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seraphim [82]
3 years ago
10

I will upvote....

Computers and Technology
1 answer:
Lady bird [3.3K]3 years ago
8 0
As we go up, the atmospheric pressure decreases constantly !

so when the balloon reaches that point when the inner pressure becomes more than outer , then the balloon starts expanding and eventually brust !
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When designing a suitable and safe flexibility training program, you should set reasonable and appropriate goals depending on wh
GalinKa [24]

Answer:

The following steps will help you design a safe and effective stretching program.

Explanation:

1. You will have to follow the ACSM's guidelines used for flexibility training.

2. evaluate your flexibility rate with the "sit-and-reach" test.

3. you have to also apply the basic principles of FITT in designing your own program.

4. have a "range-of-motion" tests performance.

5. make use of SMART guidelines when setting explicit flexibility goals.

7 0
3 years ago
Documenting Business Requirements helps developers control the scope of the system and prevents users from claiming that the new
tiny-mole [99]

Answer:

The answer is "True".

Explanation:

Industry demands documentation enables developers to monitor the system's range and protects users claims, that somehow the new system will not accomplish their business goals.

  • The main goal of this report is to offer everyone to be transparent, about what should be accomplished and when.  
  • It is the new business plan, that should be outlined in detail, that's why the given statement is true.
3 0
3 years ago
write a program that first asks the user to enter three real numbers using GUI input. Print the sum if all three numbers are pos
den301095 [7]

Answer:

Here is the complete program:

import javax.swing.JOptionPane;  //to use GUI in JAVA application

import java.util.Scanner;  //to take input from user at console

public class Main {

public static void main(String[] args) {  //start of main function body

 String num1, num2,num3;  //declare three variables to store strings of numbers for input dialog

 double number1, number2, number3, sum,product; //declare three double type variables to hold the real numbers  

 num1 = JOptionPane.showInputDialog("num1");  //shows a dialog box prompting user to enter value of first number i.e. num1

 number1 = Double.parseDouble(num1);  //scans and reads the value of first input number i.e. number1    

 num2 = JOptionPane.showInputDialog("num2");  //shows a dialog box prompting user to enter value of second number i.e. num2

 number2 = Double.parseDouble(num2);   //reads number2  

               num3 = JOptionPane.showInputDialog("num3");  //shows a dialog box prompting user to enter value of third number i.e. num3

 number3 = Double.parseDouble(num3);   //reads number3  

   if(number1>0&& number2>0 && number3>0){  //checks if all three numbers are positive

   sum = number1 + number2+ number3;  //displays the sum by adding all three positive numbers

   JOptionPane.showMessageDialog(null, "the sum is : " + sum , "Results", JOptionPane.PLAIN_MESSAGE );      }   // displays the result of sum in a message dialog box

 

    if(number1<0 && number2>0 && number3>0)      {  //checks if number2 and number3 are positive

   product = number2*number3;  //computes product

   JOptionPane.showMessageDialog(null, "the product is : " + product , "Results", JOptionPane.PLAIN_MESSAGE );      }  //displays the result

   else if (number2<0&&number1>0&&number3>0){  //checks if number1 and number3 are positive

    product = number1*number3;  //computes product

JOptionPane.showMessageDialog(null, "the product is : " + product , "Results", JOptionPane.PLAIN_MESSAGE );      }  //displays the result

   else if (number3<0 && number1>0 && number2>0)     {  //checks if number1 and number2 are positive

     product = number1*number2;  //computes product of 2 positive numbers

     JOptionPane.showMessageDialog(null, "the product is : " + product , "Results", JOptionPane.PLAIN_MESSAGE );      }  //displays the result

   else     {  /.if all the numbers are positive.

/*However this part is optional. You can just use else part for above else if statement and exclude this else part. In this way the product of all three positive numbers is not computed and product is only computed when only two numbers of the three are positive */

     product = number1*number2*number3;  //computes product of all three positive numbers

     JOptionPane.showMessageDialog(null, "the product is : " + product , "Results", JOptionPane.PLAIN_MESSAGE );      }   //displays the result

Scanner scan = new Scanner(System.in);  //creates a Scanner class object

double value1 , value2, quotient;  //declares variables to hold the values of two real numbers and the result of division of the two real numbers is stored in quotient variable

System.out.println("Enter the first real number: ");  //prompts user to enter the value of first real number from the console

value1 = scan.nextDouble();  //reads the first number from user

System.out.println("Enter the second real number : ");  //prompts user to enter the value of second real number from the console

value2 = scan.nextDouble();  //reads the second number from user

if(value1<0&&value2<0)  {  //checks if both the numbers are negative

quotient = value1/value2;  //compute the quotient

JOptionPane.showMessageDialog(null, "the quotient is : " + quotient , "Results", JOptionPane.PLAIN_MESSAGE );    }  //displays result

 else  //if both numbers are not negative

JOptionPane.showMessageDialog(null, "Both numbers are not negative" );   //displays this message in a dialog box

  }  }

Explanation:

In the above given code if you want to display the outputs in console then alter the following statements as follows:

   if(number1>0&& number2>0 && number3>0){

   sum = number1 + number2+ number3;

  System.out.println("the sum is :  " + sum); }      

 

    if(number1<0 && number2>0 && number3>0)      {

    product = number2*number3;

      System.out.println("the product is :  " + product); }          

   else if (number2<0&&number1>0&&number3>0){

          product = number1*number3;

System.out.println("the product is :  " + product);      }

   else if (number3<0 && number1>0 && number2>0)     {

     product = number1*number2;

     System.out.println("the product is :  " + product);     }

   else     {

     product = number1*number2*number3;

      System.out.println("the product is :  " + product);      }

Scanner scan = new Scanner(System.in);

double value1 , value2, quotient;

System.out.println("Enter the first real number: ");

value1 = scan.nextDouble();

System.out.println("Enter the second real number : ");

value2 = scan.nextDouble();

if(value1<0&&value2<0)  {

quotient = value1/value2;

 System.out.println("the quotient is :  " + quotient);   }

 else  

System.out.println("Both numbers are not negative");

Notice that all the JOptionPane.showMessageDialog statement are changed with System.out.println which prints the results on the console.

4 0
4 years ago
Heather is troubleshooting a computer at her worksite. She has interviewed the computer’s user and is currently trying to reprod
wlad13 [49]

Answer:

a. Establish a theory of probable cause.

Explanation:

Heather now knows the cause, and She is quite experienced. She is now studying the cause, and she is trying to come up with the theory of probable cause now. And as she will be able to come up with the theory, the practical will follow, and she will be able to remove the fault then. And hence the correct option is as mentioned in the answer section.

5 0
4 years ago
Ron is creating building blocks in Word. How can he make the building blocks that he created available?
ivann1987 [24]

Answer:

you can store those building blocks in the Normal template

Explanation:

Building Blocks are different pieces of content such as tables, lists, headers, and text boxes that can be added and used for your word document. In order to make these available, you can store those building blocks in the Normal template. This will allow you to instantly and repeatedly access them for various different projects that you may be working on.

8 0
3 years ago
Read 2 more answers
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