Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - Review Exercises: Chapter 11 - Page 773: 36

Answer

${p=\frac{9{{\pi }^{2}}}{{{N}^{2}}}}$

Work Step by Step

$N=3\pi \sqrt{\frac{1}{p}}$ This is a radical equation. Use the principle of powers and square both sides of the equation $\begin{align} & {{N}^{2}}={{\left( 3\pi \sqrt{\frac{1}{p}} \right)}^{2}} \\ & ={{3}^{2}}\cdot {{\pi }^{2}}\cdot {{\left( \sqrt{\frac{1}{p}} \right)}^{2}} \\ & ={{3}^{2}}\cdot {{\pi }^{2}}\cdot \frac{1}{p} \end{align}$ Multiply both sides by $p$ to clear fractions $\begin{align} & {{N}^{2}}\cdot p={{3}^{2}}\cdot {{\pi }^{2}}\cdot \frac{1}{p}\cdot p \\ & {{N}^{2}}\cdot p={{3}^{2}}\cdot {{\pi }^{2}} \\ \end{align}$ Divide both sides by ${{N}^{2}}$ $\begin{align} & \frac{{{N}^{2}}p}{{{N}^{2}}}=\frac{9{{\pi }^{2}}}{{{N}^{2}}} \\ & p=\frac{9{{\pi }^{2}}}{{{N}^{2}}} \end{align}$
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