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Radda [10]
4 years ago
15

Given input of a number, display that number of spaces followed by an asterisk. (1 point) Write your solution within the existin

g supplied function called problem2() and run the Zybooks test for it before continuing on to the next problem. Hint: Use the setw() command to set the field width. Alternatively, you could again use a loop.
Computers and Technology
1 answer:
Naddika [18.5K]4 years ago
7 0

Answer:

Following is the program in C++ language

#include <iostream> // header file

#include <iomanip> // header file  

using namespace std; // namespace std;

void problem2() // function definition  

{

int num ; // variable declaration  

cout << "Enter the Number: ";

cin >> num; // Read the input by user

cout << "Resulatant is:";

for(int k = 0; k <num; k++) // iterarting the loop

{

std::cout << std::setw(num); // set width

std::cout <<"*"; // print *

}

}

int main() // main function

{

problem2();// function calling

return 0;

}

Output:

Enter the Number:  3

Resulatant is:    *   *   *

Explanation:

Following is the description of program

  • Create a function problem2().
  • Declared a variable "num" of "int" type .
  • Read the input in "num" by the user .
  • Iterating the for loop and set the width by using setw() and print the asterisk.
  • In the main function  calling the problem2().
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Hello, my first time being here and I'm returning back from college and I'm now studying Java in college but I was stuck doing t
KengaRu [80]

Answer:

import java.time.LocalDate;

class TestWedding {

 public static void main(String[] args) {

   Person man1 = new Person("John", "Doe", LocalDate.parse("1990-05-23"));

   Person woman1 = new Person("Jane", "Something", LocalDate.parse("1995-07-03"));

   Person man2 = new Person("David", "Johnson", LocalDate.parse("1991-04-13"));

   Person woman2 = new Person("Sue", "Mi", LocalDate.parse("1997-12-01"));

   Couple cpl1 = new Couple(man1, woman1);

   Couple cpl2 = new Couple(man2, woman2);

   Wedding wed1 = new Wedding(cpl1, "Las Vegas", LocalDate.parse("2020-09-12"));

   Wedding wed2 = new Wedding(cpl2, "Hawaii", LocalDate.parse("2021-01-02"));  

   displayDetails(wed1, wed2);

 }

 public static void displayDetails(Wedding w1, Wedding w2) {

   System.out.println(w1.toString());

   System.out.println(w2.toString());

 }

}

---------------------------

class Couple {

 private Person person1;

 private Person person2;

 public Couple(Person p1, Person p2) {

   person1 = p1;

   person2 = p2;

 }

 public String toString() {

   return person1.toString() + " and " + person2.toString();

 }

}

----------------------------

import java.time.LocalDate;

import java.time.format.DateTimeFormatter;

class Person {

 private String firstName;

 private String lastName;

 private LocalDate birthDate;

 public Person(String first, String last, LocalDate bdate) {

   firstName = first;

   lastName = last;

   birthDate = bdate;

 }

 public String getFirstName() {

   return firstName;

 }

 public String toString() {

   DateTimeFormatter formatter = DateTimeFormatter.ofPattern("LLLL dd, yyyy");

   return String.format("%s %s born %s", this.firstName, this.lastName, birthDate.format(formatter));

 }

}

------------------------------------

import java.time.LocalDate;

import java.time.format.DateTimeFormatter;

class Wedding {

 private Couple couple;

 private String location;

 private LocalDate weddingDate;

 public Wedding(Couple c, String loc, LocalDate wDate) {

   couple = c;

   location = loc;

   weddingDate = wDate;

 }

 public String getLocation() {

       return this.location;

 }

 public String toString() {

   DateTimeFormatter formatter = DateTimeFormatter.ofPattern("LLLL dd, yyyy");

   return  

     couple.toString() +  

     " are getting married in " + location + " on "+

     weddingDate.format(formatter);

 }

}

Explanation:

I used overrides of toString to let each object print its own details. That's why this solution doesn't really require any getters. I implemented some to show how it's done, but you'll have to complete it. The solution shows how to think in an OO way; ie., let every class take care of its own stuff.

6 0
4 years ago
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

5 0
3 years ago
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Vadim26 [7]

Answer:

Hes really nice

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Chicken wing

Hot dog and baloney

Chicken and macaroni

Chillin with my homies

Chicken wings

Chicken wings

4 0
3 years ago
Read 2 more answers
The directory "a" contains its subdirectory "b" and there is no other entry in the directory "a". You want to display "the conte
Svet_ta [14]

The answer assumes that the question is about which command help accomplish display directory contents in Unix-like operating systems.

Answer:

The command is <em>ls</em>.

Explanation:

The command <em>ls</em> (short for <em>list</em>) displays a variety of important information in different ways regarding any directory contents. Because of this, it is probably one of the most used commands in Unix-like operating systems.

Any directory can contain directories and files of different sizes, recently created/modified, attributes like permissions for being accessed, and, with this command, we can see all this information by size, chronologically, by owner, and/or by many more ways.  

In the question, we can accomplish to list that the directory "a" contains its sub-directory "b" and no other entry using the next line of code:

Command-line 1: \\ ls\; a (<em>the command "says": display content of a</em>)

Result: b

The result is only the directory <em>b </em>because there is no other entry in it.

To display more information regarding <em>b</em>, we can use the many options available for the command <em>ls</em>, like <em>-a</em> (all entries), <em>-d</em> (only directories), <em>-l</em> (long listing format), and so on, e.g. ls -a  A (display all entries in directory A, included hidden files).

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3 years ago
Which of the following sorting algorithms is described by this text? "Take the item at index 1 and see if it is in order compare
trasher [3.6K]

Answer:

b. selection sort

b. 8 11 17 30 20 25

Explanation:

The options above are the correct answers to the given questions.

Selection sort is a simple comparison-based sorting algorithm.

selection sort. (algorithm) Definition: A sort algorithm that repeatedly searches remaining items to find the least one and moves it to its final location. The run time is Θ(n²), where n is the number of elements. The number of swaps is O(n).

It is this sorting algorithm that will best ne suitable in the given event.

7 0
4 years ago
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