A router, possibly. Thought I'm not for sure because I'm not in computers and technology
Answer:
answer:
#include <iostream>
#include<list>
using namespace std;
bool Greater(int x) { return x>3; } int main() { list<int>l; /*Declare the list of integers*/ l.push_back(5); l.push_back(6); /*Insert 5 and 6 at the end of list*/ l.push_front(1); l.push_front(2); /*Insert 1 and 2 in front of the list*/ list<int>::iterator it = l.begin(); advance(it, 2); l.insert(it, 4); /*Insert 4 at position 3*/ for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " "; /*Display the list*/ cout<<endl; l.erase(it); /*Delete the element 4 inserted at position 3*/ for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " "; /*Display the list*/ cout<<endl;
l.remove_if(Greater); for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " ";
/*Display the list*/
cout<<endl; return 0;
}
Answer:
C. sqrt(Math)
Explanation:
All but one of options A to E are is not a static method.
Only option C is a static method. The sqrt() is a static method of Math, that can always be used as Math.sqrt() is used;
The Math class defines all of its methods to be static. Invoking Math methods is done by using Math as a method rather than a variable of type Math; this means that sqrt(Math) doesn't rely on instance variables and don't need to be overridden, unlike others.
Lastly, sqrt(Math) is a static method because unlike other options, it is an utility method, and it is relevant to computations on primitive data types.
The purpose of the static method is in large part to offer a standard library of functions, and it doesn't need to be applied directly to an object.
Answer: Option C is correct
Explanation:
Option C is correct
