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a$b=a/a+b
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28 Apr 2021, 08:49
28 Apr 2021, 08:49
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\(a$b=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) for positive square root.
Compute 1$(4$9) with 3 decimal place accuracy
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Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Compute 1$(4$9) with 3 decimal place accuracy
Show: :: OA
0.613
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: a$b=a/a+b
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19 Aug 2021, 08:12
19 Aug 2021, 08:12
Samamammadova8888 wrote:
can anyone explain please?
\(a$b=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)
Compute 1$(4$9) with 3 decimal place accuracy
\(4$9=\frac{\sqrt{4}}{\sqrt{4}+\sqrt{9}}\)
\(4$9 = \frac{2}{2 + 3} = \frac{2}{5} = 0.40\)
Now, \(1$0.40 = \frac{\sqrt{1}}{\sqrt{1}+\sqrt{0.40}}\)
\(1$0.40 = \frac{1}{1 + 0.63} = \frac{1}{1.63} = 0.613\)
