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NEW QUESTION: 1
Your company has multiple DNS servers in the main office. You plan to install DNS on a member server in a branch office. You need to ensure that the DNS server in the branch office is able to query any DNS server in the main office, and you need to limit the number of DNS records that are transferred to the DNS server in the branch office.
What should you do?
A. Configure a stub zone on the DNS server in the branch office.
B. Configure a stub zone on the DNS server in the main office.
C. Configure a secondary zone on the DNS server in the branch office.
D. Configure a primary zone on the DNS server in the branch office.
Answer: A
Explanation:
A stub zone is a copy of a zone that contains only those resource records necessary to identify the authoritative Domain Name System (DNS) servers for that zone. A stub zone is used to resolve names between separate DNS namespaces. This type of resolution may be necessary when a corporate merger requires that the DNS servers for two separate DNS namespaces resolve names for clients in both namespaces.
http://technet.microsoft.com/en-us/library/cc779197%28v=ws.10%29.aspx

NEW QUESTION: 2
Refer to the exhibit

A business analyst needs to present the actual sales of the current year in a KPI on a dashboard. The expression in the Text field is NOT working Which two changes will correct the expression? (Select two )
A. Remove the $ sign before Year
B. Add single quotes around the whole expression
C. Add a = sign at the beginning of the expression
D. Change Representation from Text to Expression
E. Remove the $ sign between {and (
Answer: A,E

NEW QUESTION: 3
感度に基づいて資産を分類する最大の利点は何ですか？
A. 組織は、保護の適切なレベルを割り当てることができます。
Bデータを適宜割り当て資産の所有権に利用可能です。
B. スタッフは、より現実的なリスク・シナリオを作成することができます。
C. 組織は、規制ガイドラインに準拠しています。
Answer: A

NEW QUESTION: 4
Which code, inserted at line 15, generates the output "5 Bob"?
#include <iostream>
#include <string>
using namespace std;
class B;
class A {
int age;
public:
A () { age=5; };
friend void Print(A &ob, B &so);
};
class B {
string name;
public:
B () { name="Bob"; };
//insert code here
};
void Print(A &ob, B &so) {
cout<<ob.age << " " << so.name;
}
int main () {
A a;
B b;
Print(a,b);
return 0;
}
A. friend void Print(A *ob, B *so);
B. friend void Print(A &ob, B &so);
C. friend void Print(A ob, B so);
D. None of these
Answer: B