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d1i1m1o1n [39]
3 years ago
11

Write a code segment that displays the values of the integers x, y, and z on a single line, such that each value is right-justif

ied with a field width of 6.
Computers and Technology
1 answer:
Elena L [17]3 years ago
3 0

Answer:

x = int(input ("enter first number: "))

y = int(input ("enter second number: "))

z = int(input ("enter third number: "))

print('%6d %6d %6d' %(x,y,z))

Explanation:

Using python programming language we receive three integers variables (x,y,z) then using string formatting (%6) which specifies that the output should be right justified with a width of 6, the values are printed out.

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Answer:

// using c++ language

#include "stdafx.h";

#include <iostream>

#include<cmath>

using namespace std;

//start

int main()

{

 //Declaration of variables in the program

 double start_organisms;

 double daily_increase;

 int days;

 double updated_organisms;

 //The user enters the number of organisms as desired

 cout << "Enter the starting number of organisms: ";

 cin >> start_organisms;

 //Validating input data

 while (start_organisms < 2)

 {

     cout << "The starting number of organisms must be at least 2.\n";

     cout << "Enter the starting number of organisms: ";

     cin >> start_organisms;

 }

 //The user enters daily input, here's where we apply the 5.2% given in question

 cout << "Enter the daily population increase: ";

 cin>> daily_increase;

 //Validating the increase

 while (daily_increase < 0)

 {

     cout << "The average daily population increase must be a positive value.\n ";

     cout << "Enter the daily population increase: ";

     cin >> daily_increase;

 }

 //The user enters number of days

 cout << "Enter the number of days: ";

 cin >> days;

 //Validating the number of days

 while (days<1)

 {

     cout << "The number of days must be at least 1.\n";

     cout << "Enter the number of days: ";

     cin >> days;

 }

 

 //Final calculation and display of results based on formulas

 for (int i = 0; i < days; i++)

 {

     updated_organisms = start_organisms + (daily_increase*start_organisms);

     cout << "On day " << i + 1 << " the population size was " << round(updated_organisms)<<"."<<"\n";

     

     start_organisms = updated_organisms;

 }

 system("pause");

  return 0;

//end

}

Explanation:

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