College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 3 - Section 3.6 - Variation - 3.6 Exercises - Page 366: 18

Answer

$p=\dfrac{9}{8}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use $ p=k\cdot\dfrac{z^2}{r} $ and solve for the value of $k$ with the given $ p,z, $ and $ r $ values. Then use the equation of variation to solve for the value of the unknown variable. $\bf{\text{Solution Details:}}$ Since $p$ varies directly as the square of $z$ and inversely as $r,$ then $ p=k\cdot\dfrac{z^2}{r} .$ Substituting the given values, $ p=\dfrac{32}{5},z=4, $ and $ r=10 ,$ then the value of $k$ is \begin{array}{l}\require{cancel} p=k\cdot\dfrac{z^2}{r} \\\\ \dfrac{32}{5}=k\cdot\dfrac{4^2}{10} \\\\ \dfrac{32}{5}=k\cdot\dfrac{16}{10} \\\\ \dfrac{10}{16}\left(\dfrac{32}{5}\right)=\left(k\cdot\dfrac{16}{10}\right)\dfrac{10}{16} \\\\ \dfrac{10}{\cancel{16}}\left(\dfrac{\cancel{32}^2}{5}\right)=k \\\\ \dfrac{10}{\cancel{16}}\left(\dfrac{\cancel{32}^2}{5}\right)=k \\\\ \dfrac{\cancel{10}^2}{1}\left(\dfrac{2}{\cancel5}\right)=k \\\\ 2(2)=k \\\\ k=4 .\end{array} Hence, the equation of variation is given by \begin{array}{l}\require{cancel} p=4\cdot\dfrac{z^2}{r} .\end{array} If $z=3,$ and $r=32,$ then \begin{array}{l}\require{cancel} p=4\cdot\dfrac{3^2}{32} \\\\ p=4\cdot\dfrac{9}{32} \\\\ p=\cancel4\cdot\dfrac{9}{\cancel{32}^8} \\\\ p=\dfrac{9}{8} .\end{array}
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