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30 Jul 2018, 00:50
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
89% (00:50) correct
11%
(01:12)
wrong
based on 116
sessions
History
Date
Time
Result
Not Attempted Yet
If \(a\#b = a^2*\sqrt{b} - a\), where b ≥ 0, what is the value of (–4)#4?
(A) –36
(B) –28
(C) 12
(D) 28
(E) 36
(A) –36
(B) –28
(C) 12
(D) 28
(E) 36
30 Jul 2018, 02:28
Kudos
Bookmarks
Bunuel
Bunuel, can you pl check, I think there is a typo because none of the answer matches in the given function..
It should be \(a#b=a^2\sqrt{b}-a\)
In -4#4, a=-4 and b=4..
Therefore \(-4#4=(-4)^2\sqrt{4}-(-4)=4*4*2+4=32+4=36\)
E
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