Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.1 Rules for Exponents - 3.1 Exercises: 54

Answer

$\dfrac{15}{22h^{2}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \left( \dfrac{3g^5h^{-7}}{4g^2h^3} \right)\left( \dfrac{10g^{-2}h^6}{11gh^{-2}} \right) ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the Quotient Rule of the laws of exponents, which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} \left( \dfrac{3g^5h^{-7}}{4g^2h^3} \right)\left( \dfrac{10g^{-2}h^6}{11gh^{-2}} \right) \\\\ \left( \dfrac{3g^{5-2}h^{-7-3}}{4} \right)\left( \dfrac{10g^{-2-1}h^{6-(-2)}}{11} \right) \\\\ \left( \dfrac{3g^{3}h^{-10}}{4} \right)\left( \dfrac{10g^{-3}h^{8}}{11} \right) .\end{array} Using the Product Rule of the laws of exponents, which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{3g^{3}h^{-10}}{4} \right)\left( \dfrac{10g^{-3}h^{8}}{11} \right) \\\\= \dfrac{3(10)g^{3+(-3)}h^{-10+8}}{4(11)} \\\\= \dfrac{3(\cancel{10}^5)g^{0}h^{-2}}{\cancel{4}^2(11)} \\\\= \dfrac{15g^{0}h^{-2}}{22} .\end{array} Since any expression (except $0$) raised to the $0$ power is $1,$ then the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{15g^{0}h^{-2}}{22} \\\\ \dfrac{15(1)h^{-2}}{22} \\\\ \dfrac{15h^{-2}}{22} .\end{array} Using the Negative Exponent Rule of the laws of exponents, which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{15h^{-2}}{22} \\\\= \dfrac{15}{22h^{2}} .\end{array}
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