It needs an Operating System like a cable or something that will help it operate look for more and double check
Answer:
#include <stdio.h>
void interchangeCase(char phrase[],char c){
for(int i=0;phrase[i]!='\0';i++){
if(phrase[i]==c){
if(phrase[i]>='A' && phrase[i]<='Z')
phrase[i]+=32;
else
phrase[i]-=32;
}
}
}
int main(){
char c1[]="Eevee";
interchangeCase(c1,'e');
printf("%s\n",c1);
char c2[]="Eevee";
interchangeCase(c2,'E');
printf("%s\n",c2);
}
Explanation:
- Create a function called interchangeCase that takes the phrase and c as parameters.
- Run a for loop that runs until the end of phrase and check whether the selected character is found or not using an if statement.
- If the character is upper-case alphabet, change it to lower-case alphabet and otherwise do the vice versa.
- Inside the main function, test the program and display the results.
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brainly.com/question/10742732
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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.
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.
