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Real numbers x, y, and z satisfy the inequalities
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28 Jan 2022, 07:20
28 Jan 2022, 07:20
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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) \(y+z\)
(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
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28 Jan 2022, 08:10
28 Jan 2022, 08:10
Carcass wrote:
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) \(y+z\)
(A) \(y+x^2\)
(B) \(y+xz\)
(C) \(y+y^2\)
(D) \(y+2y^2\)
(E) \(y+z\)
Given:
\(-1<y<0\)
\(1<z<2\)
Since the inequality symbols are facing the same direction, we can add the two inequalities to get: \(0<y+z<2\)
So, it must be the case that \(y+z\) is positive.
Answer: E
Re: Real numbers x, y, and z satisfy the inequalities
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17 Feb 2023, 09:11
17 Feb 2023, 09:11
Hi Brent GreenlightTestPrep
How about the x in the inequalities 0<x<1 ?
How do we know only to use the other two inequalities? Could you help clarify? Thanks Brent
How about the x in the inequalities 0<x<1 ?
How do we know only to use the other two inequalities? Could you help clarify? Thanks Brent
