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strojnjashka [21]
3 years ago
13

How has the internet has changed employment

Computers and Technology
2 answers:
Alex17521 [72]3 years ago
7 0

Answer:

It provided virtural jobs instead of physical labor

Explanation:

IgorLugansk [536]3 years ago
6 0

Answer:

Freelance

The internet has also changed the types of jobs available to us. Because it makes it easier to hire and work with people from all over the world, freelancing and independent contract work has become more and more normal.

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Write a static method that implements a recursive formula for factorials. Place this method in a test program that allows the us
matrenka [14]

Answer:

Written in Java

import java.util.*;

public class Main {

  public static int fact(int n) {

     if (n == 1)

        return n;

     else

        return n * fact(n - 1);

  }

  public static void main(String[] args) {

     int num;

     Scanner input = new Scanner(System.in);

     char tryagain = 'y';

     while(tryagain == 'y'){

     System.out.print("Number: ");

     num = input.nextInt();

     System.out.println(num+"! = "+ fact(num));

     System.out.print("Try another input? y/n : ");

     tryagain = input.next().charAt(0);

}        

  }

}

Explanation:

The static method is defines here

  public static int fact(int n) {

This checks if n is 1. If yes, it returns 1

     if (n == 1)

        return n;

If otherwise, it returns the factorial of n, recursively

     else

        return n * fact(n - 1);

  }

The main method starts here. Where the user can continue executing different values of n. The program keep prompting user to try again for another number until user signals for stoppage

  public static void main(String[] args) {

This declares num as integer

     int num;

     Scanner input = new Scanner(System.in);

This initializes tryagain as y

     char tryagain = 'y';

This checks if user wants to check the factorial of a number

     while(tryagain == 'y'){

This prompts user for input

     System.out.print("Number: ");

This gets user input

     num = input.nextInt();

This passes user input to the function and also prints the result

     System.out.println(num+"! = "+ fact(num));

This prompts user to try again for another value

     System.out.print("Try another input? y/n : ");

This gets user response

     tryagain = input.next().charAt(0);

}        

Download txt
4 0
2 years ago
Write two recursive versions of the function minInArray. The function will be given a sequence of elements and should return the
Viktor [21]

Answer:

Following are the code to this question:

#include <iostream>//defining header file

using namespace std;//using namespace

int minInArray1(int arr[],int arrSize)//declaring method minInArray1

{

if(arrSize == 1)//use if block to check arrSize value is equal to 1  

{

return arr[0];//return first element of array

}

else //defining else block

{

int max= minInArray1(arr, arrSize-1);//use integer variable max to call method recursively  

if(arr[arrSize-1] < max)//use if block to check array value  

{

max = arr[arrSize-1];//use max to hold array value

}

return max;//return max variable value

}

}

int minInArray2(int arr[], int low, int high)//defining a method minInArray2  

{

if(low == high) //use if to check low and high variable value are equal

{

return arr[low];//return low variable value

}

else //defining else block

{

int minimum = minInArray2(arr, low+1, high);//defining integer variable minimum to call method minInArray2 recursively

if(arr[low] < minimum)

{

minimum = arr[low];//use minimum variable to hold array value

}

return minimum;//return minimum value

}

}

int main()//defining main method

{

int arr[10] = { 9, -2, 14, 12, 3, 6, 2, 1, -9, 15 };//defining an array arr

int r1, r2, r3, r4;//defining integer variable

r1 = minInArray1(arr, 10);//use r1 variable to call minInArray1 and hold its return value

r2 = minInArray2(arr, 0, 9);//use r1 variable to call minInArray2 and hold its return value

cout << r1 << " " << r2 << endl; //use print method to print r1 and r2 variable value

r3 = minInArray2(arr, 2, 5);//use r3 variable to call minInArray1 and hold its return value

r4 = minInArray1(arr + 2, 4); //use r4 variable to call minInArray2 and hold its return value

cout<<r3<< " " <<r4<<endl; //use print method to print r3 and r4 variable value

return 0;

}

Output:

please find the attached file.

Explanation:

In the given code two methods, "minInArray1 and minInArray2" is defined,  in the "minInArray1" it accepts two-variable "array and arrSize" as the parameter, and in the "minInArray2" method it accepts three integer variable "array, low, and high" as the parameter.

  • In the "minInArray1" method, and if the block it checks array size value equal to 1 if the condition is true it will return the first element of the array, and in the else block the max variable is defined, that calling method recursively
  • and store its value.
  • In the "minInArray2" method, an if the block it checks low and high variable value is equal. if the condition is true it will return a low array value. In the next step, the minimum value is defined, which checks the element of the array and uses a low array to store its value.
  • In the main method an array and four integer variable "r1, r2, r3, and r4" is defined, which calls two methods "minInArray1 and minInArray2" and use print method to print its value.
7 0
2 years ago
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