the levels of classification of organisms in order are: domain, kingdom, phylum, class, order, family, genus, and species. The broadest, least specific level is____________________

Answer:

hope this helps

Explanation:

Answer:

Can you add an attachment please

In this exercise we have to use the computer language knowledge in python, so the code is described as:

This code can be found in the attached image.

So the code can be described more simply below, as:

height = []

height.append([16,17,14])

height.append([17,18,17])

height.append([15,17,14])

print(height)

See more about python at brainly.com/question/26104476

Answer:

True

Explanation:

because it was correct

Answer:

The program in Python is as follows:

Length = float(input("Length: "))

Width = float(input("Width: "))

Area = Length * Width

print("Area: ",Area)

Explanation:

This gets input for length

Length = float(input("Length: "))

This gets input for width

Width = float(input("Width: "))

This calculates the area

Area = Length * Width

This prints the calculated area

print("Area: ",Area)

Answer:

Here.Hope this helps and do read carefully ,you need to make 1 .cpp file and 1 .h file

Explanation:

#include<iostream>

#include <fstream>

#include <vector>

#include <string>

#include "StatePair.h"

using namespace std;

int main() {

  ifstream inFS; // File input stream

  int zip;

  int population;

  string abbrev;

  string state;

  unsigned int i;

  // ZIP code - state abbrev. pairs

  vector<StatePair <int, string>> zipCodeState;

  // state abbrev. - state name pairs

  vector<StatePair<string, string>> abbrevState;

  // state name - population pairs

  vector<StatePair<string, int>> statePopulation;

  // Fill the three ArrayLists

  // Try to open zip_code_state.txt file

  inFS.open("zip_code_state.txt");

  if (!inFS.is_open()) {

      cout << "Could not open file zip_code_state.txt." << endl;

      return 1; // 1 indicates error

  }

  while (!inFS.eof()) {

      StatePair <int, string> temp;

      inFS >> zip;

      if (!inFS.fail()) {

          temp.SetKey(zip);

      }

      inFS >> abbrev;

      if (!inFS.fail()) {

          temp.SetValue(abbrev);

      }

      zipCodeState.push_back(temp);

  }

  inFS.close();

// Try to open abbreviation_state.txt file

  inFS.open("abbreviation_state.txt");

  if (!inFS.is_open()) {

      cout << "Could not open file abbreviation_state.txt." << endl;

      return 1; // 1 indicates error

  }

  while (!inFS.eof()) {

      StatePair <string, string> temp;

      inFS >> abbrev;

      if (!inFS.fail()) {

          temp.SetKey(abbrev);

      }

      getline(inFS, state); //flushes endline

      getline(inFS, state);

      state = state.substr(0, state.size()-1);

      if (!inFS.fail()) {

          temp.SetValue(state);

      }

      abbrevState.push_back(temp);

  }

  inFS.close();

  // Try to open state_population.txt file

  inFS.open("state_population.txt");

  if (!inFS.is_open()) {

      cout << "Could not open file state_population.txt." << endl;

      return 1; // 1 indicates error

  }

  while (!inFS.eof()) {

      StatePair <string, int> temp;

      getline(inFS, state);

      state = state.substr(0, state.size()-1);

      if (!inFS.fail()) {

          temp.SetKey(state);

      }

      inFS >> population;

      if (!inFS.fail()) {

          temp.SetValue(population);

      }

      getline(inFS, state); //flushes endline

      statePopulation.push_back(temp);

  }

  inFS.close();

  cin >> zip;

for (i = 0; i < zipCodeState.size(); ++i) {

      // TODO: Using ZIP code, find state abbreviation

  }

  for (i = 0; i < abbrevState.size(); ++i) {

      // TODO: Using state abbreviation, find state name

  }

  for (i = 0; i < statePopulation.size(); ++i) {

      // TODO: Using state name, find population. Print pair info.

  }

}

Statepair.h

#ifndef STATEPAIR

#define STATEPAIR

#include <iostream>

using namespace std;

template<typename T1, typename T2>

class StatePair {

private:

T1 key;

T2 value;

public:

// Define a constructor, mutators, and accessors

// for StatePair

StatePair() //default constructor

{}

// set the key

void SetKey(T1 key)

{

this->key = key;

}

// set the value

void SetValue(T2 value)

{

this->value = value;

}

// return key

T1 GetKey() { return key;}

// return value

T2 GetValue() { return value;}

// Define PrintInfo() method

// display key and value in the format key : value

void PrintInfo()

{

cout<<key<<" : "<<value<<endl;

}

};

#endif

In this exercise we have to use the knowledge in computational language in C++  to write the following code:

We have the code can be found in the attached image.

