Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 1 - Section 1.8 - Introduction to Variables, Algebraic Expressions, and Equations - Exercise Set - Page 83: 95

Answer

16 ft after 1 sec. 64 ft after 2 sec. 144 ft after 3 sec. 256 ft after 4 sec.

Work Step by Step

Substitute each given value and resolve. $16t^2=16(1)^2=16\times1=16$ $16t^2=16(2)^2=16\times4=64$ $16t^2=16(3)^2=16\times9=144$ $16t^2=16(4)^2=16\times16=256$
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