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It completely depends the datatype that you have assigned for the variables 'a' , 'b' , and 'c'. Check the compiler that you are using for the size of the datatype in bytes. Add them and thus you will get the answer.

Q: How many bytes are there for a plus b plus c?

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b+b+b+c+c+c+c =3b+4c

2b + 2c or 2(b + c)

And how does this relate to coins?

A+c= 2a+b

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

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b+b+b+c+c+c+c =3b+4c

b + b + b + c + c + c + c = 3b + 4c

It is impossible to give any decimal/numeric value if we are not given the values of at least one variable, so the answer is B + B + B + C + C + C.

2b + 2c or 2(b + c)

sizeof (int) will tell you (in bytes). It's often 2, 4 or 8 bytes.

And how does this relate to coins?

a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)

A+c= 2a+b

Try 'sizeof', it will return the size in bytes.

If a + b + c + d + 80 + 90 = 100, then a + b + c + d = -70.

If: a = b+c+d Then: c = a-b-d

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