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2 years ago

Ives purchases a bowl. The diameter of the bowl is 4 cm. What is the distance around the bowl?.

SAT

1 answer:

bogdanovich [222]2 years ago

7 0

Answer:

12.57cm

Explanation:

C=πd, where C is the circumference and D is the diameter.

Simply put the diameter into (d) and times it by 3.14, or by π directly if the question specifies "give your answer in terms of π".

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Is this thedrewrugrats?

kondor19780726 [428]

Answer:

no this is Patrick

Explanation:

20 characters

3 0

3 years ago

Which sentence uses the word your correctly?

lord [1]

B. Where did you finally find your missing book ?

4 0

4 years ago

Once you have your heap structure created, next you must use it as a backing structure to a priority queue. Develop a PriorityQu

saveliy_v [14]

Using the knowledge in computational language in python it is possible to write a code that Develop a PriorityQueue data structure in a file.

<h3>Writting the code in python:</h3>

import sys

class Node:

def __init__(self,key):

self.left = None

self.right = None

self.val = key

self.parent = None

class MinHeap:

def __init__(self, maxsize):

self.maxsize = maxsize

self.size = 0

self.Heap = Node(-1 * sys.maxsize )

self.FRONT = self.Heap

# A utility function to do inorder tree traversal

def isLeaf(self, node):

if node.left == None and node.right == Node:

return True

return False

# Function to swap two nodes of the heap

def swap(self, node1, node2):

node1.data, node2.data = node2.data, node1.data

# Function to heapify the node at pos

def minHeapify(self, node):

# If the node is a non-leaf node and greater

# than any of its child

if not self.isLeaf(node):

if (node.data > node.left.data or

node.data > node.right.data):

# Swap with the left child and heapify

# the left child

if node.left.data < node.right.data:

self.swap(node, node.left)

self.minHeapify(node.left)

# Swap with the right child and heapify

# the right child

else:

self.swap(node, node.right)

self.minHeapify(node.right)

# Function to insert a node into the heap

def insert(self, element):

if self.size >= self.maxsize :

return

self.size+= 1

self.bst_insert(FRONT, element)

current = FRONT

while current.parent != None and current.data < current.parent.data:

self.swap(current, current.parent)

current = current.parent

# Function to print the contents of the heap

def Print(self):

self.inorder()

# Function to build the min heap using

# the minHeapify function

def inorder(self, root):

if root:

inorder(root.left)

print(root.val)

inorder(root.right)

def bst_insert(self, root, node):

if root is None:

root = node

else:

root.next = node

self.FRONT = node

# Driver Code

if __name__ == "__main__":

r = Node(50)

bst_insert(r,Node(30))

bst_insert(r,Node(20))

bst_insert(r,Node(40))

bst_insert(r,Node(70))

bst_insert(r,Node(60))

bst_insert(r,Node(80))

# Print inoder traversal of the BST

inorder(r)

print('The minHeap is ')

minHeap = MinHeap(15)

minHeap.insert(5)

minHeap.insert(3)

minHeap.insert(17)

minHeap.insert(10)

minHeap.insert(84)

minHeap.insert(19)

minHeap.insert(6)

minHeap.insert(22)

minHeap.insert(9)

minHeap.minHeap()

minHeap.Print()

print("The Min val is " + str(minHeap.remove()))

See more about python at brainly.com/question/13437928

#SPJ1

6 0

2 years ago

At 900 ∘c,kc=0. 0108 for the reaction caco3(s)←→cao(s)+co2(g) a mixture of caco3, cao, and co2 is placed in a 10. 0-l vessel at

Mrac [35]

The concentration of CaCO3 increases in all cases because Q> K.

<h3>What is equilibrium?</h3>

A chemical reaction is said to have attained equilibrium when the rate of forward reaction is equal to the rate of reverse reaction. We are told that the Kc of the reaction is 0. 0108.

In the first case:

CaCO3 - 15.0 g/100g/mol/10 L = 0.015 M

CaO - 15.0 g/56 g/mol/10 L = 0.027 M

CO2 - 4.25 g/44 g/mol / 10 L = 0.0096 M

Q = [0.027] [0.0096]/0.015

Q = 0.017

Since Q > K, the concentration of CaCO3 increases

In the second case;

CaCO3 - 2.5 g/100g/mol/10 L =0.0025 M

CaO - 25.0 g/56 g/mol/10 L = 0.045 M

CO2 - 5.66 g/44 g/mol / 10 L = 0.013 M

Q = [0.045 ] [ 0.013]/[0.0025 ]

Q = 0.23

Q > K the concentration of CaCO3 increases

In the third case;

CaCO3 - 30.5 g/100g/mol/10 L =0.031 M

CaO - 25.5 g/56 g/mol/10 L =0.046 M

CO2 - 6.48 g//44 g/mol / 10 L = 0.015 M

Q = [0.046] [ 0.015]/[0.031]

Q = 0.022

Q> K hence the concentration of CaCO3 increases

Missing parts: At 900oC, Kc = 0.0108 for the reaction: CaCO3(s) <===> CaO(s) + CO2(g) A mixture of CaCO3, CaO, and CO2 is placed in a 10.0 Liter vessel at 900oC. For the following mixtures, will the amount of CaCO3 increase, decrease, or remain the same as the system approaches equilibrium? CaCO3 CaO CO2 At Equilibrium, CaCO3 will ?? 15.0 g 15.0 g 4.25 g Answer 2.5 g 25.0 g 5.66 g Answer 30.5 g 25.5 g 6.48 g Answer

Learn more about equilibrium: brainly.com/question/17960050

4 0

2 years ago

Which of the following is not an advantage of using data warehouse.

Deffense [45]

Answer:

Data redundancy

Explanation:

3 0

3 years ago