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vekshin1
3 years ago
15

Write a program that reads two fractions such as 1/2 and 1/4 and computes and stores the sum, the difference, the product and th

e result of dividing one by the other in four other fractions, respectively. Display the four resulting fractions. The following is a sample interaction between the user and the program: Enter two fractions: 1/2 1/4 Sum fraction: 6/8 Difference: 2/8 Product: 1/8 Quotient: 4/2 Press any key to continue. If n1/d1 and n2/d2 are the two fractions, their sum is given by: n1d2+ n2d1/d1d2 their difference is given by: n1d2-n2* d1/d1d2 Product is: n1n2/d1d2 Quotient: nl. d2/d1n2 Make sure the program pauses and waits for a key to be pressed before continuing. Put your name on top as a comment. Use comments to explain where your code may not be clear enough. After building and testing the program, submit the cpp file. There are two steps in submitting an assignment: first click Choose File and once you locate the file, click Upload to upload it. Once you have uploaded the file click Finish to submit. It will not be submitted until you click Finish. Make sure not to click Finish before uploading the files, as it will send a blank page and you can't resubmit. Each student must write his or her own program.
Computers and Technology
1 answer:
Vesnalui [34]3 years ago
5 0

Answer:

1: #include <iostream>

2: using namespace std;

3: int main()

4: {

5: int a,b,c,d;

6: cout<<"n1: ";

7: cin>>a;

8: cout<<"d1: ";

9: cin>>b;

10: cout<<"n2: ";

11: cin>>c;

12: cout<<"d2: ";

13: cin>>d;

14: int top = (a*d)+(b*c);

15: int bottom = b*d;

16: cout<<"Sum: "<<top<<"/"<<bottom<<"\n";

17: top = (a*d)-(b*c);

18: cout<<"Difference: "<<top<<"/"<<bottom<<"\n";

19: top = a*c;

20: cout<<"Product: "<<top<<"/"<<bottom<<"\n";

21: top = a*d;

22: bottom = b*c;

23: cout<<"Quotient: "<<top<<"/"<<bottom<<"\n";

24: return 0;

25: }

Explanation:

The Program is written in C++ programming language

The Program is left numbered

Line 5 was used for variable declaration.

Variables a,b represents the numerator and denominator of the first fraction

While variables c,d represent the numerator and denominator of the second fraction

Line 6 prints "n1" without the quotes which represents the numerator of the first fraction

Line 7 accepts input for the numerator of the first fraction.

Line 8 prints "d1" without the quotes which represents the denominator of the first fraction

Line 9 accepts input for the denominator of the first fraction.

Line 10 prints "n2" without the quotes which represents the numerator of the second fraction

Line 11 accepts input for the numerator of the second fraction.

Line 12 prints "d2" without the quotes which represents the denominator of the second fraction

Line 13 accepts input for the denominator of the second fraction.

Line 14 and 15 calculate the sum of the fractions which is then printed on line 16

Line 17 calculates the difference of the fractions which is then printed on line 18

Line 19 calculates the product of the fractions which is then printed on line 20

Line 21 and 22 calculates the quotient of the fractions which is then printed on line 23

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Answer:

<h2>Function 1:</h2>

#include <stdio.h> //for using input output functions

// start of the function PrintPopcornTime body having integer variable //bagOunces as parameter

void PrintPopcornTime(int bagOunces){

if (bagOunces < 3){ //if value of bagOunces is less than 3

 printf("Too small"); //displays Too small message in output

 printf("\n"); } //prints a new line

//the following else if part will execute when the above IF condition evaluates to //false and the value of bagOunces is greater than 10

else if (bagOunces > 10){

    printf("Too large"); //displays the message:  Too large in output

    printf("\n"); //prints a new line }

/*the following else  part will execute when the above If and else if conditions evaluate to false and the value of bagOunces is neither less than 3 nor greater than 10 */

else {

/* The following three commented statements can be used to store the value of bagOunces * 6 into result variable and then print statement to print the value of result. The other option is to use one print statement printf("%d",bagOunces * 6) instead */

    //int result;

    //result = bagOunces * 6;

    //printf("%d",result);

 printf("%d",bagOunces * 6);  /multiplies value of bagOunces  to 6

 printf(" seconds");

// seconds is followed with the value of bagOunces * 6

 printf("\n"); }} //prints a new line

int main(){ //start of main() function body

int userOunces; //declares integer variable userOunces

scanf("%d", &userOunces); //reads input value of userOunces

PrintPopcornTime(userOunces);

//calls PrintPopcornTime function passing the value in userOunces

return 0; }

Explanation:

<h2>Function 2:  </h2>

#include <stdio.h> //header file to use input output functions

// start of the function PrintShampooInstructions body having integer variable numCycles as parameter

void PrintShampooInstructions(int numCycles){

if(numCycles < 1){

//if conditions checks value of numCycles is less than 1 or not

printf("Too few."); //prints Too few in output if the above condition is true

printf("\n"); } //prints a new line

//else if part is executed when the if condition is false and else if  checks //value of numCycles is greater than 4 or not

else if(numCycles > 4){

//prints Too many in output if the above condition is true

printf("Too many.");

printf("\n"); } //prints a new line

//else part is executed when the if and else if conditions are false

else{

//prints "N: Lather and rinse." numCycles times, where N is the cycle //number, followed by Done

for(int N = 1; N <= numCycles; N++){

printf("%d",N);

printf(": Lather and rinse. \n");}

printf("Done.");

printf("\n");} }

int main() //start of the main() function body

{    int userCycles; //declares integer variable userCycles

   scanf("%d", &userCycles); //reads the input value into userCycles

   PrintShampooInstructions(userCycles);

//calls PrintShampooInstructions function passing the value in userCycles

   return 0;}

I will explain the for loop used in PrintShampooInstructions() function. The loop has a variableN  which is initialized to 1. The loop checks if the value of N is less than or equal to the value of numCycles. Lets say the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 1<2. So the program control enters the body of loop. The loop body has following statements. printf("%d",N); prints the value of N followed by

printf(": Lather and rinse. \n"); which is followed by printf("Done.");

So at first iteration:

printf("%d",N); prints 1 as the value of N is 1

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

1: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 2.

Now at second iteration:

The loop checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 2=2. So the program control enters the body of loop.

printf("Done."); prints Done after the above two lines.

printf("%d",N); prints 2 as the value of N is 2

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

2: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 3.

The loop again checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to false as N<numCycles  which means 3>2. So the loop breaks.

Now the next statement is:

printf("Done."); which prints Done on the screen.

So as a whole the following output is displayed on the screen:

1: Lather and rinse.

2: Lather and rinse.

Done.

The programs along with their outputs are attached.

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Answer:

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