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If x–2 – 2 = –1.96, y1/3 – 5 = –2, and xy > 0, then what is

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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 123

Kudos [?]: 1975 [0], given: 397

If x–2 – 2 = –1.96, y1/3 – 5 = –2, and xy > 0, then what is [#permalink]  18 Mar 2018, 06:40
Expert's post
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Question Stats:

87% (08:28) correct 12% (03:54) wrong based on 8 sessions
If $$x^{-2} - 2 = -1.96$$, $$y^{\frac{1}{3}} - 5 = -2$$, and xy > 0, then what is the value of xy ?

Drill 2
Question: 15
Page: 499

[Reveal] Spoiler: OA
135

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Sandy
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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 123

Kudos [?]: 1975 [0], given: 397

Re: If x–2 – 2 = –1.96, y1/3 – 5 = –2, and xy > 0, then what is [#permalink]  21 Mar 2018, 14:11
Expert's post
Explanation

First solve for x: Add 2 to both sides to yield $$x^{-2}$$ = 0.04, so $$\frac{1}{x^2}=\frac{1}{25}$$, and x =±5.

Next, add 5 to both sides of the other equation to yield $$y^{1/3} = 3$$, so $$\sqrt[3]{y}= 3$$, and y = 27.

Since y is positive, you’ll need to use the positive value of x as well, so 5 × 27 = 135.
_________________

Sandy
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Re: If x–2 – 2 = –1.96, y1/3 – 5 = –2, and xy > 0, then what is   [#permalink] 21 Mar 2018, 14:11
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