Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.4 - Formulas and Further Applications - 8.4 Exercises: 15


$r=\pm\dfrac{\sqrt{3V\pi h}}{\pi h}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ V=\dfrac{1}{3}\pi r^2h ,$ in terms of $ r ,$ use the properties of equality and the Square Root Principle to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} 3(V)=\left(\dfrac{1}{3}\pi r^2h\right)3 \\\\ 3V=\pi r^2h \\\\ \dfrac{3V}{\pi h}=r^2 \\\\ r^2=\dfrac{3V}{\pi h} .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} r=\pm\sqrt{\dfrac{3V}{\pi h}} .\end{array} Rationalizing the denominator by multiplying the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} r=\pm\sqrt{\dfrac{3V}{\pi h}\cdot\dfrac{\pi h}{\pi h}} \\\\ r=\pm\sqrt{\dfrac{3V\pi h}{\left(\pi h\right)^2}} \\\\ r=\pm\sqrt{\dfrac{1}{\left(\pi h\right)^2}\cdot3V\pi h} \\\\ r=\pm\sqrt{\left(\dfrac{1}{\pi h}\right)^2\cdot3V\pi h} \\\\ r=\pm\dfrac{1}{\pi h}\sqrt{3V\pi h} \\\\ r=\pm\dfrac{\sqrt{3V\pi h}}{\pi h} .\end{array}
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