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#2 (permalink) |
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I JUST got here.
Join Date: Aug 2005
Location: United States
Posts: 14
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Column A reduces to 1.
Since we know that a > 0 and nothing else, you cannot answer this question. Ex) let a = 1 then columna A = column B let a=0.5 then column A > column B let a=6 then column A < column B I hope this helps! ![]() |
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#3 (permalink) |
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I JUST got here.
Join Date: Mar 2005
Posts: 13
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Since c=(16 Pi a^2 b)/(9 a^2)
=(16 Pi b)/9 by cancelling a^2 => 1=(16 Pi b)/(9 c) => (+ or -)1=Sqrt[(16 Pi b)/(9 c)] Since "a" is arbitrary +ve number, and column A is either +1 or -1, So Columns A and B are not comparable. |
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