Answer:
Step by step explanation along with code and output is provided below
Explanation:
#include<iostream>
using namespace std;
// print_seconds function that takes three input arguments hours, mints, and seconds. There are 60*60=3600 seconds in one hour and 60 seconds in a minute. Total seconds will be addition of these three
void print_seconds(int hours, int mints, int seconds)
{
int total_seconds= hours*3600 + mints*60 + seconds;
cout<<"Total seconds are: "<<total_seconds<<endl;
}
// test code
// user inputs hours, minutes and seconds and can also leave any of them by entering 0 that will not effect the program. Then function print_seconds is called to calculate and print the total seconds.
int main()
{
int h,m,s;
cout<<"enter hours if any or enter 0"<<endl;
cin>>h;
cout<<"enter mints if any or enter 0"<<endl;
cin>>m;
cout<<"enter seconds if any or enter 0"<<endl;
cin>>s;
print_seconds(h,m,s);
return 0;
}
Output:
enter hours if any or enter 0
2
enter mints if any or enter 0
25
enter seconds if any or enter 0
10
Total seconds are: 8710
Answer:
It 5
Explain
number 28 is A and number 29 is 5
Answer:TRUE!!!!!!!!!!!!!!!!!!!!!
Explanation:
If 29 bits of the 32 available addressing bits are used for the subnet, then only 3 bits giving 2^3=8 combinations remain for the host addresses.
In reality, the all zeros and all ones addresses are reserved. So, 8 addresses can exist, but 6 of those are available.
The way the question is formulated it seems the answer 8 is what they're after.