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stealth61 [152]
2 years ago
11

You use the _____ sheet in the format cells dialog box to position data in a cell by centering it, for example.â

Computers and Technology
1 answer:
Murljashka [212]2 years ago
3 0
The alignment sheet

The alignment sheet in the format cells dialog box is used to position data in a cell by left or right aligning, centering, aligning at the top or bottom and rotating as well. For instance, if you choose to enter 90 in the degree area, the text will read vertically from bottom to top in the cell.


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Before buying his 12-year-old daughter her very own cell phone and laptop complete with internet access, John requires her to co
Ratling [72]

Answer:

D.

Explanation:

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1 year ago
How would you categorize the software that runs on mobile devices? Break down these apps into at least three basic categories an
I am Lyosha [343]

Mobile software is an App that can be downloaded to a mobile device and used to execute specific tasks. Once the app is initiated, it runs inside the operating system until it is been closed by the user. These are small, individual software units which can work as a browser, game, shopping app, calculator and more.


There are mainly 3 types of mobile apps:

Native applications, Hybrid applications, and Web applications.


Native Applications:

Native applications are developed to be used on a particular platform or device. They exist on the device and are accessed through icons on the device. Native apps are written for specific Operating System, which means a separate source code is written for each OS, such as iOS, Android, Linux, and Windows, this means, a native Android software will not work on a windows platform, because the Windows platform does not have the source code to operate the software, however, the advantages of native applications is that they can take the full advantage all features in the devices’


Web Application:

Web application (Web app) is a client-server software application, in which the client runs in a web browser. It is stored on a remote server and delivered over the Internet through a browser interface. It is not designed for a specific type of device or an operating system, which means it can be accessed from several devices. Web applications commonly use a combination of server-side script (ASP, PHP, etc) and client-side script (HTML, Javascript, etc.) to be developed.


Hybrid Applications:

Hybrid applications have the characteristic of both web applications and native applications. They combine the elements of both native and Web applications. They are developed using HTML, CSS, and Javascript, and then wrapped into a native application using platforms like Cordova. Since they are designed for various platforms, they don’t require different source codes like native apps, Web application development is obviously faster, simpler and rapid.

5 0
2 years ago
Read 2 more answers
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
2 years ago
Universal Containers has implemented a strict software architecture for their custom Apex code. One of the requirements is that
maxonik [38]

Answer:

The required coding for the architect to meet these requirement has been described below.

Explanation:

The following ways will help the architecture to meet the requirement.

Create a reusable SOQLQueries class; specify “With Sharing” on the methods that require user context and “Without Sharing” on the methods requiring system context.

Create a SystemSOQLQueries class and a UserSOQLQueries class; set the “With Sharing” keyword on the UserSOQLQueries class and “Without Sharing” on the SystemSOQLQueries class.

6 0
2 years ago
Prove that f(n) = 20n3 + 10nlogn + 5 is O(n3)
sp2606 [1]

Answer:

The proof is in the explanation

Explanation:

f(n) is O(n^{3}) if f(n) \leq cn^{3} for n \geq n_{0}.

So, basically, we have to solve the following inequality

f(n) \leq cn^{3}

20n^{3} + 10n\log{n} + 5 \leq cn^{3}

Dividing everything by n^{3} to simplify, we have

20 + \frac{10\log{n}}{n^{2}} + \frac{5}{n^{3}} \leq cn^{3}

I am going to use n = n_{0} = 1. So

20 + 5 \leq c

c \geq 25

There is a solution for the inequality, which proves that f(n) = 20n^{3} + 10n\log{n} + 5 is O(n^{3})

7 0
2 years ago
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