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Anarel [89]
3 years ago
11

Need help this will affect my final grade in technology and the dead line has passed!!!

Computers and Technology
1 answer:
Setler79 [48]3 years ago
6 0
The four<span> separate </span>strokes<span> are termed: ... While the piston is at T.D.C. </span><span>the compressed air-fuel mixture is ignited by a spark plug (in a </span>gasoline engine) or by heat generated through high compression<span>, forcefully returning the piston to B.D.C.

Hope this helps! : )</span>
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You have users who connect to the corporate network using their laptops. because these computers often access confidential data,
Triss [41]
<span>You implement NAP Enforcement for 802.1X</span>
7 0
3 years ago
The block of code below is supposed to display “multiple of 5” if the positive number value is in fact a multiple of 5
12345 [234]

Answer:

Answer is (value MOD 5) = 0.

Explanation:

A number A is multiple of another number B  if after dividing A by B remainder value is zero.

In mathematics modulo or (MOD) function is a function which returns remainder value after dividing two numbers.

Here in question for condition

( value MOD 5 ) = 0 code will print "Multiple of 5". For each value which is divisible by 5 we will get zero remainder implying that value is a multiple of 5.

3 0
3 years ago
Assume that getPlayer2Move works as specified, regardless of what you wrote in part (a) . You must use getPlayer1Move and getPla
kakasveta [241]

Answer:

(1)

public int getPlayer2Move(int round)

{

  int result = 0;

 

  //If round is divided by 3

  if(round%3 == 0) {

      result= 3;

  }

  //if round is not divided by 3 and is divided by 2

  else if(round%3 != 0 && round%2 == 0) {

      result = 2;

  }

  //if round is not divided by 3 or 2

  else {

      result = 1;

  }

 

  return result;

}

(2)

public void playGame()

{

 

  //Initializing player 1 coins

  int player1Coins = startingCoins;

 

  //Initializing player 2 coins

  int player2Coins = startingCoins;

 

 

  for ( int round = 1 ; round <= maxRounds ; round++) {

     

      //if the player 1 or player 2 coins are less than 3

      if(player1Coins < 3 || player2Coins < 3) {

          break;

      }

     

      //The number of coins player 1 spends

      int player1Spends = getPlayer1Move();

     

      //The number of coins player 2 spends

      int player2Spends = getPlayer2Move(round);

     

      //Remaining coins of player 1

      player1Coins -= player1Spends;

     

      //Remaining coins of player 2

      player2Coins -= player2Spends;

     

      //If player 2 spends the same number of coins as player 2 spends

      if ( player1Spends == player2Spends) {

          player2Coins += 1;

          continue;

      }

     

      //positive difference between the number of coins spent by the two players

      int difference = Math.abs(player1Spends - player2Spends) ;

     

      //if difference is 1

      if( difference == 1) {

          player2Coins += 1;

          continue;

      }

     

      //If difference is 2

      if(difference == 2) {

          player1Coins += 2;

          continue;

      }

     

     

  }

 

  // At the end of the game

  //If player 1 coins is equal to player two coins

  if(player1Coins == player2Coins) {

      System.out.println("tie game");

  }

  //If player 1 coins are greater than player 2 coins

  else if(player1Coins > player2Coins) {

      System.out.println("player 1 wins");

  }

  //If player 2 coins is grater than player 2 coins

  else if(player1Coins < player2Coins) {

      System.out.println("player 2 wins");

  }

}

3 0
3 years ago
A complete traversal of an n node binary tree is a(n)____ "operation if visiting a node is O(1)for the iterative implementation
ch4aika [34]

Answer:

B.O(n).

Explanation:

Since the time complexity of visiting a node is O(1) in iterative implementation.So the time complexity of visiting every single node in binary tree is O(n).We can use level order traversal of a binary tree using a queue.Which can visit every node in O(n) time.Level order traversal do it in a single loop without doing any extra traversal.

8 0
3 years ago
Is technology a legal discipline or law is a technological artifact.
ANEK [815]

Answer:Technology law scholars have recently started to consider the theories of affordance and technological mediation, imported from the fields of psychology, human-computer interaction (HCI), and science and technology studies (STS). These theories have been used both as a means of explaining how the law has developed, and more recently in attempts to cast the law per se as an affordance. This exploratory paper summarises the two theories, before considering these applications from a critical perspective, noting certain deficiencies with respect to potential normative application and definitional clarity, respectively. It then posits that in applying them in the legal context we should seek to retain the relational user-artefact structure around which they were originally conceived, with the law cast as the user of the artefact, from which it seeks certain features or outcomes. This approach is effective for three reasons. Firstly, it acknowledges the power imbalance between law and architecture, where the former is manifestly subject to the decisions, made by designers, which mediate and transform the substance of the legal norms they instantiate in technological artefacts. Secondly, from an analytical perspective, it can help avoid some of the conceptual and definitional problems evident in the nascent legal literature on affordance. Lastly, approaching designers on their own terms can foster better critical evaluation of their activities during the design process, potentially leading to more effective ‘compliance by design’ where the course of the law’s mediation by technological artefacts can be better anticipated and guided by legislators, regulators, and legal practitioners.

Keywords

Affordance, technological mediation, postphenomenology, legal theory, compliance by design, legal design

7 0
3 years ago
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