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anonymous
 4 years ago
Rewrite the expression xy+z' in terms of bar and the NAND operation
anonymous
 4 years ago
Rewrite the expression xy+z' in terms of bar and the NAND operation

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1334734679026:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is the NAND operation for the que above...please someone help

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0not sure what "bar" represents is this boolean algebra ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you rewrite it again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0One way could be, to draw it out, convert into NAND equivalents and form the equations.

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0im stuck of course you could be fancy and negate the NAND resulting in original AND operation > xy +z' = [(xy)']' +z'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got xy(bar).z(whole bar), sorry i dont know how to put that in latex.

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0arctic is that NAND(x,y) and z' ? it wont work for all values of 1 x=y=z=1 yields xy+z' = 1 (xy)'z' = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0noh dw:1334738031932:dw

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0ahh i see, yes that would work :) good job

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@anima Do you know what are NAND equivalents for NOT, AND and OR gates?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you dont, check this http://en.wikipedia.org/wiki/NAND_logic Then substitute all the NAND forms given to you, then form the expression again. You will yield this.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Another simpler way would be using, Boolean algebra and De Morgan's laws.
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