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22 Dec 2020, 03:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
63% (01:41) correct
37%
(01:40)
wrong
based on 277
sessions
History
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Not Attempted Yet
If \(x = \frac{1}{2}\) and \(x^2 + 2y^2 = 1\), then which of the following values is closest to the value of |y|?
(A) 0.1
(B) 0.2
(C) 0.4
(D) 0.5
(E) 0.6
(A) 0.1
(B) 0.2
(C) 0.4
(D) 0.5
(E) 0.6
28 Dec 2020, 06:19
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hero_with_1000_faces
I calculated by method of approximation
\(\frac{3}{8} = 0.37\) ~ \( 0.4\)
\(\sqrt{0.4} = \sqrt{\frac{4}{10}}\) ~ \(\sqrt{\frac{4}{9}} = \frac{2}{3} = 0.6\)
I calculated by method of approximation
\(\frac{3}{8} = 0.37\) ~ \( 0.4\)
\(\sqrt{0.4} = \sqrt{\frac{4}{10}}\) ~ \(\sqrt{\frac{4}{9}} = \frac{2}{3} = 0.6\)
General Discussion
22 Dec 2020, 04:31
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Putting x = 1/2
\(\frac{1}{4} + 2y^2 = 1\)
\(2y^2 = 1 - \frac{1}{4} = \frac{3}{4}\)
\(y^2 = \frac{3}{8}\)
\(y = \sqrt{\frac{3}{8}} = 0.6\)
Option E
Arun Kumar
\(\frac{1}{4} + 2y^2 = 1\)
\(2y^2 = 1 - \frac{1}{4} = \frac{3}{4}\)
\(y^2 = \frac{3}{8}\)
\(y = \sqrt{\frac{3}{8}} = 0.6\)
Option E
Arun Kumar
