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Llana [10]
2 years ago
11

A systems administrator is designing a directory architecture to support Linux servers using Lightweight Directory Access Protoc

ol (LDAP). The directory needs to be able to make changes to directory objects securely. Which of these common operations supports these requirements?
Computers and Technology
1 answer:
masya89 [10]2 years ago
6 0

The options available are:

Search, modify.

StartTLS, delete.

Bind, modify.

Bind, add.

Answer:

StartTLS, delete

Explanation:

StartTLS is a computer networking term for email protocol command, in which a system administrator buses it to command an email server that an email client running in a web browser, wants to turn an existing insecure connection into a secure one. Thus, during this process, StartTLS permits a client to communicate securely using LDAPv3 over TLS.

On the other hand, the DELETE operation can make a change to a directory object.

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The machine in question is only able to quickly sound the alarm because it has low latency.

In conclusion, we can say that low latency enables the machine above to sound the alarm quickly which means that low latency is important to the safety of that environment.

<em>Find out more at brainly.com/question/9337524.</em>

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2 years ago
Read 2 more answers
Input a number [1-50] representing the size of the shape and then a character [x,b,f] which represents the shape i.e. x-&gt;cros
Gnoma [55]

Answer:

C++ code given below with appropriate comments

Explanation:

pattern.cpp

#include<iostream>

using namespace std;

void printCross(int n)

{

int i,j,k;

if(n%2) //odd number of lines

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==i || j==(n-i+1))

cout<<j;

else

cout<<" ";

}

cout<<"\n";

}

}

else //even number of lines

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==i || j==(n-i+1))

{

cout<<" "<<j<<" ";

}

else

cout<<" ";

}

cout<<"\n";

}

}

}

void printForwardSlash(int n)

{

if(n%2)

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==n-i+1)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

else

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==(n-i+1))

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

}

void printBackwardSlash(int n)

{

if(n%2) // odd number of lines

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==i)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

else //even number of lines

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==i)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

}

int main()

{

int num;

char ch;

cout<<"Create a numberes shape that can be sized."<<endl;

cout<<"Input an integer [1,50] and a character [x,b,f]."<<endl;

cin>>num>>ch;

if(ch=='x' || ch=='X')

printCross(num);

else if(ch=='f' || ch=='F')

printForwardSlash(num);

else if(ch=='b' || ch=='B')

printBackwardSlash(num);

else

cout<<"\nWrong input"<<endl;

return 0;

}

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Write a setInterval() function that increases the count by 1 and displays the new count in counterElement every 100 milliseconds
GrogVix [38]

Answer:

var count = 0;

var counterElement = document.getElementById("counter");

counterElement.innerHTML = count;

var interval = setInterval(function () {

   count++;

   counterElement.innerHTML = count;

   if (count === 4) {

       clearTimeout(interval);

   }

}, 100);

Explanation:

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Answer:

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Explanation:

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Explanation:

Blackbaud's software products are specially designed to support the unique needs of nonprofit and social good organizations.

Explore our software solutions that can help you advance your organization's mission.

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