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yarga [219]
2 years ago
6

What two methods can be used to access and modify an existing program that is running on an iot device in cisco packet tracer? (

choose two.)?
Computers and Technology
1 answer:
dsp732 years ago
3 0

Cisco Packet Tracer is tool used for network simulation and visualization program.

The two methods can be used to access and modify an existing program that is running on an IoT device in Cisco packet trace are the following:

1. Click on the device then select the Programming tab.

2. Go to the registration server and login. Then select the Editor tab.

You might be interested in
The area of a square is stored in a double variable named area. write an expression whose value is length of the diagonal of the
cestrela7 [59]

Answer:

The correct answer to the following question will be "Math.sqrt(area*2)".

Explanation:

  • The Math.sqrt( ) method in JavaScript is used to find the squares root of the given figure provided to the feature as a variable.
  • Syntax Math.sqrt(value) Variables: this function takes a single variable value that represents the amount whose square root is to be determined.
  • The area of the figure is the number of squares needed to cover it entirely, like the tiles on the ground.

Area of the square = side times of the side.

Because each side of the square will be the same, the width of the square will only be one side.

Therefore, it would be the right answer.

5 0
2 years ago
A user requests an unencrypted webpage from a web server running on a computer, listening on the Internet Protocol address 10.1.
Minchanka [31]

Answer:

10.1.1.150:80

Explanation:

Socket address is the combination of an IP address and port number.

HTTP is what most unencrypted webpage use, instead of HTTPS which is the encrypted version of HTTP.

By default, webserver using Hypertext Transfer Protocol (HTTP) uses a standard port known as port 80.

The Socket address which is the combination of the IP address and standard port number will be something like this:

10.1.1.150:80

3 0
2 years ago
Write a copy assignment operator for CarCounter that assigns objToCopy.carCount to the new objects's carCount, then returns *thi
ollegr [7]

Answer:

Here is the copy assignment operator for CarCounter:

CarCounter& CarCounter::operator=(const CarCounter& objToCopy) {

carCount = objToCopy.carCount;

return *this;  }

The syntax for copy assignment operator is:

ClassName& ClassName :: operator= ( ClassName& object_name)

In the above chunk of code class name Is CarCounter and object name is objToCopy.

Assignment operator = is called which assigns objToCopy.carCount to the new objects's carCount.

This operator basically is called when an object which is already initialized is assigned a new value from another current object.

Then return *this returns the reference to the calling object

Explanation:

The complete program is:

#include <iostream>  //to use input output functions

using namespace std;   //to access objects like cin cout

class CarCounter {  //class name

public:  // public member functions of class CarCounter

CarCounter();  //constructor of CarCounter

CarCounter& operator=(const CarCounter& objToCopy);  //copy assignment operator for CarCounter

void SetCarCount(const int setVal) {  //mutator method to set the car count value

carCount = setVal;  } //set carCount so setVal

int GetCarCount() const {  //accessor method to get carCount

return carCount;  }  

private:  //private data member of class

int carCount;  };    // private data field of CarCounter

CarCounter::CarCounter() {  //constructor

carCount = 0;  //intializes the value of carCount in constructor

return; }  

// FIXME write copy assignment operator  

CarCounter& CarCounter::operator=(const CarCounter& objToCopy){

/* copy assignment operator for CarCounter that assigns objToCopy.carCount to the new objects's carCount, then returns *this */

carCount = objToCopy.carCount;

return *this;}  

int main() {  //start of main() function

CarCounter frontParkingLot;  // creates CarCounter object

CarCounter backParkingLot;   // creates CarCounter object

frontParkingLot.SetCarCount(12);  // calls SetCarCount method using object frontParkingLot to set the value of carCount to 12

backParkingLot = frontParkingLot;  //assigns value of frontParkingLot to backParkingLot

cout << "Cars counted: " << backParkingLot.GetCarCount();  //calls accessor GetCarCount method to get the value of carCount and display it in output

return 0;  }

The output of this program is:

Cars counted = 12

3 0
2 years ago
Which of the following statements are true. .ascii stores string in memory and terminate it with NULL character. There is no way
Tems11 [23]

Answer:

The true statements are:

There is an assembler directive to arrange / place floating point values in static data memory

MARS always uses setting '.set boat'

Explanation:

It is the assembler directive who arranges and places the floating point values for the static data memory. Obviously there is no such way for the MIPS assemble programming for reservation of the static data memory without having any initialization for a considerable value.

MARS would definitely use the setting set “noat” ascii would store the string in the memory and then terminate it with respect to NULL character. They cannot reserve the MIPS assembly programming for a considerable value.

4 0
2 years ago
The population of town A is less than the population of town B. However, the population of town A is growing faster than the pop
defon

Answer:

#include<iostream>

using namespace std;

void main()

{

int townA_pop,townB_pop,count_years=1;

double rateA,rateB;

cout<<"please enter the population of town A"<<endl;

cin>>townA_pop;

cout<<"please enter the population of town B"<<endl;

cin>>townB_pop;

cout<<"please enter the grothw rate of town A"<<endl;

cin>>rateA;

cout<<"please enter the grothw rate of town B"<<endl;

cin>>rateB;

while(townA_pop < townB_pop)//IF town A pop is equal or greater than town B it will break

{

townA_pop = townA_pop +( townA_pop * (rateA /100) );

townB_pop = townB_pop +( townB_pop * (rateB /100) );

count_years++;

}

cout<<"after "<<count_years<<" of years the pop of town A will be graeter than or equal To the pop of town B"<<endl;

}

Explanation:

3 0
2 years ago
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