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**The answer of a + b. A and b are both variabls which means that you only add the a and b to get your answer. Lets say you had this problem a(7h+13) you would put 7ah+13a because a goes first in the alphabet and since 13 doesn't have a letter it would be 7ah+13a.**

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Q: What is the sum of a and b?

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The sum of is the total of everything being summed; the sum total. Thus the sum of a, b and c is therefore a + b + c.

Sum of fifteen minus b

2.5

Cls input "enter two no.s ",a,b sum=a+b print "sum = ";sum end

Assuming that a and b are two non-negative numbers, then their sum is a + b and the difference is |a - b|.

b+11

(3+b)

(b+5)

The sum of b and 8 would be written b+8 or 8+b(commutative property of addition)

The expression which represents half the sum of five and a number b is (b+5)/2.

int mul (int a, int b) { int sum= 0; for (; b>0; --b) sum -= -a; for (; b<0; ++b) sum -= a; return sum; }

a = 5;b = 10;c = a + b;System.out.println("The sum is " + c);a = 5;b = 10;c = a + b;System.out.println("The sum is " + c);a = 5;b = 10;c = a + b;System.out.println("The sum is " + c);a = 5;b = 10;c = a + b;System.out.println("The sum is " + c);

You remember that 'sum' means addition so that's b+11

For example:c = a + b;(This calculates the sum of a + b, and assigns the result to variable c.)If you repeatedly want to add something to an accumulated sum:b = b + a;or better:b += a;For example:c = a + b;(This calculates the sum of a + b, and assigns the result to variable c.)If you repeatedly want to add something to an accumulated sum:b = b + a;or better:b += a;For example:c = a + b;(This calculates the sum of a + b, and assigns the result to variable c.)If you repeatedly want to add something to an accumulated sum:b = b + a;or better:b += a;For example:c = a + b;(This calculates the sum of a + b, and assigns the result to variable c.)If you repeatedly want to add something to an accumulated sum:b = b + a;or better:b += a;

3(a + b) + a = 3a + 3b + a = 4a + 3b

3a+b

1.Start 2. Input a,b,c 3. Sum = a+b+c 4. Average = sum/3 5. Output - Sum,Average 6. Stop

Let A = rolling a double Let B = sum is 11 P(A)=6/36=1/6 P(B)=2/36=1/18 since (5,6) and (6,5) produce a sum of 11. We want to find P(A/B)= P(A & B) / P(B) = 0 / P(B)=0 P(A & B) represent the event getting a double and the sum being 11.

The sum of that answer for that question is b- 4000

#include<stdio.h> main() { int a,b,sum; print f("enter two numbers"); scan f("%d%d",&a&b); sum=a+b print f(the sum of a and b is %d,sum); }

it's: a + b

3*(a + b) or 3a + 3b

Sum(All elements in B) - Sum(All elements in A)

I suppose you wanted to ask: how to do multiplication with repeated addition. int mul (int a, int b) { int sum= 0; int sign= 1; if (b<0) { sign= -1; b= -b;} for (; b; --b) sum = sum + a; return sign*sum; }

(a+b)/6

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