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Allisa [31]
2 years ago
10

How did hitles rules in nazi germany exemplify totiltarian rule?

Computers and Technology
1 answer:
ddd [48]2 years ago
7 0

Answer:

hope this helps if not srry

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Write a program that prompts the user to enter in a postive number. Only accept positive numbers - if the user supplies a negati
marusya05 [52]

Answer:

The c++ program to check prime numbers is shown below.

#include <iostream>

using namespace std;

int main() {

   int num, prime=0;

   do

   {

       cout<<"Enter a positive number."<<endl;

       cin>>num;

       if(num<1)

       {

           cout<<"Invalid number. Enter a positive number"<<endl;

           cin>>num;

       }

   }while(num<1);

   

   if(num==1 || num==2 || num==3)

       cout<<num<<" is a prime number."<<endl;

   else if(num%2 == 0)

       cout<<num<<" is not a prime number."<<endl;

   else

   {

       for(int k=3; k<num/2; k++)

       {

           if(num%k == 0)

               prime++;

       }    

   if(prime>1)

       cout<<num<<" is not a prime number."<<endl;

   else

       cout<<num<<" is a prime number."<<endl;

   }

}

OUTPUT

Enter a positive number.

-7

Invalid number. Enter a positive number

0

Enter a positive number.

79

79 is a prime number.

Explanation:

The user input is validated for positivity. A do while loop along with an if statement is implemented for verification.

do

   {

       cout<<"Enter a positive number."<<endl;

       cin>>num;

       if(num<1)

       {

           cout<<"Invalid number. Enter a positive number"<<endl;

           cin>>num;

       }

   }while(num<1);

The test for prime number is done by using multiple if else statements.

If user inputs 1, 2, or 3, message is displayed.

Else If user inputs an even number, message is displayed for not prime. This is done by taking modulo of the number upon division by 2.

Else if user inputs neither an even number nor a number less than 3, the modulus of the number is taken with divisors beginning from 3 up to half of the input number.

For this, an integer variable prime is initialized to 0. A number can be completely divisible by itself or by its factors.

If the number is divisible by any of the divisors, value of variable prime is increased by 1. If value of prime is greater than 1, this means that the user input is divisible by more than one divisor. Hence, the given number is not a prime number.

5 0
3 years ago
Which Application program saves data automatically as it is entered?
Andrei [34K]
The application program that saves data automatically as it is entered is the MS Access.
4 0
2 years ago
If you're unsure of what chart to use for a set of data, what feature does Excel include that will help you to decide? A. Sparkl
asambeis [7]
If what I found is correct it would be C, all charts... If not that then it would be B.
6 0
3 years ago
Read 2 more answers
Please add thenodes given below to construct the AVL tree show all the necessarysteps,
Paha777 [63]

Answer:

30,33,37,18,23,34,15,38,40,17

To construct the AVL tree, follow these steps and diagrams are shown in the image:

• Add 30 to the tree as the root node. Then add 33 as the right child because of 33 is greater than 30 and AVL tree is a binary search tree.

• Then add 37 as the right child of 33. Here the balance factor of node 30 becomes 0-2 = -2, unbalanced.

Use RR rotation, make node 33 the root node, 30 as the left child of 33 and 37 as the right child of 33.

• Now add 18 as the left child of 30. And 23 as the right child of 18. Here the balance factor of 30 becomes 2-0 = 2. It’s unbalanced.

Use LR rotation, make 23 the parent of 18 and 30.  

• Now add 34 as the left child of 37 and 15 as the left child of 18. Add 38 as the right child of 37. And then add 40 as the right child of 38.

• Now adding 17 as the right child of 15 makes the tree unbalanced at 18.

Use LR rotation, make 17 as the parent of 15 and 18.

Explanation:

The balance factor of a node can be either 0,1 or -1. Else the tree is called unbalanced at the node.

If the inserted node is in the left subtree of the left subtree of the unbalance node, then perform LL rotation.

If the inserted node is in the right subtree of the right subtree of the unbalance node, then perform RR rotation.

If the inserted node is in the left subtree of the right subtree of the unbalance node, then perform RL rotation.

If the inserted node is in the right subtree of the left subtree of the unbalance node, then perform LR rotation.

7 0
2 years ago
Contast between backups controlled within the SAN and backups controlled from the application server.
luda_lava [24]

Answer and Explanation:

Backup control within the SAN is one which confines the backup to the SAN and does not pass it via corporate network.

Backup traffic within the SAN are  easier to maintain and easily handled in contrast to the application server.

Backup in SAN is governed by  such applications and mechanism which allows the easy handling of the backup traffic which is not provided by the application servers.

In case of occurrence of any failure,  unlike application server, SAN provides smooth and fast migration

SAN holds data in cluster  by the use of backup techniques not used in application server.

6 0
3 years ago
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