Answer:
Quick Communication & Correspondence
Explanation:
Another main advantage of using computers in the education field is the improvement in the quality of teaching-learning process and communication between students & teachers. For this, they use Microsoft PowerPoint to prepare electronic presentations about their lectures.
Answer:
Explanation:
iname=input("Enter the file name: ")
inputfile=open(iname,'r')
lines=[]
for line in inputfile:
lines.append(line)
inputfile.close()
print("The file has ",len(lines)," lines")
while True:
linenumber=int(input("Enter the line number or 0 to quit: "))
if linenumber==0:
break
elif linenumber > len(lines):
print("Error: line number must be less than ", len(lines))
else:
print(linenumber, " : ", lines[linenumber - 1])
Answer: Depends on the power of the computer /computers and the UPS
Explanation: For someone who had a big connected network of computers and printers we would plug the computer or server into the UPS as well as the monitor and sometimes a printer sometimes the monitor was plugged into the wall
When you get a UPS you must plug it in for a certain time could be hours or even a day. A USP has a big battery in it and you must charge it (sometimes it comes semi-charged but you must plug it in to strengthen it. )
there are two connector wires that you affix to the battery and you must put a lot of strength into it to make it secure.
So in closing, you can get away with two like a desktop and a laptop.
Read your documentation that comes with the UPS
No more info??? Because I can’t think of how to answer this
Answer:
In opp friend function is a function that gives the same access to private and protected data. It is declared in class that is granting access.
Explanation:
#include<iostream.h>
using namespace std;
class Sum
{
int a, b, add;
public:
void input()
{
cout << "Enter the value of l and m:";
cin >> l>>m; taking input from users
}
friend void add(sum &t);
void display()
{
cout << "The sum is :" << z;
}
};
void add(sum & p)
{
p.add = p.a + p.b;
}
int main()
{
sum p1;
p1.input();
add(p1);
p1.display();
return false;
if(display==5)
{
return true; //true is returned if sum is equal to 5
}
else //if they are not the same
{
return false;
}
}