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anonymous

  • one year ago

Whats the trick to solve for x with this equation? 0 = (2x-2x^3) E^(1-x^2) x = ???

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  1. nincompoop
    • one year ago
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    what is this (2x-2x^3) E^(1-x^2)

  2. nincompoop
    • one year ago
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    use latex

  3. Owlcoffee
    • one year ago
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    \[(2x-2x^3)e ^{(1-x^2)}\]

  4. anonymous
    • one year ago
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    its a derivative of f[x] = x^2 E^(-x^2​ + 1) apparently I have to examine the derivative and find the zeros, without a calculator, to plot a chart.. I can see just looking at the equations that 0,-1,1 will give me the zeros.. but I can't explain why that would be true..

  5. nincompoop
    • one year ago
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    why would you need a calculator?

  6. anonymous
    • one year ago
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    sorry, whats latex ?

  7. anonymous
    • one year ago
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    why a calc, coz I'm clueless of course..

  8. nincompoop
    • one year ago
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    you probably need to use log

  9. anonymous
    • one year ago
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    Yeah I was worried about that..

  10. anonymous
    • one year ago
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    To get rid of that E^ ?

  11. Owlcoffee
    • one year ago
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    No, 1 and -1 are not zeros.

  12. nincompoop
    • one year ago
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    they are zeroes

  13. anonymous
    • one year ago
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    had me there for minute.. thought I sent the wrong equation

  14. nincompoop
    • one year ago
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    I am confused how you can "see" the values without solving it.

  15. Owlcoffee
    • one year ago
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    He may have looked at the graphical representation.

  16. anonymous
    • one year ago
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    It just looks obvious to me.. I usually plug 1 and 0 in to get an idea of the function and if 1 zeros, -1 usually does too when squares are involved.

  17. nincompoop
    • one year ago
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    then use that as your mathematical solution "it is obvious"

  18. Owlcoffee
    • one year ago
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    Nin, don't be so rude.

  19. nincompoop
    • one year ago
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    @UnkleRhaukus @UsukiDoll @rational

  20. Owlcoffee
    • one year ago
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    if you want to find the zeroes of the function: \[(2x-2x^3)e ^{1-x^2}=0\] is clearly and application of the hankelian preoperty, wich means: \[2x-2x^3=0\] \[e ^{1-x^2}= 0\]

  21. anonymous
    • one year ago
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    thats a cool one.. first time I've heard of that.. but it makes sense yes.. if a*b = 0 then a or b = 0

  22. Owlcoffee
    • one year ago
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    yes, now all you have to do is find the zeroes of those smaller and simpler parts of the function.

  23. anonymous
    • one year ago
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    silly question ... does that also apply for division?

  24. Owlcoffee
    • one year ago
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    Not really, I think you mean: \[\frac{ a }{ b }=0 <=>a=0,b=0 \] Thing is that does not work, because the expression a/b has the followin condition: \[\frac{ a }{ b }=0 <=> a \in \mathbb{R}, b \neq 0\] because you can't have a zero in the denominator.

  25. amoodarya
    • one year ago
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    hint : \[e^{something}\neq0\]

  26. Owlcoffee
    • one year ago
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    Do you mean that it does no have zeroes?

  27. anonymous
    • one year ago
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    ah amoodarya, thnx.. yea E^0 = 1

  28. nincompoop
    • one year ago
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    can you prove that \(e^{1-x^2} = 0 \)

  29. Owlcoffee
    • one year ago
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    I can.

  30. anonymous
    • one year ago
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    (2x-2x^3) is all we need though right? because if that is 0 the whole equation is 0

  31. Owlcoffee
    • one year ago
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    oh no, wait, log 0 is not possible.

  32. nincompoop
    • one year ago
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    I know it is not possible, but you said you can

  33. anonymous
    • one year ago
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    yeah, I tried to log both sides earlier.. when one side was 0, can you do that even temporarily ?

  34. Owlcoffee
    • one year ago
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    Sorry, thought there would be a constant.

