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makkiz [27]
3 years ago
6

Não basta tornar-se móvel hoje em dia. Ao se organizar para a implantação de tecnologias no contexto da mobilidade, é preciso fi

car atento a algumas características fundamentais que as aplicações móveis precisam ter. Caraceterísticas essas, que podem definir a forma de utilização dos dispositivos móveis.
Considerando as características desta apicações, faça a relação das colunas, entre o conceito e o exemplo apresentado.

1 - Riqueza da informação.
2 - Facilidade de uso.
3 - Facilidade de gerenciamento.
4 - Imediatismo.

( ) Maria Eduarda é super antenada em todos os tipos de esportes. Por isso, fica conectada o tempo todo e está sempre atualizada das últimas informações divulgadas na web.
( ) Andressa adora Moda. Por esse motivo, acompanha as tendências desse setor, através do recebimento de informações atualizadas e de diversas formas como vídeos e fotos. Isso proporciona o enriquecimento do seu conhecimento sobre a área.
( ) Eloisa é Diretora de Tecnologia em uma grande empresa do setor de aviação civil. Por ter que tomar decisões importantes todos os dias e por estar em constante deslocamento, ela acessa remotamente as aplicações da empresa para realizar os suportes necessários.
( ) Maria Clara baixou um novo aplicativo em seu celular para organizar sua agenda pessoal e profissional. Depois de consultar diversas opções, optou por uma delas em que o uso era simplificado e intuitivo, além de obedecer a comandos de voz em diversas funções.

A sequência correta desta classificação é:
Alternativas
Alternativa 1:

4 - 3 - 1 - 2.
Alternativa 2:

4 - 1 - 3 - 2.
Alternativa 3:

1 - 3 - 2 - 4.
Alternativa 4:

2 - 1 - 3 - 4.
Alternativa 5:

1 - 2 - 4 - 3.
Computers and Technology
1 answer:
Ad libitum [116K]3 years ago
4 0

Answer:

Alternativa 2:

4 - 1 - 3 - 2.

Explanation:

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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
The computer that process data that are represented in the form of discrete values are called​
Vinvika [58]

Answer:

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Explanation:

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Ratling [72]

The amplitude of a sound energy depends upon the how much energy we are giving to the wave or how much compression on the wave is taking place.

It does’t depends upon the wave length,frequency


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