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Dec 7 2015 03:29pm
I need help with a bunch of questions from a take home exam, help would be greatly appreciated it. Anyone who can walk me through these problems I can pay fg (: I will post photos of the exam in a few
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Dec 7 2015 06:14pm
wheres da efgeez?
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Dec 7 2015 07:59pm
will post photos of exam if anyone is interested in helping me out, its 10 questions can pay 300.00 fg by dec 9th when my loan is returned.

would like to be walked through each question

This post was edited by Llove on Dec 7 2015 07:59pm
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Dec 7 2015 08:00pm
are you allowed external help for your final exam?

seems odd your grade is only worth 300 fg
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Dec 7 2015 08:02pm
Quote (carteblanche @ Dec 7 2015 09:00pm)
are you allowed external help for your final exam?

seems odd your grade is only worth 300 fg


its a take home exam not a final exam. im just super busy with other school work, i just need a C in the class for preq via my schools BIO degree. if i cared about the class, i would put the time in.
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Dec 7 2015 11:18pm
Anyone ?
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Dec 8 2015 07:38am
);:
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Dec 8 2015 11:07am
1 is invalid. Counter example: c & !d
You can make the truth table yourself, just write out c d e and then under each letter go through all the combinations of true and false. Like this:

c d e
f f f
f f t
f t f
f t t

2 is invalid: !a & y


3 is invalid.

!s | a -> g
!s
therefore g
!g = m | p
since g, therefore !m & !p
Thus !p
You actually have to make a truth table per her instructions, but you'll get the same answer.

4 is valid
!(k | r)
!k & !r
(d -> a) -> r is the same as !r -> !(d -> a)
since !r therefore !(d -> a)
d & !a
!a -> b
therefore b

So we get: !r, !k, d, !a, b
b = !r, valid

This post was edited by russian on Dec 8 2015 11:23am
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Dec 8 2015 11:39am
5 is valid. You have to pick !c, in which case c -> w is always true.
If you pick c you get:
c
!d
!(c -> d)
since b -> (c -> d) and !(c -> d) therefore !b
since (!p -> r) = b and !b therefore !(!p -> r) therefore !p & !r
but we have p | r, which contradicts !p & !r
Therefore !c
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