associative? single replacement
6(ab - ac + b2 - bc)
there will be three vertex AB, BC, AC
(b + 3)(a + c)
Let ABC be a triangle. Let D and E be the mid points of AB and AC respectively. Then the mid-line theorem states that DEBC and DE = BC/2.Extend DE beyond E to F such that DE = EF. Since AE = CE, triangles ADE and CEF are equal, making CFAB (or CFBD, which is the same) because, for the transversal AC, the alternating angles DAE and ECF are equal. Also,CF = AD = BD, such that BDFC is a parallelogram. It follows that BC = DF = 2Â·DE which is what we set out to prove.Conversely, let D be on AB, E on AC, DEBC and DE = BC/2. Prove that AD = DB and AE = CE.This is because the condition DEBC makes triangles ADE and ABC similar, with implied proportion,AB/AD = AC/AE = BC/DE = 2.It thus follows that AB is twice as long as AD so that D is the midpoint of AB; similarly, E is the midpoint of AC.
It would be a straight line of length bc
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.
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
It could be a vector sum.
You could conclude that B lies between A and C.
The answer depends entirely on what AB, BC and AC are. And since you have not bothered to share that crucial bit of information, I cannot provide a more useful answer.
(a + b)(b + c)
ac is 7 if b is 3 and a is 2 a nd c is 5