Answer:
Following are code that we fill in the given in the given question
if i %2==0:# check the condition
new_list.append(elements[i])# append the list
return new_list # return the new_list
Explanation:
Missing Information :
def skip_elements(elements):# function
new_list=[]# Declared the new list
i=0 #variable declaration
for i in range(len(elements)): #iterating over the loop
if ---------# filtering out even places elements in the list
--------------# appending the list
return -------- # returning the new_list
print(skip_elements(['a','b','c','d','e','f','g'])) #display
Following are the description of the code:
- In this we declared a function skip_elements .In this function we pass the parameter into them .
- After that we declared the new_list array.
- Declared the variable "i" that are initialized with the 0 value .
- Iterating over the for loop and this loop we check the condition in the if condition.
- The if condition executed when the index is even .The statement inside the if condition we will appending the list and returning the list item .
- Finally we print the even number of index list .
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Answer:
a cell is the intersection of a column and row in an excel spreadsheet.
Answer:
True
Explanation:
Yes, the limiting condition of a linked list is the number of the elements that are present in the list. Consider a linked list contains 'n' number of elements, create an iterator which iterates over all the n elements of the linked list. So , in the limiting condition ( for loops , while loops, do while loops in all the looping conditions in fact in any conditions ) the iterator has to iterate over all the elements present in the linked list. So , the limiting condition is the number of elements in the list.
I think it would just output 3
