Question

Simplify this..giving out medals!

Answer 1

please help

Answer 2

help!

Answer 3

I factored them out and multiply for the bottom fraction's reciprocal
\[\frac{ 8x }{ (x+2)(x+2) } \times \frac{(x-2)(x+2) }{4x ^{2} }\]
\[\frac{ 8x(x-2) }{ 4x ^{2}(x+2) }\]
\[\frac{ 2(x-2) }{ x(x+2) }\]
So the answer is B ^^

Answer 4

thanks..one more please?

Answer 5

Sure

Answer 6

thank you

Answer 7

Like problem above you multiply for the reciprocal
\[\frac{ x ^{2} }{ y ^{3} } \times \frac{ y ^{6} }{ x ^{7} }\]
\[\frac{ x ^{-5} }{ y ^{-3} } = \frac{ y ^{3} }{ x ^{5} }\]
So the answer is C ^^

Answer 8

thank you.

Answer 9

Whoops I meant answer is B

Answer 10

I have 1 question..do you have time?

Answer 11

Lol yeah I'd love to help

Answer 12

= 3(2^0) + 3(2^1) + 3(2^2) + 3(2^3) + 3(2^4) + 3(2^5) + 3(2^6)
= 3(1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6)
= 3 x 1 x (2^7 -- 1) / (2 -- 1)
= 3x127
= 381

Answer 13

thanks!