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anonymous
 5 years ago
A large tank contains 100 liters of a salt solution which has a concentration of 0.25 kilograms per liter. Pure water is added to the solution at a rate of 80 liters per minute. At the same time, the wellmixed solution drains from the tank at 80 liters per minute.
Find the amount S of the salt in the solution as a function of t:
kilograms
At what time is there only 12 kilograms of salt remaining in the tank?
t= minutes
anonymous
 5 years ago
A large tank contains 100 liters of a salt solution which has a concentration of 0.25 kilograms per liter. Pure water is added to the solution at a rate of 80 liters per minute. At the same time, the wellmixed solution drains from the tank at 80 liters per minute. Find the amount S of the salt in the solution as a function of t: kilograms At what time is there only 12 kilograms of salt remaining in the tank? t= minutes

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hi dichalao, Do you still need a solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes! Or just give me some clues first :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What i have tried is to set up the first differential equation as : dS/dt= 25/(100+80t)(180t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to consider the physics of what's going on in a little bit of time,\[\delta t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and you need to consider what's actually changing. You're told 80 L/min enters the tank, and 80 L/min leaves the tank, so volume's the same. What's changing is the concentration, or even better, the mass, of salt in the solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to come up with an expression for the change in mass of salt in the tank as a function of time, then use that to get an expression for concentration and solve the system.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea i see the solution is diluting

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, let's consider what happens in that little bit of time, delta_t. No salt is entering the system, since what's coming in is pure water. Salt *is* leaving the system in the outflow. I said we need to consider mass, so, let the mass at time t be\[m(t)\] and that at time t+delta_t be\[m(t+\delta t)\]The difference between the two will be\[\delta m = m( t + \delta t)  m(t)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The change in mass is the negative of the amount leaving (because these delta quantities are always read (final)(initial), and since you have less at the end, the change will be negative). That is,\[\delta m = Q \delta t C(t)\]where Q is the flow rate (liters/min) and C(t) is concentration at time t (this is why delta_t needs to be small, otherwise C(t) won't be valid).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Notice the units on the lefthand side and the righthand side. LHS is (mass), and righthand side is (volume)/(time) x (time) x (mass)/(volume). The units equate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what we have now is\[\delta m = m(t+ \delta t)m(t)\]\[Q \delta t C(t)=m(t+\delta t)m(t)\]Divide both sides by the constant volume in the container, and by delta_t to get\[\frac{QC(t)}{V}=\frac{C(t+\delta t)C(t)}{\delta t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because \[C(t) = \frac{m(t)}{V}, C(t+\delta t)=\frac{m(t+\delta t)}{V}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That expression above (above what I *just* wrote) should look familiar. Does it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, my computer's acting up.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can now take the limit as delta_t approached zero on both sides to get,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{\delta t \rightarrow 0}\frac{QC(t)}{V}=\lim_{\delta t \rightarrow 0}\frac{C(t+ \delta t)C(t)}{\delta t}\]\[\rightarrow \frac{QC(t)}{V}=\frac{dC}{dt}\](the lefthand side is independent of delta_t, so it's just a constant in terms of taking the limit).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You now have a separable differential equation:\[\frac{dC}{C}=\frac{Q}{V}dt \rightarrow \int\limits_{}{}\frac{dC}{C}=\int\limits_{}{}\frac{Q}{V}dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks i can take it from here on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\log C=\frac{Q}{V}t+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have one more question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Suppose taht a_n>0, and lim n>infinity n^2a_n=0. Prove that sigma a_n converges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since lim n^2a_n>0, so i assume n^2a_n behave like 1/n^p where p >0, if so a_n should be have like (1/n^(2+p), a_n is a pseries, with p >1, so a_n converges. Do u think this approach is reasonable?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You might have to leave it with me. I just got up and need to get ready for uni/work. I'll get back to you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure :) Have a good one :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Before I go, I want to confirm the sequence is\[n^2a_n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have a question regarding C(t) is concentration at time t (this is why delta_t needs to be small, otherwise C(t) won't be valid).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you mean C(t)*delta_t wont be valid? Or just C(t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, C(t) is the concentration at time t, but that's always changing. We're looking at times t and t+delta_t, but only using C(t) to approximate the amount of mass that's leaving. This is valid in the limit, but if you leave everything as a finite difference (which is what you have before taking the limit), what you're really saying is\[QC(t)/V \approx m(t+ \delta t)m(t)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But because our increment in time was small, the CHANGE in C(t) over that time is assumed very small too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So then all you need to do to find the mass is multiply it by volume flow rate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know how well this is being explained. We could have used differentials instead of delta notation, but I think understanding the approximations before taking limits is useful.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a good explanation, i understand everything you did

