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# (y^2/25 + y^2/16)^(1/2) =

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Math Expert
Joined: 02 Sep 2009
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04 Mar 2018, 23:11
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35% (medium)

Question Stats:

71% (01:25) correct 29% (01:23) wrong based on 141 sessions

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$$\sqrt{\frac{y^2}{25} + \frac{y^2}{16}} =$$

A. $$\frac{2|y|}{9}$$

B. $$\frac{9|y|}{20}$$

C. $$\frac{|y|}{9}$$

D. $$\frac{|y|*\sqrt{41}}{20}$$

E. $$\frac{41|y|}{20}$$

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05 Mar 2018, 00:17
Bunuel wrote:
$$\sqrt{\frac{y^2}{25} + \frac{y^2}{16}} =$$

A. $$\frac{2|y|}{9}$$

B. $$\frac{9|y|}{20}$$

C. $$\frac{|y|}{9}$$

D. $$\frac{|y|*\sqrt{41}}{20}$$

E. $$\frac{41|y|}{20}$$

$$\sqrt{\frac{y^2}{25} + \frac{y^2}{16}}$$ = $$\sqrt{\frac{25*y^2 + 16*y^2}{25*16}}$$ = $$\sqrt{\frac{41*y^2}{400}} = \frac{\sqrt{41*y^2}}{\sqrt400}$$ = $${\frac{\sqrt41*|y|}{20}}$$ (Option D)
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25 Mar 2018, 05:54

Solution

=$$\sqrt{(\frac{y^2}{25}+ \frac{y^2}{16})}$$

= $$\sqrt{\frac{(16 y^2+25y^2)}{(25*16)}}$$

= $$\sqrt{\frac{41 y^2}{400}}$$

= $$\sqrt{\frac{41(y)^2}{(20)^2}}$$

$$\sqrt{\frac{y^2}{25}+ \frac{y^2}{16}} = |y| *\sqrt{\frac{41}{20}}$$

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Re: (y^2/25 + y^2/16)^(1/2) =  [#permalink]

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25 Mar 2018, 06:37
Bunuel wrote:
$$\sqrt{\frac{y^2}{25} + \frac{y^2}{16}} =$$

A. $$\frac{2|y|}{9}$$

B. $$\frac{9|y|}{20}$$

C. $$\frac{|y|}{9}$$

D. $$\frac{|y|*\sqrt{41}}{20}$$

E. $$\frac{41|y|}{20}$$

$$\sqrt{\frac{y^2}{25} + \frac{y^2}{16}}$$
= $$|y|\sqrt{\frac{1}{25} + \frac{1}{16}}$$
= $$|y|\sqrt{\frac{(16+25)}{25*16}}$$
= $$\frac{|y|}{20}\sqrt{(16+25)}$$
= $$\frac{|y|}{20}\sqrt{41}$$ Ans B
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Re: (y^2/25 + y^2/16)^(1/2) =   [#permalink] 25 Mar 2018, 06:37
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