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NEW QUESTION: 1
Your network consists of a single Active Directory domain. The functional level of the forest is Windows Server 2008 R2.
You need to create multiple password policies for users in your domain.
What should you do?
A. From the Schema snap-in, create multiple class schema objects.
B. From the Security Configuration Wizard, create multiple security policies.
C. From the ADSI Edit snap-in, create multiple Password Setting objects.
D. From the Group Policy Management snap-in, create multiple Group Policy objects.
Answer: C

NEW QUESTION: 2
モノの内部（IOT）ソリューションをサポートするデータベースを管理します。 データベースには、毎分1億を超えるデバイスからのメトリックが記録されています。 データベースには99.995のアップタイムが必要です。
データベースは、サイズが100ギガバイト（GB）のcheckinsという名前のテーブルを使用します。 チェックインテーブルには、デバイスからのメトリックが格納されています。 データベースには、4 TBのデータを格納するArchiveという名前のテーブルもあります。 テーブルへのすべてのアクセスにストアプロデュースを使用します。
待機タイプPAGELATCH_が大量のブロッキングを引き起こすことがわかります。
データベースのダウンタイムを最小限に抑えながら、ブロッキングの問題を解決する必要があります。
どの2つのアクションを実行しますか？ それぞれの正しい答えは解決策の一部を表しています。
A. チェックインテーブルをメモリ内のOLTPテーブルに変換します。
B. チェックインテーブルをクラスタclounmstoreインデックスに変換します。
C. checkinsテーブルにアクセスするすべてのストアドプロシージャをネイティブにコンパイルするプロシージャに変換します。
D. すべてのテーブルをクラスタ化カラムストアインデックスに変換します。
Answer: C

NEW QUESTION: 3
A customer wants to scan a serverless function as part of a build process.
Which twistcli command can be used to scan serverless functions?
A. twistcli function scan <SERVERLESS_FUNCT10N ZIP>
B. twistcli serverless scan <SERVERLESS_FUNCTION.ZIP>
C. twistcli scan serverless <SERVERLESS_FUNCTION Z1P>
D. twistcli serverless AWS <SERVERLESS_FUNCTION ZIP>
Answer: B

NEW QUESTION: 4
In an arithmetic series, the terms of the series are equally spread out. For example, in 1 + 5 + 9 + 13 + 17, consecutive terms are 4 apart. If the first term in an arithmetic series is 3, and the last term is 136, and the sum is 1,390, what are the first three terms?
A. 3, 23, 43
B. 3, 10, 17
C. 3, 139, 1251
D. 3, 36, 1/3, 70
E. 3, 69, ½, 136
Answer: B
Explanation:
Explanation/Reference:
Explanation:
If a is the 1st term and b is the common difference of an arithmetic series, then, the term of the series is a + (i − 1)b If the series has n terms, then the last term of the series is a + (n − 1)b The difference between the last term and the first term is
[a + (n − 1)b] − a = (n − 1)b = 136 − 3 = 133
Now the sum of the series is
Given, na + since (n − 1)b = 133 and a = 3
=1390, 139n = 2(1390), n = 20. Then using again (n − 1)b = 133
19b = 133, b = 7,
The first 3 terms are 3, 10, 17