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lgbasalloteBest ResponseYou've already chosen the best response.1
is that \[\huge \frac 1{2m} (4m^2  2mn)= 2m^3 + m^2n\]
 one year ago

Mathhelp346Best ResponseYou've already chosen the best response.1
yes except the m isnt in the denominator
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
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
 one year ago
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