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skihobbes · 08-02-2005, 09:28 PM

what the heck is this...when is this used??? I think what they are asking is which is greater...column a or b

windu · 08-03-2005, 05:01 AM

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!

Naseer Asif · 08-04-2005, 06:03 AM

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.

08-02-2005, 09:28 PM	#1 (permalink)
skihobbes Trying to make mom and pop proud Join Date: Aug 2005 Posts: 1	help please what the heck is this...when is this used??? I think what they are asking is which is greater...column a or b

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08-03-2005, 05:01 AM	#2 (permalink)
windu Trying to make mom and pop proud Join Date: Aug 2005 Location: United States Posts: 14	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!

08-04-2005, 06:03 AM	#3 (permalink)
Naseer Asif Trying to make mom and pop proud Join Date: Mar 2005 Posts: 13	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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