If, as is normal, ab represents a times b, etc then
ab + ab + cc = 2ab + c2 which is generally not the same as abc.
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Do you mean F = abc + abc + ac + bc + abc' ? *x+x = x F = abc + ac + bc + abc' *Rearranging F = abc + abc' + ab + bc *Factoring out ab F = ab(c+c') + ab + bc *x+x' = 1 F = ab + ab + bc *x+x = x F = bc
Commutativity.
The existence of the additive inverse (of ab).
The square root of 149 (7 squared plus 10 squared equals 149).
Area = 0.5*AB*BC*sin(ABC) = 0.5*(2x+1)*(x+2)*0.5 = 3 So, (2x+1)*(x+2) = 12 2x2 + 5x + 2 = 12 2x2 + 5x - 10 = 0 x = 1.31 (to 3 sf)
i think its ABC and if that not right ask a teacher for the answer
Do you mean F = abc + abc + ac + bc + abc' ? *x+x = x F = abc + ac + bc + abc' *Rearranging F = abc + abc' + ab + bc *Factoring out ab F = ab(c+c') + ab + bc *x+x' = 1 F = ab + ab + bc *x+x = x F = bc
You want: abc + ab Factor out the common terms which are "a" and "b" ab ( c + 1 )
yes because ab plus bc is ac
Commutativity.
The existence of the additive inverse (of ab).
The square root of 149 (7 squared plus 10 squared equals 149).
Area = 0.5*AB*BC*sin(ABC) = 0.5*(2x+1)*(x+2)*0.5 = 3 So, (2x+1)*(x+2) = 12 2x2 + 5x + 2 = 12 2x2 + 5x - 10 = 0 x = 1.31 (to 3 sf)
associative property
It can be but need not be.
15
the midpoint of AB.