Walkabout 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) \(60\sqrt{15}\)
I call these "Strange Operator" questions.
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:
[Reveal] Spoiler:
A
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