Answer:
Hackers do so for various reasons, including the challenge of it
Explanation:
I would say it is this. Plz mark brainliest Thanks:)
When a machine says function suppressed after scanning a ticket, it implies that you have won some certain amount.
<h3>What is lottery wins?</h3>
This term connote that a person has a winning ticket in a lottery that is often owned by a government.
Note that if a winning ticket is scanned, the terminal often shows a message just for you and that suppress function implies that your the ticket has won something.
Learn more about lottery from
brainly.com/question/9216200
Answer:
well, as long as there are no right or wrong answers, don't:
look to closely at the screen, as it may mess up your eyes
hold a drink above the computer, as it may spill and cause "sticky keys"
go to sites that you know will give your a computer a virus, cause they cost hundreds of dollars to repair, and some aren't able to come out
go on illegal sites/do illegal operations to the computer itself, because then you won't have a computer
Explanation:
There are two ways to convert from hexadecimal to denary gcse method. They are:
- Conversion from hex to denary via binary.
- The use of base 16 place-value columns.
<h3>How is the conversion done?</h3>
In Conversion from hex to denary via binary:
One has to Separate the hex digits to be able to know or find its equivalent in binary, and then the person will then put them back together.
Example - Find out the denary value of hex value 2D.
It will be:
2 = 0010
D = 1101
Put them them together and then you will have:
00101101
Which is known to be:
0 *128 + 0 * 64 + 1 *32 + 0 * 16 + 1 *8 + 1 *4 + 0 *2 + 1 *1
= 45 in denary form.
Learn more about hexadecimal from
brainly.com/question/11109762
#SPJ1
Answer:
#include <string>
#include <iostream>
using namespace std;
int main() {
string userInput;
getline(cin, userInput);
// Here, an integer variable is declared to find that the user entered string consist of word darn or not
int isPresent = userInput.find("darn");
if (isPresent > 0){
cout << "Censored" << endl;
// Solution starts here
else
{
cout << userInput << endl;
}
// End of solution
return 0;
}
// End of Program
The proposed solution added an else statement to the code
This will enable the program to print the userInput if userInput doesn't contain the word darn
