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anonymous

  • 4 years ago

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

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  1. asnaseer
    • 4 years ago
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    What is it that is confusing you?

  2. asnaseer
    • 4 years ago
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    are you confused as to how the left hand side is equal to the right hand side?\[40e ^{0.032(20)} \approx 75.9\]

  3. asnaseer
    • 4 years ago
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    OK, I saw from your previous posting that you don't understand how the left hand side is equal to the right hand side.

  4. asnaseer
    • 4 years ago
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    Let me try and break it down into steps for you....

  5. anonymous
    • 4 years ago
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    I don't understand how they come up with the answer

  6. asnaseer
    • 4 years ago
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    \[\begin{align} 40e^{0.032(20)}&=40*e^{0.032*20}\\ &=40*e^{0.64}\\ &=40*1.8965...\\ &=75.859...\\ &\approx75.9 \end{align}\]hope that helps...

  7. anonymous
    • 4 years ago
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    so I ignore the f(20)?

  8. asnaseer
    • 4 years ago
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    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:\[f(20)=40e^{0.032(20)}\]

  9. anonymous
    • 4 years ago
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    yes, now for e, I would use 2.718?

  10. asnaseer
    • 4 years ago
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    yes you have the correct approximate value for e.

  11. anonymous
    • 4 years ago
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    now when I take 0.64*2.718, I get 1.7344

  12. anonymous
    • 4 years ago
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    I just don't grasp it!

  13. asnaseer
    • 4 years ago
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    you made a mistake there. it is not 2.718*0.64 but \(2.718^{0.64}\)

  14. anonymous
    • 4 years ago
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    ohhhhhhhhhhhhhhhhhh

  15. asnaseer
    • 4 years ago
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    i.e. 2.718 to the power of 0.64

  16. asnaseer
    • 4 years ago
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    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.

  17. anonymous
    • 4 years ago
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    I still get the wrong answer

  18. asnaseer
    • 4 years ago
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    what do you get?

  19. anonymous
    • 4 years ago
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    I think I got it now

  20. asnaseer
    • 4 years ago
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    :-)

  21. anonymous
    • 4 years ago
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    thank you so much!

  22. asnaseer
    • 4 years ago
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    you are very welcome. I'm glad I could help you.

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spraguer (Moderator)
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