Instructor’s Solutions Manual – Algebra and Trigonometry with Corequisite Support, 5th Edition – Judith A. Beecher – ISBN 9780321981554 (Chapters 1–11 Covered)

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Algebra and Trigonometry
with Corequisite Support,
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5th Edition
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INSTRUCTOR’S
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SOLUTIONS
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MANUAL
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Judith A. Beecher
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Comprehensive Instructor’s Solutions Manual
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for Instructors and Students
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9780321981554

© Judith A. Beecher. All rights reserved. Reproduction or distribution
without permission is prohibited.



© MEDCONNOISSEUR

, TABLE OF CONTENTS

Instructor’s Solutions Manual – Algebra and Trigonometry with Corequisite Support (5th Ed.)
Author: Judith A. Beecher
ISBN: 9780321981554
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PART I: FUNCTIONS AND GRAPHS

Chapter 1: Graphs, Functions, and Models
Chapter 2: More on Functions
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Chapter 3: Quadratic Functions and Equations; Inequalities
Chapter 4: Polynomial Functions and Rational Functions
Chapter 5: Exponential Functions and Logarithmic Functions

PART II: TRIGONOMETRY
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Chapter 6: The Trigonometric Functions
Chapter 7: Trigonometric Identities, Inverse Functions, and Equations
Chapter 8: Applications of Trigonometry
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PART III: SYSTEMS, CONICS, AND DISCRETE MATHEMATICS

Chapter 9: Systems of Equations and Matrices
Chapter 10: Conic Sections
Chapter 11: Sequences, Series, and Combinatorics
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, Chapter 1
Graphs, Functions, and Models
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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).
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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
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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
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
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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.
3. To graph (4, 0) we move from the origin 4 units to the right
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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)
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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)
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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)
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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.

, 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.


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−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
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(−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
1  1  4
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10. For , 8 : y = −4x + 10
2 2 − +5 − ? 3
2 5


1

8 ? −4 · + 10 −1 − 4



2


−5
3 FALSE

−2 + 10

 1 4
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8
8 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
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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 4 3  3
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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

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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.
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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.