when you 2^10 you get 1024. Minus 32-10=22, hence this will be new subnet mask /22. How do you explain to get first 3 subnet and reserve when you have 11111111.11111111.11111100.00000000

1. Circle the different network segments found in the following figure. PC-PT PCO 192.168.2.2 262LXM 192.168.2.1 2960-2ETT Switch PC-PT PC1 192.168.2.3 192.168.1.1 PC-PT PC2 192.168.3.2 2621XM Router1 192.168.3.1 29:50-2ATT Switchi Subnet 0: AR 130.168.0.1 - 130.168.0.62 BC 130.168.0.63 Subnet 1: AR 130.168.0.65-130.168.0.126 BC 130.168.0.127 Subnet 2: AR 130.168.0.129 - 130.168.0.190 BC 130.168.0.191 PC-PT PC3 192.168.3.3 2. Given an IP address 130.168.0.0/16. Company XYZ required 1000 subnets, 2^10 (borrow 10 bits) New subnet mask /26 Subnet 0:000 Subnet 1: 001 Subnet 2: 010 a. Find the range of IP addresses and reserved addresses for the first 3 subnets. b. Determine the maximum number of hosts for each subnet. (2^6) - 2 = 62 c. Determine the new subnet mask after subnetting. /26 1111111.1111111.1111111.11 | 000000 255 255 255 192