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Eva8 [605]
3 years ago
9

The ability to adapt to increases or decreases in demand for processing or data storage is known as ________. Group of answer ch

oices scalability usability reliability portability extensibility
Computers and Technology
2 answers:
zalisa [80]3 years ago
7 0

Answer:

The correct answer is Scalability

Explanation:

Scalability is defined as the ability of a computing process to be produced in various capabilities. This means that it entails the ability to adapt to decreases or increase in demand for data storage and processing.  When a system can perform without any problems under an expanding and increasing scope, this concept is known as scalability.

Talja [164]3 years ago
6 0

Answer:

Scalability

Explanation:

<em>Scalability </em>is a generic term for the ability of a system to accommodate expansion. A system is scalable if it is able to adapt to changes in growth - either as an increase or a decrease.

<em>Extensibility</em> of a system is its ability to be easily modified. In other words, the ease of scalability of a system is a measure of its extensibility.

<em>Usability</em> of a system is the measure of how easy it is to use the system. It is the quality of the experiences of the users interacting with the system.

<em>Portability</em> of a system is its ability to be used or run in diverse environments.

<em>Reliability</em> of a system is the ability of the system to be used without failing over a specified period of time.

Therefore, the term that best describes the scenario is <em>Scalability.</em>

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Determine the distance between point (x1, y1) and point (x2, y2), and assign the result to points Distance. The calculation is:
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<u>Question:</u>

Snapshot of the question has been attached to this response.

<u>Answer:</u>

<u></u>

import java.util.Scanner;

import java.lang.Math;

public class CoordinateGeometry{

    public static void main(String []args){

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       //Declare a variable to hold the result of (x2 - x1)²

       double x;

       

       //Solve (x2 - x1)^2 and store result in the variable x

       x = Math.pow(x2 - x1, 2);

       

       //Declare a variable to hold the result of (y2 - y1)²

       double y;

       

       //Solve (y2 - y1)^2 and store result in the variable y

       y = Math.pow(y2 - y1, 2);

       

       //Now pointsDistance = SquareRootOf(x + y)

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       return;

    }

}

<u>Sample Output:</u>

Points distance:  

3.0

<u>Explanation:</u>

The above code has been written in Java and it contains comments explaining important lines of the code. Please go through the comments.

The snapshots of the program and a sample output have been attached to this response.

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