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For which of the following values of x is 3x/x^2+2x-1/5<0
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20 Oct 2022, 05:29
20 Oct 2022, 05:29
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Question Stats:
56% (01:14) correct
44% (02:02) wrong based on 25 sessions
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For which of the following values of x is \(\frac{3x}{x^2+2x}-\frac{1}{5}<0\)?
A. 10
B. 11
C. 12
D. 13
E. 14
F. 15
G. 16
A. 10
B. 11
C. 12
D. 13
E. 14
F. 15
G. 16
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GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
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GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
Re: For which of the following values of x is 3x/x^2+2x-1/5<0
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20 Oct 2022, 10:54
20 Oct 2022, 10:54
1
3x/(x^2+2x) -(1/5) < 0
3x/x(x+2)-(1/5) <0
3/(x+2)-(1/5)<0
3/(x+2) < (1/5)
Values of x that satisfy the above: Big denominators
Answers G,F,E
Posted from my mobile device
3x/x(x+2)-(1/5) <0
3/(x+2)-(1/5)<0
3/(x+2) < (1/5)
Values of x that satisfy the above: Big denominators
Answers G,F,E
Posted from my mobile device
For which of the following values of x is 3x/x^2+2x-1/5<0
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01 Mar 2024, 23:12
01 Mar 2024, 23:12
1
Simplify the expression to get \(\frac{3}{x+2}\)
\(\frac{3x}{x^2+2x} = \frac{3x}{(x+2)} = \frac{3}{x+2}\)
\(\frac{3}{x+2} - \frac{1}{5} <0\)
if \(x = 13\), \(\frac{3}{x+2} - \frac{1}{5} = \frac{3}{15} - \frac{1}{5} = 0\)
therefore the expression will be negative or less than 0 when \(x>13\)
therefore, the answers are E, F, G.
\(\frac{3x}{x^2+2x} = \frac{3x}{(x+2)} = \frac{3}{x+2}\)
\(\frac{3}{x+2} - \frac{1}{5} <0\)
if \(x = 13\), \(\frac{3}{x+2} - \frac{1}{5} = \frac{3}{15} - \frac{1}{5} = 0\)
therefore the expression will be negative or less than 0 when \(x>13\)
therefore, the answers are E, F, G.
