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: 16


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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ V=\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} \dfrac{V}{\pi h}=r^2 \\\\ r^2=\dfrac{V}{\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{V}{\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{V}{\pi h}\cdot\dfrac{\pi h}{\pi h}} \\\\ r=\pm\sqrt{\dfrac{V\pi h}{(\pi h)^2}} \\\\ r=\pm\sqrt{\dfrac{1}{(\pi h)^2}\cdot V\pi h} \\\\ r=\pm\sqrt{\left( \dfrac{1}{\pi h}\right)^2\cdot V\pi h} \\\\ r=\pm\dfrac{1}{\pi h}\sqrt{V\pi h} \\\\ r=\pm\dfrac{\sqrt{V\pi h}}{\pi h} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.