Intermediate Algebra (6th Edition)

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

Chapter 7 - Review: 13



Work Step by Step

$\sqrt (\frac{x^{12}}{36y^{2}})=\sqrt (x^{12})\times\sqrt (\frac{1}{36y^{2}})=x^{6}\times\frac{1}{6y}=\frac{x^{6}}{6y}$ We know that $\sqrt (x^{12})=x^{6}$, because $(x^{6})^{2}=x^{6\times2}=x^{12}$. We know that $\sqrt (\frac{1}{36y^{2}})=\frac{1}{6y}$, because $(\frac{1}{6y})^{2}=\frac{1}{6y}\times\frac{1}{6y}=\frac{1}{36y^{2}}$.
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