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.
I wrote my code in python 3.8:
def max_magnitude(num1, num2):
return num1 if abs(num1) > abs(num2) else num2
print(max_magnitude(-8,-2))
I hope this helps!
Using the knowledge in computational language in C code it is possible to write a code that organizes and calculates the value of the matrix of A*A and that is in up to 4 decimal places.
<h3>Writing the code in C is possible:</h3>
<em>A=[1 2 2;3 4 5;6 7 8];</em>
<em>[u ,s ,v] = svd(A);</em>
<em>k = 1;</em>
<em>A1 = u(:,1:k)*s(1:k,1:k)*v(:,1:k)'; %'</em>
<em>RMSE = rms(sqrt(mean((A - A1).^2)))</em>
See more about C code at brainly.com/question/17544466
#SPJ1
~
<h3>Types of Computer games</h3>
- Action
- Adventure
- Simulation
- Sports
- Role-playing
- Puzzlers
- Party games
<em>(</em><em>That's</em><em> </em><em>all</em><em> </em><em>i</em><em> </em><em>know</em><em>)</em><em> </em>
<h2>
</h2>
~
Answer:
You need an email and a job and to be over 18 for business ones or a legal gaurdian if you have none then ur hecced uwu :333