So in an easier way we have that the code is

#include<iostream>

#include <fstream>

#include <vector>

#include <string>

#include "StatePair.h"

using namespace std;

int main() {

 ifstream inFS;

 int zip;

 int population;

 string abbrev;

 string state;

 unsigned int i;

 vector<StatePair <int, string>> zipCodeState;

 vector<StatePair<string, string>> abbrevState;

 vector<StatePair<string, int>> statePopulation;

 inFS.open("zip_code_state.txt");

 if (!inFS.is_open()) {

     cout << "Could not open file zip_code_state.txt." << endl;

     return 1;

 }

 while (!inFS.eof()) {

     StatePair <int, string> temp;

     inFS >> zip;

     if (!inFS.fail()) {

         temp.SetKey(zip);

     }

     inFS >> abbrev;

     if (!inFS.fail()) {

         temp.SetValue(abbrev);

     }

     zipCodeState.push_back(temp);

 }

 inFS.close();

 inFS.open("abbreviation_state.txt");

 if (!inFS.is_open()) {

     cout << "Could not open file abbreviation_state.txt." << endl;

     return 1;

 }

 while (!inFS.eof()) {

     StatePair <string, string> temp;

     inFS >> abbrev;

     if (!inFS.fail()) {

         temp.SetKey(abbrev);

     }

     getline(inFS, state);

     getline(inFS, state);

     state = state.substr(0, state.size()-1);

     if (!inFS.fail()) {

         temp.SetValue(state);

     }

     abbrevState.push_back(temp);

 }

 inFS.close();

 inFS.open("state_population.txt");

 if (!inFS.is_open()) {

     cout << "Could not open file state_population.txt." << endl;

     return 1;

 }

 while (!inFS.eof()) {

     StatePair <string, int> temp;

     getline(inFS, state);

     state = state.substr(0, state.size()-1);

     if (!inFS.fail()) {

         temp.SetKey(state);

     }

     inFS >> population;

     if (!inFS.fail()) {

         temp.SetValue(population);

     }

     getline(inFS, state);

     statePopulation.push_back(temp);

 }

 inFS.close();

 cin >> zip;

for (i = 0; i < zipCodeState.size(); ++i) {

  }

 for (i = 0; i < abbrevState.size(); ++i) {

  }

 for (i = 0; i < statePopulation.size(); ++i) {

 }

}

Statepair.h

#ifndef STATEPAIR

#define STATEPAIR

#include <iostream>

using namespace std;

template<typename T1, typename T2>

class StatePair {

private:

T1 key;

T2 value;

public:

StatePair()

{}

void SetKey(T1 key)

{

this->key = key;

}

void SetValue(T2 value)

{

this->value = value;

}

T1 GetKey() { return key;}

T2 GetValue() { return value;}

void PrintInfo()

{

cout<<key<<" : "<<value<<endl;

}

};

See more about C code at brainly.com/question/19705654

Answer:

avg_owls=0.0

num_owls_zooA = int(input())

num_owls_zooB = int(input())

num_owls_zooC = int(input())

num_zoos = 3

avg_owls = (num_owls_zooA + num_owls_zooB + num_owls_zooC) / num_zoos

print('Average owls per zoo:', int(avg_owls))

Explanation:

There is a problem while you are trying the calculate the average owls per zoo.

