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The answer is 4! (4 factorial), the same as 4x3x2x1, which equals 24 combinations. The answer is 24 and this is how: A b c d A b d c A c d b A c b d A d c b A d b c B c d a B c a d B d a c B d c a B a c d B a d c C d a b C d b a C a b d C a d b C b d a C b a d D a b c D a c b D b c a D b a c D c a b D c b a
#include<stdio.h> int main() { int a,b,c,d; for(a=1; a<5; a++) { for(b=1; b<5; b++) { for(c=1; c<5; c++) { for(d=1; d<5; d++) { if(!(a==b a==c a==d b==c b==d c==d)) printf("dd\n",a,b,c,d); } } } } return 0; }
#include <stdio.h> main() { int m, n, c, d, A[10][10],temp=0; printf("\nEnter the number of rows and columns for matrix A:\n"); scanf("%d%d", &m, &n); printf("\nEnter the elements of matrix A:\n"); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < n ; d++ ) scanf("%d", &A[c][d]); printf("\nMatrix entered is:\n"); for ( c = 0 ; c < m ; c++ ) { for ( d = 0 ; d < n ; d++ ) printf("%d\t", A[c][d]); printf("\n"); } printf("\n\nMaximum element of each row is:\n"); for(c=0;c<m;c++) { for(d=0;d<n;d++) { if(A[c][d]>temp) temp=A[c][d]; } printf("\n\t\tRow %d: %d",c+1,temp); temp=0; } temp=A[0][0]; printf("\n\nMinimum element of each coloumn is:\n"); for(c=0;c<n;c++) { for(d=0;d<m;d++) { if(A[d][c]<temp) temp=A[d][c]; } printf("\n\t\tColoumn %d: %d",c+1,temp); temp=A[d][c] ; } return 0; }
int a=2, b=3, c=4, d=5; printf ("%d/%d + %d/%d = %d/%d\n", a, b, c, d, a*d+b*c, b*d);
Yes. To show the conditions on a, b, c and d given that if a/b = c/d then a+b = c+d. Suppose b != d (and that both b and d are non-zero) then: d = kb for some number k (!= 0), so c/d = c/kb = (c/k)/b so a/b = (c/k)/b => a = c/k => c = ka Thus: c + d = ka + kb = k(a + b) Which means that c + d = a + b only if k = 1. Thus if a/b = c/d then a + b = c + d only if a = c and b = d. The condition on b and d both being non-zero prevents the possibility of division by zero. If either is zero, a division by zero will occur and at least one of the fractions is infinite.