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: 13

Answer

$v=\pm\dfrac{\sqrt{kAF}}{F}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ F=\dfrac{kA}{v^2} ,$ in terms of $ v ,$ use the properties of equality and the Square Root Principle to isolate the variable. $\bf{\text{Solution Details:}}$ Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} F(v^2)=1(kA) \\\\ Fv^2=kA \\\\ v^2=\dfrac{kA}{F} .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} v=\pm\sqrt{\dfrac{kA}{F}} .\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} v=\pm\sqrt{\dfrac{kA}{F}\cdot\dfrac{F}{F}} \\\\ v=\pm\sqrt{\dfrac{kAF}{F^2}} \\\\ v=\pm\sqrt{\dfrac{1}{F^2}\cdot kAF} \\\\ v=\pm\sqrt{\left( \dfrac{1}{F} \right)^2\cdot kAF} \\\\ v=\pm\dfrac{1}{F}\sqrt{kAF} \\\\ v=\pm\dfrac{\sqrt{kAF}}{F} .\end{array}
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