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tester [92]
3 years ago
5

Using Module operator, write a java program to print odd number between 0 and 1000​

Computers and Technology
1 answer:
jolli1 [7]3 years ago
5 0

Answer:

class OddNumber

{

  public static void main(String args[])  

       {

         int n = 1000;  //Store 1000 in Variable n typed integer

         System.out.print("Odd Numbers from 1 to 1000 are:");  // Print headline of output window

         for (int i = 1; i <= n; i++)  //For loop to go through each number till end

          {

            if (i % 2 != 0)  //check if number is even or odd.Not divisible by 2 without reminder means it is odd number

             {

               System.out.print(i + " "); //print odd numbers

             }

          }

       }

}              

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Answer:

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Explanation:

i hope this helps.

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Which statement describes a difference between front-end and back-end databases?
Westkost [7]

Answer:

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Explanation:

The term front-end means interface while the term back-end means server. So imagine you go on your social media app, what you can see like your friends list and your account and the pictures you post, who liked the pictures and all of that is called the front-end and is what <em>you</em> as a user can interact with; front-end development is the programming that focuses on the visual aspect and elements of a website or app that a user will interact with its called in other words the client side. Meanwhile, back-end development focusing on the side of a website users can't see otherwise known as the server side, its what code the site uses and what it does with data and how a server runs and interacts with a user on whatever device surface they're using.

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Employees often attend trainings and read policies. They endure these, but do not internalize them. What should be your first st
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2 years ago
Question is attached as image, please help :&gt;
harkovskaia [24]

Answer:

Proved

Explanation:

Given

X = (a.\bar b)+(\bar a.b)

(a + b)\ .  \frac{}{a.b}

Required

Find out why they represent the same

To do this, we simplify either (a + b)\ .  \frac{}{a.b} or X = (a.\bar b)+(\bar a.b)

In this question, I will simplify (a + b)\ .  \frac{}{a.b}

Apply de morgan's law

(a + b)\ .  \frac{}{a.b} = (a + b) . (\bar a + \bar b)

Apply distribution property

(a + b)\ .  \frac{}{a.b} = a.\bar a + a.\bar b + \bar a . b + b . \bar b

To simplify, we apply the following rules:

a.\bar a = 0 --- Inverse law

b.\bar b = 0 --- Inverse law

So, the expression becomes

(a + b)\ .  \frac{}{a.b} = 0 + a.\bar b + \bar a.b + 0

(a + b)\ .  \frac{}{a.b}  = a.\bar b + \bar a.b

Rewrite as:

(a + b)\ .  \frac{}{a.b}  = (a.\bar b)+(\bar a.b)

From the given parameters:

X = (a.\bar b)+(\bar a.b)

This implies that:

(a + b)\ .  \frac{}{a.b} when simplified is X or (a.\bar b)+(\bar a.b)

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