Is the expression X space logical or space (Y space logical and space Z )equivalent to (X space logical or space Y )space logica
l and space (X space logical or space Z )for all possible inputs of X, Y, and Z
1 answer:
Answer:
To prove:
X+Y.Z=(X+Y).(X+Z)
Taking R.H.S
= (X+Y).(X+Z)
By distributive law
= X.X+X.Z+X.Y+Y.Z --- (1)
From Boolean algebra
X.X = X
X.Y+X.Z = X.(Y+Z)
Using these in (1)
=X+X(Y+Z)+Y.Z
=X(1+(Y+Z)+Y.Z --- (2)
As we know (1+X) = 1
Then (2) becomes
=X.1+Y.Z
=X+Y.Z
Which is equal to R.H.S
Hence proved,
X+Y.Z=(X+Y).(X+Z)
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Answer:
Slope=1
Step-by-step explanation:
Slope=y1-y2/x1-x2
Where x1,y1= (1, 1) and X2,y2=(-3, -3)
Slop=1+3/1+4
=4/4
=1
So the slope is 1.