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: 9

Answer

$r=\pm\dfrac{\sqrt{S\pi}}{2\pi}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ S=4\pi r^2 ,$ 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{S}{4\pi}=r^2 \\\\ r^2=\dfrac{S}{4\pi} .\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{S}{4\pi}} .\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{S}{4\pi}\cdot\dfrac{\pi}{\pi}} \\\\ r=\pm\sqrt{\dfrac{S\pi}{4\pi^2}} \\\\ r=\pm\sqrt{\dfrac{1}{4\pi^2}\cdot S\pi} \\\\ r=\pm\sqrt{\left(\dfrac{1}{2\pi}\right)^2\cdot S\pi} \\\\ r=\pm\dfrac{1}{2\pi}\sqrt{S\pi} \\\\ r=\pm\dfrac{\sqrt{S\pi}}{2\pi} .\end{array}
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