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adelina 88 [10]
2 years ago
9

When you propose a solution to your argument during the conflict resolution process, you should make sure you are using ______.

Computers and Technology
2 answers:
max2010maxim [7]2 years ago
7 0

None of the options is true. Correct answer: D In contrast, when you propose a solution to your argument during the conflict resolution process, you should make sure you are not using arguments from emotion, ad hominem arguments or neutral arguments. You should use strict, objective arguments and be not lead by emotions.

Sophie [7]2 years ago
4 0
The correct answer that would best complete the given statement above would be option C. <span>When you propose a solution to your argument during the conflict resolution process, you should make sure you are using neutral arguments. This is to avoid bias and further misunderstanding or conflict. Hope this answer helps.</span>
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If the input is negative, make numItemsPointer be null. Otherwise, make numItemsPointer point to numItems and multiply the value
Brilliant_brown [7]

Answer:

The code for the given statements are described below.

Explanation:

// Place code below in line 9

if(numItems < 0)  // starting of a loop

{

numItemsPointer = NULL;

}

else  

{

numItemsPointer = &numItems;

numItems = numItems * 10;   // items multiplied by 10

}   // ending of a loop

6 0
2 years ago
Write a program that first gets a list of integers from input. The input begins with an integer indicating the number of integer
Naya [18.7K]

Answer:

The program to this question as follows:

Program:

#include<iostream> //include header file.

using namespace std; //using name space.

int main() //main function.

{

int n,i,key; //define variable.

int a[n]; //define array

cout<<"Enter size of array :"; //message.

cin>>n; //input number from user.

cout<<"Enter array elements :"; //message

for(i = 0;i<n;i++) //loop

{

cin>>a[i]; //input array elements

}

cout<<"Enter a number:"; //message

cin>>key; //input number.

for(i = 0;i<n;i++) //loop

{

if(a[i]<=key) //if block

{

cout<<a[i]<<"\n "; //print array elements

}

}

return 0;

}

Output:

Enter size of array :7

Enter array elements :5

50

50

75

100

200

140

Enter a number:100

5

50

50

75

100

Explanation:

The description of the above program as follows:

  • In this program first, we include a header file and define the main method in method we define variables and array that are "n, i, key and a[]". In this function, all variable data type is "integer".
  • The variable n is used for the size of array and variable i use in the loop and the key variable is used for comparing array elements. Then we use an array that is "a[]" in the array, we use the for loop to insert elements form user input.  
  • Then we define a loop that uses a conditional statement in if block we check that array elements is less than and equal to a key variable. If this value is true so, we will print the remaining elements.  

4 0
2 years ago
This exercise shows why each pivot (in eli1nination by pivoting) must be in a different row. (a) In Example 7, make the third pi
Darya [45]

Answer:gghggggggggggggg

Explanation:

8 0
2 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
2 years ago
The following code should take a number as input, multiply it by 8, and print the result. In line 2 of the code below, the * sym
Juli2301 [7.4K]

Answer:

num = int(input("enter a number:"))

print(num * 8)

Explanation:

num is just a variable could be named anything you want.

if code was like this num = input("enter a number:")

and do a print(num * 8)

we get an error because whatever the user puts in input comes out a string.

we cast int() around our input() function to convert from string to integer.

therefore: num = int(input("enter a number:"))

will allow us to do  print(num * 8)

6 0
2 years ago
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