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Solutions Manual for College Algebra, 8th Edition by James Stewart

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Complete Solutions Manual for College Algebra, 8th Edition 8e by James Stewart, Lothar Redlin, Saleem Watson. Full Chapters Solutions are included - Chapter 1 to 9 P. PREREQUISITES. Chapter Overview P.1 Modeling the Real-World with Algebra. P.2 The Real Numbers. P.3 Integer Exponents and Scientific Notation. P.4 Rational Exponents and Radicals. P.5 Algebraic Expressions. P.6 Factoring. P.7 Rational Expressions. P.8 Solving Basic Equations. P.9 Modeling with Equations. Chapter P Review. Chapter P Test. FOCUS ON MODELING: Making the Best Decisions. 1. EQUATIONS AND GRAPHS. 1.1 The Coordinate Plane. 1.2 Graphs of Equations in Two Variables; Circles. 1.3 Lines. 1.4 Solving Quadratic Equations. 1.5 Complex Numbers. 1.6 Solving Other Types of Equations. 1.7 Solving Inequalities. 1.8 Solving Absolute Value Equations and Inequalities. 1.9 Solving Equations and Inequalities Graphically. 1.10 Modeling Variations. Chapter 1 Review. Chapter 1 Test. FOCUS ON MODELING: Fitting Lines to Data. 2. FUNCTIONS. Chapter Overview. 2.1 Functions. 2.2 Graphs of Functions. 2.3 Getting Information from the Graph of a Function. 2.4 Average Rate of Change of a Function. 2.5 Linear Functions and Models. 2.6 Transformations of Functions. 2.7 Combining Functions. 2.8 One-to-One Functions and Their Inverses. Chapter 2 Review. Chapter 2 Test. FOCUS ON MODELING: Modeling with Functions. 3. POLYNOMIAL AND RATIONAL FUNCTIONS. Chapter Overview. 3.1 Quadratic Functions and Models. 3.2 Polynomial Functions and Their Graphs. 3.3 Dividing Polynomials. 3.4 Real Zeros of Polynomials. 3.5 Complex Zeros and the Fundamental Theorem of Algebra. 3.6 Rational Functions. 3.7 Polynomial and Rational Inequalities. Chapter 3 Review. Chapter 3 Test. FOCUS ON MODELING: Fitting Polynomial Curves to Data. 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS. Chapter Overview. 4.1 Exponential Functions. 4.2 The Natural Exponential Function. 4.3 Logarithmic Functions. 4.4 Laws of Logarithms. 4.5 Exponential and Logarithmic Equations. 4.6 Modeling with Exponential Functions. 4.7 Logarithmic Scales. Chapter 4 Review. Chapter 4 Test. FOCUS ON MODELING: Fitting Exponential and Power Curves to Data. 5. SYSTEMS OF EQUATIONS AND INEQUALITIES. Chapter Overview. 5.1 Systems of Linear Equations in Two Variables. 5.2 Systems of Linear Equations in Several Variables. 5.3 Partial Fractions. 5.4 Systems of Nonlinear Equations. 5.5 Systems of Inequalities. Chapter 5 Review. Chapter 5 Test. FOCUS ON MODELING: Linear Programming. 6. MATRICES AND DETERMINANTS. Chapter Overview. 6.1 Matrices and Systems of Linear Equations. 6.2 The Algebra of Matrices. 6.3 Inverses of Matrices and Matrix Equations. 6.4 Determinants and Cramer's Rule. Chapter 6 Review. Chapter 6 Test. FOCUS ON MODELING: Computer Graphics. 7. CONIC SECTIONS. Chapter Overview. 7.1 Parabolas. 7.2 Ellipses. 7.3 Hyperbolas. 7.4 Shifted Conics. Chapter 7 Review. Chapter 7 Test. FOCUS ON MODELING: Conics in Architecture. 8. SEQUENCES AND SERIES. Chapter Overview. 8.1 Sequences and Summation Notation. 8.2 Arithmetic Sequences. 8.3 Geometric Sequences. 8.4 Mathematical Induction. 8.5 The Binomial Theorem. Chapter 8 Review. Chapter 8 Test. FOCUS ON MODELING: Modeling with Recursive Sequences. 9. COUNTING AND PROBABILITY. Chapter Overview. 9.1 Counting. 9.2 Probability. 9.3 Binomial Probability. 9.4 Expected Value. Chapter 9 Review. Chapter 9 Test.

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Solutions College Algebra, 8th Edition by James Stewart

PROLOGUE: Principles of Problem Solving

distance 1 1
1. Let r be the rate of the descent. We use the formula time  ; the ascent takes h, the descent takes h, and the
rate 15 r
2 1 1 1 1 1
total trip should take  h. Thus we have     0, which is impossible. So the car cannot go
30 15 15 r 15 r
fast enough to average 30 mi/h for the 2­mile trip.

2. Let us start with a given price P. After a discount of 40%, the price decreases to 06P. After a discount of 20%, the price
decreases to 08P, and after another 20% discount, it becomes 08 08P  064P. Since 06P  064P, a 40% discount
is better.

3. We continue the pattern. Three parallel cuts produce 10 pieces. Thus, each new cut produces an additional 3 pieces. Since
the first cut produces 4 pieces, we get the formula f n  4  3 n  1, n  1. Since f 142  4  3 141  427, we
see that 142 parallel cuts produce 427 pieces.

4. By placing two amoebas into the vessel, we skip the first simple division which took 3 minutes. Thus when we place two
amoebas into the vessel, it will take 60  3  57 minutes for the vessel to be full of amoebas.

5. The statement is false. Here is one particular counterexample:
Player A Player B
First half 1
1 hit in 99 at­bats: average  99 0 hit in 1 at­bat: average  01
Second half 1 hit in 1 at­bat: average  11 98 hits in 99 at­bats: average  98
99
Entire season 2
2 hits in 100 at­bats: average  100 99
99 hits in 100 at­bats: average  100

6. Method 1: After the exchanges, the volume of liquid in the pitcher and in the cup is the same as it was to begin with. Thus,
any coffee in the pitcher of cream must be replacing an equal amount of cream that has ended up in the coffee cup.
Method 2: Alternatively, look at the drawing of the spoonful of coffee and cream cream
mixture being returned to the pitcher of cream. Suppose it is possible to separate
the cream and the coffee, as shown. Then you can see that the coffee going into the coffee

cream occupies the same volume as the cream that was left in the coffee.


Method 3 (an algebraic approach): Suppose the cup of coffee has y spoonfuls of coffee. When one spoonful of cream
cream 1 coffee y
is added to the coffee cup, the resulting mixture has the following ratios:  and  .
mixture y1 mixture y1
1
So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing of a
y1
y
spoonful of cream and spoonful of coffee. Thus the amount of cream left in the mixture (cream in the coffee) is
y 1
1 y
1  of a spoonful. This is the same as the amount of coffee we added to the cream.
y 1 y1

7. Let r be the radius of the earth in feet. Then the circumference (length of the ribbon) is 2r. When we increase the radius
by 1 foot, the new radius is r  1, so the new circumference is 2 r  1. Thus you need 2 r  1  2r  2 extra
feet of ribbon.
1

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