EMVY42
EMVY42
29-01-2017
Mathematics
contestada
2kx+4mx−2nx−3ky−6my+3ny factor by grouping
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Xander4D
Xander4D
29-01-2017
2kx+4mx-2nx-3ky-6my+3ny = 2x(k+2m-n) -3y(k+2m-n) = (2x-3y) (k+2m-n)
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