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: 91

Answer

$\dfrac{10a^2}{b^5}$

Work Step by Step

RECALL: (i) $\dfrac{a^m}{a^n} = a^{m-n}, a \ne 0$ (ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$ Divide the coefficients by each other and the variables by each other to obtain: $=\dfrac{20}{2} \cdot a^{3-1} \cdot b^{8-13} =10 \cdot a^{2} \cdot b^{-5}$ Use rule (ii) above to obtain: $=10a^2 \cdot \dfrac{1}{b^5} \\=\dfrac{10a^2}{b^5}$
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