Question

evaluate. write your answer as a whc\\(
\frac{7^{0}}{9^{-2}} = \square
\\)

Explanation:

Step1: Simplify the numerator

Any non - zero number to the power of 0 is 1. So, for \(7^{0}\), since \(7
eq0\), we have \(7^{0}=1\).

Step2: Simplify the denominator

Using the negative exponent rule \(a^{-n}=\frac{1}{a^{n}}\) (where \(a
eq0\) and \(n\) is a positive integer), for \(9^{-2}\), we can rewrite it as \(\frac{1}{9^{2}}\). And \(9^{2}=9\times9 = 81\), so \(9^{-2}=\frac{1}{81}\).

Step3: Divide the numerator by the denominator

We have \(\frac{7^{0}}{9^{-2}}=\frac{1}{\frac{1}{81}}\). Dividing by a fraction is the same as multiplying by its reciprocal, so \(\frac{1}{\frac{1}{81}}=1\times81 = 81\).

Answer:

81