Answer:
The answer is "The first choice"
Explanation:
The circuit switches are used to provide a network that connects among the two parties, which are dedicated for the use to the parties for the length of the connection. It is a familiar technique, that is used to create a network for communication, which is used on telecalls. It also enables the hardware and circuits to share among the users, and for every user, it has direct access to the circuit during the use of the network.
Answer
You can whisper,write notes, text, e-mail, body language, and even use a sign language.
Explanation
By adjusting your communication in these ways one can decrease noise. Whispering is where you speak softly to someone where the information will be only heard by the person you are addressing and this will not make any noise. Also use of a text or an email where the message will be delivered electronically but not verbally. Then there is the body language which is a nonverbal but a physical behavior used to express a certain information or also one can use sign language.
The answer is C. double-click
Answer:
Option B, Vanishing point
Explanation:
The complete question is
Where do the projection lines converge in a perspective sketch?
A. the ground line
B. the vanishing line
C. the eye point
D. the horizon line
Solution
A point lying on the image of a perspective drawing where the drawings ( two-dimensional perspective projections) of two parallel line meet in three dimensional space is known as Vanishing point
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