  35. nincompoop
    • one year ago
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    this is beyond my skill set

  36. anonymous
    • one year ago
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    so if the zero must be in the \[0 =(2x - 2x^3)\] then \[-2x = 2x - 2x^3 -2x\] \[-2x = - 2x^3 \] \[-2x/-2 = - 2x^3 /-2 \] \[x = x^3 \] \[x/x = x^3/x \] \[1 = x^2 \] \[1/x = x^2/x \] x = 1/x Does this look right?

  37. anonymous
    • one year ago
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    lol, it cant be

  38. Owlcoffee
    • one year ago
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    Well, not really, just because : \[e ^{f(x)}=0\] Will always have positive images of x, because it is a lil' more complex than the expression: \[f:f(x)=e ^{x}\] therefore, the only part of the function that'll determine the zeroes will be: \[(2x-2x^3)\]

  39. nincompoop
    • one year ago
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    n0

  40. nincompoop
    • one year ago
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    I am interested to learn this

  41. Owlcoffee
    • one year ago
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    To find the zeroes of the expression: \[2x-2x^3=0\] You take common factor 2x: \[2x(1-x^2)\] And again, by the hankelian propety: \[2x=0 \] \[1-x^2 =0\]

  42. anonymous
    • one year ago
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    I think the solution might also be here.. x = x^3 x = x x x

  43. nincompoop
    • one year ago
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    uh...

  44. anonymous
    • one year ago
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    lol, maybe not..

  45. nincompoop
    • one year ago
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    stop trolling

  46. UsukiDoll
    • one year ago
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    ummm.... is this solve for x or find the derivative ?

  47. anonymous
    • one year ago
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    solve for x

  48. UsukiDoll
    • one year ago
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    Alright... I'm going to use what @Owlcoffee suggested earlier.

  49. UsukiDoll
    • one year ago
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    \[(2x-2x^3)e ^{1-x^2}=0\] so we're going to solve for x... more like find all x's that satisfy the equation... this feels like find the equilibrium or something. \[(2x-2x^3)=0\] \[e ^{1-x^2}=0\]

  50. nincompoop
    • one year ago
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    good luck

  51. UsukiDoll
    • one year ago
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    for the first equation we have a 2x in common \[(2x-2x^3) \rightarrow 2x(1-x^2)\] factoring again yields \[2x(1+x)(1-x)=0\] just through analysis and looking at it x = 0,1,-1

  52. UsukiDoll
    • one year ago
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    \[e ^{1-x^2}=0\] if I take the ln on both sides because e^x and ln x are inverses of each other, that e should go away ... the problem is \[e^{\ln(1-x^2)}=\ln 0 \] 1-x^2 = ln 0 and ln 0 is undefined >:/

  53. UsukiDoll
    • one year ago
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    in fact ln (negative any number) is also undefined.

  54. nincompoop
    • one year ago
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    the easy part is always the best part, yes?

  55. Owlcoffee
    • one year ago
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    Nin, what the hell? Usuki made a great job explaining it very very simple.

  56. UsukiDoll
    • one year ago
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    O____________________________________o

  57. anonymous
    • one year ago
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    would it not be okay though if \[E^(1-x^2) = 1\] because the zeros on the other term would satisfy what we need

  58. UsukiDoll
    • one year ago
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    do you mean \[e^{1-x^2} = 1\] ?

  59. anonymous
    • one year ago
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    yeah I think if x = 1,0,-1 that equation = 1 , and that's okay.

  60. anonymous
    • one year ago
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    or E

  61. nincompoop
    • one year ago
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    I still do not get E what is E

  62. anonymous
    • one year ago
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    the exponential constant = E

  63. nincompoop
    • one year ago
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    so it is not Euler's constant?

  64. UsukiDoll
    • one year ago
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    well let's supposed we are given that equation instead \[e^{1-x^2} = 1\] so taking the ln on both sides yields \[e^{\ln(1-x^2)} = \ln 1\]

  65. anonymous
    • one year ago
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    sorry, yes Eulers

  66. UsukiDoll
    • one year ago
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    ? What does Euler's Method have to do with this? That's differential equations

  67. nincompoop
    • one year ago
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    you're dazzling as always, @UsukiDoll did you get your degree yet?

  68. UsukiDoll
    • one year ago
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    that question has nothing to do with what we are trying to solve. Please delete it.

  69. nincompoop
    • one year ago
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    :)

  70. Owlcoffee
    • one year ago
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    If you know so much, Nin, why aren't you helping?

