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NEW QUESTION: 1
A. Option A
B. Option C
C. Option D
D. Option B
Answer: D
NEW QUESTION: 2
The formation of glucose from non-carbohydrates sources such as lactate and Kreb's cycle metabolites is
an anabolic process called:
A. Gluconeogenesis
B. Glycogenesis
C. Glycogenolysis
D. None of the above
E. Glycolysis
Answer: A
NEW QUESTION: 3
Your network contains an Active Directory domain named contoso.com. All domain controllers run Windows Server 2012.
In a remote site, a support technician installs a server named DC10 that runs Windows Server 2012. DC10 is currently a member of a workgroup.
You plan to promote DC10 to a read-only domain controller (RODC).
You need to ensure that a user named Contoso\User1 can promote DC10 to a RODC in the contoso.com domain. The solution must minimize the number of permissions assigned to User1.
What should you do?
A. From Active Directory Administrative Center, modify the security settings of the Domain Controllers organizational unit (OU).
B. From Active Directory Administrative Center, pre-create an RODC computer account.
C. From Active Directory Users and Computers, run the Delegation of Control Wizard on the contoso.com domain object.
D. Join DC10 to the domain. Run dsmod and specify the /server switch.
Answer: B
Explanation:
A staged read only domain controller (RODC) installation works in two discrete phases:
1.Staging an unoccupied computer account
2.Attaching an RODC to that account during promotion
Reference: Install a Windows Server 2012 Active Directory Read-Only Domain Controller (RODC)
NEW QUESTION: 4
-- Exhibit -
-- Exhibit --
Refer to the exhibit. According to the packet capture, what statements are true? (Select 2)
A. This Steelhead appliance is configured with enhanced auto-discovery.
B. The SYN packet coming from 172.16.12.99 with a destination to 172.16.11.88 carries a probe of 10 bytes + a 4 bytes notification message.
C. This Steelhead appliance is configured with enhanced peering discovery.
D. The SYN packet coming from 172.16.12.99 with a destination to 172.16.11.88 carries a probe of 10 bytes.
Answer: A,B