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I took Q as the magnitude of flow, that's why I introduced the minus sign when setting up the equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should end up with\[C(t)=\frac{1}{4}e^{\frac{4}{5}(\min^{1})t}kg/L\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which makes sense, since as t approaches infinity, C(t) approaches zero...the salt is being drained away.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and at time 0, C(0) = 1/4 kg, which is what you've stated as an initial condition.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if i neeed to find the amount S of the salt in the solution as a function of t: i just need to multiply both side by 100? since V is constant?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, (concentration) x (volume) = (mass) here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the volume flow rates were different either side, you'd have a very different situation to look at.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol it will be super complicated?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where are you from btw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, in these types of problems, look at what's actually changing, and try to come up with an expression that focuses on that. Try to formulate something you can turn into a differential equation in the end.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0UK citizen in Sydney.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where are you from? Are you male or female?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So is this school stuff?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea, i am learning sequence/series and my friend is learning 1st degree differential equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are really good at problem solving

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this for the international baccalaureate?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0international baccalaureate?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey, I just found the answer to your other problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha do you have tips on how to approach a solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the stuff you're studying is covered in the international baccalaureate  it's an international diploma that can be used to get into universities around the world.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, hang on, I'll need to check it. I was looking to apply the Dirichlet test with a series 1/n^2...but I want to see if I can get around one of the assumptions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No it is just normal school work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll think about tips on how to approach a solution and get back to you. Really, all you have to do is determine what you're being asked to find, put down the information you have before you and build a bridge.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you told that \[a_n \ge a_{n+1}>0?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, what i posted is everything given

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i wouldn't have thought about setting Qdelta_t(Ct)=m(delta_t+t)m(t) haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you ask yourself what was actually changing in the system?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Concentration's harder to come to terms with than mass, since concentration is a ratio of quantities, whereas mass is something we have an intuitive feel for.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i know that the amount of salt in the container is lessening and the amount of salts going out is also lessening

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are a genius i guess ..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like I said, experience helps

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should pass some experiences to me haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL True.. Are you still trying to find the solution to the problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but I should be showering...I have an idea, but not much time to check it out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take your time buddy, i have to go to class right now.. Thank you for helping out :) see you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Damn, I figured it out just after you left. You can use the limit comparison test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The version we need says, If \[\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=0\]then\[\sum_{n=0}^{\infty}b_n\]convergent implies\[\sum_{n=0}^{\infty}a_n\]is also convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, you need to show that the sum of a_n is convergent, and we can consider for out 'b' series, \[\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}\frac{1}{n^2}\]which is convergent (pseries, p=2>1). Therefore, taking the ratio of a_n and b_n and forming the limit gives,\[\lim_{n \rightarrow \infty}\frac{a_n}{b_n}=\lim_{n \rightarrow \infty}\frac{a_n}{1/n^2}=\lim_{n \rightarrow \infty}n^2a_n=0\]by construction (i.e. by the setup of the problem). By the theorem above, your series is convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. Whenever I see a polynomial in the variable, I think ratio test, limit comparison or Dirichlet first.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A large tank contains 450 liters of a salt solution which has a concentration of 0.65 kilograms per liter. A salt solution which has a concentration of 0.8 kilograms per liter is added to the solution at a rate of 20 liters per minute. At the same time, the solution drains from the tank at 20 liters per minute.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's going to be similar. Volume in the tank won't change again, which is useful. Try to form a mass balance like we did before.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I learned the limit comparison and ratio test just now haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah..well, that may have been why you couldn't do it ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are a couple of versions, depending on how the limit a_n/b_n behaves. I had to pick the right one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So this time, there is outflow and inflow of salt.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What does your mass balance look like?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am trying to figure it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a general formula for changes of 'stuff' in a system (for further reference):\[\Delta = io+ \gamma  \delta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Change = in  out + generated  destroyed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This isn't a chemical or nuclear reaction, so nothing is generated or destroyed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0inflow=Qdelta_t(C(t)) outflow=Q(deltat)(C(t+deltat)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your inflow isn't changing with time  it's fixed at 0.8kg/L * 20 L/min

