Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 5 - Section 5.2 - Negative Exponents and Scientific Notation - Practice: 10

Answer

$\frac{y^{2}}{81x^{6}}$

Work Step by Step

We know that $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}}=a^{n}$ (as long as a is a nonzero real number and n is an integer). Therefore, $(\frac{9x^{3}}{y})^{-2}=\frac{9^{-2}(x^{3})^{-2}}{y^{-2}}=\frac{9^{-2}x^{3\times-2}}{y^{-2}}=\frac{9^{-2}x^{-6}}{y^{-2}}=\frac{1}{9^{2}}\times\frac{1}{x^{6}}\times y^{2}=\frac{y^{2}}{81x^{6}}$.
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