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If #x is defined for all x > –2 as the square root of the nu

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If #x is defined for all x > –2 as the square root of the nu [#permalink] New post 30 Jul 2018, 09:58
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Question Stats:

88% (00:32) correct 11% (00:00) wrong based on 9 sessions
If \(#x\) is defined for all x > –2 as the square root of the number that is 2 more than x, what is the value of \(#7 - #(-1)\)?


[Reveal] Spoiler: OA
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Re: If #x is defined for all x > –2 as the square root of the nu [#permalink] New post 11 Aug 2018, 12:40
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This question isn't too involved, but it's still good form, at least for practice, to write out what #x means so you have practice for a more involved version of the question on test day.

\(#x = \sqrt{x+2}\)

Then \(#7 = \sqrt{7+2} = \sqrt{9} = 3\)

and

\(#(-1) = \sqrt{-1+2} = \sqrt{1} = 1\)

So \(#7-#(-1) = 3-1 = 2\)
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Re: If #x is defined for all x > –2 as the square root of the nu [#permalink] New post 11 Aug 2018, 16:35
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Explanation

This function defines a made-up symbol, rather than using traditional notation such as f(x).

First, translate the function:

\(#x = \sqrt{x+2}\)

The square root of any value greater than or equal to 0 is the non-negative square root of the value.

That is, the square root of 4 is just +2, not –2. Thus:
\(#7 =\sqrt{7+2} =\sqrt{9} = 3\)
\(#(-1) = \sqrt{-1+2}= \sqrt{1}= 1\)
Finally, \(#7 - #(-1) = 3 - 1 = 2\).
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Re: If #x is defined for all x > –2 as the square root of the nu   [#permalink] 11 Aug 2018, 16:35
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