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Marrrta [24]
2 years ago
13

Which protocol is often used in conjunction with ipsec to provide a remote access client vpn with user authentication?

Computers and Technology
1 answer:
amid [387]2 years ago
3 0

The L2TP or Layer 2 Tunneling Protocol is often used in conjunction with ipsec to provide a remote access client vpn with user authentication.

<h3>What is vpn?</h3>

VPN is known as a Virtual Private Network extends a private network across a public network and activates users to send and receive data across shared or public networks as if their computing devices were directly connected to the private network.

The main difference between VPN proxy and a VPN is encryption and VPNs hides all the web activities performed by the user as well as user's private IP address.

Therefore, the L2TP or Layer 2 Tunneling Protocol is often used in conjunction with ipsec to provide a remote access client vpn with user authentication.

Learn more about VPN here:

brainly.com/question/17272592

#SPJ4

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Answer:

An engine or motor is a machine designed to convert one form of energy into mechanical energy. Heat engines convert heat into work via various thermodynamic processes.

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3 years ago
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What is the process for creating a new merge document for address labels?
Archy [21]

Answer:

In Word, click Mailings > Start Mail Merge > Step-by-Step Mail Merge Wizard to start the mail merge wizard. Choose Labels, and then click Next: Starting document. Choose Label options, select your label vendor and product number, and then click OK. Click Next: Select recipients.

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Problem 2 Write a method that (a) returns an int[] , (b) takes in two sorted arrays (in descending order) and (c) returns an int
Galina-37 [17]

Answer:

Following are the Program in the Java Programming Language.

//define class

public class Main

{

//define function to merge and sort array

public static int[] sort_Array(int[] a, int[] b)

{

//set integer variable and initialize values

int x = 0, y = 0, k = 0, num = 0;

int n = a.length;

int n1 = b.length;

//set integer type array

int[] ar = new int[n+n1];

//set while loop

while (x<n && y <n1)

{

num++;

num++;

//set if conditional statement

if (a[x] > b[y])

ar[k++] = a[x++];

else

ar[k++] = b[y++];

num++;

}

//set the while loop

while (x < n)

{

num++;

ar[k++] = a[x++];

}

//set the while loop

while (y < n1)

{

num++;

ar[k++] = b[y++];

}

//print the total compares done

System.out.println("Number of compares done: " + n);

//return array

return ar;

}

//define main method

public static void main(String[] args)

{

//declaration and the initialization of array

int[] a = new int[]{18,16, 15, 6, 2};

int[] b = new int[]{13, 12, 11, 7, 1};

int[] ar;

//call function

ar = sort_Array(a, b);

//print message

System.out.println("\nThe resultant array is: \n");

//set for loop to print array

for(int x=0; x<ar.length; x++)

{

System.out.print(ar[x] + " ");

}

}

}

<u>Output</u>:

Number of compares done: 5

The resultant array is:

18 16 15 13 12 11 7 6 2 1

Explanation:

Here, we define a class "Main" inside the class, define a function "sort_Array()"and pass two integer type array argument "a" and "b".

  • Set four integer type variables "x", "y", "k", and "num" and initialize them to 0.
  • Set two integer type array "n", "n1" and initialize them the length of the array then, set integer type array variable "ar".
  • Set the while loop then, set if conditional statement then, agin set while loop to sort array in descending order.
  • Then, print the number of compares and return the value of the variable "ar".

Finally, define the main method inside it, we set two integer type array variable and initialize elements in them then, set the for loop to print the sorted elements of the array.

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olya-2409 [2.1K]
The answer to that is D.
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Eva8 [605]

Answer:

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To do this, visit discord.com and login to the web version. Once logged in using the right credentials launch the app on your PC, discord should work properly now. Ensure the system is up to date. As a last resort, uninstall and reinstall the app again.

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