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Joined: 30 Mar 2015
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In the equation 3r^{4}=(2/s), if r is doubled, then s must be
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Updated on: 14 Jun 2015, 07:35
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67% (01:21) correct 33% (01:34) wrong based on 274 sessions
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In the equation \(3r^{4}\)=\(\frac{2}{s}\), if r is doubled, the s must be multiplied by: A.\(\frac{1}{16}\) B. \(\frac{1}{8}\) C.\(\frac{1}{4}\) D. 4 E. 16
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Originally posted by petu on 14 Jun 2015, 07:26.
Last edited by petu on 14 Jun 2015, 07:35, edited 1 time in total.




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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be
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15 Jun 2015, 22:22
irus wrote: why 1/16 can someone explain pls \(3*r^4 = \frac{2}{s}\) \(s= \frac{2}{3*r^4}\) (multiply by s and divide by 3r^4) What happens when r becomes 2r. You get \(\frac{2}{3(2*r)^4} = \frac{2}{3*16*r^4}\) How is this new expression different from the old one? The new one = Old one * (1/16) Hence, s gets multiplied by 1/16 when r becomes twice.
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be
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14 Jun 2015, 10:34



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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be
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01 Nov 2015, 20:36
just plug in a number for r , like 1
Then solve for s , once you have s, all you have to do is double the value of r (which will now be: 2) , then solve for the new value of y



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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be
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23 Feb 2018, 02:47
VeritasPrepKarishma wrote: irus wrote: why 1/16 can someone explain pls \(3*r^4 = \frac{2}{s}\) \(s= \frac{2}{3*r^4}\) (multiply by s and divide by 3r^4) What happens when r becomes 2r. You get \(\frac{2}{3(2*r)^4} = \frac{2}{3*16*r^4}\) How is this new expression different from the old one? The new one = Old one * (1/16) Hence, s gets multiplied by 1/16 when r becomes twice. Hi Karishma, I have a small doubt here , I kept r to the left and s to the right and worked at it that way , below is the working : Given \(3 r^4 = \frac {2}{s}\) now r is doubled \((2r)^4 =\frac{2}{3s}\) \(16r^4= \frac{2}{3s}\) Now the above can be rearranged as \(3r^4= \frac{2}{16s}\) on comparing this with the original it seemed that s is being multiplied by 16 when r is doubled , hence I chose E. What is the error in the logic here?
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be
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24 Feb 2018, 04:17
stne wrote: VeritasPrepKarishma wrote: irus wrote: why 1/16 can someone explain pls \(3*r^4 = \frac{2}{s}\) \(s= \frac{2}{3*r^4}\) (multiply by s and divide by 3r^4) What happens when r becomes 2r. You get \(\frac{2}{3(2*r)^4} = \frac{2}{3*16*r^4}\) How is this new expression different from the old one? The new one = Old one * (1/16) Hence, s gets multiplied by 1/16 when r becomes twice. Hi Karishma, I have a small doubt here , I kept r to the left and s to the right and worked at it that way , below is the working : Given \(3 r^4 = \frac {2}{s}\) now r is doubled \((2r)^4 =\frac{2}{3s}\) \(16r^4= \frac{2}{3s}\) Now the above can be rearranged as \(3r^4= \frac{2}{16s}\) on comparing this with the original it seemed that s is being multiplied by 16 when r is doubled , hence I chose E. What is the error in the logic here? s is in the denominator. You cannot say what it multiplies by. To know "s is multiplied by ...", you need to separate out the s. s = Old expression New s = m * Old expression You need the value of m.
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be &nbs
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24 Feb 2018, 04:17






