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 - Exercise Set: 46

Answer

$\frac{8}{5a^{3}b}$

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, $\frac{-8xa^{2}b^{0}}{-5xa^{5}b}=\frac{-8}{-5}\times x^{1-1}\times a^{2-5}\times b^{0-1}=\frac{8}{5}\times x^{0}\times a^{-3}\times b^{-1}=\frac{8}{5}\times1\times\frac{1}{a^{3}}\times\frac{1}{b}=\frac{8}{5a^{3}b}$.
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