0
Still curious? Ask our experts.
Chat with our AI personalities
Vivi
Devin
MaxineAdd your answer:
Earn +20 pts
Program to find n raise to the power m?
P = 1 For K = 1 to M . P = P * N Next K PRINT "N raised to the power of M is "; P
When k is added to the expression y2-12ythe expression becomes (y p)2.find the values of p and k?
To solve this problem, we need to expand the given expression (y+k)^2 and compare it to y^2 - 12y. Expanding (y+k)^2 gives y^2 + 2yk + k^2. By comparing this to y^2 - 12y, we can see that 2yk must be equal to -12y, which implies k = -6. Additionally, comparing k^2 to 0 reveals that k = 0. Therefore, the values of p and k are p = -6 and k = 0.
How do you calculate principal when amount rate of interest and time is given?
A = P*(1+R/100)T Where A = amount P = Principal R = Interest Rate (in percentage), and T = Time Since R and T are known, you can calculate (1+R/100)T = k, say. Then A = P*k so that P = A/k
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(); }
88 P K?
Try 88 piano keys!!
