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Amiraneli [1.4K]
2 years ago
15

What job does a front-end developer perform?

Computers and Technology
1 answer:
alina1380 [7]2 years ago
5 0

Answer:

They manage making websit/es beautiful. Basically how brainly is setup? The layout is nice? Front end developers are responsible for why webpages look pretty or nicely done. <em>fun fact, my dream job is a front end developer!</em>

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You have a primary zone stored in the myzone.local.dns file. Some devices that aren't domain members are creating dynamic DNS re
Elodia [21]

Answer:

Correct answer is (C) Configure the Store the zone in Active Directory option.

Explanation:

To create dynamic DNS records in the zone, the first thing is to add an Active Directory data store by configuring the data store name, connections, credentials, and search attributes.

5 0
3 years ago
A force of 50n acts on a body of mass 5kg. calculate acceleration produced .​
guapka [62]

Answer:

Force = 50n

mass = 5kg

Force = m * acc

50 = 5 * A

A = 50/5

A = 10 m/s^2

HOPE IT HELPS!!!

Explanation:

7 0
3 years ago
A _______ record is responsible for resolving an IP to a domain name.
lianna [129]
An "Address" record also know as "host" record
3 0
3 years ago
Read 2 more answers
You will complete 3 tasks. Task 1: Determine if the input number is prime or not You will ask the user for an integer You will d
GuDViN [60]

Answer:

Program is written in C++

#include<iostream>

using namespace std;

int main()

{

//1. Prime Number

int num;

cout<<"Input Number: ";

cin>>num;

int chk = 0;

for(int i =2; i <num;i++)

{

 if(num%i==0)

 {

  chk = 1;

  break;

 }

}

if(chk == 0)

{

 cout<<num<<" is prime"<<endl;

}

else

{

 cout<<num<<" is not prime"<<endl;

}

//2. Greatest Common Factor

int num1, num2, x, y, temp, gcf;  

cout<<"Enter two numbers: ";

cin>>num1;

cin>>num2;

x = num1;

y = num2;

while (y != 0) {

   temp = y;

   y = x % y;

   x = temp;

 }

 gcf = x;

 cout<<"Greatest Common Factor: "<<gcf<<endl;

 

// 3. LCM  

cout<<"Enter two numbers: ";

cin>>num1;

cin>>num2;

x = num1;

y = num2;

while (y != 0) {

   temp = y;

   y = x % y;

   x = temp;

 }

 gcf = x;

 int lcm =(num1 * num2)/gcf;

cout<<"Least Common Multiple: "<<lcm<<endl;

return 0;

}

Explanation:

<em>I've added the full source code as an attachment where I make use of comments to explain some lines</em>

Download cpp
6 0
3 years ago
The image below shows a weather service map
USPshnik [31]

Answer:I chose A

Explanation:

7 0
3 years ago
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