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Algebra and Trigonometry with Corequisite Support 5th Edition - Solutions Manual by Judith A. Beecher, Judith A. Penna, & Marvin L. Bittinger

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This comprehensive solutions manual supports Algebra and Trigonometry with Corequisite Support, 5th Edition by Judith A. Beecher. It offers fully worked-out solutions for all odd-numbered exercises and selected review problems, aligned closely with each chapter. Covering foundational to advanced topics—including equations, inequalities, functions and their graphs, polynomial, rational, exponential, and logarithmic functions, trigonometry, systems of equations, matrices, conic sections, sequences, and series—each solution provides step-by-step guidance and detailed explanations. Designed to reinforce conceptual understanding and procedural fluency, this manual is an ideal companion for students seeking homework assistance, in-class practice, or exam preparation.

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INSTRUCTOR’S
SOLUTIONS MANUAL
ST

JUDITH A. PENNA
U

A LGEBRA & T RIGONOMETRY
VI

FIFTH EDITION
A

P RECALCULUS : A R IGHT
_A

T RIANGLE A PPROACH
FIFTH EDITION
PP
RO
Judith A. Beecher
Judith A. Penna
VE
Marvin L. Bittinger
Indiana University Purdue University Indianapolis
D
?

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

,ST
Contents
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
U
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . 57

Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . 107
VI
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 163

Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . 249

Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . 305
A
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . 357
_A
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 399

Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . 451

Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . 551
PP
Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . 635

Just-in-Time Review . . . . . . . . . . . . . . . . . . . . . 677

Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . 687
RO
VE
D
?

, Chapter 1
Graphs, Functions, and Models
ST
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)
U
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
VI
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.
A
Point E is located 5 units to the left of the y-axis and
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
_A
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
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.
PP
3. To graph (4, 0) we move from the origin 4 units to the right
y
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
2 (2, 1)
RO
To graph (−1, 4) we move from the origin 1 unit to the left
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 y
6.
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)
VE
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-
D
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.

, 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
ST
 (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
U
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
VI
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 
A
1 
8 ? −4 · + 10 −1 − 4 
 2 
 −5  3 FALSE
 −2 + 10
  1 4
8  8
_A
TRUE The equation −5 = 3 is false, so − , − is not a solu-
  2 5
1 tion.
, 8 is a solution.  3
2 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
PP
 2a + 5b = 3
 4 + 10
 3
6  14 FALSE 2·0+5· ? 3
5 
(−1, 6) is not a solution. 
0+3 
2 3 
11. To determine whether , is a solution, substitute
2 3  3 TRUE
 3
RO
3 4 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
2 3 2
6· −4· ? 1 3
3 4  3·0+4· ? 6
 2 
4−3  
 0+6 
VE
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
D
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.

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