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Serjik [45]
3 years ago
8

Brian wants to conduct an online search using a certain phrase he intends to use the word books that belong to the 1800's in his

search how should he use the word that in his search?
Computers and Technology
1 answer:
Andrews [41]3 years ago
7 0

Answer:

Use quotation marks (" ")

Explanation:

It depends on the search engine. For Google, you need to put the phrase inside of quotation (" ") marks. It also works for DogPile and Bing. I can't confirm whether it works with any other search engines, but I can confirm that it works with these three engines.

You might be interested in
Which of the following is NOT true about Parvati's web search? Group of answer choices a) She will enter the desired search text
shepuryov [24]

Answer:

The answer is letter A

Explanation:

<em>She will enter the desired search text in the query box. This statement is NOT TRUE</em>

3 0
4 years ago
Explain insert a ternary operator.?​
jek_recluse [69]

Explain insert ternary operator.?

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<h3>QUESTION;</h3>

Explain insert ternary operator.?

<h2>ANSWER;</h2>

  • <em><u>T</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>t</u></em><em><u>e</u></em><em><u>r</u></em><em><u>n</u></em><em><u>a</u></em><em><u>r</u></em><em><u>y</u></em><em><u> </u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u>r</u></em><em><u>a</u></em><em><u>t</u></em><em><u>o</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u> </u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u>r</u></em><em><u>a</u></em><em><u>t</u></em><em><u>o</u></em><em><u>r</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>a</u></em><em><u>t</u></em><em><u> </u></em><em><u>e</u></em><em><u>x</u></em><em><u>i</u></em><em><u>s</u></em><em><u>t</u></em><em><u>s</u></em><em><u> </u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>s</u></em><em><u>o</u></em><em><u>m</u></em><em><u>e</u></em><em><u> 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</u></em><em><u>t</u></em><em><u>h</u></em><em><u>a</u></em><em><u>n</u></em><em><u> </u></em><em><u>t</u></em><em><u>y</u></em><em><u>p</u></em><em><u>i</u></em><em><u>c</u></em><em><u>a</u></em><em><u>l</u></em><em><u> </u></em><em><u>o</u></em><em><u>n</u></em><em><u>e</u></em><em><u> </u></em><em><u>o</u></em><em><u>r</u></em><em><u> </u></em><em><u>t</u></em><em><u>w</u></em><em><u>o</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>a</u></em><em><u>t</u></em><em><u> </u></em><em><u>m</u></em><em><u>o</u></em><em><u>s</u></em><em><u>t</u></em><em><u> </u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u>r</u></em><em><u>a</u></em><em><u>t</u></em><em><u>o</u></em><em><u>r</u></em><em><u>s</u></em><em><u> </u></em><em><u>u</u></em><em><u>s</u></em><em><u>e</u></em><em><u>.</u></em><em> </em><em><u>l</u></em><em><u>t</u></em><em><u> 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</u></em><em><u>e</u></em><em><u>x</u></em><em><u>a</u></em><em><u>m</u></em><em><u>p</u></em><em><u>l</u></em><em><u>e</u></em><em><u>,</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>s</u></em><em><u>i</u></em><em><u>d</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>b</u></em><em><u>e</u></em><em><u>l</u></em><em><u>o</u></em><em><u>w</u></em><em><u> </u></em><em><u>J</u></em><em><u>a</u></em><em><u>v</u></em><em><u>a</u></em><em><u>S</u></em><em><u>c</u></em><em><u>r</u></em><em><u>i</u></em><em><u>p</u></em><em><u>t</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>d</u></em><em><u>e</u></em><em><u>.</u></em><em> </em><em><u>v</u></em><em><u>a</u></em><em><u>r</u></em><em><u> </u></em><em><u>n</u></em><em><u>u</u></em><em><u>m</u></em><em><u> </u></em><em><u>=</u></em><em><u>4</u></em><em><u>,</u></em><em><u> </u></em><em><u>m</u></em><em><u>s</u></em><em><u>g</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>"</u></em><em><u>"</u></em><em><u>;</u></em><em><u>i</u></em><em><u>f</u></em><em><u> </u></em><em><u>(</u></em><em><u>n</u></em><em><u>u</u></em><em><u>m</u></em><em><u> </u></em><em><u>=</u></em><em><u>=</u></em><em><u>=</u></em><em><u> </u></em><em><u>4</u></em><em><u> </u></em><em><u>)</u></em>

