Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set: 73

Answer

6 inches

Work Step by Step

Substitute the given area and solve for r. $A=\pi r^2$ Substitute the area for A. $36\pi=\pi r^2$ Divide both sides by $\pi$. $36\pi \div \pi=\pi r^2\div \pi$ Simplify. $36=r^2$ Take the square root of each side. $\sqrt{36}=\sqrt{r^2}$ Simplify. $6=r$
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