This rights protected document cannot be opened because the rights management feature has been disabled on your machine by Policy is known to be an error message.
<h3>What is the error message about?</h3>
The Error message above is known to be one that shows that your IT department has made a group policy in one's company's Active Directory. It is known to be one that tends to disables the use of ADRMS (Rights Management Feature) in all to all users.
Note that in the case above, one need to ask your IT department to be able to disable or make changes to the group policy so that it will not apply to the users who are said to require access secured Microsoft Office documents in any of Ansarada Rooms.
Another option is to look at documents off the network with the use of a personal computer or a mobile phone.
Hence, This rights protected document cannot be opened because the rights management feature has been disabled on your machine by Policy is known to be an error message.
Learn more about error message from
brainly.com/question/25671653
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Explanation:
Let, DG is the datagram so, DG= 2400.
Let, FV is the Value of Fragment and F is the Flag and FO is the Fragmentation Offset.
Let, M is the MTU so, M=700.
Let, IP is the IP header so, IP= 20.
Let, id is the identification number so, id=422
Required numbers of the fragment = ![[\frac{DG-IP}{M-IP} ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7BDG-IP%7D%7BM-IP%7D%20%5D)
Insert values in the formula = ![[\frac{2400-20}{700-20} ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B2400-20%7D%7B700-20%7D%20%5D)
Then, = ![[\frac{2380}{680} ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B2380%7D%7B680%7D%20%5D)
= ![[3.5]](https://tex.z-dn.net/?f=%5B3.5%5D)
The generated numbers of the fragment is 4
- If FV = 1 then, bytes in data field of DG=
and id=422 and FO=0 and F=1.
- If FV = 2 then, bytes in data field of DG= and id=422 and FO=85
and F=1.
- If FV = 3 then, bytes in data field of DG= and id=422 and FO=170
and F=1.
- If FV = 4 then, bytes in data field of DG=
and id=422 and FO=255
and F=0.
Answer:
This is python, I don't know what type of programming language you are learning.
import random
# Enter your code here
dice1 = random.randint(1,6)
dice2 = random.randint(1,6)
total = 0
while True:
if dice1 == 1 and dice2 == 1:
total += 1
print("Rolled: " + str(dice1) + " " + str(dice2))
break
else:
total += 1
print("Rolled: " + str(dice1) + " " + str(dice2))
dice1 = random.randint(1,6)
dice2 = random.randint(1,6)
print("It took you "+ str(total) + " rolls to get snake eyes.")
Explanation:
Answer:
All functions were written in python
addUpSquaresAndCubes Function
def addUpSquaresAndCubes(N):
squares = 0
cubes = 0
for i in range(1, N+1):
squares = squares + i**2
cubes = cubes + i**3
return(squares, cubes)
sumOfSquares Function
def sumOfSquares(N):
squares = 0
for i in range(1, N+1):
squares = squares + i**2
return squares
sumOfCubes Function
def sumOfCubes(N):
cubes = 0
for i in range(1, N+1):
cubes = cubes + i**3
return cubes
Explanation:
Explaining the addUpSquaresAndCubes Function
This line defines the function
def addUpSquaresAndCubes(N):
The next two lines initializes squares and cubes to 0
squares = 0
cubes = 0
The following iteration adds up the squares and cubes from 1 to user input
for i in range(1, N+1):
squares = squares + i**2
cubes = cubes + i**3
This line returns the calculated squares and cubes
return(squares, cubes)
<em>The functions sumOfSquares and sumOfCubes are extract of the addUpSquaresAndCubes.</em>
<em>Hence, the same explanation (above) applies to both functions</em>
Not really. xD
I only see "Not A.." and It's backwards too soooo..
