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;
}
An infinite amount of ones. 1 + 1 + 1... + 1 / 1 / 1 / 1... = n where n is any integer
By using division, multiplication, addition or subtraction
123-45-67+89??
60
3 + 4 * 5 = 23 Remember that multiplication must be carried out before addition.
Addition and subtraction are inverse operations. So you can solve addition by subtracting.
By using division, multiplication, addition or subtraction
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.
An infinite amount of ones. 1 + 1 + 1... + 1 / 1 / 1 / 1... = n where n is any integer
3-3=0
505
110 divided by 2
Problems involving the addition and subtraction of unlike fractions.
Addition and subtraction.
33*3=99 3*33=99
impossible
123-45-67+89??