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Real numbers x, y, and z satisfy the inequalities [m]0<x<1[/m], [m]-1<

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Real numbers x, y, and z satisfy the inequalities [m]0<x<1[/m], [m]-1<  [#permalink]

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New post 18 Mar 2019, 05:48
1
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

71% (01:31) correct 29% (02:00) wrong based on 21 sessions

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Real numbers x, y, and z satisfy the inequalities \(0<x<1\), \(-1<y<0\), and \(1<z<2\). Which of the following numbers is necessarily positive?

(A)\(y+x^2\)

B) \(y+xz\)

(C)\(y+y^2\)

(D)\(y+2y^2\)

E) \(y+z\)
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Re: Real numbers x, y, and z satisfy the inequalities [m]0<x<1[/m], [m]-1<  [#permalink]

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New post 18 Mar 2019, 13:13
From the given info.
X and Z are always positive.
Y can be positive or negative.

Looking at the options.
Only E will give always be positive as Z is always greater than 1 and Y is always greater than -1.

E is the answer
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Re: Real numbers x, y, and z satisfy the inequalities [m]0<x<1[/m], [m]-1<   [#permalink] 18 Mar 2019, 13:13
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Real numbers x, y, and z satisfy the inequalities [m]0<x<1[/m], [m]-1<

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