Question

Please Help

Answer 1

@BloomLocke367 @SolomonZelman

Answer 2

do you know how to begin?

Answer 3

not really @BloomLocke367

Answer 4

Okay... well you're going to want to first distribute the ^2 to each of the terms inside of the parantheses.

Answer 5

do you understand what I mean by that?

Answer 6

yes

Answer 7

okay. so what do you get after you distribute the ^2?

Answer 8

umm

Answer 9

not sure

Answer 10

ok, do we know what \(\large\color{black}{ \displaystyle {\rm a}^0 }\) is? (for any real number 'a' )

Answer 11

and do we know why?

Answer 12

I was getting ready to explain that @SolomonZelman lol. beat me to the punch XD

Answer 13

\(\large\color{black}{ \displaystyle \color{darkviolet}{\rm a}^0 = \color{darkviolet}{\rm a } ^{ \color{red}{\rm b}-\color{red}{\rm b} }}\) do you agree with this?

Answer 14

yes

Answer 15

something is going with my codes give me a sec

Answer 16

\(\large\color{black}{ \displaystyle \color{darkviolet}{\rm a}^{\color{red}{\rm b-b}}=\frac{\color{darkviolet}{\rm a}^{\color{red}{\rm b}} }{\color{darkviolet}{\rm a}^{\color{red}{\rm b}}} }\)

Answer 17

and that is equivalent to 1, because `(anything) / (anything) = 1`

Answer 18

saying when this anything is same, just like in a^b case

Answer 19

I think its b

Answer 20

close

Answer 21

A?

Answer 22

\(\Large\color{black}{ \displaystyle \left(\color{darkviolet}{\rm a}^{\color{red}{\rm b}}\right)^{\color{blue }{c}}=\color{darkviolet}{\rm a}^{(\color{red}{\rm b}\times \color{blue }{c})} }\)

Answer 23

yes A is the most accurately done answer.

Answer 24

I would not really call it correct, if 2^3 times 1= 2^3 and we don't need to write out 2^3 times 1^2, but yes, A is the correct option as far as this question is concerned.

Answer 25

Thanks @SolomonZelman

Answer 26

yw