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NEW QUESTION: 1
あなたは、C#を使用してアプリケーションを開発しています。
アプリケーションが長時間実行される処理を行うオブジェクトが含まれています。
あなたは、プロセスが完了するまで、ガベージコレクタがオブジェクトのリソースを解放しないようにする必要があります。
あなたはどのごみ収集人方法を使うべきですか?
A. WaitForPendingFinalizers()
B. WaitForFullGCApproach()
C. SuppressFinalize()
D. WaitForFullGCComplete()
Answer: C
Explanation:
You can use the SuppressFinalize method in a resource class to prevent a redundant garbage collection from being called.
Reference:
https://msdn.microsoft.com/en-us/library/system.gc.suppressfinalize(v=vs.110).aspx

NEW QUESTION: 2
A Layer 2 device that connects multiple computers within a network is a:
A. Switch
B. Repeater
C. Packet
D. Router
Answer: A
Explanation:
Explanation/Reference:
Layer 2 switching uses the media access control address (MAC address) from the host's network interface cards (NICs) to decide where to forward frames.

NEW QUESTION: 3
Die Unternehmenszentrale stellte Ihrem Büro einen Teil des Subnetzes der Klasse B zur Verfügung, das an einem neuen Bürostandort verwendet werden kann.
Gegebener Bereich: 172.30.232.0/24
* Verkauf von 57 Geräten
* HR 23 Geräte
* IT 12 Geräte
* 32 Geräte finanzieren
* Marketing 9 Geräte
ANLEITUNG
Weisen Sie die Mindestanzahl von Adressen (unter Verwendung der CIDR-Notation) zu, die für jede Abteilung erforderlich sind.
Identifizieren Sie nach der Unterbringung jeder Abteilung den nicht verwendeten Teil des Subnetzes, indem Sie auf die Frage in der Grafik antworten.
Alle Dropdowns müssen ausgefüllt sein.
Wenn Sie zu irgendeinem Zeitpunkt den Ausgangszustand der Simulation wiederherstellen möchten, klicken Sie auf die Schaltfläche Alle zurücksetzen.


Answer:
Explanation:

Explanation:
Sales network - /26 - This will provide up to 62 usable IP addresses (64-2 for subnet and broadcast IP) HR network - /27 - This will provide for up to 30usable IP's (32-2) IT - /28 - This will provide for up to 14 usable IP's (16-2) Finance - /26 - Note that a /27 is 32 IP addresses but 2 of those are reserved for the network and broadcast IP's and can't be used for hosts.
Marketing - /28
If we add up how many IP blocks are used that is 64+32+16+64+16=192.
A /24 contains 256 IP addresses, so 256-192=64.
So the last unused box should be a /26, which equates to 64 addresses

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