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swiseman6703

06.04.2022

Mathematics

Answer and Step-by-step explanation:

Alot of these can be solved with the table up above the problems.

A, B, and C.

According to the table, every time a value has an exponent of 0, it would equal to 1. Since all A B and C values follow the 0 exponent, their value is 1.

According to the table, when we have a negative exponent, the number gets turned into a fraction of $\frac{1}{x^x}$

Since our value is 9^-2, do as followed :

$\frac{1}{9^2}$

Solve what is 9^2 :

$[\frac{1}{81} ]$

$\frac{1}{x^{-5} }$

This question is different, since we have to go backwards, we have to switch the negative exponent to a positive exponent and rid the fraction.

x^5

$5a^3b^{-2}$

Since we have one negative exponent, follow the fraction rule but instead of having one as the numerator, have 5a^3.

$\frac{5a^3}{b^{2}}$

Fill in the table :

-3^4 = -3*-3*-3*-3=-81

Since -3 is in parenthesis, we can rid the negative along with the parenthesis :

(-3)^4=3^4=3*3*3=81

Follow the negative exponent rule :

$(-3)^{-4}=\frac{1}{(-3)^4} =\frac{1}{3^4} =\frac{1}{3*3*3*3}= \frac{1}{81}$

Follow the negative exponent rule yet again :

$-3^{-4}=-\frac{1}{3^4} =-\frac{1}{3*3*3*3} =-\frac{1}{81}$

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