How do you multiply 3y ^ { 0} \cdot 2x y ^ { 2}?

1 Answer
Jun 13, 2017

3y^0*2xy^2=6xy^2

Explanation:

Since any number raised to the power 0 equals 1

3y^0*2xy^2=3(1)*2xy^2

And since any number times 1 is itself

3y^0*2xy^2=3*2xy^2

Then we multiply 2xy^2 by 3

3y^0*2xy^2=6xy^2