It is - not stereospecific
, a reaction in which the stereochemistry of the reactants controls the outcome of the reaction; - bimolecular at rate-determining step
, it involves two molecules; - first order, - rates is governed by the stability of the carbocation that is formed.
The farming of maize began in Mesoamerica in 6000 B.C.
The domestication of maize made the city of Tenochtitlan possible by providing the Aztecs the ability to feed millions. Unlike the corn that is known today, the early maize had tiny cobs but dramatically became bigger and more nutritious as time passed.
Answer:
from collections import Counter
def anagram(dictionary, query):
newList =[]
for element in dictionary:
for item in query:
word = 0
count = 0
for i in [x for x in item]:
if i in element:
count += 1
if count == len(item):
newList.append(item)
ans = list()
for point in Counter(newList).items():
ans.append(point)
print(ans)
mylist = ['jack', 'run', 'contain', 'reserve','hack','mack', 'cantoneese', 'nurse']
setter = ['ack', 'nur', 'can', 'con', 'reeve', 'serve']
anagram(mylist, setter)
Explanation:
The Counter class is used to create a dictionary that counts the number of anagrams in the created list 'newList' and then the counter is looped through to append the items (tuple of key and value pairs) to the 'ans' list which is printed as output.
True
yes computers are capable of serving network servers.
The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.
int recursiveOddSum(int n) {
if(2n-1==1) return 1;
return (2n-1) + recursiveOddSum(n-1);
}
To prove the correctness of this algorithm by induction, we start from the base case as usual:
by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).
Now we can assume that 
returns indeed the sum of the first n-1 odd numbers, and we have to proof that 
returns the sum of the first n odd numbers. By the recursive logic, we have
and by induction, is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So, is the sum of the first n odd numbers, as required:
