The steps I would take are 1. Making a strong password with symbols and numbers. 2. I will set up a 2 way authentication ther for if someone would be trying to hack me I would get an alert in order to stop the attempt. 3 I will not use the same password each time.
Answer: it does – or perhaps appears to do – in the world.
Explanation:
sorry if am not right have a nice day lol
Hey Litz06,
Sorry that happened to you.
Try changing your password to something complicated so it’s more secure.
Hello <span>Endermss2118 </span>
Answer: If you do not want the border style to carry forward each time the enter key is pressed, you need to Clear formatting
Hope This Helps
-Chris
Answer:
def leap_year(y):
if y % 4 == 0:
return 1
else:
return 0
def number_of_days(m,y):
if m == 2:
return 28 + leap_year(y)
elif m == 1 or m == 3 or m == 5 or m == 7 or m == 8 or m ==10 or m == 12:
return 31
elif m == 4 or m == 6 or m == 9 or m == 11:
return 30
def days(m,d):
if m == 1:
return 0 + d
if m == 2:
return 31 + d
if m == 3:
return 59 + d
if m == 4:
return 90 + d
if m == 5:
return 120 + d
if m == 6:
return 151 + d
if m == 7:
return 181 + d
if m == 8:
return 212 + d
if m == 9:
return 243 + d
if m == 10:
return 273 + d
if m == 11:
return 304 + d
if m == 12:
return 334 + d
def days_left(d,m,y):
if days(m,d) <= 60:
return 365 - days(m,d) + leap_year(y)
else:
return 365 - days(m,d)
print("Please enter a date")
day=int(input("Day: "))
month=int(input("Month: "))
year=int(input("Year: "))
choice=int(input("Menu:\n1) Calculate the number of days in the given month.\n2) Calculate the number of days left in the given year.\n"))
if choice == 1:
print(number_of_days(month, year))
if choice == 2:
print(days_left(day,month,year))
Explanation:
Hoped this helped
