Question

Seriously need help!!(3^-1x^-1y^-1)^2(3x^-4y^3)(9x^-2y^3)0/(3x^-3y^-5)^2 simplify, step by step help!!!

Answer 1

\[
\frac{(3^{-1}x^{-1}y^{-1})^2(3x^{-4}y^3)(9x^{-2}2y^3)^0}{(3x^{-3}y^{-5})^2}
\]Is that the correct problem? Is the third factor in the numerator raised to zero?

Answer 2

yes and yes.

Answer 3

\[
\frac{(3^{-1}x^{-1}y^{-1})^2(3x^{-4}y^3)(9x^{-2}2y^3)^0}{(3x^{-3}y^{-5})^2} =
\frac{(3^{-2}x^{-2}y^{-2})*(3x^{-4}y^3)*1}{(9x^{-6}y^{-10})} = \\
\frac{3^{-2+1}x^{-2-4}y^{-2+3}}{9x^{-6}{-10}} = ?
\]Can you simplify from here on?

Answer 4

yes thank you very much!

Answer 5

you are welcome.

Answer 6

\[
\Large
\frac{3^{-2+1}x^{-2-4}y^{-2+3}}{9x^{-6}{y^{-10}}} = ?
\]

Answer 7

it would be y/3(9-10x^6)? I think

Answer 8

\(\Large
\frac{3^{-2+1}x^{-2-4}y^{-2+3}}{9x^{-6}{y^{-10}}} =
\frac{3^{-1}x^{-6}y^{1}}{9x^{-6}{y^{-10}}} =
\frac{y^{1}}{3*9*y^{-10} } = \frac{y^{1+10}}{27} = \frac{y^{11}}{27}
\)

Answer 9

Oh alright I get thank you.