Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.1 - Exponents and Scientific Notation - Exercise Set - Page 262: 34



Work Step by Step

We are given the expression $\frac{49a^{3}bc^{14}}{-7abc^{8}}$. We can simplify this expression by using the quotient rule, which holds that $\frac{a^{m}}{a^{n}}=a^{m-n}$ (where a is a nonzero real number, and m and n are integers). $\frac{49a^{3}bc^{14}}{-7abc^{8}}=(\frac{49}{-7})\times(a^{3-1})\times(b^{1-1})\times(c^{14-8})=-7a^{2}b^{0}c^{6}=-7a^{2}c^{6}$
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