Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises - Page 135: 83

Answer

She runs at 3.78 mi/h

Work Step by Step

$$v_{run} = v$$ $$v_{cycle} = v + 8$$ - Dividing the distance by the speed of each type of exercise, we will get the time (in hours). We know that the sum of both times is equal to 1 hour. $$\frac 4{v_{cycle}} + \frac{2.5}{v_{run}} = 1$$ $$\frac 4{v + 8} + \frac{2.5}{v} = 1$$ $$\frac 4{v + 8} = 1 - \frac{2.5}{v}$$ $$ 4 = (v+8)(1 - \frac{2.5}{v}) = v + 8 - 2.5 - \frac{8 \times 2.5}{v}$$ $$4 = v + 5.5 - \frac{20}{v}$$ $$\frac{20} v = v + 5.5 - 4 = v + 1.5$$ $$20 = v^2 + 1.5v $$ $$0 = v^2 + 1.5v - 20$$ $$v_1 = \frac{-(1.5) + \sqrt{1.5^2 -4\times 1 \times (-20)}}{2(1)} \approx 3.78 $$ $$v_2 = \frac{-(1.5) - \sqrt{1.5^2 -4\times 1 \times (-20)}}{2(1)} \lt 0$$ The speed can't be negative. $v_{run} = v = 3.78 \space mi/h$ Therefore, she runs at 3.78 mi/h
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