#include<stdio.h>
void main()
{
int a[3][3],b[3][3],c[3][3],i,j,k;
clrscr();
printf("Enter elements of A:");
for(i=0;i<=2;i++)
for(j=0;j<=2;j++)
scanf("%d",&a[i][j]);
printf("Enter elements of B:");
for(i=0;i<=2;i++)
for(j=0;j<=2;j++)
scanf("%d",&b[i][j]);
printf("A:");
for(i=0;i<=2;i++)
{
for(j=0;j<=2;j++)
printf("%d ",a[i][j]);
printf("");
//To change line.
}
printf("B:");
for(i=0;i<=2;i++)
{
for(j=0;j<=2;j++)
printf("%d ",b[i][j]);
printf("");
}
k=0;
while(k<=2)
{
for(i=0;i<=2;i++)
{
int sum=0;
for(j=0;j<=2;j++)
sum=sum+a[i][j]*b[j][k];
c[i][k]=sum;
}
k++;
}
printf("Result: ");
for(i=0;i<=2;i++)
{
for(j=0;j<=2;j++)
printf("%d ",c[i][j]);
printf("");
}
getch();
}
1 answer
B-J- and the Bear - 1978 B-J- and the Seven Lady Truckers Part 2 3-2 was released on:
USA: 13 January 1981
1 answer
(1) M+B=2*J
(2) M=8+B
(3) J+B=20
Three equations. Three unknowns.
Rewrite (1)
M=2*J-B
Sub into (2)
2*J-B=8+B
Solve for B
2B+8=2J or B+4=J
Sub J into (3)
2B + 4 = 20
B = 8
J = 12
M=16
1 answer
#include<stdio.h>
#include<conio.h>
void main()
{
int i=0,b,a,j;
clrscr();
while(i<=100)
{
printf("Enter the no:");
scanf("%d",&a);
b=a/2;
for(j=2;j<b;j++)
{
if(a%j==0)
break;
}
if (j==b)
i++
getch();
}
1 answer
import java.io.*;
import java.math.*;
public class addmatrices
{
public static void main(String args[])
{
int a[][]={{2,3},{4,5}};
int b[][]={{3,4},{5,2}};
int c[][]={{0,0},{0,0}};
int i,j;
System.out.println("matrix A:")
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
System.out.println(""+a[i][j]);
}
System.out.println();
}
System.out.println("matrix B:");
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
System.out.println(""+b[i][j]);
}
System.out.println();
}
System.out.println("Add Two Matrices");
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
c[i][j]=a[i][j]+b[i][j];
System.out.println(""+c[i][j]);
}
System.out.println();
}
}
}
1 answer
/* multiplication of a 3*3 matrix*/
#include<stdio.h>
main()
{
int a[3][3],b[3][3],c[3][3];
int i,j,k;
printf("enter the elements in A matrix:\n");
for(i=0;i<=2;i++)
{
for(j=0;j<=2;j++)
{
scanf("%d",&a[i][j]);
}
}
printf("enter b matrix:\n");
for(i=0;i<=2;i++)
{
for(j=0;j<=2;j++)
{
scanf("%d",&b[i][j]);
}
}
for(i=0;i<=2;i++)
{
printf("\n");
for(j=0;j<=2;j++)
{
c[i][j]=0;
for(k=0;k<=2;k++)
{
c[i][j] = c[i][j]+a[i][k] * b[k][j];
}
}
}
printf("multiplication matrix is:\n");
for(i=0;i<=2;i++)
{
for(j=0;j<=2;j++)
{
printf("%d\t",c[i][j]);
}
printf("\n");
}
}
1 answer
1 answer
#include<stdio.h>
#include<conio.h>
int main(void)
{
int a[100][100]={0}; /* initializing matrices to '0' */
int b[100][100]={0};
int c[100][100]={0}; /*matrix-c for multiplication*/
/*r1,c1 are rows and coloumns for matrix-a.r2,c2 for matrix-b.*/
int r1,c1,r2,c2;
int i,j,k;
clrscr();
printf("enter the no of ROWS and COLOUMNS for MATRIX-A\n");
