The Microsoft button was exclusively used in the 2007 versions of word, excel, powerpoint, access and outlook. It has all of the different options like to save, print, open ect. It was located in the top left hand corner of the screen. It looked a bit like this (see attachment).
Answer:
The last character in the string processed is "t".
Explanation:
The index start from 0 and the len(fruit) is 5.
When index is 0 which is less than 5; "f" is printed
When index is 1 which is less than 5; "r" is printed
When index is 2 which is less than 5; "u" is printed
When index is 3 which is less than 5; "i" is printed
When index is 4 which is less than 5; "t" is printed
When index is 5, the condition is false; so the loop content is not executed.
Therefore, the last character traversed in the string "fruit" is "t".
Answer:
<u>720</u> possible PIN can be generated.
Explanation:
To calculate different number of orders of digits to create password and PIN, we calculate permutation.
Permutation is a term that means the number of methods or ways in which different numbers, alphabets, characters and objects can arranged or organized. To calculate the permutation following formula will be used:
nPr = n!/(n-r)!
there P is permutation, n is number of digits that need to be organize, r is the size of subset (Number of digits a password contains)
So in question we need to calculate
P=?
where
n= 10 (0-9 means total 10 digits)
r= 3 (PIN Consist of three digits)
So by using formula
10P3 = 10!/(10-3)!
=10!/7!
= 10x9x8x7!/7!
= 10x9x8
= 720
Paste special allows the item being transferred to be formatted in several different ways. Paste is just going to be transferred the exact way you copied it.
Answer:
See Explaination
Explanation:
#include <iostream>
#include <string.h>
using namespace std;
char *mixem(char *s1, char *s2);
int main() {
cout << mixem("abc", "123") << endl;
cout << mixem("def", "456") << endl;
return 0;
}
char *mixem(char *s1, char *s2) {
char *result = new char[1 + strlen(s1) + strlen(s2)];
char *p1 = s1;
char *p2 = s2;
char *p = result;
while (*p1 || *p2) {
if (*p1) {
*p = *p1;
p1++;
p++;
}
if (*p2) {
*p = *p2;
p2++;
p++;
}
}
*p = '\0';
return result;
}
