Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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)?
 one year ago
 one year 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)?
 one year ago
 one year ago

This Question is Closed

mariomintchevBest ResponseYou've already chosen the best response.0
Because I get the right answer doing that with this problem...
 one year ago

KingGeorgeBest 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.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
what do you mean?
 one year ago

KingGeorgeBest 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.
 one year ago

KingGeorgeBest 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 \(20\).
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
haha i like creating my own ways to solve things.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.