anonymous
  • anonymous
I have been having major issues trying to figure this out...can someone please try and explain to me? F(20) = 40e ^0.032(20) = ~ 75.9
Mathematics
schrodinger
  • schrodinger
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asnaseer
  • asnaseer
What is it that is confusing you?
asnaseer
  • asnaseer
are you confused as to how the left hand side is equal to the right hand side?\[40e ^{0.032(20)} \approx 75.9\]
asnaseer
  • asnaseer
OK, I saw from your previous posting that you don't understand how the left hand side is equal to the right hand side.

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asnaseer
  • asnaseer
Let me try and break it down into steps for you....
anonymous
  • anonymous
I don't understand how they come up with the answer
asnaseer
  • asnaseer
\[\begin{align} 40e^{0.032(20)}&=40*e^{0.032*20}\\ &=40*e^{0.64}\\ &=40*1.8965...\\ &=75.859...\\ &\approx75.9 \end{align}\]hope that helps...
anonymous
  • anonymous
so I ignore the f(20)?
asnaseer
  • asnaseer
My guess is that you were given a function of the form:\[f(x)=40e^{0.032x}\]and were asked to evaluate the function at \(x=20\) so that you would then get:\[f(20)=40e^{0.032(20)}\]
anonymous
  • anonymous
yes, now for e, I would use 2.718?
asnaseer
  • asnaseer
yes you have the correct approximate value for e.
anonymous
  • anonymous
now when I take 0.64*2.718, I get 1.7344
anonymous
  • anonymous
I just don't grasp it!
asnaseer
  • asnaseer
you made a mistake there. it is not 2.718*0.64 but \(2.718^{0.64}\)
anonymous
  • anonymous
ohhhhhhhhhhhhhhhhhh
asnaseer
  • asnaseer
i.e. 2.718 to the power of 0.64
asnaseer
  • asnaseer
don't worry too much if you don't grasp things too quickly. as long as you keep practicing, you will slowly begin to master this subject.
anonymous
  • anonymous
I still get the wrong answer
asnaseer
  • asnaseer
what do you get?
anonymous
  • anonymous
I think I got it now
asnaseer
  • asnaseer
:-)
anonymous
  • anonymous
thank you so much!
asnaseer
  • asnaseer
you are very welcome. I'm glad I could help you.

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