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arra

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  • Gender
    Male
  • Exams
    May 2012
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    India
  1. Yes, there is a plus c, but the plus the constant of integration is on the numerator and not independent of the fraction as a whole.
  2. Yeah, that sounds about right. In the third question I got (tcost - sint)/t or something like that, I think.
  3. I think I got 300 moving to the left. Because the positive field would be repelling and the negative field would be attracting therefore they would add together and become 300, no?
  4. The Kinematics question and the one with resolution and all? Yeah, I did. For the force acting I got 3.5 x 10^4 or something like that, for the time taken to stop something like 4.2 seconds, and the distance between galaxies to the order of 10^20.
  5. 40 Volts No. The axis was to the power of -3, so it was actually .04 volts. and for the wavelength one I'm pretty sure it is 4 lambda. The one that I found reeally tricky was the force diagram. I initially had selected C, where the normal force and reaction force were equal. However, the actual answer is D, where the reaction force is less than the normal force. Because the Force at an angle is resolved and added to the reaction force to become equal
  6. If k=1, the integral test fails because we are dividing the integral by -k+1. You can rewrite the sum(u_n - L) as sum(u_n) - sum(L). Since L is a constant, the series must diverge unless L is zero. However, L is 3/2. Anyways, I hope it's okay on the third question I used x instead of t. I made sure to write "Let x=t" in the beginning of my answer. The only thing I did differently, and I think it's correct not sure, is I said that for k = 1 the integral becomes ln(lnn) which diverges to infinity. It's not 1/(k-1) because if you take lnx as u, the integral would become du/u which can't be inte
  7. If you actually understand your theory of the firm you should without doubt attempt it. It's an excellent way to score marks. If not don't touch it with a 10 foot pole.
  8. For the sequences all but the last one converge. The last one diverges because the absolute value of the sequence tends towards a specific value (3/2) and hence it alternates between the positive and negative of that value. Because this value is not 0, it converges. For the epsilon, I got N = 17 and 14500 (I'm not sure if the latter is remembered correctly, I think it is). I did not round the 17 up to 18 because N is, by definition, the "just below" the minimum necessary for epsilon to as low as the given figure (i.e. an epsilon exists for which e < (given value) for which n > N). Friend
  9. It's easy to do this by just counting the different permutations, but how can I do this explicitly/algebraically? I'm not sure what else you want me to explain. I believe my method is deductive and doesn't require listing every permutation. I also deductively found the general formula. I think what he's saying -- if he had the same problem that I had -- is that his "general" formula is still in series notation. However, with a little bit of algebraic manipulation you can - I am 100% sure -- bring this into a nice and closed equation
  10. A google search can tell you more than us. No less... His major contributions were to the neoclassical economics, i.e. what you might link with "conservative" politicians (though he liked to think of himself as the liberal in the classical sense).
  11. I sat for a while and actually managed to get a general formula that I know is right for the first four values (11 21 12 22) because I had already calculated them but it's quite complicated and I'm not even sure it's correct beyond that. Teacher is rest but probability is not her thing so not particularly sure how to check it, I mean I know that my logic is right just the formula does nottt look too pretty.
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