Answer:
The answer is "ICANN"
Explanation:
In the given question some information is missing, that is option, which can be described as follows:
A) IAB
B) ICANN
C) W3C
D) ISOC
It manages the installation and processes of multiple databases concerning the Network domain and provides a stable and safe networking service, and wrong choices were explained as follows:
- IAB, It provides a protocol to manage IETF, that's why it is wrong.
- W3C is used in web development.
- ISOC is used to provide internet accessibility.
Answer:
The answer is below
Explanation:
Elements of Server software that is built-in, in Windows workstations are:
1. Hard drives,
2. RAM (Random Access Memory)
3. Processors
4. Network adapters.
Windows Professional OS is not considered a server due to the following:
1. Windows Professional OS has a limit on the number of client connections it allowed.
2. Unlike Server, Professional OS uses less memory
2. In comparison to Server, Professional OS uses the CPU less efficiently
4. Professional OS is not built to process background tasks, unlike Server that is configured to perform background tasks.
I think that the answer wiloukd be B. For #31 and for #32 A.
Answer:
#include <iostream>
using namespace std;
class Str{ ///baseclass
public :
string super_str;
string getStr()
{
return super_str;
}
void setStr(string String)
{
super_str=String;
}
};
class str : public Str{ //inheriting Str publicly
public :
string sub_str;
string getstr()
{
return sub_str;
}
void setstr(string String)
{
sub_str=String;
}
bool notstartswith()
{
int n=sub_str.length(); //to find length of substr
bool flag=false;
for(int i=0;i<n;i++) //Loop to check beginning of Str
{
if(super_str[i]!=sub_str[i])
{
flag=true;
break;
}
}
return flag;
}
};
int main()
{
str s; //object of subclass
s.setStr("Helloworld");
s.setstr("Hey");
if(s.notstartswith()==1) //checking if str is substring of Str
cout<<"Str does not start with str";
else
cout<<"Str starts with str";
return 0;
}
OUTPUT :
Str does not start with str
Explanation:
Above program is implemented the way as mentioned. for loop is being used to check the beginning of the str starts with substring or not.
Answer:
numbers = 1:1:100;
for num=numbers
remainder3 = rem(num,3);
remainder5 = rem(num,5);
if remainder3==0
disp("Yee")
else
if remainder3 == 0 && remainder5 == 0
disp ("Yee-Haw")
else
if remainder5==0
disp("Haw")
else
disp("Not a multiple of 5 or 4")
end
end
end
end
Explanation:
- Initialize the numbers variable from 1 to 100.
- Loop through the all the numbers and find their remainders.
- Check if a number is multiple of 5, 3 or both and display the message accordingly.