As you may know, in order to calculate the average, you need to get the total number of owls in all the zoos, then divide it to the number of zoos. That is why you need to sum num_owls_zooA + num_owls_zooB + num_owls_zooC and divide the result by num_zoos

Be aware that the average is printed as integer value. That is why you need to cast the avg_owls to an int as int(avg_owls)

Answer:

Explanation:

The following function is written in Java. It takes two strings as parameters and calculates the longest substring between both of them and returns that substring length.

import java.util.*;

class Test

{

   public static void main(String[] args)

   {

       String X = "photograph";

       String Y = "tomography";

       System.out.println(X + '\n' + Y);

       System.out.println("Longest common substring length: " + longestSub(X, Y));

   }

   static int longestSub(String X, String Y)

   {

       char[] word1 = X.toCharArray();

       char[] word2 = Y.toCharArray();

       int length1 = X.length();

       int length2 = Y.length();

       int substringArray[][] = new int[length1 + 1][length2 + 1];

       int longestSubstringLength = 0;

       for (int i = 0; i <= length1; i++)

       {

           for (int j = 0; j <= length2; j++)

           {

               if (i == 0 || j == 0)

                   substringArray[i][j] = 0;

               else if (word1[i - 1] == word2[j - 1])

               {

                   substringArray[i][j]

                           = substringArray[i - 1][j - 1] + 1;

                   longestSubstringLength = Integer.max(longestSubstringLength,

                           substringArray[i][j]);

               }

               else

                   substringArray[i][j] = 0;

           }

       }

       return longestSubstringLength;

   }

}

Answer:

There are way in separating mixtures of chlorides salts such as that of sodium chloride and ammonium chloride. It can be done by crystallization, filtration or sublimation. If we want to separate the mixture, we have to heat up to 330-350 degrees Celsius and collect the gas that will be produced.

Answer: getfenv() is a type of function. Particually a envirotment function. for a lua coding.

Explanation: What this does it goes thourgh  line of code in a particular order.

This means. getfenv is used to get the current environment of a function. It returns a table of all the things that function has access to. You can also set the environment of a function to another environment.

Forgot to include examples of where this could be used. Although not very common uses, there are some. In the past Script Builders used getfenv to get the environment of a script, and setfenv to set the environment of a created script’s environment to a fake environment, so they couldn’t affect the real one. They could also inject custom global functions.

Not entirely sure if this uses getfenv or setfenv, but the use in Crazyman32’s AeroGameFramework is making the environment of each module have access to other modules without having to require them, and having access to remotes without having to directly reference them.

Answer: ummm, who knows

Explanation:

Answer:security VBA code

Explanation:

Answer:

Hope this helped you, and if it did , do consider giving brainliest.

Explanation:

import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

//   main method

  public static void main(String[] args) {

//   create a list of Comparable type

      List<Comparable> list = new ArrayList<>();

//       add elements

      list.add(1);

      list.add(2);

      list.add(3);

      list.add(4);

      list.add(5);

      list.add(6);

      list.add(7);

//       print list

      System.out.println("\nList : "+list);

//       test search method

      Comparable a = 7;

      System.out.println("\nSearch for 7 : "+search(list,a));

      Comparable b = 3;

      System.out.println("\nSearch for 3 : "+search(list,b));

      Comparable c = 9;

      System.out.println("\nSearch for 9 : "+search(list,c));

      Comparable d = 1;

      System.out.println("\nSearch for 1 : "+search(list,d));

      Comparable e = 12;

      System.out.println("\nSearch for 12 : "+search(list,e));

      Comparable f = 0;

      System.out.println("\nSearch for 0 : "+search(list,f));

  }

//   static method named search takes arguments Comparable list and Comparable parameter

  public static boolean search(List<Comparable> list, Comparable par) {

//       if list is empty or parameter is null the throw IllegalArgumentException

      if(list.isEmpty() || par == null ) {

          throw new IllegalArgumentException();

      }

//       binary search

//       declare variables

      int start=0;

      int end =list.size()-1;

//       using while loop

      while(start<=end) {

//           mid element

          int mid =(start+end)/2;

//           if par equal to mid element then return

          if(list.get(mid).equals(par) )

          {

              return true ;

          }  

//           if mid is less than parameter

          else if (list.get(mid).compareTo(par) < 0 ) {

              start=mid+1;

          }

//           if mid is greater than parameter

          else {

              end=mid-1;

          }

      }

//       if not found then retuen false

      return false;

  }

}import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

//   main method

  public static void main(String[] args) {

//   create a list of Comparable type

      List<Comparable> list = new ArrayList<>();

//       add elements

      list.add(1);

      list.add(2);

      list.add(3);

      list.add(4);

      list.add(5);

      list.add(6);

      list.add(7);

//       print list

      System.out.println("\nList : "+list);

//       test search method

      Comparable a = 7;

      System.out.println("\nSearch for 7 : "+search(list,a));

