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I was having trouble getting this picture to be the right size, but the left hand side y axis has...
Type: Posts; User: calc67x
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...
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...
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...
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...
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...