
5
6
2
2
5
5
1
5
4
2
5
3
1
5
2
2
5
1
1
2
6
6
5
5
0
2
2
1
6
2
52
26
2
52
2
6
6
6
5
0
2
or
2
5
2
5
1
38. P(2 pennies)
2
5
1
2
4
0
2
1
1
39. P(2 nickels or 2 silver-colored coins)
2
7
1
2
6
0
1
2
6
1
1
2
5
0
2
7
1
2
6
0
2
4
4
2
0
0
or
4
7
40. P(at least 1 nickel) 1 P(no nickels)
1
1
2
4
1
1
2
3
0
1
3
7
0
41. P(2 dimes or 1 penny and 1 nickel)
P(2 dimes) P(1 penny and 1 nickel)
C
C
(
(
2
9
1
,
,
2
2
)
)
C(5,
C
1
(
)
21
,
C
2
(
)
7, 1)
2
3
1
6
0
2
3
1
5
0
2
7
1
1
0
42. P(all female)
1
5
0
4
9
3
8
2
7
4
1
2
43. P(all female or all male)
P(all female) P(all male)
1
5
0
4
9
3
8
2
7
1
5
0
4
9
3
8
2
7
4
1
2
4
1
2
4
2
2
or
2
1
1
44. P(at least 3 females)
P(3 females) P(4 female)
C(5,
C
3
(
)
10
,
C
4
(
)
5, 1)
C(5,
C
4
(
)
10
,
C
4
(
)
5, 0)
2
5
1
0
0
2
5
10
2
5
1
5
0
or
4
1
2
1
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,
C
2
(
)
10
,
C
4
(
)
5, 2)
C(5,
C
3
(
)
10
,
C
4
(
)
5, 1)
1
2
0
1
0
0
2
5
1
0
0
1
2
5
1
0
0
or
5
7
46. P(word processing or playing games)
P(word processing) P(playing games) P(both)
2
5
1
3
1
4
2
6
9
0
47. P(rain or lightning)
P(rain) P(lightning) P(both)
3
5
2
5
1
5
4
5
48. P(even sum)
P(3 even cards) P(2 odd cards and 1 even card)
C(5,
C
3
(
)
9
,
C
3
(
)
4, 0)
C(4,
C
2
(
)
9
,
C
3
(
)
5, 1)
1
8
0
4
3
8
0
4
4
8
0
4
or
1
2
0
1
49. P(at least 3 women)
P(3 women) P(4 women) P(5 women)
C(6,
C
3
(
)
13
,
C
5
(
)
7, 2)
C(6,
C
4
(
)
13
,
C
5
(
)
7, 1)
C(6,
C
5
(
)
13
,
C
5
(
)
7, 0)
1
4
2
2
8
0
7
1
1
2
0
8
5
7
12
6
87
1
5
2
3
8
1
7
or
1
5
4
9
3
50. P(at least 1 doctor)
P(1 doctor) P(both doctors)
1
9
0
3
0
1
3
00
1
9
0
7
0
1
7
00
1
9
0
3
0
1
9
0
7
0
10
9
,
5
0
8
00
1
9
0
0
,0
2
0
1
0
1
9
0
9
,0
7
0
9
0
51. P(supplies or money)
P(supplies) P(money) P(both)
2
8
5
1
0
2
0
2
6
5
2
0
5
0
2
3
5
7
0
5
0
1
2
0
5
6
0
2
0
or
1
5
2
3
5
1
0
52a.
52b. P(Aor Bor C) P(A) P(B) P(C) P(Aand
B) P(Aand C) P(Band C) P(Aand Band
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)
4
7
1
2
1
5
1
2
9
1
7
1
4
4
1
4
0.93
Chapter 13 446
P
(
A
)
P
(
B
)
P
(
C
)
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).
1
3
3
6
6
6
6
3
3
6
6
5
6
3
3
6
6
4
6
...
3
3
4
6
9
6
1
1
1
P(3
3
6
6
6
6
,
18
18)
0.346
54b. 1
P(3
3
6
6
6
6
,
nn)
1
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 nis 23, Pis
about 0.51. So, 23 18 or 5 more students are
needed in the class.
55a. inclusive
55b.
55c. P(flooded engine or dead battery)
P(flooded engine) P(dead battery) P(both)
1
2
2
5
1
1
0
4
5
56. P(two threes given a sum of six)
1
5
57. (7 1)! 720
58. Let bbasket. Let mmiss.
Expand (bm)20
20
r0
r!(2
2
0
0
!
r)!
b20 rmr
Find the coefficient of the b15m5term where r5.
5!(2
2
0
0
!
5)!
b205m515,504b15m5
15,504
59. No, the spill will spread no more than 2000 meters
away.
a11200; r
1
4
2
8
0
0
0
or 0.4
s
1
a
1
r
1
1
2
0
0
.
0
04
2000
(366
3
66
3
!
48)!
36618
3
3
6
4
6
8
!
!
36618
60. 12x23x4
(x2) log 12 (x4) log 3
xlog 12 2 log 12 xlog 3 4 log 3
xlog 12 xlog 3 4 log 3 2 log 12
x(log 12 log 3) 4 log 3 2 log 12
x
x6.7549
61. Let yincome and xnumber of $1.00 increases.
income (number of customers) (cost
of a ticket)
y(400 20x)(3 x)
y1200 340x20x2
y1200 20(x217x)
y1200 1445 20(x217x72.25)
(y2645) 20(x8.5)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.
62. A
4
2
k
8270
4
(
2
7
)
2
231
,560
271.5 yards
63. xx1ta1yy1ta2
x1 t(2) y5 t(4)
x1 2ty5 4t
x1, y5t2, 4
64. 2 tan x4 0
tan x2
tan1(tan x) tan1 2
x63° 26
65. Since ADB CBD and they are alternate
interior angles, A
D
LB
C
. Simply because a45
does not mean b45, so you cannot conclude that
3bisects ABC.
The correct choice is B.
Conditional Probability
Pages 870–871 Check for Understanding
1. Sample answer: If Aand Bare independent
events, then P(AB) P(A). Thus, the formula
for conditional probability becomes P(A)
P(A
P
a
(B
nd
)
B)
or P(A) P(B) P(Aand 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.
4. P(cubes match sum greater than 5)
1
2
3
3
4
6
—
2
3
6
6
13-5
4 log 3 2 log 12
log 12 log 3
447 Chapter 13
Dead
Battery Flooded
Engine
5. P(queen face card)
1
3
6. P(all heads first coin is a head)
1
4
7. P(all heads at least 1 head)
1
7
8. P(all heads at least 2 heads)
1
4
9. P(numbers match sum greater than or equal to 9)
1
5
10. P(sum is even sum greater than or equal to 9)
2
5
11. P(numbers match or sum is even sum greater
than or equal to 9)
2
5
12a. P(disease prevented)
1
6
0
8
0
6
1
2
00
1
2
3
0
12b. P(disease prevented vaccine)
1
2
7
5
12c. P(disease prevented conventional treatment)
3
5
1
0
13a. P(legal accepted)
6
7
9
0
13b. P(rejected legal)
2
2
5
1
6
00
—
1
7
0
5
0
1
6
0
9
0
—
1
7
0
0
0
2
6
0
2
0
—
1
2
0
0
0
0
2
6
0
8
0
—
1
2
0
0
0
0
3
4
6
—
1
3
0
6
3
4
6
—
1
3
0
6
3
2
6
—
1
3
0
6
1
8
—
4
8
1
8
—
7
8
1
8
—
4
8
5
4
2
—
1
5
2
2
13c. P(not rejected counterfeit)
2
1
5
Pages 872–874 Exercises
14. P(1 head at least 1 tail)
2
3
15. P(Democrat man)
1
2
16. P(first bag first chip is blue)
2
5
17. P(girls are separated girl at an end)
3
5
18. P(number end in 52 number is even)
3
5
2
!
1
1
8
19. P(2 odd numbers sum is even)
5
8
20. P(ace black)
1
1
3
21. P(4 black)
1
1
3
22. P(face card black)
1
3
3
23. P(queen of hearts black)
0
24. P(6 of clubs black)
2
1
6
5
1
2
—
2
5
6
2
5
0
2
—
2
5
6
2
5
6
2
—
2
5
6
2
5
2
2
—
2
5
6
2
5
2
2
—
2
5
6
2
2
7
0
2
—
3
7
2
2
4 3 2 1 4 3 2 1
5!
1
2
2
4
—
2
2
0
4
1
4
6
—
1
1
0
6
1
4
2
—
1
8
2
2
4
—
3
4
1
1
00
—
1
2
0
5
0
Chapter 13 448
25. P(jack or ten black)
1
2
3
26. P(second marble is green first marble was
green)
2
7
27. P(second marble is yellow first marble was
green)
5
7
28. P(second marble is yellow first marble is yellow)
4
7
29. P(salmon bass)
C(1,
C
1
(
)
6
,
C
3
(
)
5, 2)
1
4
0
or
2
5
30. P(not walleye trout and perch)
3
4
31. P(bass and perch not catfish)
C
C
(
(
5
6
,
,
3
3
)
)
1
3
0
32. P(perch and trout neither bass nor walleye)
C
C
(
(
4
6
,
,
3
3
)
)
2
4
or
1
2
33.
P(brown eyes brown hair)
0
0
.
.
1
6
0
0
1
6
34. P(no brown hair brown eyes)
0
0
.
.
2
3
0
0
2
3
35. P(no brown eyes no brown hair)
0
0
.
.
2
4
0
0
1
2
C(1, 1)C(1, 1) (2, 1)
C(6, 3)
C(1, 1) C(1, 1) C(3, 1)
C(6, 3)
C(1, 1) C(1, 1) (4, 1)
C(6, 3)
C(1, 1) C(1, 1) C(3, 1)
C(6, 3)
C(1, 1) C(1, 1) C(4, 1)
C(6, 3)
5
8
4
7
—
5
8
3
8
5
7
—
3
8
3
8
2
7
—
3
8
5
4
2
—
2
5
6
2
36. Athe sum of the cards is 7 or less
Bat least one card is an ace
Bboth cards not an ace
P(B)
C
C
(
(
4
5
8
2
,
,
2
2
)
)
1
2
8
2
8
1
P(B) 1 P(B) 1
1
2
8
2
8
1
2
3
2
3
1
P(Aand B)
6
4
6
3
3
P(AB)
P(A
P
a
(B
nd
)
B)
4
9
3
9
37. P(sum greater than 18 queen of hearts)
1
5
9
1
38a.
38b. P(cancer smokes)
5
6
39. Aperson buys something
Bperson asks questions
P(AB) or
4
5
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.
41. P(passes studied)
4
5
P(studied)
2
3
5
4
or
5
6
42a. P(defective)
1
6
0
6
00
or
5
3
0
3
0
42b. P(chip from 3-D Images defective)
2
7
2
42c. P(functioning)
1
9
0
3
0
4
0
or
4
5
6
0
7
0
1
2
0
1
00
—
1
6
0
6
00
2
3
P(studied)
P(passes and studied)
P(studied)
1
5
2
0
0
0
—
1
5
5
0
0
0
2
2
0
5
0
—
2
3
0
0
0
5
1
2
1
5
9
1
—
5
1
2
5
5
1
1
6
4
6
3
3
—
2
3
2
3
1
C(4, 2) C(4, 1) C(20, 1)
C(52, 2)
449 Chapter 13
Brown Hair
50% 20%
20%
10%
Brown Eyes
Exercises 33–35
25
CSU
15 5
155
53. (4 2x)(3 2x) 6
12 14x4x26
4x214x6 0
2(2x1)(x3) 0
x
1
2
or x3
If x3 ft then the photo would have a negative
length and width. So, x
1
2
ft or 6 in.
54. f(x)
x25
4
(x2)
5
(x2)
f(x) is discontinuous when x2;
f(x) is undefined when x2.
55. Drop an altitude
from Bto A
E
, from
Cto ED
u
, and from
Eto B
C
. Label
the diagram as
shown.
area of unshaded region
1
2
(BC)h
area of shaded region
1
2
(AE)h
1
2
(ED)h
1
2
h(AE ED)
BC AE ED since opposite sides of a
parallelogram are equal. So, the ratio of the areas
is
1
2
h(BC)
1
2
h(BC) or 11.
The correct choice is B.
The Binomial Theorem
and Probability
Page 877 Graphing Calculator Exploration
1. S{0, 1, 2}
2. P(Bobby wins)
2
3
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.
4. P(winning 5 games) C(6, 5)
2
3
5
1
3
1
0.26
5. There is not a large enough sample of trials.
6. Increase the number of repetitions.
Page 878 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
13-6
42d. Sample answer: A chip from CyberChip Corp.
has the least probability of being defective.
P(defective from CyberChip)
5
2
0
5
0
0.05
P(defective from 3-D Images)
3
2
0
1
0
0.07
P(defective from MegaView Designs)
2
2
0
0
0
0.10
43. P(AB)
P(A
P
a
(B
n
)
dB
by definition.
So, if P(A) P(AB) then by substitution P(A)
P(A
P
a
(B
nd
)
B)
or P(Aand B) P(A) P(B). Therefore,
the events are independent.
44. P(at least 3 women)
P(3 women) P(4 women) P(5 women)
C(6,
C
3
(
)
13
,
C
5
(
)
7, 2)
C(6,
C
4
(
)
13
,
C
5
(
)
7, 1)
C(6,
C
5
(
)
13
,
C
5
(
)
7, 0)
1
4
2
2
8
0
7
1
1
2
0
8
5
7
12
6
87
1
5
2
3
8
1
7
or
1
5
4
9
3
45. C(9, 4) 126
46.
b1
3(0.5)b3(0.5)13(0.5)23(0.5)3...
1.5 0.75 0.375 ...
S
a
r
1
a11.5, r 0.5
1
0
.
.
5
5
or 3
47. They are reflections of each other over the x-axis.
48.
49. r
cos v
2
5 0
r
cos vcos
2
sin vsin
2
5
0 rsin v5
y5
50. x4ty3 2t
4
x
t
y
2
3
t
4
x
y
2
3
y
2
x
3
51. A
1
2
r2v
1
2
(8)2
9
1
8
1
7
8
1
0
54.7 ft2
52. tan 27°
2
x
5
12.7 x; 12.7 m
Chapter 13 450
y
x
2
4
42466
6
4
6
O
y
x
O
(4, 5)
(0, 3)
(4, 1)
4 ft
x
x
x
x
x
x
x
x
6 ft2
AE D
BC
h
h
h
3. First, determine P(right) and P(wrong). Second,
set up the binomial expansion (prpw)5. Third,
determine the term of the expansion. Fourth,
substitute the probability values for prand pw.
Last, compute the probability of getting exactly 2
correct answers.
4. P(only one 4) C(5, 1)
1
6
1
5
6
4
3
7
1
7
2
7
5
6
5. P(no more than two 4s)
P(no 4s) P(one 4) P(two 4s)
C(5, 0)
1
6
0
5
6
5C(5, 1)
1
6
1
5
6
4
C(5, 2)
1
6
2
5
6
3
3
7
1
7
2
7
5
6
3
7
1
7
2
7
5
6
1
7
2
7
5
7
0
6
6
6
2
4
5
8
6. P(at least three 4s)
P(three 4s) P(four 4s) P(5 fours)
C(5, 3)
1
6
3
5
6
2C(5, 4)
1
6
4
5
6
1
C(5, 5)
1
6
5
5
6
0
7
2
7
5
7
0
6
7
2
7
5
76
77
1
76
6
2
4
3
8
7. P(exactly five 4s) C(5, 5)
1
6
5
5
6
0
77
1
76
8. P(not having rain on any day) C(5, 5)
1
7
0
5
1
3
0
0
1
1
0
6
0
,
,
8
0
0
0
7
0
9. P(having rain on exactly one day)
C(5, 1)
1
3
0
1
1
7
0
4
1
3
0
6
0
,
,
0
0
1
0
5
0
or
2
7
0
2
,0
0
0
3
0
10. P(having rain no more than three days)
1 [P(rain on 4 days) P(rain on 5 days)
1
C(5, 4)
1
3
0
4
1
7
0
1C(5, 5)
1
3
0
5
1
7
0
0
1
10
2
0
8
,
3
0
5
00
10
2
0
4
,0
3
00
1
9
0
6
0
,
,
9
0
2
0
2
0
or
4
5
8
0
,
,
4
0
6
0
1
0
11. P(4 do not collapse) C(6, 4)
4
5
4
1
5
2
1
3
5
8
,6
4
2
0
5
or
3
7
1
6
2
8
5
12. P(10 stocks make money) C(13, 10)
5
8
10
3
8
3
0.1372
Pages 878–880 Exercises
13. P(never the correct color) C(4, 0)
2
3
0
1
3
4
1 1
8
1
1
8
1
1
14. P(correct at least 3 times)
C(4, 3)
2
3
3
1
3
1C(4, 4)
2
3
4
1
3
0
4
2
8
7
1
3
1
1
8
6
1
1
1
2
6
7
15. P(no more than 3 times correct)
1 P(correct 4 times)
1 C(4, 4)
2
3
4
1
3
0
1 1
1
8
6
1
1
1
1
8
6
1
6
8
5
1
16. P(correct exactly 2 times) C(4, 2)
2
3
2
1
3
2
6
4
9
1
9
2
8
7
17. P(7 correct) C(10, 7)
1
2
7
1
2
3
120
1
1
28
1
8
1
1
2
5
8
18. P(at least 6 correct)
C(10, 6)
1
2
6
1
2
4C(10, 7)
1
2
7
1
2
3
C(10, 8)
1
2
8
1
2
2C(10, 9)
1
2
9
1
2
1
C(10, 10)
1
2
10
1
2
0
210
6
1
4
1
1
6
120
1
1
28
1
8
45
2
1
56
1
4
10
5
1
12
1
2
1
10
1
24
1
1
3
0
8
2
6
4
1
5
9
1
3
2
19. P(all incorrect) C(10, 0)
1
2
0
1
2
10
1 1
10
1
24
10
1
24
20. P(at least half correct)
C(10, 5)
1
2
5
1
2
5answer to 18
252
3
1
2
3
1
2
1
3
0
8
2
6
4
1
2
0
5
2
2
4
1
3
0
8
2
6
4
3
5
1
1
9
2
21. P(4 heads) C(4, 4)
1
3
4
2
3
0
1
8
1
1
1
8
1
1
22. P(3 heads) C(4, 3)
1
3
3
2
3
1
4
2
1
7
2
3
8
8
1
23. P(at least 2 heads)
C(4, 2)
1
3
2
2
3
2answer to 22 answer to 21
6
1
9
4
9
8
8
1
8
1
1
2
8
4
1
8
9
1
2
1
7
1
24. P(6 correct answers) C(10, 6)
1
4
6
3
4
4
210
40
1
96
2
8
5
1
6
0.016
451 Chapter 13
25. P(half answers correct) C(10, 5)
1
4
5
3
4
5
252
10
1
24
1
2
0
4
2
3
4
0.058
26. P(from 3 to 5 correct answers)
C(10, 3)
1
4
3
3
4
7C(10, 4)
1
4
4
3
4
6
C(10, 5)
1
4
5
3
4
5
120
6
1
4
1
2
6
1
,3
8
8
7
4
210
2
1
56
4
7
0
2
9
9
6
252
10
1
24
1
2
0
4
2
3
4
0.25 0.15 0.06
0.46
27. P(all point up) C(10, 10)
2
5
10
3
5
0
1
9,7
1
6
0
5
2
,
4
625
1
1.049 104
28. P(exactly 3 point up) C(10, 3)
2
5
3
3
5
7
120
1
8
25
7
2
8
1
,1
8
2
7
5
0.215
29. P(exactly 5 point up) C(10, 5)
2
5
5
3
5
5
252
3
3
1
2
25
3
2
1
4
2
3
5
0.201
30. P(at least 6 point up)
C(10, 6)
2
5
6
3
5
4C(10, 7)
2
5
7
3
5
3
C(10, 8)
2
5
8
3
5
2C(10, 9)
2
5
9
3
5
1
C(10, 10)
2
5
10
3
5
0
210
15
6
,6
4
25
6
8
2
1
5
120
78
1
,
2
1
8
25
1
2
2
7
5
45
39
2
0
5
,6
6
25
2
9
5
10
1,9
5
5
1
3
2
,125
3
5
1
9,7
1
6
0
5
2
,
4
625
1
0.166
31. P(3 heads or 3 tails)
C(3, 3)
1
2
3
1
2
0C(3, 3)
1
2
3
1
2
0
1
1
8
1 1
1
8
1
1
4
32. P(at least 2 heads)
C(3, 2)
1
2
2
1
2
1C(3, 3)
1
2
3
1
2
0
3
1
4
1
2
1
1
8
1
4
8
1
2
33. P(exactly 2 tails) C(3, 2)
1
2
2
1
2
1
3
1
4
1
2
3
8
34a. The values of the function for 0 x6 are the
coefficients of the binomial expansion.
34b. Change 6 nCr X to 8 nCr X on the Y-menu.
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)
1820
1
7
0
12
1
3
0
4560
1
7
0
13
1
3
0
3
120
1
7
0
14
1
3
0
2
16
1
7
0
15
1
3
0
11
1
7
0
16
1
3
0
1
0.45 or 45%
36a. Asuccess means that a missile hits its target.
There are 6 trials and the probability of success
on each trial is 20% or
1
5
.
36b. P(between 2 and 6 missiles hit the target)
1 [P(0 missiles hit the garget) P(1 missile
hits the target)]
1
C(6, 0)
1
5
0
4
5
6C(6, 1)
1
5
1
4
5
5
1
1 1
1
4
5
0
,6
9
2
6
5
6
1
5
1
3
0
1
2
2
4
5
1
3
0
1
7
2
7
5
37. P(all men or all women)
C(10, 10)
1
4
0
10
1
6
0
0C(10, 10)
1
6
0
10
1
4
0
0
0.0062
38a. P(all carry the disease) C(20, 20)
1
1
0
20
1
9
0
0
1 1020
38b. P(exactly half have the disease)
C(20, 10)
1
1
0
10
1
9
0
10
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
C(75, 0)
1
4
00
0
1
9
0
6
0
75
C(75, 1)
1
4
00
1
1
9
0
6
0
74
0.807
40. P(less than or equal to 3 policies)
1 P(4 policies)
1
C(4, 4)
1
2
4
1
2
0
1
1
1
1
6
1
1
1
5
6
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.
P(back at her starting point) C(10, 5)
1
2
5
1
2
5
252
3
1
2
3
1
2
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.
Chapter 13 452
41c. P(exactly 20 meters from the starting point)
C(10, 6)
1
2
6
1
2
4C(10, 4)
1
2
4
1
2
6
210
6
1
4
1
1
6
210
1
1
6
6
1
4
0.41
42. P(sum less than 9 both cubes are the same)
2
3
43. P(letter is contained in house or phone)
2
5
6
2
5
6
2
3
6
2
7
6
44a. 80, 75, 70, . . .
44b. T80 5n
44c. T125
n
4
1
0
0
,0
0
0
0
0
or 40
125 g5(40)
75 g; 75° F
45. 3x16x
(x1)log 3 xlog 6
xlog 3 log 3 xlog 6
xlog 3 xlog 6 log 3
x(log 3 log 6) log 3
x
log 3
lo
g3
log 6
x0.38
46. h0, k3, a7, b5, c26
Center: (0, 3)
Foci: (26
, 3)
Vertices: major axis →(7, 3) and (7, 3)
minor axis →(0, 2) and (0, 8)
47. 2
cos
2
isin
2
2
0 2
i
0 2
i
48. WX
u
6 8, 5 (3)or 2, 8
WX
u
(6 8
)2(5
(
3))2
(2)2
82
68
or 217
49.
d27125522(71)(55) cos 106°
d101.1 cm
d27125522(71)(55) cos 74°
d76.9 cm
3
4
6
—
3
6
6
50. 4112 2162 72
432 44 72
1811 18 0
Since the remainder is 0, x4 is a factor of the
polynomial.
51.
1
6
0
3
8
1
7
2
The answer is 7/12.
Chapter 13 Study Guide and Assessment
Page 881 Understanding the Vocabulary
1. independent 2. failure
3. 14. probability
5. permutation 6. permutation with
repetitions
7. mutually exclusive 8. sample space
9. conditional 10. combinatorics
Pages 882–884 Skills and Concepts
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 21 or 720.
14. P(6, 3)
(6
6!
3)!
120
15. P(8, 6)
(8
8!
6)!
20,160
16. C(5, 3)
(5
5
3
!
)! 3!
5
2
4
1
3
3
2
2
1
1
10
17. C(11, 8)
(11
11
8
!
)! 8!
165
18.
P
P
(
(
6
5
,
,
3
3
)
)
(6
6!
3)!
(5
5!
3)!
11 10 9 8 7 6 5 4 3 2 1
3 2 1 8 7 6 5 4 3 2 1
8 7 6 5 4 3 2 1
2 1
6 5 4 3 2 1
3 21
(9 1 1)(12 1.5 1.5)
9 12
Area of page for text
Area of entire page
453 Chapter 13
55 cm
71 cm
d
106˚
55 cm
71 cm
d
106˚
74˚
2
19. C(5, 5) C(3, 2)
(5
5
5
!
)! 5!
(3
3
2
!
)! 2!
1
1
3
1
2
2
1
1
3
6 5 4 3 2 1 2 1
5 4 3 2 1 3 2 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.
2 P(5, 5) 2
(5
5!
5)!
2
543
1
21
240
21. C(3, 2) C(7, 2)
(3
3
2
!
)! 2!
(7
7
2
!
)! 2!
3
1
2
2
1
1
3 21 or 63
22.
2!
5!
2!
5
2
4
1
3
2
2
1
1
30
23.
2!
1
3
0
!
!
3!
50,400
24.
8
2
!
!
20,160
25.
3!
6!
2!
60
26.
3!
9!
2!
30,240
27. P(3 pennies)
35
5
6
1
0
1
1
1
6
28. P(2 pennies and 1 nickel)
21
5
6
4
0
1
2
3
0
29. P(3 nickels)
1
5
4
60
1
1
1
40
30. P(1 nickel and 2 dimes)
1
5
4
60
10
1
1
4
31. P(s)
1
1
6
; P(f)
1
1
5
6
odds
1
1
5
32. P(s)
2
3
0;
P(f)
1
2
7
0
odds
1
3
7
33. P(s)
1
1
40
; P(f)
1
1
3
4
9
0
odds
1
1
39
1
1
40
—
1
1
3
4
9
0
2
3
0
—
1
2
7
0
1
1
6
—
1
1
5
6
C(7, 0) C(4, 1) C(5, 2)
C(16, 3)
C(7, 0) C(4, 3) C(5, 0)
C(16, 3)
C(7, 2) C(4, 1) C(5, 0)
C(16, 3)
C(7, 3) C(4, 0) C(5, 0)
C(16, 3)
9 8 7 6 5 4 3 2 1
3 2 1 2 1
6 5 4 3 2 1
3 2 1 2 1
8 7 6 5 4 3 2 1
2 1
10 9 8 7 6 5 4 3 2 1
2 1 3 2 1 3 2 1
7 6 5 4 3 2 1
5 4 3 2 1 2 1
34. P(s)
1
1
4
; P(f)
1
1
3
4
odds
1
1
3
35. independent, P(sum of 2) P(sum of 6)
3
1
6
3
5
6
12
5
96
36. dependent, P(two yellow markets)
1
4
0
3
9
1
2
5
37. P(selecting a prime number or a multiple of 4)
P(prime number) P(a multiple of 4)
1
6
4
1
3
4
1
9
4
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)
1
7
4
1
4
4
1
2
4
1
9
4
39. P(selecting a 3 or a 4) P(3) or P(4)
1
1
4
1
1
4
1
7
40. P(selecting an 8 or a number less than 8)
P(8) P(less than 8)
1
1
4
1
7
4
4
7
41. P(sum less than 5 exactly one cube shows 1)
2
5
42. P(different numbers sum is 8)
4
5
43. P(numbers match sum is greater than or equal
to 5)
1
2
5
44. P(exactly 1 head) C(4, 1)
1
2
1
1
2
3
4
1
2
1
8
1
4
45. P(no heads) C(4, 0)
1
2
0
1
2
4
1 1
1
1
6
1
1
6
3
4
6
—
3
3
0
6
3
4
6
—
3
5
6
3
4
6
—
1
3
0
6
1
1
4
—
1
1
3
4
Chapter 13 454
46. P(2 heads and 2 tails) C(4, 2)
1
2
2
1
2
2
6
1
4
1
4
3
8
47. P(at least 3 tails)
P(3 tails) P(4 tails)
C(4, 3)
1
2
3
1
2
1C(4, 4)
1
2
4
1
2
0
4
1
8
1
2
1
1
1
6
1
1
4
1
1
6
or
1
5
6
Page 885 Applications and Problem Solving
48. C(1, 1) C(6, 4) 1 15 or 15
49.
7
2!
2520
50. P(at least 1 good chip)
1 P(both defective chips)
1
C
C
(
(
1
3
5
,
,
2
2
)
)
1
3
1
5
3
3
4
5
51a. P(female name excluding Reba)
1
7
5
51b. P(Reba’s name, then a male name)
1
1
5
1
7
4
3
1
0
Page 885 Open-Ended Assessment
1. Yes; sample answer: x
1
2
1
1
2
, so x
1
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.
Chapter 13 SAT & ACT Preparation
Page 887 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.
2. 4x
–
3
832
4x
–
3
24
x
–
3
6
(x
–
3
)262
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
3
5
or
1
5
Let xrepresent the number of green marbles
added so the probability equals 2
1
5
or
2
5
.
1
3
5
x
x
2
5
5(3 x) 2(15 x)
15 5x30 2x
3x15
x5
The correct choice is C.
4. Average
nu
s
m
um
be
o
r
f
o
t
f
e
t
r
e
m
rm
s
s
20
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
xy8. Let xz12. Let z7.
x7 12, so x5, an odd number.
5 y8, so y3, an odd number.
Statement I is false, and statement III is true.
Since yzor 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
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
is
2
3
6
0
0
0
or
1
1
3
5
. The correct choice is E.
7. ∠Aand ∠Cmust 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.
P(diamond)
1
5
3
2
or
1
4
The correct choice is C.
9. 7 different entrees are offered. 3 are chosen.
The number of combinations that can be chosen
is C(7, 3)
3!(77
!3)!
37
!4
!!
35.
The correct choice is B.
sum of five terms
5
number of green marbles
total marbles
number of green marbles
total marbles
455 Chapter 13
HHHH
HHHH
H
H
10.
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.
P
P
Por
The answer is .
1
4
1
4
4
(1)2
(2)2
area of shaded region
area of target
Chapter 13 456
1
2
The Frequency Distribution
Pages 892–893 Check for Understanding
1. Aline 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.
5a.
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:
14-1 6f. Sample answer:
6g. Sample answer: 50–60
Pages 893–896 Exercises
7a.
7b. 43220
7c. Sample answer: to determine where most of
their customers live so they can target their
advertising accordingly
8a.
8b. Sample answer: Men spend more hours driving
than women.
9a.
9b. Sales; the sales revenue is increasing, and the
rental revenue has started to decrease.
10a. range 53 4or49
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
Chapter 14 Statistics and Data Analysis
Ages Frequency
40–45 2
45–50 6
50–55 12
55–60 12
60–65 7
65–70 3
Age
70+
60-69
50-59
40-49
30-39
20-29
10-19
0-9
Percent
1900 2000
16 12 8 4 0 0481216
Frequency
Age
40050
Ages of Presidents
60 70
4
0
8
12
43026
43214 43221
43212 43220 4322945414
ZIP Codes
Age
65+
50-64
35-49
20-34
16-19
Minutes Behind the Wheel
Men Women
80 60 40 20 0 020406080
2000
2003
1997
1990
1985
Dollars (in billions)
Rental Revenue Year Sales Revenue
10 8 6 4 2 0 02468101214
10e. Sample answer:
10f. Sample answer:
10g. Sample answer: 10–20
11a. range 72 16 or 56
11b. Sample answer: 10
11c. Sample answer: 10, 20, 30, 40, 50, 60, 70, 80
11d. Sample answer: 15, 25, 35, 45, 55, 65, 76
11e. Sample answer:
11f. Sample answer:
11g. Sample answer:
12a. range 1023 404 or 619
12b. Sample answer: 100
12c. Sample answer:
12d. Sample answer:
12e. Sample answer: 400–500
13a.
Chapter 14 458
Grams of Fat Frequency
0–10 7
10–20 11
20–30 10
30–40 7
40–50 2
50–60 1
Number of Nations Frequency
10–20 2
20–30 3
30–40 8
40–50 1
50–60 1
60–70 2
70–80 1
Height (feet) Frequency
400–500 5
500–600 4
600–700 2
700–800 3
800–900 1
900–1000 2
1000–1100 3
Frequency
Grams of Fat
10020
Grams of Fat in
Fast-Food Sandwiches
30 40 50 60
4
6
2
0
8
10
Frequency
Number of Nations
10
0
20
Olympic Winter
Games
30 40 50 60 70 80
4
6
2
0
8
Frequency
Number of Nations
10
0
20
Olympic Winter
Games
30 40 50 60 70 80
4
6
2
0
8
Frequency
Height (feet)
400 600 800 1000
1
00
2
3
4
5
2003
2002
2001
2000
1999
1998
1997
1996
1995
1994
Year
American League National League
Greatest Number of Stolen Bases
for a Single Player
80 60 40 20 0 020406080
13b. Sample answer:
13c. Sample answer:
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
15.
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.
16b.
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 Sand 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%
19. (c2d)7
7
r0
r!(7
7
!
r)!
(c)7r(2d)r
To find the second term, evaluate the general term
for r1.
r!(7
7
!
r)!
(c)7r(2d)r
1!(7
7
!
1)!
(c)71(2d)1
7c62d
14c6d
20. 3.6x58.9
x ln 3.6 ln 58.9
x3.18
21.
22. (xy)2x22xy y2
(x2y2) 2(xy)
16 2(8)
32
Measures of Central Tendency
Page 903 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.
14-2
459 Chapter 14
Stolen Bases Frequency
30–40 1
40–50 6
50–60 6
60–70 4
70–80 3
Frequency
Stolen Bases
30 40 50 60 70 80
2
00
4
6
China
India
Millions
of
Tons
Wheat Rice Corn United States
300
250
200
150
100
50
0
Sales
Year
1998 1999 2000
$0
$1,200,000
$800,000
$400,000
y
x
9
xy
36
O
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.
5. X
1
4
(10 10 45 58) or 30.75
Md
10
2
45
or 27.5
Mode 10
6. X
1
1
0
(21 22 23 24 28 29 31 31
34 37)
28
Md
28
2
29
or 28.5
Mode 31
7. X
1
3
(91 94 95 98 99 105 105 107
107 107 111 111 112) 100
103.23 100
10,323
Md10,500
Mode 10,700
8a. 2 8 15 6 38 31 13 7 120
120 members
8b.
8
i1
fi120
8
i1
(fiXi) 2(3) 8(7) 15(11) 6(15)
38(19) 31(23) 13(27) 7(31)
2320
X
2
1
3
2
2
0
0
or about 19.3
8c.
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 Md17 x
3
4
8
2
x
9
x3.052631579
Md17 x
Md17 3.1
Md20.1
9a.
9b. X
4
1
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. Md21
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.
Pages 904–907 Exercises
10. X
1
4
(140 150 160 170) or 155
Md
150
2
160
or 155
Mode: none
11. X
1
5
(3 3 3 6 12) or 5.4
Md3
Mode 3
12. X
1
4
(17 19 19 21) or 19
Md19
Mode 19
13. X
1
8
(3 5 5 8 14 15 18 18)
10.75
Md
8
2
14
or 11
Mode 5 and 18
14. X
1
1
2
(54 58 62 63 64 70 76 76
87 87 98)
73.5
Md
70
2
76
or 73
Mode 87
15. X
1
1
2
(5 6 6 6 7 8 9 10 11 11
11 12)
8.5
Md
8
2
9
or 8.5
Mode 6 and 11
16a. X
1
9
(117 124 139 142 145 151
155 160 172)
145 lb
16b. Md145 lb
16c. X
1
9
(122 129 144 147 150 156
160 165 177)
150 lb
Md150 lb
Each will increase by 5 lb.
