To simplify the expression \( \frac{15^0 \cdot 7^{-2}}{(-4)^{-3}} \) using the properties of integer exponents, we proceed as follows:
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Simplify \( 15^0 \): \[ 15^0 = 1 \] So the expression simplifies to: \[ \frac{1 \cdot 7^{-2}}{(-4)^{-3}} = \frac{7^{-2}}{(-4)^{-3}} \]
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Apply the negative exponent rule, which states that \( x^{-n} = \frac{1}{x^n} \):
- For \( 7^{-2} \): \[ 7^{-2} = \frac{1}{7^2} \]
- For \( (-4)^{-3} \): \[ (-4)^{-3} = \frac{1}{(-4)^3} \]
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Rewrite the expression using these rules: \[ \frac{7^{-2}}{(-4)^{-3}} = \frac{\frac{1}{7^2}}{\frac{1}{(-4)^3}} \]
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When you divide by a fraction, you multiply by its reciprocal: \[ = \frac{1}{7^2} \cdot (-4)^3 = \frac{(-4)^3}{7^2} \]
Now, expressing everything with positive exponents, the equivalent expression is: \[ \frac{(-4)^3}{7^2} \]
From the options provided, the closest equivalent expression that matches our result is:
(−4)^3 over 7^2