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

6 0
2 years ago
Create a program which will input data into a pipe one character at a time. Count the number of characters as they are written i
LenKa [72]

Answer:

a)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

 

#define BUFSIZE 16

int main(){

   //pipe descriptors

   char msg[BUFSIZE];

   char buf[BUFSIZE];

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //pipe creation successfull

   //write four messages

   sprintf(msg , "apple");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "boy");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "cat");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "dog");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   //read

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   close(pipefd[0]);

   close(pipefd[1]);

   return 0;

}

(b)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#include<wait.h>

#define BUFSIZE 16

int main(){

   //pipe descriptors

   char msg[BUFSIZE];

   char buf[BUFSIZE];

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //pipe creation successfull

   //write four messages

   if(fork() == 0){

       //this is child

       //close unused write end

       close(pipefd[1]);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading first message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading second message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading third message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading fourth message. Content is %s\n", buf);

       close(pipefd[0]);

       //exit from child

       exit(EXIT_SUCCESS);

   }

   //this is parent process

   //close unused read end

   close(pipefd[0]);

   sprintf(msg , "apple");

   printf("This is parent process. Writing first message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "boy");

   printf("This is parent process. Writing second message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "cat");

   printf("This is parent process. Writing third message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "dog");

   printf("This is parent process. Writing fourth message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   close(pipefd[1]);

   //wait for child

   wait(NULL);

   return 0;

}

(c)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#include<sys/wait.h>

#include<sys/types.h>

#include<signal.h>

typedef struct sigaction Sigaction;

unsigned long long size = 0;

void alarmhandler(int sig){

   //alarm fired writing blocked

   printf("Write blocked after %llu characters\n", size);

   exit(EXIT_SUCCESS);

}

int main(){

   //pipe descriptors

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //install handler

   sigset_t mask , prev;

   sigemptyset(&mask);

   sigaddset(&mask , SIGALRM);

   sigprocmask(SIG_BLOCK , &mask , &prev);

   Sigaction new_action;

   sigemptyset(&new_action.sa_mask);

   new_action.sa_flags = SA_RESTART;

   new_action.sa_handler = alarmhandler;

   sigaction(SIGALRM , &new_action , NULL);

   sigprocmask(SIG_SETMASK , &prev, NULL);

   while(1){

     

       //print size on multiple of 4

       if(size != 0 && size % 1024 == 0){

           printf("%llu characters in pipe\n", size);

       }

       //reset previous alarm

       alarm(0);

       //set new alarm for 5 seconds

       alarm(5);

       //write to pipe one character

       write(pipefd[1], "A", sizeof(char));

       size++;

   }

   return 0;

}

Explanation:

Output for a, b, c are pasted accordingly

4 0
4 years ago
Suppose we perform a sequence of stack operations on a stack whose size never exceeds kk. After every kk operations, we make a c
Ainat [17]

Answer:

The actual cost of  (n) stack operations will be two ( charge for push and pop )

Explanation:

The cost of (n) stack operations involves two charges which are actual cost of operation of the stack which includes charge for pop ( poping an item into the stack) which is one and the charge for push ( pushing an item into the stack ) which is also one. and the charge for copy which is zero. therefore the maximum size of the stack operation can never exceed K because for every k operation there are k units left.

The amortized cost of the stack operation is constant 0 ( 1 ) and the cost of performing an operation on a stack made up of n elements will be assigned as 0 ( n )

7 0
3 years ago
After installing the second hard drive . what two tasks need to be done and what do they do?
Sunny_sXe [5.5K]
Create a partition using the Disk Management program and format the hard drive.

The first point will allow the hard drive to be put to use, and the second point will clear the hard drive completely so it's ready for use. 
3 0
3 years ago
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