Answered

Find answers to your most challenging questions with the help of IDNLearn.com's experts. Our Q&A platform is designed to provide quick and accurate answers to any questions you may have.

Write a complete Java program called Rooter that gets a positive integer called "start" from the user at the command line and then finds the squareroot of every number from "start" down to 0. Use a while loop to count down. Print each square root on a separate line. Include data validation to ensure the user provides a positive integer. Assume that the user enters an integer, and for validation, just check to be sure it is positive. If the validation is not passed, provide the user with suitable feedback and stay in the program to let the user try again until valid input is received. Use the Math.sqrt(double a) method to find each square root.

Sagot :

MrRoyalidn

Answer:

The program in Java is as follows:

import java.util.*;

import java.lang.Math;

public class Rooter{

public static void main(String[] args) {

    Scanner input = new Scanner(System.in);

 int start;

 System.out.print("Start: ");

 start = input.nextInt();

 while(start<=0){

     System.out.print("Number must be positive\nStart: ");

     start = input.nextInt();  }

 while(start>=0){

     System.out.println(Math.sqrt(start));

     start--;  }

}

}

Explanation:

This declares start as integer

 int start;

This prompts the user for input

 System.out.print("Start: ");

This gets input for start

 start = input.nextInt();

The following is repeated until the user input is valid i.e. positive

 while(start<=0){

     System.out.print("Number must be positive\nStart: ");

     start = input.nextInt();  }

The following while loop prints the square root of each number till 0

 while(start>=0){

     System.out.println(Math.sqrt(start));

     start--;  }

Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.