17. X
1
2
(35 2(38) 39 44 3(45) 48 2(57)
59)
45.8
Md45
Mode 45
Chapter 14 460
stem leaf
067779
13344567899
200011111338
3011124468
411127
1|313
Visits Members Cumulative Members
1–5 2 2
5–9 8 10
9–13 15 25
13–27 6 31
17–21 38 69
21–25 31 100
25–29 13 113
29–33 7 120
18. X
1
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
Md
6.8
2
6.9
or 6.85
Mode: none
19. X
1
1
5
(90 91 97 98 99 105 106
109 113 3(118) 120 2(125)) 10
1088
Md1090
Mode 1180
20. stem leaf
10555577
2000555578
300555
46
55
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
21d. X
15
1
,
0
8
0
00
or about $158
21e.
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 Md150 x
3
1
0
0
1
x
5
x5
Md150 x
Md150 5
Md155
21g. Both values represent central values of the data.
22. 7.5
1
5
(2 4 5 8 x)
37.5 19 x
18.5 x
23. 6
1
4
(x2x1 2x3x1)
24 8x
3 x
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
x8, since Md
8
2
8
or 8.
25a. X
1
1
79
[9(245) 14(275) 23(325) 30(375)
33(425) 28(475) 18(525) 12(575)
7(625) 3(675) 1(725) 1(775)]
425.6
25b.
17
2
9
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 Md400 x
3
5
3
0
13
x
.5
x20.4
5
Md400 x
Md400 20.5
Md420.5
26a. X
1
5
(3.6 3.6 3.7 3.9 4.8)
3.92
Md3.7
Mode 3.6
26b. Only the mean would change. It would increase
to 4.6.
X
1
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.
X
1
4
(3.6 3.6 3.7 3.9)
3.7
Md
3.6
2
3.7
or 3.65
Mode 3.6
27a. Sample answer: {1, 2, 2, 2, 3}
27b. Sample answer: {4, 5, 9}
27c. Sample answer: {2, 10, 10, 12}
27d. Sample answer: {3, 4, 5, 6, 9, 9}
461 Chapter 14
Cumulative
Number of
Students
200–250 9 9
250–300 14 23
300–350 23 46
350–400 30 76
400–450 33 109
450–500 28 137
500–550 18 155
550–600 12 167
600–650 7 174
650–700 3 177
700–750 1 178
750–800 1 179
Weekly Cumulative
Wages Frequency Frequency
$130–$140 11 11
$140–$150 24 35
$150–$160 30 65
$160–$170 10 75
$170–$180 13 88
$180–$190 8 96
$190–$200 4 100
Scores Number
of Students
28a.
5|3 53
28b. X
5
1
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. Md6
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.
29a. X
2
1
7
[1(170) 6(190) 10(210) 6(230)
3(250) 1(270)]
215.2
29b.
2
2
7
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 Md200 x
1
2
0
0
6
x
.5
x13
Md200 x
Md200 13
Md213
29d. X
2
1
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
Md211
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.
30a. Let X
50
9
i1
(X
fi) 0 (50 5) (50 20) (50
37) (50 44) (50 52)
(50 68) (50 71)
(50 85) (50 x)
068 x
x68
The weight should be hung 68 cm from the end.
30b. Let X
50
9
i1
(X
fi) 0 (50 5) (50 20) (50
37) (50 44) (50 52)
(50 68) (50 71)
(50 85) (50 x)
(50 x)
0118 2x
118 2x
59 x
The weight should be hung 59 cm from the end.
31a. X
1
1
0
(54 55 59 59 61 62 65 75
162 226) 1000
$87,800
31b. Md
61,000
2
62,000
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.
32a. X
1
1
00
[12(2.00) 15(2.50) 31(3.00)
37(3.50) 5(4.00)
30.4
32b.
The median class is 2.75 3.25.
58 27 31 3.25 2.75 0.50
50 27 13 Md2.75 x
0
3
.5
1
0
1
x
3
x0.2096774194
Md2.75 x
Md2.75 0.21
Md2.96
Chapter 14 462
Number of Cumulative
Goals Teams Number of Teams
160–180 1 1
180–200 6 7
200–220 10 17
220–240 6 23
240–260 3 26
260–280 1 27
Grade Point Cumulative
Averages Frequency Frequency
1.75–2.25 12 12
2.25–2.75 15 27
2.75–3.25 31 58
3.25–3.75 37 95
3.75–4.25 5 100
stem leaf
0111111122222333
334445555667778
8889999
1013335899
259
32
4
53
33. He is shorter than the mean (5 11.6) and the
median (5 11.5).
X
1
1
0
(67 68 69 69 71 72 73 74
75 78)
71.6 or 5 11.6
Md
71
2
72
71.5 or 5 11.5
34.
35. dependent;
1
3
1
1
2
0
5
3
5
36. an
3
n
n
; an1
3
nn
1
1
rlim
n→
lim
n→
3
3
n
n
(n
1
(n
1
)
)
lim
n→
3n
3
(
n
n
3n
1)
lim
n→
n
3
n
1
lim
n→
3
n
n
3
1
n
1
3
0 or
1
3
Since r1, the series is convergent.
37. FnP
(1
i
i)n1
1500
(1 0
0
.0
.0
3
3
)20 1
$40,305.56
38.
39. To find the area of the triangle use Hero’s formula:
Area s(s
a)(s
b)(s
c)
, where
s
1
2
(abc) and a10, b7, and c5.
So, s
1
2
(10 7 5)
1
2
(22) 11.
Area 11(11
10)
(11
7)(11
5)
11 1
4 6
or 264
16.25.
The correct choice is A.
n
3n
1
1
—
3
n
n
Measures of Variability
Page 914 Check for Understanding
1. The median of the data is 70, Q1is 60, and Q3is
100. The interquartile range is 40 and the semi-
interquartile 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 Q3Q1
41 25
16
Semi-interquartile range
1
2
6
8
6. X
1
8
(4.45 5.50 5.50 6.30 7.80 11.00
12.20 17.20)
8.74
MD
1
8
(4.293.24…8.46)
$3.54
1
8
(4.
292
(3.2
4)2…
8.4
62
$4.11
7. X
2
1
00
[15(5000) 30(15,000) 50(25,000)
60(35,000) 30(45,000) 15(55,000)]
30,250
13,226.39
8a. X
1
1
2
(65.7 65.9 …65.9)
70.375
Md
69.0
2
70.3
or 69.65
4.25
8b. X
1
1
2
(57.3 63.3 …57.5)
80.48
Md
77.5
2
82.1
or 79.8
17.06
(23.18)2(22.98)2…25.322
12
(4.875)2(4.475)2…6.2252
12
(25,250)215 (15,250)230 …24,750215
200
14-3
463 Chapter 14
Frequency
Speed Limit
60 65 70 75
2
00
4
6
8
10
12
14
16
18
20
22
22468
468
6
4
8
8
6
4
2
y
x
O
45403530252015
8c. Los Angeles
Las Vegas
8d. Los Angeles
8e. Los Angeles is near an ocean; Las Vegas is in a
desert.
Pages 915–917 Exercises
9. interquartile range Q3Q1
24 17
7
semi-interquartile range
7
2
or 3.5
10. interquartile range Q3Q1
21.5 12
9.5
semi-interquartile range
9
2
.5
or 4.75
11. interquartile range Q3Q1
10.5 7.6
2.9
semi-interquartile range
2
2
.9
or 1.45
12.
13. X
1
6
(152 158 …721)
381
MD
1
6
(229223…340)
211
223.14
(229)2(223)2…3402
6
Chapter 14 464
10555 65 75 85 9510090807060
10555 65 75 85 9510090807060
14 16 18 20 22 24 26 28 30
01020304050607080
5678910111213141516
14. X
1
1
0
(5.7 5.7 …3.8)
4.89
MD
1
1
0
(0.810.81…1.09)
0.672
0.73
15. X
1
1
2
(369 376 …454)
403.5
MD
1
1
2
(34.527.5…50.5)
20.25
25.31
16. X
1
5
(13 22 34 55 91)
43
Variation
774
17. X
1
1
20
[2(3) 8(7) …7(31)]
19.33
6.48
18. X
9
1
0
[3(57) 7(65) …12(97)]
81.8
9.69
19. X
8
1
5
[2(80) 11(100) …7(180)]
129.65
23.29
20a. Md259 mi
20b. Q1129 mi; Q3360 mi
20c. interquartile range Q3Q1
360 129
231 mi
20d. semi-interquartile range
23
2
1
115.5 mi
20e. An outlier would lie 231 115.5 or 346.5 mi
outside of Q1or Q3. There are no such points.
20f.
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. Md282
22b. Q142; Q3770
(49.65)22 (29.65)211 …50.3527
85
(24.8)23 (16.8)27 …15.2212
90
(16.33)22 (12.33)28 …11.6727
120
(30)2(21)2(9)2122482
5
(34.5)2(27.5)2…50.52
12
0.8120.812…(1.09)2
10
0700600500400300200100
22c. interquartile range Q3Q1
770 42
728
semi-interquartile range
72
2
8
364
22d. An outlier would lie 728 364 or 1092 points
outside of Q1or Q3. There are no such points.
22e.
22f. X
1
1
9
(22 23 …966)
404.42
22g. MD
1
1
9
(382.4381.4…561.58)
316.97
22h. Variance
118,712.56
22i. 118,71
2.56
344.55
22j. There is a great variability among the number of
teams in women’s sports.
23a. Q1$3616, Md$4125, Q3$5664
23b. interquartile range Q3Q1
5664 3616
2048
23c. An outlier would lie 2048 1024 or $3072
outside of Q1or Q3. There are two such values,
$26,954 and $27,394.
23d.
23e. X
1
1
9
(2684 2929 …27,394)
6775.95
MD
1
1
9
(4091.953846.95…20,618.05)
4463.39
23f.
7103.45
23g. The data in the upper quartile is diverse.
24a.
24b. X
4
1
2
(0 0 …635)
60.40
MD
4
1
2
(60.4060.40…574.60)
67.87
24c. Variance
14,065.48
24d. 14,06
5.48
118.60
24e. The data in the upper quartile is diverse.
(60.40)2(60.40)2…(514.60)2
42
(4091.95)2(3846.95)2…20,618.052
19
382.42381.42…561.582
19
25a. X
5
1
0
[26(9) 12(11) …2(21)]
11
25b.
2.94
26. yes; when the standard deviation is less than 1;
when both equal 0 or 1
27. See students’ work.
28a. X
3
1
5
[2(4.4) 4.9 5.4 5.5 2(6.2) 6.4
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
28b. Md8.0
28c. Mode 8.9
29a. range 68 23 or 45
29b. Sample answer: 10
29c. Sample answer: 20, 30, 40, 50, 60, 70
29d. Sample answer:
29e. Sample answer:
30.
1
9
(9!) 40,302 ways
31. x10.5(8) 1 or 3
x20.5(3) 1 or 0.5
x30.5(0.5) 1 or 0.75
32. 7 ft 7(12) or 84 in.
84 9 93 in.
9
3
3
31 in.
31 in.
3
1
1
2
or 2 ft 7 in.
The correct choice is C.
(9 11)226 (11 11)212 …(21 11)22
50
465 Chapter 14
Programs Sold Frequency
20–30 2
30–40 1
40–50 2
50–60 5
60–70 2
01000800600200 400
0100 200 300 400 500 600
025,000 30,00020,00015,0005000 10,000 Frequency
Programs
Sold
10 20 30 40 50 60 70
0
0
4
2
Page 917 Mid-Chapter Quiz
1. Sample answer: 10
2. Sample answer:
3. Sample answer:
4. stem leaf
545
62245
7156789
80245678999
902335689
5454
5. X
3
1
0
(54 55 …99)
81.1
6. Md
84
2
85
or 84.5
7. Mode 89
8.
9. MD
3
1
0
(27.126.1…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.
The Normal Distribution
Pages 922–923 Check for Understanding
1. The median, mean, and mode are the same.
2. X
1.5
14-4
3.
The second curve is less variable.
4. Sample answer:
5. 50th percentile; it contains half of the data.
6a.
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
tj70
t(35) 70
t2 → 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
tj3.5
t(2) 3.5
t1.75 →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
8a.
8b.
Chapter 14 466
Frequency
Exam Score
50 90807060
Physics Exam
100
4
2
00
6
8
10
Exam Scores Frequency
50–60 2
60–70 4
70–80 6
80–90 10
90–100 8
50 60 70 80 90 100
45 1059585756555 45 1059585756555
445 655620585550515480
64 827976737067
65 959085807570
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.
Pages 923–925 Exercises
9a.
9b. 12 1(1.5) 10.513.5
9c. 12 7.5 4.5, 16.5 12 4.5
tj4.5
t(1.5) 4.5
t3 →88.7%
9d. 12 9 3, 15 12 3
tj3
t(1.5) 3
t2 →95.5%
10a. 0.683(200) 136.6; about 137
10b. 0.955(200) 191
10c.
0.6
2
83
(200) 68.3; about 68
11a. 45% corresponds to t0.6.
82 0.6(4) 79.6 84.4
11b. 80% corresponds to t1.3
82 1.3(4) 76.8 87.2
11c. 82 76 6, 88 82 6
tj6
t(4) 6
t1.5 →86.6%
11d. 82 80.5 1.5, 83.5 82 1.5
tj1.5
t(4) 1.5
t0.375 →31.1%
12a. 25% corresponds to t0.3.
402 0.3(36) 391.2412.8
12b. 402 387 15, 417 402 15
tj15
t(36) 15
t0.416
→31.1%
12c. 402 362 40, 442 402 40
tj40
t(36) 40
t1.1
→72.9%
12d. 45% corresponds to t0.6.
402 0.6(36) 380.4 423.6
13a. X
tj150 X
tj100
140 t(20) 150 140 t(20) 100
20t10 20t40
t0.5 t2
38
2
.3%
19.15%
95
2
.5%
47.75%
19.15 47.75 66.9%
13b. X
tj180 X
tj150
140 t(20) 180 140 t(20) 150
20t40 20t10
t2t0.5
95
2
.5%
47.75%
38
2
.3%
19.15%
47.75 19.15 28.6%
13c.
t0.7 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
tj7X
tj6.5
6 t(35) 76 t(3.5) 6.5
3.5t13.5t0.5
t0.29 t0.14
23
2
.6%
11.8%
8
2
%
4%
11.8 4 7.8%
14b. X
tj 6.2 X
tj 5.5
6 t(0.35) 6.2 6 t(0.35) 5.5
0.35t0.2 0.35t0.5
t0.57 t1.43
45
2
.1%
22.55%
83
2
.8%
41.9%
22.55 41.9 64.45%
14c.
t1.3 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
15a. P(no tails)
1
2
6or
6
1
4
P(one tail) 6
1
2
1
2
5or
3
3
2
P(two tails) 15
1
2
2
1
2
4or
1
6
5
4
P(three tails) 20
1
2
3
1
2
3or
1
5
6
P(four tails) 15
1
2
4
1
2
2or
6
5
4
P(five tails) 6
1
2
5
1
2
or
3
3
2
P(six tails)
1
2
6or
6
1
4
467 Chapter 14
7.5 16.51513.51210.59
140120 180 20016010080
50%
25% 25%
6.005.65 6.70 7.056.355.304.95
80%
10% 10%
15b.
15c. X
6
1
4
[0(1) 1(6) 2(15) 3(20) 4(15)
5(6) 6(1)]
3
15d. j
1.2
15e. They are similar.
16a.
The 92nd percentile is the upper limit to 84% of
the data that is centered about the mean. 84%
corresponds to t1.4. The 92nd percentile is
1.4 standard deviations above the mean.
16b.
t0.8 corresponds to 57.6% of the data centered
about the mean.
100
2
57.6
21.2
21.2 57.6 78.8 percentile
17a. X
tj22.3
20.4 t(0.8) 22.3
0.8t1.9
t2.38 →98.4%
100
2
98.4
0.8%
17b. 100 0.8 99.2%
(0 3)2(1 3)2…(6 3)2
64
18.
96% corresponds to t2.1.
61 2.1(5) 50.5 months
19a.
70% corresponds to t1.0.
65 1.0(7) 72
19b. 65 1.0(7) 58
19c. 30% corresponds to t0.4
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.
20a. a normal distribution with a small standard
deviation
20b. a normal distribution with a large standard
deviation
20c. a distribution where values greater than the
mean are more spread out than values less than
the mean
20d. a distribution where all values occur with the
same frequency
21a. 95% corresponds to t1.96
X
tj260 X
tj250
255 1.96j260 255 1.96j250
1.96j51.96j5
j2.55 j2.55
about 2.55 mL
21b. X
tj357 X
tj353
355 t(2.55) 357 355 t(2.55)353
2.55t22.55t2
t0.78 t0.78
57.6%
22a. Md
147
2
150
or 148.5
22b. Q1110, Q3200
22c. interquartile range Q3Q1
200 110
90
22d. semi-interquartile range
9
2
0
or 45
22e.
Chapter 14 468
Frequency
Number of Tails
0123456
0
20
15
10
5
0123
t
123
84%
8% 8%
0123
t
123
57.6%
21.2% 21.2%
6156 71 76665146
96%
2% 2%
6558 79 86725144
30%
20% 20%
15% 15%
100 200 300 400 500
23. X
1
9
(19 33 42 42 45 48 55 71 79)
48.2
Md45
Mode 42
24. ysec(kc) h
k:
2
k
2
k4
c:
k
c
4
c
c4
h3
ysec(44) 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
y0.05x32.22x229.72x366.92.
25b. Sample answer:
2015 1965 50
f(50) 0.05 5032.22 50229.72 50 366.92
2553 students
26.
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, x60 m1. So, x60 90 and x30.
The correct choice is E.
Graphing Calculator Exploration:
The Standard Normal Curve
Page 926
1. The domain of the function is the set of all real
numbers. The range is the set of all real numbers
ysuch that 0 y
1
2
. The graph is symmetric
with respect to the y-axis and has the x-axis as a
horizontal asymptote. the value of f(x) approaches
0 as xapproaches .
2.
0.68268949
14-4B
3. About 68.3%; the answer for Exercise 2 can be
rounded to 0.683, which equals 68.3%.
4.
5.
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; t4 corresponds to P0.999.
8.
1; no; since the curve is approaching the x-axis
asymptotically, the area is probably not exactly
equal to 1.
Sample Sets of Data
Page 930 Check for Understanding
1. Asample 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.
5. jX
7
1
3
00
or 7.3
6. jX
3
2
.4
50
0.22
14-5
469 Chapter 14
30˚
60˚
3
12
4
5
60˚
x
˚
0.95449974
0.9973002
7. A1% confidence level is given when P99% and
t2.58.
jX
5
5
36
or 0.83
internal: X
tjX
45 2.58(0.83
)
42.8547.15
8. A5% confidence interval is given when P95%
and t1.96.
jX
5
3
.6
00
0.323316156
internal: X
tjX
55 1.96jX
54.3755.63
9a. jX
3
1
.5
50
0.29
9b. P50% corresponds to t0.7.
interval: X
tjX
27.5 0.7jX
27.3027.70 min
9c. A1% confidence level is given when P99%
and t2.58.
interval: X
tjX
27.5 2.58jX
27.7628.24 min
Pages 930–932 Exercises
10. jX
1.
8
8
1
or 0.2
11. jX
5
2
.8
50
0.37
12. jX
7
1
.8
40
0.66
13. jX
1
7
4
00
0.53
14. jX
2
1
.7
30
0.24
15. jX
13
3
.
7
5
5
0.70
16. 0.056
5.
N
6
N
100
N100,000
17. jX
5.
5
3
0
0.7495331881
interval: X
tjX
335 2.58jX
333.07336.93
18. jX
4
6
0
4
or 5
interval: X
tjX
200 2.58(5)
187.1212.9
19. jX
1
2
2
00
0.8485281374
interval: X
tjX
80 2.58jX
77.8182.91
20. jX
11
1
.
0
1
0
2
0
0.3516452758
interval: X
tjX
110 2.58jX
109.09110.91
21. P90% corresponds to t1.65.
jX
1
4
00
or 0.4
interval: X
tjX
68 1.65(0.4)
67.3468.66 in.
22. jX
2
1
.4
00
or 0.24
interval: X
tjX
24 1.96(0.24)
23.5324.47
23. jX
17
3
.
5
1
0
0.9140334473
interval: X
tjX
4526 1.96jX
4524.214527.79
24. jX
2
3
8
70
1.455650686
interval: X
tjX
678 1.96jX
675.15680.85
25. jX
0.6
8
7
0
0.0749082772
interval: X
tjX
5.38 1.96jX
5.235.53
26a. X
6
1
4
[1(4) 3(6) …2(20)]
12.375
26b. j
3.37
26c. jX
3.3
6
7
4
0.42
26d. P0.95 corresponds to t1.96.
interval: X
tjX
12.375 1.96(0.42)
11.5513.20 min
26e. tjX
1
t(0.42) 1
t2.38 →about 98.4%
27. jX
1
4
2
5
1.788854382
tjX
3
t1.68 →91.1%
100 91.1 8.9%
28a. jX
1.
5
4
0
0.1979898987 or about 0.20
28b. A5% confidence interval is given when P95%
and t1.96.
interval: X
tjX
16.2 1.96jX
15.8116.59 mm
28c. P99% corresponds to t2.58.
interval: X
tjX
16.2 2.58jX
15.6916.71 mm
28d. P0.80 corresponds to t1.3
interval: X
tjX
16.2 1.3jX
15.9416.46 mm
29a. jX
4
1
5
00
or 4.5
(4 12.375)2(6 12.375)2…(20 12.375)2
64
Chapter 14 470
29b. A1% confidence level is given when P99%
and t2.58.
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
0.4
2
128
0.2064
tjX
0.2064
2.58tjX
0.2064
jX
0.08
jX
j
N
0.08
0.
N
8
N
10
N100 packages
31a. jX
1.
1
8
0
0.57
31b. interval: X
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.
32a. jX
3.
5
2
0
0.45
32b. P50% corresponds to t0.7.
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. A5% confidence interval gives a P95% and
t1.96.
3.136
tjX
1.960jX
3.136
jX
1.6
X
tjX
753.136
X
1.96(1.6) 753.136
X
750 h
33b. jX
j
N
1.6
1
j
600
64 j; 64 h
34a. 40,000 35,000 5000, 45,000 40,000 5000
tj5000
t(500) 5000
t1 →68.3%
0.683(10,000) 6830 tires
34b. X
tj30,000
40,000 t(5000) 30,000
5000t10,000
t2 →95.5%
95
2
.5%
47.75%
0.4775(10,000) 4775 tires
753.136 746.864
34c. 50% of 10,000 0.50(10,000)
5000 tires
34d. X
tj50,000
40,000 t(5000) 50,000
5000t10,000
t2 →95.5%
100%
2
95.5%
2.25%
0.0225(10,000) 225 tires
34e. X
tj25,000
40,000 t(5000) 25,000
5000t15,000
t3 →99.7%
100%
2
99.7%
0.15%
0.0015(10,000) 15 tires
35. X
1
8
(44 49 55 58 61 68 71 72)
59.75
MD
1
8
(15.7510.75…12.25)
8.25
j
9.59
36. Sn
a1
1
a
r
1rn
n10, a1
1
1
6
, r4
S10
1
1
6
1
1
6
(4)10
1 4
(15.75)2(10.75)2…(12.25)2
8
471 Chapter 14
349
1
,
6
525
or 21,845.3125
37. xy
r cos vr sin v
1
c
s
o
in
s
v
v
1tan v
tan1(1) v
45° v
38. tan xcot x2
tan x
ta
1
nx
2
tan
t
2
an
x
x
1
2
tan2x1 2 tan x
tan2x2tan x1 0
(tan x1)20
tan x1 0
tan x1
x45°
39. *2 222(2) or 0
*1 122(1) or 1
*2 *1 0 (1) or 1
The correct choice is C.
Chapter 14 Study Guide and Assessment
Page 933 Understanding the Vocabulary
1. box-and-whisker plot
2. median
3. standard error of the mean
4. range
5. measure of central tendency
6. population
7. bimodal
8. inferential statistics
9. histogram
10. 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.
14. X
1
9
(2 4 4 4 5 5 6 7 8)
5
Md5
Mode 4
15. X
1
5
(160 200 200 240 250)
210
Md200
Mode 200
16. X
1
5
(11 13 15 16 19)
14.8
Md15
Mode: none
17. X
1
8
(5.9 6.3 6.3 6.4 6.6 6.6 6.7 6.8)
6.45
Md
6.4
2
6.6
or 6.5
Mode 6.3 and 6.6
18. X
1
8
(122 128 130 131 133 135 141
146)
133.25
Md
131
2
133
or 132
Mode: none
19. interquartile range Q3Q1
5 2
3
20. semi-interquartile range
3
2
or 1.5
21. X
1
1
0
(1 1 …6)
3.4
MD
1
1
0
(2.42.4…2.6)
1.6
22. j2.4)
2(
2.4)2
…2
.62
1.74
23. 88 78 10 98 88 10
tj10
t(5) 10
t2 →95.5%
24. 88 86 2 90 88 2
tj2
t(5) 2
t0.4 →0.311
25. 90% corresponds to t1.65.
interval: X
tj88 1.65(5)
79.7596.25
26. 0.683(150) 102.45
27. 0.955(150) 143.25
28.
0.6
2
83
(150) 51.225
29. jX
1.
9
5
0
0.16
30. jX
4
1
.9
20
0.45
31. jX
2
4
5
00
or 1.25
32. jX
1
2
8
5
or 3.6
33. jX
1
5
5
0
2.121320344
interval: X
tjX
100 2.58jX
94.53105.47
34. jX
3
1
0
5
7.745966692
interval: X
tjX
90 2.58jX
70.02109.98
35. jX
2
2
4
00
1.697056275
interval: X
tjX
40 2.58jX
35.6244.38
36. jX
0
2
.5
00
0.035
37. P0.90 corresponds to t1.65.
range: X
tjX
1.8 1.65(0.035)
1.741.86 h
38. A5% confidence level is given when P95% and
t1.96.
range: X
tjX
1.8 1.96(0.035)
1.731.87 h
39. A1% confidence level is given when P99% and
t2.58.
range: X
tjX
1.8 2.58(0.035)
1.711.89 h
40. P0.90 corresponds to t1.65.
jX
1
1
.4
00
or 0.14
range: X
tjX
4.6 1.65(0.14)
4.374.83 h
Chapter 14 472
Frequency
Weight (ounces)
91011121314
2
00
4
6
8
10
12
14
16
18
Women's Tennis Shoes
Page 937 Applications and Problem Solving
41a. stem leaf
1035679
21345
399
10 10
41b. X
1
2
(10 13 15 16 17 19 21 23
24 25 39 39)
21.75
41c. Md
19
2
21
or 20
41d. Mode 39
42. X
tj80
75 t(2) 80
2t5
t2.5 →98.8%
100%
2
98.8%
0.6%
Page 937 Open-Ended Assessment
1a. Sample answer: {2, 3, 10, 20, 40}
1b. Sample answer: 15
2. See students’ work.
Chapter 14 SAT & ACT Preparation
Page 939 SAT and ACT Practice
1. The percent increase is the ratio of the number
increase to the original amount.
Amy
1
8
0
0
1
8
Brad
3
7
0
0
3
7
Cara
2
8
0
0
1
4
Dan
3
6
0
0
1
2
Elsa
2
9
0
0
2
9
The largest fraction is
1
2
. Dan has the greatest
percent increase. The correct choice is D.
2. abbc
a
b
(b
b
bc)
b(1
b
c)
1
1
c
The correct choice is B.
3. Method 1:
0.1%(m)10%(n)mx%(10n)
0.001m0.1nm
10
x
0
10n
m100nm0.1xn
100n0.1xn
100 x
Method 2
Let mrepresent a large number such as 2000.
0.1% of m0.001(2000) or 2
10% of n2, so 0.1n2 or n20.
mx%(10n)
2000
10
x
0
(10 20)
1000 x
The correct choice is D.
4. The numbers in Sare positive numbers that are
less than 100 and the square root of each number
is an integer. So the set Scontains perfect squares
between 0 and 100.
Make a list of the
numbers, n, in set S.
From your list, you can
see that the median, or
middle value for n, is 25.
The correct choice is C.
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.
6. X
1
6
(10 20 30 35 35 50)
30
There are 3 numbers larger than 30: 35, 35, and
50. The correct choice is D.
7.
The line of best fit has a rise of 2 and a run of 5. So
the slope of the line of best fit is
2
5
. The closest
answer choice to
2
5
is
1
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.
473 Chapter 14
nn
11
42
93
16 4
25 5
36 6
49 7
64 8
81 9
y
x
O
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
mean is
1
6
5
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.
10. X
1
1
0
[820 (65) (32) 0 1 2 3
32 64 820]
1
1
0
(1 2 3)
1
6
0
The answer is 0.6, 6/10, or 3/5.
Chapter 14 474
Limits
Page 945 Graphing Calculator Exploration
1.
lim
x→0
ex
x
1
1
2.
lim
x→2
x2
x2
3
x
4
2
4
3. yis undefined when x1.
4. lim
x→2
x2
x2
3
x
4
2
lim
x→2
(
(
x
x
2
2
)
)
(
(
x
x
2
1
)
)
lim
x→2
x
x
2
1
2
2
2
1
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.
Page 946 Check for Understanding
1. Sample answer: The limit of f(x) as xapproaches a
is the number that the values of f(x) get closer to
as the values of xgets closer and closer to a.
2. Sample answer: lim
x→1f(x) is the number that the
values of f(x) approach as xapproaches 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 x1.
3. Sample answer: If f(x) is continuous at xayou
can plug ainto 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 xis to 0, the closer yis to 3. So, lim
x→0f(x)
3. However, there is a point at (0, 1), so f(0) 1.
5. lim
x→2(4x22x5) 4(2)22(2) 5
16 4 5
17
15-1 6. lim
x→0(1 x2xcos x) 1 0 20cos 0
1 1 1
1
7. lim
x→2
x
x
2
2
4
lim
x→2
(x
x
2
)(x
2
2)
lim
x→2
x
1
2
2
1
2
or
1
4
8. lim
x→0
x
x
2
3
3
4
x
x
lim
x→0
x
x
(
(
x
x
2
3
4
)
)
lim
x→0
x
x
2
3
4
0
0
2
3
4
or
3
4
9. lim
x→3
x
x
2
2
3
5
x
x
1
6
0
lim
x→3
(
(
x
x
5
3
)
)
(
(
x
x
2
2
)
)
(
(
3
3
5
3
)
)
(
(
3
3
2
2
)
)
8
6
(
(
1
5
)
)
or
1
4
5
10. lim
x→2
2x
x
2
2
5
x
x
2
2
lim
x→2
(
(
2
x
x
1
1
)
)
(
(
x
x
2
2
)
)
lim
x→2
2
x
x
1
1
2(
2
2)
1
1
or 1
11a.
11b. As the molecules get farther from the center and
closer to the pipe, ris increasing. As rincreases,
v(r) gets closer and closer to 0 in./s.
Pages 946–948 Exercises
12. The closer xis to 2, the closer yis to 1. So,
lim
x→2f(x) 1. Also f(2) 1.
13. The closer xis to 0, the closer yis to 0. So,
lim
x→0f(x) 0. However, there is point discontinuity
when x0. So f(0) is undefined.
14. The closer xis to 3, the closer yis to 4. So,
lim
x→3f(x) 4. However, there is a point at (3, 2).
So f(3) 2.
15. lim
x→2(4x23x6) 4(2)33(2) 6
16 6 6
16
16. lim
x→1(x33x24) (1)33(1)24
1 3 4
0
17. lim
x→
sin
x
x
sin
0
or 0
18. lim
x→0(xcos x) 0 cos 0
0 1 or 1
475 Chapter 15
Chapter 15 Introduction to Calculus
0.02
0.2 0.4 0.6 0.8
0.04
0.06
0.08
0.10
0.12
0.14
0.16
v
(
r
)
r
v
(
r
) 0.65(0.52
r
2)
O
19. lim
x→5
x
x
2
2
5
5
lim
x→5
(x
x
5
)(x
5
5)
lim
x→5(x5)
5 5 or 10
20. lim
x→0
2
n
n2
lim
x→02n
2(0) or 0
21. lim
x→3
x2
x2
2
x
3
x
15
lim
x→3
(x
x(x
5)
(x
3
)
3)
lim
x→3
x
x
5
3
3
5
or
3
8
22. lim
x→1
13
7
48
or
8
7
23. lim
h→2
h2
h
4h
2
4
lim
h→2
(h
h
2
)(h
2
2)
lim
h→2h2
2 2 or 0
24. lim
x→3
x3
2
2
x
x
2
2
3
x
x
6
27
1
8
1
8
9
9
or
1
2
25. lim
x→0
x
x
3
3
4
x
x
2
2
2
2
x
x
lim
x→0
x
x
(
(
x
x
2
2
4
x
x
2
2
)
)
lim
x→0
x
x
2
2
4
x
x
2
2
02
02
4(
0
0)
2
2
or 1
26. lim
x→0
x
x2
co
s
x
x
lim
x→0
x
x
(x
co
s
1
x
)
lim
x→0
x
co
sx
1
0
co
s0
1
or 1
27. lim
x→0
(x2
4
)24
lim
x→0
x24x
x
44
lim
x→0
x(x
x
4)
lim
x→0(x4)
0 4 or 4
28. lim
x→2
(x
x
1
)2
2
1
lim
x→2
x22
x
x
2
11
lim
x→2
x(
x
x
2
2)
lim
x→2x
2
29. lim
x→2
x
x
3
2
8
4
lim
x→2
lim
x→2
x2
x
2x
2
4
4
4
4
4
or 3
30. lim
x→4
x
2
3
x
6
8
4
lim
x→4
lim
x→4
x24
2
x16
424(
2
4) 16
4
2
8
or
2
1
4
2(x4)
(x4)(x24x16)
(2)22(2) 4
2 2
(x2)(x22x4)
(x2)(x2)
2(3)23(3)
332(3)23 6
133(1)24(1) 8
1 6
x33x24x8
x6
31. lim
x→1lim
x→1
lim
x→1
1
x
x
(1
1
x)
lim
x→1
1
x
1
1
or 1
32. lim
x→4lim
x→4
x
2
x
2
x4
x
2
x4
x
2
1
x
x
x1
1
x
1
x1
Chapter 15 476
lim
x→4
lim
x→4
x
2
4
2 or 4
33. lim
h→0
2h3h
h
25h
lim
h→0
h(2h2
h
h5)
lim
h→0(2h2h5)
2(0)20 5 or 5
34. lim
x→0
cos
x
(
x
)
cos
0
(
0
)
cos
()
or
35.
lim
x→0
tan
x
2x
2
36.
lim
x→1
ln (2
ln
x
x
1)
0.5
37.
lim
x→1
1
x
1
8
0.5
(x4)(x
2)
x4
38.
lim
x→0
3x
x
2s
s
in
in
x
3x
4.5
39. lim
c→0aa2c
2
aa20
2
a2
Letting capproach 0 moves the foci together, so
the ellipse becomes a circle. a2is the area of a
circle of radius a.
40.
lim
t→00.07 or 7%
41. No; the graph of f(x) sin
1
x
oscillates infinitely
many times between 1 and 1 as xapproaches 0,
so the values of the function do not approach a
unique number.
42a.
42b. yes; in the last three columns, all the decimal
places of 1
x
2
2
agree with those of cos x.
43. lim
t→2
d(t
t
)
d
2
(2)
lim
t→2
16
t
t2
2
64
lim
t→2
16(t
t
2)(
2
t2)
lim
t→216(t2)
16(2 2) or 64 ft/s
44a. As xapproaches 0, the decimals for the values of
f(x) (1 x)
1
x
approach 2.71828 ... , which is the
decimal expansion of e.
44b. He ignored the exponent. As xapproaches 0 from
the positive side,
1
x
approaches infinity. A
number close to 1 raised to a large power need
not be close to 1. If xapproaches 0 from the
negative side,
1
x
approaches negative infinity. A
number close to 1 raised to a large negative
power need not be close to 1, either.
2
1
x
0
1
t
45. When P0.99, t2.58.
Find X
.
X
1.
5
4
0
0.20
Find the range.
X
tj X
16.2 2.58(0.20)
15.68416.716 mm
46. P(not getting a 7)
1
9
0
P(never getting a 7 in five spins)
1
9
0
5
1
5
0
9
0
,
,
0
0
4
0
9
0
47. (x3y)5
5
r0
r!(5
5
!
r)!
