Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.6 - Properties of Integral Exponents - Exercise Set: 96

Answer

$\dfrac{-8}{a^3}$

Work Step by Step

RECALL: (i) $a^m \cdot a^n = a^{m+n}$ (ii) $a^{-m} = \dfrac{1}{a^m}, a \ne0$ Multiply the coefficients by each other and the variables by each other to obtain: $=[4 \cdot (-2)] \cdot a^{2+(-5)} \\=-8a^{-3}$ Use rule (ii) above to obtain: $=-8 \cdot \dfrac{1}{a^3} \\=\dfrac{-8}{a^3}$
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