From the 10.70.128.0/19 block, you need two subnets: one supporting 24 hosts and another 472 hosts, both with the longest masks possible and using the first usable address for the router. Which two interface configurations meet these requirements?

The original block is 10.70.128.0/19, which means the network address is 10.70.128.0 and the broadcast is 10.70.159.255. For 24 hosts, a /27 subnet (255.255.255.224) is needed, providing 30 usable addresses. For 472 hosts, a /23 subnet (255.255.254.0) is needed, providing 510 usable addresses. To find the correct subnets, we perform subnetting. A possible allocation is: 1. The /23 subnet (for 472 hosts) could be 10.70.148.0/23. This block ranges from 10.70.148.0 to 10.70.149.255. The first usable IP is 10.70.148.1. This matches interface vlan 1148 ip address 10.70.148.1 255.255.254.0. 2. The /27 subnet (for 24 hosts) could be 10.70.155.64/27. This block ranges from 10.70.155.64 to 10.70.155.95. The first usable IP is 10.70.155.65. This matches interface vlan 155 ip address 10.70.155.65 255.255.255.224. Other options are incorrect because their IP addresses are not the first usable in a valid subnet of the required size, or their subnet masks do not match the host requirements. For example, 10.70.147.17 is not a valid first usable IP for a /27, and 10.70.159.1/23 would overlap with the 10.70.148.0/