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: 117

Answer

$$10x^{2}y^{4}$$

Work Step by Step

$$\frac{9y^4}{x^{-2}}+ (\frac{x^{-1}}{y^2})^{-2}$$ Simplify the first term: $\frac{9y^4}{x^{-2}}$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$\frac{9y^4}{x^{-2}} = 9x^{2}y^{4}$$ Simplify the second term: $(\frac{x^{-1}}{y^2})^{-2}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(\frac{x^{-1}}{y^2})^{-2} = \frac{x^{-1\cdot-2}}{y^{2\cdot-2}} = \frac{x^{2}}{y^{-4}}$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$\frac{x^{2}}{y^{-4}} = x^{2}y^{4}$$ Rewrite the equation: $$9x^{2}y^{4} + x^{2}y^{4} = 10x^{2}y^{4}$$
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