terms = ["Bandwidth", "Hierarchy", "IPv6", "Software", "Firewall", "Cybersecurity", "Lists", "Program", "Logic",
"Reliability"]
def swap(arr, in1, in2):
w = arr[in1]
arr[in1] = arr[in2]
arr[in2] = w
print(terms)
swap(terms, 5, 1)
swap(terms, 2, 4)
swap(terms, 3, 5)
swap(terms, 5, 6)
swap(terms, 6, 8)
swap(terms, 8, 9)
print(terms)
This is how I interpreted the question. If I need to make any changes, I'll do my best. Hope this helps though.
Answer:
n = f = 7
loop f = gt(n,1) f * (n = n - 1) :s(loop)
output = f
end
Explanation:
Answer:
Following are the response to the given question:
Explanation:
The glamorous objective is to examine the items (as being the most valuable and "cheapest" items are chosen) while no item is selectable - in other words, the loading can be reached.
Assume that such a strategy also isn't optimum, this is that there is the set of items not including one of the selfish strategy items (say, i-th item), but instead a heavy, less valuable item j, with j > i and is optimal.
As 
, the i-th item may be substituted by the j-th item, as well as the overall load is still sustainable. Moreover, because 
and this strategy is better, our total profit has dropped. Contradiction.
Answer:
d) daco = new Banana;
Explanation:
Dynamically allocated variables have their memory allocated in the heap memory.We declare a dynamical variable like this:-
int *a=new int ;
It means a pointer a is created on the stack memory which hold the address of the block that hold the value of variable a in heap memory.
We already have the pointer daco. We just have to initialize with keyword new.
It will be like daco=new Banana; which matches the option d.
