Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1-8 - Cumulative Review - Page 574: 2

Answer

$$-\frac{2a^{11}}{5b^{33}}$$

Work Step by Step

We know the following rules of exponents. The list of names is on page 330. $$ (1) \ a^m\cdot a^n = a^{m+n} \\ (2) \ (ab)^m =a^mb^m \\ (3) \ (a^m)^n =a^{mn} \\ (4) \ a^{-m} = \frac{1}{a^m} \\ (5)\ \frac{a^m}{a^n} =a^{m-n} \\ (6) \ a^0=1 \\ (7) \ (\frac{a}{b})^m =\frac{a^m}{b^m}$$ Thus, we find: $$-\frac{10a^7b^{-11}}{25a^{-4}b^{22}}\\ -\frac{2a^{11}}{5b^{22-\left(-11\right)}}\\ -\frac{2a^{11}}{5b^{33}}$$
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