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Ksivusya [100]
3 years ago
12

NO LINKS<br />Write a C program called velocity.c to do the following activities.<br />• The parent process should r

ead an array (size 6) of double (frequency (1)) from the user<br />and creates a pipe to do the inter-process communication and then creates a child process<br />(Child-1).<br />• Child-1 will calculate the velocity (V) when the wavelength() is 100 m (constant) and write<br />answers to the pipe. Use the formula is given below;<br />V = fa<br />(V=Velocity, f=frequency, 1=wavelength)<br />. Then the Child-1 creates a new child process (Child-2) and the Child-2 should read values<br />from the pipe and print them to the screen.​
Computers and Technology
1 answer:
algol [13]3 years ago
3 0

Answer:

uhm.....im confused a bit here

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Drag each label to the correct location. Each label can be used more than once. Match the device to the port through which it co
son4ous [18]

Answer:

Mouse - USB

Monitor - display port, HDMI, and thunderbolt port

External hard drive - USB and thunderbolt port

Flash drive - USB

Explanation:

A computer mouse is an input device used to interact with a computer system. It connects with the computer through the USB port.

A Monitor is an output device that is used to display information on a computer system. It can be connected to the system through the display ports, HDMI, and the thunder port.

The External hard drive is a storage device that holds data for a long period of time. It has adapters to connect to the computer through the USB ports or thunderbolt ports. The flash drive is similar in function to the hard drive but small in size and storage space. It only connects with the computer through USB ports.

5 0
3 years ago
If you play video games, please answer these questions it’s for a survey for my game development class!!
yarga [219]

Answer:

Doom, fortnlte, mlnecraft, ark survival evolved, ark survival of the fittest, terraria, raft, among us, ect.

First person shooters, Third person shooters, Creative games

8-10 years

More VR, better graphics, more realistic

Maybe, not anytime soon.

Explanation:

5 0
3 years ago
Main Difference between Excel and word application<br>programs​
Sloan [31]

Answer:

See explanation

Explanation:

Word documents are mainly for typing papers and documents. This is useful for writing a report or making a book.

Excel is for writing spreadsheets and doing math equations within the program. This is useful for having budget calculators and math equations.

Hope this helped!

4 0
3 years ago
Define a class called TreeNode containing three data fields: element, left and right. The element is a generic type. Create cons
m_a_m_a [10]

Answer:

See explaination

Explanation:

// Class for BinarySearchTreeNode

class TreeNode

{

// To store key value

public int key;

// To point to left child

public TreeNode left;

// To point to right child

public TreeNode right;

// Default constructor to initialize instance variables

public TreeNode(int key)

{

this.key = key;

left = null;

right = null;

key = 0;

}// End of default constructor

}// End of class

// Defines a class to crate binary tree

class BinaryTree

{

// Creates root node

TreeNode root;

int numberElement;

// Default constructor to initialize root

public BinaryTree()

{

this.root = null;

numberElement = 0;

}// End of default constructor

// Method to insert key

public void insert(int key)

{

// Creates a node using parameterized constructor

TreeNode newNode = new TreeNode(key);

numberElement++;

// Checks if root is null then this node is the first node

if(root == null)

{

// root is pointing to new node

root = newNode;

return;

}// End of if condition

// Otherwise at least one node available

// Declares current node points to root

TreeNode currentNode = root;

// Declares parent node assigns null

TreeNode parentNode = null;

// Loops till node inserted

while(true)

{

// Parent node points to current node

parentNode = currentNode;

// Checks if parameter key is less than the current node key

if(key < currentNode.key)

{

// Current node points to current node left

currentNode = currentNode.left;

// Checks if current node is null

if(currentNode == null)

{

// Parent node left points to new node

parentNode.left = newNode;

return;

}// End of inner if condition

}// End of outer if condition

// Otherwise parameter key is greater than the current node key

else

{

// Current node points to current node right

currentNode = currentNode.right;

// Checks if current node is null

if(currentNode == null)

{

// Parent node right points to new node

parentNode.right = newNode;

return;

}// End of inner if condition

}// End of outer if condition

}// End of while

}// End of method

// Method to check tree is balanced or not

private int checkBalance(TreeNode currentNode)

{

// Checks if current node is null then return 0 for balanced

if (currentNode == null)

return 0;

// Recursively calls the method with left child and

// stores the return value as height of left sub tree

int leftSubtreeHeight = checkBalance(currentNode.left);

// Checks if left sub tree height is -1 then return -1

// for not balanced

if (leftSubtreeHeight == -1)

return -1;

// Recursively calls the method with right child and

// stores the return value as height of right sub tree

int rightSubtreeHeight = checkBalance(currentNode.right);

// Checks if right sub tree height is -1 then return -1

// for not balanced

if (rightSubtreeHeight == -1) return -1;

// Checks if left and right sub tree difference is greater than 1

// then return -1 for not balanced

if (Math.abs(leftSubtreeHeight - rightSubtreeHeight) > 1)

return -1;

// Returns the maximum value of left and right subtree plus one

return (Math.max(leftSubtreeHeight, rightSubtreeHeight) + 1);

}// End of method

// Method to calls the check balance method

// returns false for not balanced if check balance method returns -1

// otherwise return true for balanced

public boolean balanceCheck()

{

// Calls the method to check balance

// Returns false for not balanced if method returns -1

if (checkBalance(root) == -1)

return false;

// Otherwise returns true

return true;

}//End of method

// Method for In Order traversal

public void inorder()

{

inorder(root);

}//End of method

// Method for In Order traversal recursively

private void inorder(TreeNode root)

{

// Checks if root is not null

if (root != null)

{

// Recursively calls the method with left child

inorder(root.left);

// Displays current node value

System.out.print(root.key + " ");

// Recursively calls the method with right child

inorder(root.right);

}// End of if condition

}// End of method

}// End of class BinaryTree

// Driver class definition

class BalancedBinaryTreeCheck

{

// main method definition

public static void main(String args[])

{

// Creates an object of class BinaryTree

BinaryTree treeOne = new BinaryTree();

// Calls the method to insert node

treeOne.insert(1);

treeOne.insert(2);

treeOne.insert(3);

treeOne.insert(4);

treeOne.insert(5);

treeOne.insert(8);

// Calls the method to display in order traversal

System.out.print("\n In order traversal of Tree One: ");

treeOne.inorder();

if (treeOne.balanceCheck())

System.out.println("\n Tree One is balanced");

else

System.out.println("\n Tree One is not balanced");

BinaryTree

BinaryTree treeTwo = new BinaryTree();

treeTwo.insert(10);

treeTwo.insert(18);

treeTwo.insert(8);

treeTwo.insert(14);

treeTwo.insert(25);

treeTwo.insert(9);

treeTwo.insert(5);

System.out.print("\n\n In order traversal of Tree Two: ");

treeTwo.inorder();

if (treeTwo.balanceCheck())

System.out.println("\n Tree Two is balanced");

else

System.out.println("\n Tree Two is not balanced");

}// End of main method

}// End of driver class

5 0
4 years ago
You have a desktop computer that supports both IEEE 1394 and USB 2.0. You are purchasing some devices that will connect to these
zimovet [89]

Answer:

IEEE 1394 supports more devices on a single bus

Explanation:

What this means is that IEEE 1394 provides a single plug-and-socket connection on which up to 63 devices can be attached with data transfer speeds up to 400 Mbps (megabit s per second). The standard describes a serial bus( A shared channel that transmits data one bit after the other over a single wire or fiber) or pathway between one or more peripheral devices and your computer's microprocessor .

7 0
3 years ago
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