Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.7 - Quadratic Equations and Problem Solving - Exercise Set - Page 466: 14

Answer

$t=5$

Work Step by Step

We set $h$ equal to $0$: $0=-16t^{2}+400$ $16t^{2}=400$ $t^{2}=25$ $t=5$ or $t=-5$ We know that $t$ can not be negative, so the only possible value of $t$ is $5$ $t=5$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.