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

Mathhelp346
 2 years ago
Best ResponseYou've already chosen the best response.1yes except the m isnt in the denominator

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.1ooh...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
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