You've already answered it, haven't you? Data is sent through a network on bundles called packets.
Answer
B. 30
Explanation:
Assuming the code is written like this:
1. num1 = int(input())
2. num2 = 10 + num1 * 2
3. print(num2)
4. num1 = 20
5. print(num1)
Line 3 will print 30 when the number 10 is inputted at line 1.
Remember to use Order of Operations! :)
Answer:
echo %username%
whoami
Explanation:
In windows the command used to display user name which you have logged in is echo %username%.Note that it is only for windows platform only .It works on all released windows platforms.
There is another command whoami it tells the domain name also along with the username.
You have to write all these commands on the command prompt.
Main Answer:
Your primary responsibility as a system analyst concerns :b.people
Sub heading:
what does a system analyst do?
Explanation:
1. A System analyst is responsible for monitoring the effectiveness of technology systems and analysing it's efficiency for business operation and clients'needs.
2.system analyst create specific modifications
Reference Link:
https\\brainly.com
Hashtag:#SPJ4
Answer:
return value =2.
Here the function f() returns the length of the substring we traversed before we find the same character at the equal index of two substrings.
Take the inputs s= “abcd” and t= “bccd”.
• Now, p1 points to s1, i.e., p1 points to the character ‘a’ of “abcd”. And similarly, p2 points to ‘b’ of “bccd”.
• Then we compare the values at p1 and p2, are not equal, so p1 and p2 both are incremented by 1.
• Now the characters ‘b’ and ‘c’ of “abcd” and “bccd” respectively are compared. They are not equal. So both p1 and p2 both are incremented by 1.
• Now, p1 points to ‘c’ of “ abcd” that is the element at index 2 of s. And p2 points to ‘c’ of “bccd” that is the element at index 2 of t. Here value at p1 and p2 becomes equal. So the break statement is executed. We stop moving forward.
• As p1 is pointing to index 2 and s is pointing to the base that is index 0, so p1-s = 2.
Explanation:
#include<stdio.h>
int f(char *s, char *t);
void main()
{
int k = f("abcd", "bccd");
printf("%d", k);
}
int f(char *s, char *t)
{
char *p1, *p2;
for(p1 = s, p2 = t; *p1 != '\0'&& *p2 != '\0'; p1++, p2++)
{
if (*p1 ==*p2)
break;
}
return (p1-s);
}
OUPUT is given as image