      Comparable b = 3;

      System.out.println("\nSearch for 3 : "+search(list,b));

      Comparable c = 9;

      System.out.println("\nSearch for 9 : "+search(list,c));

      Comparable d = 1;

      System.out.println("\nSearch for 1 : "+search(list,d));

      Comparable e = 12;

      System.out.println("\nSearch for 12 : "+search(list,e));

      Comparable f = 0;

      System.out.println("\nSearch for 0 : "+search(list,f));

  }

//   static method named search takes arguments Comparable list and Comparable parameter

  public static boolean search(List<Comparable> list, Comparable par) {

//       if list is empty or parameter is null the throw IllegalArgumentException

      if(list.isEmpty() || par == null ) {

          throw new IllegalArgumentException();

      }

//       binary search

//       declare variables

      int start=0;

      int end =list.size()-1;

//       using while loop

      while(start<=end) {

//           mid element

          int mid =(start+end)/2;

//           if par equal to mid element then return

          if(list.get(mid).equals(par) )

          {

              return true ;

          }  

//           if mid is less than parameter

          else if (list.get(mid).compareTo(par) < 0 ) {

              start=mid+1;

          }

//           if mid is greater than parameter

          else {

              end=mid-1;

          }

      }

//       if not found then retuen false

      return false;

  }

}

Answer:

the answer is B

Explanation:

Answer:

Explanation:

The following code is written in Java and it uses nested for loops to create the array elements and then the same for loops in order to print out the elements in a pyramid-like format as shown in the question. The output of the code can be seen in the attached image below.

class Brainly {

   public static void main(String[] args) {

       int jagged[][] = new int[5][];

       int element = 1;

       for(int i=0; i<5; i++){

           jagged[i] = new int[i+1]; // creating number of columns based on current value

           for(int x = 0; x < jagged[i].length; x++) {

               jagged[i][x] = element;

               element += 1;

           }

       }

       System.out.println("Jagged Array elements are: ");

       for ( int[] x : jagged) {

           for (int y : x) {

               System.out.print(y + " ");

           }

           System.out.println(' ');

       }

       }

   }

Answer:

C

Explanation:

omputer chip

sorry btw

Answer:

2000 nibbles

Explanation:

2000 nibbles is needed for 1 kilobyte

Answer:

A

Explanation:

I feel that if everyone is with eachother, there may be a better hope to improve the next time

Answer:

Find answers below.

Explanation:

1. Worldwide Web: In computing, WWW simply means World Wide Web. The world wide web was invented by Sir Tim Berners-Lee in 1990 while working with the European Council for Nuclear Research (CERN); Web 2.0 evolved in 1999. Basically, WWW refers to a collection of web pages that are located on a huge network of interconnected computers (the Internet). Also, users from all over the world can access the world wide web by using an internet connection and a web browser such as Chrome, Firefox, Safari, Opera, etc.

In a nutshell, the World Wide Web is an information system that is generally made up of users and resources (documents) that are connected via hypertext links.

2.E-mail: An e-mail is an acronym for electronic mail and it is a software application or program designed to let users send and receive texts and multimedia messages over the internet.

It is considered to be one of the most widely used communication channel or medium is an e-mail (electronic mail).

3. Powerline networking: this is a wired networking technology that typically involves the use of existing electrical cables in a building to connect and transmit data on network devices through powerline network adapters.

4. Cable: this include network cables such as CAT-5, CAT-6 and fiber optic cables used for connecting and transmitting signals between two or more network devices.

Answer:

“hello” + “goodbye”

Explanation:

Answer:

do it! you will have to one day so why not now? hopefully she will understand and say nice things as well and be like mom ive

been thinking a lot and i really want to tell you something. Explanation:

Answer:most likely she is your parent that you live with so she might already know but has not said anything it also depends her personality but if your ready and know or think she would not hate you forever go for it

Explanation:

Solution :

Initial array = [tex]$\text{C,Q,S,A,X,B,T}$[/tex]

[tex]$n= 7$[/tex](length of the array)

[tex]$\text{1st}$[/tex] Iteration:

i = 1

  j = 1

  [tex]$\text{a[j-1]}$[/tex] = C

  a[j] = Q

  since [tex]$\text{a[j-1]}$[/tex] < a[j] , break from inner loop

Number of comparisons in 1st Iteration = 1

After 1st Iteration:

