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mariomintchev
 3 years ago
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)?
mariomintchev
 3 years ago
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)?

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

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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0what do you mean?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1For 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 \(20\).

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