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anonymous
 5 years ago
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anonymous
 5 years ago
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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits e^{(sinx)}sinx dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Make a substitution,\[u=\sin x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha i tried..by parts..and substitution..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you're looking for a solution in terms of standard integrals, I'm pretty sure it doesn't exist. You either have to write your answer in terms of the Bessel and Struve functions, or take a series solution, where you take the series expansion for sin(x), the expansion for e^(sin(x)) and then take the Cauchy product of the two series, and integrate termbyterm (which you can do since the series is uniformly convergent). Where did this problem come from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL umm someone asked this question yesterday, and i couldn't solve it.So i am asking you, the boss

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha, well, boss says that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I'm signing out...have a *lot* of work to do ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll answer your ferris wheel question later :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should chillax, it's Friday.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the difference between power and geometric series!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am chillaxing by chit chating with my roomate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Take your time on the wheel question :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A geometric series is one where there's a constant ratio between each successive terms. A power series is different in that it is not always the case that the ratio between two successive terms is constant. In a geometric series, you'd have to successive terms,\[\frac{a_{n+1}(xx_0)^{n+1}}{a_n(xx_0)^n}=\frac{a_{n+1}}{a_n}(xx_0)\]which is only constant if a_(n+1) and a_n give some number independent of n. In general, this isn't the case.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you. I will try to do a little bit more research on my ow :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL just go do your work!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have Lammy below trying to figure something out. When he/she's done, I'm out... .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright i will ask questions later if have any :) Peace~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey loki, i have a question, what test should i use to determine the convergence/divergence of the series sin(x/9)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty}\sin(9/x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think this converges. Let me check.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm distracted because some moron in my building has been playing the pellettest music I've ever heard, all day, all loud and I'm about to go break down doors ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Dude you are the boss, just order your followers to do it for you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm thinking you can set the problem up as proving it divergent using something like a limit comparison test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, had to go talk to the neighbor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problem :) i went through every single tests...but none of them worked

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To use the limit comparison, we need a series that is divergent (since we're suspicious this thing won't converge). When I think 'nonconvergence', the first series I look to to test against is the harmonic series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Should I wait for you to finish typing?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol what if i am suspicious this thing converge... then i will be in trouble

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol, yes, you would be :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Form the ratio\[\frac{1/n}{\sin \frac{9}{n}}\]and note that, in the limit, you get an indeterminate form, \[\frac{0}{0}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is great, because you now have access to L'Hopital's rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So\[\lim_{n \rightarrow \infty}\frac{1/n}{\sin 9/n}=\frac{\lim_{n \rightarrow \infty}1/n}{\lim_{n \rightarrow \infty}\sin 9/n}=\frac{\lim_{n \rightarrow \infty}\frac{1}{n^2}}{\lim_{n \rightarrow \infty}\frac{9\cos 9/n}{n^2}}=\frac{1}{9}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since cos(9/n) goes to 1 as n goes to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, since your ratio was that for a known divergent series with your own series, and because this ratio converged, you know that your series is divergent also.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So at the beginning, you have to guess whether this series diverges or converges?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Pretty much...it's okay, since you then turn towards something that can either prove or disprove your case.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A lot of mathematics is experimental.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But you yourself said you had a feeling that it was divergent...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0On that integral before, you'd have to find more than just a couple of series expansions. I oversimplified it. If you ever see something like that in practice (i.e. engineering) you would use an approximation rule, like Simpson's or something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know it behaves like pseries as n>infinity, so if i had compared it to 1/n^2, it wouldn't follow any of the outcomes for the test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/n^2 is convergent, whereas your suspicion is that sin(9/n) is divergent. It wouldn't have helped.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i used the nth term test, and the limit goes to 0, so i thought it should be convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you read the statement of the nth term test, it says:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If \[\lim_{n \rightarrow \infty}a_n \ne0 \] or if this limit does not exist as n tends to infinity, then\[\sum_{n}^{\infty}a_n \]does not converge.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The nth term test doesn't say that if your sequence tends to zero, your series converges.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It says if it DOESN'T tend to zero, your series DIVERGES.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Take 1/n for example. The limit is 0 as n goes to infinity, but the series itself does not converge.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea, so , as a normal college kid, my suspicion would be convergent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now we come to 'instinct', which is a function of experience. It means you just need practice. You're looking at good questions. Trust your teachers  they'll show you all sorts of things you need to consider.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL what my math professors have taught me the most is HUMOR

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, that *can* help..? :~S

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am stuck at number 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03..that answer is wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does this have to be submitted now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nope, but soon~~>>>> tomorrow night

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, okay. I need to go out, that's all. I'll take a look later.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH alright :) have fun!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you figure it out?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nope haha you told me to chillax

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take your time. just look at it whenever u feel like it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The first one can be 2/361. You have 1/361.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you covered asymptotic equivalence?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is that? might have..not sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know that as n> infin, some terms become negligible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, basically. The definition is this:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But how you determine which term is going to infinity faster

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, there's an order to it. You can see these orders if you make plots of the corresponding functions. Exponential functions will dominate polynomials at some point. You'll see, I'll write out how I worked out your question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you please do number 4 also :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{2^n+n^5+2}{19^{2n}+6^n+3} \iff \frac{2^n}{(19^2)^n}\]since in the numerator, the exponential dominates the quintic and 2, and in the denominator, (19^2)^n dominates 6^n+3 (19^2 > 6).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The equivalence sign should be replaced with ~. For some reason it doesn't show up in the equation editor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought n^5 is going to infinity faster

