Question

5 multiple choice 1 point evaluate: \\(\frac{(-8)^{-2}}{3^{-4}}\\) \\(\frac{3}{4}\\) \\(\frac{81}{64}\\) \\(\frac{4}{3}\\) \\(\frac{64}{81}\\) clear my selection previous next

Explanation:

Step1: Recall negative exponent rule

The negative exponent rule states that \(a^{-n}=\frac{1}{a^{n}}\) (where \(a
eq0\) and \(n\) is a positive integer). Apply this to both the numerator and the denominator.
For the numerator \((-8)^{-2}\), using the rule, we get \(\frac{1}{(-8)^{2}}\). Since \((-8)^{2}=(-8)\times(-8) = 64\), the numerator becomes \(\frac{1}{64}\).
For the denominator \(3^{-4}\), using the rule, we get \(\frac{1}{3^{4}}\). And \(3^{4}=3\times3\times3\times3 = 81\), so the denominator becomes \(\frac{1}{81}\).
Now the expression \(\frac{(-8)^{-2}}{3^{-4}}\) is equivalent to \(\frac{\frac{1}{64}}{\frac{1}{81}}\).

Step2: Divide the fractions

Dividing by a fraction is the same as multiplying by its reciprocal. So \(\frac{\frac{1}{64}}{\frac{1}{81}}=\frac{1}{64}\times\frac{81}{1}=\frac{81}{64}\).

Answer:

\(\frac{81}{64}\) (corresponding to the option: \(\boldsymbol{\frac{81}{64}}\))