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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink]
1
Carcass wrote:
If the operation @ is defined by \(x@y=\sqrt{xy}\) for all positive numbers x and y, then (5@45)@60 =

(A) 30
(B) 60
(C) 90
(D) \(30\sqrt{15}\)
(E) \(60\sqrt{15}\)


We want to evaluate: (5 @ 45) @ 60

Start with (5 @ 45)
(5 @ 45) = √[(5)(45)]
= √225
= 15


So, (5 @ 45) @ 60 = 15 @ 60
From here, 15 @ 60 = √[(15)(60)]
= √900
= 30

Answer: A
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink]
1
(5@45)@60

\(\sqrt{\sqrt{5 \times 45} \times 60}\)
\(\sqrt{\sqrt{225} \times 60}\)
\(\sqrt{900}\)
= 30
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If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink]
1
Given that \(x@y=\sqrt{xy}\) and we need to find the value of (5@45)@60

Let's start by finding the value of 5@45

To find 5@45 we need to compare 5@45 with x@y and see what is before the @ and what is after
=> x = 5 and y = 45

=> to find the value of 5@45 we need to replace x with 5 and y with 45 in \(x@y=\sqrt{xy}\)
=> \(5@45=\sqrt{5*45}\) = \(\sqrt{5*5*9}\) = \(\sqrt{5^2*3^2}\) = 5*3 = 15

=> (5@45)@60 = 15@60

Similarly, \(15@60=\sqrt{15*60}\) = \(\sqrt{15*15*4}\) = \(\sqrt{15^2*2^2}\) = 15*2 = 30

So, Answer will be A
Hope it helps!

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If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink]
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