Answer:
Time to live.
Explanation:
The Time to Live is used for the purpose of the expiration table in the rows dynamically.The Time to live is represented in the time limit to stay in the database the Time to live It is no longer possible to fetch the data that has passed its closing timeout value also it is not displayed in the store metrics.
The Time to live is available on the host in the log informing the system how long the record will stay in the database after it has been generated or last modified.
Answer:
character count
Explanation:
The answer is probably character count.
Answer:
For question a, it simplifies. If you re-express it in boolean algebra, you get:
(a + b) + (!a + b)
= a + !a + b
= b
So you can simplify that circuit to just:
x = 1 if b = 1
(edit: or rather, x = b)
For question b, let's try it:
(!a!b)(!b + c)
= !a!b + !a!bc
= !a!b(1 + c)
= !a!b
So that one can be simplified to
a = 0 and b = 0
I have no good means of drawing them here, but hopefully the simplification helped!
Answer:
print("Let's play Silly Sentences!")
print(" ")
name=input("Enter a name: ")
adj1=input("Enter an adjective: ")
adj2=input("Enter an adjective: ")
adv=input("Enter an adverb: ")
fd1=input("Enter a food: ")
fd2=input("Enter another food: ")
noun=input("Enter a noun: ")
place=input("Enter a place: ")
verb=input("Enter a verb: ")
print(" ")
print(name + " was planning a dream vacation to " + place + ".")
print(name + " was especially looking forward to trying the local \ncuisine, including " + adj1 + " " + fd1 + " and " + fd2 + ".")
print(" ")
print(name + " will have to practice the language " + adv + " to \nmake it easier to " + verb + " with people.")
print(" ")
print(name + " has a long list of sights to see, including the\n" + noun + " museum and the " + adj2 + " park.")
Explanation:
Got it right. Might be a longer version, but it worked for me.
Answer:
Explanation:
The following code is written in Python. It is a recursive function that tests the first and last character of the word and keeps checking to see if each change would create the palindrome. Finally, printing out the minimum number needed to create the palindrome.
import sys
def numOfSwitches(word, start, end):
if (start > end):
return sys.maxsize
if (start == end):
return 0
if (start == end - 1):
if (word[start] == word[end]):
return 0
else:
return 1
if (word[start] == word[end]):
return numOfSwitches(word, start + 1, end - 1)
else:
return (min(numOfSwitches(word, start, end - 1),
numOfSwitches(word, start + 1, end)) + 1)
word = input("Enter a Word: ")
start = 0
end = len(word)-1
print("Number of switches required for palindrome: " + str(numOfSwitches(word, start, end)))
