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NEW QUESTION: 1
質問をドラッグアンドドロップ
IPv4ネットワークサブネットを左側から右側の正しい使用可能なホスト範囲にドラッグアンドドロップします
Answer:
Explanation:
Explanation:
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet 172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111
1110 in 3rd octet) -> The 3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is
172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or
172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet.
Therefore the last usable host is 172.28.229.254.
NEW QUESTION: 2
A calculation view includes a rank node that uses the source data and settings shown in the graphic. Your query requests all columns of the rank node. How many rows are in the output of your query?
A. 0
B. 1
C. 2
D. 3
Answer: A
NEW QUESTION: 3
The implementation group has been using the test bed to do an IPv6 'proof-of-concept1.
After several changes to the network addressing and routing schemes, a trouble ticket has been opened indicating that the loopback address on R1 (2026::111:1) is not able to ping the loopback address on DSW2 (2026::102:1).
Use the supported commands to isolate the cause of this fault and answer the following question.
What is the solution to the fault condition?
A. Under the interface Tunnel34 configuration enter the ip address unnumbered Serial0/0/0.34 command.
B. Under the interface Tunnel34 configuration delete the tunnel mode ipv6 command.
C. Under the interface Tunnel34 configuration delete the tunnel source Serial0/0/0.34 command and enter the tunnel source 2026::34:1/122 command.
D. Under the interface Serial0/0/0.34 configuration enter the ipv6 address 2026::34:1/122 command.
Answer: B
Explanation:
As explained earlier, the problem is with route misconfigured tunnel modes on R3. R3 is using tunnel mode ipv6, while R4 is using the default of GRE. We need to remove the
"tunnel mode ipv6" command under interface Tunnel34
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