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

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Answer 1 · 2017-06-13T02:13:09

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

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 $3$

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

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