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Kamila [148]
3 years ago
10

Which of the following is an encryption tool that allows users to encrypt files and folders by simply right-clicking a given obj

ect and setting a few options?
a. PowerShellb. BitLockerc. Encrypting File System (EFS)d. Trusted PlatformModule (TPM)
Computers and Technology
2 answers:
BigorU [14]3 years ago
8 0

Answer:

C. Encrypting file system (EFS).

Explanation:

Security policies are used in networks and computer systems to protect data from the hands of cyber attackers or hackers. It promotes the confidentiality and integrity of data in the system. Encrypting a data is needed to protect the information in a document from being viewed by attackers.

The encryption hides the actual information using cryptic algorithms. Encrypting file systems are used in windows OS to protect confidential data. It encrypts and decrypts data, in the file explorer.

galben [10]3 years ago
6 0

Answer:

Option C, ENCRYPTING FILE SYSTEM (EFS).

Explanation:

Encryption is one of the best protection methods available to keep all one's digital assets safe.

Encrypting file system (EFS) is a feature of windows 2000 operating system that allows any file or folder be stored in an encrypted form and decrypted only by an individual user and an authorized recovery agent.

EFS provides filesystem-level encryption. This technology enables files to be transparently encrypted to protect confidential data from attackers with physical access to the computer.

To encrypt or a file using EFS;

Open windows explorer

Right-click the file or folder and select properties

Click advanced in the general tab

Check encrypt contents to secure data

Click ok.

Therefore, option C, ENCRYPTING FILE SYSTEM, is an encryption tool that allows users to encrypt files and folders by simply right-clicking a given object and setting a few options.

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Type the correct answer in the box. Spell all words correctly.
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4 years ago
Write a static method called bothStart that allows the user to input two Strings and returns the String that is the longest subs
marishachu [46]

Answer:

  1.    public static String bothStart(String text1, String text2){
  2.        String s = "";
  3.        if(text1.length() > text2.length()) {
  4.            for (int i = 0; i < text2.length(); i++) {
  5.                if (text1.charAt(i) == text2.charAt(i)) {
  6.                    s += text1.charAt(i);
  7.                }else{
  8.                    break;
  9.                }
  10.            }
  11.            return s;
  12.        }else{
  13.            for (int i = 0; i < text1.length(); i++) {
  14.                if (text1.charAt(i) == text2.charAt(i)) {
  15.                    s += text1.charAt(i);
  16.                }else{
  17.                    break;
  18.                }
  19.            }
  20.            return s;
  21.        }
  22.    }

Explanation:

Let's start with creating a static method <em>bothStart()</em> with two String type parameters, <em>text1 </em>&<em> text2</em> (Line 1).  

<em />

Create a String type variable, <em>s,</em> which will hold the value of the longest substring that both inputs start with the same character (Line 2).

There are two possible situation here: either <em>text1 </em>longer than<em> text2 </em>or vice versa. Hence, we need to create if-else statements to handle these two position conditions (Line 4 & Line 13).

If the length of<em> text1</em> is longer than <em>text2</em>, the for-loop should only traverse both of strings up to the length of the <em>text2 </em>(Line 5). Within the for-loop, we can use<em> charAt()</em> method to extract individual character from the<em> text1</em> & <em>text2 </em>and compare with each other (Line 15). If they are matched, the character should be joined with the string s (Line 16). If not, break the loop.

The program logic from (Line 14 - 20) is similar to the code segment above (Line 4 -12) except for-loop traverse up to the length of <em>text1 .</em>

<em />

At the end, return the s as output (Line 21).

5 0
3 years ago
Can someone smart help me
nalin [4]

Answer:

Answer

Explanation:

The only thing that is wrong with this code is the fact that the link is not completed (should be https://[link name].com/)

Other than that the code is fine

7 0
3 years ago
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