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: 88

Answer

$\frac{10}{x^{8}z^{15}}$

Work Step by Step

We are given the expression $\frac{30x^{-7}yz^{-14}}{3xyz}$. 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). $(\frac{30}{3})\times(x^{-7-1})\times(y^{1-1})\times(z^{-14-1})=10\times x^{-8}\times y^{0}\times z^{-15}$ Note that $y^{0}=1$. $10\times x^{-8}\times z^{-15}$ 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). $10\times x^{-8}\times z^{-15}=\frac{10}{x^{8}z^{15}}$
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