Answer:
By being a part of a block to a block chain and each block in the block chain is precisely connected to the block that was before it.
Explanation:
I Hope This Helps You
Answer:
The most probable reason for this is A.
Explanation:
B should be wrong because the TV being connected to a different network segment still means that it is on the Wi-Fi network, it is just using a different switch or a repeater.
C should be wrong because the brightness of the TV does not have anything to do with being able to connect to the phone.
D should be wrong because in the question itself it says that "you tap the mirroring option on your device" which clearly means that the mobile device supports display mirroring.
The answer should be A, if the device and the TV are not on the same Wi-Fi network, than they simply can not connect and the mirroring can not be done.
I hope this answer helps.
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.
</span>
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.
Answer:
The function is as follows:
def divisible_by(listi, n):
mylist = []
for i in listi:
if i%n == 0:
mylist.append(i)
return mylist
pass
Explanation:
This defines the function
def divisible_by(listi, n):
This creates an empty list
mylist = []
This iterates through the list
for i in listi:
This checks if the elements of the list is divisible by n
if i%n == 0:
If yes, the number is appended to the list
mylist.append(i)
This returns the list
return mylist
pass
