To find an equivalent expression for \(\frac{\sqrt{49}}{7^{-3}}\), we'll start by simplifying both the numerator and the denominator.
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Calculate the square root in the numerator: \[ \sqrt{49} = 7 \]
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Now, substitute this into the expression: \[ \frac{\sqrt{49}}{7^{-3}} = \frac{7}{7^{-3}} \]
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Use the quotient property of exponents, which states that \(\frac{a^m}{a^n} = a^{m-n}\):
- Here, we have \(7^1\) in the numerator and \(7^{-3}\) in the denominator. \[ \frac{7^1}{7^{-3}} = 7^{1 - (-3)} = 7^{1 + 3} = 7^{4} \]
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Convert to rational exponents:
- The expression \(7^{4}\) can also be represented as: \[ 7^{4} = 7^{\frac{4}{1}} \]
So the rational exponent for the new expression is \(\frac{4}{1}\).
Thus, the final answer is: The rational exponent for the new expression is 4.