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: 95

Answer

$\dfrac{-6}{a^2}$

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: $=[2 \cdot (-3)] \cdot a^{5+(-7)} \\=-6a^{-2}$ Use rule (ii) above to obtain: $=-6 \cdot \dfrac{1}{a^2} \\=\dfrac{-6}{a^2}$
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