Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.2 - More Work with Exponents and Scientific Notation - Exercise Set - Page 268: 47

Answer

$\dfrac{16a^{2}b^{9}}{9}$

Work Step by Step

Using laws of exponents, the given expression simplifies to \begin{align*} & \dfrac{3^{-2}a^{-5}b^6}{4^{-2}a^{-7}b^{-3}} \\\\&= \dfrac{4^{2}a^{-5-(-7)}b^{6-(-3)}}{3^{2}} \\\\&= \dfrac{16a^{2}b^{9}}{9} .\end{align*}
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