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NEW QUESTION: 1
You have a server named Server1 that runs Windows Server 2012. You promote Server1 to domain controller. You need to view the service location (SVR) records that Server1 registers on DNS.
What should you do on Server1?
A. Open the Srv.sys file
B. Run ipconfig/displaydns
C. Run Get-DnsServerDiagnostics
D. Open the Netlogon.dns file
Answer: D
Explanation:
A. Timestamp server driver
B. Netlogon service creates a log file that contains all the locator resource records stored in netlogon.
C. used to display current resolver cache content
D. Gets DNS event logging details http://technet.microsoft.com/en-us/library/cc959303.aspx http://technet.microsoft.com/en-us/library/jj649883(v=wps.620).aspx

NEW QUESTION: 2
Which two qualities of PKI key exchange overcome asymmetric cryptography scalability issues? (Choose two.)
A. the trusted introducer uses the signed certificates of the endpoints that need to communicate
B. only the public key of the introducer has to be initially known and verified by all other entities
C. the introducer digitally signs the public key of the user with the public key of the introducer to generate a signed certificate
D. pki uses only a single trusted introducer
Answer: B,D
Explanation:
Explanation/Reference:
Chapter 14: Understanding Cryptographic Fundamentals and PKI pg. 374 A PKI does not eliminate the need for authenticity when you exchange public keys in an asymmetric encryption environment, but it solves the scalability issues associated with that process.
A PKI uses the concept of a single, trusted introducer. Instead of securely exchanging all public keys among all devices, only the public key of the trusted introducer has to be securely distributed to all devices.
This exchange is usually done by downloading the public key and then verifying it out of band.

NEW QUESTION: 3
Given:

Which expression when added at line 1 will produce the output of 1.17?
A. float z = Math.round((float)x/y*100)/(float)100;
B. float z = (float)(Math.round((float)x/y*100)/100);
C. float z = Math.round((int)(x/y),2);
D. float z = Math.round((float)x/y,2);
Answer: A
Explanation: