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mixer [17]
3 years ago
11

Knowledge of html and css is considered essential for the job of a(n _______.

Computers and Technology
1 answer:
7nadin3 [17]3 years ago
4 0
<span>Knowledge of html and css is considered essential for the job of a web-designer. Html and css are useful if you want to build an informational web-site or landing page. Html is used for page structure and basic style. Css used for advanced styles, it helps you to make your web-site look prettier. If you want to make the advanced functions on your web-site, you should use javascript or php.</span>
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Write an algorithm and draw a flowchart for switching off a machine after it has made 500 glass bottles. use an appropriate cond
masha68 [24]

The algorithm is as follows:

1. Start

2. bottles = 0

3. While bottles != 500

3.1 Create bottle

3.2 bottle = bottle + 1

4. Switch off machine

5. Stop

Explanation:

This begins the algorithm

1. Start

This initializes bottles to 0

2. bottles = 0

The loop is repeated until 500 bottles is created

3. While bottles != 500

This creates a new bottle

3.1 Create bottle

This increments the number of bottles by 1

3.2 bottle = bottle + 1

This switches of the machine after all bottles are created

4. Switch off machine

End algorithm

5. Stop

THANKS

4 0
3 years ago
What is true about the dilation?
madreJ [45]

Hey hey hey! I recently took the test and the answer is D | (• ◡•)|

7 0
2 years ago
Read 2 more answers
Create a Binary Expressions Tree Class and create a menu driven programyour program should be able to read multiple expressions
solong [7]

Answer:

Explanation:

Program:

#include<iostream>

#include <bits/stdc++.h>

using namespace std;

//check for operator

bool isOperator(char c)

{

switch(c)

{

case '+': case '-': case '/': case '*': case '^':

return true;

}

return false;

}

//Converter class

class Converter

{

private:

string str;

public:

//constructor

Converter(string s):str(s){}

//convert from infix to postfix expression

string toPostFix(string str)

{

stack <char> as;

int i, pre1, pre2;

string result="";

as.push('(');

str = str + ")";

for (i = 0; i < str.size(); i++)

{

char ch = str[i];

if(ch==' ') continue;

if (ch == '(')

as.push(ch);

else if (ch == ')')

{

while (as.size() != 0 && as.top() != '('){

result = result + as.top() + " ";

as.pop();

}

as.pop();

}

else if(isOperator(ch))

{

while (as.size() != 0 && as.top() != '(')

{

pre1 = precedence(ch);

pre2 = precedence(as.top());

if (pre2 >= pre1){

result = result + as.top() + " ";

as.pop();

}

else break;

}

as.push(ch);

}

else

{

result = result + ch;

}

}

while(as.size() != 0 && as.top() != '(') {

result += as.top() + " ";

as.pop();

}

return result;

}

//return the precedence of an operator

int precedence(char ch)

{

int choice = 0;

switch (ch) {

case '+':

choice = 0;

break;

case '-':

choice = 0;

break;

case '*':

choice = 1;

break;

case '/':

choice = 1;

break;

case '^':

choice = 2;

default:

choice = -999;

}

return choice;

}

};

//Node class

class Node

{

public:

string element;

Node *leftChild;

Node *rightChild;

//constructors

Node (string s):element(s),leftChild(nullptr),rightChild(nullptr) {}

Node (string s, Node* l, Node* r):element(s),leftChild(l),rightChild(r) {}

};

//ExpressionTree class

class ExpressionTree

{

public:

//expression tree construction

Node* covert(string postfix)

{

stack <Node*> stk;

Node *t = nullptr;

for(int i=0; i<postfix.size(); i++)

{

if(postfix[i]==' ') continue;

string s(1, postfix[i]);

t = new Node(s);

if(!isOperator(postfix[i]))

{

stk.push(t);

}

else

{

Node *r = nullptr, *l = nullptr;

if(!stk.empty()){

r = stk.top();

stk.pop();

}

if(!stk.empty()){

l = stk.top();

stk.pop();

}

t->leftChild = l;

t->rightChild = r;

stk.push(t);

}

}

return stk.top();

}

//inorder traversal

void infix(Node *root)

{

if(root!=nullptr)

{

cout<< "(";

infix(root->leftChild);

cout<<root->element;

infix(root->rightChild);

cout<<")";

}

}

//postorder traversal

void postfix(Node *root)

{

if(root!=nullptr)

{

postfix(root->leftChild);

postfix(root->rightChild);

cout << root->element << " ";

}

}

//preorder traversal

void prefix(Node *root)

{

if(root!=nullptr)

{

cout<< root->element << " ";

prefix(root->leftChild);

prefix(root->rightChild);

}

}

};

//main method

int main()

{

string infix;

cout<<"Enter the expression: ";

cin >> infix;

Converter conv(infix);

string postfix = conv.toPostFix(infix);

cout<<"Postfix Expression: " << postfix<<endl;

if(postfix == "")

{

cout<<"Invalid expression";

return 1;

}

ExpressionTree etree;

Node *root = etree.covert(postfix);

cout<<"Infix: ";

etree.infix(root);

cout<<endl;

cout<<"Prefix: ";

etree.prefix(root);

cout<<endl;

cout<< "Postfix: ";

etree.postfix(root);

cout<<endl;

return 0;

}

3 0
3 years ago
A class that ______________ an interface must contain methods for all abstract methods in the interface. Otherwise, the class mu
WINSTONCH [101]

Answer:  A class that implement an interface must contain methods for all abstract methods in the interface. Otherwise, the class must be declared as abstract.

Explanation:

It is necessary to implement all the abstract method that are present in the interface. Basically, this is one of the rule of abstract method. As, abstract method is define as without any implementation.

If any class contain abstract method then, it must be declare as abstract. Therefore, if a class are not implemented an interface method then, it should be declare as abstract.    

8 0
2 years ago
The aim of this activity is to implement an algorithm that returns the union of elements in a collection. Although working with
shusha [124]

Answer:

The function in Python is as follows:

def unite_lists(A, B):

   union = []

   for elem in A:

       if elem in B:

           union.append(elem)

   return union

Explanation:

This defines the function

def unite_lists(A, B):

This initializes an empty list for the union list

   union = []

This iterates through list A

   for elem in A:

This checks if the item in list A is in list B

       if elem in B:

If it is present, the item is appended to the union list

           union.append(elem)

This returns the union of the two lists

   return union

3 0
3 years ago
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