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kogti [31]
3 years ago
13

Consider the following code snippet:

Computers and Technology
1 answer:
Schach [20]3 years ago
6 0

Answer:

c. The Auto class overloads the setVehicleClass method.

Explanation:

In this snippet of code the Auto class overloads the setVehicleClass method. This is because if a super-class and a sub-class have the same method, the sub-class either overrides or overloads the super-class method. In this case the sub-class Auto is overloading the setVehicleClass method because the parameters are different. The Auto class methods parameter are of type int while the super-class methods parameter are of type double. Therefore, it will overload the method if an int is passed as an argument.

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pickupchik [31]

Answer:

<u>Python 3</u>

Explanation:

The language's latest iteration, Python 3.9, was released on October 5, 2020. It includes even more new features such as relaxed grammar restrictions, flexible function and variable annotations, and new string methods to remove prefixes and suffixes.

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Does anyone else realize how the only tutor is for math never any other subjects?​
Slav-nsk [51]

Answer:

yes I do and I wonder why ??

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If the task is to write firewall specifications for the preparation of a(n) __________, the planner would note that the delivera
USPshnik [31]

Answer:

RFP

Explanation:

If the task is to write firewall specifications for the preparation of a(n)<u> RFP</u> , the planner would note that the deliverable is a specification document suitable for distribution to vendors.

RFP means request for proposal.

3 0
3 years ago
Create a function void process(char ch, int x, int y)- to accept an arithmetic operator (+,-./, in argum
saul85 [17]

Answer:

The program is as follows:

import java.util.*;

public class Main{

   public static void process(char ch, int x, int y){

if(ch == '+'){

    System.out.print(x+y);  }

else if(ch == '-'){

    System.out.print(x-y);  }

else if(ch == '*'){

    System.out.print(x*y);  }

else if(ch == '/'){

    if(y!=0){

        double num = x;

        System.out.print(num/y);      }

    else{

        System.out.print("Cannot divide by 0");     } }

else{

    System.out.print("Invalid operator"); }

 }

public static void main(String[] args) {

 Scanner input = new Scanner(System.in);

 int x, y;

 char ch;

 System.out.print("Enter two integers: ");

 x = input.nextInt(); y = input.nextInt();

 System.out.print("Enter operator: ");

 ch = input.next().charAt(0);  

 process(ch,x, y);

}

}

Explanation:

The function begins here

   public static void process(char ch, int x, int y){

If the character is +, this prints the sum of both integers

<em> if(ch == '+'){</em>

<em>     System.out.print(x+y);  }</em>

If the character is -, this prints the difference of both integers

<em> else if(ch == '-'){</em>

<em>     System.out.print(x-y);  }</em>

If the character is *, this prints the product of both integers

<em> else if(ch == '*'){</em>

<em>     System.out.print(x*y);  }</em>

If the character is /, this prints the division of both integers.

<em> else if(ch == '/'){</em>

<em>     if(y!=0){</em>

<em>         double num = x;</em>

<em>         System.out.print(num/y);      }</em>

<em>     else{</em>

<em>This is executed if the denominator is 0</em>

<em>         System.out.print("Cannot divide by 0");     } }</em>

Invalid operator is printed for every other character

<em>else{</em>

<em>     System.out.print("Invalid operator"); }</em>

<em> </em> }

The main begins here

public static void main(String[] args) {

 Scanner input = new Scanner(System.in);

This declares both integers

 int x, y;

This declares the operator

 char ch;

Get input for both integers

<em>  System.out.print("Enter two integers: ");</em>

<em>  x = input.nextInt(); y = input.nextInt();</em>

Get input for the operator

<em>  System.out.print("Enter operator: ");</em>

<em>  ch = input.next().charAt(0);   </em>

Call the function

 process(ch,x, y);

}

}

6 0
3 years ago
Can you believe everything you read online?
Tasya [4]

Answer:

no

Explanation:

bcoz some time it can be a rumour

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