Precalculus (6th Edition)

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

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 72

Answer

$r = + \sqrt {\frac{A}{\pi}}, -\frac{A}{\pi}$

Work Step by Step

$A = \pi r^2$ $\frac{A}{\pi} = r^2$ $r = + \sqrt {\frac{A}{\pi}}, -\frac{A}{\pi}$
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