## Algebra 1

Published by Prentice Hall

# Chapter 7 - Exponents and Exponential Functions - Mid-Chapter Quiz - Page 439: 6

#### Answer

$\frac{64m^6}{9}$

#### Work Step by Step

We start with the given expression: $(3^2)^{-1}(4m^2)^3$ To raise a power to a power, we multiply the exponents: $(3^{-2})(4m^2)^3$ The negative exponent rule states that for every nonzero number $a$ and integer $n$, $a^{-n}=\frac{1}{a^n}$. We use this rule to rewrite the expression: $\frac{(4m^2)^3}{3^2}=\frac{(4m^2)^3}{3\times3}=\frac{(4m^2)^3}{9}$. Now, we will simplify the numerator. To raise a product to a power, we raise each factor to the power and multiply: $\frac{4^3(m^2)^3}{9}$ To raise a power to a power, we multiply the exponents: $\frac{4^3m^6}{9}$ We expand and simplify the constant power: $\frac{4\times4\times4m^6}{9}=\frac{64m^6}{9}$

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