Answer:
A. 255.255.255.192
B. 62 address
C. First 130.56.0.1 and last 130.56.0.62
D. First 130.56.255.193 and last 130.56.255.254
Explanation:
A. We have a class b address, we know that because end in 130.56.0.0/16
We want 1024 subsets this is equal 2^10 = 1024.
Then we sum both decimal number 10 + 16 = 26, we can represent in subnet mask like 255.255.255.192.
B. We already have used 26 bits, in total 32, we must use the rest of the bits for the address
32 - 26 = 6
2^6 = 64, but we use two per subnet cannot be allocated and subnet mask.
We have in total 62 address.
C. We have 62 address for logic the last address is 130.56.0.62, and the first one is 130.56.0.1, because we cannot use the 130.56.0.0
D. We could represent these address from this mask 255.255.255.192 where First address is 130.56.255.193 and the last 130.56.255.254.
