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: 87

Answer

$\frac{7}{x^{3}z^{5}}$

Work Step by Step

We are given the expression $\frac{14x^{-2}yz^{-4}}{2xyz}$. To simplify, we can separate like terms and then use 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). $(\frac{14}{2})\times(x^{-2-1})\times(y^{1-1})\times(z^{-4-1})=7\times x^{-3}\times y^{0}\times z^{-5}$ Note that $y^{0}=1$. $7\times x^{-3}\times z^{-5}$ To simplify this into only positive exponents, we know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer). $7\times x^{-3}\times z^{-5}=\frac{7}{x^{3}z^{5}}$
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