Array : C,Q,S,A,X,B,T

2nd Iteration:

i = 2

  j = 2

  a[j-1] = Q

  a[j] = S

  since a[j-1] < a[j], break from inner loop

Number of comparisons in 2nd Iteration = 1

After 2nd Iteration:

Array : C,Q,S,A,X,B,T

3rd Iteration:

i = 3

  j = 3

  a[j-1] = S

  a[j] = A

  since a[j-1] > a[j], exchange a[2] with a[3]

  Array : C,Q,A,S,X,B,T

  j = 2

  a[j-1] = Q

  a[j] = A

  since a[j-1] > a[j], exchange a[1] with a[2]

  Array : C,A,Q,S,X,B,T

  j = 1

  a[j-1] = C

  a[j] = A

  since a[j-1] > a[j], exchange a[0] with a[1]

  Array : A,C,Q,S,X,B,T

  j = 0, break from inner loop

Number of comparisons in 3rd Iteration = 3

After 3rd Iteration:

Array : A,C,Q,S,X,B,T

4th Iteration:

i = 4

  j = 4

  a[j-1] = S

  a[j] = X

  since a[j-1] < a[j], break from inner loop

Number of comparisons in 4th Iteration = 1

After 4th Iteration:

Array : A,C,Q,S,X,B,T

5th Iteration:

i = 5

  j = 5

  a[j-1] = X

  a[j] = B

  since a[j-1] > a[j], exchange a[4] with a[5]

  Array : A,C,Q,S,B,X,T

  j = 4

  a[j-1] = S

  a[j] = B

  since a[j-1] > a[j], exchange a[3] with a[4]

  Array : A,C,Q,B,S,X,T

  j = 3

  a[j-1] = Q

  a[j] = B

  since a[j-1] > a[j], exchange a[2] with a[3]

  Array : A,C,B,Q,S,X,T

  j = 2

  a[j-1] = C

  a[j] = B

  since a[j-1] > a[j], exchange a[1] with a[2]

  Array : A,B,C,Q,S,X,T

  j = 1

  a[j-1] = A

  a[j] = B

  since a[j-1] < a[j], break from inner loop

Number of comparisons in 5th Iteration = 5

After 5th Iteration:

Array : A,B,C,Q,S,X,T

6th Iteration:

i = 6

  j = 6

  a[j-1] = X

  a[j] = T

  since a[j-1] > a[j], exchange a[5] with a[6]

  Array : A,B,C,Q,S,T,X

  j = 5

  a[j-1] = S

  a[j] = T

  since a[j-1] < a[j], break from inner loop

Number of comparisons in 6th Iteration = 2

After 6th Iteration:

Array : A,B,C,Q,S,T,X

Sorted Array : A B C Q S T X  

Answer:

C.

Explanation:

Internet of things

The Basic premise of the Internet of Things is that objects can be tagged, tracked, and monitored through a local network, or across the Internet option (c) is correct.

The term "Internet of things" refers to actual physical items that have sensors, computing power, software, and other innovations and can link to those other systems and gadgets via the Internet or other communication networks and share files with them.

As we know,

Using a local network or the Internet, things may be identified, tracked, and monitored is the fundamental tenet of the Internet of Things.

Thus, the Basic premise of the Internet of Things is that objects can be tagged, tracked, and monitored through a local network, or across the Internet option (c) is correct.

Learn more about the IoT here:

https://brainly.com/question/25703804

#SPJ2

Answer:

- the usernameMaker function

Explanation:

When you look at the code, you can see it's located alongside the usernameMaker. It it only utilized then, therefore being it's scope. For reassurance, it's also correct on Edge.

I hope this helped!

