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NEW QUESTION: 1
展示を参照してください。

エンジニアは、R1がDHCPからIP構成を取得するトポロジを展開します。スイッチとDHCPサーバーの構成が完全で正しい場合。タスクを完了するために、R1とR2で構成する必要があるコマンドの2つのセットはどれですか。 （2つ選択してください）
A. R1(config)# interface fa0/0
R1(config-if)#ip helper-address 192.0.2.2
B. R2(config)# interface gi0/0
R2(config-if)#ip address dhcp
C. R1(config)# interface fa0/0
R1(config-if)#ip helper-address 198.51.100.100
D. R1(config)# interface fa0/0
R1(config-if)#ip address dhcp
R1(config-if)# no shutsown
E. R2(config)# interface gi0/0
R2(config-if)#ip helper-address 198.51.100.100
Answer: D,E

NEW QUESTION: 2
複数のデータベースを持つMicrosoft SQL Server 2014インスタンスを管理します。 2ノードのSQL Serverフェールオーバークラスターがあります。クラスターはストレージエリアネットワーク（SAN）を使用します。あなたはI / Oの問題を発見しました。 SANは最大容量であり、追加のディスクを追加することはできません。
最小限のコストでSAN上のI / Oワークロードを削減する必要があります。
あなたは何をするべきか？
A. tempdbファイルをローカルディスクに移動します
B. tempdbのデータファイルとログファイルを展開します。
C. テーブル変数を使用するようにアプリケーションコードを修正する
D. ユーザーデータベースをローカルディスクに移動します。
Answer: A
Explanation:
Explanation
The use of local disks for TempDB allows us to have more flexibility when configuring for optimal performance. It is a common performance recommendation to create the TempDB database on the fastest storage available. With the capability to utilize local disk for TempDB placement we can easily utilize disks that are larger, have a higher rotational speed or use SSD disks.
References:https://www.mssqltips.com/sqlservertip/2817/sql-server-2012-cluster-with-tempdb-on-local-disk/

NEW QUESTION: 3
Which changes introduced independently will allow the code to compile and display 0 0 1 1 8 8 9 9 (choose all that apply)?
#include <iostream>
#include <set>
#include <vector>
using namespace std;
class A {
int a;
public:
A(int a):a(a){}
int getA() const { return a;}
/* Insert Code Here 1 */
};
/* Insert Code Here 2*/
int main(){
A t[] ={ 3, 4, 2, 1, 6, 5, 7, 9, 8, 0 };
set<A>s(t, t+10);/* Replace Code Here 3 */
multiset<A> s1(s.begin(),s.end());/* Replace Code Here 4 */
s1.insert(s.begin(),s.end());
s1.erase(s1.lower_bound(2),s1.upper_bound(7));
multiset<A>::iterator i=s1.begin();/* Replace Code Here 5 */
for( ;i!= s1.end(); i++)
{
cout<<i?>getA()<<" ";
}
cout<<endl;
return 0;
}
A. bool operator < (const A & b) const { return a<b.a;} inserted at Place 1
B. bool operator < (const A & b) const { return b.a<a;} inserted at Place 1
C. struct R { bool operator ()(const A & a, const A & b) { return a.getA()<b.getA();} }; inserted at Place 2 replacing line marked 3 with set<A, R>s(t, t+10); replacing line marked 4 with multiset<A,R> s1(s.begin(),s.end()); replacing line marked 5 with multiset<A,R>::iterator i=s1.begin();
D. operator int() const { return a;} inserted at Place 1
Answer: A,C,D