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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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15 Jun 2016, 00:09
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If p ≠ 0 and \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\), then r = A) p + 1 B) 2p – 1 C) \(p^2+ 1\) D) \(2p^2 – 1\) E) \(p^2 + p – 1\) OG 2017 New Question
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Last edited by AbdurRakib on 22 Jun 2016, 12:19, edited 1 time in total.



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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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15 Jun 2016, 06:22
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AbdurRakib wrote: If p ≠ 0 and p – {\(\frac{(1 – p2 )}{p}\)} = \(\frac{r}{p}\), then r = A) p + 1 B) 2p – 1 C) \(p^2\)+ 1 D) 2\(p^2\) – 1 E) \(p^2\) + p – 1 OG 2017 New Question \(p –\frac{(1 – p^2 )}{p} = \frac{p^2 –(1 – p^2 )}{p} = \frac{p^2 –1 + p^2}{p} = \frac{2p^2 –1}{p} = \frac{r}{p}\)................. \(r= 2p^21\) D
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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22 Jun 2016, 12:12
AbdurRakib equation is missing an indication of power for "p2". I actually thought it was a typo for 2p.



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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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22 Jun 2016, 12:20
arthurmatsuda wrote: AbdurRakib equation is missing an indication of power for "p2". I actually thought it was a typo for 2p. edited thanks for your remarks
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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05 Dec 2016, 17:34
AbdurRakib wrote: If p ≠ 0 and p – {\(\frac{(1 – p^2 )}{p}\)} = \(\frac{r}{p}\), then r =
A) p + 1 B) 2p – 1 C) \(p^2\)+ 1 D) 2\(p^2\) – 1 E) \(p^2\) + p – 1 We are given that p  [(1p^2)/p] = r/p, and we must isolate r. We can start by multiplying the entire equation by p and then simplifying: p^2  (1  p^2) = r p^2  1 + p^2 = r 2p^2  1 = r Answer: D
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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19 Feb 2017, 09:28
Isolating r, we get r = p [ p  {(1p^2)/p} ]
r= p^2  (1p^2)
r= p^2  1 + p^2
r= 2p^2  1 = answer is D.



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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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04 Apr 2017, 21:31
\(p\cfrac { \left( 1{ p }^{ 2 } \right) }{ p } =\cfrac { r }{ p } \\ \left[ p\left( p\cfrac { \left( 1{ p }^{ 2 } \right) }{ p } =\cfrac { r }{ p } \right) \right] \\ { p }^{ 2 }\left( 1{ p }^{ 2 } \right) =r\\ { p }^{ 2 }1+{ p }^{ 2 }=r\\ 2{ p }^{ 2 }1=r\) PEMDAS in the 3rd line got me thinking whether I should multiply that negative to \(\left( 1{ p }^{ 2 } \right)\).
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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26 Jul 2017, 02:23
BunuelHow to know if 1p^2 is in bracket or not? I found r=1 because of this error.
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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26 Jul 2017, 03:14



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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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17 Sep 2017, 22:46
Bunuel. Once we multiply both sides by p, p in the numerator becomes p^2, but what happens to the p in the denumerator? I understand that it gets cancel, but I don't see how.



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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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17 Sep 2017, 22:52
Valentina86 wrote: If p ≠ 0 and \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\), then r =
A) p + 1
B) 2p – 1
C) \(p^2+ 1\)
D) \(2p^2 – 1\)
E) \(p^2 + p – 1\)
Bunuel. Once we multiply both sides by p, p in the numerator becomes p^2, but what happens to the p in the denumerator? I understand that it gets cancel, but I don't see how. \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\); Multiply by p: \(p*(p – \frac{(1 – p^2 )}{p}) =p*\frac{r}{p}\); Expand the left hand side and reduce by p in the right hand side: \(p^2 – p*\frac{(1 – p^2 )}{p} =r\); Reduce by p in the second term on the left hand side: \(p^2 –(1 – p^2 ) =r\); Open the brackets: \(p^2 –1 + p^2 =r\); \(2p^21=r\) Answer: D. Hope it's clear.
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