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maria [59]
4 years ago
9

Game Design Help please

Computers and Technology
1 answer:
valina [46]4 years ago
4 0
This game on cool math
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Can i have help for a ggogle class room
andrew11 [14]

Answer:

or tell ur teacher to add u in ur class

Explanation:

3 0
3 years ago
Read 2 more answers
How do you change the name on your brainly account?
serg [7]

Answer:

For changing your username you can contact to administrator of the site. If you are using app then go to message option. Then search "rishilaugh" then you can message him. Hope it helps you!

8 0
4 years ago
Write a static method middleValue that takes three int parameters, and returns a int . It should return the middle value of the
anygoal [31]

We use if-else structure to check the each possible scenario and return the median accordingly in the middleValue() method. The main is also provided so that you can test the method.

Comments are used to explain the each line.

You may see the output in the attachment.

public class Main

{

public static void main(String[] args) {

   

    //call the method for different scenarios

    System.out.println(middleValue(1, 2, 3));

    System.out.println(middleValue(1, 3, 2));

    System.out.println(middleValue(2, 1, 3));

    System.out.println(middleValue(2, 3, 1));

    System.out.println(middleValue(3, 1, 2));

    System.out.println(middleValue(3, 2, 1));

 

}

       //method that takes three int and returns an int

public static int middleValue(int n1, int n2, int n3) {

    //set the median as n1

    int median = n1;

   

    //check the situation where the n1 is the highest

    //if n2 is greater than n2 -> n1 > n2 > n3

    //if not -> n1 > n3 > n2

    if(n1 > n2 && n1 > n3){

        if(n2 > n3)

            median = n2;

        else

            median = n3;

    }

   

    //check the situation where the n2 is the highest

    //if n3 is greater than n1 -> n2 > n3 > n1

    //if not -> n2 > n1 > n3

    //note that we set the median as n1 by default, that is why there is no else part

    else if(n2 > n1 && n2 > n3){

        if(n3 > n1)

            median = n3;

    }

   

    //otherwise, n3 is the highest

    //if n2 is greater than n1 -> n3 > n2 > n1

    //if not -> n3 > n1 > n2

    //note that we set the median as n1 by default, that is why there is no else part

    else{

        if(n2 > n1)

            median = n2;

    }

   

    return median;

}

}

You may see another if-else question at:

brainly.com/question/13428325

6 0
3 years ago
Write a C++ program to find all numbersless than 1000 which are:
Elden [556K]

Answer:

The c++ program is shown below.

#include <iostream>

using namespace std;

int main() {    

   cout<<"Numbers less than 1000 which are divisible by 7"<<endl;    

   for(int k=1; k<1000; k++)

   {

       // number should give remainder 0 which show complete divisibility

       if(k%7 == 0)

           cout<<k<<"\t";

   }    

   cout<<endl<<"Numbers less than 1000 which are divisible by 11"<<endl;    

   for(int k=1; k<1000; k++)

   {

       if(k%11 == 0)

           cout<<k<<"\t";

   }    

   cout<<endl<<"Numbers less than 1000 which are divisible by 7 but not divisible by 11"<<endl;    

   for(int k=1; k<1000; k++)

   {

       // for 11, number should not give remainder 0 which shows incomplete divisibility

       if(k%7 == 0 && k%11 != 0)

           cout<<k<<"\t";

   }    

   cout<<endl<<"Numbers less than 1000 which are divisible by 7 and divisible by 11"<<endl;

   

   for(int k=1; k<1000; k++)

   {

       if(k%7 == 0 && k%11 == 0)

           cout<<k<<"\t";

   }    

   cout<<endl<<"Numbers less than 1000 which are not divisible by 7 and not divisible by 11"<<endl;    

   for(int k=1; k<1000; k++)

   {

       if(k%7 != 0 && k%11 != 0)

           cout<<k<<"\t";

   }    

   return 0;

}

Explanation:

The test for divisibility is done by using the modulus operator which is used as a condition inside the if statement. This test is done inside for loop.

All the numbers from 1 to 999, less than 1000, are divided by 7 and/ or 11 depending on the sub question. Only the numbers which are completely divisible are displayed. Divisible numbers give remainder 0 always.

The divisibility test by 7 is shown below.

cout<<"Numbers less than 1000 divisible by 7"<<endl;    

   for(int k=1; k<1000; k++)

   {

       if(k%7 == 0)

           cout<<k<<"\t";

   }    

In other words, all the numbers divisible by 7 are same as the numbers in the table of 7.

The same logic shown above is applied for other sub questions to test for divisibility by 11 and combination of 7 and 11.

To improve readability, tabs and new lines are inserted at appropriate places.

8 0
3 years ago
A device which is not connected to the cpu is known as?​
Dvinal [7]

Answer:

would be called an Off-line device

3 0
3 years ago
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