(x)5r(3y)r
To find the third term, evaluate the general term
for r2.
r!(5
5
!
r)!
(x)5r(3y)r
2!(5
5
!
2)!
(x)52(3y)2
2!
5!
3!
(x3)(9y2)
10(x3)(9y2)
90x3y2
48. (16y8)
3
4
416
3y8
3
4
23y6
8y6
49. center
5
2
5
,
5
2
1
(5, 2)
The foci are on the x-axis.
a
9
2
1
or 4
b
1
2
(5)
or 3
(x
425)2
(y
322)2
1 →
(x
16
5)2
(y
9
2)2
1
50.
51. WX
u
3 4, 6 0
7, 6
WX
u
(3
4)2
(6
0)2
49
36
85
52. C
25
1
in.
3
1
6
y
i
d
n.
17
1
6
m
0y
i
d
0.0012395804 mi
v
6
C
5
52437.09741
w
v
t
360
v
0s
14.6 rps
477 Chapter 15
f(x)
xy
10.06697
0.1 0.06908
0.1 0.06956
10.07177
2
1
x
0
1
x
x10.50.10.01 0.001
cos x0.540302 0.877583 0.995004 0.999950 1.000000
1
x
2
2
0.5 0.875 0.995 0.99995 1.000000
1234
0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
53. csc 270°
sin
1
270°
1
y
1
1
or 1
54. Use a graphing calculator to
find the rational roots at
1
4
and
4
3
.
55.
y→as x→, y→as x→
56.
(1)(6) (2)(3)
6 (6)
12
57. yes; opposite sides have the same slope
mAB
1
8
mDC
1
8
mBC
1
4
mAD
1
4
58. If 2n8, then n3.
3n2332
35or 243
Graphing Calculator Exploration:
The Slope of a Curve
Page 950
1-6. Exact answers are given. Accept all reasonable
approximations.
1. 4
2. 1
15-2A
2
6
1
3
3. 0.5
4. 1
5. 1
6. 3
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.
9. At (0, 1),
d
d
y
x
1. For other points on the curve, the
values for yand
d
d
y
x
are approximately the same.
Derivatives and Antiderivatives
Pages 957–958 Check for Understanding
1. 4x3is the derivative of x4. x4is an antiderivative
of 4x3.
2. Letting n1 in the expression
n
1
1
xn1
results in
x
0
0
, which is undefined.
15-2
Chapter 15 478
y4x52x24
xy
100 4 100
10 4 105
16
10 399804
100 4 1010
y
x
A
D
C
(0, 3)
(2, 5) (10, 4)
(8, 4)
B
y
x
(0, 1)
270˚
3. f(xh) means substitute the quantity xhinto
the function. On the other hand, f(x) hmeans
substitute xinto the function, then add hto the
result. Using f(x) hinstead of f(xh) in the
definition of the derivative results in:
lim
h→0
f(x)h
h
f(x)
lim
h→0
h
h
lim
h→01
1
You would always get 1.
4. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
3(xh) 2 (3x2)
h
13. C(x) 1000 10x0.001x2
C(x) 0 10 1x110.001(2)x21
C(x) 10 0.002x
C(1000) 10 0.002(1000)
8
The marginal cost is $8.
Pages 958–960 Exercises
14. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
2(xh
h
)2x
lim
h→0
2x2
h
h3x
lim
h→0
2
h
h
2
15. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
7(xh) 4 (7x4)
h
479 Chapter 15
lim
h→0
lim
h→0
3
h
h
3
5. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
(xh)2xh(x2x)
h
3x3h2 3x2
h
lim
h→0
lim
h→0
2xh
h
h2h
lim
h→0
h(2x
h
h1)
lim
h→02xh1
2x0 1 or 2x1
6. f(x) 2x23x5
f(x) 2 2x213 1x110
4x3
7. f(x) x32x23x6
f(x) 1 3x312 2x213 1x110
3x24x3
8. f(x) 3x42x33x2
f(x) 3 4x412 3x313 1x110
12x36x23
9. yx22x3
d
d
y
x
2 1x212 lx110
d
d
y
x
2x2
f(1) 2(1) 2
4
10. f(x) x2
F(x)
2
1
1
x21C
1
3
x3C
11. f(x) x34x2x3
F(x)
3
1
1
x314
2
1
1
x21
1
1
1
x11
3xC
1
4
x4
4
3
x3
1
2
x23xC
12. f(x) 5x52x3x24
F(x) 5
5
1
1
x512
3
1
1
x31
2
1
1
x21
4xC
5
6
x6
1
2
x4
1
3
x34xC
x22xh h2xhx2x
h
lim
h→0
lim
h→0
7
h
h
7
16. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
lim
h→0
3x
h
3h3x
lim
h→0
h
3h
3
17. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
lim
h→0
lim
h→0
h
4h
4
18. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
lim
h→0
lim
h→0
4xh 2
h
h25h
lim
h→0
h(4x
h
2h5)
4x2(0) 5 or 4x5
19. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
(xh)35(xh)26 (x35x26)
h
2x24xh 2h25x5h2x25x
h
2(xh)25(xh) (2x25x)
h
4x4h9 4x9
h
4(xh) 9 (4x9)
h
3(xh) (3x)
h
7x7h4 7x4
h
lim
h→0
x33x2h3xh2h35x210xh 5h26 x35x26
–
––––––––––––
––
h
lim
h→0
3x23x(0) 0210x5(0)
3x210x
h(3x23xh h210x5h)
h
20. f(x) 8x
f(x) 8 1x11
8
21. f(x) 2x6
f(x) 2 1x110
2
22. f(x)
1
3
x
4
5
f(x)
1
3
1x110
1
3
23. f(x) 3x22x9
f(x) 3 2x212 1x110
6x2
24. f(x)
1
2
x2x2
f(x)
1
2
2x211 1x110
x1
25. f(x) x32x25x6
f(x) 3x312 2x215 1x110
3x24x5
26. f(x) 3x47x32x27x12
f(x)
3 4x417 3x312 2x217 1x110
12x321x24x7
27. f(x) (x23)(2x7)
2x37x26x21
f(x) 2 3x317 2x216 1x110
6x214x6
28. f(x) (2x4)2
4x216x16
f(x) 4 2x2116 1x110
8x16
29. f(x) (3x4)3
27x3108x2144x64
f(x)27 3x31108 2x21144 1x110
81x2216x144
30. f(x)
2
3
x3
1
3
x2x9
f(x)
2
3
3x31
1
3
2x211 1x110
2x2
2
3
x1
31. yx3
d
d
y
x
3x2
f(1) 3(1)2
3
32. yx37x24x9
d
d
y
x
3x214x4
f(1) 3(1)214(1) 4
7
33. y(x1)(x2)
x2x2
d
d
y
x
2x1
f(1) 2(1) 1
1
34. y(5x27)2
25x470x249
d
d
y
x
100x3140x
f(1) 100(1)3140(1)
240
35. f(x) x6
F(x)
6
1
1
x61C
1
7
x7C
36. f(x) 3x4
F(x) 3
1
1
1
x114xC
3
2
x24xC
37. f(x) 4x26x7
F(x) 4
2
1
1
x216
1
1
1
x117xC
4
3
x33x27xC
38. f(x) 12x26x1
12
2
1
1
x216
1
1
1
x11xC
4x33x2xC
39. f(x) 8x35x29x3
F(x)
8
3
1
1
x315
2
1
1
x219
1
1
1
x113xC
2x4
5
3
x3
9
2
x23xC
40. f(x)
1
4
x4
2
3
x24
F(x)
1
4
4
1
1
x41
2
3
2
1
1
x214xC
2
1
0
x5
2
9
x34xC
41. f(x) (2x3)(3x7)
6x25x21
F(x) 6
2
1
1
x215
1
1
1
x1121xC
2x3
5
2
x221xC
42. f(x) x4(x2)2
x64x54x4
F(x)
6
1
1
x614
5
1
1
x514
4
1
1
x41C
1
7
x7
2
3
x6
4
5
x5C
43. f(x)
x34
x
x2x
x24x1
F(x)
2
1
1
x214
1
1
1
x11xC
1
3
x32x2xC
44. f(x)
2x2
x
5x
3
3
2x1
F(x) 2
1
1
1
x11xC
x2xC
45. Any function of the form F(x)
1
6
x6
1
4
x4
1
3
x3
xC, where Cis a constant.
46a. v(12) 15 4(12)
1
8
(12)2
81 ft/s
Chapter 15 480
46b. v(t) 15 4t
1
8
t2
v(t) 0 4 1t11
1
8
2t21
4
1
4
t
v(12) 4
1
4
(12)
7 ft/s2
46c. When t12 the car’s velocity is increasing at a
rate of 7 ft/s per second.
46d. v(t) 15 4t
1
8
t2
s(t) 15t4
1
1
1
t11
1
8
2
1
1
t21C
15t2t2
2
1
4
t3C
When t0, s(t) should equal 0, so C0.
s(t) 15t2t2
2
1
4
t3
46e. s(12) 15(12) 2(12)2
2
1
4
(12)3
540 ft
47. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
x
1
h
1
x
h
49a. h(t) 3 80t16t2
h(t) v(t) 0 80 1t1116 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
50. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
exh
h
ex
lim
h→0
exe
h
hex
lim
h→0
ex(e
h
h1)
exlim
h→0
eh
h
1
Acalculator indicates that lim
h→0
eh
h
1
1, so
f(x) ex1 ex. i. e. exis its own derivative.
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)
100p2p2
r(p) 100 1p112 2p21
100 4p
0 100 4p
100 4p
25 p; 25 cents
52. lim
x→3
x2
x
3x
3
3
lim
x→3
(x
x
3
)(x
3
1)
lim
x→3x1
3 1 or 4
53a.
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.
P(sum 7 given that the numbers match)
3
6
1
2
55. ana1rn1
a69
1
3
61
2
9
43
or
2
1
7
56. y136e0.06(30) 74
96°
481 Chapter 15
lim
h→0
x(x
x
h)
x(
x
x
h
h)
h
lim
h→0
lim
h→0
x(x
1
h)
x(x
1
0)
x
1
2
48a. When y2010, 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).
Find the slope of the tangent line at (2010, 2.75).
T(2010)
0.19
In 2010 the amount spent on health care will be
increasing at a rate of about $190 billion per
year.
2.75 – 2.56
2010 – 2009
x(x
h
h)
h
Health Care Spending
(Trillions of Dollars)
$3.5
0
0.5
1.0
2.5
2.0
1.5
3.0
2010
Source: Centers for Medicare & Medicaid Services
20132007
Projections
20042001
170 190 210 230 250180 200 220 240 260
Page 960 Mid-Chapter Quiz
1. lim
x→3(2x24x6) 2(3)24(3) 6
18 12 6
36
2. lim
x→2
x
2
2
x2
9x
7x
14
6
lim
x→2
(
(
2
x
x
7
3
)
)
(
(
x
x
2
2
)
)
lim
x→2
2
x
x
7
3
2(
2
2)
7
3
or 5
3. lim
x→0
sin
x
2x
2
4. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
(xh)23 (x23)
h
Chapter 15 482
x
sin
x
x
10.9093
0.1 1.9867
0.01 1.9999
0.01 1.9999
0.1 1.9867
10.9093
lim
h→0
lim
h→0
2xh
h
h2
lim
h→0
h(2x
h
h)
2x0 or 2x
5. f(x)
f(x) 0is a constant.
6. f(x) 3x25x2
f(x) 3 2x215 1x110
6x5
7. R(M) M2
C
2
M
3
C
2
M2
1
3
M3
R(M)
C
2
2M21
1
3
3M32
CM M2
8. f(x) x27x6
F(x) 1
2
1
1
x217
1
1
1
x116xC
1
3
x3
7
2
x26xC
9. f(x) 2x3x28
F(x) 2
3
1
1
x31
2
1
1
x218xC
1
2
x4
1
3
x38xC
10. f(x) 2x46x32x5
F(x)
2
4
1
1
x416
3
1
1
x312
1
1
1
x115xC
2
5
x5
3
2
x4x25xC
x22xh h23 x23
h
y
y
3 sin( 45˚)
180˚270˚360˚450˚
90˚
O
57. x2y2Dx Ey F0
22(1)22DEF0
→2DEF5 0
(3)2023DF0 →3DF9 0
1242D4EF0 →D4EF17 0
2DEF5 03DF90
(3DEF9) 0(D4EF17) 0
5DE4 04D4E80
20D4E16 04
1
3
4E8 0
(4D4E8) 04E
2
3
8
24D80E
7
3
D
1
3
3
1
3
F9 0
F8
The solution of the system is D
1
3
, E
7
3
, and
F8
x2y2
1
3
x
7
3
y8 0
x
1
6
2
y
7
6
2
1
1
6
8
9
58. 5
cos
5
6
isin
5
6
5
2
3
i
1
2
5
2
3
5
2
i
59. xx1ta1→x8t3
yy1ta2→y3t2
60.
61.
d2253231922(253)(319) cos 42° 12
d165,7
77
161,4
14 cos
42° 1
2
d214.9 m
62.
p
q
:
1
2
, 1,
3
2
, 3
1 2383
253
3
253
0
63
1
2
21
0
1
2
0
The rational roots are 3,
1
2
, and 1.
63. 90 xyz360
xyz270
The correct choice is D.
42˚12
253 m 319 m
H
2
H
1
A
d
Area Under a Curve
Page 966 Check for Understanding
1. Sample answer: yx4
2. Sample answer: Subdivide the interval from ato b
into nequal 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 napproaches 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.
4.
2
0
x2dx lim
n→
n
i 1
2
n
i
2
n
2
lim
n→
n
8
3
n(n1)
6
(2n1)
lim
n→
4
3
2n3
n
3
3
n2n
lim
n→
4
3
2
n
3
n
1
2
8
3
units2
5.
3
1
x2dx
3
0
x2dx
1
0
x2dx
lim
n→
n
i1
3
n
i
2
n
3
lim
n→
n
i1
n
i
2
n
1
lim
n→
2
n
7
3
n(n1)
6
(2n1)
lim
n→
n
1
3
n(n1)
6
(2n1)
lim
n→
9
2
2n3
n
3
3
n2n
lim
n→
1
6
2n3
n
3
3
n2n
lim
n→
9
2
2
n
3
n
1
2
lim
n→
1
6
2
n
3
n
1
2
9
1
3
or
2
3
6
unit2
6.
1
0
x3dx lim
n→
n
i1
n
i
3
n
1
lim
n→
n
1
4
n2(n
4
1)2
lim
n→
1
4
n2
n
2
2
n1
lim
n→
1
4
1
n
2
n
1
2
1
4
unit2
7.
6
0
x2dx lim
n→
n
i1
6
n
i
2
n
6
lim
n→
2
n
1
3
6
n(n1)
6
(2n1)
lim
n→36
2n3
n
3
3
n2n
lim
n→36
2
n
3
n
1
2
72
15-3 8.
3
0
x3dx lim
n→
n
i1
3
n
i
3
n
3
lim
n→
8
n
1
4
n2(n
4
1)2
lim
n→
8
4
1
n2
n
2
2
n1
lim
n→
8
4
1
1
n
2
n
1
2
8
4
1
9a.
6
0
32tdt lim
n→
n
i1
32
6
n
i
n
6
lim
n→
11
n
5
2
2
n(n
2
1)
lim
n→576
n2
n
2
n
lim
n→576
1
n
1
576 ft
9b.
10
0
32tdt lim
n→
n
i1
32
1
n
0i
1
n
0
lim
n→
32
n
0
2
0
n(n
2
1)
lim
n→1600
n2
n
2
n
lim
n→1600
1
n
1
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.
Pages 966-968 Exercises
10.
2
0
(x1)dx lim
n→
n
i1
2
n
i
1
n
2
lim
n→
n
2
2(
n
1)
1
2(
n
2)
1
. . .
2
n
n
1
lim
n→
n
2
n
n
2
(1 2 ... n)
lim
n→
n
2
n
n
2
n(n
2
1)
lim
n→
n
2
(2n1)
lim
n→
4
n
n
n
2
4 units2
11.
3
0
x2dx lim
n→
n
i1
3
n
i
2
n
3
lim
n→
2
n
7
3
n(n1)
6
(2n1)
lim
n→
9
2
2n3
n
3
3
n2n
lim
n→
9
2
2
n
3
n
1
2
9 units2
483 Chapter 15
12.
2
1
x2dx
0
1
x2dx
2
0
x2dx
1
0
x2dx
2
0
x2dx By symmetry
lim
n→
n
i1
n
i
2
n
1
lim
n→
n
i1
2
n
i
2
n
2
lim
n→
n
1
3
n(n1)
6
(2n1)
lim
n→
n
8
3
n(n1)
6
(2n1)
lim
n→
1
6
2n3
n
3
3
n2n
lim
n→
4
3
2n3
n
3
3
n2n
lim
n→
1
6
2
n
3
n
1
2
lim
n→
4
3
2
n
3
n
1
2
1
3
8
3
or 3 units2
13.
3
1
xdx
3
0
xdx
1
0
xdx
lim
n→
n
i1
3
n
i
n
3
lim
n→
n
i1
n
i
n
1
lim
n→
n
9
2
n(n
2
1)
lim
n→
n
1
2
n(n
2
1)
lim
n→
9
2
n2
n
2
n
lim
n→
1
2
n2
n
2
n
lim
n→
9
2
1
n
1
lim
n→
1
2
1
n
1
9
2
1
2
or 4 units2
14.
5
0
x2dx lim
n→
n
i1
5
n
i
2
n
5
lim
n→
1
n
2
3
5
n(n1)
6
(2n1)
lim
n→
12
6
5
2n3
n
3
3
n2n
lim
n→
12
6
5
2
n
3
n
1
2
12
3
5
units2
15.
5
1
2x2dx
5
0
2x3
1
0
2x3
lim
n→
n
i1
2
5
n
i
3
n
5
lim
n→
n
i1
2
n
i
3
n
1
lim
n→
2
n
5
4
0
n2(n
4
1)2
lim
n→
n
2
4
n2(n
4
1)2
lim
n→
62
2
5
n2
n
2
2
n1
lim
n→
1
2
n2
n
2
2
n1
lim
n→
62
2
5
1
n
2
n
1
2
lim
n→
1
2
1
n
2
n
1
2
62
2
5
1
2
or 312 units2
16.
5
0
x4dx lim
n→
n
i1
5
n
i
4
n
5
lim
n→
31
n
2
5
5
lim
n→
62
6
5
6
1
n
5
1
n
0
2
n
1
4
625 units2
6n515n410n3n
30
17.
4
0
(x26x)dx lim
n→
n
i1
4
n
i
26
4
n
i
n
4
lim
n→
n
4
4
n
1
26
4
n
1
4
n
2
26
4
n
2
. . .
4
n
n
26
4
n
n
lim
n→
n
4
1
n
6
2
(1222. . . n2)
2
n
4
(1 2 . . . n)
lim
n→
n
4
1
n
6
2
n(n1)
6
(2n1)
2
n
4
n(n
2
1)
lim
n→
6
n
4
3
2n33
6
n2n
9
n
6
2
n2
2
n
lim
n→
3
3
2
(2
n
3
n
1
2
) 48
1
n
1
6
3
4
48 or
20
3
8
units2
18.
3
0
(x2x1)dx
lim
n→
n
i1
3
n
i
2
3
n
i
1
n
3
lim
n→
n
3
3
n
1
2
3
n
1
1
3
n
2
2
3
n
2
1
. . .
3
n
n
2
3
n
n
1
lim
n→
n
3
n
9
2
(1222. . . n2)
n
3
(1 2 . . . n) n
lim
n→
n
3
n
9
2
n(n1)
6
(2n1)
n
3
n(n
2
1)
n
lim
n→
2
n
7
3
2n23
6
n2n
n
9
2
n2
2
n
3
lim
n→
9
2
2
n
3
n
1
2
9
2
1
n
1
3
9
9
2
3
12
9
2
or
1
2
5
units2
19. lim
n→
n
i1
sin i
n
n
20.
2
0
8xdx lim
n→
n
i1
8
2
n
i
n
2
lim
n→
3
n
2
2
n(n
2
1)
lim
n→16
n2
n
2
n
lim
n→16
1
n
1
16
Chapter 15 484
21.
4
1
(x2)dx lim
n→
n
i1
1
3
n
i
2
n
3
lim
n→
n
3
n
i1
3
3
n
i
lim
n→
n
3
3
3
n
1
3
3
n
2
. . .
3
3
n
n
lim
n→
n
3
3n
n
3
(1 2 . . . n)
lim
n→
n
3
3n
n
3
n(n
2
1)
lim
n→9
n
9
2
n2
2
n
lim
n→9
9
2
1
n
1
9
9
2
or
2
2
7
22.
4
0
x2dx lim
n→
n
i1
4
n
i
2
n
4
lim
n→
6
n
4
3
(122232. . . n2)
lim
n→
6
n
4
3
n(n1)
6
(2n1)
lim
n→
3
3
2
2n3
n
3
3
n2n
lim
n→
3
3
2
2
n
3
n
1
2
6
3
4
23.
5
3
8x3dx
5
0
8x3dx
3
0
8x3dx
lim
n→
n
i1
8
5
n
i
3
n
5
lim
n→
n
i1
8
3
n
i
3
n
3
lim
n→
50
n
0
4
0
n2(n
4
1)2
lim
n→
6
n
4
4
8
n2(n
4
1)2
lim
n→1250
n2
n
2
2
n1
lim
n→162
n2
n
2
2
n1
lim
n→1250
1
n
2
n
1
2
lim
n→162
1
n
2
n
1
2
1250 162 or 1088
24.
4
1
(x24x2)dx
4
0
(x24x2)dx
1
0
(x24x2)dx
lim
n→
n
i1
4
n
i
24
4
n
i
2
n
4
lim
n→
n
i1
n
i
24
n
i
2
n
1
lim
n→
n
4
4
n
1
24
4
n
1
2
4
n
2
2
4
4
n
2
2
...
4
n
n
24
4
n
n
2
lim
n→
n
1
n
1
24
n
1
2
n
2
2
4
n
2
2
. . .
n
n
24
n
n
2
lim
n→
n
4
1
n
6
2
(1222. . . n2)
1
n
6
(1 2 . . . n) 2n
lim
n→
n
1
n
1
2
(1222. . . n2)
n
4
(1 2 . . . n) 2n
lim
n→
6
n
4
3
n(n1)
6
(2n1)
6
n
4
2
n(n
2
1)
8
lim
n→
n
1
3
n(n1)
6
(2n1)
n
4
2
n(n
2
1)
2
lim
n→
3
3
2
2n3
n
3
3
n2n
32
n2
n
2
n
8
lim
n→
1
6
2n3
n
3
3
n2n
2
n2
n
2
n
2
lim
n→
3
3
2
2
n
3
n
1
2
32
1
n
1
8
lim
n→
1
6
2
n
3
n
1
2
2
1
n
1
2
6
3
4
32 8
1
3
2 2
24
6
3
3
or 45
25.
2
0
(x5x2)dx lim
n→
n
i1
2
n
i
5
n
2
2
n
i
2
n
2
lim
n→
n
2
n
i1
2
n
i
5
2
n
i
2
lim
n→
n
2
2
n
1
5
2
n
1
2
2
n
2
5
2
n
2
2
. . .
2
n
n
5
2
n
n
2
lim
n→
n
2
3
n
2
5
(1525. . . n5)
n
4
2
(1222. . . n2)
lim
n→
6
n
4
6
n
8
3
n(n1)
6
(2n1)
lim
n→
1
3
6
4
3
2n3
n
3
3
n2n
lim
n→
1
3
6
2
n
6
n
5
2
n
1
4
4
3
2
n
3
n
1
2
3
3
2
8
3
or
4
3
0
26.
5
0
x3dx lim
n→
n
i1
5
n
i
3
n
5
lim
n→
6
n
2
4
5
n2(n
4
1)2
lim
n→
62
4
5
n2
n
2
2
n1
lim
n→
62
4
5
1
n
2
n
1
2
62
4
5
2n66n55n4n2
n6
2n66n55n4n2
12
485 Chapter 15
27.
3
0
1
2
x3dx lim
n→
n
i1
1
2
3
n
i
3
n
3
lim
n→
2
8
n
1
4
n2(n
4
1)2
lim
n→
8
8
1
n2
n
n
2
1
lim
n→
8
8
1
1
n
1
n
1
2
8
8
1
or 10.125 ft2
28a. f(20)80 2(20)
$40
28b.
40
20
(80 2x)dx
lim
n→
n
i1
80 2
20
2
n
0i
2
n
0
lim
n→
2
n
0
n
i1
40
4
n
0i
lim
n→
2
n
0
40
4
n
1
40
4
n
2
...
40
4
n
n
lim
n→
2
n
0
40n
4
n
0
(1 2 . . . n)
lim
n→
800
8
n
0
2
0
n(n
2
1)
lim
n→
800 400
n2
n
2
n
lim
n→
800 400
1
n
1
800 400 or $400
29.
10
0
(6 0.06x2)dx
lim
n→
n
i1
6 0.06
1
n
0i
2
1
n
0
lim
n→
1
n
0
6 0.06
10
n
1
2
6 0.06
10
n
2
2
. . .
6 0.06
10
n
n
2
lim
n→
1
n
0
6n
n
6
2
(1222. . . n2)
lim
n→
60
6
n
0
3
n(n1)
6
(2n1)
lim
n→
60 10
2n3
n
3
3
n2n
lim
n→
60 10
2
n
3
n
1
2
60 20 or 40
10
10
6 0.06x2dx 2(40) or 80 By symmetry
To make a tunnel 100 ft long, multiply 80 by 100.
80(100) 8000 ft3
30. Setting the two functions equal
to each other and solving for x,
we find that the curves cross
when x0 and x1. xx2
for 0 x1, so we can find
the desired area by subtracting
the area between the graph of
yx2and the x-axis from the
area between the graph of
yxand the x-axis. These areas are
1
3
unit2and
1
2
unit2, respectively, so the answer is
1
2
1
3
1
6
unit2.
31a.
31b.
12
0
(50 36t3t2)dt
lim
n→
n
i1
50 36
1
n
2i
3
1
n
2i
2
1
n
2
lim
n→
1
n
2
50 36
12
n
1
3
12
n
1
2
50 36
12
n
2
3
12
n
2
2
...
50 36
12
n
n
3
12
n
n
2
lim
n→
1
n
2
50n
4
n
32
(1 2 ... n)
4
n
3
2
2
(1222... n2)
lim
n→
600
51
n
8
2
4
n(n
2
1)
51
n
8
3
4
n(n1)
6
(2n1)
lim
n→
600 2592
n2
n
2n
864
2n3
n
3
3
n2n
lim
n→
600 2592
1
n
1
864
2
n
3
n
1
2
600 2592 1728
$1464
31c.
$1
1
4
2
64
$122
32a.
32b.
10
0
(3.5t0.25t2)dt
lim
n→
n
i1
3.5
1
n
0i
0.25
1
n
0i
2
1
n
0
lim
n→
1
n
0
3.5
10
n
1
0.25
10
n
1
2
3.5
10
n
2
0.25
10
n
2
2
...
3.5
10
n
n
0.25
10
n
n
2
lim
n→
1
n
0
3
n
5
(1 2 ... n)
2
n
5
2
(1222
... n2)
lim
n→
3
n
5
2
0
n(n
2
1)
2
n
5
3
0
n(n1)
6
(2n1)
lim
n→
175
n2
n
2
n
12
3
5
2n3
n
3
3
n2n
lim
n→
175
1
n
1
12
3
5
2
n
3
n
1
2
175
25
3
0
or
27
3
5
m
Chapter 15 486
y
x
y
x
2
y
x
1
1
2
O
r
(
t
)
t
r
(
t
) 50 36
t
3
t
2
160
40
80
120
O
v
(
t
)
t
v
(
t
) 3.5
t
0.25
t
2
v
(
t
) 1.2
t
0.03
t
2
10
15
5
12345678910
O
10
0
(1.2t0.03t2)dt
lim
n→
n
i1
1.2
1
n
0i
0.03
1
n
0i
2
1
n
0
lim
n→
1
n
0
1.2
10
n
1
0.03
10
n
1
2
1.2
10
n
2
0.03
10
n
2
2
...
1.2
10
n
n
0.03
10
n
n
2
lim
n→
1
n
0
1
n
2
(1 2 ... n)
n
3
2
(1222... n2)
lim
n→
1
n
2
2
0
n(n
2
1)
3
n
0
3
n(n1)
6
(2n1)
lim
n→
60
n2
n
2
n
5
2n3
n
3
3
n2n
lim
n→
60
1
n
1
5
2
n
3
n
1
2
60 10 or 70 m
The first one results in a greater distance
covered.
33. The equation yr2x
2
can be rearranged to
obtain x2y2r2, which is the circle centered at
the origin of radius r.Inthe equation yr2x
2
,
ymust be nonnegative, so the graph is only the
top half of the circle. Therefore, the value of the
integral is
1
2
the area of a circle of radius r,
or
1
2
r2.
34. f(x) 3x2x27x
f(x) 3 3x311 2x217 1x11
9x22x7
35. lim
x→2
x
x
2
2
2
2
2
2
0
4
or 0
36. log
1
3
x3
x
1
3
3
x32or 27
37. u
u
2, 5, 33, 4, 7
2 (3), 5 4, 3 (7)
1, 1, 10
i
u
j
u
10k
u
38. cos 2v1 2 sin2v
1 2
3
5
2
1
1
2
8
5
2
7
5
39. y
1
2
sin 10v
Amplitude
1
2
Period
2
1
0
or
5
40. Use a graphing calculator to find the maximum
width of x5.4 cm.
[5, 10] scl1 by [30, 90] scl1
41. 6282102
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
ABC
1
2
(6)(8) or 24 square inches. If the
rectangle has a width of 3 inches, then it has a
length of
2
3
4
or 8 inches, since Aw. The perimeter
of the rectangle 2(3) 2(8) or 22 inches. The
correct choice is C.
Page 969 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.
The Fundamental Theorem of
Calculus
Pages 972–973 Check for Understanding
1. f(x)dx represents all of the function that have
f(x) as their derivative.
b
a
f(x)dx is a number; it
gives the area under the graph of yf(x) from
xato xb.
2. Sample answer: Let a0, b1, f(x) x, and
g(x) x. Then
b
a
f(x)g(x)dx
1
0
x2dx
1
3
, but
b
a
f(x)dx
b
a
g(x)dx
1
0
xdx
1
0
xdx
1
2
1
2
1
4
.
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.
5. (2x24x3)dx 2
1
3
x34
1
2
x23xC
2
3
x32x23xC
6. (x33x1)dx
1
4
x43
1
2
x2xC
1
4
x4
3
2
x2xC
15-4
487 Chapter 15
7.
0
2
(4 x2)dx 4x
1
3
x3
0
2
40
1
3
03
4(2)
1
3
(2)3
0
8
8
3
1
3
6
units2
8.
2
0
x4dx
1
5
x5
2
0
1
5
25
1
5
05
3
5
2
0 or
3
5
2
units2
9.
1
1
(x24x4)dx
1
3
x34
1
2
x24x
1
1
1
3
x32x24x
1
1
1
3
132 124 1
1
3
(1)32 (1)24 (1)
1
3
9
7
3
or
2
3
6
units2
10.
3
1
2x3dx 2
1
4
x4
3
1
1
2
x4
3
1
1
2
34
1
2
14
8
2
1
1
2
or 40
11.
4
1
(x2x6)dx
1
3
x3
1
2
x26x
4
1
1
3
43
1
2
426 4
1
3
13
1
2
126 1
1
3
12
3
6
5
22
6
4
3
6
5
or
6
2
3
12.
2
0
(2x23x2)dx 2
1
3
x33
1
2
x22x
2
0
2
3
x3
3
2
x22x
2
0
2
3
23
3
2
222 2
2
3
0
3
2
0 2 0
1
3
4
0 or
1
3
4
13.
4
2
(x3x6)dx
1
4
x4
1
2
x26x
4
2
1
4
44
1
2
426 4
1
4
24
1
2
226 2
48 (6) or 54
14.
10
1
cm
10
1
0
m
cm
0.1 m
0.1
0
500xdx 500
1
2
x2
0.1
0
250x2
0.1
0
(250 (0.1)2) (250 02)
2.5 0 or 2.5 J
Pages 973–976 Exercises
15. x5dx
1
6
x6C
16. 6x7dx 6
1
8
x8C
3
4
x8C
17. (x22x4)dx
1
3
x32
1
2
x24xC
1
3
x3x24xC
18. (3x2x6)dx 3
1
3
x31
1
2
x26xC
x3
1
2
x26xC
19. (x42x23)dx
1
5
x52
1
3
x33xC
1
5
x5
2
3
x33xC
20. (4x56x37x28)dx
4
1
6
x66
1
4
x47
1
3
x38xC
2
3
x6
3
2
x4
7
3
x38xC
21. (x26x3)dx
1
3
x36
1
2
x23xC
1
3
x33x23xC
22.
3
0
2x2dx 2
1
3
x3
3
0
2
3
x3
3
0
2
3
33
2
3
03
18 0 or 18 units2
23.
3
2
(x22)dx
1
3
x32x
3
2
1
3
332 3
1
3
232 2
3
4
3
or
1
3
3
units2
24.
2
0
(4xx3)dx 4
1
2
x21
1
4
x4
2
0
2x2
1
4
x4
2
0
222
1
4
24
202
1
4
04
4 0 or 4 units2
25.
4
0
x3dx
1
4
x4
4
0
1
4
44
1
4
04
64 0, or 64 units2
26.
1
1
3x6dx 3
1
7
x7
1
1
3
7
x7
1
1
3
7
17
3
7
(1)7
3
7
3
7
or
6
7
unit2
27.
0
2
(x22x)dx
1
3
x32
1
2
x2
0
2
1
3
x3x2
0
2
1
3
0302
1
3
(2)3(2)2
0
2
3
0
or
2
3
0
units2
Chapter 15 488
28.
3
1
(x22x3)dx 1
1
3
x32
1
2
x23x
3
1
1
3
x3x23x
3
1
1
3
33323 3
1
3
13123 1
9
1
3
1
or
1
3
6
units2
29.
1
0
(x3x)dx
1
4
x4
1
2
x2
1
0
1
4
14
1
2
12
1
4
04
1
2
02
3
4
0 or
3
4
unit2
30.
3
1
(x38x10)dx
1
4
x48
1
2
x210x
3
1
1
4
x44x210x
3
1
1
4
344 3210 3
1
4
(1)44 (1)210 1
34
4
5
2
4
3
or 92 units2
31.
7
0
6x2dx 6
1
3
x3
7
0
2x3
7
0
(2 73) (2 03)
686 0 or 686
32.
4
2
3x4dx 3
1
5
x5
4
2
3
5
x5
4
2
3
5
45
3
5
25
30
5
72
9
5
6
or
29
5
76
33.
3
1
(x4)dx
1
2
x24x
3
1
1
2
3243
1
2
(1)241
3
2
3
7
2
or 20
34.
5
1
(3x22x1)dx 3
1
3
x32
1
2
x2x
5
1
x3x2x
5
1
(53525) (13121)
105 1 or 104
35.
3
1
(x3x2)dx
1
4
x4
1
3
x3
3
1
1
4
34
1
3
33
1
4
14
1
3
13
4
4
5
1
1
2
1
1
3
2
5
1
1
2
or
3
3
4
36.
1
0
(x42x21)dx
1
5
x52
1
3
x3x
1
0
1
5
x5
2
3
x3x
1
0
1
5
15
2
3
131
1
5
05
2
3
030
2
1
8
5
0 or
2
1
8
5
37.
0
1
(x4x3)dx
1
5
x5
1
4
x4
0
1
1
5
05
1
4
04
1
5
(1)5
1
4
(1)4
0
2
9
0
or
2
9
0
38.
2
0
(x3x1)dx
1
4
x4
1
2
x2x
2
0
1
4
24
1
2
222
1
4
04
1
2
020
8 0 or 8
39.
5
2
(x23x8)dx
1
3
x33
1
2
x28x
5
2
1
3
x3
3
2
x28x
5
2
1
3
53
3
2
528 5
1
3
(2)3
3
2
(2)28 2
26
6
5
7
3
4
26
6
5
14
6
8
or
41
6
3
40.
3
1
(x3)(x1)dx
3
1
(x22x3)dx
1
3
x32
1
2
x23x
3
1
1
3
x3x23x
3
1
1
3
33323 3
1
3
13123 1
9
5
3
or
3
3
2
41.
