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Solution Manual For College Algebra, 8th Edition by James Stewart Lothar RedlinSaleem Watson Chapter

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PROLOGUE: Principles of Problem Solving distance 1 1 ; the ascent takes h, the descent takes h, and the rate 15 r 1 1 1 1 1 2  h. Thus we have     0, which is impossible. So the car cannot go total trip should take 30 15 15 r 15 r fast enough to average 30 mi/h for the 2­mile trip.

1. Let r be the rate of the descent. We use the formula time 

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 First half Second half Entire season

1 1 hit in 99 at­bats: average  99 1 hit in 1 at­bat: average  11

2 2 hits in 100 at­bats: average  100

Player B 0 hit in 1 at­bat: average  01

98 hits in 99 at­bats: average  98 99

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 1 coffee y cream  and  . is added to the coffee cup, the resulting mixture has the following ratios: mixture y1 mixture y1 1 of a So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing 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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