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 - Page 536: 20

Answer

$d=\dfrac{k^2}{F^2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ F=\dfrac{k}{\sqrt{d}} ,$ in terms of $ h ,$ square both sides. Then use the laws of exponents and the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Squaring both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} (F)^2=\left(\dfrac{k}{\sqrt{d}}\right)^2 \\\\ F^2=\left(\dfrac{k}{\sqrt{d}}\right)^2 .\end{array} Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^m}{z^p} \right)^q=\dfrac{x^{mq}}{z^{pq}},$ the equation above is equivalent to \begin{array}{l}\require{cancel} F^2=\dfrac{(k)^2}{(\sqrt{d})^2} \\\\ F^2=\dfrac{k^2}{d} .\end{array} Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} d(F^2)=\left(\dfrac{k^2}{d}\right)d \\\\ dF^2=k^2 \\\\ d=\dfrac{k^2}{F^2} .\end{array}
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