A+bc equals ac+ b
yes because ab plus bc is ac
6(ab - ac + b2 - bc)
there will be three vertex AB, BC, AC
If the greatest common factor/divisor of A and B is 1 then they are coprime - they do not share any prime factors. Multiplying both through by C means, obviously, that each number now divides by C. In fact, C is their greatest common divisor, since AC and BC do not have further common factors after C is taken out. Hence the GCF of AC and BC is not merely a factor of C - it is C. (The question makes sense only if A, B and C are integers.)
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 = 2·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.
associative? single replacement
A+BC+AC+B=A+BC+AC+B unless any of these variables has an assigned value.
yes because ab plus bc is ac
This is a 'Sngle Displacement' reaction ( A + BC --> AC + B
It is possible, depending on what on earth AC and BC are!
A+bc---> b+ac
A+bc---> b+ac
(a + b)(b + c)
never A+ :))
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.
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 would be a straight line of length bc