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x3_drummerchick

  • one year ago

can someone explain to me how to do this problem? will give medals! (problem is in picture below)

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  1. x3_drummerchick
    • one year ago
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  2. nincompoop
    • one year ago
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    your problem is very interesting how do you propose solving it if you were to figure it out?

  3. Jhannybean
    • one year ago
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    ok so \[f(a) = \frac{4a}{a-7}\]\[f(a+h) = \frac{4a+4h}{a+h+7}\] therefore... \[\frac{f(a+h)-f(a)}{h} = \frac{\dfrac{4a+4h}{a+h+7} - \dfrac{4a}{a-7}}{h}\]

  4. Jhannybean
    • one year ago
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    So forget the \(h\) at the bottom for now and concentrate on finding an LCD between \(a+h+7\) and \(a-7\)

  5. Jhannybean
    • one year ago
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    sorry, typo.

  6. Jhannybean
    • one year ago
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    \[ \frac{f(a+h)-f(a)}{h} = \frac{\dfrac{4a+4h}{a+h-7} - \dfrac{4a}{a-7}}{h}\]

  7. x3_drummerchick
    • one year ago
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    ohh okay i see where this is going, the initial start confused me, i forgot which way to plug it in. oh and jhannybean. youre always to my rescue! you helped me understand things well enough yto get to precalc, so thanks!!

  8. Jhannybean
    • one year ago
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    Woot no problem :D

  9. nincompoop
    • one year ago
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    thank you, beanerwoman

  10. x3_drummerchick
    • one year ago
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    you should be a teacher

  11. Jhannybean
    • one year ago
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    ...

  12. nincompoop
    • one year ago
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    she shouldn't like for real ;) she should be the CEO of openstudy

  13. Jhannybean
    • one year ago
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    HAHA

  14. Jhannybean
    • one year ago
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    So let's finish this badboy.

  15. Jhannybean
    • one year ago
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    What did you get as your LCD?

  16. x3_drummerchick
    • one year ago
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    one sec

  17. nincompoop
    • one year ago
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    don't finish me :(

  18. x3_drummerchick
    • one year ago
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    cant you multiply whole thing by (a+h-7)and(a-7)? ?

  19. Jhannybean
    • one year ago
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    Yup, That's what I got \[\frac{(a-7)(4a+4h)-4a(a+h-7)}{(a-7)(a+h-7)}\]

  20. x3_drummerchick
    • one year ago
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    then it all cancels, well most of it, right?

  21. x3_drummerchick
    • one year ago
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    oh wait nvm

  22. Jhannybean
    • one year ago
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    No no, it doesn't all cancel. we have to expand it

  23. x3_drummerchick
    • one year ago
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    yeah im lost again :(

  24. x3_drummerchick
    • one year ago
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    buut doesnt the denominator go away?

  25. Jhannybean
    • one year ago
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    No, we're trying to find an LCD by multiplying both of the denominators of the two fractions with eachother, we can't simply cancel them otherwise there would be no point of finding the LCD in the first place!

  26. nincompoop
    • one year ago
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    show that it goes away

  27. Jhannybean
    • one year ago
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    And remember, we're still dividing EVERYTHING by h, so we need to expand our function to see which h values are still left :D

  28. x3_drummerchick
    • one year ago
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    okay

  29. x3_drummerchick
    • one year ago
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    is this right?

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  30. Jhannybean
    • one year ago
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    I'llcompare it with mine :) one minute!

  31. x3_drummerchick
    • one year ago
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    okie dokie

  32. x3_drummerchick
    • one year ago
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    i think i did somehing wrong

  33. x3_drummerchick
    • one year ago
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    then i distribute negative here? :

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  34. x3_drummerchick
    • one year ago
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    then i cancel out terms?

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  35. Jhannybean
    • one year ago
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    Yeah I found where I had messed up too haha

  36. nincompoop
    • one year ago
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    perform the operation instead of using the word cancel

  37. Jhannybean
    • one year ago
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    \[\frac{(a-7)(4a+4h)-4a(a+h-7)}{(a-7)(a+h-7)}\]\[=\frac{4a^2+4ah-28a-28h-4a^2-4ah+28a}{a^2+ah-7a-7a-7h+49}\]\[=\frac{\cancel{4a^2}\cancel{+4ah}\cancel{-28a}-28h\cancel{-4a^2}\cancel{-4ah}\cancel{+28a}}{a^2+ah-7a-7a-7h+49}\]\[=\frac{-28h}{a^2+ah-14a-7h+49}\]

  38. Jhannybean
    • one year ago
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    So now all of this is divided by h.

  39. x3_drummerchick
    • one year ago
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    holy sheesh

  40. Jhannybean
    • one year ago
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    \[=\dfrac{\dfrac{-28h}{a^2+ah-14a-7h+49}}{h}\]

  41. Jhannybean
    • one year ago
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    That means we're multiplying in h throughout the entire denominator.

  42. x3_drummerchick
    • one year ago
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    why are we multiplying it through the denominator only?

  43. Jhannybean
    • one year ago
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    Its a rule in dividing fractions: \(\dfrac{\dfrac{a}{b}}{c} \iff \dfrac{a}{bc} \)

  44. x3_drummerchick
    • one year ago
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    oh wow i didnt know that

  45. Jhannybean
    • one year ago
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    And actually, I just realized we don't eve have to multiply through the denominator, we can just leave it alone.

  46. Jhannybean
    • one year ago
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    It would look like: \[\dfrac{\dfrac{-28h}{a^2+ah-14a-7h+49}}{h} \longrightarrow \frac{-28h}{h(a^2+ah-14a-7h+49)}\]

  47. x3_drummerchick
    • one year ago
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    so what do we do with the denominator now? thi is super intense 0_0

  48. Jhannybean
    • one year ago
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    And therefore, because we're multiplying both the numerator AND denominator by h, we can eliminate the h altogether. \[\frac{-28\cancel{h}}{\cancel{h}(a^2+ah-14a-7h+49)} =\boxed{ -\frac{28}{a^2+ah-14a-7h+49}}\]

  49. x3_drummerchick
    • one year ago
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    holy rap girl. youre a genius... im gonna have to study this one.

  50. Jhannybean
    • one year ago
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    So now we can state that \[\boxed{\frac{f(a+h)-f(a)}{h} = \frac{\dfrac{4a+4h}{a+h-7} - \dfrac{4a}{a-7}}{h} = -\frac{28}{a^2+ah-14a-7h+49}}\]

  51. nincompoop
    • one year ago
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    jhan, that looks pretty

  52. Jhannybean
    • one year ago
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    Lol ty.

  53. x3_drummerchick
    • one year ago
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    i would have never gotten this...

  54. Jhannybean
    • one year ago
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    More importantly, do you understand my steps?

  55. x3_drummerchick
    • one year ago
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    yes, and ive screenshotted all of them. im going to rewrite them all in my notes with your steps, i forgot about the division rule!!

  56. Jhannybean
    • one year ago
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    Yeah that's important \(\checkmark\)

  57. x3_drummerchick
    • one year ago
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    pretty much, we plugged in those functions in for x, got lcd common denominators, combined like terms, divided all by h in which we innstead, multiplied, then canceled out terms (h in num + denom)

  58. Jhannybean
    • one year ago
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    yes, good job.

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