The options are missing from the question.
Below are the options.
A) 192.168.15.0
B) 192.168.15.16
C) 192.168.15.32
D) 192.168.15.64
Answer: The correct option to the question is option D
192.168.15.64
Explanation:
The Network is: 192.168.15.64/26 11000000.10101000.00001111.01000000
Then the Broadcast is: 192.168.15.127 11000000.10101000.00001111.01111111
We see the HostMin as: 192.168.15.65 11000000.10101000.00001111.01000001
And the HostMax as: 192.168.15.126 11000000.10101000.00001111.01111110
Answer:
All flags are On ( c, z , N )
Explanation:
Given data:
4-bit operation
Assuming 2's complement representation
<u>Determine status flags that are on after performing </u> 1010+0110
1 1
1 0 1 0
0 1 1 0
1 0 0 0 0
we will carry bit = 1 over
hence C = 1
given that: carry in = carry out there will be zero ( 0 ) overflow
hence V = 0
also Z = 1
But the most significant bit is N = 1
Answer:
Explanation:
develloppemetn de grace en option est logiquement une biologiste experimenter des annes 1256 a toulouse en france
Answer:
The buffer has room for 499 characters (you always have to reserve 1 for the terminating \0 character).
The code copies all characters passed in the commandline (argv[1]) into this buffer without checking. If there are more than 499 characters supplied, these will be copied into a memory location that is not intended for it. This will probably cause your program to crash, but if the extra data is somehow executed by the processor as if it were a program, this could be a way to sneak a virus into your computer.
So, while copying data, it is important to always limit the maximum amount to the allocated space.
