Q: What is a lots of 1 k?

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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

If you're talking about (K-0)=1, then the answer is most definitely 1.

#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(); }

K is two times m add 1 k = (2 m ) + 1 k=2m+1

%%%fim1 is our image%%% [ r c ] = size(fim1); even=zeros(r,(c/2)); %first level decomposition %one even dimension for j = 1:1:r a=2; for k =1:1:(c/2) even(j,k)=fim1(j,a); a=a+2; end end %one odd dim odd=zeros(r,(c/2)); for j = 1:1:r a=1; for k =1:1:(c/2) odd(j,k)=fim1(j,a); a=a+2; end end [ lenr lenc ]=size(odd) ; %one dim haar for j = 1:1:lenr for k =1:1:lenc fhigh(j,k)=odd(j,k)-even(j,k); flow(j,k)=even(j,k)+floor(fhigh(j,k)/2); end end %2nd dimension [len2r len2c ]=size(flow); for j = 1:1:(len2c) a=2; for k =1:1:(len2r/2) %even separation of one dim leven(k,j)=flow(a,j); heven(k,j)=fhigh(a,j); a=a+2; end end %odd separtion of one dim for j = 1:1:(len2c) a=1; for k =1:1:(len2r/2) lodd(k,j)=flow(a,j); hodd(k,j)=fhigh(a,j); a=a+2; end end %2d haar [ len12r len12c ]=size(lodd) ; for j = 1:1:len12r for k =1:1:len12c %2nd level hh f2lhigh(j,k)=lodd(j,k)-leven(j,k); %2nd level hl f2llow(j,k)=leven(j,k)+floor(f2lhigh(j,k)/2); %2nd level lh f2hhigh(j,k)=hodd(j,k)-heven(j,k); %2nd level ll f2hlow(j,k)=heven(j,k)+floor(f2hhigh(j,k)/2); end end % level=level-1;

Related questions

K-Mart,Big lots.,

Chose any non-zero integer k. Then 1/1 = k/k.Chose any non-zero integer k. Then 1/1 = k/k.Chose any non-zero integer k. Then 1/1 = k/k.Chose any non-zero integer k. Then 1/1 = k/k.

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;

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

K

It is 1, since k^1 = k.

-k = -1*k, so the coefficient is minus 1

warehouse k-mart lots and lots of place's [only in nz]

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

0

K=1

(k - 1)(k + 1)(k - 2)(k + 2)