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Inessa05 [86]
3 years ago
10

Which of the following would allow for the QUICKEST restoration of a server into a warm recovery site in a case in which server

data mirroring is not enabled?a. Full backupb. incremental backupc. Differential back upd. Snapshot
Computers and Technology
1 answer:
adelina 88 [10]3 years ago
5 0

Answer: C. Differential backup

Explanation: There are several ways od ensuring the preservation and storage of data even cases of disaster, one of such ways is data data mirroring which allows data to be replicated or copied in real time and several backup options. In cases where there there is need to restore a server, the warm recovery site provides a data or disaster recovery option used to mitigate the effect of data loss on organization. In the absence of data mirroring, differential backup option, provides the quickest recovery option as it only requires changes in the data stored after the last full backup. These speed experieced should be expected due to the relatively low data been dealt with rather than the entire data.

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What type of IPv6 address should you use when you have multiple routers on a subnet and want hosts to use the nearest router for
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2 years ago
Remove gray from RGB Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers repres
sergejj [24]

Answer:  

Here is the C++ program:  

#include <iostream>   //to use input output functions    

using namespace std;   //to identify objects cin cout    

int main() {   //start of main method    

int red,green,blue,smallest;   //declare variables to store integer values of red,green, blue and to store the smallest value    

cout<<"Enter value for red: ";  //prompts user to enter value for red    

cin>>red;  //reads value for red from user    

cout<<"Enter value for green: ";  //prompts user to enter value for green  

cin>>green; //reads value for green from user    

cout<<"Enter value for blue: "; //prompts user to enter value for blue    

cin>>blue;   //reads value for blue from user    

//computes the smallest value  

if(red<green && red<blue) //if red value is less than green and blue values    

smallest=red;   //red is the smallest so assign value of red to smallest    

else if(green<blue)   //if green value is less than blue value    

smallest=green; //green is the smallest so assign value of green to smallest  

else //this means blue is the smallest    

smallest=blue;  //assign value of blue to smallest    

//removes gray part by subtracting smallest from rgb  

red=red-smallest; //subtract smallest from red    

green=green-smallest; //subtract smallest from green    

blue=blue-smallest; //subtract smallest from blue    

cout<<"red after removing gray part: "<<red<<endl;  //displays amount of red after removing gray    

cout<<"green after removing gray part: "<<green<<endl; //displays amount of green after removing gray  

cout<<"blue after removing gray part: "<<blue<<endl;  } //displays amount of blue after removing gray  

Explanation:  

I will explain the program using an example.    

Lets say user enter 130 as value for red, 50 for green and 130 for blue. So  

red = 130    

green = 50

blue = 130  

First if condition if(red<green && red<blue)   checks if value of red is less than green and blue. Since red=130 so this condition evaluate to false and the program moves to the else if part else if(green<blue) which checks if green is less than blue. This condition evaluates to true as green=50 and blue = 130 so green is less than blue. Hence the body of this else if executes which has the statement: smallest=green;  so the smallest it set to green value.    

smallest = 50    

Now the statement: red=red-smallest; becomes:    

red = 130 - 50    

red = 80    

the statement: green=green-smallest;  becomes:    

green = 50 - 50    

green = 0    

the statement: blue=blue-smallest; becomes:    

blue = 130 - 50    

blue = 80    

So the output of the entire program is:    

red after removing gray part: 80                                                                                                 green after removing gray part: 0                                                                                                blue after removing gray part: 80    

The screenshot of the program along with its output is attached.

5 0
2 years ago
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