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valina [46]
3 years ago
5

If you had three extra siblings, what would be your birth order and what personalities would you like them to have?

Computers and Technology
2 answers:
Goshia [24]3 years ago
8 0

Answer:

I already have 4 step brothers, 2 real brothers, 1 step sister, and a half sister, so 3 more people would be a lot. If I got to pick their personalities and gender, I would want 1 more sister and 2 more brothers. I would watn a more alt brother, a more fem brother, and an indie sister. Kinda just an aesthetic type for them.

Annette [7]3 years ago
3 0

Answer:

mixed personalities l

that will be cool

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Which statement best describes one reason why assembly language is easier
viktelen [127]

Answer:

machine language uses binary code and assembly language uses mnemonic codes to write a program.

Explanation:

In a nutshell, machine language uses binary code, which is almost impossible for humans to decipher, whereas assembly language uses mnemonic codes to write a program. Mnemonic codes make it simpler for humans to understand or remember something, and so make the language a bit easier for humans to use than machine code.

6 0
2 years ago
We need goku and naruto fight story someone
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Answer:

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Explanation:

8 0
3 years ago
Read 2 more answers
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

5 0
3 years ago
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NemiM [27]
Possibly rephrase or rewrite the heading or whatever else you decide to repeat. You should never say the exact thing twice. 
Hope I helped :)
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3 years ago
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Answer:

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Explanation:

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