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Ket [755]
2 years ago
9

Edie wants to visit her university's website. What software application should she use?

Computers and Technology
2 answers:
grandymaker [24]2 years ago
6 0
For PLATO it's C-Browser
MatroZZZ [7]2 years ago
3 0

Answer:

C. browser; I am 100% sure

Explanation:

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All animations on the world wide web are flash animations
natima [27]

Not every animation on the web is Flash but currently the majority is. There exists simple GIF based animations as well as other technologies for animation such as Silverlight and HTML5. Most believe HTML5 will eventually replace Flash.

6 0
3 years ago
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Pharming involves redirecting users to a fraudulent Web site even when the user has typed in the correct address in the Web brow
Anna35 [415]

Answer:

Pharming involves:

redirecting users to a fraudulent website even when the user has typed in the correct address in the web browser.          

Explanation:

  • When a hacker steals your confidential and private information by installing a malicious code on your computer which misdirects you to fake and fraudulent websites, this is called pharming.
  • Hacker installs virus on your computer which manipulates the host files on your computer.
  • The hacker uses DNS cache poisoning to attack DNS services on your computer which makes many users to enter fraudulent site, unknowingly, as they type the URL of the real website.
  • These fraudulent website developed by hacker can install virus on your computer and steal confidential information from your computer.
  • Pharming is done by hackers in order to steal sensitive data from a person's computer.
8 0
2 years ago
Is it possible for the front and rear references in a circular array implementation to be equal?
azamat

Answer:

Yes it is possible for the following cases:-

  1. When the queue is full.
  2. When the queue is empty.

Explanation:

When the queue is full the the front and the rear references in the circular array implementation are equal because after inserting an element in the queue we increase the rear pointer.So when inserting the last element the rear pointer will be increased and it will become equal to front pointer.

When the queue is empty the front and rear pointer are equal.We remove an element from queue by deleting the element at front pointer decreasing the front pointer when there is only one element and we are deleting that element front and rear pointer will become equal after deleting that element.

7 0
2 years ago
What are two best practices for creating ads? (Choose two.) Select All Correct Responses Implement one ad per ad group per keywo
salantis [7]

The two best practices for creating ads are:

  • Include three to five ads and at least three extensions in each ad group.
  • Optimize the campaign’s ad rotation for clicks or conversion actions.

<h3>How do one Create one responsive search ad per ad group?</h3>

In any ad group, one need to have at least a single responsive search ad.

Note that this will help to optimize one's performance, and as such, the two best practices for creating ads are:

  • Include three to five ads and at least three extensions in each ad group.
  • Optimize the campaign’s ad rotation for clicks or conversion actions.

Learn more about ads from

brainly.com/question/1658517

#SPJ1

5 0
1 year ago
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

5 0
3 years ago
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