Please help i will give you brainelst answer part B thank you!!!!!!!!!!!!!1
1 answer:
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Answer:
A and C
Explanation:
Option A:
In IPv6 there is a rule to reduce an IPv6 address when there are two or more consecutive segments of zeros just one time. This rule says that you can change the consecutive zeros for “::”
Here is an example
How to reduce the following IPv6 address?
ff02:0000:0000:0000:0000:0000:0000:d500
Ans: ff02::d500
Example 2:
2001:ed02:0000:0000:cf14:0000:0000:de95
Incorrect Answer -> 2001:ed02::cf14::de95
Since the rule says that you can apply “::” just one time, you need to do it for a per of zero segments, so the correct answer is:
Correct Answer -> 2001:ed02::cf14:0:0:de95
Or
2001:ed02:0:0:cf14::de95
Option C:
Since in IPv6 there are available addresses which means 340.282.366.920.938.463.463.374.607.431.768.211.456 (too many addresses), there is no need of NAT solution, so each device can have its own IP address by the same interface to have access through the internet if needed. If not, you can block the access through internet by the firewall.
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;
}
