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: 71

Answer

$\frac{x^{10}}{2y}$

Work Step by Step

$\sqrt (\frac{x^{20}}{4y^{2}})=\frac{x^{10}}{2y}$, because $(\frac{x^{10}}{2y})^{2}=\frac{x^{10}}{2y}\times\frac{x^{10}}{2y}=\frac{x^{10+10}}{(2\times2)\times y^{1+1}}=\frac{x^{20}}{4y^{2}}$
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