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
To complete an addition pattern using 3, you can rearrange the addition statements by using subtraction. For example, if the pattern is 3, 6, 9, you can see that each number increases by 3. To find the previous number in the sequence, you can subtract 3 from the current number: 9 - 3 = 6 and 6 - 3 = 3. This shows that subtraction can help identify the consistent step in an addition pattern.
123-45-67+89??
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
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.
To complete an addition pattern using 3, you can rearrange the addition statements by using subtraction. For example, if the pattern is 3, 6, 9, you can see that each number increases by 3. To find the previous number in the sequence, you can subtract 3 from the current number: 9 - 3 = 6 and 6 - 3 = 3. This shows that subtraction can help identify the consistent step in an addition pattern.
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??