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3V0-21.25

Exam Code: 3V0-21.25

Exam Name: Advanced VMware Cloud Foundation 9.0 Automation

Version: V22.75

Q & A: 580 Questions and Answers

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NEW QUESTION: 1
You need to calculate the number of days from 1st January 2019 until today.
Dates are stored in the default format of DD-MON-RR.
Which two queries give the required output? (Choose two.)
A. SELECT SYSDATE - TO_DATE('01-JANUARY-2019') FROM DUAL;
B. SELECT TO_CHAR(SYSDATE,'DD-MON-YYYY') - '01-JAN-2019' FROM DUAL;
C. SELECT ROUND(SYSDATE - TO_DATE('01/JANUARY/2019')) FROM DUAL;
D. SELECT TO_DATE(SYSDATE, 'DD/MONTH/YYYY') - '01/JANUARY/2019' FROM DUAL;
E. SELECT ROUND(SYSDATE - '01-JAN-2019') FROM DUAL;
Answer: A,C

NEW QUESTION: 2
You are developing an application. The application converts a Location object to a string by using a
method named WriteObject. The WriteObject() method accepts two parameters, a Location object and an
XmlObjectSerializer object.
The application includes the following code. (Line numbers are included for reference only.)

You need to serialize the Location object as a JSON object. Which code segment should you insert at line 20?
A. New XmlSerializer(typeof(Location))
B. New NetBataContractSenalizer {}
C. New DataContractSerializer(typeof(Location))
D. New CataConcractJsonSerializer(typeof(Location))
Answer: D

NEW QUESTION: 3
暗号化アルゴリズムでは、鍵長はどのような情報を提供しますか?
A. 順列の数
B. パケットサイズ
C. 暗号ブロックサイズ
D. ハッシュブロックサイズ
Answer: A
Explanation:
In cryptography, an algorithm's key space refers to the set of all possible permutations of a keys.
If a key were eight bits (one byte) long, the keyspace would consist of 28 or 256 possible keys. Advanced Encryption Standard (AES) can use a symmetric key of 256 bits, resulting in a key space containing 2256 (or 1.1579*1077) possible keys.
Source: https://en.wikipedia.org/wiki/Key_space_(cryptography)

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