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sergiy2304 [10]
2 years ago
12

A folder has been shared with other users and set to read-only. What does this mean for users?

Computers and Technology
2 answers:
skelet666 [1.2K]2 years ago
8 0

Answer:

it means they can not make edits or changes to said folder

Explanation:

wariber [46]2 years ago
3 0

Answer:

Users cannot add new folders or files.

Explanation:

So adding new folders or files would be editing it but if its read only they cant edit.

I just took the test ;)

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Write a 5 – 7+ sentence paragraph about Thanksgiving that DOES NOT use the letter “t.”
Norma-Jean [14]

A period in all of our lives which is recognizing a major bond including family and friends. When summer concludes, a new season comes around bringing a cold, and chilly breeze which is fall. Fall is a special season because occurrences such as leaves falling from conifers, looking for a good pumpkin in the pumpkin field, or simply enjoying  your family and friends company. Drinking warm drinks such as coffee and baking cookies can help you embrace such an amazing season. When fall rolls around, Always be aware of nearby occurrences which can bring you joy during a cold season such as fall.

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Each of the following are advanced strategies you may use to help you conduct a search on the Internet except _____. using a wil
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i cant see anything tho theres nothing there

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Read 2 more answers
Which statement about synchronous communication is true?
tangare [24]
In general, synchronous communication means you have to wait for the answer all the time. The programming logic is simpler, but the cost that you spend a lot of time waiting.

If the options are:

<span>a. The people communicating don't need to be online at the same time.
b. There is lag time in the communication.
c. The communication occurs in real time.

a is false, you do need to be online to receive the message
b is true, typically you continue only after an acknowledgement
c is true, you wait for acknowledgement that occurs in real time (not necessarily fast though)</span>
3 0
3 years ago
Analyze the following code. public class Test { public static void main(String[] args) { double radius; final double PI= 3.15169
sergiy2304 [10]

Answer:

The output of the given code as follows:

Output:

Area is: 12.60676

Explanation:

In the given code some information is missing so, the correct code to this question can be described as follows:

Program:

public class Test //defining class  

{

   public static void main(String[] args)//defining the main method

   {

       double radius= 2; //defining double variable radius

       final double PI= 3.15169; //defining double variable PI

       double area = radius * radius * PI; //defining double variable area that calculates values

       System.out.println("Area is: " + area); //print values

   }  

}

Explanation:

  • In the given java code a class "Test" is defined, in which a double variable "radius" is defined, which holds a value, that is 2.
  • In the next step, a double constant variable, that is PI is defined, that holds a value, that is "3.15169".
  • Then another double variable area is defined, that calculates the area value, and prints its value.
4 0
3 years ago
consider Java and explore its scope rules. One aspect of the scope rules of any language is when data or methods can have the sa
navik [9.2K]

Answer and Explanation:

Abstract class Point {

   int x = 1, y = 1;

   void move(int dx, int dy) {

       x += dx;

       y += dy;

       alert();

   }

   abstract void alert();

}

abstract class Colored Point extends Point {

   int color;

}

class Simple Point extends Point {

   void alert() { }

}

Here, a class Point is proclaimed that must be declared abstract, in light of the fact that it contains an assertion of a unique strategy named alert. The subclass of Point named Colored Point acquires the dynamic technique alert, so it should likewise be proclaimed theoretical. Then again, the subclass of Point named Simple Point gives a usage of alarm, so it need not be dynamic.

The statement:

Point p = new Point();

would bring about an aggregate time mistake; the class Point can't be launched in light of the fact that it is theoretical. Be that as it may, a Point variable could accurately be instated with a reference to any subclass of Point, and the class Simple Point isn't digest, so the statement:

Point p = new Simple Point();

would be correct. Instantiation of a Simple Point causes the default constructor and field initializers for x and y of Point to be executed.

5 0
3 years ago
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