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Old Jun 23, 2003 | 07:48 PM
  #41  
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http://www.math.com/school/subject1/lesson.../S1U1L10GL.html



wow i was rite
Old Jun 23, 2003 | 07:58 PM
  #42  
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I'm e to the x.
Old Jun 23, 2003 | 08:05 PM
  #43  
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i made it home!



mike
Old Jun 23, 2003 | 08:19 PM
  #44  
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Congrats.
Old Jun 23, 2003 | 08:24 PM
  #45  
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somebody lock this thread
Old Jun 23, 2003 | 08:50 PM
  #46  
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MATH!
Old Jun 23, 2003 | 09:03 PM
  #47  
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Originally Posted by TyresmokinRx7' date='Jun 23 2003, 08:27 AM
Hire you to do what? What do you want to do.
I have a degree from hopkins in mechanical engineering, I'll do anything at this point its getting lean!
Old Jun 23, 2003 | 10:11 PM
  #48  
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Originally Posted by ROTARYROCKET7' date='Jun 23 2003, 10:00 AM
AN interger is non negative because it lies in the set of real numbers. If you go into negative numbers with intergers you are now entering the set of imaginary numbers. Those problems are brain teasers and the way to solve it by trial and error using odd and even numbers. I just failed my math test for Differential Equations.



SOlve this problem:



x^2 means x squared okay..

y' means first derivative.. '' means second and so on

e= eulers equation = 2.7

ln= natural log

cos t^2 + sin t^2 =1 ( identity)

cot = cotangent

sec= secant



IIGht



Find a Particular Solution for



y'' + 4y= 3 csc t ( csc is the cosecant so its 1/sin (t) )



SO the general solution is yc(t) = C1 cos 2t + C2 sin2t



The general solution is off this form because the discriminat b2-4ac ( part of the quadratic equation of the form -b +- (b2-4ac)^(1/2) / 2(a) ). is less than 0. so then the general equation is y= e ^ lambda (t) ( c1 cos u(t) + c2 sin u(t) ) and the discrimant therfore gives us lambda = 0 so e raised to the zero is 1 and then we get C1 cos 2t + C2 sin2t.





Find the coefficiants c1 and c2 byusing variation of parameters where the first condition that needs to be satisfied is



1) v1'y1 + v2'y2= 0

2) v1'y1' + v2'y2' = csc (t) by using trigonometric substituition. Engineers of NP.com gimme this answer!







The answer is y= 3 sint + 3/2 ln | csc t - cot t|sin2t + c1 cos2t + c2 sin2t
WTF? That **** confused the hell out of me!
Old Jun 24, 2003 | 11:40 AM
  #49  
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Yo freestyle reaction i got something for you. Or anyone else solve these basic math problems.







1). iight what is e raised to the natural log of x squared?



2). whats the Intergral of sin x?

3) what is the derivative of (4x sqaured + 5) squared?

4.) what is the half angle formula for cos squared x.?



These are basic ****. YEah u can look up these identities behind your math book. lol.



Yeah intergers are negative my bad but i dont consider them negative cause when ever we use intergers we never really use em with the negatives well at least in my math study. we just say non-zero or something to denote that they are all positives.



I hate imaginary numbers which are the +- of a square root..





***********BONUS*************

For all you smartees

What is the Partial Derivative of F(x,Y) = 3x^2 + xy + 2x^2y^3..

find Fx, Fy, and Fxy....
Old Jun 24, 2003 | 01:55 PM
  #50  
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1. x=2e ???

2. 0 or -cosx i am pretty sure zero

3. 2(4x+5)(8x) = 64x^2 + 80x

4. can't remember



does the bonus want implicit differentiation? i haven't seen a question worded like that. i only took high school calculus

anyways implicit differentiation gives us y' = (-6x - 6x2y^3) / (x + 6y^2 2x^2)



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