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 - Review Exercises: 98

Answer

$-\dfrac{a^8}{2b^5}$

Work Step by Step

RECALL: (i) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$ (ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$ Divide the numerical coefficients then use rule (i) above to obtain: $=\frac{-10}{20} \cdot a^{5-(-3)}b^{6-11} \\=-\frac{1}{2} \cdot a^{5+3}b^{-5} \\=-\frac{1}{2} a^8b^{-5}$ Use rule (ii) above to obtain: $=-\dfrac{1}{2} \cdot a^8 \cdot \dfrac{1}{b^5} \\=-\dfrac{a^8}{2b^5}$
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