scanf("%d %d",&r1,&c1);
printf("enter the no of ROWS and COLOUMNS for MATRIX-B\n");
scanf("%d %d",&r2,&c2);
if(c1==r2)
{
printf("matrix multiplication possible\n\nenter numbers in MATRIX-A\n");
/* to enter numbers in matrix-a*/
for(i=0;i<r1;i++)
for(j=0;j<c1;j++)
scanf("%d",&a[i][j]);
printf("enter numbers in MATRIX-B\n");
/* to enter numbers in matrix-b*/
for(i=0;i<r2;i++)
for(j=0;j<c2;j++)
scanf("%d",&b[i][j]);
/*for matrices multiplication*/
for(i=0;i<r1;i++)
for(j=0;j<c2;j++)
for(k=0;k<c1;k++)
*(*(c+i)+j)+=*(*(a+i)+j)*(*(*(b+k)+j));
printf("result of multiplication of matrices is \n");
/*to display as matrix format*/
for(i=0;j<r1;i++)
{
printf("\n");
for(j=0;j<c2;j++)
printf("%d\t",*(*(c+i)+j));
}
} //if
else
printf("MATRIX multiplication not Possible\n");
getch();
return 0;
}
/* TWINKLING TWINS */
2 answers
No. Title Lyrics Music Length
1. "Introduction to Cardiology" Benji Madden B. Madden 0:47
2. "Let the Music Play" B. Madden, Joel Madden, Don Gilmore B. Madden 4:12
3. "Counting the Days" B. Madden, J. Madden B. Madden, John Feldmann 2:51
4. "Silver Screen Romance" B. Madden, J. Madden, Gilmore B. Madden 3:10
5. "Like It's Her Birthday" B. Madden, J. Madden, Gilmore B. Madden 3:30
6. "Last Night" B. Madden, J. Madden, Gilmore B. Madden, Gilmore 3:40
7. "Sex on the Radio" B. Madden, J. Madden, Sam Hollander, Dave Katz B. Madden 3:16
8. "Alive" B. Madden, J. Madden, Gilmore B. Madden 3:14
9. "Standing Ovation" B. Madden, J. Madden, Feldmann B. Madden, Feldmann 3:39
10. "Harlow's Song (Can't Dream Without You)" J. Madden, B. Madden, Luke Walker J. Madden 3:34
11. "Interlude: The Fifth Chamber" B. Madden, J. Madden, Gilmore 1:29
12. "1979" B. Madden, J. Madden B. Madden, Matt Squire 2:59
13. "There She Goes" B. Madden, J. Madden, Noel Fisher B. Madden, Fisher 3:22
14. "Right Where I Belong" B. Madden, J. Madden B. Madden, Squire 3:52
15. "Cardiology" B. Madden, J. Madden, Walker B. Madden 2:56
1 answer
Let Joe's amount be represented as ( J ) and Bill's amount as ( B ). According to the problem, ( B = \frac{1}{2}J ) and ( B + J = 60 ). Substituting the first equation into the second gives ( \frac{1}{2}J + J = 60 ), which simplifies to ( \frac{3}{2}J = 60 ). Solving for ( J ) gives ( J = 40 ), and substituting back, ( B = \frac{1}{2} \times 40 = 20 ). Therefore, Bill has 20 dollars.
1 answer
B. J. Surhoff debuted on April 8, 1987, playing for the Milwaukee Brewers at County Stadium; he played his final game on October 2, 2005, playing for the Baltimore Orioles at Oriole Park at Camden Yards.
1 answer
B. J. Upton debuted on August 2, 2004, playing for the Tampa Bay Devil Rays at Tropicana Field; he played his final game on September 29, 2013, playing for the Atlanta Braves at Turner Field.
