Algebra help! please!!?

Question

how do you do..

( -x^2y^4) ( -x^3y^100

--------------------------------

x^9y

^=power, ---= divide

thanks!

Answer 1

= (x^5y^104)

   ---------------

    x^9y

=y^103

  ---------

    x^4

In the numerator you add exponents together.  The negative x cancels out because a negative times a negative is positive. Then you do

x^9 - x^5 so you can get all your x'es in the denominator. And then you do y^104 - y

Answer 2

( -x^2y^4) ( -x^3y^100

--------------------------------  = [( -x^2y^4) ( -x^3y^100)]/(x^9y)

x^9y

=   +(x^5y^104)/(x^9y)

=   y^103/x^4 = (y^103)(x^-4)

Remember that (x^a)(x^b) = x^(a+b)

and (x^a)/(x^b) = x^(a-b) = (x^a)(x^-b)

Answer 3

( -x^2y^4) ( -x^3y^100)/x^9y

+(x^5y^104)/(x^9y)

+y^103/x^4

Answer 4

( -x^2y^4) ( -x^3y^100

--------------------------------

x^9y

is

- - (x^-9 x^2x^4)(y^4y^100y^-1) == x^-2y^103= (y^103)/x^2