3
2
(x1)3dx
3
2
(x33x23x1)dx
1
4
x43
1
3
x33
1
2
x2x
3
2
1
4
x4x3
3
2
x2x
3
2
1
4
3433
3
2
323
1
4
2423
3
2
222
1
4
5
0 or
1
4
5
42.
1
0
x2
x
x
2
2
dx
1
0
(x
x
2
)(x
2
1)
dx
1
0
(x1)dx
1
2
x2x
1
0
1
2
121
1
2
020
3
2
43.
2
0
x(4x21)dx
2
0
(4x3x)dx
4
1
4
x4
1
2
x2
2
0
x4
1
2
x2
2
0
24
1
2
22
04
1
2
02
18 0 or 18
489 Chapter 15
44.
1
1
(x1)(3x2)dx
1
1
(3x25x2)dx
3
1
3
x35
1
2
x22x
1
1
x3
5
2
x22x
1
1
13
5
2
122 1
(1)3
5
2
(1)22 1
1
2
1
1
2
or 6
45a.
20.5
0
x3dx
1
4
x4
20.5
0
1
4
20.54
1
4
04
44,152.52 0 or 44,152.52
20
i1
i3
202(20
4
1)2
400
4
(4
0
41)
or 44,100
45b.
100.5
0
x2dx
1
3
x3
100.5
0
1
3
100.53
1
3
03
338,358.38 0 or 338,358.38
100
i1
i2
100(10
6
1)(201)
or 338,350
46.
2
0
490,000 xdx 490,000
1
2
x2
2
0
245,000 x2
2
0
(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 xis
positive, so each term in the sum
n
i1
f(xi)xis
negative. Therefore, the sum is negative. since
b
a
f(x)dx is a limit of negative sums, it is also
negative.
47b.
2
0
(x25)dx
1
3
x35x
2
0
1
3
235 2
1
3
035 0
2
3
2
47c. Since the function is negative, the integral in
part b gives the opposite of the area of the
region. The area is
2
3
2
.
48. The integral represents the
area of a right triangle. The
value is
1
2
(3)(9)
2
2
7
.
100(100 1)(2 100 1)
6
49a.
49b.
6
1
0
6
0
75 8x
1
2
x2
dx
1
6
75x8
1
2
x2
1
2
1
3
x3
6
0
1
6
75x4x2
1
6
x3
6
0
1
6
75 6 4 62
1
6
63
75 0 4 02
1
6
03
1
6
(558 0)
$93
49c.
12
1
6
12
6
75 8x
1
2
x2
dx
1
6
75x8
1
2
x2
1
2
1
3
x3
12
6
1
6
75x4x2
1
6
x3
12
6
1
6
75 12 4 122
1
6
122
75 6 4 62
1
6
63
1
6
(1188 558)
$105
50.
R
R
(R2x2)dx R2x
1
3
x3
R
R
R2R
1
3
R3
R2R
1
3
(R)3
2
3
R3
2
3
R3
4
3
R3
51a. 1000 k(6.4 106)2
1000 k
40.96
1
1012
4.1 1016 k; 4.1 1016Nm2
51b.
3.8 108
6.4 106
4.1 1016x2dx
4.1 1016 (1)x1
3.8
6.4
108
106
4.1
x
1016
3.8
6.4
108
106
4
3
.
.
1
8
1
1
0
0
1
8
6
4
6
.
.
1
4
1
1
0
0
1
6
6
1.1 10864.1 108
6.3 109J
Chapter 15 490
y
x
y
3
x
6
O
2
2
4
6
4
6
8
10
12345
f
(
x
)
x
f
(
x
) 75 8
x
x
2
O
1
2
10
20
30
40
50
60
70
80
90
100
110
52.
2
0
1
2
x2dx lim
n→
n
i1
1
2
2
n
i
2
n
2
lim
n→
4
2n
3
2
n(n1)
6
(2n1)
lim
n→
4
6
2n3
n
3
3
n2n
lim
n→
2
3
2
n
3
n
1
2
4
3
53. f(x) 2x63x22
f(x) 2 6x613 2x210
12x56x
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)
2
5
5
0
2
4
4
9
2
4
3
8
2
4
2
7
4
2
6
5
0
3
6
56. APert
600e0.06(15)
$1475.76
57. h6; k1
hp3
6 p3 or p3
(yk)24p(xh)
(y1)212(x6)
58. r
2
2
2
or 2
v
2
3
3
or
3
2
cos
3
isin
3
; 2
1
2
i
2
3
1 3
i
59. sin 52°
4
v
5
y
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.
Chapter 15 Study Guide and Assessment
Page 977 Understanding the Vocabulary
1. false; sometimes 2. true
3. false; indefinite 4. true
5. false; secant 6. false; tangent
7. false; derivative 8. true
9. false; rate of change 10. false; derivative
Pages 978–980 Skills and Concepts
11. There is a point at (2, 1) so f(2) 1. However,
the closer xis to 2, the closer yis to 3.
So, lim
x→2f(x) 3.
12. lim
x→2(x3x25x6) (2)3(2)25(2) 6
8 4 10 6
4
13. lim
x→0(2xcos x) 2(0) cos 0
0 1
1
14. lim
x→1
x
x
2
3
6
6
lim
x→1
(x
x
6
)(x
1
6)
lim
x→1x6
1 6 or 5
15. lim
x→0
5
2
x
x
2
lim
x→0
5
2
x
5
2
(0)
0
16. lim
x→4
x2
x2
3
x
2
x
10
lim
x→44
(x
x(x
5)
(x
2
)
2)
lim
x→4
x
x
5
4
4
5
or 4
17. lim
x→0(xsin x) 0 sin 0
0
18. lim
x→0
x2
2
x
x
cos x
lim
x→0
x(x
2x
cos x)
lim
x→0
x
2
cos x
0
2
cos 0
1
2
19. lim
x→2lim
x→2
(x
x
2
2
)
(x2
4
4)
lim
x→2x2
2 2 or 4
20. lim
x→0
(x
2
3
x
)29
lim
x→0
x26x
2
x
99
lim
x→0
x(x
2
x
6)
lim
x→0
x
2
6
0
2
6
or 3
21. lim
x→5
x2
x29
x
5
x
20
lim
x→5
(x
x(x
5)
(x
5
)
4)
lim
x→5
x
x
4
5
5
4
or
1
5
22. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
lim
h→0
lim
h→0
2
h
h
2
2x2h1 2x1
h
2(xh) 1 (2x1)
h
x32x24x8
x24
491 Chapter 15
52˚
45 ft/s
Vy
Vx
23. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
4(xh)23(xh) 5 (4x23x5)
h
35. f(x)
1
2
x32x23x2
F(x)
1
2
3
1
1
x312
2
1
1
x21
3
1
1
1
x112xC
1
8
x4
2
3
x3
3
2
x22xC
36. f(x) x45x32x6
F(x)
4
1
1
x415
3
1
1
x312
1
1
1
x11
6xC
1
5
x5
5
4
x4x26xC
37. f(x) (x4)(x2)
x22x8
F(x)
2
1
1
x212
1
1
1
x11 8xC
1
3
x3 x2 8xC
38. f(x)
x2
x
x
x1
F(x)
1
1
1
x11xC
1
2
x2xC
39.
2
0
2x dx lim
n→
n
i1
2
2
n
i
n
2
lim
n→
n
8
2
n(n
2
1)
lim
n→4
n2
n
2n
lim
n→4
1
n
1
4 units2
40.
1
0
x3dx lim
n→
n
i1
n
i
3
n
1
lim
n→
n
1
4
n2(n
4
1)2
lim
n→
1
4
n2
n
22
n1
lim
n→
1
4
1
n
2
n
1
2
1
4
unit2
41.
4
3
x2dx
4
0
x2dx
3
0
x2dx
lim
n→
n
i1
4
n
i
2
n
4
lim
n→
n
i1
3
n
i
2
n
3
lim
n→
6
n
4
3
n(n1)
6
(2n1)
lim
n→
2
n
7
3
n(n1)
6
(2n1)
lim
n→
6
6
4
2n3
n
3
3
n2n
lim
n→
2
6
7
2n3
n
3
3
n2n
lim
n→
6
6
4
(2
n
3
n
1
2
)lim
n→
2
6
7
2
n
3
n
1
2
6
3
4
2
3
7
3
3
7
units2
Chapter 15 492
lim
h→0
4x28xh 4h23x3h5 4x23x5
h
lim
h→0
8xh 4
h
h23h
lim
h→0
h(8x
h
4h3)
lim
h→08x4h3
8x3
24. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
lim
h→0
x33x2h3xh2 h3 3x 3h x33x
h
(xh)33(xh) (x33x)
h
lim
h→0
lim
h→0
lim
h→03x23xh h23
3x23
25. f(x) 2x6
f(x) 2 6x61
12x5
26. f(x) 3x7
f(x) 3 1x110
3
27. f(x) 3x25x
f(x) 3 2x215 1x11
6x5
28. f(x)
1
4
x2x4
f(x)
1
4
2x211 1x110
1
2
x1
29. f(x)
1
2
x42x3
1
3
x4
f(x)
1
2
4x41 2 3x31
1
3
1x110
2x36x2
1
3
30. f(x) (x3)(x4)
x27x12
f(x) 2x217 1x110
2x7
31. f(x) 5x3(x43x2)
5x715x5
f(x) 5 7x7115 5x51
35x675x4
32. f(x) (x2)3
x36x212x8
f(x) 3x316 2x2112 1x110
3x212x12
33. f(x) 8x
F(x) 8
1
1
1
x11C
4x2C
34. f(x) 3x22
F(x) 3
2
1
1
x212xC
x32xC
h(3x23xh h23)
h
3x2h3xh2h3 3h
h
42.
2
1
6x2dx
2
0
6x2dx
1
0
6.x2dx
lim
n→
n
i1
6
2
n
i
2
n
2
lim
n→
n
i1
6
n
i
2
n
1
lim
n→
4
n
8
3
n(n1)
6
(2n1)
lim
n→
n
6
3
n(n1)
6
(2n1)
lim
n→
4
6
8
2n3
n
3
3
n2n
lim
n→
6
6
2n3
n
3
3
n2n
lim
n→8(2
n
3
n
1
2
) lim
n→1
2
n
3
n
1
2
16 2 or 14 units2
43.
4
2
6xdx 6
1
2
x2
4
2
3x2
4
2
(3 42) (3 22)
48 12 or 36
44.
2
3
3x2dx 3
1
3
x3
2
3
x3
2
3
23(3)2
8 (27) or 35
45.
2
2
(3x2x3)dx 3
1
3
x31
1
2
x23x
2
2
x3
1
2
x23x
2
2
(23
1
2
223 2)
(2)3
1
2
(2)23(2)
12 (16) or 28
46.
4
0
(x2)(2x3)dx
4
0
(2x27x6)dx
2
1
3
x37
1
2
x26x
4
0
2
3
x3
7
2
x26x
4
0
2
3
43
7
2
426 4
2
3
03
7
2
026 0
36
3
8
0 or
36
3
8
47. 6x4dx 6
1
5
x5C
6
5
x5C
48. (3x22x)dx 3
1
3
x32
1
2
x2C
x3x2C
49. (x25x2)dx
1
3
x35
1
2
x22xC
1
3
x3
5
2
x22xC
50. (3x54x47x)dx 3
1
6
x64
1
5
x57
1
2
x2C
1
2
x6
4
5
x5
7
2
x2C
Page 981 Applications and Problem Solving
51. lim
t→100
1
2
m
1
5
0
t2
2
1
2
m
1
5
1
0
002
2
0.0000125 m
52. c(x) 9x5135x310,000
c(x) 9 5x51135 3x310
45x4405x2
c(2.6) 45(2.6)4405(2.6)2or $681.41
53a.
6
1
0
h
m
r
i
52
1
8
m
0
i
ft
3
1
60
h
0
r
s
88 ft/s
a
88
5
f
s
t/s
17.6 ft/s2
53b. v(t)
t
0
17.6dx
17.6x
t
0
17.6t17.6(0)
17.6t
53c. d(t)
t
0
17.6xdx
17.6
1
1
1
x11
t
0
8.8x2
t
0
8.8t28.8(0)2
8.8t2
Page 981 Open-Ended Assessment
1. Sample answer: f(x) x22x2;
lim
x→1(x22x2) 122(1) 2
5
2. Sample answer: g(x) 16x3;
1
0
16x3dx 4x4
1
0
4 144 04
4
Chapter 15 SAT & ACT Preparation
Page 983 SAT and ACT Practice
1.
1
2
1
3
1
1
2
1
3
493 Chapter 15
1
3
6
2
6
6
The correct choice is A.
1
1
6
2. Let xrepresent the length of the second side of the
triangle. Then the first side has length 2x.
Clearly the perimeter must be greater than 3x, so
eliminate answer choices A and B. Use the
Triangle Inequality Theorem.
x2x?Px2x3x
3x?P6x
? x2xPx2xx
? xP4x
Since the perimeter cannot equal 6xor 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.
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
height and base must be
6
2
. The area of one
triangle is
1
2
6
2
6
2
or 9. The area of both
triangles is 2 9 or 18. The correct choice is B.
5. x2x6
x2x6 0
(x3)(x2) 0
At x3 and x2,
the inequality equals
0. Test values that are
greater than and less
than 3 and 2 to
determine for which
values the inequality
is less than 0.
For values of xthat are between 2 and 3, the
inequality is true.
The correct choice is C.
6. Since 5 is prime, 3(5) or 15.
5
Chapter 15 494
x2x6 0
xy
36
20
06
30
46
x
2
x
?
Since 16 is composite,
1
2
(16) or 8.
16
15 8 or 23.
165
Now, you need to determine which of the choices is
equal to 23. Calculate each one.
10.5 69 63
23 34.5
The correct choice is B.
7. The nth term of an arithmetic sequence with first
term a1and common difference dis given by
ana1(n – 1)d. a1 4, and d3, 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} x21.
So {{x}} {x21}.
And {x21} (x21)21
(x42x21) 1
x42x2
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.
The answer is 5.
16
20
6946
312321
Lesson 1-1
Page A26
1. D{2, 1, 2}; R{4, 2, 4}; no
2. D{3, 0.5, 0.5, 3}; R{0.5, 3}; yes
3. D{1, 0, 2, 5, 7}; R{1, 2, 3, 5, 7}; yes
4. D{2, 2.3, 3.2}; R{4, 1, 3, 4}; no
5. 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
7. h(1.5)
2(1
3
.5)
3
3
or 1
8. k(5m) 3(5m)23
3(25m2) 3
75m23
Lesson 1-2
Page A26
1. f(x) g(x) 2x1 x23x1
x25x2
f(x) g(x) 2x1 x23x1
x2x
f(x) g(x) (2x1)(x23x1)
2x35x25x1
g
f
(x)
g
f(
(
x
x
)
)
x2
2
x
3
x
1
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 24x4x2
3. [f g](x) f(g(x))
f(x9)
1
3
(x9) 1
1
3
x2
[g f](x) g (f(x))
g
1
3
x1
1
3
x1 9
1
3
x8
495 Extra Practice
Extra Practice
f
(
x
)
x
O
f
(
x
)
x
2
x
O
f
(
x
) 3
x
4
f
(
x
)
f
(
x
)
x
O
f
(
x
) 1
x
O
f
(
x
) 4
x
f
(
x
)
4. [f g](x) f(g(x))
f(2x3x2x1)
2(2x3x2x1)
4x32x22x2
[g f](x) g(f(x))
g(2x)
2(2x)3(2x)2(2x) 1
16x34x22x1
Lesson 1-3
Page A26
1. x2 0
x2
2. 3x4 0
3x4
x
4
3
3. 1 0, false 4. 4x0
none x0
5. 2x1 0
2x1
x
1
2
5. m
1
3
or
1
3
y6
1
3
(x(2))
3y18 x2
x3y16 0
6. x10 is a horizontal line; perpendicular slope is
undefined.
y15 or y15 0
7. m
2
5
or
2
5
y(7)
2
5
(x3)
5y35 2x6
2x5y29 0
Lesson 1-6
Page A27
1.
2. Sample answer: y0.6091x1171.6
m
2
4
0
7
0
.
1
2
–
–
1
4
9
0
9
.5
0
0.6091
y40.5 0.6091(x1990)
y0.6091x1171.6
3. Sample answer: y0.6125x1178; r0.99
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
Extra Practice 496
6. x5 0
x 5
x5
Lesson 1-4
Page A26
1. ymx b→y2x1
2. y2 1(x1)
y2 x1
yx3
3. ymx b→y
1
4
x3
4. y(4) 0(x(2))
y4 0
y4
5. m
3
2
1
2
2
4
or
1
2
y1
1
2
(x2)
y1
1
2
x1
y
1
2
x2
6. m
0
6
(
0
1)
6
1
or 6
ymx b→y6x6
7. y0
8. y0 1.5(x10)
y1.5x15
Lesson 1-5
Page A27
1. None of these; the slopes are neither the same nor
opposite reciprocals.
2. y(2) 1(x0)
y2 x
xy2 0
3. y3 2(x1)
y3 2x2
2xy1 0
4. y1 is a vertical line; parallel slope is undefined.
y12 or y12 0
x
O
f
(
x
) 2
x
1
f
(
x
)
x
O
f
(
x
)
x
5
f
(
x
)
Enrollment
(millions)
Year
199019911992199319941995 19971998199920002001
1996
0
43
44
42
41
40
0
48
45
46
47
y
x
497 Extra Practice
f
(
x
)
x
O
x
O
g
(
x
) |
x
2|
g
(
x
)
x
O
h
(
x
)
x
O
f
(
x
)
x
O
g
(
x
) |5 2
x
|
g
(
x
)
x
O
k
(
x
)
y
x
O
y
2
y
x
O
x
y
3
y
x
O
y
x
1
y
x
O
4
x
2
y
6
y
x
O
1
x
y
4
y
x
O
y
|
x
|
Lesson 1-7
Page A27
1. 2.
3. 4.
5. 6.
Lesson 1-8
Page A27
1. 2.
3. 4.
y
x
O
3
y
6
x
3
2
x
y
1 0
y
x
O
y
3
x
8
x
y
4
(1, 5)
y
x
O
3
x
y
1
2
y
6
x
4
5. 6.
Lesson 2-1
Page A28
1.
y2x1; consistent and dependent
2.
(1, 5); consistent and independent
3.
no solution; inconsistent
4. 5x2y1x2y5
x2y51 2y5
4x42y6
x1y3
(1, 3)
5. 2x4y82x3y8
2x3y82x3(0) 8
y02x8
x4
(4, 0)
6
1
5
5
6. 8x2y2→16x4y4
3x4y23 3x4y23
19x19
x1
8x2y2
8(1) 2y2
2y10
y5
(1, 5)
Lesson 2-2
Page A28
1. xy6yz8
xz2yz2
yz82y10
y5
xy6xz2
x5 61 z2
x1z3
(1, 5, 3)
2. 2x2yz6→4x4y2z12
xy2z6x y2z6
3x5y18
x2yz7
2x2yz6
x4y13
3x5y18 →3x5y18
x4y13 3x12y39
7y21
y3
x4y13 xy2z6
x4(3) 13 1 (3) 2z6
x12z4
z2
(1, 3, 2)
3. 2x3yz1
xyz4
3x2y3
xyz4→2x2y2z8
3x2y2z 33x2y2z3
5x5
x1
3x2y3xyz4
3(1) 2y31 0 z4
2y0z3
y0z3
(1, 0, 3)
Lesson 2-3
Page A28
1. 2xy1x2y1
y1 2xx2(1 2x) 1
3x2 1
3x3
x1
y1 2(1)
1
(1, 1)
2. x2y5x2y5
2x2y21 2y5
3x32y4
x1y2
(1, 2)
3. 3xy74y5x
3x7 y4(3x7) 5x
12x28 5x
7x28
x4
y3(4) 7
5
(4, 5)
4. AC
1 (5)
5 1
6 4
4 7
1 0
2 8
Extra Practice 498
6
6
10
11
1
10
5. DE
1 (5)
3 2
4 1
2 (3)
6. 4B
4(3)
4(2)
4(0)
4(3)
4(2)
4(4)
7. impossible
8. 2C3A
3(1)
3(5)
3(6)
3(4)
3(1)
3(2)
2(5)
2(1)
2(4)
2(7)
2(0)
2(8)
10 (3)
2 15
8 18
14 12
0 3
16 6
9. impossible
13
17
26
26
3
22
10. ED
1
3
4
2
5
2
1
3
14
3
14
16
1(1) (5)(3)
3(1) 2(3)
1(4) (5)(2)
3(4) 2(2)
12
8
0
12
8
16
11. BC
5
1
4
7
0
8
3
2
0
3
2
4
4.
2
0
5
1
6
2
3
3
0
1
1
0
499 Extra Practice
2(5) 0(1) (3)(4)
4(5) (3)(1) 2(4)
2(7) 0(0) (3)(8)
4(7) (3)(0) 2(8)
22
15
10
44
BC D
22 1
15 3
10 (4)
44 2
Lesson 2-4
Page A28
1. 0.5
1
3
1
2
2
1
2
1
1
2
2
6
2
4
5
3
23
18
6
42
A
2
1
2
, 1
1
2
, B(1, 2), C(1, 3)
2.
J(4, 5), K(5, 8), L(6, 4), M(1, 2)
1
2
6
4
5
8
4
5
2
3
2
3
2
3
2
3
1
5
4
7
3
5
6
2
3.
N(5, 2), P(3, 8), Q(0, 4)
0
4
3
8
5
2
0
4
3
8
5
2
0
1
1
0
y
x
O
L
(4, 7)
J
(6, 2)
K
(3, 5)
M
(1, 5)
L
(6, 4)
M
(1, 2)
J
(4, 5)
K
(5, 8)
4
4
4
8
8
848
y
x
O
Q
(0, 4)
P
(3, 8)
N
(5, 2)
P
(3, 8)
Q
(0, 4)
N
(5, 2)
4
4
4
8
848
W(3, 3), X(6, 2), Y(5, 1), Z(2, 0)
5. Rot90 Ryx
1
0
0
1
1
0
0
1
2
0
5
1
6
2
3
3
0
1
1
0
F(4, 2), G(2, 3), H(6, 7)
Lesson 2-5
Page A29
1.
3(2) (11)(7)
83
2.
3(2) (7)(5)
29
3.
5(6)
1
2
(0)
30
4.
2
2
3
0
1
1
1
3
5
0
6
5
1
2
5
2
3
7
7
2
3
11
6
7
2
3
4
2
6
7
2
3
4
2
0
1
1
0
1
0
2
1(5)
0(19) 2(2)
1
5.
2
1
4
3
2
3
1
4
3
1
1
3
5
2
3
3
5
2
3
1
1
1
3
2
2
3
4
3
1
4
4
3
1
4
2
3
y
x
O
F
(4, 2)
G
(2, 3)
H
(6, 7)
F
(4, 2)
G
(2, 3)
H
(6, 7)
4
8
4
4
8
848
y
x
O
X
(6, 2)
Y
(5, 1)
Z
(2, 0)
W
(3, 3)
X
(6, 2)
Y
(5, 1)
Z
(2, 0)
W
(3, 3)
4
8
4
4
8
848
1(11) 3(19) 2(6)
34
y
x
O
A
B
(2, 4)
B
(1, 2)
C
(1, 3)
A
(5, 3)
C
(2, 6)
2, 1
()
1
21
2
6.
1
6
2
0
3
5
4
5
2
Extra Practice 500
4
0
(1)
4(36) 0(22) 1(19)
163
7.
1
5
1
2
3
1
1
2
3
1
1
21
15
3
5
5
2
6
2
5
2
6
2
3
5
8.
4
1
0
0
10
4
5
0
10
4
5
1
10 0
54
1
5
2
5
3
5
1
5
9.
1
2
6
5
4
3
6
5
4
3
1
56
34
0
1
4
1
1
0
1
8
3
5
2
2
3
2
10.
3
1
3
5
3
1
6
5
3
1
6
1
35
61
3
5
3
1
1
1
3
1
3
1
2
1
11.
22
6
x
y
2
2
3
1
1
8
2
3
2
1
2
3
2
1
1
32
12
1
8
1
8
22
6
2
3
2
1
x
y
2
2
3
1
2
3
2
1
(4, 5)
12.
6
7
x
y
2
1
4
3
4
5
x
y
1
1
0
2
4
1
3
2
4
1
3
1
42
31
1
1
0
1
1
0
6
7
2
4
1
3
x
y
2
1
4
3
2
4
1
3
(2, 1)
13.
4
13
x
y
1
3
2
4
2
1
x
y
1
1
0
1
2
3
4
1
2
3
4
1
21
43
1
1
0
1
1
0
4
13
1
2
3
4
x
y
1
3
2
4
1
2
3
4
5
2
, 1
5
2
1
x
y
y
x
O
(3, 7)
(3, 1)
(0, 1)
y
x
O
(0, 3)
(0, 2) (4, 2)
(4, 7)
y
x
O
(1, 6)
(1, 2)
(1, 1)
(1, 1)
Lesson 2-6
Page A29
1.
vertices: (3, 1), (0, 1), (3, 7)
f(x, y) 4x3y
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.
vertices: (0, 3), (4, 7), (4, 2), (0, 2)
f(x, y) 2xy
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
3.
vertices: (1, 1), (1, 6), (1, 2), (1, 1)
f(x, y) xy
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
Lesson 2-7
Page A29
1. Let xthe number of cars.
Let ythe number of buses.
6x30y600
xy60
x0
y0
C(x, y) 3x8y
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 xnumber of gallons of black walnut.
Let ynumber of gallons of chocolate mint.
y2x
y20 thousand
xy45 thousand
C(x, y) 2.95x2.95y
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 athe number of company A’s cards.
Let bthe number of company B’s cards.
ab90
a40
b25
C(a, b) 0.30a0.32b
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.
4. Let xthe number of video store hours.
Let ythe number of landscaping company
hours.
y0
x4
xy15
C(x, y) 5x7y
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.
Lesson 3-1
Page A30
1. f(x) 4x
f(x) 4(x)f(x) (4x)
f(x) 4xf(x) 4x
yes
2. f(x) x23
f(x) (x)23f(x) (x23)
f(x) x23f(x) x23
no
3. f(x)
3
1
x3
f(x)
3(
1
x)3
f(x)
3
1
x3
f(x)
3
1
x3
f(x)
3
1
x3
yes
4. xy 2→ab 2
x-axis a(b) 2
ab 2
ab 2; no
y-axis (a)b2
ab 2
ab 2; no
yx(b)(a) 2
ab 2; yes
yx(b)(a) 2
ab 2; yes
yx, yx
5. yx23→ba23
x-axis (b) a23
ba23; no
y-axis b(a)23
ba23; yes
yx(a) (b)23
ab23; no
yx(a) (b)23
ab23; no
y-axis
501 Extra Practice
O
y
x
(0, 20) (50, 10)
(60, 0)
(0, 0)
x
y
60
6
x
30
y
600
20
20
40
60
40 60
x
0
y
0
Ox
y
(4, 11)
(15, 0)
(4, 0)
x
y
15
84
4
8
12
16
12 16
y
0
x
4
x
y
(15, 30)
(10, 20)
(25, 20)
x
y
45
2010
10
20
30
40
30 40
y
2
x
Oa
b
(40, 50)
(40, 25) (65, 25)
a
b
90
4020
20
40
60
80
60 80
b
25
a
40
6. y2
2
7
x2
1→b2
2
7
a2
1
x-axis (b)2
2
7
a2
1
b2
2
7
a2
1; yes
y-axis b2
2(
7
a)2
1
b2
2
7
a2
1; yes
yx(a)2
2(
7
b)2
1
a2
2
7
b2
1; no
yx(a)2
2(
7
b)2
1
a2
2
7
b2
1; no
x-axis, y-axis
7. x4y→a4b
x-axis a4(b)
a4b; no
y-axis (a)4b
a4b; yes
yx(b)4(a)
b4a; no
yx(b)4(a)
b4a; no
y-axis
8. y3x→b3a
x-axis (b) 3a
b3a; no
y-axis b3(a)
b3a; no
yx(a) 3(b)
a3b; no
yx(a) 3(b)
a3b
a3b; no
none of these
9. y
x21
→ba21
x-axis (b) a21
ba21
ba21
; yes
y-axis b
(a)2
1
ba21
; yes
yx(a)
(b)2
1
a(b)2
1
; no
yx(a)
(b)2
1
ab21
ab21
; no
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 502
4. g(x) is a translation of f(x) right 1 unit and
compressed vertically by a factor of 2.
5. y
1
2
x
5
Lesson 3-3
Page A30
1. 2.
3. 4.
5. 6.
7. Case 1 Case 2
x23x23
(x2) 3x2 3
x2 3x5
x1
{x1 x5}
8. Case 1 Case 2
4x218 4x218
(4x2) 18 4x2 18
4x2 18 4x20
4x16 x5
x4
{xx4 or x5}
9. Case 1 Case 2
5 2x95 2x9
(5 2x) 95 2x9
5 2x92x4
2x14 x2
x7
{x2 x7}
y
x
O
y
|
x
4|
y
x
O
y
x
2 1
y
x
O
y
2|
x
1|
y
x
O
y
x
2
y
x
O
y
(
x
2)2 1
y
x
O
y
|
x
| 2
6. f(x) x26
yx26
xy26
x6 y2
yx6
f1(x) x
6; No, it is not a function.
7. f(x) (x2)2
y(x2)2
x(y2)2
x
y2
y2 x
f1(x) 2 x
; No, it is not a function.
8. f(x)
2
x
y
2
x
x
2
y
2xy
y2x
f1(x) 2x; Yes, it is a function.
9. f(x)
x
1
4
y
x
1
4
x
y
1
4
y4
1
x
y
1
x
4
f1(x)
1
x
4; Yes, it is a function.
10. f(x) x28x2
yx28x2
xy28y2
x2 y28y
x2 16 (y4)2
x18
y4
4 x18
y
f1(x) 4 x18
; No, it is not a function.
11. f(x) x34
yx34
xy34
x4 y3
3x4
y
f1(x)
3x4
; Yes, it is a function.
12. f(x)
(x
3
1)2
y
(x
3
1)2
x
(y
3
1)2
(y1)2
3
x
y1
3
x
y1
3
x
f1(x) 1
3
x
; No, it is not a function.
503 Extra Practice
x
O
f
(
x
)
f
1(
x
)
f
(
x
)
x
O
f
(
x
)
f
1(
x
)
f
(
x
)
x
O
f
1(
x
)
f
(
x
)
f
(
x
)
10. Case 1 Case 2
x13 1x13 1
(x1) 3 1(x1) 3 1
(x1) 4x3
x1 4
x5
{xx5 or x3}
11. Case 1 Case 2
2x327 2x327
(2x3) 27 2x3 27
2x3 27 2x24
2x30 x12
x15
{x15 x12}
12. Case 1 Case 2
3x43x03x43x0
(3x4) 3x0(3x4) 3x0
6x4 04 0; true
6x4
x
2
3
all real numbers
Lesson 3-4
Page A30
1. 2.
3.
4. f(x) 4x5
y4x5
x4y5
x5 4y
y
x
4
5
f1(x)
x
4
5
; Yes, it is a function.
5. f(x) 2x2
y2x2
x2y2
x2 2y
y
x
2
2
f1(x)
x
2
2
; Yes, it is a function.
Lesson 3-5
Page A31
1. Yes; the function approaches 1 as xapproaches 2
from both sides.
2. No; the function is undefined when x3.
3. No; the function is undefined when x1.
4. Yes; the function approaches 1 as xapproaches 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→
Lesson 3-6
Page A31
1. abs. max.: (1, 2)
2. rel. min.: (3, 0), rel max.:
(1, 3), abs. min.: (2, 1)
Lesson 3-7
Page A31
1. x2y
x
3
x
2
y
y
as x→, y→3; y3
2. x3y
x
2
x2
3
y
2
x
x
2
2
——
x
x
2
x
3
2
3
1
2
x
3
x
x
—
x
x
2
x
4. yes yx5
x5
x3x2
2x
1
→yx5
x
1
6
3
x23xAs x→,
x
1
6
3
→0.
5x1So, the graph of f(x)
5x15 will approach that of
16 yx5.
Lesson 3-8
Page A31
1. ykx y 4x
8 k(2) y4(9)
4 ky36
2. gkw g
1
3
0
w
10 k(3) 4
1
3
0
w
1
3
0
k
6
5
w
3. t
k
r
t
r
84
6
1
k
4
rt 84
84 kr(7) 84
r12
4. ykxz y 3xz
60 k(5)(4) y3(5)(10)
3 ky150
5. y
x
k
2
y
2
x
4
2
3
27
(3
k
)2
yx2243
243 ky(5)2243
y9.72
6. akbc3a1.5bc3
36 k(3)(2)3a1.5(5)(3)3
1.5 ka202.5
Lesson 4-1
Page A32
1. yes; f(x) x37x22x40
f(2) (2)37(2)22(2) 40
8 28 4 40
0
2. no; f(x) x37x22x40
f(1) (1)37(1)22(1) 40
1 7 2 40
36
3. no; f(x) x37x22x40
f(2) (2)37(2)22(2) 40
8 28 4 40
24
4. yes; f(x) x37x22x40
f(5) (5)37(5)22(5) 40
125 175 10 40
0
Extra Practice 504
y
no horizontal
asymptotes since as
x→, yis undefined
3. h(x)
x2
x
6x
5
5
y
x2
x
6x
5
5
y
x5, x1
x
x
2
x
5
2
——
x
x
2
2
6
x
x
2
x
5
2
x5
(x5)(x1)
2
1
x
x
3
2
y
as x→, y→0; y0
1
x
x
5
2
——
1
6
x
x
5
2
5. (x3)(x4) 0
x27x12 0; even; 2
6. (x(2))(x(1))(x2) 0
(x2)(x1)(x2) 0
(x2)(x2x2) 0
x3x24x4 0; odd; 3
7. (x(1.5))(x(1))(x1) 0
(x1.5)(x1)(x1) 0
(x1.5)(x21) 0
x31.5x2x1.5 0; odd; 3
8. (x(2))(x(i))(xi) 0
(x2)(xi)(xi) 0
(x2)(x21) 0
x32x2x2 0; odd; 1
9. (x(3i))(x3i)(x(i))(xi) 0
(x3i)(x3i)(xi)(xi) 0
(x29)(x21) 0
x410x29 0; even; 0
10. (x(1))(x1)(x2)(x3) 0
(x1)(x1)(x2)(x3) 0
(x21)(x25x6) 0
x45x35x25x6 0; even; 4
Lesson 4-2
Page A32
1. x24x5 0
x24x5
x24x4 5 4
(x2)29
x2 3
x2 3x2 3
x5x1
2. x26x8 0
x26x8
x26x9 8 9
(x3)21
x3 1
x3 1x3 1
x2x4
3. m23m20
m23m2
m23m
9
4
2
9
4
m
3
2
2
1
4
7
m
3
2
2
17
m
3
2
2
17
4. 2a28a6 0
a24a3 0
a24a3
a24a4 3 4
(a2)27
a2 7
a2 7
5. h212h4
h212h36 4 36
(h6)240
h6 210
h6 210
6. x29x1 0
x29x1
x29x
8
4
1
1
8
4
1
x
9
2
2
7
4
7
x
9
2
2
77
x
9
2
2
77
7. b24ac (3)24(4)(7) or 121; 2 real
x
(3)
2
(4
)
121
x
3
8
11
x
7
4
or x1
8. b24ac (2)24(1)(10) or 44; 2 real
w
2
2
(1
)
44
w1 11
9. b24ac (5)24(12)(6) or 263; 2 imaginary
t
t
5i
2
4
263
10. b24ac (6)24(1)(13) or 88; 2 real
x
6
2
(1
)
88
x 3 22
11. b24ac (4)24(4)(1) or 0; 1 real
n
(
2
4
(
)
4
)
0
n
1
2
12. b24ac (6)24(4)(15) or 276; 2 real
x
6
2(4
)
276
x
6
8
269
x
3
4
69
Lesson 4-3
Page A32
1. 2110 8
216
188
x8, R8
2. 11341
122
1221
x22x2, R1
3. 11035
112
1123
x2x2, R3
(5) 263
2(12)
505 Extra Practice
4. 412734
4844
12110
x32x2x1
5. f(x) x22x8
f(4) (4)22(4) 8
16 8 8
0; yes
6. f(x)x312
f(1) (1)312
1 12 or 13; no
7. f(x) 4x32x26x1
f(1) 4(1)32(1)26(1) 1
4 2 6 1
7; no
8. f(x)x44x216
f(4) (4)44(4)216
256 64 16
208; no
Lesson 4-4
Page A32
1. p: 1, 2, 3, 6
q: 1
p
q
: 1, 2, 3, 6
rational roots: 3, 1, 2
2. p: 1
q: 1, 2
p
q
: 1,
1
2
rational root:
1
2
3. p: 1, 2
q: 1
p
q
: 1, 2
rational root: 1
4. p: 1, 2, 4, 8
q: 1, 2, 3, 6
p
q
: 1, 2, 4, 8,
1
2
,
1
3
,
2
3
,
4
3
,
8
3
,
1
6
rational roots:
2
3
,
1
2
5. 2 or 0 positive
f(x) x34x2x4
1 negative
x25x4 0
(x4)(x1) 0
x4, x1
rational zeros: 1, 1, 4
6. 2 or 0 positive
f(x) x4x33x25x10
2 or 0 negative
rational zeros: none
7. 2 or 0 positive
f(x) 4x37x3
1 negative
4x26x2 0
(4x2)(x1) 0
x
1
2
, x1
rational zeros:
3
2
,
1
2
, 1
Extra Practice 506
r1256
11328
11160
21430
210516
31510 24
31120
r212 31
1213 0 1
1235 89
1
2
202 20
1
2
223
9
2
1
4
3
r110 2
1122 0
1100 2
2136 10
2112 6
r6122 48
16729 33 25
1652723 15
2
3
632412 0
1
2
6424 16 0
r1414
11540
r113 510
1125 0 10
10 1 993935 9360
r4073
3
2
4620
8. 3 or 1 positive
f(x) x4x34x4
1 negative
rational zero: 1
Lesson 4-5
Page A33
1.
