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I would spend my money on getting a decent third-grade spelling and grammar book. Improve your present life and then see about the past ones.
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∙ 15y ago
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Who is the famest perso N n in the world?
Jj
When was Lost N Found - JJ Lin album - created?
Lost N Found - JJ Lin album - was created in 2011.
What dress did neil arnstrong wore in the space?
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C program for displaying perfect numbers using nested loop?
for (n=1; n<1000; ++n) { for (sum=0, k=1; k<=n/2; ++k) if (n%k==0) sum += k; if (sum==n) printf ("%d\n", n); }
How do you write it in numbers k is a function of n and 7 times n equals k?
k = f(n) = 7n
C program for optimal merge pattern?
#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(); }
Sum of 1 plus 2 plus 3 plus 4 plus .. plus n?
n(n+1)/2 You can see this from the following: Let x=1+2+3+...+n This is the same as x=n+(n-1)+...+1 x=1+2+3+...+n x=n+(n-1)+...+1 If you add the corresponding terms on the right-hand side of the two equations together, they each equal n+1 (e.g., 1+n=n+1, 2+n-1=n+1, ..., n+1=n+1). There are n such terms. So adding the each of the left-hand sides and right-hand sides of the two equations, we get: x+x=(n+1)+(n+1)+...+(n+1) [with n (n+1) terms on the right-hand side 2x=n*(n+1) x=n*(n+1)/2 A more formal proof by induction is also possible: (1) The formula works for n=1 because 1=1*2/2. (2) Assume that it works for an integer k. (3) Now show that given the assumption that it works for k, it must also work for k+1. By assmuption, 1+2+3+...+k=k(k+1)/2. Adding k+1 to each side, we get: 1+2+3+...+k+(k=1)=k(k+1)/2+(k+1)=k(k+1)/2+2(k+1)/2=(k(k+1)+2(k+1))/2=((k+2)(k+1))/2=(((k+1)+1)(k+1))/2=((k+1)((k+1)+1)/2
Write a program in c language to print a diamond?
// // THIS IS A MACH SIMPLER SOLUTION: // void Diamond(int n) { for (int i=0;i<=2*n;i++,printf("\n")) for (int j=0;j<=2*n;j++) (abs(i-n)+abs(j-n)<=n ? printf("*") : printf(" ")); } //============================================= #include<stdio.h> main() { int i,j,k,n,a,b,c,x; printf("enter the # of rows of graphical output"); scanf("%d",&n); /* UPPER HALF OF KITE */ for(i=1;i<=n;i++) { printf("\t"); for (k=1;k<=(n-i);k++) { printf(" "); } for(j=0;j<i;j++) { printf("*"); printf(" "); } for(k=1;k<=(n-i-1);k++) { printf(" "); } printf("\n"); } /* LOWER PART OF KITE */ for(i=(n-1);i>0;i--) { printf("\t"); for (k=(n-i);k>0;k--) { printf(" "); } for(j=i;j>0;j--) { printf("*"); printf(" "); } for(k=(n-i-1);k>0;k--) { printf(" "); } printf("\n"); } getch(); }
If a fair coin is tossed eight times what is the probability of getting exactly 6 heads?
You need a formula for this. If the probability (in one toss) of getting head is "p", then the probability of getting exactly k heads out of n tosses is: (n,k) p^k (1-p)^(n-k) where (n,k) denotes the number of combinations of k elements among n. You should also know that (n,k) = n! / (( n-k)! k! ) so here, with n=8, k=6, and p=.5 you have (n,k) = 8*7 / 2 = 28 and your probability is : 28 * 1/2^6 * 1/2^2 = 28 / 256 = 7 / 64
How do you draw a cycle in a c program?
/*PROGRAM TO IMPLEMENT GAUSS-jordan method.#include#define MX 20main(){float a[MX] [MX+1],m,p;int i,j,k,n;puts("\n how many equations?:");scanf("%d",&n);for(i=0;i
What has the author N K Nezhdanova written?
N. K. Nezhdanova has written: '\\'
What has the author N K Sugimura written?
N. K. Sugimura has written: '\\'
