If, as is normal, ab represents a times b, etc then
ab + ab + cc = 2ab + c2 which is generally not the same as abc.
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).
associative property
The square root of 149 (7 squared plus 10 squared equals 149).
the midpoint of AB.
C minus B equals AB
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)
It can be but need not be.