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If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/

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If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/ [#permalink]

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New post 03 Sep 2017, 06:03
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Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/ [#permalink]

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New post 03 Sep 2017, 07:59
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Bunuel wrote:
If \(\sqrt{x} =20\) and \(y = \sqrt{16}\), then \(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=\)


A. \(9\sqrt{41}\)

B. 2

C. \(\frac{9\sqrt{41}}{41}\)

D. 1

E. 9/41



Hi Bunuel,

Is there a mistake with the question? Or am i doing something wrong?
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Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/ [#permalink]

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New post 03 Sep 2017, 08:04
pushpitkc wrote:
Bunuel wrote:
If \(\sqrt{x} =20\) and \(y = \sqrt{16}\), then \(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=\)


A. \(9\sqrt{41}\)

B. 2

C. \(\frac{9\sqrt{41}}{41}\)

D. 1

E. 9/41


Hi Bunuel,

Is there a mistake with the question? Or am i doing something wrong?


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Edited. Thank you.
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Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/ [#permalink]

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New post 03 Sep 2017, 08:13
Bunuel wrote:
If \(\sqrt{x} =20\) and \(\sqrt{y} =16\), then \(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=\)


A. \(9\sqrt{41}\)

B. 2

C. \(\frac{9\sqrt{41}}{41}\)

D. 1

E. 9/41



Hi

\(\sqrt{x} =20.....x=400\) and \(\sqrt{y} =16,.....y=256\),
\(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}=\frac{20+16}{√(400+256)}=\frac{36}{√656}=\frac{36}{4√41}=\frac{9√41}{41}\)

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Re: If x^(1/2) = 20 and y^(1/2) = 16, then (x^(1/2) + y^(1/2))/   [#permalink] 03 Sep 2017, 08:13
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