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It is not possible to answer the question as it appears because it makes no sense whatsoever.

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0The sum of two positive integers is never zero. The sum of two numbers a and b can only be zero if a=-b, or a=0 and b=0. Since 0 is not a positive integer, and a and b cannot both be positive integers if a=-b, then it is impossible for the sum of two positive integers to be zero. _______________________________________________________________ The above answer is correct. Here is another way to say it: An integer is any whole number including negative numbers, positive numbers and zero. However, a "positive integer" is a whole number greater than zero. The "sum of two positive integers" means you are adding two numbers greater than zero together. Therefore, the sum of two positive integers can never be a negative integer, and can never be zero. Example: 1 + 1 = 2

int a = 1; int b = 2; int c = a + b; // Sum

addition a + b

The flowchart to read 10 positive integers K>10 Start A N K=1 Sum = 0 Sum = Sum + K2 B Is Y Print K > 100? sum K=k+1 End B A

Sine(A+ B) = Sine(A)*Cosine(B) + Cosine(A)*Sine(B).

On a multiple choice test with answers of A, B, C, and D, the answer of 'B' would only be the correct answer if all other choices are incorrect.

641

There are infinitely many answers: 1+a/b and 2-a/b for any pair of positive integers a and b, with a<b.

The associative property states that, for the sum of three or more integers the order in which the summation in carried out does not make a difference to the answer. Thus, for any three integers, A, B and C: (A + B) + C = A + (B + C) and so, without ambiguity, we can write either as A + B + C. Note that A + B need not be the same as B + A. The order of the integers DOES matter. It is the order of the summing that does not.

I assume you meant "sum". The answer is 60. 60=4 (integers) * 15 (average value of the integers). average of 4 integers = 15 --> (a + b + c + d)/4 = 15 --> (a + b + c + d) = 15 * 4 --> (a + b + c + d) = 60

If a and b are integers, then a times b is an integer.

cls rem a program to sum two number read a,b data 10,30 sum=a+b print "the result of sum=";sum end

Given:a + b = 2a - b = 8Then:a = b + 8∴ (b + 8) + b = 2∴ 2b = -6∴ b = -3∴ a = 5

A. formulaii B. The above is incorrect. "Formulae" and "formulas" are correct.

#include<stdio.h> #include<conio.> void main() { int a,b,sum; clrscr(); cin<<"enter value of a"; cout>>a; cin<<"enter value of b"; cout<<b; sum=a+b; cout<<"sum of two numbers=" getch(); }

Because a is rational, there exist integers m and n such that a=m/n. Because b is rational, there exist integers p and q such that b=p/q. Consider a+b. a+b=(m/n)+(p/q)=(mq/nq)+(pn/mq)=(mq+pn)/(nq). (mq+pn) is an integer because the product of two integers is an integer, and the sum of two integers is an integer. nq is an integer since the product of two integers is an integer. Because a+b equals the quotient of two integers, a+b is rational.

Call the numbers "a" and "b": a + b > a - b b > -b 2b > 0 b > 0 In other words, when the second number is greater than zero.

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

accourding to novanet the correct answer (the sum of A and B).

625*16=10000625+16=641

Look at it the other way - by reverting the operation. The reason it is not a whole number is because if it where, then the subtraction of two integers would be a fraction! If a + b = c (a is a non-integer fraction, b and c are integers), then c - b = a. You would have a fraction as a result of subtracting two integers. However, adding or subtracting two integers always gives you an integer.

On a multiple choice test with answers of A, B, C, and D, the answer of 'A' would only be the correct answer if all other choices are incorrect.

If b is positive, then reduce each of a and b by subtracting 1, repeatedly until b is zero. If b is negative, repeatedly add 1 to both until b is zero. The final value of a is the result you ask for. If a and b are two integers, their sum is a+b, not a-b.

x^2-8x+3 This expression cannot be factored as (x+a)(x+b) where a and b are integers. This is because you cannot find two integers such that their sum is -8 and product 3.

Let a and b be the two numbersThe data we have is:a - b = -18 ........ equation (1)a + b = 2 ........ equation (2)Adding equations (1) and (2)(a - b) + (a + b) = -18 + 22a = -16a = -8Using a = -8 in equation (2),-8 + b = 2b = 2 + 8b = 10

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