1 answer
Although it is set in 1912, Priestley wrote the play in 1945
2 answers
#include<stdio.h>
int main(){
int a[2][2],b[2][2],c[2][2],i,j;
int m1,m2,m3,m4,m5,m6,m7;
printf("Enter the 4 elements of first matrix: ");
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",&a[i][j]);
printf("Enter the 4 elements of second matrix: ");
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",&b[i][j]);
printf("\nThe first matrix is\n");
for(i=0;i<2;i++){
printf("\n");
for(j=0;j<2;j++)
printf("%d\t",a[i][j]);
}
printf("\nThe second matrix is\n");
for(i=0;i<2;i++){
printf("\n");
for(j=0;j<2;j++)
printf("%d\t",b[i][j]);
}
m1= (a[0][0] + a[1][1])*(b[0][0]+b[1][1]);
m2= (a[1][0]+a[1][1])*b[0][0];
m3= a[0][0]*(b[0][1]-b[1][1]);
m4= a[1][1]*(b[1][0]-b[0][0]);
m5= (a[0][0]+a[0][1])*b[1][1];
m6= (a[1][0]-a[0][0])*(b[0][0]+b[0][1]);
m7= (a[0][1]-a[1][1])*(b[1][0]+b[1][1]);
c[0][0]=m1+m4-m5+m7;
c[0][1]=m3+m5;
c[1][0]=m2+m4;
c[1][1]=m1-m2+m3+m6;
printf("\nAfter multiplication using \n");
for(i=0;i<2;i++){
printf("\n");
for(j=0;j<2;j++)
printf("%d\t",c[i][j]);
}
return 0;
}
Sample output:
Enter the 4 elements of first matrix: 1
2
3
4
Enter the 4 elements of second matrix: 5
6
7
8
The first matrix is
1 2
3 4
The second matrix is
5 6
7 8
After multiplication using
19 22
43 50
1 answer
// transpose for the sparse matrix void main() { clrscr(); int a[10][10],b[10][10]; int m,n,p,q,t,col; int i,j; printf("enter the no of row and columns :\n"); scanf("%d %d",&m,&n); // assigning the value of matrix for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("a[%d][%d]= ",i,j); scanf("%d",&a[i][j]); } } printf("\n\n"); //displaying the matrix printf("\n\nThe matrix is :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("%d\t",a[i][j]); } printf("\n"); } t=0; printf("\n\nthe non zero value matrix are :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { // accepting only non zero value if(a[i][j]!=0) { t=t+1; b[t][1]=i; b[t][2]=j; b[t][3]=a[i][j]; } } } printf("a[0 %d %d %d\n",m,n,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,b[i][1],b[i][2],b[i][3]); } a[0][1]=n; a[0][2]=m; a[0][3]=t; int s[10],u[10]; if(t>0) { for(i=1;i<=n;i++) { s[i]=0; } for(i=1;i<=t;i++) { s[b[i][2]]=s[b[i][2]]+1; } u[1]=1; for(i=2;i<=n;i++) { u[i]=u[i-1]+s[i-1]; } for(i=1;i<=t;i++) { j=u[b[i][2]]; a[j][1]=b[i][2]; a[j][2]=b[i][1]; a[j][3]=b[i][3]; u[b[i][2]]=j+1; } } printf("\n\n the fast transpose matrix \n\n"); printf("a[0 %d %d %d\n",n,m,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,a[i][1],a[i][2],a[i][3]); } getch(); }
