I think u mean by comparing them so basically interest assessments are when u have interest in them and the value ones are when they actually mean to u and u take them serious thats pretty much i know and please leave a thanks
Answer:
a. A hardware and software solution for remote workers, providing users with a data-encrypted gateway through a firewall and into a corporate network
Explanation:
VPN is the initial for Virtual Private Network. it is a highly secured channel that is encrypted end to end for for exchanging information via a public network like the internet.
Answer:
1)Select the cells you want to shade.
2)On the Home tab of ribbon select Conditional Formatting > New Rule
3)Select 'Use a formula to determine which cells to format' > enter your formula in the 'Edit the Rule Description' field.
4)Click the Format button and from the Format Cells dialog box select the Fill tab > choose your weapon (colour, pattern, fill effect etc.):
Explanation:
Answer:
File Or Folder I think it is
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