College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 180: 106

Answer

$2.9$ feet

Work Step by Step

Plug the given time into the formula. Multiply the denominator on both sides, and then square both sides to get the vertical distance. $t=\frac{\sqrt d}{2}$ $0.85=\frac{\sqrt d}{2}$ $0.85*2=\sqrt d$ $1.7=\sqrt d$ $d\approx2.9$
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