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Bess [88]
3 years ago
11

How to keep tools and equipment safeuh​

Computers and Technology
2 answers:
ivanzaharov [21]3 years ago
8 0

Answer:

Use pelican, or similar heavy duty cases

Explanation:

Pelican is a brand btw. 10/10 would recommend

7nadin3 [17]3 years ago
5 0
Use a tool box like handy Mandy hope this helps lol yay
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Hamilton is the best
san4es73 [151]

Answer:

Yes he is

Explanation:

6 0
3 years ago
Alice is adding a chart in a spreadsheet. Which option will allow you her to give the chart a suitable title and subtitle?
Anon25 [30]

Chart elements is the answer.


8 0
3 years ago
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Write two methods: encrypt and decrypt. encrypt should #take as input a string, and return an encrypted version #of it according
Harman [31]

Answer:

The code is given below with appropriate comments

Explanation:

CIPHER = (("D", "A", "V", "I", "O"),

         ("Y", "N", "E", "R", "B"),

         ("C", "F", "G", "H", "K"),

         ("L", "M", "P", "Q", "S"),

         ("T", "U", "W", "X", "Z"))

# Add your code here!

def encrypt(plaintext):

   theList = []

   for char in plaintext:

       if char.isalpha():

           char = char.upper()

           if char == "J":

               char = "I"

           theList.append(char)

   if len(theList) % 2 == 1:

       theList.append("X")

   for i in range(0, len(theList), 2):

       if theList[i] == theList[i + 1]:

           theList[i + 1] = "X"

       findex = [-1, -1]

       sindex = [-1, -1]

       for j in range(len(CIPHER)):

           for k in range(len(CIPHER)):

               if theList[i] == CIPHER[j][k]:

                   findex = [j, k]

               if theList[i + 1] == CIPHER[j][k]:

                   sindex = [j, k]

       # same row

       if (findex[0] == sindex[0]):

           findex[1] += 1

           sindex[1] += 1

           if findex[1] == 5:

               findex[1] = 0

           if sindex[1] == 5:

               sindex[1] = 0

           theList[i] = CIPHER[findex[0]][findex[1]]

           theList[i + 1] = CIPHER[sindex[0]][sindex[1]]

       # same column

       elif (findex[1] == sindex[1]):

           findex[0] += 1

           sindex[0] += 1

           if findex[0] == 5:

               findex[0] = 0

           if sindex[0] == 5:

               sindex[0] = 0

           theList[i] = CIPHER[findex[0]][findex[1]]

           theList[i + 1] = CIPHER[sindex[0]][sindex[1]]

       else:

           theList[i] = CIPHER[findex[0]][sindex[1]]

           theList[i + 1] = CIPHER[sindex[0]][findex[1]]

   return "".join(theList)

def decrypt(ciphertext):

   theString = ""

   findex = [-1, -1]

   sindex = [-1, -1]

   for i in range(0, len(ciphertext), 2):

       for j in range(len(CIPHER)):

           for k in range(len(CIPHER)):

               if ciphertext[i] == CIPHER[j][k]:

                   findex = [j, k]

               if ciphertext[i + 1] == CIPHER[j][k]:

                   sindex = [j, k]

       if (findex[0] == sindex[0]):

           findex[1] -= 1

           sindex[1] -= 1

           if findex[1] == -1:

               findex[1] = 4

           if sindex[1] == -1:

               sindex[1] = 4

           theString += CIPHER[findex[0]][findex[1]]

           theString += CIPHER[sindex[0]][sindex[1]]

       # same column

       elif (findex[1] == sindex[1]):

           findex[0] -= 1

           sindex[0] -= 1

           if findex[0] == -1:

               findex[0] = 4

           if sindex[0] == -1:

               sindex[0] = 4

           theString += CIPHER[findex[0]][findex[1]]

           theString += CIPHER[sindex[0]][sindex[1]]

       else:

           theString += CIPHER[findex[0]][sindex[1]]

           theString += CIPHER[sindex[0]][findex[1]]

   return theString

# Below are some lines of code that will test your function.

# You can change the value of the variable(s) to test your

# function with different inputs.

#

# If your function works correctly, this will originally

# print: QLGRQTVZIBTYQZ, then PSHELXOWORLDSX

print(encrypt("PS. Hello, worlds"))

print(decrypt("QLGRQTVZIBTYQZ"))

5 0
3 years ago
I found a random Nether Portal in my Minecraft Singleplayer world. Why is it there?
Mkey [24]

Answer:

This is a known bug in earlier versions of the game.

Explanation:

An easy way to solve this that I have seen work before is to completely destroy the 'new' portal that appears in the overworld. Once done, you should be able to enter your portal and exit from the same portal.

To help avoid this, you should also never install a portal in your world in an area that is at all enclosed. The portal system seems to dislike portals appearing in places they should 'not be' based on the initial construction of the world.

Also avoid dying, as it can cause random nether portals to form in parts of your world.

6 0
3 years ago
Read 2 more answers
________is a nickname for social media tools which allows users to contribute content easily and to easily connect to each other
Nata [24]

Answer:

Web 2.0

Explanation:

The term that is being defined here is known as Web 2.0. As mentioned in the question this term refers to any and all social platform websites that emphasize user-generated content, ease of use, participatory culture, and interoperability for end users. Some examples of Web 2.0 features would be social media sites like FB, blogs, wikis, video sharing sites, image sharing sites, etc.

4 0
4 years ago
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