Good luck <3

Answer:

hope this helps,if it did pls mark my ans as brainliest

Explanation:

using System;

class Printer{ string companyName; int pagesOutput;

double price;

public Printer(string companyName,int pagesOutput,double price){

this.companyName=companyName; this.pagesOutput=pagesOutput; this.price=price;

}

public virtual void Display(){ Console.WriteLine("companyName: "+companyName+" pagesOutput: "+pagesOutput+" price: "+price);

}

}

class PrinterCumScanner:Printer{ int imageDpi;

public PrinterCumScanner(string companyName,int pagesOutput,double price,int imageDpi):base(companyName,pagesOutput,price){ this.imageDpi=imageDpi; }

public override void Display(){ base.Display(); Console.WriteLine("imageDpi: "+imageDpi);

}

}

public class Program { static void Main(string[] args) { const int NUM_PRINTERS=3;

Printer []stock=new Printer[NUM_PRINTERS];

stock[0]=new Printer("HP",40,89.50);

stock[1]=new PrinterCumScanner("HP",40,215.0,320); stock[2]=new PrinterCumScanner("Cannon",30,175.0,240);

foreach(Printer aPrinter in stock){ aPrinter.Display();

}

}

}

Answer:

Design

Explanation:

seems the most correct one..

Answer:

C code

Explanation:

#include <stdio.h>

void histrogram(int counters[])

{

   int i,j;

   int count;

   for(i=0;i<26;i++)

   {

       count=counters[i];

       printf("%c ",i+97);

       for(j=0;j<count;j++)

       {

           printf("="); //= is used

       }

       printf("\n");

   }

}

int main()

{

   FILE* fp;

   int i;

   int arr[26];

   char c;

   int val;

   // Open the file

fp = fopen("story.txt", "r");

if (fp == NULL) {

printf("Could not open file ");

return 0;

}

else

{

   for(i=0;i<26;i++)

   arr[i]=0;

   for (c = getc(fp); c != EOF; c = getc(fp))

   {

       if(c>='a' && c<='z')

       {

           val = c-97;

           //printf("%d ",val);

           arr[val]++;

           }

       }

       histrogram(arr);

   }

}

Answer:

I think it's B correct me if I'm wrong

Explanation:

Answer:

B

Explanation:

I say it is the right one although in this 2021 there is no other game but it is its 10th anniversary

Answer:

1 First Generation

The period of first generation: 1946-1959. Vacuum tube based.

2 Second Generation

The period of second generation: 1959-1965. Transistor based.

3 Third Generation

The period of third generation: 1965-1971. Integrated Circuit based.

4 Fourth Generation

The period of fourth generation: 1971-1980. VLSI microprocessor based.

5 Fifth Generation

The period of fifth generation: 1980-onwards. ULSI microprocessor based.

The phrase "generation" refers to a shift in the technology that a computer is/was using. The term "generation" was initially used to describe different hardware advancements. These days, a computer system's generation involves both hardware and software.

The Japan Ministry of International Trade and Industry (MITI) launched the Fifth Generation Computer Systems (FGCS) initiative in 1982 with the goal of developing computers that use massively parallel computing and logic programming.

The generation of computers is determined by when significant technological advancements, such as the use of vacuum tubes, transistors, and microprocessors, take place inside the computer. There will be five computer generations between now and 2020, with the first one starting around 1940.

Learn more about generations of computers here:

https://brainly.com/question/9354047

#SPJ2

Answer:

brainliest if it did help you

Explanation:

Given array : A[0...n-1]Now, create a count array Cnt[0...n-1]then, initialize Cnt to zero

for(i=0...n-1)

Cnt[i]=0 ---------O(n) time

Linearly traverse the list and increment the count of respected number in count array.

for(i=0...n-1)

Cnt[A[i]]++; ---------O(n) time

Check in the count array if duplicates exist.

for(i=0...n-1){

if(C[i]>1)

output (i); -----------O(n) time

}

analysis of the above algorithm:

Algorithm takes = O(1 + n + n + n)

                           = O(n)          

//java code

public class SortIntegers {

  public static void sort(int[] arr) {

    int min = arr[0];

    int max = arr[0];

    for (int i = 0; i < arr.length; ++i) {

      if (arr[i] > max) {

        max = arr[i];

      }

      if (arr[i] < min) {

        min = arr[i];

      }

    }

    int counts[] = new int[max - min + 1];

    for (int i = 0; i < arr.length; ++i) {

      counts[arr[i] - min]++;

    }

    for (int i = 0; i < counts.length; ++i) {

      if (counts[i] > 1) {

        System.out.print((i + min) + " ");

      }

    }

    System.out.println();

  }

  public static void main(String[] args) {

    sort(new int[] { 5, 4, 10, 2, 4, 10, 5, 3, 1 });

  }

}

brainlest   pl     jk  mmmmmmmmmmmmm