Answer:
#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n)
{
for(int j=2;j<=n-1;j++) //loop to check prime..
{
if(n%j==0)
return false;
}
return true;
}
int main(){
int n;
cout<<"Enter the integer"<<endl;//taking input..
cin>>n;
if(isPrime(n))//printing the message.
{
cout<<"The number you have entered is prime"<<endl;
}
else
{
cout<<"The number is not prime"<<endl;
}
return 0;
}
Output:-
Enter the integer
13
The number you have entered is prime
Explanation:
The above written program is in C++.I have created a function called isPrime with an argument n.I have used a for loop to check if the number is prime or not.In the main function I have called the function isPrime for checking the number is prime or not.
Answer:
A: Backstage view > Account Settings
Explanation:
Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
Answer:
a. A only
Explanation:
When air, is in excess of allowable amounts, it can cause the system to operate at pressures that are higher than normal. This means that the recycled refrigerant is operating at high pressure. Thus Technician A is correct.
However, Technician B is incorrect because unlike reclaimed refrigerants where the process strip the refrigerant of impurities making it to meet the standards of a new refrigerant, the recycled refrigerant is not as pure as it's contaminants are only reduced.
Therefore, only technician A is correct.
Answer:
Explanation: What is that word you typed?