1 and 0, 2 and 3
2.
1 and 2
3.
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
Lesson 4-6
Page A33
1.
6
x
x5
6 x25x
x25x6 0
(x2)(x3) 0
x2 0orx3 0
x2x3
2.
y
7
1
4
y
y
y
1
7y4(y1) y2
3y4 y2
y23y4 0
(y4)(y1) 0
y4 0ory1 0
y4y1
3.
r
5
1
r
4
1
r2
1
1
5(r1) 4(r1) 1
5r5 4r4 1
r10
4. 2
2
1
t
t
4
2
2
t
4
2
t
1
2
2(t2) 4 1
2t4 3
2t7
t3.5
5.
3
1
w
5
4
w
1
1
5
5 12 w
17 w; w0
Test w1:
3(
1
1)
5(
4
1)
?
1
1
5
1
3
4
5
?
1
1
5
1
1
7
5
?
1
1
5
; true
Test w1:
3(
1
1)
5(
1
1)
?
1
1
5
1
3
1
5
?
1
1
5
1
8
5
?
1
1
5
; false
Test w18:
3(
1
18)
5(
1
18)
?
1
1
5
5
1
4
9
1
0
?
1
1
5
1
4
35
?
1
1
5
; true
w0 or w17
507 Extra Practice
r11044
1100 4 0
4152076 300
r245
22811
1261
0245
1227
2205
3221
r100 5
2124 13
1111 6
0100 5
1111 4
2124 3
r1014 2
212326
11104 2
01014 2
11104 6
212310 22
6.
x
x
2
x
x
4
6
(x6)(x2) x(x4)
x28x12 x24x
12 4x
3 x; x0 or 6
Test x1:
(1
)
1
2
?
(
(
1
1
)
)
4
6
3
?
5
7
; false
Test x1:
1
1
2
?
1
1
4
6
1
?
3
5
; true
Test x4:
4
4
2
?
4
4
4
6
1
2
?0; false
Test x7:
7
7
2
?
7
7
4
6
5
7
?3; true
0 x3 or x6
Lesson 4-7
Page A33
1. 2 3
t
4Check: 2 3
1
3
4
4
2 3t16 16
4
3t14 4 4
t
1
3
4
2. 4 x2
1Check: 4 11 2
1
x2
34 3 1
x2 91 1
x11
3.
3y7
10 2Check:
3505
7
10 2
3y7
88 10 2
y7 512 2 2
y505
4. a1
5 a6
a1 10a1
25 a6
10a1
30
a1
3
a1 9
a10
Check: 10
1
5 10
6
3 5 2
2 2
no real solution
5. 2x3
2
2x3 4
2x1
x
1
2
2x3 0
2x3
x
3
2
Test x2: 2(2)
3
?2
1
?2; meaningless
Test x0: 2(0)
3
?2
3
?2; true
Test x1: 2(1)
3
?2
5
?2; false
Solution:
3
2
x
1
2
6.
46a
2
4
6a2 256
6a258
a43
6a2 0
6a2
a
1
3
Test a0:
46(0)
2
?4
42
?4; meaningless
Test a1:
46(1)
2
?4
44
?4; false
Test a44:
46(44)
2
?4
4262
?4; true
Solution: a43
Lesson 4-8
Page A33
1. f(x) 0.75x2
2. f(x) x3x2x2
3. Sample answer: f(x) 0.51x20.02x0.79
4a. Sample answer: y1.632x99.275
4b. Sample answer: about 122.123 thousand
f(x) 1.632x99.275
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° 4312
75° 43 12
3. 29.44° 29° (0.44 60)
29° 26.4
29° 26(0.4 60)
29° 2624
29° 26 24
4. 87.81° 87° (0.81 60)
87° 48.6
87° 48(0.6 60)
87° 4836
87° 48 36
5. 144° 12 30144° 12
6
1
0
°
30
36
1
0
°
0
144.208°
6. 38° 15 10
38° 15
6
1
0
°
10
36
1
0
°
0
38.253°
7. 107° 12 45
107° 12
6
1
0
°
45
36
1
0
°
0
107.213°
8. 51° 14 3251° 14
6
1
0
°
32
36
1
0
°
0
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°
Lesson 5-2
Page A34
1. cot v
ta
1
nv
cot vor
6
5
1
—
5
6
2. sin v
cs
1
cv
sin v
2
1
.5
or 0.4
3. (AC )2(BC )2(AB)2
142162(AB)2
452 (AB)2
452
AB; 452
or 2113
sin A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin A
2
1
1
6
13
or
8
11
1
3
13
cos A
2
1
1
4
13
or
7
11
1
3
13
tan A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan A
1
1
6
4
or
8
7
csc A
2
1
1
6
13
or
1
8
13
sec A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec A
2
1
1
4
13
or
1
7
13
cot A
1
1
4
6
or
7
8
4. (AC )2(BC )2(AB)2
252(BC )2282
BC2159
BC 159
sin A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin A
2
1
8
59
cos A
2
2
5
8
tan A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan A
2
1
5
59
csc A
2
1
8
59
or
28
1
5
1
9
59
sec A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec A
2
2
8
5
cot A
2
1
5
59
or
25
1
5
1
9
59
5. (AC)2(BC)2(AB)2
9262 (AB)2
117 (AB)2
117
AB; 117
or 313
sin A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin A
3
6
13
or
2
13
13
cos A
3
9
13
or
3
13
13
tan A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc A
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan A
6
9
or
2
3
csc A
3
6
13
or
2
13
sec A
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot A
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec A
3
9
13
or
3
13
cot A
9
6
or
3
2
Lesson 5-3
Page A34
1. tan v
y
x
Since tan v0, y0.
cot v
x
y
0
x
cot vis undefined.
2. Sample answers: 90°, 270°
cos v
x
r
Since cos v0, x0.
On the unit circle, x0 when v90° or v270°.
509 Extra Practice
3. rx2y
2
(1)2
(2
)2
5
sin v
y
r
cos v
x
r
sin v
2
5
or
2
5
5
cos v
1
5
or
5
5
tan v
y
x
csc v
y
r
tan v
2
1
or 2 csc v
2
5
sec v
x
r
cot v
x
y
sec v
1
5
or 5
cot v
1
2
or
1
2
4. rx2y
2
(2)2
(2)2
22
sin v
y
r
cos v
x
r
sin v
2
2
2
or
2
2
cos v
2
2
2
or
2
2
tan v
y
x
csc v
y
r
tan v
2
2
or 1csc v
2
2
2
or 2
sec v
x
r
cot v
x
y
sec v
2
2
2
or 2
cot v
2
2
or 1
5. rx2y
2
(5)2
(2)2
29
sin v
y
r
cos v
x
r
sin v
2
29
or
2
29
29
cos v
5
29
or
5
29
29
tan v
y
x
csc v
y
r
tan v
2
5
csc v
2
29
sec v
x
r
cot v
x
y
sec v
5
29
cot v
5
2
6. rx2y
2
(4)2
(3)2
5
sin v
y
r
cos v
x
r
sin v
3
5
cos v
4
5
tan v
y
x
csc v
y
r
tan v
3
4
csc v
5
3
sec v
x
r
cot v
x
y
sec v
5
4
cot v
4
3
Lesson 5-4
Page A34
1. tan A
a
b
tan 38°
1
a
5
15 tan 38° a
11.7 a
2. cos B
a
c
cos 87°
1
a
9
19 cos 87° a
1.0 a
3. cos B
a
c
cos 65.4°
16
c
.5
c
cos
16
6
.
5
5
.4°
c39.6
4. tan B
a
b
tan 42.5°
1
a
2
a
tan
1
4
2
2.5°
a13.1
5. sin B
b
c
sin 75°
5
b
.8
5.8 sin 75° b
5.6 b
6. tan 48°
20
x
b
x
t
2
a
0
n
48
b
°
tan 42°
b
x
tan 42°
20 tan 42° btan 42° btan 48°
20 tan 42° b(tan 48° tan 42°)
b
85.7 b; 85.7 ft
Lesson 5-5
Page A35
1. Let Aarcsin
3
4
. Then sin A
3
4
.
sin
arcsin
3
4
3
4
2. Let Acos1
1
2
. Then cos A
1
2
.
sec A
co
1
sA
sec Aor 2
sec
cos
3. Let Atan11. Then tan A1.
tan(tan11) 1
4. tan A
a
b
tan A
3
2
8
5
Atan1
3
2
8
5
A56.7°
1
1
2
20 tan 42°
(tan 48° tan 42°)
b
––
t
2
a
0
n
48
b
°
Extra Practice 510
b
x
20 feet
48˚
42˚
5. sin B
b
c
sin B
1
1
7
9
Bsin1
1
1
7
9
B63.5°
6. cos B
a
c
cos B
2
3
4
0
Bcos1
2
3
4
0
B36.9°
7. A180° (60° 75°) or 45°
K
1
2
a2
sin
s
B
in
s
A
in C
K
1
2
(8)2
sin 6
si
0
n
°
4
si
5
n
°
75°
K37.9 units2
8. K
1
2
bc sin A
K
1
2
(16)(12) sin 43°
K65.5 units2
Lesson 5-7
Page A35
1. Since 145° 90°, consider Case II.
5 10; no solution
2. Since 25° 90°, consider Case I.
bsin A10 sin 25°
4.226182617
9 4.226182617; 2 solutions
sin
a
A
sin
b
B
sin
6
25°
si
1
n
0
B
6 sin B10 sin 25°
Bsin1
10 si
6
n25°
B44.77816685
B180° 44.8° or 135.2°
Solution 1
C180° (25° 44.8°) or 110.2°
sin
a
A
sin
c
C
sin
6
25°
sin 1
c
10.2°
c
6s
s
i
i
n
n
1
2
1
5
0
°
.2°
c13.32398206
B44.8°, C110.2°, c13.3
Solution 2
C180° (25° 135.2°) or 19.8°
sin
a
A
sin
c
C
sin
6
25°
sin 1
c
9.8°
c
6s
s
i
i
n
n
1
2
9
5
.
°
8°
c4.809133219
B135.2°, C19.8°, c4.8
3. Since 56° 90°, consider Case I.
Csin B50 sin 56°
41.45187863
34 41.5; no solution
511 Extra Practice
7. cos B
a
c
cos B
1
9
2
.
.
2
6
Bcos1
1
9
2
.
.
2
6
B43.1°
8. tan A
a
b
tan A
2
3
8
6
.
.
4
5
Atan1
2
3
8
6
.
.
4
5
A37.9°
Lesson 5-6
Page A35
1. C180° (75° 50°) or 55°
sin
a
A
sin
b
B
sin
a
A
sin
c
C
sin
7
75°
bsin 50°
sin
7
75°
sin
c
55°
b
7
s
s
in
in
7
5
5
0
°
°
c
7
s
s
in
in
7
5
5
5
°
°
b5.551472956 c5.936340197
C55°, b5.6, c5.9
2. B180° (97° 42°) or 41°
sin
c
C
sin
b
B
sin
c
C
sin
a
A
sin
12
42°
sin
b
41°
sin
12
42°
sin
a
97°
b
12
si
s
n
in
42
4
°
1°
a
12
si
s
n
in
42
9
°
7°
b11.76557801 a17.80004338
B41°, a17.8, b11.8
3. A180° (49° 32°) or 99°
sin
a
A
sin
b
B
sin
a
A
sin
c
C
sin
10
99°
sin
b
49°
sin
10
99°
sin
c
32°
b
10
si
s
n
in
99
4
°
9°
c
10
si
s
n
in
99
3
°
2°
b7.641171301 c5.365247745
A99°, b7.6, c5.4
4. B180° (22° 41°) or 117°
sin
b
B
sin
a
A
sin
b
B
sin
c
C
sin
2
1
5
17°
sin
a
22°
sin
2
1
5
17°
sin
c
41°
a
2
s
5
in
si
1
n
1
2
7
2
°
°
c
2
s
5
in
si
1
n
1
4
7
1
°
°
a10.51077021 c18.40780654
B117°, a10.5, c18.4
5. K
1
2
bc sin A
K
1
2
(12)(6) sin 34°
K20.1 units2
6. C180° (87° 56.8°) or 36.2°
K
1
2
c2
sin
s
A
in
s
C
in B
K
1
2
(6.8)2
sin 8
si
7
n
°
3
si
6
n
.2
5
°
6.8°
K32.7 units2
4. C180° (45° 85°) or 50°
sin
a
A
sin
c
C
sin
a
45°
sin
15
50°
a
15
si
s
n
in
50
4
°
5°
a13.8459352
sin
b
B
sin
c
C
sin
b
85°
sin
15
50°
b
15
si
s
n
in
50
8
°
5°
b19.50659731
C50°, a13.8, b19.5
Lesson 5-8
Page A35
1. a2b2c22bc cos A
a262822(6)(8) cos 62°
a254.93072997
a7.411526831
sin
a
A
sin
b
B
sin
a
62°
sin
6
B
sin B
6sin
a
62°
Bsin1
6sin
a
62°
B45.62599479
C180° (62° 45.6°) or 72.4°
a7.4, B45.6°, C72.4°
2. a2b2c22bc cos A
92721222(7)(12) cos A
112 168 cos A
Acos1
1
1
6
12
8
A48.1896851
sin
a
A
sin
b
B
sin
9
A
sin
7
B
sin B
7si
9
nA
Bsin1
7si
9
nA
B35.43094469
C180° (48.2° 35.4°) or 96.4°
A48.2°, B35.4°, C96.4°
3. b2a2c22ac cos B
b21421822(14)(18) cos 48°
b2182.7581744
b13.51880817
sin
a
A
sin
b
B
si
1
n
4
A
sin
b
48°
sin A
14 si
b
n48°
Asin1
14 si
b
n48°
A50.31729382
C180° (50.3° 48°) or 81.7°
b13.5, A50.3°, C81.7°
4. c2a2b22ab cos C
c2(14.2)2(24.5)22(14.2)(24.5) cos 85.3°
c2744.8771857
c27.29243825
sin
a
A
sin
c
C
s
1
i
4
n
.2
A
sin 8
c
5.3°
sin A
14.2 si
c
n 85.3°
Asin1
14.2 si
c
n 85.3°
A31.23444201
B180° (31.2° 85.3°) or 63.5°
c27.3, A31.2°, B63.5°
5. s
1
2
(abc)
s
1
2
(4 7 10)
s10.5
Ks(s
a)(s
b)(s
c)
K10.5(1
0.5
4)(10
.5 7
)(10.5
10
)
K119.43
75
K10.9 units2
6. s
1
2
(abc)
s
1
2
(4 6 5)
s7.5
Ks(s
a)(s
b)(s
c)
K7.5(7.
5 4
)(7.5
6)(7.5
5)
K98.43
75
K9.9 unit2
7. s
1
2
(abc)
s
1
2
(12.4 8.6 14.2)
s17.6
Ks(s
a)(s
b)(s
c)
K17.6(1
7.6
12.4)
(17.6
8.6
)(17.6
14
.2)
K2800.
512
K52.9 units2
8. s
1
2
(abc)
s
1
2
(150 124 190)
s232
Ks(s
a)(s
b)(s
c)
K232(2
32
150)(2
32
124)(2
32
190)
K86,29
2,864
K9289.4 units2
Extra Practice 512
Lesson 6-2
Page A36
1. 5 210or about 31.4 radians
2. 3.8 27.6or about 23.9 radians
3. 14.2 228.4or about 89.2 radians
4. 2.1 24.2
q
v
t
q
4.
5
2
q2.6 radians/s
5. 1.5 23
q
v
t
q
3
2
q4.7 radians/min
6. 15.8 231.6
q
v
t
q
31
1
.
8
6
q5.5 radians/s
7. 140 2280
q
v
t
q
28
2
0
0
q44.0 radians/min
8. q
v
t
q
2
3
0
qabout 0.2 radian/s
Lesson 6-3
Page A36
1. 1
2. 0
3. 1
4.
5.
9a.
d22823522(28)(35) cos 110°
d22679.359481
d51.8 in.
9b. Area 2
1
2
(28)(35) sin 110°
920.9 in2
Lesson 6-1
Page A36
1. 120° 120°
18
0°
2
3
2. 280° 280°
18
0°
14
9
3. 440° 440°
18
0°
22
9
4. 150° 150°
18
0°
5
6
5.
8
3
8
3
18
0°
480°
6.
5
1
2
5
1
2
18
0°
75°
7. 2 2
18
0°
114.6°
8. 10.5 10.5
18
0°
601.6°
9. reference angle:
5
6
6
; Quadrant II
sin
5
6
1
2
10. reference angle:
4
3
3
; Quadrant III
sin
4
3
2
3
11.
9
4
is coterminal with
4
; Quadrant I
cos
9
4
2
2
12.
2
3
is coterminal with
2
cos
3
2
0
13. If the diameter 10 in., the radius 5 in.
80° 80°
1
80
4
9
srv
s5
4
9
s7.0 in.
35 in.
28 in.
110˚
513 Extra Practice
432
1
1
y
x
O
y
sin
x
43
1
1
y
x
O
y
cos
x
Lesson 6-4
Page A36
1. 22;
2
1
2
2. 33;
0
2
.
5
4
3.
1
2
1
2
; 8
4. A0.5
2
k
6
A0.5 k
2
6
or
1
3
y0.5 sin
3
v
5. A2
2
k
3
A2kor 6
y2 sin 6v
6. A
3
5
2
k
4
A
3
5
k
2
4
or
1
2
y
3
5
cos
2
v
7. A0.25
2
k
8
A0.25 k
2
8
or
4
y0.25 cos
4
v
2
3
2
1
4
Lesson 6-5
Page A37
1.
k
c
2
or
2
2.
k
c
2
1
or 2
3.
k
c
or
4. A2
2
k
2
1
c
h1
A2kor 1 c
y2 sin (v) 1
5. A0.5
2
k
4
8
c
0h3
A0.5 k
2
4
or 8 c0
y0.5 sin 8v3
6. A20
2
k
2
4
c
2h4
A20 kor 4 c8
y20 cos (4v8) 4
7. A
3
4
2
k
10 0h
1
2
c
5
2
2
2
2
2
1
2
Extra Practice 514
23
1
2
2
1
y
O
y
2 cos
23
1
2
3
2
3
1
y
O
y
3 sin 0.5
2468
1
1
y
O
y
4
1
2cos
23
1
1
y
O
y
sin(2 )
23
2
2
y
O
y
2 cos( 2)
23
1
1
y
O
y
sin
2
2
()
A
3
4
k
2
1
0
or
5
c0
y
3
4
cos
5
v
1
2
Lesson 6-6
Page A37
1a. 12.1 2.7 9.4 h
1b. 12.1 2.7 14.8 h
1c. m10 represents the middle of October.
d2.7 sin (0.5(10) 1.4) 12.1
d10.9 h
2. A
6
2
or 3;
2
1
4
7
y3 cos
7
t
Lesson 6-7
Page A37
1.
2.
3.
Lesson 6-8
Page A37
1. Let vCos10.
Cos v0
v
2
4. If ytan
3
4
, then y1.
Cos1
tan
3
4
Cos1y
Cos1(1)
5. Let aCos1
1
2
and bSin10.
Cos a
1
2
Sin b0
a
3
b0
sin
Cos1
1
2
Sin10
sin (ab)
sin
3
0
sin
3
2
3
6. Let vsin1
2
3
.cos
2 Sin1
2
3
cos (2v)
sin v
2
3
cos
2
3
v
3
cos
2
3
1
2
Lesson 7-1
Page A38
1. csc v
si
1
nv
1
1
co
s2v
1
1
1
4
2
515 Extra Practice
23
4
8
4
8
y
O
y
cot
()
4
23
2
4
y
O
y
sec 2
23
2
4
2
4
y
O
y
csc(2 2)
2. Let vArcsin 0.
Sin v0
v0
3. Let vTan11. cos(Tan11) cos
4
Tan v1
2
2
v
4
1
1
1
5
6
3
6
6
6
2
6
3. cos
13
6
cos
13
6
2
cos
6
4. tan (315°)
c
s
o
in
s
(
(
3
3
1
1
5
5
°
°
)
)
c
s
o
in
s
4
4
5
5
°
°
tan 45°
sin (45° (360°))
cos (45° (360°))
4
15
1
1
5
5
4
15
15
2. tan v
co
1
tv
1
3
6
1
4
15
5. csc (930°)
sin (
1
930°)
1
sin (360°(2) 210°)
5. Sample answer: cos x1
c
c
s
o
c
tx
x
1
1
cos x1
6. Sample answer: cot x2
2 tan xsin x2 cos xcsc x
2
c
s
o
in
s
x
x
sin xcos x
sin
1
x
2
sin2x
co
sx
cos2x
sin
1
x
2
co
1
sx
sin
1
x
2
c
s
o
in
s
x
x
2 cot x
Lesson 7-3
Page A38
1. cos 75° cos (45° 30°)
cos 45° cos 30° sin 45° sin 30°
2
2
2
3
2
2
1
2
6
4
2
2. sin 105° sin (60° 45°)
sin 60° sin 45° cos 60° cos 45°
2
3
2
2
1
2
2
2
6
4
2
3. tan
1
2
tan
4
6
tan
4
tan
p
6
1 tan
4
tan
6
c
s
o
in
s
x
x
sin
1
x
Extra Practice 516
sin
1
(210°)
sin
1
(30°)
sin
1
30°
csc 30°
6.
t
s
a
in
n
v
v
sin v
c
s
o
in
s
v
v
co
1
sv
sec v
7. cot vtan vsin vsec v
c
s
o
in
s
v
v
c
s
o
in
s
v
v
sin v
co
1
sv
1
c
s
o
in
s
v
v
1 tan v
8. (1 sin x)(1 sin x) 1 sin2x
cos2x
9.
c
c
s
o
c
tx
x
s
co
in
s
x
x
c
s
o
in
s
x
x
sin x
sin
1
x
cos x
cos x
c
s
o
in
s
x
x
sin x
Lesson 7-2
Page A38
1. csc2vcot2vsin vcsc v
csc2vcot2vsin v
si
1
nv
csc2vcot2v1
csc2vcsc2v
2.
se
c
c
sc
v
v
se
c
c
sc
v
v
sin vcos v
csc
s
v
ec
se
v
cv
csc
c
v
sc
se
v
cv
sin vcos v
cs
1
cv
se
1
cv
sin vcos v
sin vcos vsin vcos v
3. sin2xcos2xsec2xtan2x
1 tan2x1 tan2x
1 1
4. sec Acos Atan Asin A
co
1
sA
cos Atan Asin A
co
1
sA
c
c
o
o
s
s
2
A
A
tan Asin A
1
co
c
s
o
A
s2A
tan Asin A
s
c
i
o
n
s
2
A
A
tan Asin A
c
s
o
in
s
A
A
sin Atan Asin A
tan Asin Atan Asin A
1
3
3
1 1
3
3
2 3
4. tan
7
1
2
tan
3
4
tan
3
tan
4
1 tan
3
tan
4
1
3
3
1
3
3
1
3
3
1
3
3
1
3
1
1
1
3
1
2 3
1 3
3
1
5. sec
2
1
9
2
sec
5
1
2
1
cos
4
6
9. If cot x
4
3
, then tan x
3
4
.
sec2y 1 tan2y
5
4
21 tan2y
1
9
6
tan2y
3
4
tan y
tan (xy)
1
ta
n
t
x
an
x
ta
ta
n
n
y
y
3
4
3
4
——
1
3
4
3
4
517 Extra Practice
1
cos
4
cos
6
sin
4
sin
6
1
2
2
2
3
2
2
1
2
6
4
2
6
6
2
2
6
2
6. cot 375° cot 15°
tan (45
1
°30°)
1 tan 45° tan 30°
1
2 3
7. sin x1
2
5
2
cos y1
3
4
2
2
2
1
5
1
7
6
2
5
1
7
4
sin (xy) sin xcos ycos xsin y
5
21
4
7
2
5
3
4
6
2
7
0
3
8. 122
52
119
122
112
23
sin x
1
1
2
19
sin y
1
2
2
3
cos (xy) cos xcos ysin xsin y
1
5
2
1
1
2
1
1
1
2
19
1
2
2
3
55
1
44
2737
1
3
3
——
1
3
3
1
3
3
——
1
3
3
1
3
3
1 1
3
3
1
tan 45° tan 30°
1
6
42
2
7
4
10. 726
2
85
; cos x
6
85
85
; sin x
7
85
85
csc y
8
5
825
2
39
sin
1
y
8
5
cos y
8
39
sin y
5
8
sec (xy)
cos (x
1
y)
1
cos xcos ysin xsin y
6
4
—
1
7
6
1
6
85
85
8
39
7
85
85
5
8
63315
3585
63315
3585
680
63315
3585
Lesson 7-4
Page A39
1. sin 15° sin
3
2
0°
1c
2
os 30°
2
2
3
Since 15° is in Quadrant I, sin 15°
2
2
3
.
2. cos 75° cos
15
2
0°
1c
o
2
s150
°
2
2
3
Since 75° is in Quadrant I, cos 75°
2
2
3
.
1
2
3
2
1
2
3
2
483315
28085
179
3. tan
1
2
tan
1 cos
—
6
—
1 cos
—
6
—
6
—
2
7. 722
2
35
; sin v
3
7
5
; tan v
3
2
5
sin 2v2 sin vcos v
2
3
7
5
2
7
12
4
9
5
cos 2vcos2vsin2v
2
7
2
3
7
5
2
4
4
1
9
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
3
2
5
1
3
2
5
2
12
4
1
5
8. 322
2
5
; cos v
3
5
, tan v
2
5
5
sin 2v2 sin vcos v
2
2
3
3
5
4
9
5
cos 2vcos2vsin2v
3
5
2
2
3
2
1
9
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
2
5
5
1
2
5
5
2
45
9. (3)2
(1)2
10
; sin v
3
10
10
; cos v
1
1
0
0
sin 2v2 sin vcos v
2
3
10
10
1
1
0
0
3
5
cos 2vcos2vsin2v
1
1
0
0
2
3
10
10
2
4
5
tan 2v
1
2
t
t
a
a
n
n
v
2v
1
2
(
(
3
3
)
)2
3
4
Extra Practice 518
2
2
3
3
2
2
3
3
(2
1
3
)2
or
2 3
Since
1
2
is in Quadrant I, tan
1
2
2 3
.
4. cos 22.5° cos
4
2
5°
1c
2
os 45°
1
2
3
1
2
3
1
2
2
2
2
2
2
Since 22.5° is in Quadrant I, cos 22.5°
2
2
2
.
5. sin
5
1
2
sin
5
6
—
2
1 cos
5
6
2
2
2
3
Since
5
1
2
is in Quadrant I, sin
5
1
p
2
2
2
3
.
6. tan 112.5° tan 225°
1
1
c
c
o
o
s
s
2
2
2
2
5
5
°
°
1
2
2
1
2
2
1
2
3
2
2
2
2
2
2
2
2
2
(2
2
2
)2
or
1 2
Since 112.5° is in Quadrant II,
tan 112.5° 1 2
.
10. csc v
3
2
32(
2)2
5
si
1
nv
3
2
cos v
3
5
sin v
2
3
tan v
2
5
5
sin 2v2 sin vcos v
2
2
3
3
5
4
9
5
cos 2vcos2vsin2v
3
5
2
2
3
2
1
9
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
2
5
5
1
2
5
5
2
45
Lesson 7-5
Page A39
1. 4 cos2x2 0
cos2x
1
2
cos x
2
2
x45°, 135°, 225°, 315°
2. sin2xcsc x1 0
sin2x
s
1
in
x1 0
sin x1
x90°
3. 3
cos x2 cos x
3
c
s
o
in
s
x
x
2 cos x
3
c
s
o
in
s
x
x
2 cos x0
cos x
s
in
3
x
2
0
cos x0or
s
in
3
x
2 0
x90°, 270°
s
in
3
x
2
2
3
sin x
x60°, 120°
4. 3 cos2x6 cos x3
3 cos2x6 cos x30
3(cos2x2 cos x1) 0
3(cos x1)(cos x1) 0
cos x1 0
cos x1
x0°
Lesson 7-6
Page A39
1. xcos 30° ysin 30° 12 0
2x
3
1
2
y12 0
3
xy24 0
2. xcos
3
ysin
3
2 0
1
2
x
2
3
y2 0
x3
y4 0
3. xcos 150° ysin 150°
1
2
0
2
3
x
1
2
y
1
2
0
3
xy 1 0
4. A2
B2
421
02
229
2
4
x
29
1
2
0
y
29
2
1
0
29
0
2
29
29
x
5
29
29
y
5
29
29
0
p
5
29
29
sin f
5
29
29
, cos f
2
29
29
tan f
5
2
fArctan
5
2
180°
f248°
5. A2
B2
12(
1)2
2
x
2
y
2
2
2
0
2
2
x
2
2
y2
0
p2
sin f
2
2
, cos f
2
2
tan f1
fArctan (1) 360°
f315°
6. A2
B2
223
2
13
2
13
x
3
13
y
1
1
2
3
0
2
13
13
x
3
13
13
y
12
1
3
13
0
p
12
1
3
13
sin f
3
13
13
, cos f
2
13
13
tan f
3
2
fArctan
3
2
f56°
519 Extra Practice
10. cos 70°
1
h
.7
sin 70°
1
x
.7
1.7 cos 70° h1.7 sin 70° v
0.58 h1.60 v
11. cos 76°
2
h
.1
sin 76°
2
v
.1
2.1 cos 76° h2.1 sin 76° v
0.51 h2.04 v
12. cos 30°
3
h
.6
sin 30°
3
v
.6
3.6 cos 30° h3.6 sin 30° v
3.12 h1.8 v
Lesson 8-2
Page A40
1. AB
u
4 3, 1 6
1, 5
AB
u
12(
5)2
26
2. AB
u
2 (1), 2 3
1, 1
AB
u
(1)2
(1
)2
2
3. AB
u
1 0, 8 (4)
1, 4
AB
u
(1)2
(4
)2
17
4. AB
u
3 1, 9 10
2, 19
AB
u
22(
19)2
365
5. AB
u
3 (6), 6 0
3, 6
AB
u
32(
6)2
45
35
6. AB
u
0 4, 7 (5)
4, 12
AB
u
(4)2
122
160
410
7. 5, 6526
2
61
5i
u
6j
u
8. 2, 4(2)2
42
20
25
2i
u
4j
u
9. 10, 5(10)
2(
5)2
125
55
10i
u
5j
u
10. 2.5, 6(2.5)2
62
42.25
6.5
2.5i
u
6j
u
Extra Practice 520
Lesson 7-7
Page A39
1.
3
(2)
3
2
2(
(
1
)
2
)2
2
2
13
2
13
13
2.
2(3
)
22
4
(0)
4
2
2
4
20
2
4
5
2
5
5
3.
5
10
; d
5
10
4. y
2
3
x2 →2x3y6 0
2(
4)
2
2
3
(
(
2)
3
)2
6
8
13
8
13
13
5. (0, 1)
d
0
2
1
(
2
1)
2
2
4
5
6
6
5
5
; d
6
5
5
6. (0, 3)
d
3
(3)
3
2
2(
(
0
)
2
)2
7
2
13
2
13
13
7. (2, 1)
d
2(2)
2
5
2
(
1
5
)
2
4
2
5
9
0.9; d0.9 unit
Lesson 8-1
Page A40
1. 2.
3.
4. 3.6 cm; 89° 5. 2.6 cm; 23°
6. 3.7 cm; 357° 7. 1.2 cm; 342°
8. 7.2 cm; 330° 9. 8.8 cm; 340°
2
10
3(1) (4) 1
(3)2
12
1.7 cm
70˚
2.1 cm
104˚
3.6 cm
330˚
11. 2, 622(
6)2
40
210
2i
u
6j
u
12. 15, 12(15)
2(
12)2
369
341
15i
u
12j
u
Lesson 8-3
Page A40
1. p
u
21, 2, 13(4, 3, 0
2, 4, 212, 9, 0
10, 5, 2
2. p
u
1, 2, 1
1
2
2, 2, 44, 3, 0
1, 2, 11, 1, 24, 3, 0
2, 2, 3
3. p
u
22, 2, 44, 3, 0
4, 4, 84, 3, 0
0, 7, 8
4. p
u
3
4
4, 3, 021, 2, 1
3,
9
4
, 02, 4, 2
1, 1
3
4
, 2
5. 2, 4, 15, 4, 3v
u
0, 0, 0
7, 0, 4v
u
0, 0, 0
v
u
0 7, 0 0, 0 4
v
u
7, 0, 4
Lesson 8-4
Page A41
1. 3, 42, 53 2 4 5
26; no
2. 3, 24, 63 4 2 6
0; yes
3. 5, 32, 35 2 3 (3)
19; no
4. 8, 62, 38 (2) 6 (3)
34; no
5. 3, 4, 04, 3, 63 4 4 (3) 0 6
0; yes
6. 4, 5, 11, 2, 34 (1) 5 (2) 1 3
11; no
7. 1, 0, 31, 1, 2
i
u
j
u
k
u
3i
u
j
u
k
u
3, 1, 1
3, 1, 1(1, 0, 33 1 1 0 1 3
0; yes
3, 1, 11, 1, 23 1 1 1 1 2
0; yes
3
2
0
1
1
1
8. 3, 0, 41, 5, 2
i
u
j
u
k
u
20i
u
10j
u
15k
u
20, 10, 15
20, 10, 153, 0, 420 3 (10) 0
15 4
0; yes
20, 10, 151, 5, 220 (1) (10)
5 15 2
0; yes
9. 1, 1, 02, 1, 3
i
u
j
u
k
u
3i
u
3j
u
3k
u
3, 3, 3
3, 3, 31, 1, 03(1) 311(3) 0
0; yes
3, 3, 3(2, 1, 33 2 3 1 (3) 3
0; yes
10. 1, 3, 26, 1, 2
i
u
j
u
k
u
8i
u
10j
u
19k
u
8, 10, 19
8, 10, 191, 3, 28 (1) 10 (3) 19 2
0; yes
8, 10, 196, 1, 28610 (1) 19 (2)
0; yes
Lesson 8-5
Page A41
1. magnitude 2002
2202
297.32 N
direction: tan v
2
2
0
2
0
0
v42.3°
2. r
u
25025022(50)(50) cos 120°
r
u
27500
r
u
86.60 mph; 30°
3. r
u
2350228022(350)(280) cos 135°
r
u
339,492.9291
r
u
582.66 N
s
5
in
82
1
.
