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Lina20 [59]
3 years ago
5

the development of modern computer system have been evolutionary, discuss this phenomena and further explain how current trends

in computing would impact future computer systems develpoment
Computers and Technology
1 answer:
Volgvan3 years ago
5 0

Answer: Why are there so many of these?

The modern computer system has been evolutionary. Modern computers have allowed for us to complete hard tasks rather easily. With modern computers, we are able to calculate advanced equations with the click of a button. Trends in computing like RTX can rapidly improve performance, rendering, and tons of other graphical heavy actions. In September of 2018, we got the first RTX card, now in 2021, RTX has become exponentially better than what it was in 2018. With the modern computers comes quantum computers. Quantum computing is estimated to be 100 million times faster than any modern computer. IBM, the company to make the first computer stated that we are in the decade of quantum computing. And who knows? Maybe quantum computing will come sooner rather than later.

Explanation:

You probably should modify this a little so you don't get in trouble. I researched each topic I put into this for about an hour so I hope this is what you need. If it isn't, please don't hesitate to tell me what I didn't add or what went wrong.

Have a good day :)

You might be interested in
Write a program that takes a point (x,y) from theuser and find where does the point lies. The pointcan
agasfer [191]

Answer:

C++ Program .

#include<bits/stdc++.h>

using namespace std;

int main()

{

int x,y;//declaring two variables x and y.

string s;//declaring string s..

cout<<"enter x and y"<<endl;

cin>>x>>y;//taking input of x and y..

if(x>=0 &&y>=0) //condition for 1st quadrant..

cout<<"the point lies in 1st Quadrant"<<endl;

else if(x<=0 &&y>=0)//condition for 2nd quadrant..

cout<<"the point lies in 2nd Quadrant"<<endl;

else if(x>=0 &&y<=0)//condition for 3rd quadrant..

cout<<"the point lies in 3rd Quadrant"<<endl;

else //else it is in  4th quadrant..

cout<<"the point lies in 4th Quadrant"<<endl;

cout<<"enter n to terminate the program"<<endl;

while(cin>>s)//if the user has not entered n the program will not terminate..

{

   if(s=="n")

   {

       cout<<"the program is terminated"<<endl;

       exit(0);

   }

   cout<<"you have not entered n please enter n to terminate the program<<endl;

}

}

Explanation:

The above written program is for telling the point lies in which quadrant.I am first declaring two variables x and y.Then after that taking input of x and y after that checking in which quadrant the point lies.

Taking input of the string s declared earlier for program termination the program will keep running until the user enters n.

4 0
3 years ago
Topic: Video Games.
Butoxors [25]

Answer:

Fire Emblem

Explanation:

4 0
3 years ago
A manufacturing company has several one-off legacy information systems that cannot be migrated to a newer OS due to software com
Digiron [165]

The resiliency technique which would provide the aforementioned capabilities is: D. Full backups.

An operating system (OS) can be defined as a system software which is pre-installed on a computing device, so as to manage or control software application, computer hardware and user processes.

In this scenario, a manufacturing company cannot migrate its several one-off legacy information systems (IS) to a newer operating system (OS), due to software compatibility issues.

Resiliency can be defined as a measure of the ability of a network, server, storage system, computing system or data center, to recover quickly and continue operating when it experience adverse conditions such as:

  • Power failure.
  • Equipment failure.
  • Data loss.

In Computers and Technology, there are four (4) main resiliency technique and these include:

I. Redundancy.

II. RAID 1+5.

III. Virtual machines.

IV. Full backups.

Full backup is a resiliency technique which create backups of the systems for recovery and it allows operating system (OS) patches to be installed on computer systems.

