Existen 2 clases comunes:
sistema de lazo abierto: la salida no efecto sobre el sistema
sistema de lazo cerrado: la toma de desiciones del sistema de la entrada si no también de la salida
Answer:
The program is as follows:
<em>5 INPUT A,B</em>
<em>6 PROD = A * B</em>
<em>7 PRINT PROD</em>
<em>8 TOTAL = A + B</em>
<em>9 PRINT TOTAL</em>
<em>10 DIFF = A - B</em>
<em>11 PRINT DIFF</em>
<em>12 END</em>
Explanation:
This gets input for the two numbers
<em>5 INPUT A,B</em>
This calculates the product
<em>6 PROD = A * B</em>
This prints the calculated product
<em>7 PRINT PROD</em>
This calculates the sum
<em>8 TOTAL = A + B</em>
This prints the calculated sum
<em>9 PRINT TOTAL</em>
This calculates the difference
<em>10 DIFF = A - B</em>
This prints the calculated difference
<em>11 PRINT DIFF</em>
This ends the program
<em>12 END</em>
Answer:
first off you need to take of the screws make sure its unpluged and you open up to see the mother bored
Explanation:
Answer:
your computer will not allow it
Explanation:
because it is not one of the main dyonostics
Answer:
For question a, it simplifies. If you re-express it in boolean algebra, you get:
(a + b) + (!a + b)
= a + !a + b
= b
So you can simplify that circuit to just:
x = 1 if b = 1
(edit: or rather, x = b)
For question b, let's try it:
(!a!b)(!b + c)
= !a!b + !a!bc
= !a!b(1 + c)
= !a!b
So that one can be simplified to
a = 0 and b = 0
I have no good means of drawing them here, but hopefully the simplification helped!