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Type: Posts; User: calc67x

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  1. instantaneous velocity/average velocity from looking at position graph

    https://www.freemathhelp.com/forum/attachment.php?attachmentid=9154&d=1518795876&thumb=1&stc=1

    I was having trouble getting this picture to be the right size, but the left hand side y axis has...
  2. finally got it!

    I tried to post earlier but apparently did not press the correct button.
    I did finally notice that e-2x(K+P0(e2x -1))would equal 0+P0*1 -0.
    This would equal KP0 in the numerator and P0 in...
  3. Actually, I did notice in the denominator, we...

    Actually, I did notice in the denominator, we have K+P0(e2x-1)
    This would be K +P0*e2x-P0
    when multiplied by e-2x we would get 0 +P0*1-P0*0 = P0
    Then the numerator would be KP0 and the...
  4. Dr. Peterson: I corrected the parentheses as you...

    Dr. Peterson:
    I corrected the parentheses as you suggested
    It appears that if you multiply e2x with e-2x (e to negative exp power) you would get 1.
    in that case, KP0 would remain in the numerator...
  5. Can't understand this concept: lim_{x->infty} P(x)=(KP_0 e^{2x})/(K+P_0(e^{2x}-1))

    I had a math problem that I need help with. I already have the answer but I can't arrive at the explanation.

    It is the following:

    P(x)=(KP0e2x)/(K+P0(e2x-1))
    limx -- infinity

    They want us...
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