A. browser
Web browsers are programs like internet explorer, safari, chrome. You use them to access web pages on the internet.
Answer:
sorry for the wait but the answer is b
Explanation:
The python program that creates a Bankaccount class for a Bank ATM that is made up of the customers and has a deposit and withdrawal function is given below:
<h3>Python Code</h3>
# Python program to create Bankaccount class
# with both a deposit() and a withdraw() function
class Bank_Account:
def __init__(self):
self.balance=0
print("Hello!!! Welcome to the Deposit & Withdrawal Machine")
def deposit(self):
amount=float(input("Enter amount to be Deposited: "))
self.balance += amount
print("\n Amount Deposited:",amount)
def withdraw(self):
amount = float(input("Enter amount to be Withdrawn: "))
if self.balance>=amount:
self.balance-=amount
print("\n You Withdrew:", amount)
else:
print("\n Insufficient balance ")
def display(self):
print("\n Net Available Balance=",self.balance)
# Driver code
# creating an object of class
s = Bank_Account()
# Calling functions with that class object
deposit()
s.withdraw()
s.display()
Read more about python programming here:
brainly.com/question/26497128
#SPJ1
An Average of 10 times faster.
Hope this helps!
Answer: provided in the explanation section
Explanation:
Given that:
Assume D(k) =║ true it is [1 : : : k] is valid sequence words or false otherwise
now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.
So, from here we have that D(k) is given by the following recorance relation:
D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1 : : : k] ∦ array of boolean variable.
for a ← 1 to m
do ;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1 : : : a])
d (a) ← True
(b). Algorithm Output
if D[k] = = True
stack = temp stack ∦stack is used to print the strings in order
c = k
while C > 0
stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c
P = W (p) - 1
output stack
= 0 =
cheers i hope this helps !!!
