Algorithm to find the sum and average of 3 numbers: Start. Input three numbers (let's call them A, B, and C). Calculate the sum (Sum = A + B + C). Calculate the average (Average = Sum / 3). Output the sum and average. End. Flowchart: [Start] ↓ [Input A, B, C] ↓ [Sum = A + B + C] ↓ [Average = Sum / 3] ↓ [Output Sum, Average] ↓ [End]
3+c
"the same as the others added together" For example: a=b+c+d a is equal to the sum of b plus c plus d. 9=2+3+4
c + n
It is simply: a+c
3+c
1.Start 2. Input a,b,c 3. Sum = a+b+c 4. Average = sum/3 5. Output - Sum,Average 6. Stop
//sum and product of 3 nos #include #include void main() { int a,b,c; printf("enter the 3 nos"); scanf("%d%d%d",&a,&b,&c); printf("sum of 3 nos",a+b+c); printf("product of 3 nos",a*b*c); getch(); }
matrix
Sum means to add.. so the equation would look like 3(c+d) or (c+d)*3. You could also add them separately. 3c+3d Hope that helps!
Um, x2+3x-5=0? This is ax2+bx+c where a=1, b=3, and c=-5. The sum of the roots is -b/a so that means the sum of the roots is -3. Also, product of the roots is c/a. That means the product of the roots is -5. -3+(-5)= -8. There you have it.
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.
J C Sum was born in 1976.
"the same as the others added together" For example: a=b+c+d a is equal to the sum of b plus c plus d. 9=2+3+4
As a product of its prime factors: 3*3*3*3 = 81 and 4*3 = 12
To solve for the sum of the first n number of cubes, use (1+2+3+4+5...+n)2. So for n=100, (1+2+3+4+5...+100)2 = c (cubes) (101 * 50)2 = c 50502 = c c = 25502500
c + n