3
6
5
6
°
s
2
i
8
n
0
v
sin v
280
58
si
2
n
.6
1
6
35°
sin v0.34
v19.9°
4. r
u
902
1102
r
u
142.13 N
2
2
3
1
1
6
0
3
1
1
1
2
4
2
0
5
3
1
521 Extra Practice
110 N
90 N
Lesson 8-6
Page A41
1. x2, y3t1, 0
x2 ty3 0
x2 ty3
2. x(1), y(4)t5, 2
x1, y4t5, 2
x1 5ty4 2t
x1 5ty4 2t
3. x(3), y6t2, 4
x3, y6t2, 4
x3 2ty6 4t
x3 2ty6 4t
4. x3, y0t0, 1
x3 0y0 t
x3yt
5. x3t→t
3
x
y2 t→ty2
y2
3
x
y
1
3
x2
6. x1 2t→t
x
2
1
y4t→t
4
y
4
y
2
x
1
2
y2x2
7. x3t10 →t
x
3
10
yt1→ty1
y1
1
3
x
1
3
0
y
1
3
x
7
3
Lesson 8-7
Page A41
1. vy70 sin 34°
39.14 yd/s
vx70 cos 34°
58.03 yd/s
2a. xtv
u
cos v
75tcos 25°
ytv
u
sin v
1
2
gt2h
75tsin 25°
1
2
(32)t25
5 75tsin 25° 16t2
2b. y0 when t?
t
t
0.1468595989 or t2.127882701
t2.13 s
x75tcos 25°
75(2.13) cos 25°
145 ft
75 sin 25° (75 si
n 25°)
24(
16)(
5
)
2(16)
Extra Practice 522
70 yd/s
34˚
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
K
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
M
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
N
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
P
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234 1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234 0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
2c. y5 75tsin 25° 16t2
5 75
2.
2
13
sin 25° 16
2.
2
13
2
20.6 ft
Lesson 8-8
Page A41
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
Page A42
1. 2.
3. 4.
5. 6.
7. 8.
6. x5 cos 240° y5 sin 240°
5
1
2
5
2
3
5
2
or 2.5
5
2
3
or about 4.33
(2.5, 4.33)
523 Extra Practice
7. x2
rcos v2
r
co
2
sv
r2 sec v
8. y6
rsin v6
r
si
6
nv
r6 csc v
9. x2y236
(rcos v)2(rsin v)236
r2(cos2vsin2v) 36
r236
r6 or r6
10. x2y23y
(rcos v)2(rsin v)23rsin v
r2(cos2vsin2v) 3rsin v
r23rsin v
r3 sin v
11. r4
r216
x2y216
12. r4 cos v
r24rcos v
x2y24x
Lesson 9-4
Page A42
1. A2
B2
62(
5)2
61
6
61
x
5
61
y
6
61
0
cos f
6
61
, sin f
5
61
, p
6
61
61
fArctan
5
6
180°
140°
6
61
61
rcos (v140°)
2. A2
B2
329
2
310
3
3
10
x
3
9
10
y
3
90
10
0
cos f
1
1
0
0
, sin f
3
10
10
, p310
fArctan 3
72°
310
rcos (v72°)
3. 8 rcos (v30°)
0 r(cos vcos 30° sin vsin 30°) 8
0
2
3
rcos v
1
2
rsin v8
0
2
3
x
1
2
y8
0 3
x2y16
4. 1 rcos (v)
0 r(cos vcos sin vsin ) 1
0 rcos v0 1
0 x1
x1
9. r5
or r5
Lesson 9-2
Page A42
1. 2.
circle spiral of Archimedes
3. 4. Sample answer:
rsin 5v
cardioid
Lesson 9-3
Page A42
1. r12(
1)2
vArctan
1
1
2
2
7
4
2
,
7
4
2. r320
2
vArctan
0
3
9
or 3 0
(3, 0)
3. r22(
2
)2
vArctan
2
2
6
or about 2.45 0.62
(2.45, 0.62)
4. x2 cos
4
y2 sin
4
2
2
2
2
2
2
2
2
2
, 2
5. x
1
4
cos
2
y
1
4
sin
2
1
4
(0)
1
4
(1)
0
1
4
0,
1
4
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
5101520
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
5.
Lesson 9-5
Page A43
Extra Practice 524
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
1. i10 (i4)3i2
1 3(1)
1
2. i17 (i4)4i
14i
i
3. i1000 (i4)250
1250
1
5. (4 i) (3 5i) (4 (3)) (i5i)
1 4i
6. (6 6i) (2 4i) (6 (2))(6i(4i))
4 2i
7. (3 i)(5 3i) 15 4i3i2
18 4i
8. (2 5i)2(2 5i)(2 5i)
4 20i25i2
21 20i
9. (1 2
i)(38
i)38
i32
i16
i2
3 22
i32
i4i2
7 2
i
10.
4
1
i
i
4
1
i
i
1
1
i
i
4
1
5
i
i
2
i2
3
2
5i
3
2
5
2
i
11.
6
2
2i
i
6
2
2i
i
2
2
i
i
12
4
10
i
i
2
2i2
10
5
10i
2 2i
12.
(
4
i
2
2
)
i
2
i2
4
4i
2
i
4
3
4
4
2
i
i
4
4
2
2
i
i
12
16
2
2i
4
i2
8i2
4
20
22i
1
5
1
1
0
1
i
4. i12 i4(i4)3 (i4)1
1311
2
i
O
(4, 1)
i
O
(0, 5)
Lesson 9-6
Page A43
1. 4x6yi14 12i
4x14 6y12
x
1
4
4
y
1
6
2
x3.5 y2
2. 3.
z421
2
z02(
5)2
17
25
or 5
4. z22(
3
)
2
7
5. r424
2
vArctan
4
4
32
or 42
4
4 4i42
cos
4
isin
4
6. r(2)2
12
vArctan
1
2
5
2.68
2 i5
(cos 2.68 isin 2.68)
7. r42(
2
)
2
vArctan
4
2
2
18
or 32
5.94
4 2
i32
(cos 5.94 isin 5.94)
8a. 5(cos 0.9 isin 0.9) 3.11 3.92j
8(cos 0.4 jsin 0.4) 7.37 3.12j
8b. (3.11 3.92j) (7.37 3.12j)
(3.11 7.37) (3.92j3.12j)
10.48 7.04johms
8c. r(10.48
)2(7
.04)2
vArctan
1
7
0
.
.
0
4
4
8
12.63 0.59
10.48 7.04j12.63 (cos 0.59 jsin 0.59) ohms
Lesson 9-7
Page A43
1. r6 4 or 24 v
2
4
3
4
24
cos
3
4
isin
3
4
24
2
2
i
2
2
122
122
i
i
O
(2, 3)
2. ror 6 v
3
4
3
1
2
3. d(x2
x1)2
(y2
y1)2
d(rr
)2(
2 6
)2
d02(
8)2
d64
or 8
x1
2
x2
,
y1
2
y2
r
2
r
,
6
2
(2)
(r, 2)
4.
6
2
x2
,
2
2
y2
(5, 8)
6
2
x2
5
2
2
y2
8
6 x210 2 y216
x216 y214
Then Ahas coordinates (16, 14).
5. Aquadrilateral 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.
6a.
0
2
50
,
0
2
40
(25, 20)
6b. d(x2
x1)2
(y2
y1)2
d(25
0)2
(20
0)2
d252
202
d1025
d541
or about 32 ft
Lesson 10-2
Page A44
1. (xh)2(yk)2r2
[x(2)]2(y2)2
2
2
(x2)2(y2)22
2. (xh)2(yk)2r2
(x0)2(y(4))242
x2(y4)216
525 Extra Practice
7
4
6(cos
7
4
isin
7
4
) 6
2
2
i
2
2
32
32
i
3. r5 2 or 10 v135° 45°
180°
10(cos 180° isin 180°) 10(1 i(0))
10
Lesson 9-8
Page A43
1. 44
cos (4)
2
isin (4)
2
256(cos 2isin 2)
256(1 i(0))
256
2. r122
(5)2
vArctan
12
5
13 1.965587446
133(cos (3)(v) isin (3)(v)) 2035 828i
3. r121
2
vArctan
1
1
2
4
2
1
3
cos
1
3
4
isin
1
3
1
2
2
1
3
cos
1
2
isin
1
2
1.08 0.29i
4. r(1)2
02
v
1
1
1
5
cos
1
5
() isin
1
5
()
1
cos
5
isin
5
0.81 0.59i
Lesson 10-1
Page A44
1. d(x2
x1)2
(y2
y1)2
d(4 (
2))2
(5
2)2
d623
2
d45
or 35
x1
2
x2
,
y1
2
y2
2
2
4
,
2
2
5
(1, 3.5)
2. d(x2
x1)2
(y2
y1)2
d(8 (
3))2
(1
6)2
d112
(7)2
d170
x1
2
x2
,
y1
2
y2
3
2
8
,
6
2
(1)
(2.5, 2.5)
y
x
O
(
x
2)2 (
y
2)2 2
y
x
O
x
2 (
y
4)2 16
Ox
y
(5, 10)
(5, 2)
(2, 5)
(2, 8)
3. x249 y2→x2y249
4. x2y26x8y18 0
(x26x9) (y28y16) 18 9 16
(x3)2(y4)27
5. x2y2Dx Ey F0
22(2)22D2EF0→2D2EF8
02(4)20(D)4EF0→4EF16
(2)2(2)22D2EF0
→2D2EF8
2D2EF82(2 0 2EF) 2(8)
2D2EF84EF)16
4D0F0
D0
4E(0) 16
4E16
E4
x2y24y0center: (0, 2)
x2(y24y4) 0 4radius: 2
x2(y2)24
6. x2y2Dx Ey F0
(1)232D3EF0→D3EF10
(4)2624D6EF0
→4D6EF52
(7)2327D3EF0
→7D3EF58
2(D3EF) 2(10) 4D 6EF)2(52
4D6EF)2(52 2(7D3EF) 2(58)
2DF)2(32 10DF)2(64
2DF32 (8) 3EF10
10DF64 4(8) 6EF52
12D96 24)3E42
D83E18
E6
(8)3(6) F10
26 F10
F16
x2y28x6y16 0
(x28x16) (y26y9) 16 16 9
(x4)2(y3)29
center: (4, 3)
radius: 3
7. r(4 2
)2(0
(
5))2
r225
2
r29
; r229
(x4)2(y0)229
(x4)2y229
Lesson 10-3
Page A44
1. center: (h, k) (0, 0)
a
1
2
0
or 5
b
6
2
or 3
(x
a2
h)2
(y
b2k)2
1
(x
52
0)2
(y
320)2
1
2
x
5
2
y
9
2
1
c2a2b2
c225 9
c216
c4
foci: (4, 0)
2. center: (h, k) (2, 1)
a
8
2
or 4
b
4
2
or 2
(y
a2
k)2
(x
b2
h)2
1
(y
42
1)2
[x
2
(
2
2)]2
1
(y
16
1)2
(x
4
2)2
1
c2a2b2
c216 4
c212
c12
or 23
foci: (2, 1 23
)
3. The major axis contains the foci and it is located
on the x-axis.
center: (h, k) (4, 1)
c2a2b2
c264 16
c280
c80
or 45
foci: (hc, k) (4 45
, 1)
major axis vertices: (ha, k) (4 8, 1)
(12, 1) and (4, 1)
minor axis vertices: (h, kb) (4, 1 4)
(4, 5) and (4, 3)
Extra Practice 526
y
x
O
x
2
y
2 49
4884
4
4
8
8
y
x
O
(
x
3)2 (
y
4)2 7
y
x
O
(
x
4)2
64
4
4812
4
4
8(
y
1)2
16 1
Lesson 10-5
Page A45
1. vertex (h, k) (0, 0)
4p4
p1
focus: (hp, k) (0 1, 0) or (1, 0)
directrix: xhp
x0 1
x1
axis of symmetry: yk
y0
2. x24x4 12y12
(x2)212(y1)
vertex (h, k) (2, 1)
4p12
p3
focus: (h, kp) (2, 1 3) or (2, 4)
directrix: ykp
y1 3
y2
axis of symmetry: xh
x2
3. vertex: (h, k) (2, 3)
focus: (hp, k) (0, 3)
hp0, k3
2 p0
p2
(yk)24p(xh)
(y3)24(2)[x(2)]
(y3)28(x2)
4. directrix: ykp3
focus: (h, kp) (0, 2)
kp32.5 p2
kp2p0.5
2k5
k2.5
(xh)24p(yk)
(x0)24(0.5)[y(2.5)]
x22(y2.5)
527 Extra Practice
y
x
O
y
2 4
x
y
x
O
x
2 4
x
4 12
y
12
Lesson 10-4
Page A45
1.
2.
3. center: (h, k) (4, 3)
(x
a2
h)2
(y
b2
k)2
1
[x
3
(
2
4)]2
(y
22
3)2
1
(x
9
4)2
(y
4
3)2
1
4. transverse axis: xh2
foci: (h, kc) (2, 7) kc7
(h, kc) (2, 3) kc3
2k4
k2; c5
vertices: (h, ka) (2, 5) 2 a5
(h, ka) (2, 1) a3
a2b2c2
32b252
b216
b4
(y
a2
k)2
(x
b2
h)2
1
(y
32
2)2
(x
42
2)2
1
(y
9
2)2
(x
16
2)2
1
8
4
4
484
y
x
O
(
x
1)
y
2 7
3(
x
1)
(1, 2)
58)(1, 2
58)(1, 2
y
2 7
3
8
8
4
4
4884
y
x
O
xy
16
Lesson 10-6
Page A45
1. A1, C1; since AC, the conic is a circle.
x2y28x2y13 0
(x28x16) (y22y1) 13 16 1
(x4)2(y1)24
2. A1, C4; since Aand Chave different
signs, the conic is a hyperbola.
x24y210x16y5
(x210x25) 4(y24y4) 5 25 16
(x5)24(y2)24
(x
4
5)2
(y
1
2)2
1
3. A0, C1; since A0, the conic is a parabola.
y25x6y9 0
(y26y9) 5x9 9
(y3)25x
4. A1, C2; since Aand Chave the same sign
x22y22x8y15
(x22x1) 2(y24y4) 15 1 8
(x1)22(y2)224
(x
24
1)2
(y
12
2)2
1
Lesson 10-7
Page A45
1. B24AC 024(1)(1)
0 4 or 4
Since 4 0 and AC, the graph is a circle.
x2y29
(x1)2[y(1)]29
(x1)2(y1)29
x22x1 y22y1 9
x2y22x2y7 0
2. B24AC 024(4)(1)
0 16 or 16
Since 16 0 and AC, the graph is an ellipse.
4x2y216
4[x(3)]2[y(2)]216
4(x3)2(y2)216
4(x26x9) (y24y4) 16
4x2y224x4y24 0
3. B24AC 024(49)(16)
0 3136 or 3136
Since 3136 0, the graph is a hyperbola.
49x216y2784
49
xcos
4
ysin
4
216
xsin
4
ycos
4
2784
49
2
2
x
2
2
y
216
2
2
x
2
2
y
2784
49
1
2
(x)2xy
1
2
(y)2
16
1
2
(x)2xy
1
2
(y)2
784
4
2
9
(x)249xy
4
2
9
(y)28(x)216xy
8(y)2784
49(x)298xy49(y)216(x)232xy
16(y)21568
33(x)2130xy33(y)21568 0
4. B24AC 024(4)(25)
0 400 or 400
Since 400 0, the graph is a hyperbola.
4x225y264
4(xcos 90° ysin 90°)225(xsin 90°
ycos 90°)264
4(0 y)225(x0)264
4(y)225(x)264 0
5. B24AC
22
24(1)(2)
8 8 or 0
parabola
tan 2v
A
B
C
tan 2v
1
2
2
2
tan 2v22
2v70.52877937°
v35, 35°
6. B24AC 524(15)(5)
25 300 or 275
Since 275 0 and AC, the graph is an ellipse.
tan 2v
A
B
C
tan 2v
15
5
5
tan 2v
1
2
2v26.56505118°
v13°
Lesson 10-8
Page A45
1. xy 3x2y28
y
3
x
x2
3
x
28
x2
x
9
2
8
x49 8x2
x48x29 0
(x29)(x21) 0
x29 0orx21 0
x29x21
x3x1
or i
If x3, then y
(
3
3)
or 1.
If x3, then y
(
3
3)
or 1.
Since x1
is an imaginary number, disregard
this solution.
(3, 1), (3, 1)
Extra Practice 528
y
x
O
(3, 1)
(3, 1)
2. xy4x210y210
xy4(y4)210y210
y28y16 10y210
9y28y6 0
y(8) (8)2
4(9
)(6)
11. ((3f)2)3(3f)6
(3
1
f)6
36
1
f6
72
1
9f6
12.
c
c
4
3
a
a
2
c
c
8
6
a
a
c14a
c1
1
4a
13. (2n
1
3
3n
1
2
)626n
6
3
36n
6
2
64n2799n3
46,656n5
14.
216
h
h
6
3
1
3
2
h
1
9
6
1
3
h
9
3
216
1
3
529 Extra Practice
y
8
1
2
8
70
y
4
9
70
y
4
9
70
or y
4
9
70
y1.4 y0.5
If y1.4, then x(1.4) 4 or 5.4.
If y0.5, then x(0.5) 4 or 3.5.
(5.4, 1.4), (3.5, 0.5)
Lesson 11-1
Page A46
8
8
4
4
4884
y
x
O
(5.4, 1.4)
(3.5, 0.5)
1. (12)2
(1
1
2)2
1
1
44
2. 122
1
1
22
1
1
44
3. (4 6)34363
64 216
13,824
4.
2
3
4
2
3
4
4
1
8
6
1
5. 16
(42)
1
2
16
16
1
2
1
4
6
or 4
6. 27
1
2
20
1
2
(323)
1
2
(225)
1
2
(32)
1
2
3
1
2
(22)
1
2
5
1
2
3 2 15
1
2
615
7.
4625
2625
2
4
625
1
2
625
or 25
8. 1
15
6
3
1
3(15)6
1
1
52
2
1
25
9. (2a4)222(a4)2
4a8
10. (x4)3x5x12 x5
x17
3
h
2
3
16
h
6
3
15.
3
z4(z4)
1
2
3z4z2
3z6
z
6
3
z2
16. (4r2t5)(16r4t8)
1
4
4r2t5
(16
1
4
r
4
4
t
8
4
)
(4r2t5)(2rt2)
8r3t7
17. a3b5
a
3
2
b
5
2
18.
364m9n
6
64
1
3
m
9
3
n
6
3
(43)
1
3
m3n2
4m3n2
19. 15
3r12t2
15r
1
3
2
t
2
3
15r4t
2
3
20.
8256x2
y16
256
1
8
x
2
3
y
1
8
6
(28)
1
8
x
1
4
y2
2x
1
4
y2
Lesson 11-2
Page A46
1. 2.
216
1
3
h3
y
x
O
y
3
x
y
x
O
y
3
x
9. log36 6 x
36x6
(62)x6
62x61
2x1
x
1
2
10. log3 y4
34y
81 y
11. log5rlog58
r8
12. log535 log5dlog55
log5
3
d
5
log55
3
d
5
5
7 d
13. log44
x
log44
1
2
x
1
2
log44 x
1
2
x
14. log4(2x3) log415
2x3 15
2x12
x6
15. 4 log82
1
3
log827 log8a
log824log827
1
3
log8a
log816 log83 log8a
log848 log8a
48 a
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 104log 8
4 log 10 log 8
4 0.9031
3.0969
3. log 0.14 log (0.01 14)
log 102log 14
2 log 10 log 14
2 1.1461
0.8539
4. log381
l
l
o
o
g
g
8
3
1
4
5. log612
l
l
o
o
g
g
1
6
2
1.3869
6. log529
l
l
o
o
g
g
2
5
9
2.0922
Extra Practice 530
y
x
O
y
3
x
1
3.
Lesson 11-3
Page A46
1. p(100 a)ebt a
p(100 18)e0.6(2) 18
42.7%
2. yaekt c
y140e0.01(10) 70
197° F
3a. y6.7e
4
t
8.1
y6.7e
4
1
8
5
.1
y0.271292 millions of cubic feet
y271,292 ft3
3b. y6.7e
4
t
8.1
y6.7e
4
5
8
0
.1
y2.560257 millions of cubic feet
y2,256,275 ft3
4. Continuously Semiannually
APert AP
1
n
r
nt
A5000e0.058(20) A5000
1
0.0
2
58
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 11-4
Page A47
1. 16
1
4
2
2.
1
2
38
3. 41
1
4
4. log8x2
5. logx32 5
6. log
1
4
16 2
7. log5
1
5
x
5x
1
5
5x51
x1
8. log327 x
3x27
3x33
x3
7. 3x45
xlog 3 log 45
x
l
l
o
o
g
g
4
3
5
x3.4650
8. 6x2x1
xlog 6 (x1) log 2
xlog 6 xlog 2 log 2
xlog 6 xlog 2 log 2
x (log 6 log 2) log 2
x
log
6
l
og
l
2
og 2
x0.6309
9. 5 log ylog 32
log y5log 32
y532
y525
y2
Lesson 11-6
Page A47
1. 3.5553
2. 0.5763
3. 3.4398
4. log15 10
l
l
n
n
1
1
0
5
0.8503
5. log314
l
l
n
n
1
3
4
2.4022
6. log8350
ln
ln
35
8
0
2.8171
7. 5x90
xln 5 ln 90
x
ln
ln
9
5
0
x2.7959
8. 7x25.25
(x2) ln 7 ln 5.25
xln 7 2 ln 7 ln 5.25
xln 7 ln 5.25 2 ln 7
x
ln 5.25
ln
7
2ln7
x1.1478
9. 4x43
xln 4 ln 43
x
ln
ln
4
4
3
x1.3962
10. 6ex48
ex8
xln eln 8
x2.0794
11. 50.2 e0.2x
ln 50.2 0.2xln e
ln
0
5
.2
0.2
x
x19.5801
12. 16 10(1 ex)
1.6 1 ex
0.6 ex
ln 0.6 xln e
0.5108 x
Lesson 11-7
Page A47
1. t
0
l
.
n
04
2
5
15.40 yr
2. t
0
ln
.0
2
6
11.55 yr
3. t
0.0
ln
81
2
25
8.53 yr
4a. y5.2449(1.5524)x
4b. y5.2449(eln 1.5524)x
y5.2449e0.4398x
4c. Use t
ln
k
2
; k0.4398
t
0.
l
4
n
3
2
98
1.58 hr
Lesson 12-1
Page A48
1. d3 7 or 4
1 (4) 5, 5 (4) 9,
9 (4) 13, 13 (4) 17
5, 9, 13, 17
2. d1 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. d8 (14) or 6
2 6 4, 4 6 10,
10 6 16, 16 6 22
4, 10, 16, 22
4. d2.8 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. dx4xor 5x
6x(5x) 11x, 11x(5x) 16x,
16x(5x) 21x, 21x(5x) 26x
11x, 16x, 21x, 26x
6. d(2y2) (2y4)
2y2y(2) (4)
2
2y2, 2y2 2 2y4,
2y4 2 2y6, 2y6 2 2y8
2y2, 2y4, 2y6, 2y8
531 Extra Practice
7. ana1(n1)d
a16 2 (16 1)5
77
8. 20 6 (n1)(2)
26 2n2
28 2n
14 n
9. 42 a1(12 1)4
42 a144
2 a1
10. 30 7 (13 1)d
23 12d
1
1
1
2
1
d
11. d10 10.5 or 0.5
a24 10.5 (24 1)(0.5)
a24 1
12. Sn
n
2
[2a1(n1)d]; d2.8 2 or 0.8
S12
1
2
2
[2 2 (12 1)(0.8)]
6(4 8.8)
76.8
13. Sn
n
2
[2a1(n1)d]
80
n
2
[2 (4) (n1)4]
160 n(8 4n4)
160 12n4n2
0 4(n23n40)
0 4(n8)(n5)
n8 or n5
Since ncannot be negative, n8.
Lesson 12-2
Page A48
1. r
1
7
4
or 0.5
3.5(0.5) 1.75, 1.75(0.5) 0.875,
0.875(0.5) 0.4375
1.75, 0.875, 0.4375
2. r
4
2
or 2
8(2) 16, 16(2) 32, 32(2) 64
16, 32, 64
3. ror
3
4
3
8
3
4
3
9
2
,
3
9
2
3
4
1
2
2
7
8
,
1
2
2
7
8
3
4
5
8
1
1
2
3
9
2
,
1
2
2
7
8
,
5
8
1
1
2
4. r
10
5
or 0.5
2.5(0.5) 1.25, 1.25(0.5) 0.625,
0.625(0.5) 0.3125
1.25, 0.625, 0.3125
5. r
8
8
2
or 2
16
2
162
,162
2
32, 32
2
322
162
, 32, 322
3
8
1
2
6. r
a
a
1
8
0
or
a
1
2
a6
a
1
2
a4, a4
a
1
2
a2, a2
a
1
2
1
a4, a2, 1
7. ana1rn1
a69(2)61
288
8. 100 a1(4)81
100 16,384 a1
4
2
0
5
96
a1
9. 10 a1
1
2
51
10
1
1
6
a1
160 a1
a2160
1
2
or 80
a380
1
2
or 40
160, 80, 40
10. 256 4r41
64 r3
4 r
4(4) 16, 16(4) 64
4, 16, 64, 256
11. Sn
a1
1
a
r
1rn
; r
9
3
or 3
S6
3
1
3(
3
3)6
3
2
2
187
1092
12a. There are four 15-minute periods in an hour, so
r24or 16.
btb016t
12b. b412 164
786,432
Lesson 12-3
Page A49
1. lim
n→
4
3n
2n
lim
n→
3
4
n
lim
n→
2
3
n
n
0
2
3
2
3
2. Limit does not exist. lim
n→
n4
n
3
3n
lim
n→
n
n
3
2
lim
n→n. But as napproaches infinity, nbecomes
increasingly large, so there is no limit.
3. lim
n→
8n2
4n
6
2
n2
lim
n→
8
4
n
n
3
2
lim
n→
4
6
n
n
2
lim
n→
4
n
2
2
2 0 0
2
4. lim
n→
4n2
n
2
2n
2
1
lim
n→
4
1
0
0
0
4
4
n
n
2
2
2
n
n
2
n
1
2
––––
n
n
2
2
n
2
2
Extra Practice 532
5. lim
n→
n3
5
n
2
2
n
3
4
lim
n→
n
n
3
3
n
n
2
3
n
4
3
n
5
3
2
n
n
3
3
2. an
n
3
, an1
n
3
1
rlim
n→
lim
n→
n
n
1
lim
n→
n
n
lim
n→
n
1
1 0
1
test provides no information
3. an
3n(n
1
1)3
, an11
3n1(n1 1)3
n
3
1
—
n
3
533 Extra Practice
1
0
0
2
0
1
2
6. Limit does not exist.
lim
n→
—
2
2
nn
n
—
lim
n→
2n
—
2
n
n
—
lim
n→
2n
lim
n→(2n1) or lim
n→2nAs napproaches infinity, 2n
becomes increasingly large, so there is no limit.
7. 0.0
9
—
1
9
00
—
—
10,
9
000
—
...
a1
—
1
9
00
—
, r
—
1
1
00
—
Sn
1
9
00
—
1
—
1
1
00
—
1
—
n
2
—
1
—
1
1
1
—
8. 0.13
1
1
0
1
3
00
10
3
00
...
a1
1
3
00
, r
1
1
0
Sn
1
1
0
1
3
00
—
1
1
1
0
1
2
5
9. 7.4
0
7
7
1
4
0
0
0
7
0
1,0
4
0
0
0
7
,000
...
a1
1
4
0
0
0
7
0
, r
10
1
00
Sn7
1
4
0
0
0
7
0
——
1
10
1
00
7
2
1
7
1
10. a1
2
1
0
, ror
1
2
4
1
0
—
2
1
0
Sn
2
1
0
—
1
1
2
2
2
0
or
1
1
0
11. Sndoes 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. an(2n)2, an1(2n1)2
22n22n2
rlim
n→
2
2
2n
2
n
2
lim
n→
22
2
n
2
n22
4
divergent
rlim
n→
3n1(n
1
2)3
––––
3n(n
1
1)3
lim
n→
3n(n33n23n1)
3n3(n36n212n8)
lim
n→
n
n
3
3
3
n
n
3
2
3
n
n
3
n
1
3
———
3
n
n
3
3
6
n
n
3
2
1
n
2
3
n
n
8
3
1
3
0
0
0
0
0
0
1
3
convergent
4. an
2
4
n
, an1
2n
4
1
rlim
n→
2n
4
1
—
2
4
n
lim
n→
2n
2n
2
1
2
convergent
5. The general term is
6n
7
1
.
6n
7
1
n
1
for all n, so divergent
6. The general term is
(2
1
n)2
or
4
1
n2
.
4
1
n2
n
1
for all n, so convergent.
7. an
2n1
1
, an1
2n1
1
1
rlim
n→
2n1
1
1
——
2n1
1
lim
n→
2n
2n
2
1
1
lim
n→
2
2
n
n
2
1
n
——
2n
2n
2
2
1
n
1
2
0
0
1
2
convergent
8. an
n
2
n
2
, an1
(n
2n
1
)
1
2
rlim
n→
n
2n
1
3
—
n
2
n
2
Lesson 12-6
Page A50
1. (2 x)416 32x24x28x3x4
2. (nm)5n55n4m10n3m210n2m4
5nm4m5
3. (4ab)364a348a2b12ab2b3
4. (m3)6m6(3)06 m5(3)1
65
2
m
4
1
(3)2
6 5 4 3 m2(3)4
4 3 2 1
6 5 4 m3(3)3
3 2 1
Extra Practice 534
lim
n→
2n
2n
(
2
n
(n
3)
2)
lim
n→
2
n
n
3
2
lim
n→
2
n
n
n
2
—
n
n
n
3
2
1
0
0
2
divergent
Lesson 12-5
Page A49
1.
5
n1
(3n1) (3 1 1) (3 2 1)
(3 3 1) (3 4 1)
(3 5 1)
2 5 8 11 14
40
2.
6
k3
4a4 3 4 4 4 5 4 6
12 16 20 24
72
3.
7
k 3
(k22) (322) (422) (522)
(62 2) (722)
7 14 23 34 47
125
4.
8
j4
j
j
3
4
4
3
5
5
3
6
6
3
7
7
3
8
8
3
4
7
5
8
6
9
1
7
0
1
8
1
3
2
9
6
2
8
4
3
0
5.
4
p0
3p3031323334
1 3 9 27 81
121
6.
n1
2
3
4
n2
3
4
12
3
4
22
3
4
3...
2
3
4
1
1
2
1
1
8
2
3
7
2
... 2
3
4
S
6
7.
4
n1
(3n2) 8.
10
r2
4r
9.
9
m0
4m110.
w1
w!
1
1
2
—
1
3
4
6 5 4 3 2 m1(3)5
5 4 3 2 1
m618m5135m4540m3
1215m21458m729
5. (2rs)4(2r)4(s)04(2r)3(s)1
43
2
(2
r
1
)2(s)2
43
3
2
2
(2
r
1
)1(s)3
16r432r3s24r2s28rs3s4
6. (5x4y)3(5x)3(4y)03(5x)2(4y)1
32(5
2
x)
1
1
(4y)2
125x3300x2y240xy264y3
7.
5!(8
8
!
5)!
x85y5x3y5
56x3y5
8.
4!(7
7
!
4)!
b7434
b39
315b3
9.
2!(1
1
0
0
!
2)!
(4z)102(w)2
(4z)8(w)2
45 65,536z8w2
2,949,120z8w2
10.
7!(1
1
2
2
!
7)!
(2h)127(k)7
(2h)5(k)7
792 32h5(k)7
25,344h5k7
Lesson 12-7
Page A50
1. ln (3) ln (1) ln (3)
i1.0986
2. ln (4.6) ln (1) ln (4.6)
i1.5261
3. ln (0,75) ln (1) ln (0.75)
i0.2877
4. e1.2 1 1.2
(1
2
.2
!
)2
(1
3
.2
!
)3
(1
4
.2
!
)4
1 1.2 0.72 0.288 0.0864
3.29
5. e0.7 1 (0.7)
(0
2
.
!
7)2
(0
3
.
!
7)3
(0
4
.
!
7)4
1 0.7 0.245 0.057 0.010
0.50
12 11 10 9 8 7 6 5 4 3 2 1
7 6 5 3 2 1 5 4 3 2 1
10 9 8 7 6 5 4 3 2 1
2 1 8 7 6 5 4 3 2 1
7 6 5 4 3 2 1
4 3 2 1 3 2 1
8 7 6 5 4 3 2 1
5 4 3 2 1 3 2 1
3 2 1(5x)0(4y)3
3 2 1
4 3 2 1(2r)0(s)4
4 3 2 1
6 5 4 3 2 1 m0(3)6
6 5 4 3 2 1
6. e3.65 1 3.65
(3.
2
6
!
5)2
(3.
3
6
!
5)3
(3.
4
6
!
5)4
1 3.65 6.661 8.105 7.395
26.81
7. cos x1
x
2
2
!
x
4
4
!
x
6
6
!
x
8
8
!
cos
4
cos 0.7854
1
(0.7
2
8
!
54)2
(0.7
4
8
!
54)4
(0.7
6
8
!
54)6
(0.7
8
8
!
54)8
1
0.6
2
169
0.3
2
8
4
05
0.
7
2
2
3
0
47
0
4
.
0
1
,
4
3
4
2
8
0
0.7071
actual value: cos
4
2
2
0.7071
8. sin xx
x
3
3
!
x
5
5
!
x
7
7
!
x
9
9
!
sin
6
sin 0.5236
0.5236
(0.5
3
2
!
36)3
(0.5
5
2
!
36)5
(0.5
7
2
!
36)7
(0.5
9
2
!
36)9
0.5236
0.1
6
435
0.
1
0
2
3
0
94
0
5
.0
0
1
4
0
0
8
3
0
6
.
2
0
,
0
8
3
8
0
0
0.5000
actual value: sin
6
0.5
9. cos
3
1
x
2
2
!
x
4
4
!
x
6
6
!
x
8
8
!
cos
3
cos 1.0472
1
(1.0
2
4
!
72)2
(1.0
4
4
!
72)4
(1.0
6
4
!
72)6
(1.0
8
4
!
72)8
1
1.0
2
966
1.2
2
0
4
26
1.
7
3
2
1
0
88
1
4
.
0
4
,
4
3
6
2
2
0
0.5000
actual value: cos
3
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) 42or 16
f(16) 162or 256
f(256) 2562or 65,536
f(65,536) 65,5362or 4,294,967,296
16, 256, 65,536, 4,294,967,296
3. z02i
z10.5(2i) i
2i
z20.5(2i) i
2i
z30.5(2i) i
2i
2i, 2i, 2i
4. z04 4i
z10.5(4 4i) i
2 2ii
2 3i
z20.5(2 3i) i
1 1.5ii
1 2.5i
z30.5(1 2.5i) i
0.5 1.25ii
0.5 2.25i
2 3i, 1 2.5i, 0.5 2.25i
5. p1p0rp0
p14000 (0.054)4000
4216
p24216 (0.054)4216
4443.66
p34443.66 (0.054)4443.66
4683.62
p44683.62 (0.054)4683.62
4936.54
p54936.54 (0.054)4936.54
5203.11
$4216, $4443.66, $4683.62, $4936.54, $5203.11
Lesson 12-9
Page A50
1. Step 1: Verify that the formula is valid for n1.
Since S12 and 1(1 1) 2, the formula is valid
for n1.
Step 2: Assume the formula is valid for nkand
show that it is also valid for nk1.
2 4 6 ... 2kk(k1)
Derive the formula for nk1 by adding
2(k1) to each side.
2 4 6 ... 2k2(k1)
k(k1) 2(k1)
2 4 6 ... 2k2(k1)
(k1)(k2)
Apply the original formula for nk1.
Sk1(k1)((k1) 1) or (k1)(k2)
The formula gives the same result as adding the
(k1) term directly. Thus, if the formula is valid
for nk, it is also valid for nk1. Since Snis
valid for n1, it is also valid for n2, n3,
and so on indefinitely. Hence the formula is valid
for all positive integral values of n.
535 Extra Practice
2. Step 1: Verify that the formula is valid for n1.
Since S11 and
1(1 1
6
)(1 2)
1, the formula is
valid for n1.
Step 2: Assume the formula is valid for nkand
show that it is also valid for nk1.
1 3 6 ...
k(k
2
1)
k(k1
6
)(k2)
Derive the formula for nk1 by adding
(k1)
2
(k2)
to each side.