  71. UsukiDoll
    • one year ago
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    ok so e is the mathematical exponent.... it's also Euler's number... and it's default value is 2.71... but enough with that.. let's go back to what we're doing...this is going off topic and it's not helping hughfuve

  72. UsukiDoll
    • one year ago
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    so assuming we have \[e^{\ln(1-x^2)} = \ln 1\] since ln 1 -> 0 \[1-x^2=0\] \[(1+x)(1-x) = 0\] \[x=1,-1\]

  73. anonymous
    • one year ago
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    Is it a valid technique to solve for that first half of the equation... (2x - 2x^3) and then take the 3 zeros that you found.. and plug them into the second part of the equation. Or would that be frowned upon?

  74. UsukiDoll
    • one year ago
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    I'm trying to think if this is going to work... if I plug in x = 1 on the original equation... I will have \[e^{\ln 0} = 0\] I know ln 0 is undefined.. but \[e^x \] and \[\ln x \] are inverses of each other so I can use that to make the equation hold true. wait let me see that suggestion..

  75. UsukiDoll
    • one year ago
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    our zeros from the first equation x = 0,1,-1 x=1 is going to be a problem if I put it in the second equation

  76. UsukiDoll
    • one year ago
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    \[e ^{1-x^2}=0\] let x = 1 \[e ^{1-(1)^2}=0\] \[e ^{1-1}=0\] \[e ^{0}=0\] \[1 \neq 0\]

  77. UsukiDoll
    • one year ago
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    x = -1 will produce the same issue to because (-1)^2 - > (-1)(-1) = 1 Same thing will occur

  78. Owlcoffee
    • one year ago
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    Thing is, if you plug the zeroes you found, say for instance, x=0: \[f(0)=(2(0)-2(0)^3)e ^{1-0^2}\] you'll end up having: \[f(0)=0.e^1\] which is just : \[f(0)=0\] And that'll happen for 1, and -1. it will make the whole function a "0", but it will not make e^(1-x^2) equal zero.

  79. UsukiDoll
    • one year ago
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    and we can't use x = 0 either you will end up with \[e^1 = 0 \rightarrow 2.71 \neq 0 \]

  80. anonymous
    • one year ago
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    but does E^(1-x^2) have to equal zero? if the full equation is \[0 == (2x-2x^3) (E^1-x^2)\] then even if E^(n) = 1 or E, is our equation not still a valid 0?

  81. anonymous
    • one year ago
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    oops on my equation.. its supposed to have everything after E in the exponent.

  82. UsukiDoll
    • one year ago
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    the zero's we found for x =0,-1,1 is not going to make \[e^{(1-x^2)}=0\] true for 2 reasons. you will either have \[e^0 \neq 0 \rightarrow 1 \neq 0\] or \[e \neq 0 \rightarrow 2.71 \neq 0 \]

  83. UsukiDoll
    • one year ago
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    but I think if you combine those two together... the left side of the equation the one with the 2x should make it go 0... yeah let's try that MAHAHAHAH!

  84. anonymous
    • one year ago
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    I guess I'm not understanding why we are searching for 0 on that portion. If A * B = 0 right, then only A OR B needs to be zero.

  85. Owlcoffee
    • one year ago
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    Thing is the understanding you have for the function: \[f:f(x)= e^x\] if you look at the graph, you'll see that it has no zeroes.

  86. UsukiDoll
    • one year ago
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    \[0 =(2x-2x^3) (e^{(1-x^2)})\]

  87. UsukiDoll
    • one year ago
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    so if I substituted x = 0 \[0 =(2(0)-2(0)^3) (e^{(1-(0)^2)})\] \[0 =(2(0)-2(0)) (e^{(1-0)})\] \[0 =(0-0)) (e^{(1)})\] \[0=(0)(e)\] e is approximately 2.71 0=(0)(2.71) 0=0

  88. UsukiDoll
    • one year ago
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    it's like we need that \[(2x-2x^3)\] portion to help us get rid of the e ... otherwise we're stuck because x = 1 yields (2-2) -> (0)

  89. anonymous
    • one year ago
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    totally, cant escape it hey

  90. UsukiDoll
    • one year ago
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    I think .. splitting the problem up is only necessary to find the zeros... but we need the whole problem together and have x = 0,1,-1 in order for that equation to hold true. If we keep the split up version... something isn't going to cancel and will lead to something false.

