The view that perpetual processes can be thought of in terms of a software/hardware metaphor is known as the: information processing view.
<h3>What is the Information Processing View?</h3>
Information processing view is explained by the cognitive theory to explain how the brain encodes information and how information are filtered from what we pay attention to in a particular moment. This also determines what is stored in the short-term or in our long-term memory.
Therefore, the view that perpetual processes can be thought of in terms of a software/hardware metaphor is known as the: information processing view.
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Solution :
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Below, I believe are the multiple choices attached to this question
A. Power, Structural, and Technical Systems.
B. Construction.
C. Design/ Pre-Construction.
D. Maintenance/ Operations
The answer is A: Power, Structural, and Technical Systems.
The career pathways in the Architecture and Construction deal with all aspects of designing, planning, and maintaining all kinds of structures we live or work in. It is organized into 3 career pathways; Construction, Design/ PreConstruction, and Maintenance/ Operations. This field also covers the servicing of equipment such as plumbing, electrical wiring, escalators, and elevators.
Greek word origins be helpful at a doctors offics
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
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As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
