Respuesta :

mariamikayla
[tex]1) \ 2xy^2* (-3x^2y)\to \boxed{-6x^3y^3} \\\\\\ 2) \ \frac{1}{3x^2y}*(-9xy^3)= \frac{1}{x}*(-3y^2})\to\boxed{-\frac{3y^2}{x}}[/tex]
Multiply the numbers and letters...
Let's separate them...

2 * -3 = -6
x * x^2 (add your exponents) = x^(1+2) = x^3
y^2 * y (add exp) = y^(2+1) = y^3

Put it all together!

-6x^3y^3

Do the same for no. 2.

(1/3)*-9 = -9/3 = -3
x^2 * x = x^(2+1) = x^3
y * y^3 = y^(1+3) = y^4

Put it together...

-3x^3y^4

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