INSTRUCTOR’S
SOLUTIONS MANUAL
JUDITH A. PENNA
A LGEBRA & T RIGONOMETRY
@
FIFTH EDITION
Ap
P RECALCULUS : A R IGHT
T RIANGLE A PPROACH
lu
FIFTH EDITION
ss
tu
Judith A. Beecher
vi
Judith A. Penna
a
Marvin L. Bittinger
Indiana University Purdue University Indianapolis
Boston Columbus Hoboken Indianapolis New York San Francisco
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
@Aplusstuvia
, Contents
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . 57
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . 107
@
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 163
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . 249
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . 305
Ap
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . 357
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 399
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . 451
lu
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . 551
ss
Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . 635
Just-in-Time Review . . . . . . . . . . . . . . . . . . . . . 677
Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . 687
tu
vi
a
@Aplusstuvia
,Chapter 1
Graphs, Functions, and Models
4. y
Exercise Set 1.1
4 (1, 4)
1. Point A is located 5 units to the left of the y-axis and 2
(5, 0) (4, 0)
4 units up from the x-axis, so its coordinates are (−5, 4). 4 2 2 4 x
Point B is located 2 units to the right of the y-axis and 2
(4, 2)
2 units down from the x-axis, so its coordinates are (2, −2). 4 (2, 4)
Point C is located 0 units to the right or left of the y-axis
@
and 5 units down from the x-axis, so its coordinates are
(0, −5). 5. To graph (−5, 1) we move from the origin 5 units to the
left of the y-axis. Then we move 1 unit up from the x-axis.
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3, 5). To graph (5, 1) we move from the origin 5 units to the right
of the y-axis. Then we move 1 unit up from the x-axis.
Point E is located 5 units to the left of the y-axis and
Ap
4 units down from the x-axis, so its coordinates are To graph (2, 3) we move from the origin 2 units to the right
(−5, −4). of the y-axis. Then we move 3 units up from the x-axis.
Point F is located 3 units to the right of the y-axis and To graph (2, −1) we move from the origin 2 units to the
0 units up or down from the x-axis, so its coordinates are right of the y-axis. Then we move 1 unit down from the
(3, 0). x-axis.
To graph (0, 1) we do not move to the right or the left of
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2. G: (2, 1); H: (0, 0); I: (4, −3); J: (−4, 0); K: (−2, 3); the y-axis since the first coordinate is 0. From the origin
L: (0, 5) we move 1 unit up.
3. To graph (4, 0) we move from the origin 4 units to the right
y
ss
of the y-axis. Since the second coordinate is 0, we do not
move up or down from the x-axis.
4
To graph (−3, −5) we move from the origin 3 units to the (2, 3)
2
left of the y-axis. Then we move 5 units down from the (5, 1) (0, 1) (5, 1)
x-axis. 4 2 4 x
tu
To graph (−1, 4) we move from the origin 1 unit to the left 2 (2, 1)
of the y-axis. Then we move 4 units up from the x-axis. 4
To graph (0, 2) we do not move to the right or the left of
the y-axis since the first coordinate is 0. From the origin
vi
6. y
we move 2 units up.
To graph (2, −2) we move from the origin 2 units to the 4
right of the y-axis. Then we move 2 units down from the (5, 2)
a
2
x-axis. (5, 0) (4, 0)
4 2 2 4 x
y 2
4 (4, 3)
(1, 4) 4 (1, 5)
2 (0, 2)
(4, 0) 7. The first coordinate represents the year and the second co-
4 2 2 4 x ordinate represents the number of Sprint Cup Series races
2 (2, 2)
in which Tony Stewart finished in the top five. The or-
(3, 5) 4 dered pairs are (2008, 10), (2009, 15), (2010, 9), (2011, 9),
(2012, 12), and (2013, 5).
8. The first coordinate represents the year and the second
coordinate represents the percent of Marines who are
women. The ordered pairs are (1960, 1%), (1970, 0.9%),
(1980, 3.6%), (1990, 4.9%), (2000, 6.1%), and (2011, 6.8%).
Copyright
c 2016 Pearson Education, Inc.
@Aplusstuvia
, 2 Chapter 1: Graphs, Functions, and Models
9. To determine whether (−1, −9) is a solution, substitute 12. For (1.5, 2.6): x2 + y 2 = 9
−1 for x and −9 for y.
(1.5)2 + (2.6)2 ? 9
y = 7x − 2
2.25 + 6.76
−9 ? 7(−1) − 2 9.01 9 FALSE
−7 − 2 (1.5, 2.6) is not a solution.