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First, in a unit of time, delta t, how much salt is entering the system?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00.8(20)(delta t) (delta t in min of time)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just set it equal to some constant, a, or something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You don't have to, but when you have numbers floating around, it can stuff the algebra if you're not careful.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it will just be aoutflow=m(t+delta t)m(t) and then everything goes the same way?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0give a round of applause to loki haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but what about the outflow...what's your expression?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0might want to save the round of applause for a bit =D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You might want to write your inflow as\[a \delta t Q\]where a is your constant concentration and Q is the inflow rate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0outflow= 20(delta t)(Constantly changing concentration) :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You might want to write your inflow as\[a \delta t Q\]where a is your constant concentration and Q is the inflow rate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your outflow is write. Don't be :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL you need more sleep

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0concentration = mass over volume mass=(mass initial+mass inflowmass outlfow)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, what does\[m(t+ \delta t)m(t)\]equal?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(constant inflow)a  (changing outflow), b delta t(C(t+delta t) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, you're giving me a headache...I'll write it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry.. i am not floating with you :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[m(t+\delta t)m(t)=a \delta t QC(t) \delta t Q=(aC(t))Q \delta t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the change is just *in*  *out*, as per that Delta equation I put up.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0give me some time to digest haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a is constant concentration for inflow, while C(t) is changing concentration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0delta t is intended to cancel the time component in Q?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got it now i think .. haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got to go grab food now...otherwise i have to eat myself tonite .see you ~~thanks !!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Divide both side by V, the volume of the system to obtain\[C(t+\delta t)C(t)=\frac{Q}{V}(aC(t))\delta t\]and divide both sides by delta t\[\frac{C(t+\delta t)C(t)}{\delta t}=\frac{Q}{V}(aC(t))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Take the limit as delta_t approaches zero on both sides to get\[\frac{dC}{dt}=\frac{Q}{V}(aC(t)) \rightarrow \frac{dC}{aC}=\frac{Q}{V}dt \rightarrow \int\limits_{}{}\frac{dC}{aC}=\int\limits_{}{}\frac{Q}{V}dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should get\[C(t)=aC(0)e^{\frac{Qt}{V}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh wait...I forgot about the salt already in there! This is what happens when you do it on the fly. Go eat.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll write something later.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, this is much easier when you do it on paper. This site needs something like a whiteboard. I'll write it up and scan.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow! you are the most amazing person in the world haha! So for the first question do i also need to add the original amount of salt to the equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know, I've forgotten the first question already. I don't think so since if you put t=0 into the solution, you should get the initial amount in the tank. Just check.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Always check the initial and terminating conditions (t=0, t> infinity) to see if your answer makes sense.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL i am too lazy to do so when there is a smart guy in the house :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, flattery will get you everywhere.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am just fooling around ~~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is there a way that i can use to find square root super fast

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you mean? By hand?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea or better in mind

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0NewtonRaphson method is a numerical method for extracting solutions from equations. It has something called, "quadratic convergence" which makes it very fast. Plus, it's easy to use.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I explained it to someone last week. I'll see if I can find it. If not, the result for square root is \[x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You start with an initial guess, x_0, and the number it spits out on the LHS will be x_1, which will be closer to the solution. But that means that this is a BETTER guess than before, so we can plug the result into the RHS to get an EVEN BETTER guess on the LHS and so on...you repeat for infinity, but in reality, it only takes a couple of cycles to get something close.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, that's only for the square root of 2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For\[y=\sqrt{a}\]it would be\[x_{n+1}=\frac{x_n}{2}+\frac{a}{2x_n}=\frac{1}{2}\left( x_n+\frac{a}{x_n} \right)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LHS = lefthand side.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, start with an initial guess for sqrt(2) of 1.5. Plug 1.5 into the RHIS of the equation and you should get an answer, 1.4166666...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Plug that answer into the RHS again and you get on the LHS: 1.414215686...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Already it's getting close.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this something similar to newton's method?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My professor talked a little bit about it last quarter