1 3 6 ...
k(k
2
1)
(k1)
2
(k2)
k(k1
6
)(k2)
(k1)
2
(k2)
1 3 6 ...
k(k
2
1)
(k1)
2
(k2)
k(k1)(k2) 3(k1)(k2)
6
4. dependent
5. dependent
6. P(5, 5)
(5
5!
5)!
543
1
21
120
7. P(8, 3)
(8
8!
3)!
336
8. P(4, 1)
(4
4!
1)!
4
3
3
2
2
1
1
4
9. P(10, 9)
(10
1
0!
9)!
3,628,800
10. P(9, 6)
(9
9!
6)!
60,480
11. P(7, 3)
(7
7!
3)!
210
12.
P
P
(
(
5
2
,
,
2
1
)
)
5
3
!
!
1
2
!
!
10
13.
P
P
(
(
8
7
,
,
6
4
)
)
8
7
!
!
3
2
!
!
24
14.
P(5,
P
2
(1
)
0
,
P
1
(8
)
,4)
8 7 6 5 4 3 2 1 3 2 1
7 6 5 4 3 2 1 2 1
(8
8!
6)!
—
(7
7!
4)!
5 4 3 2 1 1
3 2 1 2 1
(5
5!
2)!
—
(2
2!
1)!
7 6 5 4 3 2 1
4 3 2 1
9 8 7 6 54 3 2 1
3 2 1
10 9 8 7 6 5 4 3 2 1
1
8 7 6 5 4 3 2 1
5 4 3 2 1
Extra Practice 536
1 3 6 ...
k(k
2
1)
(k1)
2
(k2)
(k1)(k2)(k3)
6
1
9
0
!
!
8
4
!
!
5
3
!
!
3360
15. C(4, 2)
(4
4
2
!
)! 2!
4
2
3
1
2
2
1
1
6
16. C(10, 7)
(10
10
7
!
)! 7!
120
10 9 8 7 6 5 4 3 2 1
3 2 1 7 6 5 4 3 2 1
9 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 54 3 2 1
10 9 8 7 6 5 4 3 2 1 4 3 2 1 3 2 1
(5
5!
2)!
(8
8!
4)!
———
(10
1
0!
1)!
Apply the original formula for nk1.
Sk1
or
The formula gives the same result as adding the
k1 term directly. Thus, if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2,
n3, and so on indefinitely. Hence, the formula
is valid for all positive integral values of n.
3. Sn: 5n1 2rfor some integer r.
Step 1: Verify that Snis valid for n1.
S1511 or 4. Since 4 2 2, Snis valid for
n1.
Step 2: Assume that Snis valid for nkand show
that it is also valid for nk1.
Sk→5k1 2rfor some integer t
Sk1→5k11 2tfor some integer t
5k1 2r
5(5k1) 5(2r)
5k15 10r
5k11 10r4
5k11 2(5r2)
Thus, 5k11 2t, where t(5r2) is an
integer. Thus if Skis valid, then Sk1is also
valid. Since Snis valid for n1, it is also valid for
n2, n3, and so on indefinitely. Hence, 5n1
is even for all positive integral values of n.
Lesson 13-1
Page A51
1. Using the Basic Counting Principle,
6 6 6 216.
2. P(8, 8)
(8
8!
8)!
40,320
3. independent
8 7 6 5 4 3 2 1
1
(k1)(k2)(k3)
6
(k1)((k1) 1)((k1) 2)
6
17. C(6, 5)
(6
6
5
!
)! 5!
6
18. C(4, 3) C(7, 3)
(4
4
3
!
)! 3!
(7
7
3
!
)! 3!
4
1
3
3
2
2
1
1
140
19. C(3, 1) C(8, 7)
(3
3
1
!
)! 1!
(8
8
7
!
)! 7!
24
20. C(9, 5) C(4, 3)
(9
9
5
!
)! 5!
(4
4
3
!
)! 3!
504
Lesson 13-2
Page A51
1.
5040
2.
10,080
3.
840
4.
1,814,400
5.
4
2
!
!
43
2
2
1
1
12
6. circular; (4 1)! or 6
7. circular; since the bracelet can be turned over
there are
(9
2
1)!
or 20,160 permutations
8. linear; 5! or 120
Lesson 13-3
Page A51
1. P(ace)
5
4
2
or
1
1
3
2. P(a card of 5 or less)
2
5
0
2
or
1
5
3
3. P(a red face card)
5
6
2
or
2
3
6
4. P(not a queen) 1 P(a queen)
1
5
4
2
4
5
8
2
or
1
1
2
3
5. P(blue)
8
2
42
1
2
4
or
1
7
4 4 4 4 4
52
10 9 8 7 6 5 4 3 2 1
2 1
10!
2!
7 6 5 4 3 2 1
3 2 1
7!
3!
8 7 6 5 4 3 2 1
2 1 2 1
8!
2! 2!
7 6 5 4 3 2 1
1
7!
0!
4 3 2 1
1 2 1
9 8 7 6 5 4 3 2 1
4 3 2 1 5 4 3 2 1
8 7 6 5 4 3 2 1
1 7 6 5 4 3 2 1
321
2 1 1
7 6 5 4 3 2 1
4 3 2 1 3 2 1
6 5 4 3 2 1
1 5 4 3 2 1
6. P(green)
8
4
42
1
4
4
or
2
7
7. P(not red)
8
4
4
2
2
1
6
4
or
3
7
8. P(red or green)
8
8
4
4
2
1
1
2
4
or
6
7
9. P(s)
1
4
6
1
3
5
P(f) 1 P(s)
2
1
0
1
2
1
0
or
1
2
9
0
odds or
1
1
9
2
1
0
—
1
2
9
0
537 Extra Practice
10. P(s)
1
6
6
1
5
5
P(f) 1 P(s)
1
8
1
1
8
or
7
8
odds or
1
7
11. P(s)
C(2,
C
1
(
)
16
,
C
2
(
)
4, 1)
P(f) 1 P(s)
2
12
0
4
1
1
1
5
—
1
1
5
—
1
1
4
5
odds or
1
1
4
12. P(s) P(1 white, 1 yellow) P(1 yellow, 1 red)
P(1 white, 1 red)
C(2,
C
1
(
)
16
,
C
2
(
)
4, 1)
C(4,
C
1
(
)
1
6
C
,
(
2
1
)
0, 1)
C(2,
C
1
(
)
1
6
C
,
(
2
1
)
0, 1)
2
12
0
4
4
1
2
1
0
0
2
1
2
1
0
0
1
1
5
1
3
1
6
1
3
7
0
P(f)1 P(s)
1
1
3
7
0
or
1
3
3
0
odds or
1
1
7
3
Lesson 13-4
Page A52
1. dependent;
6
8
2
7
1
3
4
2. independent;
1
1
0
8
1
1
0
8
2
8
5
1
3. independent;
1
2
1
6
0
1
3
0
4. inclusive;
1
6
1
6
3
1
6
3
1
6
1
5. exclusive;
5
4
2
5
4
2
5
8
2
or
1
2
3
1
3
7
0
—
1
3
3
0
1
1
5
—
1
1
4
5
1
8
—
7
8
Lesson 13-5
Page A52
1. P(second clip is bluefirst clip was red)
1
8
2
1
4
1
—
1
8
2
4. P(exactly 1 hit) C(5, 1)
1
2
0
0
0
0
0
1
1
8
0
0
0
0
0
4
5
1
5
2
6
5
2
6
5
2
6
5
2
6
5
or 0.4096
5. P(exactly 3 hits) C(5, 3)
1
2
0
0
0
0
0
3
1
8
0
0
0
0
0
2
10
1
1
25
1
2
6
5
6
3
2
2
5
or 0.0512
6. P(at least 4 hits)
P(4 hits) P(5 hits)
C(5, 4)
1
2
0
0
0
0
0
4
1
8
0
0
0
0
0
1C(5, 5)
1
2
0
0
0
0
0
5
1
8
0
0
0
0
0
0
5
6
1
25
4
5
1
31
1
25
1
3
2
1
1
25
or 0.00672
Lesson 14-1
Page A53
1. range 70 22 or 48
2. Sample answer: 10
3. Sample answer: 20, 30, 40, 50, 60, 70, 80
4. Sample answer: 25, 35, 45, 55, 65, 75
5.
6.
7. Sample answer: 40–50
Extra Practice 538
1
4
1
2. P(second clip is bluefirst clip was blue)
1
4
2
1
3
1
—
1
4
2
1
3
1
3. P(numbers on dice matchsum is greater than 7)
3
3
6
—
1
3
5
6
1
3
5
or
1
5
4. P(sum is greater than 7numbers match)
3
3
6
—
3
6
6
3
6
or
1
2
5. P(ball is from second boxball is white)
P(2nd box and white)
P(white)
1
2
3
8
——
1
2
4
7
1
2
3
8
2
5
1
3
Lesson 13-6
Page A52
1. P(all heads) C(3, 3)
1
2
3
1
2
0
1
1
8
1
1
8
2. P(exactly 2 tails) C(3, 2)
1
2
2
1
2
1
3
1
4
1
2
3
8
3. P(at least 2 heads)
P(2 heads) P(3 heads)
C(3, 2)
1
2
2
1
2
1C(3, 3)
1
2
3
1
2
0
3
1
4
1
2
1
1
8
1
3
8
1
8
4
8
or
1
2
Grams of Fat Frequency
20-30 6
30-40 7
40-50 8
50-60 3
60-70 5
70-80 1
Frequency
Grams of Fat
10020
Grams of Fat Consumed
by Adults
30 40 50 60 70 80
4
6
2
0
8
Lesson 14-2
Page A53
1. X
1
4
(130 150 180 190)
162.5
Md
150
2
180
or 165
Mode: none
2. X
1
6
(15 16 17 18 18 19)
17.2
Md
17
2
18
or 17.5
Mode 18
3. X
1
6
(25 28 30 36 38 42)
33.2
Md
30
2
36
or 33
Mode: none
4. X
1
1
0
(1 2 3 4 5 5 6 9 9 10)
5.4
Md5
Mode 5 and 9
5. X
1
1
2
(2.3 2.5 4 2(5.6) 6 6.4
6.5 2(7) 8 10)
5.9
Md
6
2
6.4
or 6.2
Mode 5.6 and 7
6. X
1
1
2
(14 2(15) 16 20 21 24 27
28 36 2(39))
24.5
Md
21
2
24
or 22.5
Mode 15 and 39
7. X
1
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
Md5.7
Mode 5.4 and 5.6
8. X
1
1
4
(800 820 830 890 960 970 980
1040 2(1050) 1080 1110 1170
1180)
995
Md
980
2
1040
or 1010
Mode 1050
9. stem leaf
1279
23446899
3445679
4024
55
12 12
10. 8
1
6
(4 5 6 9 10 x)
48 34 x
14 x
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
7.5
7
2
x
15 7 x
8 x
Lesson 14-3
Page A54
1. interquartile range Q3Q1
58 39
19
semi-interquartile range
1
2
9
or 9.5
2. interquartile range Q3Q1
7.65 2.75
4.9
semi-interquartile range
4
2
.9
or 2.45
3. X
1
9
(150 175 180 180 195 200 212
220 250)
195.7
MD
1
9
(45.820.815.8
... 54.2)
21.98
j
27.56
4. X
1
1
1
(1.4 2 2.4 2.9 3 3.5 3.7 4.2
4.6 5.3 5.5)
3.5
MD
1
1
1
(2.11.51.1...2)
1.05
j
1.26
5a. Md18
5b. Q1
16
2
18
or 17
Q320
5c. interquartile range Q3Q1
20 17
3
5d. semi-interquartile range
3
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.
(2.1)2(1.5)2(1.1)2... 22
11
(45.8)2(20.8)2... 54.22
9
539 Extra Practice
40 50 60 70 80
12345678910
5f.
Lesson 14-4
Page A54
1a. 25% corresponds to t0.3.
10 0.3(2) 9.4 10.6
1b. 10 8 2, 14 10 4
tj2tj4
t(2) 2t(2) 4
t1t2
68
2
.3%
34.15%
95
2
.5%
47.75%
34.15% 47.75% 81.9%
1c. 10 7 3, 10 10 0
tj3tj0
t(2) 3t(2) 0
t1.5 t0
86
2
.6%
43.3%
1d. 80% corresponds to t1.3.
10 1.3(2) 7.4 12.6
2a. 0.683(400) 273.2
2b. 0.955(400) 382
2c.
0.6
2
83
(400) 136.6
Lesson 14-5
Page A54
1. jX
1.
9
2
0
or about 0.13
2. jX
3
1
.4
00
or 0.34
3. jX
12
2
.
4
4
0
or about 0.80
4. A1% confidence level is given when P99% and
t2.58.
jX
4.
4
2
0
0.6640783086
interval: X
tjX
150 2.58jX
148.29 151.71
5. A1% confidence level is given when P99% and
t2.58.
jX
1
7
0
8
1.132277034
interval: X
tjX
320 2.58jX
317.08 322.92
Lesson 15-1
Page A55
1. lim
x→4(x22x2) 422(4) 2
22
2. lim
x→1(x4x32x1) (1)4(1)32(1) 1
1
3. lim
x→0(xsin x) 0 sin 0
0
4. lim
x→4
x
x
2
1
4
6
lim
x→4
(x
x
4
)(x
4
4)
lim
x→4(x4)
4 4
8
5. lim
x→2
x
x
2
2
5
x
x
2
6
lim
x→2
(
(
x
x
3
1
)
)
(
(
x
x
2
2
)
)
lim
x→2
x
x
3
1
2
2
3
1
1
3
6. lim
x→2
x2
3x
5
x
9
24
lim
x→2
(x
3(x
8)
(x
3
)
3)
lim
x→2
x
3
8
2
3
8
3
6
or
1
2
Lesson 15-2
Page A55
1. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
5(xh
h
)5x
lim
h→0
5x5
h
h5x
lim
h→0
5
h
h
5
2. f(x) lim
h→0
f(xh
h
)f(x)
lim
h→0
lim
h→0
lim
h→0
9
h
h
9
3. f(x)
1
2
x
2
3
f(x)
1
2
1x110
1
2
4. f(x) x24x8
f(x) 2x214 1x110
2x4
9x9h2 9x2
h
9(xh)2(9x2)
h
Extra Practice 540
15 16 17 18 19 20 21 22 23 24
5. f(x) x5
F(x)
5
1
1
x51C
1
6
x6C
6. f(x) 2x28x2
F(x) 2
2
1
1
x218
1
1
1
x112xC
2
3
x34x22xC
7. f(x)
1
5
x3
3
4
x1
F(x)
1
5
3
1
1
x31
3
4
1
1
1
x11xC
2
1
0
x4
3
8
x2xC
8. f(x)
x32
x
x2x
x22x1
F(x)
2
1
1
x212
1
1
1
x11xC
1
3
x3x2xC
Lesson 15-3
Page A55
1.
3
0
5xdxlim
n→
n
i1
5
3
n
i
n
3
lim
n→
4
n
5
2
n(n
2
1)
lim
n→
2
4
n
5
2
(n2n)
lim
n→
4
2
5
n
n
2
2
lim
n→
4
2
5
n
n
2
4
2
5
0
4
2
5
2.
5
1
(x1)dx
5
0
(x1)dx
1
0
(x1)dx
lim
n→
n
i1
5
n
i
1
n
5
lim
n→
n
i1
n
i
1
n
1
lim
n→
n
5
5
n
1
1
5
n
2
1
...
5
n
n
1
lim
n→
n
1
n
1
1
n
2
1
...
n
n
1
lim
n→
n
5
n
n
5
(1 2 ... n)
lim
n→
n
1
n
n
1
(1 2 ... n)
lim
n→
n
5
n
n
5
n(n
2
1)
lim
n→
n
1
n
n
1
n(n
2
1)
lim
n→
n
5
7
2
n
5
2
lim
n→
n
1
3
2
n
1
2
lim
n→
3
2
5
2
2
n
5
lim
n→
3
2
2
1
n
3
2
5
0
3
2
0
3
2
2
or 16
3.
2
0
(x24x4)dx
lim
n→
n
i1
2
n
i
24
2
n
i
4
n
2
lim
n→
n
2
2
n
1
24
2
n
1
4
2
n
2
2
4
2
n
2
4
...
2
n
3
24
2
n
3
4
lim
n→
n
2
n
4
2
(1222... n2)
n
8
(1 2 ... n) 4n
lim
n→
n
2
n
4
2
n(n1)
6
(2n1)
n
8
n(n
2
1)
4n
lim
n→
n
8
3
2n33
6
n2n
1
n
6
2
n2
2
n
8
lim
n→
4
3
2
n
3
n
1
2
8
1
n
1
8
8
3
8 8
5
3
6
4.
2000
1500
(6 0.002x)dx
lim
n→
n
i1
6 0.002
1500
50
n
0i
5
n
00
lim
n→
5
n
00
n
i1
3
n
i
lim
n→
5
n
00
3
n
1
3
n
2
...
3
n
n
lim
n→
5
n
00
3n
n
1
(1 2 ... n)
lim
n→
1500
5
n
0
2
0
n(n
2
1)
lim
n→
1500 250
1
n
1
1500 250 or $1250
Lesson 15-4
Page A55
1. x6dx
1
7
x7C
2. 5x4dx 5
1
5
x5C
x5C
3. (x2x5)dx
1
3
x3
1
2
x25xC
4. (4x4x26)dx 4
1
5
x5
1
3
x36xC
4
5
x5
1
3
x36xC
5.
2
2
14x6dx 14
1
7
x7
2
2
2x7
2
2
(2 27) (2 (2)7)
256 (256)
512
541 Extra Practice
6.
6
0
(x2)dx
1
2
x22x
6
0
1
2
622 6
1
2
022 0
30 0
30
7.
4
2
(x24)dx
1
3
x34x
4
2
1
3
434 4
1
3
234 2
1
3
6
1
3
6
3
3
2
8.
5
4
(x4)(x2)dx
5
4
(x22x8)dx
1
3
x32
1
2
x28x
5
4
1
3
x3x28x
5
4
1
3
53528 5
1
3
43428 4
7
3
0
8
3
0
1
3
0
Extra Practice 542
543 Chapter Tests
Chapter Tests
Chapter 1 Test
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(a2) a2 3(a2)2
a2 3(a24a4)
a2 3a212a12
3a211a10
6a. E
4
P
d2
4
1
(
2
0
0
.9
0
)2
117.89 Im/m2
6b. dcannot be negative or zero.
7. f(x) g(x) x27 x3
x2x4
f(x) g(x) x27 (x3)
x2x10
f(x) g(x) (x27)(x3)
x33x27x21
g
f(
(
x
x
)
)
x
x
2
3
7
8. [fg](x) f(g(x))
f(4x5)
4x5 1
4x4
[g f](x) g(f(x))
g(x1)
4(x1) 5
4x4 5
4x1
9. [fg](x) f(g(x))
f(2x26)
5(2x26)
10x230
[gf](x) g(f(x))
g(5x)
2(5x)26
50x26
10. 11.
y
x
O
y
4
x
12
2
2
4
6
8
10
12
14
16
21123 45345
4
y
x
O
y
|
x
| 2
12. 13.
15. m
8
2
0
4
8
6
or
3
4
y4
3
4
(x0)
y4
3
4
x
y
3
4
x4
14. y3
5
3
(x(1))
y3
5
3
(x1)
y3
5
3
x
5
3
y
5
3
x
1
3
4
16. parallel: y2 4(x0)
y2 4x
4xy2 0
perpendicular: y2
1
4
(x0)
y2
1
4
x
4y8 x
x4y8 0
17. x5y3 ⇒m
1
5
parallel: y2
1
5
(x(1))
y2
1
5
(x1)
5y10 x1
x5y11 0
perpendicular: y2
5
1
(x(1))
y2 5(x1)
y2 5x5
5xy3 0
18. 19.
y
x
O
2
x
y
1
y
x
O
y
3
x
6
x
O
g
(
x
)
x
1
g
(
x
)
f
(
x
)
x
O
20a.
20b. Sample answer: Using (47, 67) and (95, 88),
y0.44x46.44
m
8
9
8
5
6
4
7
7
0.4375
y67 0.4375(x47)
y67 0.4375x20.5625
y0.4375x46.4375
y0.44x46.44
Chapter 2 Test
Page A57
1.
2. 3xy7→6x2y14
5x2y12 5x2y12
x2
3xy7
3(2) y7
y1
(2, 1)
3. 4x5y2→12x15y 6
3x2y13 12x8y52
23y46
y2
4x5y2
4x5(2) 2
4x12
x3
(3, 2)
4. 4x6y3z20 →4x6y3z20
x5yz15 3x15y3z45
7x9y25
x5yz15 →2x10y2z30
7xy2z17xy2z1
9x11y31
7x9y25 →63x81y225
9x11y31 63x77y217
4y 8
y2
7x9y25 x5yz15
7x9(2) 25 1 5(2) z15
7x711 z15
x1z4
(1, 2, 4)
5. y12 2x3x4 2y
3x4 2(12 2x)
3x4 24 4x
7x28
x4
y12 2(4)
4
(4, 4)
6. AB
1 3
2 0
5 (2)
4 (3)
Chapter Tests 544
Economics
Grade
Statistics Grade
20 40 60 80 1000
40
60
20
80
100
y
x
O
x
y
4
2
x
y
1
(1, 3)
4
2
7
1
7. 3C
3 (3)
3(4)
3(0)
3(2)
3(1)
3 (3)
9
12
0
6
3
9
8. 2B
2(3)
2(0)
2(2)
2(3)
6
0
4
6
2BA
6 1
0 2
4 (5)
6 4
5
2
1
10
9. BC
3
4
0
2
1
3
3
0
2
3
2(3) 3(4)
3(3) 0(4)
2(0) 3(2)
3(0) 0(2)
2(1) 3(3)
3(1) 0(3)
18
9
6
0
11
3
10. 1.5
A(1.5, 4.5), B(6, 4.5), and C(9, 3)
9
3
6
4.5
1.5
4.5
6
2
4
3
1
3
6
6
2
4
6
8
4
4
2246810
y
x
O
A
A
BC
C
B
11.
F(3, 2), G(1, 5), H(6, 1), J(3, 5)
12.
5(6) 3(7)
9
13.
2(9) (10)(4)
58
14.
1
2
(1)
1(1) 2(10) 1(3)
16
15.
1
5
1
2
3
1
1
2
3
1
1
21
13
0
1
3
4
1
2
3
4
1
2
0
1
1
1
2
2
0
1
1
3
4
4
9
2
10
7
6
5
3
3
5
6
1
1
5
3
2
3
5
6
1
1
5
3
2
0
1
1
0
545 Chapter Tests
y
x
O
J
J
H
H
G
G
F
F
3
5
1
5
1
5
2
5
16.
1
1
0
4
3
2
4
4
3
2
4
1
34
42
1
5
2
5
2
5
1
3
0
17. The inverse does not exist since
0.
18.
7
3
x
y
1
1
2
3
4
12
5
15
1
5
1
2
1
3
1
2
1
3
1
21
31
1
5
1
5
2
3
x
y
7
3
1
2
1
3
x
y
1
1
2
3
1
2
1
3
O
y
x
(2, 4)
(2, 2)
(6, 0)
(
4
3
1
2, 0
y
y
3
x
4
x
3
)
y
0
x
2
19.
f(x, y) 2xy
f(6, 0) 2(6) 0 or 12 cminimum
f(2, 4) 2(2) 4 or 0
f(2, 2) 2(2) 2 or 2
f
4
3
, 0
2
4
3
0 or
8
3
cmaximum
20. Let xnumber of ads.
Let ynumber of commercial minutes.
x3
y4
100x200y1300
f(x, y) 12,000x16,000y
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.
Chapter 3 Test
Page A58
1. y2x1→b2a1
x-axis b2a1
b2a1no
y-axis b2(a) 1
b2a1no
yxa2b1
b
a
2
1
no
yxa2(b) 1
a2b1
b
a
2
1
no
none of these
O
y
x
(3, 5)
(3, 4)
(5, 4)
100
x
200
y
1300
y
4
x
3
9. f(x) 5x4
y5x4
x5y4
x4 5y
y
x
5
4
f1(x)
1
5
x
4
5
Yes, it is a function.
10. f(x)
x
3
2
y
x
3
2
x
y
3
2
y2
3
x
y
3
x
2
f1(x)
3
x
2; Yes, it is a function.
11. f(2)
2
2
2
2
4
0
No; the function is undefined when x2.
12. Yes; the function is defined when x0, the
function approaches 1 as xapproaches 0 from both
sides, and f(0) 1.
13. y→as x→, y→as x→
14. At x4, yis at a minimum
value, since f(4) f(3) and
f(4) f(5).
15. At x1, yis at a point of
inflection, since
f(0) f(1) f(2).
16. x1y
x
4
x
1
y
4
x
x
x
x
1
x
Chapter Tests 546
2. y
x
2
2
→b
a
2
2
x-axis b
a
2
2
b
a
2
2
no
y-axis b
(a
2
)2
b
a
2
2
yes
yxa
b
2
2
b
a
2
no
yxa
(
2
b)2
a
b
2
2
b
a
2
no; y-axis
3. xy23→ab23
x-axis a(b)23
ab23yes
y-axis ab23
ab23no
yxba23
ab3
no
yxb(a)23
ba23
ab
3
no; x-axis
4. xy 5→ab 5
x-axis a(b) 5
ab 5no
y-axis (a)b5
ab 5no
yxba5
ab 5yes
yx(b)(a) 5
ab 5yes
yx, yx
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.
7.
8.
x
O
f
(
x
) 5
x
4
f
(
x
)
f
1(
x
) 1
54
5
x
xy
311
412
511
xy
01
12
23
y
as x→, y→4; y4
4
1
1
x
17. x24 0
(x2)(x2) 0
x2
y
x2
x
4
y
x
x
2
x
x
2
2
x
4
2
y
1
x
1
x
4
2
as x→, y→0; y0
y
x
O
y
|
x
4|
y
x
O
y
2
x
2 3
5
123452354
10
15
20
25
30
35
40
45
50
18a. y
0
0
.
.
2
2
5
5
0
0
.
.
0
0
0
0
4
1
x
x
y
0.
x
25
0.0
x
04x
0.
x
25
0.0
x
01x
6. 11513 53 60
16760
167600
x36x27x60
7. f(x) x38x22x11
f(2) (2)38(2)22(2) 11
8 32 4 11
9; no
8. f(x) 4x42x2x3
f(1) 4(1)42(1)21 3
4 2 2
0; yes
9. 1 positive
f(x) 6x311x23x2
2 or 0 negative
rational zeros: 2,
1
3
,
1
2
10. 1 positive
f(x) x4x39x217x8
3 or 1 negative
rational zero: 1
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
upper bound: 2
f(x) x33x25x9
lower bound: 5
547 Chapter Tests
y
as x→, y→
0
0
.
.
0
0
0
0
4
1
or 4; y4
18b. As the amount of 4 molar solution added
increases, the molarity of the mixture
approaches 4.
19. ykx y 0.25x
0.5 k(2) y0.25(10)
0.25 ky2.5
20. y
x
k
2
y
7
x
2
2
8
(3
k
)2
18
7
x
2
2
72 kx
24
x2
Chapter 4 Test
Page A59
1. (x4)(xi)(x(i) (x4)(xi)(xi)
(x4)(x2i2)
(x4)(x21)
x34x2x4
2. b24ac (5)24(1)(4)
9
Since b24ac 0, there are 2 real roots.
n
n
(5
2
)
(1
)
9
n
5
2
3
n
5
2
3
or n
5
2
3
n4n1
3. b24ac (7)24(1)(3)
61
Since b24ac 0, there are 2 real roots.
z
z
(7
2
)
(
1)
61
z
7
2
61
4. b24ac (5)24(2)(4)
7
Since b24ac 0, there are 2 imaginary roots.
x
x
(5
2
)
(
2)
7
x5
4
7
i
5. 22334
42 10
2156
2x2x5, R6
bb24
ac
2a
bb24
ac
2a
bb24
ac
2a
0.
x
25
0.004
0.
x
25
0.001
r61132
1
2
614 4 0
1
3
6960
26110
r11917 8
81 963 487 3888
81747393 3136
11 2724 32
11 0980
r1359
114110
215 5 1
r1359
11272
2117 5
3105 6
41115
512534
14. Sample answer: 1; 2
upper bound: 1
f(x) 2x43x3x2x1
lower bound: 2
15.
8
1
0
a
1
1
1
0
a
1
8
7
0
a
8
7
0
16.
x
4
2
x2
3
4
1
4
x
4
2
(x2)
3
(x2)
1
4
4(x2) 3
(x2)
4
(x2)
16(x2) 12 (x2)(x2)
16x32 12 x24
x216x16 0
x
x
16
2
83
x8 43
17.
x
5
2
5
x
3
2
x
15x15(x2) 2(x2)
15x15x30 2x4
2x34
x17; x2 or 0
Test x18:
18
5
2
?
5
18
3(
2
18)
1
5
6
?
1
5
8
2
1
7
0.3125
?0.3148 true
Test x3:
3
5
2
?
5
3
3(
2
3)
5
?
5
3
2
9
5
?1
8
9
false
Test x1:
1
5
2
?
5
1
3(
2
1)
5
?5
2
3
5
?5
2
3
true
Test x1:
1
5
2
?
5
1
3(
2
1)
5
3
?5
2
3
1
2
3
?5
2
3
false
x17, 2 x0
18. y2
3 0Check: 11 2
3 0
y2
39
3 0
y2 93 3 0
y11 0 0
16 162
4(1)(1
6)
2(1)
19. 2x2
3x5
2x2 3x5
7 x
Check: 2(7)
2
3(7)
5
16
16
20. 11 1
0m
9
11 10m81
10m70
m7
11 10m0
10m11
m
1
1
0
1
Test m8: 11 1
0(8)
?9
91
?9
9.54
?9true
Test m0: 11 1
0(0)
?9
11
?9
3.32
?9false
Solution: m7
21.
2z
5
2
z
z
1
1
6
(z
5z
2)
(2z
11
3)
2z
5
2
z
z
1
1
6
z
A
2
2z
B
3
5z11 A(2z3) B(z2)
Let z
3
2
.
5
3
2
11 A
2
3
2
3
B
3
2
2
7
2
7
2
B
1 B
Let z2.
5(2) 11 A(2 (2) 3) B(2 2)
21 7A
3 A
2z
5
2
z
z
1
1
6
z
3
2
2z
1
3
22.
(
7
x
x
2
2
1
1
)
8
(x
x
2
1
)
(
7
x
x
2
2
1
1
)
8
(x
x
2
1
)
x
A
1
x
B
2
x
C
1
7x218x1 A(x1)(x2) B(x1)(x1)
C(x1)(x2)
Let x1.
7(1)218(1) 1A(1 1)(1 2) B(1 1)(1 1)
C(1 1)(1 2)
24 6C
4 C
Let x2.
7(2)218(2) 1 A(2 1)(2 2)
B(2 1)(2 1)
C(2 1)(2 2)
9 3B
3 B
Let x1.
7(1)218(1) 1 A(1 1)(1 2)
B(1 1)(1 1)
C(1 1)(1 2)
12 2A
6 A
(
7
x
x
2
2
1
1
)
8
(x
x
2
1
)
x
6
1
x
3
2
x
4
1
7x218x1
(x1)(x1)(x2)
Chapter Tests 548
r23111
125 456
r23111
121232
22 1 1 13
23. quadratic
24. Let xthe height.
Then 6xthe length and 6x7 the width.
Vx(6x)(6x7)
120 36x342x2
0 6x37x220
Use a graphing calculator to graph the related
function.
V(x) 6x37x220
It appears as if the zero is at x2.
Use the Factor Theorem to check
V(2) 6(2)37(2)220 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 sthe speed of the freight train.
Then 30 sthe speed of the car.
Car: 500 (30 s)t⇒t
30
50
0
s
Train: 350 st ⇒t
35
s
0
30
50
0
s
35
s
0
500 s10,500 350s
150s10,500
s70 km/h
Chapter 5 Test
Page A60
1.
9
3
9
6
5
0
°
°
2.76
a360°(2) 995°
a720° 995°
x275°; IV
2.
3
2
6
3
0
4
°
°
0.65
a360°(1) 234°
a360° 234°
a126°; II
3.
4
3
1
6
0
0
°
°
1.14
a360°(1) 410°
a360° 410°
a50°; I
4.
3
1
6
2
0
4
°
5°
3.46
a360° (4) 1245°
a1440° 1245°
a195°; III
5. (RS)2(ST )2(RT )2
(RS)2122152
RS2225 144
RS 81
or 9 in.
sin R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
cos R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
sin R
1
1
2
5
or
4
5
cos R
1
9
5
or
3
5
tan R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
csc R
s
h
id
y
e
po
o
t
p
e
p
n
o
u
s
s
it
e
e
tan R
1
9
2
or
4
3
csc R
1
1
5
2
or
5
4
sec R
s
h
id
y
e
po
a
t
d
e
j
n
a
u
ce
s
n
e
t
cot R
s
s
i
i
d
d
e
e
a
op
d
p
ja
o
c
s
e
i
n
te
t
sec R
1
9
5
or
5
3
cos R
1
9
2
or
3
4
6. tan 60°
y
x
tan 60°
1
3
or 3
7. sec 270°
x
r
sec 270°
1
0
; undefined
8. sin (405°) sin (45°)
sin (45°) y
sin (45°)
2
2
9. rx2y
2
r325
2
r34
sin v
y
r
cos v
x
r
tan v
y
x
sin v
5
34
cos v
3
34
tan v
5
3
sin v
5
34
34
cos v
3
34
34
csc v
y
r
sec v
x
r
cot v
x
y
csc v
5
34
sec v
3
34
cot v
3
5
10. rx2y
2
r(4)2
22
r20
or 25
sin v
y
r
cos v
x
r
tan v
y
x
sin v
2
2
5
cos v
2
4
5
tan v
2
4
sin v
5
5
cos v
2
5
5
tan v
1
2
csc v
y
r
sec v
x
r
cot v
x
y
csc v
2
2
5
sec vcot v
2
4
csc v5
sec v
2
5
cot v2
11. rx2y
2
r02(
3)2
r9
or 3
sin v
y
r
cos v
x
r
tan v
y
x
sin v
3
3
cos v
0
3
tan v
0
3
sin v1cos v0undefined
csc v
y
r
sec v
x
r
cot v
x
y
csc v
3
3
sec v
3
0
cot v
0
3
csc v1undefined cot v0
25
4
549 Chapter Tests
12. cos A
b
c
cos °
4
c
2
c
cos
42
77°
c186.7
13. tan B
a
b
tan 27°
1
b
3
b13 tan 27°
b6.6
14. sin A
a
c
sin 32° 17
1
a
4
a14 sin 32° 17°
a7.5
15. cos B
a
c
cos B
2
3
3
7
Bcos1
2
3
3
7
B51.6°
16. tan A
a
b
tan A
1
3
1
Atan1
1
3
1
A15.3°
17. Let hthe height.
sin 70°
6
h
5
h65 sin 70°
h61.1 m
18. B180° 36° 87° or 57°
K
1
2
a2
sin
s
B
in
s
A
in C
K
1
2
(24)2
sin 5
si
7
n
°
3
si
6
n
°
87°
K410.4 units2
19. K
1
2
bc sin A
K
1
2
(56.4)(92.5) sin 58.4°
K2221.7 units2
20.
