A Peer-to-peer network architecture would work best for Edward and Matthew.
Although they have the option of creating a Client-Server type of network architecture, the number of people that they are currently working with would cost them a lot more than a peer-to-peer architecture.
One of the advantages of using a Peer-to-peer network architecture for a small network would be the access of files between all computers in the network. Even if one of the computers will fail, the other computers will still have access to the different files and information they will need to keep the company moving forward.
<u>Paired programming:</u>
Paired programming is a new technique where one person gets one another to<em> write the code</em>. Here one will write the code and other will start verifying as the type.
So we might think it is waste to employ two person for same task but it is not so. There are many <em>benefits in incorporating this technique</em>. Let us see those benefits one by one.
- <em>Inter-personal skill gets improved
</em>
- <em>Lower amount of coding mistake
</em>
- <em>Learn from each other
</em>
- <em>Peer review enhance collaboration
</em>
Challenges expect to arise during pair-programming
-
Two heads are superior to one. On the off chance that the driver experiences a <em>hitch with the code</em>, there will be two of them who'll take care of the issue.
- Builds up your <em>staff's relational aptitudes</em>. Working together on a solitary venture encourages your group to welcome the estimation of <em>correspondence and collaboration. </em>
Overcome Method
The most ideal approach to move <em>toward blending</em> is to accomplice two software engineers and have them share a PC. The pair ought to have the option to choose how to part the work, and it is prudent that they should <em>switch jobs frequently.</em>
This is in the wrong caterigore and yes the answer is 8.0 × 103 N
Answer:
see explaination
Explanation:
void insertion( int e,int *x, int start, int end)
{
if (e >= x[end])
x[end+1] = e;
else if (start < end)
{
x[end+1] = x[end];
insertion(e, x, start, end-1);
}
else
{
x[end+1] = x[end];
x[end] = e;
}
}
void insertion_recurssion(int *b, int start, int end)
{
if(start < end)
{
insertion_sort_recur(b, start, end-1);
insertion(b[end], b, start, end-1);
}
}
void main()
{
insertion_recurssion(x,0,5);
}
