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muminat
3 years ago
15

The fundamental building block in every Hypertext Markup Language (HTML) document is the _____ tag, which marks a component in t

he document.
Computers and Technology
2 answers:
AVprozaik [17]3 years ago
6 0

Answer:

<div></div>

Explanation:

The <div></div> tag allows you to build containers to the text that you want to include in your html

dolphi86 [110]3 years ago
5 0

Answer:

The fundamental building block in every Hypertext Markup Language (HTML) document is the element tag.

Explanation:

Element describes the structure of any HTML document or a web page. An element will contain “Start and End tags”. There are few elements which contains start tag but does not contain an end tag.

There are elements to represent heading, paragraph, a body tag to define a body, anchor tag to link one link page to another. These tag contains both start and end tag. There are few tags where there is a start tag and no end tag.

Eg. Break tag which includes an empty line in the web page.

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Your task is to write a C program that measures the latencies of various system calls. In particular, you want to know 1) the co
tensa zangetsu [6.8K]

Answer and Explanation:

#include <stdio.h>

#include<fcntl.h>

#include <sys/time.h>

#include<time.h>

#define MAX 1000

int main()

{

int pid;

int i,fd ;

char c[12];

FILE *fp;

struct timeval start,end;

double time1,time2,time3;

//open file for writing

fp=fopen("output.txt","w");

 

if(!fp)

{

printf("Not able to open the file output.txt\n");

return -1;

}

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system(pid = getpid());

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time1 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

//to get micro second , divide multiply time by 1000000 , to get nano multiply time by 1000000000

printf("getpid(): %.10f %.10f\n",time1*1000000,time1*1000000000);

fprintf(fp,"getpid():%.10f %.10f\n",time1*1000000,time1*1000000000);

//in similar way execute other two commands ,open and read

 

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system(open("/dev/null", O_RDONLY ));

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time2 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

printf("open(): %.10f %.10f\n",time2*1000000,time2*1000000000);

fprintf(fp,"open():%.10f %.10f\n",time2*1000000,time2*1000000000);

//in similar way execute other two commands ,open and read

fd = open("/dev/dev",O_RDONLY );

//printf("fd = %d\n",fd);

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system( read(fd,c,10));

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time3 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

printf("read(): %.10f %.10f\n",time3*1000000,time3*1000000000);

fprintf(fp,"read(): %.10f %.10f\n",time3*1000000,time3*1000000000);

}

----------------------------------------------------------

//output

//I have written output to standard output also , you can remove that

getpid(): 0.1690000000 169.0000000000    

open(): 0.1890000000 189.0000000000    

read(): 3.1300000000 3130.0000000000

------------------------------------------------------

//Makefile content

prob2.o : prob2.c    

         gcc -c  prob2.c                                                                                                                                      

prob2 : prob2.o                                                                                                                                                

       gcc -o prob2 prob2.o                                                                                                                                    

all   :                                                                                                                                                        

       gcc -o prob2 prob2.c                                                                                                                                    

clean:                                                                                                                                                          

       rm -rf prob2.o  

---------------------------------------

use

$make all

then execute as below

$./prob2

3 0
3 years ago
Define operating system?explain the types of operating system on the basis of use.​
GalinKa [24]

Answer:

An operating system is a software that helps the any computers basic needs or functions.  For example, the windows on a computer is a operating system.

Explanation:

4 0
2 years ago
Sino ang pinaka matalino sa math sa buong mundo​
Svetllana [295]

Answer:

lahat ng tao ay matalino kasama ka na dun kaya lang di natin ito inilalabas dahil natatamad tayo

6 0
3 years ago
Read 2 more answers
[30 points, will mark Brainliest] Which of the following is the lowest hexadecimal value? Explain why. Options to chose; F2, 81,
larisa86 [58]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

First, we need to convert these hexadecimal numbers into decimal numbers, then we can easily identify which one is the lowest hexadecimal.

The hexadecimal numbers are F2, 81, 3C, and 39.

F2 = (F2)₁₆ = (15 × 16¹) + (2 × 16⁰) = (242)₁₀

81 = (81)₁₆ = (8 × 16¹) + (1 × 16⁰) = (129)₁₀

3C = (3C)₁₆ = (3 × 16¹) + (12 × 16⁰) = (60)₁₀

39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀

The 39 is the lowest hexadecimal number among the given numbers.

Because 39 hex is equal to 57 decimal.

39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀

5 0
3 years ago
_____________ describes the abstraction of web-based computers, resources, and services that system developers can utilize to im
Vika [28.1K]

Answer:

c. Cloud computing

Explanation:

Cloud computing -

It refers to the on - time need of the computer resource , in order to store data , computing power by the user , is referred to as cloud computing.

It refers to the availability of the data centers to the user .

This method enables the sharing of resources .

Hence, from the information of the question,

The correct option is c. Cloud computing .

6 0
3 years ago
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