1 answer
#include
#include
void main()
{
char reg[20];
int q[20][3],i,j,len,a,b;
clrscr();
for(a=0;a<20;a++)
{
for(b=0;b<3;b++)
{
q[a][b]=0;
}
}
printf("Regular expression: \n");
scanf("%s",reg);
len=strlen(reg);
i=0;
j=1;
while(i
{
if(reg[i]=='a'&®[i+1]!='/'&®[i+1]!='*')
{
q[j][0]=j+1;
j++;
}
if(reg[i]=='b'&®[i+1]!='/'&®[i+1]!='*')
{
q[j][1]=j+1;
j++;
}
if(reg[i]=='e'&®[i+1]!='/'&®[i+1]!='*')
{
q[j][2]=j+1;
j++;
}
if(reg[i]=='a'&®[i+1]=='/'&®[i+2]=='b')
{
q[j][2]=((j+1)*10)+(j+3);
j++;
q[j][0]=j+1;
j++;
q[j][2]=j+3;
j++;
q[j][1]=j+1;
j++;
q[j][2]=j+1;
j++;
i=i+2;
}
if(reg[i]=='b'&®[i+1]=='/'&®[i+2]=='a')
{
q[j][2]=((j+1)*10)+(j+3);
j++;
q[j][1]=j+1;
j++;
q[j][2]=j+3;
j++;
q[j][0]=j+1;
j++;
q[j][2]=j+1;
j++;
i=i+2;
}
if(reg[i]=='a'&®[i+1]=='*')
{
q[j][2]=((j+1)*10)+(j+3);
j++;
q[j][0]=j+1;
j++;
q[j][2]=((j+1)*10)+(j-1);
j++;
}
if(reg[i]=='b'&®[i+1]=='*')
{
q[j][2]=((j+1)*10)+(j+3);
j++;
q[j][1]=j+1;
j++;
q[j][2]=((j+1)*10)+(j-1);
j++;
}
if(reg[i]==')'&®[i+1]=='*')
{
q[0][2]=((j+1)*10)+1;
q[j][2]=((j+1)*10)+1;
j++;
}
i++;
}
printf("Transition function \n");
for(i=0;i<=j;i++)
{
if(q[i][0]!=0)
printf("\n q[%d,a]-->%d",i,q[i][0]);
if(q[i][1]!=0)
printf("\n q[%d,b]-->%d",i,q[i][1]);
if(q[i][2]!=0)
{
if(q[i][2]<10)
printf("\n q[%d,e]-->%d",i,q[i][2]);
else
printf("\n q[%d,e]-->%d & %d",i,q[i][2]/10,q[i][2]%10);
}
}
getch();
}
4 answers
Only 2 of the 50 states begin with B, E, J or K. Therefore 4% of the states begin with B, E, J or K. The states are as follows:
· Kansas
· Kentucky
2 answers
Hold [Shift] and type [J][M][T][B][0][2]
2 answers
B-J- and the Bear - 1978 Fly a Wild Horse 2-11 was released on:
USA: 8 December 1979
1 answer
B-J- and the Bear - 1978 The Friendly Double Cross 2-21 was released on:
USA: 29 March 1980
1 answer
B-J- and the Bear - 1978 Cain's Son-in-Law 2-5 was released on:
USA: 27 October 1979
1 answer
B-J- and the Bear - 1978 Through the Past Darkly 2-15 was released on:
USA: 26 January 1980
1 answer
A.) j(a) = a^2 - 2a + 4
B.) j(3) = (3)^2 - 2(3) + 4 = 9 - 6 + 4 = 7
C.) j(x^2) = (x^2)^2 - 2(x^2) + 4 = x^4 - 2x^3 + 4
D.) j(x+3) = (x + 3)^2 - 2(x + 3) + 4 = x^2 +6x + 9 - 2x - 6 + 4 = x^2 + 4x + 7
E.) j(x+h) = (x + h)^2 - 2(x + h) + 4 = x^2 + 2hx + h^2 - 2x - 2h + 4
1 answer
A.) j(a) = a^2 - 2a + 4
B.) j(3) = (3)^2 - 2(3) + 4 = 9 - 6 + 4 = 7
C.) j(x^2) = (x^2)^2 - 2(x^2) + 4 = x^4 - 2x^3 + 4
D.) j(x+3) = (x + 3)^2 - 2(x + 3) + 4 = x^2 +6x + 9 - 2x - 6 + 4 = x^2 + 4x + 7
E.) j(x+h) = (x + h)^2 - 2(x + h) + 4 = x^2 + 2hx + h^2 - 2x - 2h + 4
1 answer