Answer:
The Following are the solution to this question:
Explanation:
In Option a:
In the point (i) 
is transitive, which means it converts one action to others object because if 
indicates 
. It's true by definition, that becomes valid. But if 
, which implies 
. it's a very essential component. If 
. They 
will also be 
.
In point (ii), The value of 
is convergent since the 
. It means they should be dual a and b constant variable, therefore 
could only be valid for the constant variable, that is 
.
In Option b:
In this algorithm, the input size value is equal to 1 object, and the value of A is a polynomial-time complexity, which is similar to its outcome that is 
. It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop
. All internal loops operate on a total number of 
generations and therefore the final time complexity is .
Answer:
Round-Robin DNS
Explanation:
According to my experience in Information Technology and Networking it can be said that based on the information provided the best solution would be to set up a Round-Robin DNS. This term refers to a technique used to balance the load on a server, where a client request is sent to each server one at a time, and then the system repeats the process from the top of the request list. This prevents the server from being drowned in a sea of simultaneous requests.
If you have any more questions feel free to ask away at Brainly.
Answer:
I believe the answer is B.
Answer:
In general the number of bit registers in Intel 80x86 CPU design when combined together forms a 16 - bit register
An example of the -bit registers are AH, AL, BH, BL, CH, CL, DH, and DL
Explanation:
Solution
The 8086 CPU design has a total of eight 8-bit registers and these register can be integrated together to make 16- bit register as well.
The 16-bit data is stored by breaking the data into a low-order byte and high order byte.
The name of the 8 bit registers is shown below:
AH, AL, BH, BL, CH, CL, DH, and DL
Answer:
Code is given below and output is attached as an image.
Explanation:
#include <iostream>
#include <fstream>
using namespace std;
bool isPalindrome(int n)
{
// Find reverse of n
int rev = 0;
for (int i = n; i > 0; i /= 10)
rev = rev * 10 + i % 10;
// If n and rev are same,then n is a palindrome
return (n == rev);
}
int main()
{
int min = 1; // Lower Bound
int max = 200; // Upper Bound
ofstream myfile;
myfile.open("palindrome.txt");
for (int i = min + 1; i < max; i++)
if (isPalindrome(i))
myfile << i << endl;
myfile.close();
return 0;
}