−9 −9 TRUE For (−3, 0): x2 + y 2 = 9
The equation −9 = −9 is true, so (−1, −9) is a solution. (−3)2 + 02 ? 9
To determine whether (0, 2) is a solution, substitute 0 for
9+0
x and 2 for y.
9 9 TRUE
y = 7x − 2
(−3, 0) is a solution.
2 ? 7 · 0 − 2 1 4
13. To determine whether − , −
0−2 is a solution, substitute
2 5
2 −2 FALSE 1 4
− for a and − for b.
@
The equation 2 = −2 is false, so (0, 2) is not a solution. 2 5
2a + 5b = 3
10. For
1
, 8 : y = −4x + 10 1 4
2 2 − +5 − ? 3
2 5
1
Ap
8 ? −4 · + 10 −1 − 4
2
−5 3 FALSE
−2 + 10
1 4
8 8 TRUE The equation −5 = 3 is false, so − , − is not a solu-
2 5
1 tion.
, 8 is a solution. 3
2
lu
To determine whether 0, is a solution, substitute 0 for
5
For (−1, 6): y = −4x + 10 3
a and for b.
5
6 ? −4(−1) + 10
2a + 5b = 3
ss
4 + 10
3
6 14 FALSE 2·0+5· ? 3
5
(−1, 6) is not a solution.
0+3
2 3
2 3 3 TRUE
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11. To determine whether , is a solution, substitute
3 4 3 3
3 The equation 3 = 3 is true, so 0, is a solution.
for x and for y. 5
4
6x − 4y = 1 3
14. For 0, : 3m + 4n = 6
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2 3 2
6· −4· ? 1 3
3 4 3·0+4· ? 6
2
4−3
0+6
a
1 1 TRUE
2 3 6 6 TRUE
The equation 1 = 1 is true, so , is a solution. 3
3 4 0, is a solution.
3 2
To determine whether 1, is a solution, substitute 1 for 2
2
3 For ,1 : 3m + 4n = 6
x and for y. 3
2 2
6x − 4y = 1 3·
+4·1 ? 6
3
3
6·1−4· ? 1 2+4
2
6 6 TRUE
6−6 2
0 1 FALSE The equation 6 = 6 is true, so
3
, 1 is a solution.
3
The equation 0 = 1 is false, so 1, is not a solution.
2
Copyright
c 2016 Pearson Education, Inc.
@Aplusstuvia
SOLUTIONS MANUAL
JUDITH A. PENNA
A LGEBRA & T RIGONOMETRY
@
FIFTH EDITION
Ap
P RECALCULUS : A R IGHT
T RIANGLE A PPROACH
lu
FIFTH EDITION
ss
tu
Judith A. Beecher
vi
Judith A. Penna
a
Marvin L. Bittinger
Indiana University Purdue University Indianapolis
Boston Columbus Hoboken Indianapolis New York San Francisco
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
@Aplusstuvia
, Contents
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . 57
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . 107
@
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 163
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . 249
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . 305
Ap
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . 357
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 399
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . 451
lu
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . 551
ss
Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . 635
Just-in-Time Review . . . . . . . . . . . . . . . . . . . . . 677
Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . 687
tu
vi
a
@Aplusstuvia
,Chapter 1
Graphs, Functions, and Models
4. y
Exercise Set 1.1
4 (1, 4)
1. Point A is located 5 units to the left of the y-axis and 2
(5, 0) (4, 0)
4 units up from the x-axis, so its coordinates are (−5, 4). 4 2 2 4 x
Point B is located 2 units to the right of the y-axis and 2
(4, 2)
2 units down from the x-axis, so its coordinates are (2, −2). 4 (2, 4)
Point C is located 0 units to the right or left of the y-axis
@
and 5 units down from the x-axis, so its coordinates are
(0, −5). 5. To graph (−5, 1) we move from the origin 5 units to the
left of the y-axis. Then we move 1 unit up from the x-axis.
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3, 5). To graph (5, 1) we move from the origin 5 units to the right
of the y-axis. Then we move 1 unit up from the x-axis.
Point E is located 5 units to the left of the y-axis and
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4 units down from the x-axis, so its coordinates are To graph (2, 3) we move from the origin 2 units to the right
(−5, −4). of the y-axis. Then we move 3 units up from the x-axis.
Point F is located 3 units to the right of the y-axis and To graph (2, −1) we move from the origin 2 units to the
0 units up or down from the x-axis, so its coordinates are right of the y-axis. Then we move 1 unit down from the
(3, 0). x-axis.
To graph (0, 1) we do not move to the right or the left of
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2. G: (2, 1); H: (0, 0); I: (4, −3); J: (−4, 0); K: (−2, 3); the y-axis since the first coordinate is 0. From the origin
L: (0, 5) we move 1 unit up.
