Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.1 - Radicals and Radical Functions - Exercise Set: 4

Answer

$\frac{3}{5}$

Work Step by Step

$\sqrt \frac{9}{25}=\frac{3}{5}$, because $(\frac{3}{5})^{2}=\frac{3}{5}\times\frac{3}{5}=\frac{9}{25}$
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