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01022013, 10:35 AM  #9181 
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Join Date: Oct 2008
Location: UK/NO

How do you rearrange:
v = 1 + bx + cx² to get a straight line?
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01072013, 06:25 PM  #9182 
brought dipping sticks
Join Date: Apr 2011
Location: Onterrible

When doing linear equations, how would you write the equation for a vertical line?

01072013, 07:25 PM  #9183  
I'm too old for this ****
Join Date: Oct 2007
Location: OH

Quote:
x=something
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01072013, 07:28 PM  #9184  
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Join Date: Feb 2007

Quote:
x = constant, much like y = constant is a horizontal line!
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01122013, 03:27 AM  #9185 
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Join Date: Sep 2009
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I have just started integration and I'm still a little confused... how would you integrate this?
(square root of x  1/x)^2
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01122013, 03:31 AM  #9186 
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Is the entire square root being squared?

01122013, 06:55 AM  #9187  
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Join Date: Sep 2009
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Quote:
Ok yeah that wasn't very clear... it's [(square root of x)(1/x)]^2
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Signatureless. And this doesn't count. Because I said so. Last edited by sherry07 : 01122013 at 06:57 AM. 

01122013, 07:07 AM  #9188 
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Join Date: Mar 2007
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So (√(x)  1/x)^2.
Foil and simplify: (√(x)  1/x)(√(x)  1/x) = √(x)^2  2(√(x)/x) + 1/x^2 = x  2x^(1/2) + x^2. So for the indefinite integral we have ∫ x  x^(1/2) + x^2 dx = ∫x dx  ∫2x^(1/2) dx + ∫x^2dx = (1/2)x^2  4x^1/2  x^1 + C Last edited by MakinLattes : 01122013 at 07:08 AM. 
01132013, 06:37 AM  #9189 
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How do you find a dy/dx expression when:
sin(xy) = y + x ?
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01132013, 06:58 AM  #9190  
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Join Date: Feb 2006

Quote:
Implicity differentiation dy/dx=Fx/Fy, so take the sin to the right and simply partial differentiste to x and y.
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Quote:
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01132013, 07:31 AM  #9191 
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Thank you!
The answer key just said: (to a simlar question; there's no answer key for exam sets) 1. sin(xy) = x 2. [x(dy/dx) + y] cos(xy) = 1 3. (dy/dx)=[(cos(xy))^1  y] / x Though the method you mentioned yielded the same answer on that question. I'm curious about what happens between step 1. and 2. here...
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Last edited by sfaune92 : 01132013 at 07:39 AM. 
01132013, 08:49 PM  #9192  
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Join Date: Jun 2007
Location: Bangor, Norn Iron/Manchester

Quote:
cos(xy)[xdy/dx+y]=dy/dx+1 dy/dx(xcos(xy)1)=1ycos(xy) dy/dx=(1ycos(xy))/(xcos(xy)1) 

01142013, 08:50 AM  #9193 
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Join Date: Sep 2008
Location: London

Hey guys, doing a computing report and was wondering if there's a way to get the convergence of this series for different values of a? Is there a way to express this limit in terms of a? I'm still not great with just plucking limits out of thin air
Any help is appreciated!
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01142013, 09:50 AM  #9194 
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Join Date: Sep 2006
Location: Glasgow.

Sorry, this is a really stupid question but I'm having a hard time visualising this question in my head. Can anyone explain it to me a little better?
A pendulum with a cord length, r=0.5m, swing on a vertical plane. When the pendulum is in the 2 horizontal positions of theta=90' and theta=270', its speed is 5.00ms^1 
01142013, 11:20 AM  #9195  
° ͜ ͡°
Join Date: Sep 2008
Location: London

Quote:
Imagine the pendulum upside down, completely vertical, with the weight at the top  this is with θ = 0°. Now rotate it clockwise 90°, that's the first horizontal position, and the other is, yeah, 270°. It's saying that each time θ=90°, v (instantaneous speed) is 5m/s. ...what's the actual question though? :p Quote:
Dayum that looks pretty fascinating... Are you in 3rd year or something? I've seen some stuff about this in IOP news, I never understand anything in physics news though
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When I think of the perpetual journey through life When it always feels like autumn The wind moves slowly to the north And the flowers die Rain falls in my dreams         Last edited by papershredder : 01142013 at 11:23 AM. 

01162013, 06:43 AM  #9196 
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Location: UK/NO

I've got a simple stats question...
So I've calculated mean, variance, standard deviation, and standard error of the mean. How do I find "the 70% confidence limit for the true value"? EDIT: Another question... How do you linearise this: t² = (d² + 4h²) / v² Given that t: dependent variable d: independent variable h: constant v: constant (Or is it possible to have v² as a function of d²?)
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Last edited by sfaune92 : 01162013 at 12:36 PM. 
01162013, 01:15 PM  #9197  
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Location: London

Quote:
I was taught something about a 67% confidence level, I'm assuming this is the same thing  if you have your standard deviation s then the error on your average value (at the 67% confidence level) should be s/sqrt(N) where N is the number of values used to calculate the standard deviation. And I'm not really sure what that second question is asking is that something to do with taking the first order of the Taylor expansion?
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When I think of the perpetual journey through life When it always feels like autumn The wind moves slowly to the north And the flowers die Rain falls in my dreams         Last edited by papershredder : 01162013 at 01:17 PM. 

01162013, 01:30 PM  #9198 
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Location: UK/NO

Thank you.
Also, the second part seems to be asking of t as a function of d. I don't think series has anything to do with it as it is not included in that module. I tried using logs as well, but it didn't go over to well because of that annoying plus on top of the denominator. Regarding your question on the previous page, is it based on a binomial series?
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01162013, 11:20 PM  #9199 
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Join Date: Sep 2008
Location: London

Well can't you just square root it? t = (d+2h)/v ?
which is essentially t = (1/v)d + (2h/v) in straight line form and uh, my question was finding the limit/convergence of that series, in terms of a :/ still stuck on it lol, need to finish writing the report by sunday :'(
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When I think of the perpetual journey through life When it always feels like autumn The wind moves slowly to the north And the flowers die Rain falls in my dreams         
01162013, 11:39 PM  #9200 
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Join Date: Sep 2008
Location: Tampa

You can't square root it because those values are squared independently and then added. To linearize it you have to get both the dependent and independent variables to a power of 1, but I'm not actually sure how to do that for that problem (been through calc 3 and math methods, I'm ashamed of myself).
What class is that question for? 
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