void math(int*, int*, int*, int*)
void main()
{
int a, b, c, d;
puts("ENTER VALUES TO A & B");
math(&a,&b,&c,&d);
printf("sum= %d \n diff= %d", c,d);
getch();
}
void math( int*a, int*b, int*c, int*d)
{
*c= *a+*b;
*d= *a-*b;
}
By using division, multiplication, addition or subtraction
An infinite amount of ones. 1 + 1 + 1... + 1 / 1 / 1 / 1... = n where n is any integer
123-45-67+89??
3 + 4 * 5 = 23 Remember that multiplication must be carried out before addition.
60
Addition and subtraction are inverse operations. So you can solve addition by subtracting.
By using division, multiplication, addition or subtraction
An infinite amount of ones. 1 + 1 + 1... + 1 / 1 / 1 / 1... = n where n is any integer
Polynomials are the simplest class of mathematical expressions. The expression is constructed from variables and constants, using only the operations of addition, subtraction, multiplication and non-negative integer exponents.
3-3=0
110 divided by 2
505
Problems involving the addition and subtraction of unlike fractions.
33*3=99 3*33=99
impossible
123-45-67+89??
Addition and subtraction.