To simplify the expression \( \frac{15^0 \cdot 7^{-2}}{(-4)^{-3}} \) and express it with only positive exponents, let's go step by step using the properties of integer exponents.
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Simplify \( 15^0 \):
- \( 15^0 = 1 \)
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The expression now becomes: \[ \frac{1 \cdot 7^{-2}}{(-4)^{-3}} = \frac{7^{-2}}{(-4)^{-3}} \]
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Apply the property of exponents for division: \[ \frac{a^{-m}}{b^{-n}} = \frac{b^n}{a^m} \] Thus, we have: \[ = 7^{-2} \cdot (-4)^3 \]
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Next, since we want to express everything with positive exponents, we can rewrite \( 7^{-2} \): \[ 7^{-2} = \frac{1}{7^2} \] Therefore, the expression becomes: \[ = (-4)^3 \cdot \frac{1}{7^2} = \frac{(-4)^3}{7^2} \]
Now that we have simplified the expression, let's compare it with the options provided:
- \( \frac{1}{7^2} \cdot (-4)^{-3} \) (has negative exponent)
- \( \frac{4^3}{7^2} \) (not the same as \((-4)^3\))
- \( \frac{7^2}{(-4)^3} \) (inverted)
- \( \frac{(-4)^3}{7^2} \) (this matches!)
Thus, the equivalent expression is: \[ \boxed{\frac{(-4)^3}{7^2}} \]