Question

Pls help me, I don't know how to do it?:(​

Answer 1

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.

D.

According to the table, when we have a negative exponent, the number gets turned into a fraction of
`(1)/(x^x)`

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

`(1)/(9^2)`

Solve what is 9^2 :

`[(1)/(81) ]`

E.

`(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`

F.

`5a^3b^(-2)`

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

`(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)=(1)/((-3)^4) =(1)/(3^4) =(1)/(3*3*3*3)= (1)/(81)`

Follow the negative exponent rule yet again :

`-3^(-4)=-(1)/(3^4) =-(1)/(3*3*3*3) =-(1)/(81)`