Answer:
The program in C++ is as follows:
#include<bits/stdc++.h>
using namespace std;
class Complex {
public:
int rl, im;
Complex(){ }
Complex(int Real, int Imaginary){
rl = Real; im = Imaginary;
}
};
int main(){
int real, imag;
cout<<"Real: "; cin>>real;
cout<<"Imaginary: "; cin>>imag;
Complex ComplexNum(real, imag);
cout<<"Result : "<< ComplexNum.rl<<" + "<<ComplexNum.im<<"i"<<endl;
}
Explanation:
See attachment for explanation
The answer is true.
Explanation:
The scanner class's methods are the methods in java.util, which allows the user to read values of various types. If the nextLine is issued after a numeric read and the numeric value is at the end of the line, nextLine returns the empty string.
The problem occurs when you click the enter key which is a new line \n character. nextInt() has only the integer but skips the new line \n.
To solve this problem, you have to add the input.nextLine() after reading the int it will consume the \n.
Hence, make input.nextLine(); call after input.nextint(); which reads till end of life.
Answer:
E. 172.16.18.255 255.255.252.0
Explanation:
Oh goodie, this is my home turf.
The answer is E) 172.16.18.255 255.255.252.0
This is because your subnet network ID includes mask \22, which means the ending with 255 255.255.252.0, which is standard for Class B. Only option E falls as an adequate host due to the network being 172.16.16 and broadcasting 16.19.
Answer:
Problem-solving. Why is problem-solving so valued? Companies face a lot of obstacles. Those better able to cope
Explanation:
