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ch4aika [34]
3 years ago
7

You type a web address in your web browser. order the steps, which describe the internet name resolution process for the web add

ress.

Computers and Technology
1 answer:
padilas [110]3 years ago
6 0
First of all let's make it clearer, what is meant by "domain name resolution"?
"Domain name resolution" is the process of translating your web address into its corresponding IP address. e.g. translating (www.brainly.com to 104.16.32.180)
Computers cannot understand the human language they only can understand numbers that's why we need this process.

So, the process (in a very simple way) is divided into four steps:
1) Lookup "root nameservers"
2) Lookup "Top Level Domain TDL nameservers"
3) Lookup "domain nameservers"
4) Lookup "subdomain nameservers" [if exist]

If you still confused please check out this video https://www.youtube.com/watch?v=BCjUbpIzRs8&index=10&list=PL4bq-KmCeyBmMvBB3-RT6ikdp8ljV9GeT

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Changing the configuration of a database falls under which category of databaseâ administration?

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3 years ago
If someone said to you, “Pseudocode is a waste of time! It’s just more work because you end up writing the same program twice,”
myrzilka [38]

Answer:

It allows the designer to focus on main logic without being distracted by programming languages syntax.

* Since it is language independent, it can be translated to any computer language code.

* It allows designer to express logic in plain natural language.

* It is easier to write actual code using pseudocode.

* Unlike algorithms, pseudocodes are concise so pseudocodes are more readable and easier to modify.

Explanation:

8 0
3 years ago
It’s been six months since the disk crash at CSM Tech Publishing, and the owner is breathing a little easier because you install
Semmy [17]

Answer:

Storage Spaces

Explanation:

The feature that would best accommodate his needs would be the Storage Spaces feature that has been implemented in the Windows and Windows Server systems. This feature basically takes various hard drives and combines them together, this ultimately acts as copies of each other which protect the data of each one in the unfortunate case that one of the drives tends to fail. This feature also allows you to add more drives to the existing ones with very minimal effort without restricting configuration. Which would solve the needs that are needed in this scenario.

5 0
3 years ago
Compare Fibonacci (recursion vs. bottom up)
ipn [44]

Answer:

C++ code explained below

Explanation:

#include<bits/stdc++.h>

#include <iostream>

using namespace std;

int FiboNR(int n)

{

int max=n+1;

int F[max];

F[0]=0;F[1]=1;

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

{

F[i]=F[i-1]+F[i-2];

}

return (F[n]);

}

int FiboR(int n)

{

if(n==0||n==1)

return n;

else

return (FiboR(n-1)+FiboR(n-2));

}

int main()

{

long long int i,f;

double t1,t2;

int n[]={1,5,10,15,20,25,30,35,40,45,50,55,60,65,70,75};

cout<<"Fibonacci time analysis ( recursive vs. non-recursive "<<endl;

cout<<"Integer FiboR(seconds) FiboNR(seconds) Fibo-value"<<endl;

for(i=0;i<16;i++)

{

clock_t begin = clock();

f=FiboR(n[i]);

clock_t end = clock();

t1=double(end-begin); // elapsed time in milli secons

begin = clock();

f=FiboNR(n[i]);

end = clock();

t2=double(end-begin);

cout<<n[i]<<" "<<t1*1.0/CLOCKS_PER_SEC <<" "<<t2*1.0/CLOCKS_PER_SEC <<" "<<f<<endl; //elapsed time in seconds

}

return 0;

}

7 0
3 years ago
The following problem illustrates the way memory aliasing can cause unexpected program behavior. Consider the following function
laila [671]

Answer:

A

Explanation:

The value will be the original stored in the int variables. This is because the method swap(int xp, int yp) accepts xp and yp parameters by value (They are passed by value and not by reference). Consider the implementation below:

public class Agbas {

   public static void main(String[] args) {

   int xp = 2;

   int yp = xp;

   swap(xp,yp); //will swap and print the same values

       System.out.println(xp+", "+yp); // prints the original in values 2,2

 int xp = 2;

   int yp = 5;

   swap(xp,yp); // will swap and print 5,2

       System.out.println(xp+", "+yp); // prints the original in values 2,5

   }

   public static void swap(int xp, int yp){

       int temp = xp;

       xp = yp;

       yp = temp;

       System.out.println(xp+", "+yp);

   }

}

5 0
3 years ago
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