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That factors to (b + 3)(a + c)

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12y ago

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Can you factor this ab plus 3a-bc-3c?

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Factor 3a plus ab plus 3c plus bc?

a(3+b)+c(3+b) * * * * * This is easy to finish: . . . = (a + c)(3 + b).


If ab plus bc equals ac then ac equals ab plus bc?

yes because ab plus bc is ac


Ab plus bc equals bc plus ab what property is this?

Commutativity.


Use Boolean algebra to simplify the logic function and realize the given function and minimized function using discrete gates. f equals ab c plus abc plus ac plus bc plus abC.?

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


How do you factor ab plus bc plus b2 plus ac?

(a + b)(b + c)


What is 15ac plus 20bc plus 6ad plus 8bd?

15 ac + 20 bc + 6 ad + 8 bd =3a (5c + 2d) + 4b (5c + 2d) =(3a + 4b) (5c + 2d)


If point C is between points A and B then AC plus CB .?

The real answer is Bc . Hate these @


What would a triangle look like if ab plus ac equals bc?

It would be a straight line of length bc


What is AB plus BC equals AC an example of?

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.


What does AB plus BC equals AC mean?

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.


If a over b equals c over d then is it possible that a plus b equals c plus d If not then give Mathematical Reason for that?

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 :)