Answer:
Internal:
#CPU; That retrieves &execute instructions.
#Modem; Modulates& demodulates electric signals.
#RAM;Gives application a place to store &access data on a short time periods.
External:
#Mouse; Transmits commands and controlling movements.
#Moniter; Device used to display video output from computer.
#Printer; Accepts text, graphics to the paper.
Explanation:
Hope this will help you.
I think it might be the USD but I'm not sure.
B, C, and D are the correct answers.
<span>If a attachment is not reliable to open, terrible effects can happen, peradventure it may have a virus or even malware that can destroy a computers software.
To avoid this and stay on the safe side, try the following:-
- Open it in protected view
- Do not save the attachment on your computer
- Look at the author and read the message carefully to make sure it is not biased.
- Open it on a flash-drive </span>
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
