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Mademuasel [1]
3 years ago
13

Which of the following are acceptable to share? Check all of the boxes that apply.

Computers and Technology
1 answer:
Burka [1]3 years ago
8 0

Answer:

B. files that you have permission to share  

D. works that a Creative Commons license says you can use

You might be interested in
What is the difference between "call by value" and "call by reference"? Explain with example in c language.
Serjik [45]

Answer:

<u>Call by reference</u>

In an function if the variables are passed as reference variables this means that the variables are pointing to the original arguments.So the changes made in the function on the reference variables will be reflected back on the original arguments.

For example:-

#include<stdio.h>

void swap(&int f,&int s)

{

   int t=f;

   f=s;

  s =temp;  

}

int main()

{

int n,m;

n=45;

m=85;

swap(n,m);

printf("%d %d",m,n);

return 0;  

}

the values of m and n will get swapped.  

<u> Call by value</u>

In this program the values of m and n will not get swapped because they are passed by value.So duplicate copies of m and n will be created and manipulation will be done on them.

#include<stdio.h>

void swapv(int f,int s)

{

   int t=f;  

   f=s;  

   s=temp;

}

int main()

{  

int n,m;

n=45;

m=85;

swapv(n,m);

printf("%d %d",n,m);

return 0;

}

7 0
3 years ago
The process known as "bitmapping" is defined as information in _____. <br><br> finish the sentence.
valentina_108 [34]

graphics is the answer

4 0
3 years ago
Hi, Everybody i have a question it is almost my B-day i want this lego set
salantis [7]

Answer:

You can look at different websites of look on an app for people selling it, or something :p

Explanation:

3 0
3 years ago
Suppose we compute a depth-first search tree rooted at u and obtain a tree t that includes all nodes of g.
Temka [501]

G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

<h3>What are DFS and BFS?</h3>

An algorithm for navigating or examining tree or graph data structures is called depth-first search. The algorithm moves as far as it can along each branch before turning around, starting at the root node.

The breadth-first search strategy can be used to look for a node in a tree data structure that has a specific property. Before moving on to the nodes at the next depth level, it begins at the root of the tree and investigates every node there.

First, we reveal that G exists a tree when both BFS-tree and DFS-tree are exact.

If G and T are not exact, then there should exist a border e(u, v) in G, that does not belong to T.

In such a case:

- in the DFS tree, one of u or v, should be a prototype of the other.

- in the BFS tree, u and v can differ by only one level.

Since, both DFS-tree and BFS-tree are the very tree T,

it follows that one of u and v should be a prototype of the other and they can discuss by only one party.

This means that the border joining them must be in T.

So, there can not be any limits in G which are not in T.

In the two-part of evidence:

Since G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

The complete question is:

We have a connected graph G = (V, E), and a specific vertex u ∈ V.

Suppose we compute a depth-first search tree rooted at u, and obtain a tree T that includes all nodes of G.

Suppose we then compute a breadth-first search tree rooted at u, and obtain the same tree T.

Prove that G = T. (In other words, if T is both a depth-first search tree and a breadth-first search tree rooted at u, then G cannot contain any edges that do not belong to T.)

To learn more about  DFS and BFS, refer to:

brainly.com/question/13014003

#SPJ4

5 0
2 years ago
php Exercise 3: Function Write a function named word_count that accepts a string as its parameter and returns the number of word
Bingel [31]

Answer:

<?php  

function word_count($string){   // function that takes a string a parameter and returns the number of words in that string

   $space = 0;   // this variable is used in order to detect space, new line and tab spaces in a string

   $words = 1; // this variable is used to identify word presence in string

   $include = $space;  //used to maintain the state of $words and $space

   $counter = 0;  // this counts the number of words in the string

   $i = 0;  //this moves through each character of the string

   while ($i < strlen($string)){  // iterates through every character of string until the last character of string is reached

       if ($string[$i] == " "||$string[$i] == "\n" || $string[$i] == "\t")  //if the space or new line or tab space is identified in the string

           $include = $space;  //set the state of include as space if the next character is a space newline or a tab space

       else if ($include == $space) {  //if next character is a word and current state i.e. $include holds $space

           $include = $words;   //  then set the state i.e. $include as $words

           ++$counter; } //increments i to move to next character at each iteration

       ++$i;  } //returns the number of words in a string

   return $counter; }

$str = "Hello, how are you ";  //sample string

echo "Words: " . word_count($str);  // calls word_count function to return number of words in str.

?>  

Explanation:

The program has a function word_count that accepts a string as its parameter and returns the number of words in the string. In the function there are three variables $space and $words and $include used as state variables. For instance $space identifies space, new line and tab spaces in the $string so it specifies that a space has occurred. $words identifies words in the $string so it specifies that a word has occurred. $include holds this state information. $i moves through every character of the $string and when a space occurs in the $string then the state $include is set to $space. When a word occurs then state $include is set to $words. $counter variable, which is used to count the number of words is incremented to 1 when previous state is $space and next character is a word character. Every time a word is seen in the string, this variable is incremented to 1 to keep track of number of words in the string. When $i reaches the end of the string then this loop stops and counter returns the number of words in the $string.

5 0
3 years ago
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