Please help and explain

Answer: A,B
Step-by-step explanation:
[tex](\frac{3^2*3^-^2}{3^3})^2[/tex]
You can distribute the exponent.
[tex]\frac{3^2^*^2*3^-^2^*^2}{3^3^*^2} =\frac{3^4*3^-^4}{3^6}[/tex]
or you can simplify first and then square.
[tex](\frac{3^2*3^-^2}{3^3})^2[/tex]
[tex](\frac{3^2^+^(^-^2^)}{3^3})^2[/tex]
[tex](\frac{3^0}{3^3})^2=(\frac{1}{3^3} )^2=\frac{(1)^2}{(3^3)^2}=\frac{1}{3^6}[/tex]
Answer:
A and B.
Step-by-step explanation:
If we simplified this equation, we could begin by dealing with the exponents inside of the parenthesis. Remember, when multiplying you will need to add exponents.
This gives us:
[tex](\frac{3^{2}*3^{-2} }{3^{3}} )^{2}[/tex] --> [tex](\frac{3^{0}}{3^{3}} )^{2}[/tex]
Simplifying this gets us:
[tex](\frac{1}{3^{3}} )^{2}[/tex]
Now, multiply the exponent outside of the parenthesis with the exponent inside resulting in:
[tex]\frac{1}{3^{6}}[/tex]
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Reasoning for answer A:
The step of multiplying the outside exponents with those inside was done first:
[tex](\frac{3^{2}*3^{-2} }{3^{3}} )^{2}[/tex] --> [tex]\frac{3^{4}*3^{-4} }{3^{6}}[/tex]
Therefore, we get answers A and B.