There is no way to define the 'best' site. Each site offers a unique experience along with the fact it might not display the same information as the other ones. The APEX site (As far as I know, it is a school test site) Won't really be able to help unless you are willing to do a few tests. I would go for Scholarshipexperts.com
The answers are as follows:
a) F(A, B, C) = A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC
= A'(B'C' + B'C + BC' + BC) + A((B'C' + B'C + BC' + BC)
= (A' + A)(B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC
= B'(C' + C) + B(C' + C) = B' + B = 1
b) F(x1, x2, x3, ..., xn) = ∑mi has 2n/2 minterms with x1 and 2n/2 minterms
with x'1, which can be factored and removed as in (a). The remaining 2n1
product terms will have 2n-1/2 minterms with x2 and 2n-1/2 minterms
with x'2, which and be factored to remove x2 and x'2, continue this
process until the last term is left and xn + x'n = 1
Answer:
Explanation:
The following code is written in Python. It creates a method for each one of the questions asked and then tests all three with the same test case which can be seen in the picture attached below.
def alternating_list(lst1, lst2):
lst3 = []
for x in range(len(lst1)):
lst3.append(lst1[x])
try:
lst3.append(lst2[x])
except:
pass
if len(lst2) > len(lst1):
lst3.extend(lst2[len(lst1):])
return lst3
def reverse_alternating(lst1, lst2):
lst3 = []
if len(lst1) == len(lst2):
for x in range(len(lst1) - 1, -1, -1):
lst3.append(lst1[x])
lst3.append(lst2[x])
return lst3
def alternating_list_no_extra(lst1, lst2):
lst3 = []
max = 0
if len(lst1) > len(lst2):
max = len(lst2)
else:
max = len(lst1)
for x in range(max):
lst3.append(lst1[x])
try:
lst3.append(lst2[x])
except:
pass
return lst3
False because the internet connects with multiple communication networks to exchange information
A ( micro ) is usually a smaller version of a data warehouse