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If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a

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Intern
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Joined: 26 May 2015
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If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a [#permalink]

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New post 17 Jul 2017, 11:57
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E

Difficulty:

  25% (medium)

Question Stats:

72% (00:50) correct 28% (00:57) wrong based on 43 sessions

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If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =

a. -1
b. 0
c. 1
d. 9
e. 25
[Reveal] Spoiler: OA
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Senior Manager
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Joined: 24 Apr 2016
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Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a [#permalink]

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New post 17 Jul 2017, 14:20
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-1∆(1∆ -1) = -1∆[(-1-1)^2]/1 = -1∆4 = [(4+1)^2]/1 = 25
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Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a [#permalink]

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New post 17 Jul 2017, 17:55
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Esguitar wrote:
If the operation ∆ is defined by a∆b = \((b-a)^2/a^2\) for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =
a. -1
b. 0
c. 1
d. 9
e. 25

Or, the longish way (not too bad - 54 seconds)

If a∆b = \(\frac{(b-a)^2}{a^2}\), then

-1∆(1∆ -1) =

1. Parentheses first:

(1∆ -1) = \(\frac{(-1-1)^2}{(1)^2}\) = \(\frac{(-2)^2}{1}\) = \(\frac{4}{1}\) = 4

4 becomes b -->

2. New expression is -1 ∆ 4

\(\frac{(4-(-1))^2}{(-1)^2}\) = \(\frac{(5)^2}{1}\) = \(\frac{25}{1}\) = 25

Answer E
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Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a [#permalink]

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New post 18 Jul 2017, 01:13
Thanks a lot, I got it.
I could not see how the expression (1∆ -1) becomes (−1−1)^2/(1)^2, I assume that it is plugin number where a=1 and b=-1.
Thanks again.
Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a   [#permalink] 18 Jul 2017, 01:13
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