j + 4
four more than j or j+4 or j--4 or 2j+(4-j) or (0.0j + 0.04) *100 BOGIF buy one get infinity free
Designate the number of wrong sums for each person by the first letter of their names, c, j, and w. From the statement of the problem, c = j + 4, w = c + 4 = j + 4 + 2 = j +6, and c + j + w = 16. Substituting for c and w in terms of j: (j + 4) + j + (j + 6) = 16, or 3j + 10 = 16, or 3j = 6, or j = 6/3 = 2.
//to multiply two matrices... #include<stdio.h> main() { int a[4][4]; int b[4][4]; int c[4][4]; int i,j,k; printf("\n Enter elements into the first matix....\n"); for(i=0;i<4;i++) { for(j=0;j>4;j++) { scanf("%d",&a[i][j]); } } printf("\n Enter elements into the second matrix.....\n"); for(i=0;i<4;i++) { for(j=0;j<4;j++) { scanf("%d",&b[i][j]); } } printf("\n 1st Matrix.....\n"); for(i=0;i<4;i++) { for(j=0;j<4;j++) { printf("\t%d",a[i][j]); } printf("\n"); } printf("\n 2nd Matrix......\n); for(i=0;i<4;i++) { for(j=0;j<4;j++) { printf("\t%d",b[i][j]); } printf("\n"); } for(i=0;i<4;i++) { for(j=0;j<4;j++) { for(k=0;k<4;k++) { c[i][j]+=(a[i][k]*b[k][j]); } } } printf("\n Resultant Matrix...\n"); for(i=0;i<4;i++) { for(j=0;j<4;j++) { printf("\t%d",c[i][j]); } printf("\n"); } }
{(J/4)-8}<4 if j<48
void main() { int arr[4][4]; int i,j,a,b,f; printf("\nInput numbers to 4*4 matrix"); for(i=0;i<4;i++) { for(j=0;j<4;j++) { printf("\nKey in the [%d][%d]) value",i+1,j+1); scanf("%d",&arr[i][j]); } } for(i=0;i<4;i++) { for(j=0,f=0;j<4;j++) { if(i!=j&&f==0) continue; a=arr[i][j]; b=arr[j][i]; arr[i][j]=b; arr[j][i]=a; f=1; } } for(i=0;i<4;i++) { for(j=0;j<4;j++) printf("%d ",arr[j][i]); printf("\n"); } }
#include <iostream> int main() { int i,j; for(i=1;i<6;i++) // Change 6 to a higher number to get more lines of output { for(j=i;j<i+4;j++) // Change 4 to a higher number to get more numbers in each line cout<<j; cout<<"\n"; } return 0; }
j-4
It's just like solving algebra: [(J/4)-8] < (4) (J/4) < (12) (J) < (12*3) J < 48 The inequality is true when J is less than 48.
#include<stdio.h> #include<conio.h> main() { int a[4][4],b[4][4],c[4][4],i,j; printf("enter the elements of matrix a"); for(i=0;i<=3;i++) for(j=0;j<=3;j++) scanf("%d",&a[i][j]); printf("the first matris is"); for(i=0;i<=3;i++) { printf("\n"); for(j=0;j<=3;j++) printf("%d",a[i][j]); } printf("Enter the elements of second matrix"); for(i=0;i<=3;i++) for(j=0;j<=3;j++) scanf("%d",&b[i][j]); printf("the second matrix is"); for(i=0;i<=3;i++) { printf("\n"); for(j=0;j<=3;j++) printf("%d",b[i][j]); } for(i=0;i<=4;i++) for(j=0;j<=4;j++) c[i][j]=a[i][j] + b[i][j]; printf("the addition of matrix is"); for(i=0;i<=3;i++) { for(j=0;j<=3;j++) printf("%d\t",c[i][j]); printf("\n"); } getch(); }
V/2 + 4
The power for the first scenario is 10 W (200 J / 20 s = 10 W), while the power for the second scenario is 12.5 W (50 J / 4 s = 12.5 W). Therefore, the second scenario has more power.
4 elephants on a jumbo jet