Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.6 - Properties of Integral Exponents - Exercise Set - Page 80: 97

Answer

$$=\frac{3z^3}{x^{7}y^{6}}$$

Work Step by Step

$(-\frac{1}{4}x^{-4}y^{-5}z^{-1})(-12x^{-3}y^{-1}z^{4})$ Recall the product rule: $a^{m}\cdot a^{n} = a^{m+n}$ Thus, $$(-\frac{1}{4}\cdot -12)(x^{-4-3})(y^{-5-1})(z^{-1+4})$$ $$=3(x^{-7})(y^{-6})(z^{3})$$ Recall the negative exponent rule: $a^{-n} = \frac{1}{a^n}$ Hence, $$=\frac{3z^3}{x^{7}y^{6}}$$
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