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True [87]
3 years ago
11

When cleaning a computer, you need only worry about what you can see.

Computers and Technology
1 answer:
almond37 [142]3 years ago
4 0

Answer:

False because the inside could be dirty and cause problems.

Explanation:

You might be interested in
Technician A says that generic scan tools must be able to read all generic OBD-II codes. Technician B says that all generic scan
Likurg_2 [28]

Answer:

A. Technicians A and B

Explanation:

When we're talking about generic scanners and about all OBD-II codes, in this case, both technician A and B is the correct answer. Because we can scan all OBD-II codes with a generic scan.

But the technician A just says generic tools must be able to read all generic OBD-II codes and technicians B just says generic scan tools must be able to read manufacture OBD-II code, both are correct.

4 0
3 years ago
For this lab you will write a class to create a user-defined type called Shapes to represent different shapes, their perimeters
maw [93]
<h2>Answer:</h2>

<u />

========= Shape.java  ===========

//import the Scanner class

import java.util.Scanner;

public class Shape{

   //required fields

  private String shape;

   private double area;

   private double perimeter;

   //default constructor

  public Shape(){

       this.shape = "unknown";

       this.area = 0.0;

       this.perimeter = 0.0;

   }

   //constructor with one parameter

   public Shape(String shape){

       this.setShape(shape);

       this.area = 0.0;

       this.perimeter = 0.0;

   }

   //accessors and mutators

  public void setShape(String shape){

       this.shape = shape;

   }

  public String getShape(){

       return this.shape;

   }

   public double getPerimeter(){

       return this.perimeter;

   }

  public double getArea(){

       return this.area;

   }

  public void setPerimeter(Scanner scr){

       if(this.getShape().equals("circle")){

           System.out.println("Enter the radius of the circle");

           double radius = scr.nextDouble();

           this.perimeter = 2 * 3.142 * radius;

       }

       else if(this.getShape().equals("rectangle")){

           System.out.println("Enter the width");

           double width = scr.nextDouble();

           System.out.println("Enter the height");

           double height = scr.nextDouble();

           this.perimeter = 2 * (width + height);

       }

       else if(this.getShape().equals("square")){

           System.out.println("Enter the height or width");

           double height = scr.nextDouble();

           this.perimeter = 4 * height;

       }

       else if(this.getShape().equals("unknown")){

           System.out.println("You must define a shape first before calculating perimeter");

           this.perimeter = 0.0;

       }

       else {

           System.out.println("You must define a shape first before calculating perimeter");

           this.perimeter = 0.0;

       }

   }

   public void setArea(Scanner scr){

       if(this.getShape().equals("circle")){

           System.out.println("Enter the radius of the circle");

           double radius = scr.nextDouble();

           this.area = 3.142 * radius * radius;

       }

       else if(this.getShape().equals("rectangle")){

           System.out.println("Enter the width");

           double width = scr.nextDouble();

           System.out.println("Enter the height");

           double height = scr.nextDouble();

           this.area = width * height;

       }

       else if(this.getShape().equals("square")){

           System.out.println("Enter the height or width");

           double height = scr.nextDouble();

           this.area = height * height;

       }

       else if(this.getShape().equals("unknown")){

           System.out.println("You must define a shape first before calculating area");

           this.area = 0.0;

       }

       else {

           System.out.println("You must define a shape first before calculating area");

           this.area = 0.0;

       }

   }

   //Own methods

   //1. Method to show the properties of a shape

   public void showProperties(){

       System.out.println();

       System.out.println("The properties of the shape are");

       System.out.println("Shape : " + this.getShape());

       System.out.println("Perimeter : " + this.getPerimeter());

       System.out.println("Area : " + this.getArea());

   

   }

   //2. Method to find and show the difference between the area and perimeter of a shape

   public void getDifference(){

       double diff = this.getArea() - this.getPerimeter();

       System.out.println();

       System.out.println("The difference is " + diff);

   }

}

========= ShapeTest.java  ===========

import java.util.Scanner;

public class ShapeTest {

   public static void main(String [] args){

       Scanner scanner = new Scanner(System.in);

       // create an unknown shape

       Shape shape_unknown = new Shape();

       //get the shape

       System.out.println("The shape is " + shape_unknown.getShape());

       //set the area

       shape_unknown.setArea(scanner);

       //get the area

       System.out.println("The area is " + shape_unknown.getArea());

       //set the perimeter

       shape_unknown.setPerimeter(scanner);

       //get the perimeter

       System.out.println("The perimeter is " + shape_unknown.getPerimeter());

       // create another shape - circle

       Shape shape_circle = new Shape("circle");

       //set the area

       shape_circle.setArea(scanner);

       //get the area

       System.out.println("The area is " + shape_circle.getArea());

       //set the perimeter

       shape_circle.setPerimeter(scanner);

       //get the area

       System.out.println("The perimeter is " + shape_circle.getArea());

       //get the properties

       shape_circle.showProperties();

       //get the difference between area and perimeter

       shape_circle.getDifference();

   }

}

<h2>Sample output:</h2>

The shape is unknown

You must define a shape first before calculating area

The area is 0.0

You must define a shape first before calculating perimeter

The perimeter is 0.0

Enter the radius of the circle

>> 12

The area is 452.448

Enter the radius of the circle

>> 12

The perimeter is 452.448

The properties of the shape are

Shape : circle

Perimeter : 75.408

Area : 452.448

The difference is 377.03999999999996

<h2>Explanation:</h2>

The code above is written in Java. It contains comments explaining important parts of the code. Please go through the code for more explanations. For better formatting, the sample output together with the code files have also been attached to this response.

Download java
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
57aa40cd91dceaa5454e65fc5d209315.png
5 0
3 years ago
Two negative reviews and no positive reviews is enough to consider the website to have a negative reputation.
Bas_tet [7]

Two negative reviews and no positive reviews is enough to consider the website to have a negative reputation: False.

<h3>What is reputation?</h3>

In Computer technology, reputation can be defined as a metric which is used to determine the quality of a particular website, especially based on the opinions and ratings by its end users.

In this context, we can infer that two negative reviews and no positive reviews isn't an enough metric to consider a website as having a negative reputation.

Read more on negative reputation here: brainly.com/question/2343924

#SPJ1

4 0
2 years ago
When you use the tilde operator (~ in a url for the attribute of a server control, it represents the _____ directory of the webs
LuckyWell [14K]
~ usually represents the home directory( or the directory from which each user's repository is immediately accessible.
none of the options are necessarily correct
6 0
3 years ago
To use an automated teller machine (ATM), a customer must enter his or her four-digit Personal Identification Number (PIN). How
schepotkina [342]

Answer:

10000 possible PINs.

Explanation:

Considering the fact that only numbers are permitted (0 - 9) which gives 10 characters and also that the digits can be repeated, the total number of possible PINs that are possible is 10 x 10 x 10 x 10 = 10000

6 0
3 years ago
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