That factors to (b + 3)(a + c)
The GCF is b.
The GCF is b.
It depends on whether or not a and c are coprime (have no factors in common). Suppose x is the greatest common factor of a and c, where x = 1 if a and c are coprime. ie GCF(a, c) = x then let a = x*m and c = x*n where m and n are coprime then LCM(ab, bc) = LCM(x*m*b, x*n*b) = x*m*n*b
The answer depends on whether or not a is a factor of c.
Recall distributivity a(b + c) = ab + ac = (b + c)a and associativity (ab)c = a(bc) (a + b) + c = a + (b + c) as well as commutativity ab = ba a + b = b + a we are gonna need those. See for yourself when I applied each to learn the trick: ax - bx - ay + yb = (ax - bx) + (-ay + yb) = x(a - b) + -y(a - b) = (x - y)(a - b)
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a(3+b)+c(3+b) * * * * * This is easy to finish: . . . = (a + c)(3 + b).
yes because ab plus bc is ac
Commutativity.
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
(a + b)(b + c)
We can use the identity ((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)). Given that (a + b + c = 12) and (a^2 + b^2 + c^2 = 64), we can substitute these values into the identity: [ 12^2 = 64 + 2(ab + ac + bc). ] Calculating (12^2) gives us 144, so: [ 144 = 64 + 2(ab + ac + bc). ] Subtracting 64 from both sides gives us: [ 80 = 2(ab + ac + bc). ] Dividing by 2, we find: [ ab + ac + bc = 40. ] Thus, the value of (ab + ac + bc) is 40.
15 ac + 20 bc + 6 ad + 8 bd =3a (5c + 2d) + 4b (5c + 2d) =(3a + 4b) (5c + 2d)
The real answer is Bc . Hate these @
It would be a straight line of length bc
AB plus BC equals AC is an example of the Segment Addition Postulate in geometry. This postulate states that if point B lies on line segment AC, then the sum of the lengths of segments AB and BC is equal to the length of segment AC. It illustrates the relationship between points and segments on a line.
If point b is in between points a and c, then ab +bc= ac by the segment addition postulate...dont know if that was what you were looking for... but that is how i percieved that qustion.