<span>This statement is true.
When you are making a right turn you will turn into the right lane of the roadway that you entered into. In the countries that drive on the right side, you will always stay on the right side even when turning. If you do not you will enter the lane of the vehicles that are going in the opposite direction which can lead to some pretty serious, life threatening situations. </span>
Answer:
The program completed in a third of the time with six computers versus one computer.
Explanation:
The speedup of a parallel solution is measured in the time it took to complete the task sequentially divided by the time it took to complete the task when done in parallel (or put succinctly, speedup = sequential/parallel).A speedup of 3 means that the program took three times as long on one computer, which is the same thing as saying it took a third of the time with all six computers working in parallel.
Hey there,
Your question states: <span>Which careers have the highest minimum experience requirement?
I believe that your correct answer would be "</span>network administrator". Now the reason why this would be the answer is not just because I say that, it's because being a administrator does not need high experience requirement. It's a basic thing to do. This is why being a "network administrator" would be <span>the highest minimum experience requirement.
_____________________________________________________________
~Jurgen</span>
Answer:
Explanation:
#include <iostream>
using namespace std;
int main()
{
string cwords[8]={"ten", "fading", "post", "card", "thunder", "hinge", "trailing", "batting"};
string temp="";
for (int i=0; i<8; i++)
{
temp=cwords[i];
int j=temp.length();
if (temp[j-3]=='i' && temp[j-2]=='n' && temp[j-1]=='g')
{
cout<<temp<<endl;
}
}
return 0;
}
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