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densk [106]
3 years ago
7

: Each individual data items of record is called a

Computers and Technology
1 answer:
musickatia [10]3 years ago
6 0
Each individual data items of record is called field (letter A).
The client may communicate with Proxy server to use a protocol to proxy the communication between the client and the DBMS. Proxy servers lets you hide your real Ip address and replaces it with a new IP obtained from proxy server sites.

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There are two kinds of emotions: positive and negative. True False
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False

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can we get a picture of the problem ?

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Which RFC reserves three ranges of IP addresses for private use - a single Class A (10.0.0.0-10.255.255.255), 16 Class Bs (172.1
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Answer:

The answer is "1918".

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Array Challenge Have the function ArrayChallenge (arr) take the array of numbers stored in arr and return the string true if any
vampirchik [111]

Answer:

The code is given as follows,

Explanation:

Code:

#include <stdio.h>

#include <string.h>  

int n; //to store size of array  

char* ArrayChallenge(int arr[]) //function returns string true or false

{

  int i, j;

  int sum = 0;

  for(i = 0; i < n; i++)

  {

      sum = sum + arr[i]; //count sum

  }  

  for(i = 0; i < n; i++)

  {

      for(j = i+1; j < n; j++) //loop for every two elements in array

      {

          if(arr[i]*arr[j] > 2*sum) //check if proudct of two elements > 2 times sum

          {

              printf("\n%d x %d = %d, 2xSum = %d\n", arr[i], arr[j], arr[i]*arr[j], 2*sum);

              return "true"; //If proudct of two elements > 2 times sum. return true

          }

      }

  }

  return "false"; // If proudct of two elements < 2 times sum. return false

}  

int main()

{  

  printf("\nEnter size of array: ");

  scanf("%d", &n); //read size of array

  int A[n]; //array of size n

  printf("\nEnter array elements: ");

  int i;

  for(i = 0; i < n; i++)

  {

      scanf("%d", &A[i]); //read array from stdin

  }

  printf("%s\n",ArrayChallenge(A)); //ccall function and print answer

 

  return 0;

}

6 0
3 years ago
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