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notka56 [123]
3 years ago
6

What command would you use to see how many concurrent telnet sessions you can run on the IFT MAIN router and how many could can

you have on that router
Computers and Technology
1 answer:
Mariulka [41]3 years ago
6 0

"Computer, I demand information about how many concurrent telnet sessions I can run on the IFT MAIN router and how many I could have on that router, quickly!"

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The first mechanical computer design in by Charles Babbage was Called​
White raven [17]

Answer:

<h2>Analytical Engine</h2>

Explanation:

Analytical Engine, generally considered the first computer, designed and partly built by the English inventor Charles Babbage in the 19th century (he worked on it until his death in 1871).

7 0
3 years ago
Read 2 more answers
What are some other ways to program a robot to navigate a complicated environment other than straight paths and right angle (90
In-s [12.5K]

Answer:

The most popular method of robot programming is probably the teach pendant. ... To program the robot, the operator moves it from point-to-point, using the buttons on the pendant to move it around and save each position individually. When the whole program has been learned, the robot can play back the points at full speed.

8 0
3 years ago
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What type of computer is likely to use so-dimms, have an internal power supply, and use a desktop processor socket?
Gekata [30.6K]
Obviously, it is a laptop. So-Dimms are about half as long as Dimms, and are meant for laptops.
8 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
Compare the elements of the basic Software Development Life Cycle with 2 other models. How are they similar? How are they differ
Artemon [7]

Answer:

Explanation:

One of the basic notions of the software development process is SDLC models which stands for Software Development Life Cycle models. SDLC – is a continuous process, which starts from the moment, when it’s made a decision to launch the project, and it ends at the moment of its full remove from the exploitation. There is no one single SDLC model. They are divided into main groups, each with its features and weaknesses. The most used, popular and important SDLC models are given below:

1. Waterfall model

2. Iterative model

3. Spiral model

4. V-shaped model

5. Agile model

Stage 1. Planning and requirement analysis

Each software development life cycle model starts with the analysis, in which the stakeholders of the process discuss the requirements for the final product.

Stage 2. Designing project architecture

At the second phase of the software development life cycle, the developers are actually designing the architecture. All the different technical questions that may appear on this stage are discussed by all the stakeholders, including the customer.  

Stage 3. Development and programming

After the requirements approved, the process goes to the next stage – actual development. Programmers start here with the source code writing while keeping in mind previously defined requirements. The programming by itself assumes four stages

• Algorithm development

• Source code writing

• Compilation

• Testing and debugging

Stage 4. Testing

The testing phase includes the debugging process. All the code flaws missed during the development are detected here, documented, and passed back to the developers to fix.  

Stage 5. Deployment

When the program is finalized and has no critical issues – it is time to launch it for the end users.  

SDLC MODELS

Waterfall – is a cascade SDLC model, in which development process looks like the flow, moving step by step through the phases of analysis, projecting, realization, testing, implementation, and support. This SDLC model includes gradual execution of every stage completely. This process is strictly documented and predefined with features expected to every phase of this software development life cycle model.

ADVANTAGES  

Simple to use and understand

DISADVANTAGES

The software is ready only after the last stage is over

ADVANTAGES

Management simplicity thanks to its rigidity: every phase has a defined result and process review

DISADVANTAGES

High risks and uncertainty

Iterative SDLC Model

The Iterative SDLC model does not need the full list of requirements before the project starts. The development process may start with the requirements to the functional part, which can be expanded later.  

ADVANTAGES                                        

Some functions can be quickly be developed at the beginning of the development lifecycle

DISADVANTAGES

Iterative model requires more resources than the waterfall model

The paralleled development can be applied Constant management is required

Spiral SDLC Model

Spiral model – is SDLC model, which combines architecture and prototyping by stages. It is a combination of the Iterative and Waterfall SDLC models with the significant accent on the risk analysis.

4 0
3 years ago
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