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I totally agree with you. I mess them up to and about did this one. I saw that if I make 2r = 1, then r=1/2 so to the power of 3 doesn't matter, it's still 1/1. Then I did r =1 (double it) so then V = 1/8. So I was thinking oh, go from 1 to 1/8 multiply by 1/8...but I had worked the problem backwards. There are tricks the authors throw at us, and then there are tricks we bring to the test ourselves. We have to eliminate our own tricks. Easier said than done!

mbawaters wrote:

I always mess up these kinds... Say r=2 => V1 = 1/64 when r=1; V2 = 1/8 V2 = 8*V1. B?

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi [#permalink]

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07 Apr 2015, 20:28

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