Question

1. (2xy^2)(-3x^2y)

2.(1/3x^2y)(-9xy^3)=

Answer 1

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

Answer 2

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