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: 10

Answer

$d=\pm\dfrac{\sqrt{skw}}{kw}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ s=kwd^2 ,$ in terms of $ d ,$ 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}{kw}=d^2 \\\\ d^2=\dfrac{s}{kw} .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} d=\pm\sqrt{\dfrac{s}{kw}} .\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} d=\pm\sqrt{\dfrac{s}{kw}\cdot\dfrac{kw}{kw}} \\\\ d=\pm\sqrt{\dfrac{skw}{(kw)^2}} \\\\ d=\pm\sqrt{\dfrac{1}{(kw)^2}\cdot skw} \\\\ d=\pm\sqrt{\left( \dfrac{1}{kw}\right)^2\cdot skw} \\\\ d=\pm\dfrac{1}{kw}\sqrt{skw} \\\\ d=\pm\dfrac{\sqrt{skw}}{kw} .\end{array}
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