Answer:
\(9^{-2}\\\\\dfrac{9^{8}}{9^{10}}\\\\9^{5}\times9^{-7}\)
Step-by-step explanation:
Let's evaluate the answers.
Think of numbers with negative exponents as numbers on the wrong side of the fraction. So if you have a negative exponent, you put it at the bottom of the fraction or up and then you remove the negative sign.
So let's take the first answer:
\(9^{-2} \:is\:actually\:equal\:to\: \dfrac{1}{9^{2}}\\\\\dfrac{1}{9^{2}} = \dfrac{1}{81}}\)
Now the next one, according to the law of exponents, when dividing we subtract the exponents. (But this only applies if the coefficients are the same!)
\(\dfrac{m^{x}}{m^{y}} = m^{x-y}\)
Applying this rule to the second answer:
\(\dfrac{9^{8}}{9^{10}} = 9^{8-10} = 9^{-2}=\dfrac{1}{9^{2}} = \dfrac{1}{81}\)
The third one:
\(9^5\times9^{-7} = \dfrac{9^5}{9^{7}} = 9^{5-7} = 9^{-2} = \dfrac{1}{9^2} = \dfrac{1}{81}\)
If you apply the same rules to the rest of the choices, you wil find that they are not equivalent to 1/81.
