Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 3 - Polynomial and Rational Functions - 3.6 Variation - 3.6 Exercises - Page 387: 14

Answer

$p=9/8=1.125$

Work Step by Step

"p varies directly as z squared and inversely as r" can be written as: $p=\frac{kz^2}{r}$ We can calculate the constant $k$ based on the given set of value of p, z and r. $\frac{32}{5}=\frac{k*4^2}{10}$ $\frac{32}{5}=\frac{8k}{5}$ $8k=32$ $k=4$ Then, we use this $k=4$ value to find $p$ in the second set. $p=\frac{4*3^2}{32}=36/32=9/8$
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