Answer:
The goal of enterprise software is to enable the activities of large organizations, supporting a wide range of different user roles.
Answer is A : Cache memory
Answer:
#include<stdio.h>
#include<conio.h>
int m=0,n=4;
int cal(int temp[10][10],int t[10][10])
{
int i,j,m=0;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(temp[i][j]!=t[i][j])
m++;
}
return m;
}
int check(int a[10][10],int t[10][10])
{
int i,j,f=1;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
if(a[i][j]!=t[i][j])
f=0;
return f;
}
void main()
{
int p,i,j,n=4,a[10][10],t[10][10],temp[10][10],r[10][10];
int m=0,x=0,y=0,d=1000,dmin=0,l=0;
clrscr();
printf("\nEnter the matrix to be solved,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&a[i][j]);
printf("\nEnter the target matrix,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&t[i][j]);
printf("\nEntered Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",a[i][j]);
printf("\n");
}
printf("\nTarget Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",t[i][j]);
printf("\n");
}
while(!(check(a,t)))
{
l++;
d=1000;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(a[i][j]==0)
{
x=i;
y=j;
}
}
//To move upwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=0)
{
p=temp[x][y];
temp[x][y]=temp[x-1][y];
temp[x-1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move downwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x+1][y];
temp[x+1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move right side
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x][y+1];
temp[x][y+1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move left
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=0)
{
p=temp[x][y];
temp[x][y]=temp[x][y-1];
temp[x][y-1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
printf("\nCalculated Intermediate Matrix Value :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",r[i][j]);
printf("\n");
}
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
a[i][j]=r[i][j];
temp[i][j]=0;
}
printf("Minimum cost : %d\n",d);
}
getch();
}
Explanation:
The answer is 2 frequencies.
A Full-Duplex QAM 64 K Modem uses two frequencies. One frequency is used for upstream and the other for downstream. A variety of common QAM forms are available and include, 16 QAM, 32 QAM, 64 QAM, and 256 QAM. For example, for domestic broadcast use, 64 and 256 QAM are used for cable modem and digital cable TV. The modulation scheme for this modem uses both amplitude and phase.
