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{(3^{-1} + 4^{-1})^{-2}}

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{(3^{-1} + 4^{-1})^{-2}}  [#permalink]

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New post Updated on: 15 Feb 2016, 06:50
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Question Stats:

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\(\frac{(3^{-1} + 4^{-1})^{-2}}{\frac{4^{2}}{7^{2}}}\) =

A. \(\frac{12}{9}\)

B. \(\frac{49}{144}\)

C. \(\frac{12}{7}\)

D. \(7\)

E. \(9\)

Originally posted by Ekland on 15 Feb 2016, 06:47.
Last edited by ENGRTOMBA2018 on 15 Feb 2016, 06:50, edited 1 time in total.
Renamed the topic
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Re: {(3^{-1} + 4^{-1})^{-2}}  [#permalink]

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New post 15 Feb 2016, 07:00
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1
Nez wrote:
\(\frac{(3^{-1} + 4^{-1})^{-2}}{\frac{4^{2}}{7^{2}}}\) =

A. \(\frac{12}{9}\)

B. \(\frac{49}{144}\)

C. \(\frac{12}{7}\)

D. \(7\)

E. \(9\)


\(\frac{(3^{-1} + 4^{-1})^{-2}}{\frac{4^{2}}{7^{2}}} =\) = [(1/3 + 1/4)^-2]/[16/49]
= [(4/12 + 3/12)^-2]/[16/49]
= [(7/12)^-2]/[16/49]
= [(12/7)^2]/[16/49]
= [144/49]/[16/49]
= [144/49][49/16]
= 144/16
= 9
= E

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{(3^{-1} + 4^{-1})^{-2}}  [#permalink]

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New post 01 Aug 2016, 10:44
=(49/16)*(1/(1/3+1/4)^2)
=49*12*12/(16*7*7)
=144/16
=9
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{(3^{-1} + 4^{-1})^{-2}}  [#permalink]

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New post 01 Aug 2016, 11:53
Nezdem wrote:
\(\frac{(3^{-1} + 4^{-1})^{-2}}{\frac{4^{2}}{7^{2}}}\)


\((3^{-1} + 4^{-1})^{-2} = (\frac{1}{3} + \frac{1}{4})^{-2} = \frac{7}{12}^{-2} = \frac{12}{7}^2 = \frac{12^2}{7^2}\)

\(\frac{\frac{12^2}{7^2}}{\frac{4^{2}}{7^{2}}} = \frac{12^2}{7^2} \times \frac{7^2}{4^2} = \frac{12^2 \times 7^2}{4^2 \times 7^2} = \frac{12^2}{4^2} = \frac{12}{4}^2 = 3^2 = 9\)


E. \(9\)
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Re: {(3^{-1} + 4^{-1})^{-2}}  [#permalink]

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New post 28 Oct 2017, 20:22
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Re: {(3^{-1} + 4^{-1})^{-2}}   [#permalink] 28 Oct 2017, 20:22
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