Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.2 Applications and Modeling with Linear Equations - 1.2 Exercises - Page 102: 28

Answer

$6.2$ mph.

Work Step by Step

Distance traveled = (average speed)(time of travel). Let $x$ mph be the wind speed. Against the wind, the average speed is $(180-x)$ mph, and time =3 h, so distance = $3(180-x)$ miles. With the wind, the average speed is $(180+x)$ mph, and time =$2.8$ h, so distance = $2.8(180+x)$ miles. The distances discussed are equal. $3(180-x)=2.8(180+x)$ $ 540-3x=504+2.8x\qquad$ ... add $3x-504$ $36=5.8x$ $x=\displaystyle \frac{36}{5.8}\approx 6.2$ The wind speed is $6.2$ mph.
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