cyount isn't a word. Check your spelling and try again.
63° 2' n, 29° 55' w
I think that you're doing the same thing that I did for IB. Here's what I think you're asking for : 0, 2, 6, 12, 20, 30, 42... c₁ = 0 c₂ = 2 c₃ = 6 (c₂ + 4 = 2 + 4 = 6) c₄ = 12 (c₃ + 6 = 2 + 4 + 6 = 12) c₅ = 20 (c₄ + 8 = 2 + 4 + 6 + 8 = 20) c₆ = 2 + 4 + 6 + 8 + 10 etc... dn = (n/2) <2c₁ + (n-1) 2> dn = (n/2) <2 (2) + (n-1) 2> dn = (n/2) (4 + 2n - 2) dn = (n/2) (2 + 2n) dn = (2n/2) + (2n²/2) dn = n + n²
#include<stdio.h> int main (void) { int i ,c,n; for(i=100;i<=200;i++) { for(n=1;n<=i;n++) { if(i%n==0) { c++; } if(c==2) } printf("the numbers =%2d",c); } }
That is an unsolvable problem, because there are an infinite number of even numbers. However, with limits, you could say... int NMax = ...; int Sum = 0; for (N=2; N<=NMax N+=2) Sum += N;
#include<iostream.h> #include<conio.h> void main() { clrscr(); int i,k,a[10],c[10],n,l; cout<<"Enter the no. of elements\t"; cin>>n; cout<<"\nEnter the sorted elments for optimal merge pattern"; for(i=0;i<n;i++) { cout<<"\t"; cin>>a[i]; } i=0;k=0; c[k]=a[i]+a[i+1]; i=2; while(i<n) { k++; if((c[k-1]+a[i])<=(a[i]+a[i+1])) { c[k]=c[k-1]+a[i]; } else { c[k]=a[i]+a[i+1]; i=i+2; while(i<n) { k++; if((c[k-1]+a[i])<=(c[k-2]+a[i])) { c[k]=c[k-1]+a[i]; } else { c[k]=c[k-2]+a[i]; }i++; } }i++; } k++; c[k]=c[k-1]+c[k-2]; cout<<"\n\nThe optimal sum are as follows......\n\n"; for(k=0;k<n-1;k++) { cout<<c[k]<<"\t"; } l=0; for(k=0;k<n-1;k++) { l=l+c[k]; } cout<<"\n\n The external path length is ......"<<l; getch(); }
Well, darling, the antiderivative of 70x is 35x^2 + C, where C is the constant of integration. It's as simple as that, honey. Just integrate like there's no tomorrow and don't forget that little "+ C" at the end.
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
int(X) = X^n+1/n+1 so X^2/2 + C or, my preference in form 1/2X^2 + C
The answer is Centillion
The electric field at that location is 20 N/C. This can be determined using the formula E = F/q, where E is the electric field, F is the force, and q is the charge. Given that F = 40 N and q = 2 C, we can plug in the values to find E = 40 N / 2 C = 20 N/C.
To balance the reaction Ni + C₄H₈N₂O₂ → Ni(C₄H₈N₂O₂)₂, the coefficients would be 1 for Ni and 2 for C₄H₈N₂O₂. This results in the balanced equation: 1 Ni + 2 C₄H₈N₂O₂ → 1 Ni(C₄H₈N₂O₂)₂.
long factorial(int); int main() { int i, n, c; printf("Enter the number of rows you wish to see in pascal triangle\n"); scanf("%d",&n); for ( i = 0 ; i < n ; i++ ) { for ( c = 0 ; c