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 Post subject: Problem solving OG 11 # 244
PostPosted: Thu Jun 11, 2009 2:48 pm 
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Joined: Sun Jun 07, 2009 7:57 pm
Posts: 3
Hello,

Can someone help, I have no idea how do to this problem. It is PS #244 from OG 11. Thanks!


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 Post subject: Re: Problem solving OG 11 # 244
PostPosted: Tue Jun 16, 2009 1:25 pm 
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Joined: Mon Apr 06, 2009 5:44 pm
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The first thing you want to do is create whole numbers in the ratio.

lets convert .0015 to 15. We move the decimal over 4 places to the right but we have to balance this out on the 10 power. So what we have is 15 x 10^-4. Now, lets do the same for the denominator. We want to turn 0.03 to 3. So we move the decimal to the right two decimal points and we adjust our 10 power. So we have 3 x 10^-2. So we can rewrite the ratio
[15 x 10^-4 x 10^m / .03 x 10^-2 x 10^k] = [15 x 10^(m-4)] / [3 x 10^(k-2)] = 5 x 10^7. So we have
5 x [10^(m-4)-(k-2)] = 10^7 = 5 x 10^(m-k-2) = 5 x 10^7 so we have

m-k-2 = 7 adding 2 to both sides we have m-k = 9.


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 Post subject: Re: Problem solving OG 11 # 244
PostPosted: Tue Jun 16, 2009 1:31 pm 
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Here is what we have to do. First rewrite the numerator as [15 x 10^-4] x 10^m and rewrite the numerator as [3x 10^-2] x 10^k and set this equal to 5 x 10^7. Now, lets group our ten powers together. So in the numerator we have: 15 x 10^(m-4) / 3 x 10^(k-2) = 5 x 10^7. Now lets divide our ratio through and subtract the 10 power in the denominator from the 10 power in the numerator (number properties rule). So we have 5 x 10^[(m-4)-(k-2)] = 5 x 10^7. So, lets equate our powers.

(m-4)-(k-2) = 7 => m - 4 - k +2 = 7 => m-k = 9


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