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August 13th, 2011, 11:49 AM   #1
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mean proportions

Hello all!

I can find the mean proportion of ab and bc (it was the first, and simplest, problem) as b squared = ac or b = the sq. root of ac.

Whoops! I just saw that the text offers the answer as b sq. x root of ac.... i.e., b ac with the square root bracket, no = sign.

And the next problem is find the mean proportion of 8a squared and 2b squared (only the a and the b are "squared").

I'm hung up on whether I should put this into an equation as 8/a squared = 2/b squared which would result in (I think)

8b sq = 2 a sq

Am I on the right track or way off? I think I'm off track here!

If you have a clue, then it would be appreciated.

Thanks!
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August 14th, 2011, 05:53 AM   #2
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Re: mean proportions--I got it!

OK, I'm a little slow but I finally got it.

ab to bc means ab/x = x/bc;

thus, ac (b squared) = x squared, take the square root and one has x = (b)the square root of ac.

And the other problem follows similarly....

8asquared/x = x/2bsq.; thus x squared = 16 a squared b squared, take the square root and the result is x = 4ab.

Voila!
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