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svp [43]
3 years ago
14

Which Creative Commons license type allows others to use and build upon work non-commercially, provided that they credit origina

l author and maintain the same licensing?
Computers and Technology
1 answer:
Vera_Pavlovna [14]3 years ago
8 0

Answer:

Creative Commons license type allows others to use and build upon work non-commercially, provided that they credit original author and maintain the same licensing is described below in detail.

Explanation:

Attribution-Non financially-ShareAlike

This permission lets others adapt, remix, and develop upon your work non-financially, as long as they charge you and license their new inventions under identical times. There are six separate license classes, scheduled from most to least licensed. the material in any mechanism or arrangement, so long as attribution is given to the originator.

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Assume that x is a char variable that has been declared and already given a value. Write an expression whose value is true if an
maria [59]

Answer:

^[A-Fa-f0-9]+$

Explanation:

The hexadecimal (Base 16) can contain 0-9, A-F, a-f and any of these can be the part of the Hexadecimal base 16 code.

In python we have the re.match function, and the syntax is as below:

re.match(pattern, string, flag=0). And we can use this function to check whether the entered char variable is a regex or not. Like:

re.match(pattern,x, flag=0)

8 0
4 years ago
How are overhead projectors being used to improve teaching and learning in the classroom?
baherus [9]
<span>Projects are easier to organize in programs like PowerPoint. Also it is definitely a lot less Complicated than a chalkboard or a whiteboard.</span>
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3 years ago
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Write a program that demonstrates the skills we've learned throughout this quarter. This type of project offers only a few guide
UNO [17]

This type of project offers only a few guidelines, allowing you to invest as much time and polish as you want, as long as you meet the program requirements described below.

The code given below is tic-tac-toe

Explanation:

import java.util.Arrays;

import java.util.InputMismatchException;

import java.util.Scanner;

public class TicTacToe {

static Scanner in;

static String [ ] board;

static String turn;

public static void main(String[] args) {

 in = new Scanner(System.in);

 board = new String[9];

 turn = "X";

 String winner = null;

 populateEmptyBoard();

 System.out.println("Welcome to 2 Player Tic Tac Toe.");

 System.out.println("");

 printBoard();

 System.out.println("X's will play first. Enter a slot number to place X in:");

while (winner == null) {

  int numInput;

  try {

   numInput = in.nextInt();

   if (!(numInput > 0 && numInput <= 9)) {

    System.out.println("Invalid input; re-enter slot number:");

    continue;

   }

  } catch (InputMismatchException e) {

   System.out.println("Invalid input; re-enter slot number:");

   continue;

  }

  if (board[numInput-1].equals(String.valueOf(numInput))) {

   board[numInput-1] = turn;

   if (turn.equals("X")) {

    turn = "O";

   } else {

    turn = "X";

   }

   printBoard();

   winner = checkWinner();

  } else {

   System.out.println("Slot already taken; re-enter slot number:");

   continue;

  }

 }

 if (winner.equalsIgnoreCase("draw")) {

  System.out.println("It's a draw! Thanks for playing.");

 } else {

  System.out.println("Congratulations! " + winner + "'s have won! Thanks for playing.");

 }

}

static String checkWinner() {

 for (int a = 0; a < 8; a++) {

  String line = null;

  switch (a) {

  case 0:

   line = board[0] + board[1] + board[2];

   break;

  case 1:

   line = board[3] + board[4] + board[5];

   break;

  case 2:

   line = board[6] + board[7] + board[8];

   break;

  case 3:

   line = board[0] + board[3] + board[6];

   break;

  case 4:

   line = board[1] + board[4] + board[7];

   break;

  case 5:

   line = board[2] + board[5] + board[8];

   break;

  case 6:

   line = board[0] + board[4] + board[8];

   break;

  case 7:

   line = board[2] + board[4] + board[6];

   break;

  }

  if (line.equals("XXX")) {

   return "X";

  } else if (line.equals("OOO")) {

   return "O";

  }

 }

 for (int a = 0; a < 9; a++) {

  if (Arrays.asList(board).contains(String.valueOf(a+1))) {

   break;

  }

  else if (a == 8) return "draw";

 }

 System.out.println(turn + "'s turn; enter a slot number to place " + turn + " in:");

 return null;

}

static void printBoard() {

 System.out.println("/---|---|---\\");

 System.out.println("| " + board[0] + " | " + board[1] + " | " + board[2] + " |");

 System.out.println("|-----------|");

 System.out.println("| " + board[3] + " | " + board[4] + " | " + board[5] + " |");

 System.out.println("|-----------|");

 System.out.println("| " + board[6] + " | " + board[7] + " | " + board[8] + " |");

 System.out.println("/---|---|---\\");

}

static void populateEmptyBoard() {

 for (int a = 0; a < 9; a++) {

  board[a] = String.valueOf(a+1);

 }

}

}

5 0
3 years ago
What displays the columns dialog box?
kkurt [141]

More Columns Command displays the Column Dialog Box.  Our column choices aren't limited only to the drop-down menu. Select More Columns at the bottom of the menu so as  to access the Columns dialog box.

3 0
3 years ago
Miguel's boss asks him to distribute information about a new dress code as quickly as possible to the entire staff. There are fi
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Figure out how many people are that need the dress code and how many people that don't need the dress code. He should give the people that need dress code the dress codes.


I hope I helped you. :)
4 0
3 years ago
Read 2 more answers
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