Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - Quiz - Page 123: 10

Answer

$$r=\frac{\pm\sqrt{2A\theta}}{\theta}$$

Work Step by Step

To solve for r, isolate $r^2$ on one side of the equation $A=\frac{1}{2}r^2\theta$ $(2)A=\frac{1}{2}(2)r^2\theta$ $2A=r^2\theta$ $\frac{2A}{\theta}=r^2$ apply the square root property: $x^2=k$, $x=\pm\sqrt{k}$ $r=\pm\sqrt{\frac{2A}{\theta}}$ $r=\pm\sqrt{\frac{2A}{\theta}\times\frac{\theta}{\theta}}$ $r=\pm\sqrt{\frac{2A\theta}{(\theta)^2}}$ $$r=\frac{\pm\sqrt{2A\theta}}{\theta}$$
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