Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapters 1-9 Cumulative Review Problem Set: 18

Answer

$-\frac{3a^{3}}{b}$

Work Step by Step

Since all terms are multiplied by each other, we can collect constants and variables separately in their own parentheses. We then use the rules $a^{m}\times a^{n}=a^{m+n}$ and $\frac{1}{x^2}= x^{-2}$ to simplify the expression: $\frac{-12a^{-2}b^{3}}{4a^{-5}b^{4}}$ =$(\frac{-12}{4})\times(\frac{a^{-2}}{a^{-5}})\times(\frac{b^{3}}{b^{4}})$ =$(-3)\times(a^{-2-(-5)})\times(b^{3-4})$ =$(-3)\times(a^{-2+5})\times(b^{-1})$ =$(-3)\times(a^{3})\times(b^{-1})$ =$(-3)\times(a^{3})\times(\frac{1}{b})$ =$-\frac{3a^{3}}{b}$
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