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P (4 and C)?
The answer depends on what P represents and also what 4 and C are.
650 p c in the you k?
The questions should read '650 p c in the UK? Answer: 650 parliamentary constituencies in the UK!
How do you divide by 5 and have a remainder with 3 and 8 and a remainder of 2 and 9 and a remainder of 4?
The smallest positive integer, p, that satisifesp = 5a + 3 = 8b + 2 = 9c + 4 (where a, b and c are integers) is 58.So the solution set is 58 + 360k where k is an integer.The smallest positive integer, p, that satisifesp = 5a + 3 = 8b + 2 = 9c + 4 (where a, b and c are integers) is 58.So the solution set is 58 + 360k where k is an integer.The smallest positive integer, p, that satisifesp = 5a + 3 = 8b + 2 = 9c + 4 (where a, b and c are integers) is 58.So the solution set is 58 + 360k where k is an integer.The smallest positive integer, p, that satisifesp = 5a + 3 = 8b + 2 = 9c + 4 (where a, b and c are integers) is 58.So the solution set is 58 + 360k where k is an integer.
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(); }
4 C P of the C?
four central points of the compass