3. To graph (4, 0) we move from the origin 4 units to the right
y
ss
of the y-axis. Since the second coordinate is 0, we do not
move up or down from the x-axis.
4
To graph (−3, −5) we move from the origin 3 units to the (2, 3)
2
left of the y-axis. Then we move 5 units down from the (5, 1) (0, 1) (5, 1)
x-axis. 4 2 4 x
tu
To graph (−1, 4) we move from the origin 1 unit to the left 2 (2, 1)
of the y-axis. Then we move 4 units up from the x-axis. 4
To graph (0, 2) we do not move to the right or the left of
the y-axis since the first coordinate is 0. From the origin
vi
6. y
we move 2 units up.
To graph (2, −2) we move from the origin 2 units to the 4
right of the y-axis. Then we move 2 units down from the (5, 2)
a
2
x-axis. (5, 0) (4, 0)
4 2 2 4 x
y 2
4 (4, 3)
(1, 4) 4 (1, 5)
2 (0, 2)
(4, 0) 7. The first coordinate represents the year and the second co-
4 2 2 4 x ordinate represents the number of Sprint Cup Series races
2 (2, 2)
in which Tony Stewart finished in the top five. The or-
(3, 5) 4 dered pairs are (2008, 10), (2009, 15), (2010, 9), (2011, 9),
(2012, 12), and (2013, 5).
8. The first coordinate represents the year and the second
coordinate represents the percent of Marines who are
women. The ordered pairs are (1960, 1%), (1970, 0.9%),
(1980, 3.6%), (1990, 4.9%), (2000, 6.1%), and (2011, 6.8%).
Copyright
c 2016 Pearson Education, Inc.
@Aplusstuvia
, 2 Chapter 1: Graphs, Functions, and Models
9. To determine whether (−1, −9) is a solution, substitute 12. For (1.5, 2.6): x2 + y 2 = 9
−1 for x and −9 for y.
(1.5)2 + (2.6)2 ? 9
y = 7x − 2
2.25 + 6.76
−9 ? 7(−1) − 2 9.01 9 FALSE
−7 − 2 (1.5, 2.6) is not a solution.
−9 −9 TRUE For (−3, 0): x2 + y 2 = 9
The equation −9 = −9 is true, so (−1, −9) is a solution. (−3)2 + 02 ? 9
To determine whether (0, 2) is a solution, substitute 0 for
9+0
x and 2 for y.
9 9 TRUE
y = 7x − 2
(−3, 0) is a solution.
2 ? 7 · 0 − 2 1 4
13. To determine whether − , −
0−2 is a solution, substitute
2 5
2 −2 FALSE 1 4
− for a and − for b.
@
The equation 2 = −2 is false, so (0, 2) is not a solution. 2 5
2a + 5b = 3
10. For
1
, 8 : y = −4x + 10 1 4
2 2 − +5 − ? 3
2 5
1
Ap
8 ? −4 · + 10 −1 − 4
2
−5 3 FALSE
−2 + 10
1 4
8 8 TRUE The equation −5 = 3 is false, so − , − is not a solu-
2 5
1 tion.
, 8 is a solution. 3
2
lu
To determine whether 0, is a solution, substitute 0 for
5
For (−1, 6): y = −4x + 10 3
a and for b.
5
6 ? −4(−1) + 10
2a + 5b = 3
ss
4 + 10
3
6 14 FALSE 2·0+5· ? 3
5
(−1, 6) is not a solution.
0+3
2 3
2 3 3 TRUE
tu
11. To determine whether , is a solution, substitute
3 4 3 3
3 The equation 3 = 3 is true, so 0, is a solution.
for x and for y. 5
4
6x − 4y = 1 3
14. For 0, : 3m + 4n = 6
vi
2 3 2
6· −4· ? 1 3
3 4 3·0+4· ? 6
2
4−3
0+6
a
1 1 TRUE
2 3 6 6 TRUE
The equation 1 = 1 is true, so , is a solution. 3
3 4 0, is a solution.
3 2
To determine whether 1, is a solution, substitute 1 for 2
2
3 For ,1 : 3m + 4n = 6
x and for y. 3
2 2
6x − 4y = 1 3·
+4·1 ? 6
3
3
6·1−4· ? 1 2+4
2
6 6 TRUE
6−6 2
0 1 FALSE The equation 6 = 6 is true, so
3
, 1 is a solution.
3
The equation 0 = 1 is false, so 1, is not a solution.
2
Copyright
c 2016 Pearson Education, Inc.
@Aplusstuvia
