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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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 Updated on: 22 Jun 2016, 12:19
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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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Originally posted by AbdurRakib on 15 Jun 2016, 00:09.
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 =
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15 Jun 2016, 06:22
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 \(p –\frac{(1 – p^2 )}{p} = \frac{p^2 –(1 – p^2 )}{p} = \frac{p^2 –1 + p^2}{p} = \frac{2p^2 –1}{p} = \frac{r}{p}\)................. \(r= 2p^2-1\) D
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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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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Director
Status: I don't stop when I'm Tired,I stop when I'm done
Joined: 11 May 2014
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GPA: 2.81
WE: Business Development (Real Estate)
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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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 =
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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 - [(1-p^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 =
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19 Feb 2017, 09:28
Isolating r,
we get
r = p [ p - {(1-p^2)/p} ]
r= p^2 - (1-p^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 =
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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 =
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26 Jul 2017, 02:23
BunuelHow to know if 1-p^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 =
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26 Jul 2017, 03:14
Shiv2016 wrote: BunuelHow to know if 1-p^2 is in bracket or not? I found r=-1 because of this error. In \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\), it does not matter whether 1 - p^2 is in brackets.
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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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 =
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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: \(p^2 –1 + p^2 =r\); \(2p^2-1=r\) Answer: D. Hope it's clear.
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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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01 Mar 2018, 11:39
chetan2u wrote: 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 \(p –\frac{(1 – p^2 )}{p} = \frac{p^2 –(1 – p^2 )}{p} = \frac{p^2 –1 + p^2}{p} = \frac{2p^2 –1}{p} = \frac{r}{p}\)................. \(r= 2p^2-1\) D hi chetan2u, can you please explain the last step of getting this \([m]r= 2p^2-1\) from this \(\frac{2p^2 –1}{p} = \frac{r}{p}\) thanks!
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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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01 Mar 2018, 11:53
dave13 wrote: chetan2u wrote: 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 \(p –\frac{(1 – p^2 )}{p} = \frac{p^2 –(1 – p^2 )}{p} = \frac{p^2 –1 + p^2}{p} = \frac{2p^2 –1}{p} = \frac{r}{p}\)................. \(r= 2p^2-1\) D hi chetan2u, can you please explain the last step of getting this \([m]r= 2p^2-1\) from this \(\frac{2p^2 –1}{p} = \frac{r}{p}\) thanks! Hey dave13If you multiply or divide both the sides of a fraction by a variable/constant, the value remains the same Another way to do this is as follows \(\frac{2p^2 –1}{p} = \frac{r}{p}\) -> \(2p^2 –1= \frac{r}{p}*p\) (by cross multiplying) -> r = 2\(p^2\) - 1 Here p on the right-hand side cancel each other out(because in a fraction, common number/variables in the numerator and denominator cancel each other out) Hope this helps you!
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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11 May 2019, 22:09
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) \(2p^2 – 1\) E) \(p^2 + p – 1\) OG 2017 New Question Simplify the expression, P^2-1+p^2/p=r/p =>2p^2-1=r [multiply both side by 'P'] Answer is D Posted from my mobile device
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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05 Jul 2020, 12:15
\(p – \frac{(1 – p^2 )}{p} =\frac{ r}{p}\)
\(r=p(p-\frac{(1 – p^2 )}{p} )\)
\(r=p^2-(1-p^2)\)
\(r=2p^2-1\)
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =
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05 Jul 2020, 12:15
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