Verified Answer :

mariamikayla

\(1) \ 2xy^2* (-3x^2y)\to \boxed{-6x^3y^3} \\\\\\ 2) \ \frac{1}{3x^2y}*(-9xy^3)= \frac{1}{x}*(-3y^2})\to\boxed{-\frac{3y^2}{x}}\)

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

Related Questions