which option is equivalent to $\frac{(9)^{-4}}{(-3)^{-8}}$?\na. 1\nb. -1\nc. $\frac{1}{3^{4}}$\nd…

Explanation:

Step1: Rewrite 9 as 3²

$$\frac{(9)^{-4}}{(-3)^{-8}}=\frac{(3^{2})^{-4}}{(-3)^{-8}}$$

Step2: Apply power - of - a - power rule ((a^{m})^{n}=a^{mn})

For ((3^{2})^{-4}), we have ((3^{2})^{-4}=3^{2\times(-4)} = 3^{-8}). And ((-3)^{-8}=[(-1)\times3]^{-8}=(-1)^{-8}\times3^{-8}), since ((ab)^{n}=a^{n}b^{n}), and ((-1)^{-8}=\frac{1}{(-1)^{8}} = 1), so ((-3)^{-8}=3^{-8}).

Step3: Simplify the fraction

$$\frac{3^{-8}}{3^{-8}}=1$$

Answer:

A. 1