A. /lib
B. /var
C. /home
D. /opt
E. /etc
Answer: C
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NEW QUESTION: 1
Which command shows the connection table in human readable format?
A. fw tab -t connections -h
B. fw tab -t connections -u
C. fw tab -t connections -f
D. fw tab -t connections -s
Answer: C
NEW QUESTION: 2
A. /lib
B. /var
C. /home
D. /opt
E. /etc
Answer: C
NEW QUESTION: 3
Which two IPv6 address pairs can be assigned on a point-to-point link with a /125 mask? (Choose two.)
A. 2001:DB8::1 and 2001:DB8::2
B. 2001:DB8::10 and 2001:DB8::17
C. 2001:DB8::9 and 2001:DB8::10
D. 2001:DB8::1 and 2001:DB8::18
Answer: A,B
Explanation:
2001:DB8::x/125 = 2001:0DB8:0000:0000:0000:0000:0000:00xx
1- The 0's disapear because the 3 zero ommiting rules: Leading zeros can be omitted Groups of four zeros can be abbreviated to a zero Multiple contiguous groups (2-byte groups) can be abbreviated to double colon (::) but only once. 2- IPv6 is made by 128bits, therefore 128-125 = 3 This means that the last 3 bits represent the hosts, just like in IPv4 3- IPv6 is represented in Hex numbers: Ex: 2001 = 00100001 4- Then you can determine the answer just like in IPv4 (or at least in a very similar way) 2001:DB8::1 and 2001:DB8::2 belong to the /125 mask because of the 3 last bits 2001:DB8::9 and 2001:DB8::10 are obviously not in the same /125 subnet (remember 10 in hex is much different thatn 10 in decimal) 2001:DB8::10 and 2001:DB8::17 fall into the /125 subnet 2001:DB8::1 and 2001:DB8::18 it goes beyond the /125 subnet 5- In case you didn't understand why: 2001:DB8::1 - 2001:DB8::9 = First Subnet's Host 2001:DB8::10 - 2001:DB8::18 = Second Subnet's Hosts