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Fynjy0 [20]
3 years ago
7

on which two interfaces or ports can security be improved by configuring executive timeouts? (choose two.)

Computers and Technology
1 answer:
ELEN [110]3 years ago
5 0

Executive timeouts are configured using ports such as Console port, Aux port, VTY port.

Executive time outs are used to automatically disconnect devices which have been idle for a specific period of time.

In IOS devices, the executive timeout automatically disconnect users after 10 minutes of inactivity. These users can either be console or VTY user. The executive timeout can be set using line mode command.

Executive timeouts are configured using ports such as Console port, Aux port, VTY port.

Find out more using: brainly.com/question/13025617

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Write a program that reads in an integer value for n and then sums the integers from n to 2 * n if n is nonnegative, or from 2 *
devlian [24]

Answer:

Following are the program in c++

First code when using for loop

#include<iostream> // header file  

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n  

cin>>n;

if(n > 0) // check if n is nonnegative

{

for(i=n; i<=(2*n); i++)

{

sum1+= i;

}

}

else //  check if n  is negative

{

for(i=(2*n); i<=n; i++)

{

sum1+= i;

}

}

cout<<"The sum is:"<<sum1;

return 0;

Output

Enter the number: 1

The sum is:3

second code when using while loop

#include<iostream> // header file  

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n  

cin>>n;

if(n > 0) // check if n is nonnegative

{

int i=n;  

while(i<=(2*n))

{

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}

else //  check if n  is negative

{

int i=n;  

while(i<=(2*n))

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}

}

cout<<"The sum is:"<<sum1;

return 0;

}

Output

Enter the number: 1

The sum is:3

Explanation:

Here we making 2 program by using different loops but using same logic

Taking a user input in "n" variable.

if n is nonnegative then we iterate the loops  n to 2 * n and storing the sum in "sum1" variable.

Otherwise iterate a loop  2 * n to n and storing the sum in "sum1" variable.  Finally display  sum1 variable .

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Answer:

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Explanation:

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Answer:

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