180°
2
36°
72°
sin
22
36°
sin
x
72°
xsin 36° 22 sin 72°
s
22
si
s
n
in
36
7
°
2°
x35.6 cm
Perimeter 22 35.6 35.6
93.2 cm
21. Since 98° 90°, consider Case II.
ca; one solution
sin
90
98°
si
6
n
4
A
64 sin 98° 90 sin A
Asin1
64 s
9
in
0
98°
A44.8°
B180° 98° 44.8° or 37.2°
sin
b
37.2
sin
90
98°
90 sin 37.2° bsin 98°
b
90
s
s
i
i
n
n
9
3
8
7
°
.2°
b54.9
A44.8°, B37.2°, b54.9
22. Since 31° 90°, consider Case I.
absin A
9 20 sin 31°
9 10.3; none
23. a2b2c22bc cos A
132721522(7)(15) cos A
169 274 210 cos A
105 210 cos A
Acos1
1
2
0
1
5
0
A60°
sin
13
60°
sin
7
B
7 sin 60° 13 sin B
Bsin1
7si
1
n
3
60°
B27.8°
C180° 60° 27.8° or 92.2°
A60°, B27.8°, C92.2°
24. b2a2c22ac cos B
b202
242
2(20)(
24) co
s 47°
b17.92432912
sin
b
47°
si
2
n
4
C
24 sin 47° bsin C
Csin1
24 si
b
n47°
C78.3°
A180° 47° 78.3° or 54.7°
b17.9, C78.3°, A54.7°
25. Let xthe distance between the transmitters.
x270° 13022(70)(130) cos 130°
x21,80
0 1
8,200
cos 1
30°
x183.0 miles
Chapter Tests 550
70˚
65 m
h
72˚
36˚
22 cm
xx
70 miles 130 miles
130˚
x
14. yArccsc x
xArccsc y
Csc xyor yCsc x
15. ytan x
xtan y
arctan xyor yarctan x
16. sin
Arccos
1
2
sin 60°
2
3
17. tan
sin1
1
2
tan
6
tan
7
6
1
3
or
3
3
18. vr
v
t
2300 240,000
2
t
t
240
2
,0
3
0
0
0
0
(2)
t656 hours or 27.3 days
19. A
73
2
21
h
73
2
21
2
k
12
A26 h47 k
2
1
2
A26 k
6
y26 sin
6
tc
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
Sample answer: y26 sin
6
t
2
3
47
551 Chapter Tests
y
x
O
y
sin
x
2
2
1
1
Chapter 6 Test
Page A61
1. 225° 225°
18
0°
5
4
2. 480° 480°
18
0°
8
3
3. reference angle:
5
6
6
; Quadrant 2
sin
5
6
1
2
4. reference angle:
5
4
4
; Quadrant 3
tan
5
4
1
5. vr
v
t
v12
7
1
.1
v85.2 cm/s
6. 7.
8. 33;
2
4
or
2
9. 22; or
4
3
10. A4
2
k
4
A4k
2
4
or 0.5
y4 sin 0.5v2
11. A0.5
2
k
2
4
c
4
A0.5 kor 4 c
y0.5 cos (4v) 1
12. 13.
2
2
2
3
2
y
x
O
y
cos
x
2
2
1
1
y
O
y
3 cos
22
2
3
3
y
O
y
tan (2 )
4
3
2
5
4
1
3
2
1
4
5
4
4
2
2
y
x
O
y
Csc
x
y
Arccsc
x
2
2
2
y
arctan
x
y
tan
x
y
x
O
2
2
2
2
20.
6
1
5
h
m
o
i
u
le
r
s
52
1
8
m
0
i
f
l
e
e
et
360
1
0
h
se
o
c
u
o
r
nds
95.3
ft/s
tan v
r
v
g
2
vtan1
1
(
2
9
0
5
0
.3
(3
)2
2)
v0.23 radians
Chapter 7 Test
Page A62
1. sin2vcos2v1
1
3
2cos2v1
cos2v
8
9
cos v
2
3
2
2. tan2v1 sec2v
tan2v1 (2)2
tan2v3
tan v3
3. sin2vcos2v1
4
5
2cos2v1
cos2v
2
9
5
cos v
3
5
sec v
co
1
sv
sec vor
5
3
4. sin v
cs
1
cv
sin vor
3
5
sin2vcos2v1
3
5
2cos2v1
cos2v
1
2
6
5
cos v
4
5
5. tan (420°)
c
s
o
in
s
(
(
4
4
2
2
0
0
°
°
)
)
sin (360° 60°)
cos (360° 60°)
1
5
3
1
3
5
8.
s
s
e
in
c
x
x
c
s
o
in
s
x
x
cot x
cot x
sec xcos xsin2x
sin xcos x
Chapter Tests 552
c
s
o
i
s
n
6
6
0
0
°
°
tan 60°
6. tan v(cot vtan v) sec2v
tan vcot vtan2vsec2v
1 tan2vsec2v
sec2vsec2v
7. sin2Acos2A(1 sin A)(1 sin A)
sin2A
c
s
o
in
s2
2
A
A
(1 sin A)(1 sin A)
cos2A(1 sin A)(1 sin A)
1 sin2A(1 sin A)(1 sin A)
(1 sin A)(1 sin A) (1 sin A)(1 sin A)
s
1
in
x
s
c
in
o
2
s
x
x
cot x
c
s
o
in
s
x
x
cot x
cot xcot x
9.
1
cos
si
x
nx
1
cos
si
x
nx
2 sec x
2 sec x
cos xcos xsin xcos xsin xcos x
1 sin2x
1
2
c
s
o
i
s
n
x
2x
2 sec x
2
co
c
s
o
2
s
x
x
2 sec x
co
2
sx
2 sec x
2 sec x2 sec x
10. csc (AB) sec B
sin Acos Atan B
csc (AB)
co
1
s B
sin Acos A
c
s
o
in
s
B
B
csc (AB) 1
sin Acos Bcos Asin B
csc (AB)
sin (A
1
B)
csc (AB) csc (AB)
11. cot 2v
1
2
cot v
1
2
tan v
cos 2v
2
c
s
o
i
s
n
v
v
2
s
c
in
os
v
v
cot 2v
co
2
s
s
2
i
v
n
vc
s
o
in
s
2
v
v
cot 2v
c
s
o
in
s
2
2
v
v
cot 2vcot 2v
12. sin 255° sin (225° 30°)
sin 225° cos 30° cos 225° sin 30°
2
2
2
3
2
2
1
2
6
4
2
2
4
6
13. tan
5
1
2
tan
6
4
tan
6
tan
4
1 tan
6
tan
4
1
3
1
1
1
3
1
1
1
3
1
1
3
1
1
3
1
1
3
1
2
3
1
3
1
1
3
3
2
4
3
2
3
3
2 3
553 Chapter Tests
14. tan v
c
s
o
in
s
v
v
3
7
sin 2v2 sin vcos v
2
4
7
3
4
3
8
7
cos 2vcos2vsin2v
3
4
2
4
7
2
1
9
6
1
7
6
1
2
6
or
1
8
tan 2v
1
2
t
t
a
a
n
n
v
2v
2
3
7
1
3
7
2
4
7
3
4
sin2vcos2v1
sin2v
3
4
21
sin2v
1
7
6
sin v
4
7
20. yx3
xy3 0
A2
B2
(1)2
12
or 2
1
2
x
1
2
y
3
2
0
2
2
x
2
2
y
3
2
2
0
sin f
2
2
, cos f
2
2
, p
3
2
2
; Quadrant II
tan for 1
ftan1(1)
135°
21. 5 5y10x
0 10x5y5
A2
B2
102
(5)2
or 55
5
1
0
5
x
5
5
5
y
5
5
5
0
2
5
5
x
5
5
y
5
5
0
sin f
5
5
, cos f
2
5
5
, p
5
5
; Quadrant IV
tan for
1
2
5
5
2
5
5
2
2
2
2
553 Chapter Tests
37
15. cos (22.5°) cos
4
2
5°
1c
2
os 45°
1
2
2
2
2
3
7
2
9
2
4
2
2
2
2
16. tan2x3
tan x
tan2x3
tan x0
tan x(tan x3
) 0
tan x0ortan x3
0
x0° tan x3
x60°
17. cos 2xcos x0
2 cos2xcos x1 0
(2 cos x1)(cos x1) 0
2 cos x1 0orcos x1 0
cos x
1
2
cos x1
x120° x0°
18. sin xcos x0
sin xcos x
c
s
o
in
s
x
x
1
tan x1
x45° or x225°
19. 2 cos2x3 sin x3
2(1 sin2x) 3 sin x3 0
2 sin2x3 sin x1 0
(2 sin x1)(sin x1) 0
2 sin x1 0 or sin x1 0
sin x
1
2
sin x1
x30°, 150° x90°
ftan1
1
2
2
333°
22. 2xy6 →2xy6 0
d
Ax
1
A2
B
y1
B
2
C
d
2(
5)
2
2
1(8
1
)
2
6
d
8
5
or
8
5
5
8
5
5
23. d
A
x1
A
B
2
y
1
B2
C
d
3(
6
)
32
4
(8)
4
2
2
d
1
5
6
1
5
6
24. 5x2y7 →5x2y7 0
y
3
4
x1 →3x4y4 0
d1
5x
1
52
2
y1
2
2
7
d2
3
x1
3
2
4y1
4
2
4
5x1
2
2
y
9
17
3x1
5
4y14
25x110y135 329
x1429
y1429
(25 329
)x(10 429
)y
35 429
0
25. R
v
g
02
sin 2v
R
v
g
02
2 sin vcos v
R
8
3
8
2
2
2
3
5
4
5
R232.32 ft
Chapter 8 Test
Page A63
1. 2.5 cm, 60° 2. 1.6 cm, 25°
3.
3.9 cm, 46°
4.
1.8 cm, 334°
5. AB
u
1 3, 9 6
4, 3
AB
u
(4)2
(3)2
25
or 5
6. AB
u
3 (2), 10 7
5, 3
AB
u
523
2
34
7. AB
u
9 2, 3 (4), 7 5
7, 1, 2
AB
u
721
222
54
or 36
8. AB
u
8 (4), 10 (8), 2 (2)
4, 2, 4
AB
u
(4)2
(2
)24
2
36
or 6
9. r
u
s
u1 4, 3 3, 4 (6)
5, 0, 10
10. 3s
u3 4, 3 3, 3 6
12, 9, 18
2r
u
2 (1), 2 3, 2 4
2, 6, 8
3s
u2r
u
12 (2), 9 6, 18 8
14, 3, 26
11. r
u
3s
u1 12, 3 9, 4 (18)
11, 12, 14
12. r
u
(1)2
32
42
26
13. s
u423
2(
6)2
61
14. r
u
i
u
3j
u
4k
u
15. s
u4i
u
3j
u
6k
u
Chapter Tests 554
3.9 cm
46˚
a
b
a b
1.8 cm
334˚
a
2b
2b a
40˚
55˚
60 lb
125 lb
125˚
165˚
x
16. r
u
s
u1(4) 3(3) 4(6)
19
17. r
u
s
u
i
u
j
u
k
u
30 i
u
10 j
u
15 k
u
30, 10, 15
18. No; since r
u
s
u19 and not 0.
19.
x212526022(125)(60) cos 55°
x19,22
5 1
5,000
cos 5
5°
x103.06 lb
s
1
i
0
n
3
5
.0
5
6
°
si
6
n
0
v
103.06 sin v60 sin 55°
vsin1
60
10
si
3
n
.0
5
6
5°
v28.48°
v40° 28.48° 40° or 68.48°
20. xx1ta1yy1ta2
x3 t(2) y11 t(5)
x2t3y5t11
21. x2t3yt1
x
2
3
ty1 t
y1
x
2
3
y
1
2
x
1
2
22. vx100 cos 2° vy100 sin 2°
vx99.94 mph vy3.49 mph
23. The figure is four times the original size and
reflected over the yz-plane.
24. AB
u
1.5 cos 60°, 0, 1.5 sin 60°
0.75, 0, 0.753
F
u
0, 0, 110
T
u
AB
u
F
u
i
u
j
u
k
u
0.75 0 0.753
00 110
i
u
u
j
k
u
0i
u
82.5j
u
0k
u
T
u
028
2.52
02
82.5 lb-ft
0
0
0.75
0
0.753
110
0.75
0
0.753
110
0
0
3
3
1
4
4
6
1
4
4
6
3
3
1.5 ft
110 lb
60˚
555 Chapter Tests
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
A
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
B
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
C
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234
0
6
11
2
6
3
2
3
3
4
6
5
3
2
6
7
3
5
1234 2468 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
1234 0˚
330˚
180˚
90˚
30˚
60˚
270˚
240˚
150˚
120˚
210˚
300˚
25. ytv
u
sin v
1
2
gt2xtv
u
cos v
y28tsin 35°
1
2
(32)t2x28tcos 35°
0 4t(7 sin 35° 4t)x23.02 feet
4t0 or 7 sin 35° 4t0
t0t
7s
in
4
35°
t1.003758764
Chapter 9 Test
Page A64
1. 2.
3. 4.
5. 6.
7.
8. r222
2
vArctan
2
2
8
or 22
4
22
,
4
9. r(6)2
02
36
or 6
Since x0 and y0, v.
(6, )
14. y3
rsin v3
r
s
in
3
v
r3 csc v
15. x2y23x
r23rcos v
r3 cos v
16. r7
r249
x2y249
17. 5 rcos (v45°)
0 rcos (v45°) 5
0 r (cos vcos 45° sin vsin 45°) 5
0
2
2
rcos v
2
2
rsin v5
0
2
2
x
2
2
y5
0 2
x2
y10
18. A2
B2
523
2
34
5
34
x
3
34
y
3
34
0
cos f
5
34
34
, sin f
3
34
34
, p
3
34
34
fArctan
3
5
180°
211°
prcos (vf)
3
34
34
rcos (v211°)
19. A2
B2
22(
4)2
20
or 25
2
2
5
x
2
4
5
y
2
1
5
0
cos f
5
5
, sin f
2
5
5
, p
10
5
fArctan
1
2
360°
297°
prcos (vf)
10
5
rcos (v297°)
20. i93 (i4)23 i
123 i
i
10. r(2)2
(3
)2
vArctan
3
2
13
4.12
3.61
(3.61, 4.12)
11. xrcos vyrsin v
3 cos
5
4
3 sin
5
4
3
2
2
3
2
2
3
2
2
3
2
2
3
2
2
,
3
2
2
12. xrcos vyrsin v
2 cos
7
6
2 sin
7
6
2
2
3
2
1
2
3
1
(3
, 1)
13. xrcos vyrsin v
4 cos 1.4 4 sin 1.4
0.68 3.94
(0.68, 3.94)
21. (2 5i) (2 4i) (2 (2)) (5i4i)
0 (i)
i
22. 6i(3 2i) 6i3 2i
8i3
23. (3 5i)(3 2i) 9 9i10i2
19 9i
24. (1 3i)(2 i)(1 2i) (2 7i3i2)(1 2i)
(1 7i)(1 2i)
1 9i14i2
13 9i
25.
6
2
2
i
i
6
2
2
i
i
2
2
i
i
12
4
1
0i
i
2
2i2
10
5
10i
2 2i
26. r(4)2
42
vArctan
4
4
32
or 42
3
4
42
cos
3
4
isin
3
4
27. r(5)2
02
vArctan
0
5
25
or 5
5(cos isin )
28. r4 3v
3
2
4
12
6
4
or
7
4
12
cos
7
4
isin
7
4
12
2
2
i
2
2
62
62
i
29. r
2
3
3
v
2
3
6
2
3
6
or
2
2
cos
2
isin
2
2(0 i)
2i
30. r12(
1)2
vArctan
1
1
2
2
7
4
(1 i)8(2
)8
cos (8)
7
4
isin (8)
7
4
16 (cos 14isin 14)
16 (1 0)
16
31. r02(
27)2
v
2
729
or 27
327i
(0 27i)
1
3
27
1
3
cos
1
3
2
isin
1
3
2
3
cos
6
i sin
6
3
2
3
i
1
2
3
2
3
3
2
i
32. x3i0 →x3i
Find the cube roots of i.
r021
2
v
2
1
or 1
(0 i)
1
3
1
cos
2
2n
isin
2
2n
1
3
1
1
5
cos
6
2n
3
isin
6
2n
3
cos
6
4n
isin
6
4n
x1cos
6
isin
6
2
3
1
2
i
x2cos
5
6
isin
5
6
2
3
1
2
i
x3cos
9
6
isin
9
6
i
33. EIZ
8 (cos 307° jsin 307°)
20 (cos 115° jsin 115°)
8 20 [cos (307° 115°) jsin (307° 115°)]
160 (cos 422° jsin 422°)
160 (cos 62° jsin 62°)
Chapter 10 Test
Page A65
1. d(x2
y1)2
(y2
y1)2
d3 (
1))2
(1
2)2
d42(
1)2
d17
x1
2
x2
,
y1
2
y2
1
2
3
,
2
2
1
1,
3
2
2. d(x2
y1)2
(y2
y1)2
d(2k
3k)2
(k
1 (k
1))
2
d(k)2
(2
)2
dk24
x1
2
x2
,
y1
2
y2
3k
2
2k
,
k1
2
k1
5
2
k, k
3. r(8
(6)2
(3
(4
))2
r(2)2
(7)2
r53
(xh)2(yk)2r2
(x(8)2(y3)2
53
2
(x8)2(y3)253
Chapter Tests 556
i
i
O
213 2
i
213 2
i
1
1
1
1
4. center: (h, k) (0, 0)
a210 b26ca2b
2
a10
b6
c10
6
or 2
foci: (h, kc) (0, 2)
major axis vertices: (h, ka)
0, 10
minor axis vertices: (hb, k) (6
, 0)
5. e
a
c
1
2
(x
a2
h)2
(y
b2
k)2
1
If c
1
2
, then a1.
(x
1
0)2
1
ca2b
2
(y0)2
3
4
10. A1, C0; since C0, the conic is a parabola.
x26x8y7 0
x26x9 8y7 9
(x3)28y16
(x3)28(y2)
11. A4, C1; since Aand Chave opposite signs,
the conic is a hyperbola.
4x2y21
y
1
2
1
12. A1, C1; since AC, the conic is a circle.
x2y24x12y36 0
x24x4 y212y36 4
(x2)2(y6)24
x2
1
4
557 Chapter Tests
1
2
12b
2
1
4
1 b2
b2
3
4
6. center: (h, k) (4, 2)
a216 b27b2c2a2
a4b7
7 c216 →c23
foci: (h, kc)
4, 2 23
vertices: (h, ka) (4, 2 4)
(4, 6) and (4, 2)
asymptotes: yk
a
b
(xh)
y2
4
7
(x(4))
y2
4
7
7
(x4)
7. foci: (h, kc) ⇒h5e
3
2
a
c
kc4b2c2a2
kc2b23222
2k2b25
k1; c3
(y
a2
k)2
(x
b2
h)2
1
(y
22
1)2
(x(
5
5))2
1
(y
4
1)2
(x
5
5)2
1
8. vertex: (h, k) (0, 3)
4p8 ⇒p2
focus: (hp, k) (0 2, 3) (2, 3)
directrix: xhp
x0 2
x2
axis of symmetry: yk
y3
9. 2kp5foci: (h, kp) (3, 5)
2kp2directrix: ykp2
2kp7
k
7
2
, p
3
2
(xh)24p(yk)
(x3)24
3
2
y
7
2
(x3)26
y
7
2
y
x
O
y
x
O
y
x
O
x2
4
3
y2
1
13. A3, C16; since Aand Chave opposite
signs, the conic is a hyperbola.
3x216y218x128y37 0
3(x26x9) 16(y28y16) 37 27 256
3(x3)216(y4)2192
(y
12
4)2
(x
64
3)2
1
14. A9, C1; since Aand Chave opposite signs,
the conic is a hyperbola.
9x2y290x8y200 0
9(x210x25) (y28y16) 200 225 16
9(x5)2(y4)29
(x
1
5)2
(y
9
4)2
1
15. A2, C13; since Aand Chave opposite
signs, the conic is a hyperbola.
2x213y25 0
2x213y25
13
5
y2
2
5
x2
1
1
x2
5
2
y2
1
5
3
16. A0, C1; since A0, the conic is a parabola.
y22x10y27 0
y210y25 2x27 25
(y5)22x2
(y5)22(x1)
17. A1, C2; since Aand Chave the same sign
and AC, the conic is an ellipse.
x22y22x12y11 0
(x22x1) 2(y26y9) 11 1 18
(x1)22(y3)28
(x
8
1)2
(y
4
3)2
1
18. y2x2x
19. x2 cos ty2 sin t
2
x
cos t
2
y
sin t
cos2tsin2t1
2
x
2
2
y
21
x
4
2
y
4
2
1
x2y24
20. B24AC 024(4)(1)
16
AC; ellipse
4(x1)2(y3)236
4(x1 3)2(y3 (5))236
4(x2)2(y2)236
4x216x16 y24y4 36
4x2y216x4y16 0
Chapter Tests 558
y
x
O
2
2
2
4
6
8
10
12
14
4
6
24681012 1446
y
x
O
y
x
O
yx
O
y
x
O
21. B24AC 024(2)(1)
8
hyperbola
Replace xwith xcos 60° ysin 60° or
1
2
x
2
3
y.
Replace ywith xsin 60° ycos 60° or
2
3
x
1
2
y.
2
1
2
x
2
3
y
2
2
3
x
1
2
y
28
2
1
4
(x)2
2
3
xy
3
4
(y)2
3
4
(x)2
2
3
xy
1
4
(y)2
8
1
2
(x)23
xy
3
2
(y)2
3
4
(x)2
2
3
xy
1
4
(y)28
(x)263
xy5(y)232 0
(x)263
xy5(y)232 0
22. (x1)2y21x24y24
x22x1 y21x24(x22x) 4
x2y22x0x24x28x4
y2x22x3x28x4 0
(3x2)(x2) 0
x
2
3
or x2
If x
2
3
, y
2
3
2
2
2
3
0.9.
If x2, y
(2)2
2(2
)
0.
(2, 0), (0.7, 0.9)
23.
24. x2y2Dx Ey F0
0212D(0) E(1) F0 ⇒EF1
(2)232D(2) E(3) F0
⇒2D3EF13
4252D(4) E(5) F0
⇒4D5EF41
4D6E2E26 3E3F63
4D5EF41 11E3F67
11E3F67 8E64
E8
EF12D3EF13
8 F12D3(8) 7 13
F72D4
D2
x2y22x8 y7 0
x22x1 y28y16 7 1 16
(x1)2(y4)210
center: (h, k) (1, 4)
radius 10
If a person is walking northeast, the person will
first hit the motion detector at (3, 9).
Chapter 11 Test
Page A66
559 Chapter Tests
y
x
O
1. 343
2
3
3343
2
72
49
2. 64
1
3
3
1
64
1
4
3. ((2a)3)2(2a)6
(2
1
a)6
26
1
a6
64
1
a6
4.
x
3
2
y2z
5
4
4x
1
2
2
y8z
2
4
0
x6y8z5
5.
327a6b
12
(33)
1
3
a
6
3
b
1
3
2
3a2b4
6. m
1
2
n
2
3
m
3
6
n
4
6
6m3n4
7.
8.
9. 4
1
2
210.
1
6
3216
11. log5625 412. log8m5
13. logx32 5
x532
x
1
5
32
3
1
2
x5
1
2
5x5
1
2
x
y
x
O
25. x2y290
x2(2x3)29
x24x212x9 90
5x212x81 0
(5x27)(x3) 0
x
2
5
7
or x3
y2x3
y2
2
5
7
3 or
3
5
9
y2(3) 3 or 9
y
x
O
14. log5(2x) log5(3x4)
2x3x4
4 x
15. 3.6x72.4
xlog 3.6 log 72.4
x
l
l
o
o
g
g
7
3
2
.
.
6
4
x3.3430
16. 6x182x
(x1) log 6 (2 x) log 8
xlog 6 log 6 2 log 8 xlog 8
xlog 6 xlog 8 2 log 8 log 6
x(log 6 log 8) 2 log 8 log 6
x
2
lo
lo
g
g
6
8
lo
lo
g
g
8
6
17. log415
l
l
o
o
g
g
1
4
5
1.9534
18. log30.9375
log
lo
0
g
.9
3
375
0.0587
19. log81 3
l
l
o
o
g
g
8
3
1
1
4
20. log 542 2.7340 21. ln 0.248 1.3943
22. antiln (1.9101) 0.1481
23. t
ln
k
2
t
0
l
.
n
05
2
4
t12.84 yr
24. Let xthe original number of bacteria.
3xxek(6)
3 e6k
ln 3 6k
k
ln
6
3
k0.1831020481
8xxe0.1831t
8 e0.1831t
ln 8 0.1831t
t
0.
l
1
n
8
8
31
t11.3568626 hours
0.3568626(60) 21 minutes
11 hours, 21 minutes
25. Q(t) Qe
B
t
C
1 106(5 106)e
3(2
t
106)
1
5
e
6
t
106
ln 0.2
6
t
106
tln 0.2(6 106)
t9.66 106s
Chapter 12 Test
Page A67
1. d4.5 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. d1 (6) or 5
a24 6 (24 1)5
a24 109
3. 8 4 (5 1)d
12 4d
3 d
4 3 1, 1 3 2, 2 3 5
4, 1, 2, 5, 8
4. 345
n
2
(2(12) (n1)5)
690 24n5n25n
0 5n219n690
0 (5n69)(n10)
n
6
5
9
or n10
Since there cannot be a fractional number of
terms, n10.
5. ror
2
5
1
1
0
1
4
Chapter Tests 560
2
1
5
2
5
1
2
25
,
1
2
25
2
5
6
4
25
,
6
4
25
2
5
31
8
25
1
2
25
,
6
4
25
,
31
8
25
6. 1 16r51
1
1
6
r4
1
2
r
16
1
2
8, 8
1
2
4, 4
1
2
2
16, 8, 4, 2, 1
7. ror 2
5
5
2
S10
5
2
5
2
(2)10
1 2
5
2
51
2
20
1
51
2
15
8. does not exist;
lim
n→
3
n
n
3
2
3
1
lim
n→, lim
n→33, and lim
n→
n
1
2
0,
so the denominator approaches 3. As napproaches
infinity, the nterm in the numerator makes the
whole numerator approach infinity, so the entire
fraction has no limit.
9. lim
n→
2n
n
3
3
3
4
n
lim
n→
1
2
0
0
1
2
n
n
3
3
n
4
3
2
n
n
3
3
3
n
n
3
n
n
3
2
3
n
1
2
10. The general term is
3
1
n2
.
3
1
n2
n
1
for all n, so convergent
11. The series is arithmetic, so it is divergent.
12. Sample answer:
19
k
15k
13. Sample answer:
k1
6
3
2
k1
14. (2a3b)5
(2a)5(3b)05(2a)4(3b)1
54(2
2
a)
3
1
(3h)2
5 4 3 2 (2a)1(3b)4
4 3 2 1
5 4 3(2a)2(3b)3
3 2 1
Evaluate the original formula for nk1.
The formula gives the same result as adding the
(k1) term directly. Thus, if the formula is valid
for nk, it is also valid for nk1. Since the
formula is valid for n1, it is also valid for n2.
Since it is valid for n2, it is also valid for n3,
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, n40. Since the interest is
compounded quarterly, the rate per period is
0.
4
08
or 0.02. The common ratio ris 1.02.
a1200(1.02) 204
Sn
a1
1
a
r
1rn
S40
S40 $12,322.00
Chapter 13 Test
Page A68
1. P(6, 2)
(6
6!
2)!
30
2. P(7, 5)
(7
7!
5)!
2520
3. C(8, 3)
(8
8
3
!
)! 3!
56
4. C(5, 4)
(5
5
4
!
)! 4!
5
1
4
4
3
3
2
2
1
1
5
5. Using the Basic Counting Principle,
5 4 3 2 1 120.
6. P(5, 3)
(5
5!
3)!
54
2
3
1
21
60
7. (7 1)! 6! 720
8. C(3, 1) C(12, 8)
(3
3
1
!
)! 1!
(12
12
8
!
)! 8!
3
2
2
1
1
1
1485
9. C(4, 2) C(6, 3)
(4
4
2
!
)! 2!
(6
6
3
!
)! 3!
4
2
3
1
2
2
1
1
120
6 5 4 3 2 1
3 2 1 3 2 1
12 11 10 9 8 7 6 5 4 3 2 1
4 3 2 1 8 7 6 5 4 3 2 1
8 7 6 5 4 3 2 1
5 4 3 2 1 3 2 1
7 6 5 4 3 2 1
2 1
6 5 4 3 2 1
4 3 2 1
204 204(1.02)40
1 1.02
(k1)(k2)(4k9)
3
(k1)[(k1) 1][4(k1) 5]
3
561 Chapter Tests
32a5240a4b720a3b21080a2b3
810ab4243b5
15.
5!(1
1
0
0
!
5)!
a10525a532
8064a5
16.
4!(8
8
!
4)!
(3x)84(y)4
(3x)4(y)4
8 7 65 4 3 2 1
4 3 2 1 4 3 2 1
10 9 8 7 6 5 4 3 2 1
5 4 3 2 1 5 4 3 2 1
5 4 3 2 1(2a)0(3b)5
5 4 3 21
k(k1)
3
(4k5)
2(k1)(2k3)
k(k1)
3
(4k5)
6(k1)
3
(2k3)
k(k1)(4k5) 6(k1)(2k3)
3
(k1)(k2)(4k9)
3
(k1)(4k217k18)
3
(k1)(k(4k5) 6(2k3)
3
70(81x4)(y4)
5670x4y4
17. r(2)2
22
or 22
vArctan
2
2
or
3
4
2 2i22
cos
3
4
isin
3
4
22
ei
3
4
18. z02i
z13(2i) (2 i)
6i2 i
2 5i
z23(2 5i) (2 i)
6 15i 2 i
8 14i
z33(8 14i) (2 i)
24 42i2 i
26 41i
19. Step 1: Verify that the formula is valid for n1.
Since 2 1(2 1 1) 6 and 6,
the formula is valid for n1.
Step 2: Assume that the formula is valid for nk
and prove that it is valid for nk1.
2 3 4 5 6 7 ... 2k(2k1)
k(k1)
3
(4k5)
234567...2k(2k1)
2(k1)(2k3)
1(1 1)(4 1 5)
3
10. P(2 white)
1
6
0
5
9
1
3
11. P(s)
1
4
P(f) 1 P(s)
1
1
4
or
3
4
odds or
1
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)
4
1
5
3
2
1
5
2
1
5
1
0
1
1
4
0
9
4
9
8
16
3
,6
3
60
13. P(sum of 8) P(sum of 4)
3
5
6
3
3
6
1
1
2
5
96
or
4
5
32
14. P(3 odd digits)
1
2
1
2
1
2
1
8
15. P(all red or all blue) P(all red) P(all blue)
1
3
2
1
2
1
1
1
0
1
5
2
1
4
1
1
3
0
2
1
20
2
1
2
2
1
0
16. P(ace or black card)
P(ace) P(black card) P(black ace)
5
4
2
2
5
6
2
5
2
2
2
5
8
2
or
1
7
3
17. P(3’stogetheroddnumber)
1
4
3
4
4. Sample answer: 1.5, 4.5, 7.5, 10.5, 13.5, 16.5, 19.5
5. Sample answer:
6.
7. X
8
1
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.
Md
10
2
10
or 10
9. Order the values from least to greatest. Since
there are 15 terms, the median is the
15
2
1
or
8th term.
Md2.344
10. Q12.341; Q32.347
11. interquartile range Q3Q1
2.347 2.341
0.006
semi-interquartile range
0.0
2
06
0.003
12.
13. X
1
1
5
(2.334 2.338 ... 2.350)
2.34393333
MD
1
1
5
(2.34393333 2.334...
2.34393333 2.350)
0.0025
14. j
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 t1.65.
24 1.65(2.8) 19.3828.62
(2.34393333 2.334)2... (2.34393333 2.350)2
15
Chapter Tests 562
P(3’s together and odd number)
P(odd number)
1
5
8
18. P(both eveneven product)
2
6
6
or
1
3
3
19. P(no more than two heads)
P(0 heads) P(1 head) P(2 heads)
C(5, 0)
2
3
0
1
3
5C(5, 1)
2
3
1
1
3
4
C(5, 2)
2
3
2
1
3
3
2
1
43
2
1
4
0
3
2
4
4
0
3
2
5
4
1
3
or
1
8
7
1
20. C(7, 4)
4
5
4
1
5
3
1
1
5
7
,6
9
2
2
5
or 0.114688
Chapter 14 Test
Page A69
1. range 20 1 or 19
2. Sample answer: 3
3. Sample answer: 0, 3, 6, 9, 12, 15, 18, 21
P(both even and even product)
P(even product)
Number
of Tallies Frequency
Absences
0–3 ||| 3
3–6 ||||| 6
6–9 |||||||||||| | 1 6
9–12 |||| |||| |||| |||| |||| 2 5
12–15 |||| |||| |||| || 1 7
15–18 |||| |||8
18–21 |||| 5
2.338 2.340 2.342 2.344 2.346 2.348 2.350
Frequency
Number of Absences
306912151821
12
18
6
3
0
24
27
9
15
21
17. 24 18.4 5.6 32.4 24 8.4
tj5.6 tj8.4
t(2.8) 5.6 t(2.8) 8.4
t2t3
95
2
.5%
47.75%
99
2
.7%
49.85%
47.75% 49.85% 97.6%
10. f(x) 6x42x230
f(x) 6 4x412 2x210
24x34x
11. f(x) 2x54x3
2
5
x26
f(x) 2 5x514 3x31
2
5
2x210
10x412x2
4
5
x
12. f(x) 2x4(x33x2)
2x76x6
f(x) 2 7x716 6x61
14x636x5
13. f(x) (x3)2
x26x9
f(x) 2x216 1x110
2x6
14. f(x) 2x6
F(x)
1
2
1
x116xC
x26xC
15. f(x) x34x2x4
F(x)
3
1
1
x31
2
4
1
x21
1
1
1
x114xC
1
4
x4
4
3
x3
1
2
x24xC
16. f(x)
1
2
x3
2
7
x5
F(x)
1
2
3
1
1
x31
2
7
1
1
1
x115xC
1
8
x4
1
2
4
x25xC
1
8
x4
1
7
x25xC
17. f(x)
x34
x
x2x
x24x1
F(x)
2
1
1
x21
1
4
1
x11xC
1
3
x32x2xC
18.
2
0
x3dx lim
n→
n
i1
2
n
i
3
n
2
lim
n→
1
n
6
4
n2(n
4
1)2
lim
n→4
n2
n
2
2
n1
lim
n→4
1
n
2
n
1
2
4 units2
19.
3
1
3x2dx 3
1
3
x3
3
1
x3
3
1
3313
26 units2
20.
1
0
(2x3)dx 2
1
2
x23x
1
0
x23x
1
0
[123(1)] [023(0)]
4
563 Chapter Tests
18. 29.6 24 5.6
tj5.6
t(2.8) 5.6
t2
95
2
.5%
47.75%
19. jX
j
N
jX
3
4
.6
00
jX
0.18
20. A5% level of confidence is given when P95%.
95% corresponds to t1.96.
X
tjX
57 1.96(0.18)
56.6557.35
Chapter 15 Test
Page A70
1. The closer xis to 1, the closer yis to 1.
So, lim
x→1f(x) 1. Also, f(1) 1.
2. The closer xis to 2, the closer yis to 4.
So, lim
x→2f(x)4. However, there is a point at (2, 1).
So f(2) 1.
3. lim
x→0
x2
x
3x
lim
x→0(x3)
3
4. lim
x→1
x2
x
3x
1
2
lim
x→1
(x
x
2
)(x
1
1)
lim
x→1(x2)
1
5. lim
x→3lim
x→3
lim
x→3
x2
x
3x
3
9
32
3
3
(3
3
)9
2
6
7
or
2
9
6. lim
x→1
x2
3
x2
2
x
5
3
12
3
(1
2
)2
(1
)
5
3
2
2
or 1
7. f(x) lim
h→0
f(xh
h
)f(x)
(x3)(x3)
(x3)(x23x9)
x29
x327
lim
h→0
(xh)22(xh) (x22x)
h
lim
h→0
lim
h→0
2xh
h
h22h
lim
h→0(2xh2)
2x2
8. f(x)
1
2
x27x1
f(x)
1
2
2x217 1x110
x7
9. f(x) 4x34
f(x) 4 3x310
12x2
x22xh h22x2hx22x
h
21.
3
1
(x2x6) dx
1
3
x3
1
2
x26x
3
1
1
3
(3)3
1
2
(3)26(3)
1
3
(1)3
1
2
(1)26(1)
2
2
7
3
6
7
2
3
2
22. (1 2x)dx x
2
2
x2C
xx2C
23. (3x24x7)
3
3
x3
4
2
x27xC
x32x27xC
24. h(t) 3 95t16t2
h(t) v(t) 0 95 1t1116 2t21
95 32t
h(2) v(2) 95 32(2)
31 ft/s
25. V2
r
0
b
r
x2hx
dx
2
h
r
1
3
x3h
1
2
x2
r
0
2
3
h
r
r3
h
2
r2
3
h
r
03
h
2
02
2
h
3
r2
h
2
r2
V(h, r) V(2, 3) 2
2
3
32
2
2
32
6units3
Chapter Tests 564
