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A+bc equals ac+ b

Q: What kind of reaction is A plus BC æ B plus AC?

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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 = 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.

Related questions

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+ :))

It would be a straight line of length bc

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

hello