You should get up and move every 30 minutes
Hope this helps
-scav
Answer:
breaking a problem into smaller parts
Explanation:
In language of computer science, decomposition is the process in which complex problems are divided into simpler parts. These simpler parts helps in the solving of the complex problems. They are made easier and comprehensible for the user to understand. Each simple part is further analyzed separately and the solution of the complex issues are derived. The process becomes lengthy but the solution is found.
Answer:
You need to explain the entire network layout first.
Explanation:
Bringing on new IT Staff can be time consuming. But depending on the possession you need to explain to them how the domain lay out is.
Answer:
writing code for a software program.
Explanation:
Software development life cycle (SDLC) can be defined as a strategic process or methodology that defines the key steps or stages for creating and implementing high quality software applications.
Some of the models used in the software development life cycle (SDLC) are; waterfall model, incremental model, spiral model, agile model, big bang model, and V-shaped model.
The five (5) standard stages of development in software development are;
I. Analysis.
II. Design.
III. Implementation (coding).
IV. Testing.
V. Maintenance.
A typical day in programming and software development would involve writing code for a software program.
Generally, software applications or programs require a code containing series of sequential instruction to perform specific tasks, commands and processes. These sets of code are typically written by a software developer (programmer).
Answer:
boolean isEven = false;
if (x.length % 2 == 0)
isEven = true;
Comparable currentMax;
int currentMaxIndex;
for (int i = x.length - 1; i >= 1; i--)
{
currentMax = x[i];
currentMaxIndex = i;
for (int j = i - 1; j >= 0; j--)
{
if (((Comparable)currentMax).compareTo(x[j]) < 0)
{
currentMax = x[j];
currentMaxIndex = j;
}
}
x[currentMaxIndex] = x[i];
x[i] = currentMax;
}
Comparable a = null;
Comparable b = null;
if (isEven == true)
{
a = x[x.length/2];
b = x[(x.length/2) - 1];
if ((a).compareTo(b) > 0)
m = a;
else
m = b;
}
else
m = x[x.length/2];
