Intermediate Algebra (6th Edition)

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

Chapter 7 - Review: 69

Answer

$\frac{p^{8}\sqrt p}{11}$

Work Step by Step

$\sqrt(\frac{p^{17}}{121})=\sqrt(p^{16}\times p)\times\sqrt(\frac{1}{121})=\sqrt(p^{16})\times \sqrt p\times\sqrt(\frac{1}{121})=\frac{p^{8}\sqrt p}{11}$ We know that $\sqrt (p^{16})=p^{8}$, because $(p^{8})^{2}=p^{8\times2}=p^{16}$ We know that $\sqrt(\frac{1}{121})=\frac{1}{11}$, because $(\frac{1}{11})^{2}=\frac{1}{121}$
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