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Shkiper50 [21]
3 years ago
13

1

Computers and Technology
1 answer:
Sergio [31]3 years ago
3 0

Answer:

The answer to this question is given below in the explanation section

Explanation:

The ctrl+shift+e command is used to select track changes in a document.

Track Changes is a feature that is used in MS word or in wordprocessing application to track the changes inserted or modified by different users in a document. This helps the user to know where new text inserted or modified. Track changes help users to know new updates being inserted by other users in the document.

Furthermore, you can access the track changes features in the word file under the review tab. where you can find other related settings and features related to the track changes.

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When creating a spreadsheet, there's no need to worry about design and how it will be organized; the program will take care of t
Rainbow [258]
False: a computer program do many things, but it can't read your mind. It doesn't know what kind of formatting you need for your spreadsheet. There are so many potential layouts of a spreadsheet, that the computer couldn't decide what to lay it out for you. Eventually the computer can see what you're trying to lay it out as and can help that way, but it needs to e started first. Having a uniform sheet that is well organized by you, is much easier to read than gobbledegook that has been spewed everywhere.

I hope this was helpful!
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3 years ago
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Need help ASPA!!!!!!!!!
professor190 [17]
Answer is D the internet
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3 years ago
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Create a class called Hokeemon that can be used as a template to create magical creatures called Hokeeemons. Hokeemons can be of
Viefleur [7K]

Answer:

See Explaination

Explanation:

/*****************************************Hokeemon.java********************************/

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class Hokeemon {

private String name;

private String type;

private int age;

public Hokeemon(String name, String type, int age) {

super();

this.name = name;

this.type = type;

this.age = age;

}

public Hokeemon() {

this.name = "";

this.type = "";

this.age = 0;

}

public String getName() {

return name;

}

public void setName(String name) {

this.name = name;

}

public String getType() {

return type;

}

public void setType(String type) {

this.type = type;

}

public int getAge() {

return age;

}

public void setAge(int age) {

this.age = age;

}

public String liveIn() {

if (this.type.equalsIgnoreCase("dwarf")) {

return "Mountain";

} else if (this.type.equalsIgnoreCase("elf")) {

return "Dale";

} else if (this.type.equalsIgnoreCase("fairy")) {

return "Forest";

} else {

return "Shire";

}

}

public boolean areFriends(Hokeemon other) {

if (this.type.equalsIgnoreCase(other.type)) {

return true;

} else if ((this.type.equalsIgnoreCase("dwarf") && other.type.equalsIgnoreCase("elf"))

|| (this.type.equalsIgnoreCase("elf") && other.type.equalsIgnoreCase("dwarf"))) {

return true;

} else if ((this.type.equalsIgnoreCase("hobbit") && other.type.equalsIgnoreCase("fairy"))

|| (this.type.equalsIgnoreCase("fairy") && other.type.equalsIgnoreCase("hobbit"))) {

return true;

} else {

return false;

}

}

public static Hokeemon[] getData(String file) {

Hokeemon[] hokeemons = new Hokeemon[8];

int i = 0;

try {

Scanner scan = new Scanner(new File(file));

while (scan.hasNextLine() && i < hokeemons.length) {

String line = scan.nextLine();

String[] data = line.split("\\s+");

String name = data[0];

String type = data[1];

int age = Integer.parseInt(data[2]);

hokeemons[i] = new Hokeemon(name, type, age);

i++;

}

scan.close();

} catch (FileNotFoundException e) {

System.err.println(e);

}

return hokeemons;

}

public static String getBio(Hokeemon[] hokeemons) {

String s = "";

for (Hokeemon hokeemon : hokeemons) {

s += "I am " + hokeemon.getName() + ": Type " + hokeemon.getType() + ": Age=" + hokeemon.getAge()

+ ", I live in " + hokeemon.liveIn() + "\n";

s += "My friends are: ";

for (Hokeemon hokeemon2 : hokeemons) {

if (hokeemon.areFriends(hokeemon2) && !hokeemon.equals(hokeemon2)) {

s += (hokeemon2.getName() + " ");

}

}

s += "\n";

}

return s;

}

atOverride

public String toString() {

return "Name " + name + ": Type " + type + ": Age=" + age + "\n";

}

}

/*******************************HokeemonMain.java**************************/

import java.util.Arrays;

public class HokeemonMain {

public static void main(String[] args) {

Hokeemon[] hokeemons = Hokeemon.getData("data.txt");

System.out.println(Arrays.toString(hokeemons));

System.out.println(Hokeemon.getBio(hokeemons));

}

}

/**********************************data.txt********************/

Noddy Dwarf 4

Legolas Elf 77

Minerva Fairy 33

Samwise Hobbit 24

Merry Hobbit 29

Warhammer Dwarf 87

Ernedyll Elf 19

Frodo Hobbit 18

3 0
3 years ago
Print "Censored" if userInput contains the word "darn", else print userInput. End with newline. Ex: If userInput is "That darn c
barxatty [35]

Answer:

#include <string>

#include <iostream>

using namespace std;

int main() {

string userInput;

getline(cin, userInput);

// Here, an integer variable is declared to find that the user entered string consist of word darn or not

int isPresent = userInput.find("darn");

if (isPresent > 0){

cout << "Censored" << endl;

// Solution starts here

else

{

cout << userInput << endl;

}

// End of solution

return 0;

}

// End of Program

The proposed solution added an else statement to the code

This will enable the program to print the userInput if userInput doesn't contain the word darn

6 0
3 years ago
A sample member of the list data is a1 = ['male', True] where the second item is True if the person is on the phone.
Ira Lisetskai [31]
Answer:

0

Explanation:
In the lists, indexation starts from 0, because of that in if statement we compare item[0] which is 'male' and 'male', and then males is itterated within for loop.
8 0
3 years ago
Read 2 more answers
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