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NEW QUESTION: 1
-= is a _________________ type of operator.
A. Unary
B. Arithmetic
C. Assignment
D. Logical
Answer: C
NEW QUESTION: 2
A network has four routers running IS-IS with the NET addresses below:
R1 - 49.1234.1111.1111.1111.00
R2 - 49.1234.2222.2222.2222.00
R3 - 49.4321.1111.1111.1111.00
R4 - 49.4321.2222.2222.2222.00
How can IS-IS authentication be enabled between all routers?
A. domain authentication with the command(config-rtr)#domain-passwordword
B. area authentication with the command (config-if)#isisarea-passwordword
C. interface authentication with the command (config-if)#isis passwordword
D. domain authentication with the command (config-if)#isis domain-passwordword
Answer: A
NEW QUESTION: 3
Your storage administrator informs you about some disk corruption issues in the disk subsystem where your Oracle Cluster Registry (OCR) is located. Your storage administrator has resolved these corruption issues by applying a patch to the storage subsystem, but to be sure that the OCR is not affected by this problem the administrator wants you to check the integrity of the OCR. How would you do this?
A. You dump the OCR contents into a text file by using the OCRDUMP command, and if that succeeds without errors you know that your OCR is not corrupted.
B. You create a physical backup of the OCR by using the OCRCONFIG command, and if this succeeds without errors you know that the OCR is not corrupted.
C. You create a logical export from the OCR by using the OCRCONFIG command and analyze the export file for errors.
D. There is no easy way to check your OCR for corruptions.
E. You can check whether the OCR is not corrupted by using the OCRCHECK command.
Answer: E
NEW QUESTION: 4
A. Option D
B. Option C
C. Option E
D. Option B
E. Option A
Answer: A
Explanation:
The IP address 192.168.1.17 255.255.255.0 specifies that the address is part of the 192.168.1.0/24 subnet
24 mask bits = 255.255.255.0 28 mask bits = 255.255.255.240 192.168.1.0/24 subnet has a host range of 192.168.1.1 to 192.168.1.254 (0 being network and 255 being broadcoast) 192.168.1.17/28 subnet has a host range of 192.168.1.17 to 192.168.1.30 (16 being network and 31 being broadcast) 192.168.1.65/28 subnet has a host range of 192.168.1.65 - 192.168.1.78 (64 being network and 79 being broadcast)
if fa0/0 was left as /24, you can see that the host range includes the host range of 192.168.1.64/28 which conflicts.
Simply speaking, you can't overlap the subnets.
By changing the subnet mask of fa0/0 to 255.255.255.240, these networks would no longer overlap.