Intermediate Algebra (12th Edition)

Published by Pearson

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises - Page 278: 148

Answer

$\dfrac{15}{8y^{5}z^{3}}$

Work Step by Step

Using the laws of exponents, the given expression, $\dfrac{(-5y^3z^4)^2(2yz^5)^{-2}}{10(y^4z)^{3}(3y^3z^2)^{-1}} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{(-5)^2y^{3(2)}z^{4(2)}(2)^{-2}y^{-2}z^{5(-2)}}{10y^{4(3)}z^{3}(3)^{-1}y^{3(-1)}z^{2(-1)}} \\\\= \dfrac{(-5)^2(3)^{1}y^{6}z^{8}y^{-2}z^{-10}}{10(2)^{2}y^{12}z^{3}y^{-3}z^{-2}} \\\\= \dfrac{25(3)y^{6+(-2)}z^{8+(-10)}}{10(4)y^{12+(-3)}z^{3+(-2)}} \\\\= \dfrac{75y^{4}z^{-2}}{40y^{9}z^{1}} \\\\= \dfrac{\cancel{5}(15)y^{4-9}z^{-2-1}}{\cancel{5}(8)} \\\\= \dfrac{15y^{-5}z^{-3}}{8} \\\\= \dfrac{15}{8y^{5}z^{3}} .\end{array}

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