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-Dominant- [34]
3 years ago
14

"The pkill command terminates _________."

Computers and Technology
1 answer:
Helen [10]3 years ago
4 0

Answer:

d. all instances of a process with the same name matched by a regular expression

Explanation:

In Linux OS there is <em>kill</em> command that you can use to force applications to shut down. When you execute <em>kill</em> command, you are actually sending a signal to the system to force it to terminate the incorrectly behaving application.

The syntax for <em>kill</em> is:

$ kill [signal or option] PID

You have to know <em>PID</em> (Process IDentification number) of the desired process to complete this command.

The <em>pkill</em> command allows you to use advanced <em>regex</em> patterns and other matching criteria. Instead of using the <em>PID</em>, you can now terminate the application by entering the name of its process. For example, to shut down <em>Firefox</em> simply enter the command:

$ pkill firefox

Since it matches the regular expression pattern, you can also enter the name only partially, for example:

$ pkill fire

You might be interested in
¿como la imagen organiza la realidad?
Troyanec [42]

Answer:

Las imágenes son las percepciones visuales que las personas tienen respecto de la realidad que los rodea. Así, a través de la visión, las personas pueden interpretar el contexto en el cual se encuentran inmersos, organizando los distintos componentes de la realidad en la cual desarrollan sus vidas, para poder comprender entonces de qué modo proceder ante las diferentes eventualidades de la vida.

Es decir que, a través de las imágenes, y en conjunto con las demás percepciones sensoriales, los seres humanos pueden contextualizarse en un entorno en el cual se desenvuelven, organizando su vida y su realidad a futuro.

7 0
2 years ago
Which of the following are considered transactions in an information system?
Marrrta [24]

Answer: (C) All of them.

Explanation:

 All the given options are example of the transaction in the information system.

As, the money deposited in the bank account is the process that take place computerized for transaction purpose. Now a days we can easily done transaction through wire transfer at anywhere and anytime by using the information system technology.  

Students can easily study online and also record their answers in the online test by using the information system technology.  

Customers can also doing shopping online by adding various products and items in the online shopping cart by using various e-commerce websites like amazon, flip-cart etc.  

5 0
3 years ago
Write a function with two parameters, prefix (a string, using the string class from ) and levels (an unsigned integer). The func
Allisa [31]

Answer:

Here is the program:

#include <iostream>  //to use input output functions

#include <string>  // to use functions to manipulate strings

using namespace std;  //to identify objects cin cout

void function(string prefix, unsigned int levels){  // function that takes two parameters, a string, using the string class and levels an unsigned integer

   if (levels == 0) {  //if number of levels is equal to 0

       cout << prefix << endl;  //displays the value of prefix

       return;    }  

  for (int i = 1; i <=9 ; i++){  //iterates 1 through 9 times

       string sections = (levels == 1 ? "" : ".");  //if the number of levels is equal to 1 then empty space in sections variable otherwise stores a dot

       string output = prefix +  std::to_string(i) + sections;   // displays the string prefix followed by the section numbers. Here to_string is used to convert integer to string

       function(output, levels - 1);   } }   //calls function by passing the resultant string and levels-1  recursively to print the string prefix followed by section numbers

int main() {  // start of main function

   int level = 2;  //determines the number of levels

   function("BOX", level);  } //calls function by passing the string prefix and level value

Explanation:

The program has a function named function() that takes two parameters, prefix (a string, using the string class from ) and levels (an unsigned integer). The function prints the string prefix followed by "section numbers" of the form 1.1., 1.2., 1.3., and so on. The levels argument determines how many levels the section numbers have. If the value of levels is 0 means there is 0 level then the value of prefix is printed. The for loop iterates '1' through '9' times for number of digits in each level. If the number of levels is 1 then space is printed otherwise a dot is printed. The function() calls itself recursively to print the prefix string followed by section numbers.

Let us suppose that prefix = "BOX" and level = 1

If level = 1 then the loop for (int i = 1; i <=9 ; i++) works as follows:

At first iteration:

i = 1

i <=9 is true because value of i is 1

string sections = (levels == 1 ? "" : "."); this statement checks if the levels is equal to 1. It is true so empty space is stored in sections variable so,

sections = ""

Next, string output = prefix +  std::to_string(i) + sections; statement has prefix i.e BOX plus value of i which is 1 and this int value is converted to string by to_string() method plus sections has an empty space. So this statement becomes

string output = BOX + 1  

So this concatenates BOX with 1 hence output becomes:

output = BOX1

At second iteration:

i = 2

i <=9 is true because value of i is 2

string sections = (levels == 1 ? "" : "."); is true so

sections = ""

Next, string output = prefix +  std::to_string(i) + sections; statement becomes

string output = BOX + 2  

So this concatenates BOX with 1 hence output becomes:

output = BOX2

At third iteration:

i = 3

i <=9 is true because value of i is 3

string sections = (levels == 1 ? "" : "."); is true so

sections = ""

Next, string output = prefix +  std::to_string(i) + sections; statement becomes

string output = BOX + 3  

So this concatenates BOX with 1 hence output becomes:

output = BOX3

Now at each iteration the prefix string BOX is concatenated and printed along with the value of i. So at last iteration:

At last iteration:

i = 9

i ==9 is true because value of i is 9

string sections = (levels == 1 ? "" : "."); is true so

sections = ""

Next, string output = prefix +  std::to_string(i) + sections; statement becomes

string output = BOX + 9  

So this concatenates BOX with 1 hence output becomes:

output = BOX9

After this the loop breaks at i = 10 because the condition i<=9 becomes false. So the output of the entire program is:

BOX1                                                                                                                                          BOX2                                                                                                                                          BOX3                                                                                                                                          BOX4                                                                                                                                          BOX5                                                                                                                                          BOX6                                                                                                                                          BOX7                                                                                                                                          BOX8                                                                                                                                          BOX9  

The program along with the output is attached.

4 0
3 years ago
Compare the time spent on each of the above operations, and describe your observations. If an operation is too fast, you may wan
Artist 52 [7]

Here's the complete question below that clarifies what you need to do

<u>Explanation</u>:

"In this task, we will study the performance of public-key algorithms. Please prepare a file ( message.txt) that contains a 16-byte message. Please also generate an 1024-bit RSA public/private key pair. Then, do the following:

1)Encrypt message.txt using the public key; save the the output in message_enc.txt.

2)Decrypt message_enc.txt using the private key.

3)Encrypt message.txt using a 128-bit AES key.

<em><u>Compare the time spent on each of the above operations, and describe your observations. If an operation is too fast, you may want to repeat it for many times, i.e., 5000 times, and then take an average.</u></em>

<em><u> After you finish the above exercise, you can now use OpenSSL's speed command to do such a benchmarking. Please describe whether your observations are similar to those from the outputs of the speed command?</u></em>

3 0
3 years ago
Richard owns a chain of hardware stores. He wants to update the store website and write a numbered list of all the items in his
Digiron [165]
Use ordered list
  1. tag and put each item inside of a list item
  2. tag.
8 0
3 years ago
Read 2 more answers
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