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by LalaB » Tue Jan 31, 2012 10:09 pm
C is the asnw

P/(P-1)/(P/(P-1)-1)=P(P-1)/(P-1)=P
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by arpitad » Tue Jan 31, 2012 10:21 pm
I understand that P#= P/P-1, but what do you do with the second #?
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by arpitad » Tue Jan 31, 2012 10:22 pm
I understand that P#= P/P-1, but what do you do with the second #?
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by LalaB » Tue Jan 31, 2012 10:36 pm
P# = P/P-1

(P#)#= P/(P-1)#= P/(P-1)/(P/(P-1)-1)
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by pemdas » Tue Jan 31, 2012 10:38 pm
P# = P/(P-1)

divide into two stages, p# as given and p##=p/(p-1) : ((p/(p-1)-1)
simplify to get p/(p-1) : (p-p+1)/(p-1)= p/(p-1) * (p-1)/1=p

c
arpitad wrote:If, P# = P/P-1 what is the value of P## ?


(A) p/p-1
(B) 1/p
(C) p
(D) 2-p
(E) p-1
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