  91. anonymous
    • one year ago
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    I think that will be an adequate solution..

  92. anonymous
    • one year ago
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    factor out the first part of the problem, and claim the 2nd term as basically irrelevant.

  93. UsukiDoll
    • one year ago
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    yeah.. because by setting the second equation to 0 , you will run into problems.. because ln 0 is undefined. If that's the case, just stick with the first part of the equation.

  94. UsukiDoll
    • one year ago
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    I've done this before when I had to find the equilibrium for Mathematical biology... if we come across something that's undefined...ignore it and focus on the problem that will give us the 0's .. in this case it was the first part of the equation which gave us x =0,1, -1.. plugging those x values back into the whole equation, will make your equation 0=0 which is true

  95. anonymous
    • one year ago
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    Sorry, I guess I should have mentioned, I just realized it might be important to know, there is a fixed domain too by the way.. it's +2 .. -2 so we dont need to worry about - infinity

  96. UsukiDoll
    • one year ago
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    ok no problem.. the domain lies in the x-axis

  97. UsukiDoll
    • one year ago
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    |dw:1434530843604:dw|

  98. UsukiDoll
    • one year ago
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    |dw:1434530877368:dw|

  99. anonymous
    • one year ago
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    So I am thinking, within the domain there is no way for E^anything to become 0

  100. UsukiDoll
    • one year ago
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    well e itself is 2.71 and e^0 is 1

  101. anonymous
    • one year ago
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    So I can claim then as there is no way for E^(1-x^2) within the domain (-2,2) to become 0, then the zero is to be found in the other part of the equation.

  102. UsukiDoll
    • one year ago
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    yes the zeros are found in the other part of the equation .

  103. anonymous
    • one year ago
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    Thanks Usuki.. you're awesome. and thanks to everyone who came to help.. much appreciated.

  104. UsukiDoll
    • one year ago
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    and if we have x = 0 we end up with e which is 2.711 and for x =1,-1 we have e^0 which is 1... I am unable to have that equation go to 0 .

  105. UsukiDoll
    • one year ago
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    I think I did way more than the Qualified Helper... if you've paid for this question... you got ripped off :(

  106. UsukiDoll
    • one year ago
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    actually it's @Owlcoffee and me doing all the work

  107. UsukiDoll
    • one year ago
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    but still if I had to pay for qualified help, I would expect to have extraordinary result and expert explanations, not chit chat :/

  108. anonymous
    • one year ago
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    lol, well, you saved the day then :)

  109. Owlcoffee
    • one year ago
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    Sorry, i wish i could've helped better :(

  110. anonymous
    • one year ago
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    true, sometimes the help is not so helpful.. it costs. 10 a question I think

  111. UsukiDoll
    • one year ago
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    sorry 10 owlbucks is $2 I think... you didn't pay that much did you? I notice your question was highlighted in yellow

  112. UsukiDoll
    • one year ago
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    @Owlcoffee you did a good job too.

  113. anonymous
    • one year ago
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    owlcoffee, yes you did great.

  114. UsukiDoll
    • one year ago
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    but my main concern is whether or not you have already paid for qualified help... because if you did, I would ask for a refund and re-evaluation of the qualified helper in question

  115. anonymous
    • one year ago
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    actually. that email address they say you get a refund on.. doesnt work.. it bounces.

  116. UsukiDoll
    • one year ago
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    :(... but that highlighted yellow question meant that you've paid for qualified help right?

  117. Owlcoffee
    • one year ago
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    You paid for a qualified helper!?

  118. UsukiDoll
    • one year ago
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    his question was highlighted in yellow... unusual since free questions are in grey

  119. anonymous
    • one year ago
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    yes, I just figured it was too late at night and bad timing.

  120. UsukiDoll
    • one year ago
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    =( get a refund...pm a moderator and explained what happened. You've been ripped off

  121. anonymous
    • one year ago
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    I pay for most my questions.. they're not always handled too well. but often I get a very conscientious helper..

  122. UsukiDoll
    • one year ago
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    not for this one... this qh was chit chatting and barely did anything.

