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You can write "the sum of..." with a plus sign; in this case:b + 11.

Q: The sum of b and 11?

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The sum of 11 and 11 is 22.

9, 11, 13, 15 The solution equation is A + B = 24 where B = A +2 (the consecutive odd integer) 2A +2 = 24 A = 11, B = 13

(b+5)

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

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

Related questions

b+11

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.

a + b = 35 a - b = 11 2b + 11 = 35 2b = 24 b = 12 a = 23

Add together all the digits in the odd positions in the number. Sum = A Add together all the digits in the even positions in the number. Sum = B If A-B is 0 or if it divisible by 11 (positive or negative), then the original number is divisible by 11.

No, thanks.

The sum of 11 and 11 is 22.

If a + b = 25 and a - b = 11 as asked, we can solve like this: a - b = 11 => a = 11 + b [Now we have expressed a in terms of b. Let's substitute that information in the first equation.] a + b = 25 => (11 + b) + b = 25 => 11 + b + b = 25 => 11 + 2b = 25 => 2b = 14 => b = 14/2 & b = 7 Substitute the value for b (the 7) into either equation and solve like this: a + b = 25 => a + 7 = 25 => a = 25 - 7 & a = 18 Lastly, substitute your answers into either equation and see if it checks.

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

The sum of addends 11 and 15 is: 11 + 15 = 26

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.

The sum of 88 and 11 is 99.

The sum of 11 and 7 is 18.