Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 446: 89

Answer

$r=\dfrac{\sqrt{A\pi}}{2\pi}$

Work Step by Step

Multiplying the denominator by an expression that will make it a perfect root of the index, then the rationalized-denominator form of the given equation, $ r=\sqrt{\dfrac{A}{4\pi}} ,$ is \begin{array}{l}\require{cancel} r=\sqrt{\dfrac{A}{4\pi}\cdot\dfrac{\pi}{\pi}} \\\\ r=\sqrt{\dfrac{A\pi}{4\pi^2}} \\\\ r=\sqrt{\dfrac{A\pi}{(2\pi)^2}} \\\\ r=\dfrac{\sqrt{A\pi}}{2\pi} .\end{array}
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