  123. UsukiDoll
    • one year ago
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    right @Owlcoffee ?

  124. Owlcoffee
    • one year ago
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    I believe qualified helpers should be evaluated by the help given and not smartscore. Nincoops is sadly, not a "helper". Glad you came @UsukiDoll I was about to just leave this question here.

  125. UsukiDoll
    • one year ago
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    thanks :) on top of that nin confessed to not knowing and had the nerve to tag me and 2 99ers who I am fans with. Like, if you're a qualified helper... you should know what the question is already talking about and guide the person through the whole process.

  126. UsukiDoll
    • one year ago
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    qualified helper <--- know what's going on already...and be able to explain and clarify, not omg I have no idea what'a going on ... let me just tag someone else. That's not being a qualified helper. That's a qualified FRAUDULENT helper.

  127. Owlcoffee
    • one year ago
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    so, regardless of that, if you ever need help you can tag me or @UsukiDoll and we'll help you as soon as we can. @hughfuve

  128. anonymous
    • one year ago
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    guys thats's awesome thank you, I kind of dont want to rock the boat around here.. I really need the help right now.. I'm like 4 chapters into a 15 chapter course and I'm already hitting the halfway timer.

  129. UsukiDoll
    • one year ago
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    it depends on the question...there are some topics that I've forgot though since it has been a very long time .

  130. anonymous
    • one year ago
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    got about 5 weeks to get through the rest of it.

  131. UsukiDoll
    • one year ago
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    is this a summer school course?

  132. anonymous
    • one year ago
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    yeah

  133. UsukiDoll
    • one year ago
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    I remember having to re-take college algebra during the summer.. it was crazy intense 6 weeks, but I already knew the material so it wasn't that bad. I had to re-take because the professor I had during the spring semester was tough as nails.

  134. anonymous
    • one year ago
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    its a disaster of a curriculum for calculus using mathematica, but someone has butchered it up, and everything is out of order..

  135. anonymous
    • one year ago
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    at least my professor is pretty cool.. having an retriceof a teacher would be awful

  136. UsukiDoll
    • one year ago
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    is it calculus 1,2,3, or 4?

  137. anonymous
    • one year ago
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    I think it';s an entry level its 234 business calc

  138. UsukiDoll
    • one year ago
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    o-------o. Oh I'm horrible at Business Calculus . I used to be a business major.

  139. UsukiDoll
    • one year ago
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    It's strange... Business Calculus makes me look like an idiot, but regular Calculus is easy.. I passed all 4 sections with above average grades.

  140. anonymous
    • one year ago
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    but its just poorly explained.. if I didnt have calc videos, I'd be in bad trouble. they just throw the equations at you.. say.. this is what mathetmatica did.. now explain it..

  141. UsukiDoll
    • one year ago
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    Mathematica is a computer program.. I had to use it for Calculus II

  142. anonymous
    • one year ago
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    like they're throwing derivatives at you.. in chapter 3, and asking you what they are, when they dont even introduce them until chapter 4.

  143. UsukiDoll
    • one year ago
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    hmmm that could be the definition of the derivative... I had to use that long beast before using the real derivative techniques.

  144. UsukiDoll
    • one year ago
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    \[\frac{f(x+h)-f(x)}{h}\]

  145. Owlcoffee
    • one year ago
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    \[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\]

  146. anonymous
    • one year ago
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    yes, that part was ugly.

  147. anonymous
    • one year ago
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    lol, man was that ugly

  148. UsukiDoll
    • one year ago
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    that looks like mean value theorem xD or the rate of change formula

  149. Owlcoffee
    • one year ago
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    it is the definition of derivative. just a variation of it, there are 3, I think.

  150. anonymous
    • one year ago
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    is the other the slope formula?

  151. Owlcoffee
    • one year ago
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    By definition, the derivative gives you the slope of the tangent line on any point of a function.

  152. anonymous
    • one year ago
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    I guess thats uski's difference quotient up there too = slope

  153. anonymous
    • one year ago
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    okay.. time to ask another Q ...

  154. UsukiDoll
    • one year ago
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    I might be away from keyboard. I have to do something.

  155. anonymous
    • one year ago
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    starting a new thread.. all good.

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