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<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em>
Answer:
//Here is the for loop in C.
for(n=10;n>0;n--)
{
printf("count =%d \n",n);
}
Explanation:
Since C is a procedural programming language.Here if a loop that starts with n=10; It will run till n becomes 0. When n reaches to 0 then loop terminates otherwise it print the count of n.
// here is code in C++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{ // variables
int n;
// for loop that runs 10 times
// when n==0 then loop terminates
for(n=10;n>0;n--)
{
cout<<"count ="<<n<<endl;
}
return 0;
}
Output:
count =10
count =9
count =8
count =7
count =6
count =5
count =4
count =3
count =2
count =1
Answer:
The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.
![rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)](https://tex.z-dn.net/?f=rank%28A%29%3Drank%5Cleft%20%28%20%5Cleft%20%5B%20A%7CB%20%5Cright%20%5D%20%5Cright%20%29%5C%3Aand%5C%3An%3Drank%28A%29)
Then satisfying this theorem the system is consistent and has one single solution.
Explanation:
1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.
![rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)](https://tex.z-dn.net/?f=rank%28A%29%3Drank%5Cleft%20%28%20%5Cleft%20%5B%20A%7CB%20%5Cright%20%5D%20%5Cright%20%29%5C%3Aand%5C%3An%3Drank%28A%29)

Then the system is consistent and has a unique solution.
<em>E.g.</em>

2) Writing it as Linear system


3) The Rank (A) is 3 found through Gauss elimination


4) The rank of (A|B) is also equal to 3, found through Gauss elimination:
So this linear system is consistent and has a unique solution.