Algebra: A Combined Approach (4th Edition)

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

Chapter 5 - Section 5.1 - Exponents - Exercise Set - Page 344: 62

Answer

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

Work Step by Step

Based on the quotient rule for exponents, we know that $\frac{a^{m}}{{a}^{n}}=a^{m-n}, a\ne0$ (where $m$ and $n$ are positive integers and $a$ is a real number). Therefore, $\frac{9a^{4}b^{7}}{27ab^{2}}=\frac{9}{27}\times a^{4-1}\times b^{7-2}=\frac{1}{3}\times a^{3}\times b^{5}=\frac{a^{3}b^{5}}{3}$.
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