Question

Please help and explain

Answer 1

Answer: A,B

Step-by-step explanation:

$(\frac{3^2*3^-^2}{3^3})^2$

You can distribute the exponent.

$\frac{3^2^*^2*3^-^2^*^2}{3^3^*^2} =\frac{3^4*3^-^4}{3^6}$

or you can simplify first and then square.

$(\frac{3^2*3^-^2}{3^3})^2$

$(\frac{3^2^+^(^-^2^)}{3^3})^2$

$(\frac{3^0}{3^3})^2=(\frac{1}{3^3} )^2=\frac{(1)^2}{(3^3)^2}=\frac{1}{3^6}$

Answer 2

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:

$(\frac{3^{2}*3^{-2} }{3^{3}} )^{2}$ --> $(\frac{3^{0}}{3^{3}} )^{2}$

Simplifying this gets us:

$(\frac{1}{3^{3}} )^{2}$

Now, multiply the exponent outside of the parenthesis with the exponent inside resulting in:

$\frac{1}{3^{6}}$

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Reasoning for answer A:

The step of multiplying the outside exponents with those inside was done first:

$(\frac{3^{2}*3^{-2} }{3^{3}} )^{2}$ --> $\frac{3^{4}*3^{-4} }{3^{6}}$

Therefore, we get answers A and B.