Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set - Page 633: 78


6 seconds

Work Step by Step

Substitute the given distance and solve for t. $d=16t^2$ Substitute the distance for d. $576=16t^2$ Divide both sides by 16. $576 \div 16=16t^2\div16$ Simplify. $36=t^2$ Take the square root of each side. $\sqrt{36}=\sqrt{t^2}$ Simplify. $6=t$
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