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 [tex]\frac{1}{x^x}[/tex]
Since our value is 9^-2, do as followed :
[tex]\frac{1}{9^2}[/tex]
Solve what is 9^2 :
[tex][\frac{1}{81} ][/tex]
E.
[tex]\frac{1}{x^{-5} }[/tex]
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.
[tex]x^5[/tex]
F.
[tex]5a^3b^{-2}[/tex]
Since we have one negative exponent, follow the fraction rule but instead of having one as the numerator, have 5a^3.
[tex]\frac{5a^3}{b^{2}}[/tex]
Fill in the table :
[tex]-3^4 = -3*-3*-3*-3=-81[/tex]
Since -3 is in parenthesis, we can rid the negative along with the parenthesis :
[tex](-3)^4=3^4=3*3*3=81[/tex]
Follow the negative exponent rule :
[tex](-3)^{-4}=\frac{1}{(-3)^4} =\frac{1}{3^4} =\frac{1}{3*3*3*3}= \frac{1}{81}[/tex]
Follow the negative exponent rule yet again :
[tex]-3^{-4}=-\frac{1}{3^4} =-\frac{1}{3*3*3*3} =-\frac{1}{81}[/tex]