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, if you plot n^5 and 2^n, you'll see 2^n overtake pretty quickly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But listen, you can always check domination by using the definition. I'll scan in something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You asked for tricks in math  I love asymptotic equivalence when it comes to limits  along with L'Hopital's rule. It saves your butt.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{2^n/(19^2)^n}{r^n}=\left( \frac{2}{361r} \right)^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then choose r = 2/361 to ensure a convergent limit that's nonzero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can show that the original expression is asymptotically equivalent to the one we use in the end by taking the ratio between the two and showing the ratio approaches 1 as n approaches infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does exponential function always dominate quadratic, quintic ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a^n>n^m\] for all a >1 and positive exponent. You have positive exponent because of the problem, and here you have the numbers 2, 6, 19. It is the case,\[19^n>6^n>2^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Remember, when you're dealing with limits, the definition of limits makes use of "for *large* n...so you have to be thinking about the general properties of functions once they go through the motions and their domains start heading to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:P can you prove a^n >n^m

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but I'll show you a plot...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They're all monotonic increasing functions, so when it gets to a point where the rate of one outstrips the other, the one with the slower rate of increase is dead in the water.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that a free program haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do i input multiple graphs

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just type them in on the one page. y=x^2, y=x^4, etc. They should show up.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know if you've done the last one, but I have \[r=\left( \frac{2^{7/10}}{19} \right)^3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what you up to right now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I should be doing my work...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Seriously going offline...it's getting dark and I've done nothing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.youtube.com/watch?v=C3v3aThfFA just when you have time loki, check out the electric car my friends from the university i wanna go built, it won xprize 2nd place

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL when is your presentation ? i forgot~~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@ oktal Nice car ~ haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll check it out  I saved the link. And presentations on Wed and Fri >< And to both of you  there's a site you should check out if you haven't: wolframalpha.com. You can check out math questions, like integrals and such.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, when I found out they were builing it, i could not believe it, and when it won google xprize 2nd place, I was stunned

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, i will see it, thanks bye

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol you still have few days to prepare for it no worries. you should chill with us

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, let him work, he has a life to manage...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL he has troublesome neighbors to manage as matter of fact :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and actually it belongs to the university now, as it funded the building of it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's pretty cool. they should at least be able to take it out on weekends ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"Hey ladies, check out my revolutionary design..." ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol. oktal, check out a nerd link above for free plotting software, GeoGebra.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they did it few times, drove around town :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could have used it to check out that absolute value problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I'm out...seriously need to get sorted. Later!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Laters ~~ i am going to to bed too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nice meeting you okatal :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, see you, nice to meet you too dichalao

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hi Loki, can you help me out with this problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What did you enter before?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0brb...when's this due?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, are these your physics questions that aren't due till the end of the week or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is due on tuesday before lecture

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your value is what you get when you work out magnitude, since \[a=\frac{v^2}{r}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.011.7 is what i got from that equation..but it is wrong..probably becasue this is not uniform circular motion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a vector problem, and you're going to have to subtract an initial vector from a final vector...no, this is uniform circular motion...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So,\[a=\frac{\Delta v}{\Delta t}=\frac{v_2v_1}{t_2t_1}\]where the a and v's are vectors.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The velocity in uniform circular motion is\[v=(r \omega \sin \theta, r \omega \cos \theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why is there a negative sign for V_x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we know, from definition of angular speed and uniform circular motion that,\[\omega = \frac{\theta }{t} \rightarrow \theta= \omega t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So \[v=(r \omega \sin (\omega t), r \omega \cos (\omega t))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a minus sign from when we take the derivative of the xcomponent of displacement with respect to time. That component is a cosine.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since\[v = r \omega \rightarrow \omega = \frac{v}{r}\]so\[V=(v \sin \frac{vt}{r}, v \cos \frac{vt}{r})=v(\sin \frac{vt}{r}, \cos \frac{vt}{r})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should be able to work out two vectors at different times and then subtract.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that making sense? I hope it is 'cause I just got up and I'm doing the onthefly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in fact, we haven't "learned" angular speed .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a minus sign on the last expression outside sine...typo.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It goes like this: let the arc of a circle of radius r have length (distance) l. Then that distance is\[l=r \theta\]when theta s in radians. Take the derivative with respect to time on both sides to get\[\frac{dl}{dt}=\frac{d(r \theta)}{dt}=r \frac{d \theta }{dt}+ \theta \frac{dr}{dt}=r \frac{d \theta }{dt}:=r \omega\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dr/dt is zero by construction of the problem > the radius is fixed and therefore unchanging in time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Angular speed is the rate of change in the angle.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there something wrong with this site..i didn't reply"kk"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do i just plug in t=0,and t=0.6?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it came through...and there's always something wrong with this site.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would plug 0 and 0.6.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me check something before you waste a turn.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there a way solving this problem without using angular speed?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Not off the top of my head.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\Delta v = 5.3(\sin (1.325), \cos (1.325)1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a=\frac{\Delta v}{\Delta t}\approx \frac{1}{0.6}(5.14,4.01) \approx (8.57, 6.68)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am going to ask my professor... see if there is another way to do it..since we haven't covered angular velocity....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. At least it worked. I hate online submission. I used to hate it as an undergrad.; you can never argue a point if it's wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anyways, I have to sign out and get some work done. Happy mathing :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL you meant physicsing?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright thanks for helping!! have a great day...still have an essay to do..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0At least they run you into the ground.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL true~ see ya ~buddy
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