Answer:
The name of a .java file should <u>always match the name of the class inside.</u>
Explanation:
In Java programming the program written in code editor is first saved with .java extension. The name of this .java file should be same as that of the class declared inside the file.
This .java file is then compiled and converted to .class file which contains the java bytecode. This bytecode can then be executed by java virtual machine(JVM).
However it is not always necessary that name of .java files should be same as that of class inside it. The name should be same only when the class inside is declared as public.
In case it is not declared as public one can name .java file different than the actual class name.
Answer:A) Syslog
Explanation: Syslog is the log for the messaging in the computing field.It acts as a separator for the different task that is related to messaging. The task usually are storing of message, production of the message through software,analyzing message, reporting it etc.
Other options given in the question such as WORM storage is for the storage technology,UTM is for the management for the threat situation and firewall logging is related with log/tables for firewall.
Thus, the correct option is option (A).
the answer is E. ............
Any image that helps you, the reader, understand the text that the visual aid is accompanied with is referred to as a visual graphic or graphic aid.
Too frequently, readers lazily scan or entirely ignore graphs, diagrams, charts, and tables. Grid graphs, tables, bar charts, flow charts, maps, pie diagrams, and drawings and sketches are the most popular. Relationships are displayed using grid graphs. A visual aid should always be used in conjunction with preparation to interest the audience, improve their comprehension of your message, elicit an emotional response, and assist you in communicating it effectively. Charts, diagrams, graphs, maps, flashcards, posters, images, photos, booklets, folders, pamphlets, cartoons, and comics are examples of graphic aids.
Learn more about graphic here-
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Answer:
user_age = int(input())
if user_age > 17 and user_age != 25:
print("Eligible")
else:
print("Not eligible")
Explanation: