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26 Mar 2015, 04:42
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Difficulty:
35%
(medium)
Question Stats:
67% (01:35) correct
33%
(01:49)
wrong
based on 362
sessions
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Date
Time
Result
Not Attempted Yet
In the xy-coordinate system, the distance between points \((2\sqrt{3}, \ -\sqrt{2})\) and \((5\sqrt{3}, \ 3\sqrt{2})\) is approximately
A. 4.1
B. 5.9
C. 6.4
D. 7.7
E. 8.1
Kudos for a correct solution.
A. 4.1
B. 5.9
C. 6.4
D. 7.7
E. 8.1
Kudos for a correct solution.
hsbinfy
Joined: 02 Mar 2012
Last visit: 13 Nov 2017
Posts: 196
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Given Kudos: 4
Schools: Schulich '16
26 Mar 2015, 07:27
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D
distance =sqrt(x2-x1)2 +(y2-y1)2 (2 outside brackets stands for sqaure)
distance=sqrt32+27
=sqrt59
approx 7.7
distance =sqrt(x2-x1)2 +(y2-y1)2 (2 outside brackets stands for sqaure)
distance=sqrt32+27
=sqrt59
approx 7.7
26 Mar 2015, 11:32
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The distance is
\(\sqrt{(5\sqrt{3}-2\sqrt{3})^2+(3\sqrt{2}+\sqrt{2})^2}=\sqrt{(3\sqrt{3})^2+(4\sqrt{2})^2}=\sqrt{9\cdot 3+16\cdot 2}=\sqrt{59}\)
since \(7^2=49\) and \(8^2=64\), \(7<\sqrt{59}<8\). Only D. gives a possible value and is the correct answer
\(\sqrt{(5\sqrt{3}-2\sqrt{3})^2+(3\sqrt{2}+\sqrt{2})^2}=\sqrt{(3\sqrt{3})^2+(4\sqrt{2})^2}=\sqrt{9\cdot 3+16\cdot 2}=\sqrt{59}\)
since \(7^2=49\) and \(8^2=64\), \(7<\sqrt{59}<8\). Only D. gives a possible value and is the correct answer
