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nikklg [1K]
2 years ago
9

Assume that a gallon of paint covers about 350 square feet of wall space. Create an application with a main() method that prompt

s the user for the length, width, and height of a rectangular room. Pass these three values to a method that does the following: Calculates the wall area for a room Passes the calculated wall area to another method that calculates and returns the number of gallons of paint needed Displays the number of gallons needed Computes the price based on a paint price of $32 per gallon, assuming that the painter can buy any fraction of a gallon of paint at the same price as a whole gallon Returns the price to the main() method The main() method displays the final price. For example:
Computers and Technology
1 answer:
Makovka662 [10]2 years ago
5 0

Answer:

<em>C++</em>

#include <iostream>

#include <cmath>

using namespace std;

///////////////////////////////////////////////////////////////////////

int noOfGallons(int wallArea) {

   int gallon = 350;

   int gallon_price = 32.0;

   float total_price = 0.0;

   ///////////////////////////////////////////

   float noOfGallons = (float)wallArea/(float)gallon;

   printf("No of gallons needed %.2f", noOfGallons);

   cout<<endl;

   //////////////////////////////////////////

  total_price = noOfGallons*gallon_price;

   return round(total_price);

}

int wallArea(int length, int width, int height) {

   int wall_area = length*width;

   int total_price = noOfGallons(wall_area);

}

///////////////////////////////////////////////////////////////////////

int main() {

   int length, width, height;

   ///////////////////////////////////////////

   cout<<"Enter length: ";

   cin>>length;

   

   cout<<"Enter width: ";

   cin>>width;

   

   cout<<"Enter height: ";

   cin>>height;

   

   cout<<endl;

   ///////////////////////////////////////////

   int total_price = wallArea(length, width, height);

  cout<<"Total price: $"<<total_price;

   ///////////////////////////////////////////

   return 0;

}

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kodGreya [7K]

The correct answer is; She can purchase OEM codes from e-commerce sites such as Amazon or eBay.

Further Explanation:

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5 0
2 years ago
The __________ of a desktop computer is the case that houses the computerâs critical parts, such as the processing and storage d
Phantasy [73]
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4 0
2 years ago
When creating a table in the relational database design from an entity in the extended e-r model, the attributes of the entity b
larisa86 [58]
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8 0
2 years ago
C# Write, compile, and test a program named PersonalInfo that displays a person’s name, birthdate, work phone number, and cell p
Soloha48 [4]

Answer:

#include <stdio.h>

#Include<iostream>

using namespace std;

//Define required function

PersonalInfo( )   {

    printf("Name   : Robert Josh\n");  

    printf("DOB    : July 14, 1975\n");  

    printf("Work Phone : 00-00000000\n");  

    printf("Cell Phone : 11-777777777\n");  

    return(0);  

 }

int main()   {

    //Call function

 PersonalInfo( );

 return 0;

 }

Explanation:

Its just a simple code to print required details when the function PersonalInfo is called.

5 0
2 years ago
Import java.util.scanner; public class sumofmax { public double findmax(double num1, double num2) { double maxval; // note: if-e
jeyben [28]

Here you go,


Import java.util.scanner

public class SumOfMax {

   public static double findMax(double num1, double num2) {

       double maxVal = 0.0;

       // Note: if-else statements need not be understood to

       // complete this activity

       if (num1 > num2) { // if num1 is greater than num2,

           maxVal = num1; // then num1 is the maxVal.

       }

       else { // Otherwise,

           maxVal = num2; // num2 is the maxVal.

       }

       return maxVal;

   }

   public static void main(String[] args) {

       double numA = 5.0;

       double numB = 10.0;

       double numY = 3.0;

       double numZ = 7.0;

       double maxSum = 0.0;

       /* Your solution goes here */

       maxSum = findMax(numA, numB); // first call of findMax

       maxSum = maxSum + findMax(numY, numZ); // second call

       System.out.print("maxSum is: " + maxSum);

       return;

   }

}

/*

Output:

maxSum is: 17.0

*/

6 0
2 years ago
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