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How do you find the LCM of k to the 2nd power k to the 2nd power-1 and k to the 2nd power minus 2k plus 1?
Factor them. k2 = k x k k2 - 1 = (k - 1)(k + 1) k2 - 2k + 1 = (k - 1)(k - 1) Combine the factors, eliminating duplicates. k2(k + 1)(k - 1)(k - 1) = k5 - k4 - k3 + k2, the LCM
What is k in parenthesis K minus 0 then equals 1?
If you're talking about (K-0)=1, then the answer is most definitely 1.
Gaussian elimination in c?
#include<stdio.h> #include<stdlib.h> #include<math.h> #include<conio.h> void main(void) { int K, P, C, J; double A[100][101]; int N; int Row[100]; double X[100]; double SUM, M; int T; do { printf("Please enter number of equations [Not more than %d]\n",100); scanf("%d", &N); } while( N > 100); printf("You say there are %d equations.\n", N); printf("From AX = B enter elements of [A,B] row by row:\n"); for (K = 1; K <= N; K++) { for (J = 1; J <= N+1; J++) { printf(" For row %d enter element %d please :\n", K, J); scanf("%lf", &A[K-1][J-1]); } } for (J = 1; J<= N; J++) Row[J-1] = J - 1; for (P = 1; P <= N - 1; P++) { for (K = P + 1; K <= N; K++) { if ( fabs(A[Row[K-1]][P-1]) > fabs(A[Row[P-1]][P-1]) ) { T = Row[P-1]; Row[P-1] = Row[K-1]; Row[K-1] = T; } } if (A[Row[P-1]][P-1] 0) { printf("The matrix is SINGULAR !\n"); printf("Cannot use algorithm --- exit\n"); exit(1); } X[N-1] = A[Row[N-1]][N] / A[Row[N-1]][N-1]; for (K = N - 1; K >= 1; K--) { SUM = 0; for (C = K + 1; C <= N; C++) { SUM += A[Row[K-1]][C-1] * X[C-1]; } X[K-1] = ( A[Row[K-1]][N] - SUM) / A[Row[K-1]][K-1]; } for( K = 1; K <= N; K++) printf("X[%d] = %lf\n", K, X[K-1]); getch(); }
What equation shows that k is one more than twice m?
K is two times m add 1 k = (2 m ) + 1 k=2m+1
How much does 1 k equal?
1 k equals 1 thousand, commonly used ingame.
Write a program to s wap kth and k plus 1th element?
Assuming the elements are integer type... a[k] ^= a[k+1]; a[k+1] ^= a[k]; a[k] ^= a[k+1]; ...but if they are not integer type... temp = a[k]; a[k] = a[k+1]; a[k+1] = temp;
How do you find the LCM of k to the 2nd power k to the 2nd power-1 and k to the 2nd power minus 2k plus 1?
Factor them. k2 = k x k k2 - 1 = (k - 1)(k + 1) k2 - 2k + 1 = (k - 1)(k - 1) Combine the factors, eliminating duplicates. k2(k + 1)(k - 1)(k - 1) = k5 - k4 - k3 + k2, the LCM
What is the power of k?
It is 1, since k^1 = k.
What is the coefficient of k in -k?
-k = -1*k, so the coefficient is minus 1
What k -k equals 2?
k and -k right? -k x -1 =k k+k= 2 k= 1 unless you mean multiply then that would be -k x-1 =k k x k= 2 1.4142 rounded to the nearest ten thousandth
Which of these is the smallest k k plus 1 kr br-1?
0
Which value of k makes 5 - k + 12 = 16 a true statement?
K=1
Factor k4 - 5k2 plus 4?
(k - 1)(k + 1)(k - 2)(k + 2)
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
Is 'k equals k plus 1' same as 'k plus equals 1' in Java?
Yes, they are exactly the same, both of them increment k in 1.
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(); }
Why is the sum of the reciprocals of all of the divisors of a perfect number equal to 2?
Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)which implies that 1/f(r) = f(k+1-r)/NThen 1/f(1) + 1/(f(2) + ... + 1/f(k)= f(k)/N + f(k-1)/N + ... + f(1)/N= [f(k) + f(k-1) + ... + f(1)] / N= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N
