Solutions for College Algebra: Graphs and Models, 6th edition by Marvin Bittinger, Judith Beecher, David Ellenbogen, Judith Penna, All Chapters

SOLUTIONS
COLLEGE ALGEBRA: GRAPHS
AND MODELS
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6th Edition




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, INSTRUCTOR’S
SOLUTIONS MANUAL
DANIEL S. MILLER
Niagara County Community College




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COLLEGE ALGEBRA

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EIGHTH EDITION
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Robert Blitzer
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Miami Dade College
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, Instructor Solutions
College Algebra 8th edition


Table of Contents

Chapter P Fundamental Concepts of Algebra........................................................................1
Chapter 1 Equations and Inequalities ..................................................................................75
Chapter 2 Functions and Graphs .......................................................................................215
Chapter 3 Polynomial and Rational Functions ..................................................................359




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Chapter 4 Exponential and Logarithmic Functions ...........................................................515




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Chapter 5 Systems of Equations and Inequalities .............................................................593
Chapter 6 Matrices and Determinants ...............................................................................729




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Chapter 7 Conic Sections ..................................................................................................845
Chapter 8 Sequences, Induction, and Probability ..............................................................905
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, Chapter P
Fundamental Concepts of Algebra

Section P.1 6. a. 1− 2
Check Point Exercises Because 2 ≈ 1.4, the number inside the
absolute value bars is negative. The absolute
1. 8 + 6( x − 3)2 = 8 + 6(13 − 3)2 value of x when x < 0 is –x. Thus,
= 8 + 6(10) 2 (
1 − 2 = − 1 − 2 = 2 −1 )
= 8 + 6(100)
= 8 + 600 b. π −3
= 608 Because π ≈ 3.14, the number inside the
absolute value bars is positive. The absolute
2. a. Since 2016 is 16 years after 2000, substitute 16




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value of a positive number is the number itself.
for x. Thus,
T = − x 2 + 361x + 3193




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π − 3 = π − 3.
= −(16) 2 + 361(16) + 3193
= 8713 x
c.
The average cost of tuition and fees at public x




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U.S. colleges for the school year ending in Because x > 0, x = x.
2016 was $8713.
x x
Thus, = =1
b. The formula underestimates the actual answer
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by $65.
7. −4 − (5) = −9 = 9
3. The elements common to {3, 4, 5, 6, 7} and
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{3, 7, 8, 9} are 3 and 7. The distance between –4 and 5 is 9.
{3, 4,5, 6, 7} ∩ {3, 7,8,9} = {3, 7}
8. 7(4 x 2 + 3x) + 2(5 x 2 + x)
4. The union is the set containing all the elements of = 7(4 x 2 + 3 x) + 2(5 x 2 + x)
either set.
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{3, 4,5, 6, 7} ∪ {3, 7,8,9} = {3, 4,5, 6, 7,8,9} = 28 x 2 + 21x + 10 x 2 + 2 x
= 38 x 2 + 23 x
 π 
5.  −9, − 1.3, 0, 0.3, , 9, 10 
9. 6 + 4[7 − ( x − 2)]
 2 
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= 6 + 4[7 − x + 2)]
a. Natural numbers: 9 because 9 =3 = 6 + 4[9 − x]
= 6 + 36 − 4 x
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b. Whole numbers: 0, 9 = 42 − 4 x

c. Integers: −9, 0, 9
Concept and Vocabulary Check P.1
d. Rational numbers: −9, − 1.3, 0, 0.3, 9
C1. expression
π C2. b to the nth power; base; exponent
e. Irrational numbers: , 10
2
C3. formula; modeling; models
π
f. Real numbers: −9, − 1.3, 0, 0.3, , 9, 10 C4. intersection; A ∩ B
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C5. union; A ∪ B



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