(B/C) + (B/A) = 1
(1/B) = (1/A) + (1/C)
B = AC/(A + C)
This is the familiar lensmakers formula, or two resistors or inductors in parallel,
or two capacitors in series.
If 'A', 'B', and 'C' are supposed to be the vertices of a triangle, then the triangle is
degenerate. It looks just like a straight line with the letters 'A' and 'C' at its ends
and the letter 'B' sticking out somewhere between them for no apparent reason.
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Commutativity.
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
It could be a vector sum.
You could conclude that B lies between A and C.
yes because ab plus bc is ac
Commutativity.
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.
36
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.
12
5
ac is 7 if b is 3 and a is 2 a nd c is 5