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh do you have any cool math tricks that you can show me haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'd have to think... Some tricks, you can add zero or multiply by 1 and not change a number...so you can extend this idea when it comes to solving a lot of problems since 0 and 1 can be written in many convenient forms. Example \[\int\limits_{}{}\frac{u}{1+u}du=\int\limits_{}{}\frac{u+0}{1+u}du \int\limits_{}{}\frac{u+11}{1+u}du\]\[=\int\limits_{}{}1\frac{1}{1+u}du=u\log (1+u) + c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why multiplying two negatives gives a positive...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cuz ..mathematicians say so LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2 \times 3 \]\[= 2 \times 3 +0 \]\[=2 \times 3 + 2 \times 0 \]\[=2 \times 3+2 \times (3+3)\]\[=2 \times 3 + 2 \times 3 + 2 \times 3\]\[=3 \times (2+2)+ 2 \times 3\]\[=3 \times 0 + 2 \times 3\]\[=0+2\times 3\]\[=2 \times 3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Everyone takes it for granted. Hardly anyone ever checks or bothers to figure out why.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea this is how education been conducted these days

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Were you math major? i hear that only math majors actually get into this kind of proofs, for example proof 1+1=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes. pure mathematics and physics.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0COOL so you can help me with physics too haha!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why do you have classes so late?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I had no choice, i picked my class on the last day of classes registration because i didn't have many credits when i entered this school.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you go to University of Sydney ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Postgrad. stuff at Oxford

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i had to fight for classes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Isn't it a school though?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL that explained why you have 129 fans here haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you heard of University of California, Davis?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't get the American system.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i am attending Davis ..since i couldnt get into UC berkerley

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is annoying and difficult to deal with..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can't you try and transfer, or do postgraduate stuff...or even try for another course?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can transfer after my sophmore year , but it is really hard to transfer from one college to another college at the same level

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is much easier if you try to transfer from a city college to a University

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your system's quite weird. It seems almost punitive. In the Commonwealth, it's so much easier. We can transfer easily and get exchange placements.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea, i had an internship at a company last summer. One of my colleagues is an immigrant from UK. He told me that the school system here is much more complicated than the one in UK

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it has to do with the fact all universities in the Commonwealth (well, most) are public, and there are links between schools that act as 'feeder schools', although you can choose where you go.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The governments ensure as much as possible there is enough money to allow people places in courses they want (assuming the academics think they can handle the work).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you actually go to lower division schools, elementary, middle and high school, that help you to get into the university you want?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We have nothing like that here. NOw they are evening cutting classes because of the recent budget cut

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, it's not like that. Because the government funds everything, the standards of all the universities have to be the same, which means most people are happy studying at a university that is close geographically to their home. There's no (as far as I'm aware) discrimination in selection.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you want to do research, then you have to get yourself into one of the major ones here. At Sydney, I was taught and worked with people who were at Oxford, Cambridge, Yale, etc.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think there was a mathematician who worked at Stanford too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where did you do your undersgraduate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a guy in my old physics school there who worked at NASA after graduating. He's back now. I think he discovered something to do with black holes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are you going to do when you finish your studies? I get the impression with the US system that you guys study for a general degree and then use it (if you want) to get into a professional school.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OMG. You are surrounded by ...genius man so jealous

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont know...what i am gonna do.. I am majoring in mechanical engineering.. i dont know what i can do with it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well here, mechanical engineers make a lot of dosh.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maybe you could focus on making moolah?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL you mean i design and produce my own moolah?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea that s what i meant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LITTERALLY producing moolah

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, subtleties are lost online.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why not, your Government does.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How about you .what you going to do after post grad

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I'm taking a break and teaching mathematics at a grammar school in Sydney. I'll get back into research a bit later.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you think the earth's magnetic field is gonna turn 180% around

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i asked about the international baccalaureate before because that's part of what I'm teaching.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Re. magnetic field, I think it's supposed to flip, and can do so anytime 'soon', but 'soon' is in astronomical terms, so it could be like, forever.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What kind of research are you working on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gtg i will ttyl man..Thanks for all the help~:)
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