If a figure consists of 4 sides that are all congruent, then it is a parallelogram. Figure ABCD is a parallelogram. Therefore, ABCD has 4 congruent sides.
Determine if the statement is valid.
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OpenStudy (anonymous):
A.No,p-> q , q:p is an invalid argument.
B.No,p -> q ,~p:~q is an invalid argument.
C.Yes,p->q is true, therefore q is true. It is a valid argument.
jimthompson5910 (jim_thompson5910):
you got what?
OpenStudy (anonymous):
i got C.
jimthompson5910 (jim_thompson5910):
so you're saying that if p->q is true, then so is q->p ?
OpenStudy (anonymous):
yes.
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OpenStudy (anonymous):
?
OpenStudy (anonymous):
@jim_thompson5910 Am i wrong?
jimthompson5910 (jim_thompson5910):
unfortunately you are, it's invalid to say that the converse is equivalent to the original conditional
OpenStudy (anonymous):
:/ Darn it.
jimthompson5910 (jim_thompson5910):
think about it: If I told you that "Figure ABCD is a parallelogram", does this mean that it has all 4 sides congruent? the answer is "no" because some parallelograms do not have all 4 sides congruent
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OpenStudy (anonymous):
So if its not C. then is No, p->q , is an invalid arguement.
jimthompson5910 (jim_thompson5910):
so which one, A or B?
OpenStudy (anonymous):
B?
OpenStudy (anonymous):
They both look the same to me though.
jimthompson5910 (jim_thompson5910):
they're not
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jimthompson5910 (jim_thompson5910):
A is q -> p
jimthompson5910 (jim_thompson5910):
B is ~q -> ~p
OpenStudy (anonymous):
So my answer was A?
jimthompson5910 (jim_thompson5910):
one is the converse, the other is the contrapositive
jimthompson5910 (jim_thompson5910):
yeah it's A
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OpenStudy (anonymous):
so its the converse.A.
jimthompson5910 (jim_thompson5910):
oh wait, they've formatted it differently, one sec
jimthompson5910 (jim_thompson5910):
A is still correct, I was just using different notation