Answer:
Hi, for this exercise we have two laws to bear in mind:
Morgan's laws
NOT(А).NOT(В) = NOT(A) + NOT (B)
NOT(A) + NOT (B) = NOT(А).NOT(В)
And the table of the Nand
INPUT OUTPUT
A B A NAND B
0 0 1
0 1 1
1 0 1
1 1 0
Let's start!
a.
Input OUTPUT
A A A NAND A
1 1 0
0 0 1
b.
Input OUTPUT
A B (A NAND B ) NAND (A NAND B )
0 0 0
0 1 0
1 0 0
1 1 1
C.
Input OUTPUT
A B (A NAND A ) NAND (B NAND B )
0 0 0
0 1 1
1 0 1
1 1 1
Explanation:
In the first one, we only need one input in this case A and comparing with the truth table we have the not gate
In the second case, we have to negate the AND an as we know how to build a not, we only have to make a nand in the two inputs (A, B) and the make another nand with that output.
In the third case we have that the OR is A + B and we know in base of the morgan's law that:
A + B = NOT(NOT(А).NOT(В))
So, we have to negate the two inputs and after make nand with the two inputs negated.
I hope it's help you.
Answer:
open the header. hope this helps
<span>To the computer f, this answer is because when reading the statement I assume that there is no type of connection and / or communication between the computer a and c; therefore to be profitable the computer c should return the package sending it back to computer f.</span>
Maybe in atosave, Computers mostly save what your working on :3
