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  • Exam Number/Code : FCSS_NST_SE-7.4
  • Exam Name : FCSS - Network Security 7.4 Support Engineer
  • Questions and Answers : 260 Q&As
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NEW QUESTION: 1
ネットワークエンジニアは、このタスクを達成するために、Ciscoネットワーク環境ソリューションを管理および構成する必要がありますか?
A. 構成を各IPSにプッシュするcisco適応型セキュリティデバイスマネージャー
B. SIGHTマネージャーを起動して、インストール済みのIPSユニットに設定をバンドルおよびプッシュします
C. Cisco IPSは、エクスプレスおよびプッシュ構成をipsユニットに管理します
D. cisco security 4.5以降および構成バンドルを各,,,,,にプッシュする
Answer: B

NEW QUESTION: 2

A. Option B
B. Option D
C. Option E
D. Option A
E. Option C
Answer: D,E

NEW QUESTION: 3
Given:

What is the result?
A. false
\sales\quarter\ . . \qtrlreport.txt
B. true
.. \ . . \ . . \ annualreport.txt
C. true
\. . \ . . \annualreport.txt
D. false
\quarter\ . . \qtrlreport.txt
Answer: A
Explanation:
Explanation/Reference:
Explanation:
( richtig !! import java.nio.file.Path; import java.nio.file.Paths; ) original-Aufgabe war ( falsch >>
import java.io.file.Path; import java.io.file.Paths; )
The relativize method that can be used to construct a relative path between two paths.
relativize
Path relativize(Path other)
Constructs a relative path between this path and a given path.
Parameters: other - the path to relativize against this path
Returns: the resulting relative path, or an empty path if both paths are equal
Note:
Relativization is the inverse of resolution. This method attempts to construct a relative path that when
resolved against this path, yields a path that locates the same file as the given path. For 18 example, on
UNIX, if this path is "/a/b" and the given path is "/a/b/c/d" then the resulting relative path would be "c/d".
Where this path and the given path do not have a root component, then a relative path can be constructed.
A relative path cannot be constructed if only one of the paths have a root component. Where both paths
have a root component then it is implementation dependent if a relative path can be constructed. If this
path and the given path are equal then an empty path is returned.
For any two normalized paths p and q, where q does not have a root component, p.relativize(p.resolve
(q)).equals(q)
When symbolic links are supported, then whether the resulting path, when resolved against this path,
yields a path that can be used to locate the same file as other is implementation dependent. For example,
if this path is "/a/b" and the given path is "/a/x" then the resulting relative path may be "../x". If "b" is a
symbolic link then is implementation dependent if "a/b/../x" would locate the same file as "/a/x".

 

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