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Otrada [13]
3 years ago
12

You have three users who travel to four branch offices often and need to log on to the RODCs at these offices. The branch office

s are connected to the main office with slow WAN links. You don’t want domain controllers at the main office to authenticate these four users when they log on at the branch offices. What should you do that requires the least administrative effort yet adheres to best practices?
Computers and Technology
1 answer:
liberstina [14]3 years ago
8 0

Answer:

Configure Caching on RODCs using PRP.

Explanation

Read only domain controller is a domain controller that allows active directories for read only purposes. An administrator can setup RODCs for branches of a company and control the domain controller of the database server, updating the database which can be viewed as a read only partition in RODC.

Authenticating users in a RODC rely on the domain controller to forward the credentials to the read only domain controller. This can impose to high traffic on a slow WAN network link, especially when many users are trying to authenticate, resulting to high bandwidth.

To avoid this, credential caching by configuring password replication policy (PRP) on the RODC is vital.  

When PRP is configure on RODC, the user only get to authenticate once, then the user password is replicated and encrypted for subsequent use.

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In mathematics, "quadrant I" of the cartesian plane is the part of the plane where x and y are both positive. Given a variable,
lbvjy [14]

Answer:

#include <iostream>

using namespace std;

struct Cartesian {

double x;

double y;

};

int main() {

// creating a pointer p of type struct Cartesian

struct Cartesian *p = new Cartesian ;

cout << " Enter x: ";

// accessing the structure variable x by arrow "->"

cin >> p->x;

cout << "Enter y: ";

// accessing the structure variable y by arrow "->"

cin >> p->y;

// expression to check whether x and y lie in Quadrant 1

if (p->x > 0 && p->y > 0) {

 cout << "X and Y are in Quadrant 1 ";

}

else

{

 cout << "x And y are not in Quadrant 1";

}

// deleting the struct pointer p

delete p;

return 0;

}

Explanation:

in order to locate memory in heap, keyword "new" is used in C++ so,

struct Cartesian *p = new Cartesian ;

in order to access data members of the structure using pointer we always use an arrow "->". otherwise a dot oprerator is used to access data members.

in order to check whether x and y lie in 1st quadrent, we must use && operator in our if condition. so

if (p->x > 0 && p->y > 0)

&& will return true iff x and y both are +ve.

deleting the struct pointer p is important beacuse we are allocating memory in heap so heap memory should be used a resource and must be release when not needed otherwise it can cause memory leakage problems. so  

delete p;

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3 years ago
An array subscript can be an expression, but only as long as the expression evaluates to what type?
densk [106]

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Write a program named Deviations.java that creates an array with the deviations from average of another array. The main() method
Westkost [7]

Answer:

hello your question is incomplete attached is the complete question and solution

answer : The solution is attached below

Explanation:

Below is a program named Derivations.java that creates an array with the deviations from average of another array.

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3 years ago
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