I bought mine at k-mart.......
The unscrambled word "Y A N E K" rearranges to form the country "Kenya."
D-K-N-Y is pronounced as individual letters: "D-K-N-Y".
monika did. [k][ii][n][k][y][s][e][k][z]
#include#includevoid main(){ int i, n, j, k, l ;float xo, y[20], f[10][10],X[10],Y[10],h,u,p;clrscr();printf("Enter the value of n(No.of data pairs - 1) : \n");scanf("%d" ,&n);printf("Enter the initial value of x :\n ");scanf("%f" ,&xo);printf("Enter the step size h :\n ");scanf("%f",&h);printf("Enter the values of y\n");for(i=0;i
E k o n o m i y a
K-c-a-j-y-e-k-n-o-m.
No country can be spelled using only these letters.
In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. Its simplest version reads (x+y)^n=\sum_{k=0}^n{n \choose k}x^ky^{n-k}\quad\quad\quad(1) whenever n is any non-negative integer, the numbers {n \choose k}=\frac{n!}{k!(n-k)!} are the binomial coefficients, and n! denotes the factorial of n. This formula, and the triangular arrangement of the binomial coefficients, are often attributed to Blaise Pascal who described them in the 17th century. It was, however, known to Chinese mathematician Yang Hui in the 13th century. For example, here are the cases n = 2, n = 3 and n = 4: (x + y)^2 = x^2 + 2xy + y^2\, (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\, (x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4.\, Formula (1) is valid for all real or complex numbers x and y, and more generally for any elements x and y of a semiring as long as xy = yx.
1
This is beyond sick. Legal advisors for the K-Mart corporation, which it is, evolving out of the old Kresge chain, should take decisive legal action! A woman having a bab y in a discount store- this is horrible.
#include<stdio.h> #include<math.h> int main() { float x[10],y[10][10],sum,p,u,temp; int i,n,j,k=0,f,m; float fact(int); printf("\nhow many record you will be enter: "); scanf("%d",&n); for(i=0; i<n; i++) { printf("\n\nenter the value of x%d: ",i); scanf("%f",&x[i]); printf("\n\nenter the value of f(x%d): ",i); scanf("%f",&y[k][i]); } printf("\n\nEnter X for finding f(x): "); scanf("%f",&p); for(i=1;i<n;i++) { k=i; for(j=0;j<n-i;j++) { y[i][j]=(y[i-1][j+1]-y[i-1][j])/(x[k]-x[j]); k++; } } printf("\n_____________________________________________________\n"); printf("\n x(i)\t y(i)\t y1(i) y2(i) y3(i) y4(i)"); printf("\n_____________________________________________________\n"); for(i=0;i<n;i++) { printf("\n %.3f",x[i]); for(j=0;j<n-i;j++) { printf(" "); printf(" %.3f",y[j][i]); } printf("\n"); } i=0; do { if(x[i]<p && p<x[i+1]) k=1; else i++; }while(k != 1); f=i; sum=0; for(i=0;i<n-1;i++) { k=f; temp=1; for(j=0;j<i;j++) { temp = temp * (p - x[k]); k++; } sum = sum + temp*(y[i][f]); } printf("\n\n f(%.2f) = %f ",p,sum); return 0; }