Mathhelp346
Does 1/2m(4m^22mn)=2m^3+m^2n?



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lgbasallote
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is that \[\huge \frac 1{2m} (4m^2  2mn)= 2m^3 + m^2n\]

Mathhelp346
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yes except the m isnt in the denominator

lgbasallote
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ooh...well then
\[\huge \frac 12 m ( 4m^2  2mn)\]
distribute
\[\huge \implies \frac{4m^2 \times m}2  \frac{2mn \times m}2\]
multiply..
\[\huge \implies 2m^3 + m^2 n\]
so yeah you're right

Mathhelp346
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ok thanks

lgbasallote
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welcome

Mathhelp346
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lol thanks