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No, not necessarily. In theory you can assign different names to variables, and it doesn't always follow that they are alphabetical. So a divided by b could be x or t or n or any letter you like. Equally, you could use different letters in the initial calculation resulting in c, so x divided by y could be c. Having said that, letters used are commonly used in sequence, so a divided by b = c would be common. Having a, b and c together or x, y and z together is common. In certain situations though, certain letters are given special designations as in the famous E = mc² equation.

Q: Is a divided by b c?

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Because there is no way to define the divisors, the equations cannot be evaluated.

b divided by 2

I think I know how to do this but I'm not sure...It's something like:A, B & C are consecutive so:A=A, B=A+/-1, C=A+/-2So:A/(A+/-1)/(A+/-2)=0.3I think it can be worked out from there but I'm not sure how...

No it is not a, b, c or d.

2: if a*b=c, then (1) c/a = b and (2) c/b = a hope this answers your question

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Because there is no way to define the divisors, the equations cannot be evaluated.

b divided by 2

67 plus 13 = 80 80 divided by 8 = 10

I think I know how to do this but I'm not sure...It's something like:A, B & C are consecutive so:A=A, B=A+/-1, C=A+/-2So:A/(A+/-1)/(A+/-2)=0.3I think it can be worked out from there but I'm not sure how...

You add 2 fractions with the same denominator [c], so the sum is the sum of the numerators divided by the denominator: a/c + b/c = (a+b)/c

No. is it not a, b,c, or d.

No it is not a, b, c or d.

2: if a*b=c, then (1) c/a = b and (2) c/b = a hope this answers your question

AB/8Cas divided by 8/A can be reversed to multiplied by A/8so B/C x A/8 = AB/8CExample: If A=4 B=6 C=3then 6/3 divided by 8/4 = 2/2 = 1also AB/8C = 24/24 = 1Q.E.D.

yes

take a/b divided by c/d, this is the same as a/b X d/c So you multiply by the reciprocal

The remainder is 0.If A has a remainder of 1 when divided by 3, then A = 3m + 1 for some integer mIf B has a remainder of 2 when divided by 3, then B = 3n + 1 for some integer n→ A + B = (3m + 1) + (3n + 2)= 3m + 3n + 1 + 2= 3m + 3n + 3= 3(m + n + 1)= 3k where k = m + n + 1 and is an integer→ A + B = 3k + 0→ remainder when A + B divided by 3 is 0-------------------------------------------------------------------------From this, you may be able to see that:if A when divided by C has remainder Ra; andif B when divided by C has remainder Rb; then(A + B) divided by C will have remainder equal to the remainder of (Ra + Rb) divided by C