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mariomintchev
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For #9, will I always get the right answer if I take the volume (32), divide it by the pressure (4), and multiply it by the rate (2)?
 2 years ago
 2 years ago
mariomintchev Group Title
For #9, will I always get the right answer if I take the volume (32), divide it by the pressure (4), and multiply it by the rate (2)?
 2 years ago
 2 years ago

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mariomintchev Group TitleBest ResponseYou've already chosen the best response.0
Because I get the right answer doing that with this problem...
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
I don't think so. I'm pretty sure that's an artifact of \(2^2=4\). Or something similar.
 2 years ago

mariomintchev Group TitleBest ResponseYou've already chosen the best response.0
what do you mean?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Well, if we do the actual problem out, we get that \(c=128\). Then, we have that \[dV=\;{128 \over P^2} dP\]We know that \(dP =2\) and \(P=4\). Solving, we find that \(dV =16\). So really, it seems like it's because everything is just a power of two.
 2 years ago

mariomintchev Group TitleBest ResponseYou've already chosen the best response.0
i see
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
For a counterexample, let \(V=40\). Then we get that \(dV =35/2\) using the method I wrote out. Using your hypothetical method, we get that it would be \(20\).
 2 years ago

mariomintchev Group TitleBest ResponseYou've already chosen the best response.0
haha i like creating my own ways to solve things.
 2 years ago
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