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

1 Answer
256
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#

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