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dangina [55]
3 years ago
15

A client wishes to update their legacy system even though there have been no

Computers and Technology
1 answer:
gulaghasi [49]3 years ago
8 0

Answer:

A client wishes to update their legacy system even though there have been no security breaches since its implementation five years ago. If the client has not suffered any attacks, why is it still necessary to update their system? ... Because new security threats emerge all the time.

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What are new technologies, products, or services that eventually surpass the existing dominant technology or product in a market
kicyunya [14]

Answer:

Disruptive innovation

Explanation:

Disruptive innovation are considered those new technologies, products or services whose application can greatly impact the manner in which an industry or a market functions. This is because the surpass the existing dominant product or technology. One example of a disruptive innovation is the internet. The internet altered the manner in which business was done by companies, but negatively affected those who refused to adopt it.

4 0
3 years ago
The National Institute of Standards and Technology (NIST) guidelines list four different states a mobile device can be in when y
Ahat [919]

Answer:

quiescent

Explanation:

The National Institute of Standards and Technology (NIST) guidelines list four different states a mobile device can be in when you extract data. The quiescent mode is a dormant mode that conserves battery life while maintaining user data and performing other background functions.

4 0
3 years ago
Which is the fastest method of sharing public folders?
Rasek [7]

Answer:

I'm stuck on this but the correct answers I would at-least pick is A and C I need more information about what operating system you have on the computer.

3 0
3 years ago
In Java.Use a single for loop to output odd numbers, even numbers, and an arithmetic function of an odd and even number. Use the
rosijanka [135]

Answer:

<em>This program is written using java programming language</em>

<em>Difficult lines are explained using comments (See Attachment for Source file)</em>

<em>Program starts here</em>

import java.util.*;

import java.lang.Math;

public class oddeven{

public static void main(String [] args)

{

 double even,odd;//Declare variables even and odd as double

 //The next iteration prints odd numbers

 for(int i = 1;i<=173;i+=2)

 {

  odd = i;

  System.out.format("%.4f",odd);//Round to 4 decimal places and print

System.out.print("\t");

 }

 System.out.print('\n');//Start printing on a new line

 //The next iteration prints even numbers

 for(int i = 0;i<=173;i+=2)

 {

  even=i;

  System.out.format("%.4f",even);//Round to 4 decimal places and print

System.out.print("\t");

 }

 System.out.print('\n');//Start printing on a new line

 double  ssqrt;//Declare ssqrt to calculate the square root of sum of even and odd numbers

 for(int i = 0;i<=173;i+=2)

 {

  ssqrt = Math.sqrt(2*i+1);//Calculate square root here

  System.out.format("%.4f",ssqrt); //Round to 4 decimal places and print

System.out.print("\t");

 }

}

}

Explanation:

Libraries are imported into the program

import java.util.*;

import java.lang.Math;

The following line declares variables even and odd as double

double even,odd;

The following iteration is used to print odd numbers  with in the range of 0 to 173

for(int i = 1;i<=173;i+=2)  {

odd = i;

System.out.format("%.4f",odd); This particular line rounds up each odd numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

The following code is used to start printing on a new line

System.out.print('\n');

The following iteration is used to print even numbers  with in the range of 0 to 173

for(int i = 0;i<=173;i+=2)

{

even=i;

System.out.format("%.4f",even); This particular line rounds up each even numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

The following code is used to start printing on a new line

System.out.print('\n');

The next statement declares ssqrt as double to calculate the square root of the sum of even and odd numbers

double  ssqrt;

for(int i = 0;i<=173;i+=2)

{

ssqrt = Math.sqrt(2*i+1);This line calculates the square root of sum of even and odd numbers

System.out.format("%.4f",ssqrt); This particular line rounds up each even numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

Download java
7 0
3 years ago
¿Cuáles son los 5 Procesadores de textos on line más usados?
labwork [276]

Answer:

5 Procesadores de textos

Explanation:

Google Docs.

Textilus - Edición Microsoft Word.

TextEdit.

Kingsoft Office Writer.

Páginas.

8 0
3 years ago
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