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 11 months ago
simplify the following using Boolean algebra.
(xy)'+xz'+yz
a)y+z'
b)x+y
c)X'+y
d)xz'
 11 months ago
simplify the following using Boolean algebra. (xy)'+xz'+yz a)y+z' b)x+y c)X'+y d)xz'

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ybarrap
 11 months ago
Best ResponseYou've already chosen the best response.1$$ \Large{ (xy)'+xz'+yz\\ =[(xy)(xz')'(yz)']'\\ =[(xy)(x'+z)(y'+z')]'\\ =[((xyx')+(xyz))(y'+z')]'\\ =[0+xyzy'+xyzz']'\\ =[0+0+0]'\\ =[0]'\\ =1 } $$ This is a tautology, because it is always true. Here is another method using a Truth Table dw:1395590669537:dw This also shows a tautology

ybarrap
 11 months ago
Best ResponseYou've already chosen the best response.1Does this make sense?

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.1(xy)' + xz' + yz We know that (xy)' = x' + y' [De morgan's laws] x' + y' + xz' + yz x' + z' + y' + z x' + y' + 1 = 1 None of the options are as 1 @ybarrap Is truth table the only method to find that out?

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.1@moli1993 I guess you have to see which of the options also show Tautology [Always true].

ybarrap
 11 months ago
Best ResponseYou've already chosen the best response.1The truth table is authoritative. This is a tautology. None of the options are a tautology so they are all incorrect.

moli1993
 11 months ago
Best ResponseYou've already chosen the best response.1I am so sorry , wrote the wrong question here : it is (xz)'+xz'+yz and the options are same as given earlier !

ybarrap
 11 months ago
Best ResponseYou've already chosen the best response.1It's the same process

moli1993
 11 months ago
Best ResponseYou've already chosen the best response.1process is same but it's not a tautology i suppose !

dumbcow
 11 months ago
Best ResponseYou've already chosen the best response.0(xz)' = x' + z' z' +xz' = z'(1+x) = z' > x' + z' + yz = x' +z' +z = x' + 1 = 1

moli1993
 11 months ago
Best ResponseYou've already chosen the best response.1how x' + z' + yz = x' +z' +z

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.1(xz)'+xz'+yz We know that (xz)' = x' + z' [De morgan's laws] x' + z' + xz' + yz x' + (z' + xz') + yz [ In z' + xz' take z' common, z'(1+x) = z'(1) = z' ] x' + z' + yz [yz = zy] z' + zy + x' Using simplification law [A + A'B = A+B] we get, z' + y + x' Final answer should be x' + y + z'

moli1993
 11 months ago
Best ResponseYou've already chosen the best response.1but it doesnt matches any of the options given above

AkashdeepDeb
 11 months ago
Best ResponseYou've already chosen the best response.1(xz)' + y I am pretty sure, this is the final answer. :/ @ybarrap @dumbcow ?

moli1993
 11 months ago
Best ResponseYou've already chosen the best response.1me too I think u r correct !

dumbcow
 11 months ago
Best ResponseYou've already chosen the best response.0yep http://www.wolframalpha.com/input/?i=~%28x+and+z%29+or+%28x+and+~z%29+or+%28y+and+z%29

moli1993
 11 months ago
Best ResponseYou've already chosen the best response.1Thank you all for helping me !
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