College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises - Page 69: 105

Answer

$-12$ degrees Fahrenheit

Work Step by Step

Given: Wind chill = $35.74+.6215*T -35.75*V^{.16} +.4275*T*V^{.16}$ $T=10$ degrees Fahrenheit $V=30$ mph wind Substitute the given values if $V$ and $T$ into the formula above to obtain $\text{Wind chill temperature} = 35.74+0.6215\cdot 10 -35.75\cdot 30^{0.16} +0.4275\cdot10\cdot30^{0.16}$ $\text{Wind chill temperature} = 35.74 + 6.215 – 35.75\cdot1.72321474 + 4.275\cdot1.72321474$ $\text{Wind chill temperature} = -12.28318394$ $\text{Wind chill temperature} \approx -12$
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