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hichkok12 [17]
2 years ago
10

What are three ways of verifying legitimate right of access to a computer system?

Computers and Technology
1 answer:
dezoksy [38]2 years ago
7 0

Something you know (such as a password)

Something you have (such as a smart card)

Something you are (such as a fingerprint or other biometric method)

<h3>What are the 3 methods of authentication?</h3>
  • The three authentication factors are Knowledge Factor – something you know, e.g. password.
  • Possession Factor – something you have, e.g.mobile phone. Inherence Factor – something you are, e.g., fingerprint.

To learn more about it, refer

to brainly.com/question/22654163

#SPJ4

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To maintain your body temperature, your body converts chemical potential energy into thermal energy T or F
PilotLPTM [1.2K]
True
This is physics btw
CPE --> TE
4 0
3 years ago
Help me
denis23 [38]

Answer:

a = 6, b = 0

Explanation:

The loop ran 3 times before b == 0. "while ((b != 0)" is essentially saying: 'While b is not equal to 0, do what's in my loop'. Same general thing with "&& ((a / b) >= 0)". The "&&" is specifying that there should be another loop condition, while the rest states: 'as long as a ÷ b is greater than 0, do what's in my loop'. If both of these conditions are met, the loop will run, It will continue this until the conditions are not met. Hope that helped! :)

7 0
3 years ago
A brief communication used in businesses is called a _____.
Bad White [126]

Answer:

C. Memo

Explanation:

Hope this helps :)

6 0
3 years ago
Read 2 more answers
In the original UNIX operating system, a process executing in kernel mode may not be preempted. Explain why this makes (unmodifi
Elena L [17]

Answer:

the preemption is -> The ability of the operating

system to preempt or stop a currently

scheduled task in favour of a higher priority

task. The scheduling may be one of, but not

limited to, process or 1/0 scheduling etc.

Under Linux, user-space programs have always

been preemptible: the kernel interrupts user

space programs to switch to other threads,

using the regular clock tick. So, the kernel

doesn't wait for user-space programs to

explicitly release the processor (which is the

case in cooperative multitasking). This means

that an infinite loop in an user-space program

cannot block the system.

However, until 2.6 kernels, the kernel itself was

not preemtible: as soon as one thread has

entered the kernel, it could not be preempted to

execute an other thread. However, this absence

of preemption in the kernel caused several

oroblems with regard to latency and scalability.

So, kernel preemption has been introduced in

2.6 kernels, and one can enable or disable it

using the cONFIG_PREEMPT option. If

CONFIG PREEMPT is enabled, then kernel code

can be preempted everywhere, except when the

code has disabled local interrupts. An infinite

loop in the code can no longer block the entire

system. If CONFIG PREEMPT is disabled, then

the 2.4 behaviour is restored.

So it suitable for real time application. Only

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5 0
3 years ago
The program that solves problem 2 (a) on p.275 should be named as proj4_a.c. It must include and implement the following functio
avanturin [10]

Answer:

#include <iostream>

#include <cstring>

using namespace std;

bool isAPalindrome(char* palindrome);

int main()

{

   char palindrome[30];

   bool palindrome_check;

   cout << "Please enter an word or phrase.\n";

   cin.getline(palindrome, 30);

   palindrome_check = isAPalindrome(palindrome);

   if (palindrome_check = true)

   {

       cout << "Input is a palindrome\n";

   }

   else

   {

       cout << "Inputis not a palindrome\n;";

   }

system("pause");

return 0;

}

bool isAPalindrome(char* palindrome)

{

   char* front;  

   char* rear;  

front = palindrome;// starts at the left side of the c string

rear = (palindrome + strlen(palindrome)) - 1;//starts at the right side of the c-string. adds the c string plus the incriment value of s

while (front <= rear)

{

 if (front = rear)

 {

  front++;

  rear--;

 }

 else

 {

  return false;

 }

}

   return true;

}

7 0
3 years ago
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