Read more: brainly.com/question/17586013

3 0
3 years ago
In JAVA please:
blagie [28]

Answer:

import java.util.Scanner;

public class ArraysKeyValue {

public static void main (String [] args) {

final int SIZE_LIST = 4;

int[] keysList = new int[SIZE_LIST];

int[] itemsList = new int[SIZE_LIST];

int i;

keysList[0] = 13;

keysList[1] = 47;

keysList[2] = 71;

keysList[3] = 59;

itemsList[0] = 12;

itemsList[1] = 36;

itemsList[2] = 72;

itemsList[3] = 54;  

/* Your solution goes here */

for ( i = 0; i < SIZE_LIST; i++){

 if (keysList[i]>50){

  System.out.println(itemsList[i] + " ");  }  }

System.out.println("");

}

}

Explanation:

I will explain the whole program flow.

  • There are two arrays here
  • keysList and itemsList
  • The first list (keysList) contains the following elements:

13 element at first position of the array (0th index)

47 element at second  position of the array (1st index)

71 element at third position of the array (2nd index)

59 element at fourth position of the array (3rd index)

  • The other list (itemsList) contains the following elements:

12 element at first position of the array (0th index)

36 element at second  position of the array (1st index)

72 element at third position of the array (2nd index)

54 element at fourth position of the array (3rd index)

  • The size of the array elements is fixed which is 4 and is stored in the variable SIZE_LIST.
  • Then the loop starts. The loop contains a variable i which is initialized to 0. First it checks if the value of i is less than the size of the list. It is true as SIZE_LIST=4 and i=0.
  • So the program control enters the body of the loop.
  • In first iteration, IF condition checks if the i-th element of the keysList is greater than 50. As i=0 So the element at 0th index of the keysList is 13 which is not greater than 50 so the body of IF statement will not execute as the condition evaluates to false. The value of i increments by 1 so now i becomes 1.
  • In next iteration loop again checks if the value of i is less than the size of the list which is true again so the body of the loop executes.
  • IF condition checks if the i-th element of the keysList is greater than 50. As i=1 So the element at 1st index of the keysList is 47 which is not greater than 50 so the body of IF statement will not execute as the condition evaluates to false. The value of i is incremented by 1 so now i becomes 2.
  • In next iteration loop again checks if the value of i is less than the size of the list which is true again as i= 2 which is less than SIZE_LIST so the body of the loop executes.
  • IF condition checks if the i-th element of the keysList is greater than 50. As i=2 So the element at 2nd index of the keysList is 71 which is greater than 50 so the body of IF statement is executed as the condition evaluates to true. So in the body of the IF statement there is a print statement which prints the i-th element of the itemsList. As i = 2 so the value at the index 2 of the itemsList is displayed in the output which is 72. Next value of i is incremented by 1 so now i becomes 3.
  • In next iteration loop again checks if the value of i is less than the size of the list which is true again as i= 3 which is less than SIZE_LIST so the body of the loop executes.
  • IF condition checks if the i-th element of the keysList is greater than 50. As i=3 So the element at 3rd index of the keysList is 59 which is greater than 50 so the body of IF statement is executed as the condition evaluates to true. So in the body of the IF statement there is a print statement which prints the i-th element of the itemsList. As i = 3 so the value at the index 3 of the itemsList is displayed in the output which is 54. Next value of i is incremented by 1 so now i becomes 4.
  • In next iteration loop again checks if the value of i is less than the size of the list which is now false as i=4 which is equal to the SIZE_LIST= 4. So the loop breaks.
  • So the output of the above program is:

72

54

5 0
3 years ago
Analyze the following code. Is count &lt; 100 always true, always false, or sometimes true or sometimes false at Point A, Point
GaryK [48]

Answer:

Point A: Always True

Point B: Sometimes false

Point C: Always False

Explanation:

In the given code snippet. Point A is the first statement within the While loop the statement System.out.println("Welcome to Java!"); will only be executed if the while condition evaluates to true.

At Point B, The statement count++ increases the value of the counter at every iteration, while it will be true for most occasions, at the last increament, this statement will be false that is at count=100, The condition will be false at this point just before program execution breaks out of the loop

Point C is outside of the loop, this happens when the given condition is no longer true.

8 0
3 years ago
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