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Nimfa-mama [501]
3 years ago
14

For C++ (please make sure it runs properly)

Computers and Technology
1 answer:
Jlenok [28]3 years ago
7 0

Answer:

The C code is below.

Explanation:

#define amplitude 1

02     #define b 1

03     #define c 200

04     class Sinewave

05     {

06     protected:

07

08                 double freq;

09                 int y;

10

11     public:

12                 Sinewave()

13                 {

14

15                 }              

16                 void generateSinewave()

17                 {

18                     COLORREF yellow = RGB(255,255,0);

19                     COLORREF lightblue = RGB(173,216,230);

20

21                     // make sure the names match

22                     SetConsoleTitle(L"ConGraphics");

23                     HWND hWnd = FindWindow(NULL, L"ConGraphics");

24                     HDC hDC = GetDC(hWnd);

25

26                     //for(int x = 0; x < freq; x++)

27                     for(int x = 0;; x++)

28                     {

29                         // center at y = 200 pixels

30                          

31                         y = amplitude*(int)(sin(x/100.0)*100 + 150);

32                         SetPixel(hDC, x, y, lightblue);

33                          

34                          

35                     }

36                      

37                 }

38 };

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Define a method pyramidVolume with double parameters baseLength, baseWidth, and pyramidHeight, that returns as a double the volu
marysya [2.9K]
<h2>Question:</h2>

Define a method pyramidVolume with double parameters baseLength, baseWidth, and pyramidHeight, that returns as a double the volume of a pyramid with a rectangular base. Relevant geometry equations:

Volume = base area x height x 1/3

Base area = base length x base width.

(Watch out for integer division)

import java.util.Scanner;

public class CalcPyramidVolume {

   /* Your solution goes here */

   public static void main (String [] args) {

     System.out.println("Volume for 1.0, 1.0, 1.0 is: " + pyramidVolume(1.0, 1.0,   1.0));

     return;

    }

}

<h2>Answer:</h2><h2></h2>

import java.util.Scanner;

public class CalcPyramidVolume {

   /* Your solution goes here */

   public static void main (String [] args) {

       System.out.println("Volume for 1.0, 1.0, 1.0 is: " + pyramidVolume(1.0, 1.0, 1.0));

       return;

   }

   

   //Begin method definition

   public static double pyramidVolume(double baseLength, double baseWidth, double pyramidHeight){

       

       //First, calculate the base area of the pyramid

       //store the result in a double variable

       double baseArea = baseLength * baseWidth;

       

       //Then, calculate the volume of the pyramid

       //using the base area and the base width

       double volume = 1 / 3.0 * baseArea * pyramidHeight;

       

       //return the volume

      return volume;

   }

}

<h2>Output:</h2>

Volume for 1.0, 1.0, 1.0 is: 0.3333333333333333

<h2>Explanation:</h2>

The code above contains comments explaining the important lines of the code. The output of the code has also been provided above.

The parts of the code that are worth noting are:

(i) The return type of the method pyramidVolume should be a <em>double</em> since calculations are done using <em>double</em> values. i.e the method header should be written as

<em>public static double pyramidVolume(){</em>

<em>}</em>

<em />

(ii) The method requires three(3) parameters of type double: baseLength, baseWidth and pyramidHeight. These should be included in the method. Therefore, the complete header definition should be:

<em>public static double pyramidVolume(double baseLength, double baseWidth, double pyramidHeight){</em>

<em />

<em>}</em>

(iii) The formula for calculating the volume of the pyramid should be:

volume = 1 / 3.0 * baseArea * baseWidth;

Rather than;

volume = 1 / 3 * baseArea * baseWidth;

This is because integer division yields integer result. If 1 / 3 is evaluated, the result will be 0 since the decimal part will be truncated thereby making the result of the volume = 1 / 3 * baseArea * baseWidth will be 0.

Therefore, the work around for that should be to write 1 / 3.0 or 1.0 / 3.0 or 1.0 / 3.

(iv) After the calculation, the result of volume should be returned by the method. This will enable a proper call by the main method for execution.

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