That factors to (b + 3)(a + c)
The GCF is b.
The GCF is 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)
That will depend on what type of quadrilateral it is and other information such as its area.
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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)
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
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.
a/b=c/d =>ad=bc =>a =bc/d b =ad/c c =ad/b d =bc/a so if a+b=c+d is true => (bc/d)+(ad/c)=(ad/b)+(bc/a) => (bc2+ad2)/dc=(da2+cb2)/ab => ab(bc2+ad2)=dc(da2+cb2) and since ad=bc, => ab(adc+add) =dc(ada+adc) => abadc+abadd =dcada + dcadc => abadc-dcadc =dcada-abadd => (ab-dc)adc =(dc-ab)add ad cancels out => (ab-dc)c =(dc-ab)d => -(dc-ab)c =(dc-ab)d => -c = d so there's your answer :)
You cannot prove it since it is not true for a general quadrilateral.