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NEW QUESTION: 1
ストレージ設計を作成するときに考慮する必要があるトピックは何ですか？
A. Application I/O requirements
B. All of the above
C. Latency
D. Growth rate
Answer: B
NEW QUESTION: 2
You have a server named Server1 that runs Windows Server 2016. Server1 has the Windows Application Proxy role service installed.
You need to publish Microsoft Exchange ActiveSync services by using the Publish New Application Wizard. The ActiveSync services must use preauthentication.
How should you configure Server1? To answer, select the appropriate options in the answer area.
Answer:
Explanation:
NEW QUESTION: 3
You have a Hyper-V host named Server1 that runs Windows Server 2012 R2.
Server1 hosts a virtual machine named VM1 that runs Windows Server 2012 R2.
VM1 has several snapshots.
You need to modify the snapshot file location of VM1.
What should you do?
A. Shut down VM1, and then modify the settings of VM1.
B. Right-click VM1, and then click Export...
C. Delete the existing snapshots, and then modify the settings of VM1.
D. Pause VM1, and then modify the settings of VM1.
Answer: C
Explanation:
you will need to navigate to the Hyper-V Management snap-in
(C:\ProgramData\Microsoft\Windows\Hyper-V) and from there access the Snapshot file Location tab where you can
change the settings for the VM1 snapshot file location. However, since there are already several snapshots in
existence, you will need to delete them first because you will not be able to change the location of the snapshot file
while there is an existing snapshot and you need to modify the snapshot file location of VM1.
NEW QUESTION: 4
What will happen when you attempt to compile and run the following code?
# include <iostream>
# include <set>
# include <vector>
using namespace std;
int main(){
int t[] ={ 3, 4, 2, 1, 6, 5, 7, 9, 8, 0 };
vector<int>v(t, t+10);
multiset<int> s1(v.begin(),v.end());
multiset<int, greater<int> > s2(v.begin(), v.end());
for(multiset<int, greater<int> >::iterator i=s2.begin();i!= s2.end(); i++) { cout<<*i<<" ";
}
for(multiset<int>::iterator i=s1.begin();i!= s1.end(); i++) {
cout<<*i<<" ";
}
cout<<endl;
return 0;
}
The output will be:
A. 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0
B. 0 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 0
C. 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9
D. 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
Answer: C
