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NEW QUESTION: 1
The forwarding table on the leaf switch is divided between local and global entries. Which two options are contained in the global station table? (Choose two.)
A. Default router to the spine proxy
B. All hosts attached to the fabric
C. Outside routes
D. Local cache of fabric endpoints (not directly attached)
E. Directly attached endpoints
Answer: A,D
Explanation:
http://www.cisco.com/c/en/us/solutions/collateral/data-centervirtualization/application-centric-infrastructure/white-paper-c07-732033.html

NEW QUESTION: 2
Your network contains a single Active Directory domain. All servers run Windows Server 2008 R2. You deploy a new server that runs Windows Server 2008 R2. The server is not connected to the internal network.
You need to ensure that the new server is already joined to the domain when it first connects to the internal network.
What should you do?
A. From a domain-joined computer, run djoin.exe and specify the /requestodj parameter. From the new server, run djoin.exe and specify the /provision parameter.
B. From a domain-joined computer, run djoin.exe and specify the /provision parameter. From the new server, run djoin.exe and specify the /requestodj parameter.
C. From a domain controller, run sysprep.exe and specify the /generalize parameter. From the new server, run sysprep.exe and specify the /oobe parameter.
D. From a domain controller, run sysprep.exe and specify the /oobe parameter. From the new server, run sysprep.exe and specify the /generalize parameter.
Answer: B

NEW QUESTION: 3
What happens when you attempt to compile and run the following code?
# include <deque>
# include <iostream>
# include <algorithm>
using namespace std;
template<class T>struct Out {
ostream & out;
Out(ostream & o): out(o){}
void operator() (const T & val ) { out<<val<<" "; } };
struct Sequence { int start;
Sequence(int start):start(start){}
int operator()() {return 10*(1+(start++ %3));}
};
int main() {
deque<int> d1(10);
generate(d1.begin(), d1.end(), Sequence(1));
sort(d1.begin(), d1.end());
pair<deque<int>::iterator, deque<int>::iterator > result = equal_range(d1.begin(), d1.end(),
20);
for_each(result.first, result.second, Out<int>(cout));cout<<endl;
return 0;
}
Program outputs:
A. 20 20 20 20
B. 10 20 20 20 20
C. 10 10 10 20 20 20 20 30 30 30
D. 20 20 20 20 30
E. 10 20 20 20 20 30
Answer: A