Algebra: A Combined Approach (4th Edition)

Published by Pearson

Appendix A - Practice: 5

Answer

Practice 5 (Solution) $(\,\frac{4x^3}{3y^{-1}})\,^3(\,\frac{y^{-2}}{3x^{-1}})\,^{-1}$ =$\frac{(\,4x^3)\,^3}{(\,3y^{-1})\,^3}\cdot\frac{(\,y^{-2})\,^{-1}}{(\,3x^{-1})\,^{-1}}$ (Power of a quotient) =$\frac{4^3x^9y^2}{3^3y^{-3}\cdot3^{-1}x^1}$ (Power rule) =$\frac{64x^{9-1}y^{2-(-3)}}{3^{3+(-1)}}$ (Quotient rule, Product rule) =$\frac{64x^8y^5}{9}$

Work Step by Step

Practice 5 (Solution) $(\,\frac{4x^3}{3y^{-1}})\,^3(\,\frac{y^{-2}}{3x^{-1}})\,^{-1}$ =$\frac{(\,4x^3)\,^3}{(\,3y^{-1})\,^3}\cdot\frac{(\,y^{-2})\,^{-1}}{(\,3x^{-1})\,^{-1}}$ (Power of a quotient) =$\frac{4^3x^9y^2}{3^3y^{-3}\cdot3^{-1}x^1}$ (Power rule) =$\frac{64x^{9-1}y^{2-(-3)}}{3^{3+(-1)}}$ (Quotient rule, Product rule) =$\frac{64x^8y^5}{9}$

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