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anonymous

  • one year ago

help! fan and medal! a ball is thrown vertically upward from the top of a 100-foot tower, with an initial velocity of 20ft/sec. its position function is s(t)=-16t^2+20t+100. what is its velocity in ft/sec when t=1 second? a. -12 b.-44 c.100 d.-32 what is the average rate of change of y with respect to x over the interval [-2,5] for the function y=3+2? a. -9 b.3 c.1/3 ( wrong answer ) d.-1

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  1. Astrophysics
    • one year ago
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    What's stopping you from plugging t = 1 for the first problem?

  2. anonymous
    • one year ago
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    im getting a choice that is not presented. I don't know what im doing wrong see: -16(1)^2+20(1)+100 I get 104

  3. Astrophysics
    • one year ago
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    Oh I see, you want the velocity, what you have there is the position, take the derivative and plug t = 1.

  4. anonymous
    • one year ago
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    what would be the derivative?

  5. Astrophysics
    • one year ago
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    Have you done calculus?

  6. anonymous
    • one year ago
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    I am taking the class. lol I am stuck on this lesson. what would be the answer to my questions ?would you be able to work the problems out and show me how to do it. im confused.

  7. Astrophysics
    • one year ago
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    \[s'(t) = -32t-20\] would be your velocity

  8. Astrophysics
    • one year ago
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    First derivative of displacement is velocity, the second is acceleration.

  9. anonymous
    • one year ago
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    where are you he 32 from o.o

  10. Astrophysics
    • one year ago
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    Mhm, I just used the power rule

  11. Astrophysics
    • one year ago
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    \[\frac{ d }{ dx } x^n = nx^{n-1}\]

  12. anonymous
    • one year ago
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    more confused lol. ok show me how to solve this. o.o

  13. Astrophysics
    • one year ago
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    Ok it seems you haven't got far at all in calculus yet lol

  14. anonymous
    • one year ago
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    yup

  15. Astrophysics
    • one year ago
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    Ok, so can you tell me what you know?

  16. anonymous
    • one year ago
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    pretty much what I first posted while trying to find an answer to my questions lol. I have limited time so im fussing over what I am doing wrong lol. o.) here is my guess: 1- -44 2- -9 am I right?

  17. Astrophysics
    • one year ago
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    You can use other methods such as the kinematic equations I suppose, but this question requires you to understand calculus, I gave you the derivative already, so you should be able to figure it out :)

  18. anonymous
    • one year ago
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    so what I am getting now for my first answer is -32. did I do that right?

  19. Astrophysics
    • one year ago
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    No, the derivative is s'(t) = -32t-20, so you still have to plug in t=1 in this equation.

  20. Astrophysics
    • one year ago
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    And for your second question average rate of change is \[\frac{ f(b)-f(a) }{ b-a }\]

  21. anonymous
    • one year ago
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    ok I plugged t=1 in and got -52. not one of my choices.

  22. Astrophysics
    • one year ago
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    How did you get that?

  23. Astrophysics
    • one year ago
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    \[s(t) = -16t^2+20t+100,~~~s'(t) = -32t+20\] plug t = 1, into s'(t), so find s'(1).

  24. Astrophysics
    • one year ago
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    Like I said s(t) is your position, but s'(t) is your velocity

  25. Astrophysics
    • one year ago
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    And you're looking for velocity when t = 1

  26. anonymous
    • one year ago
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    1=-32(1)+20 1= -32+20 1=-12 then what? t=-12?

  27. Astrophysics
    • one year ago
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    Well you leave s'(1) alone, but yes, - 12.

  28. anonymous
    • one year ago
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    I said -12 earlier lol

  29. Astrophysics
    • one year ago
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    You said -32 and -52

  30. Astrophysics
    • one year ago
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    But do you understand how that is?

  31. Astrophysics
    • one year ago
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    You will need to read your book/ youtube videos I think, then you will really understand it.

  32. anonymous
    • one year ago
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    yes. thanks. um on the second question a would be -2 and b is 5. right?

  33. Astrophysics
    • one year ago
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    Yup

  34. Astrophysics
    • one year ago
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    a = -2, b = 5

  35. Astrophysics
    • one year ago
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    y=2+3? That's just a horizontal line

  36. Astrophysics
    • one year ago
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    is there suppose to be an x there somewhere haha

  37. anonymous
    • one year ago
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    y=3x+2 im supposed to look for the average rate of y with respect to x over the interval [-2,5] for the function y=3x+2. so this is basically like my slope formula? m=y2-y1/x2-x1?

  38. Astrophysics
    • one year ago
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    Yes, similarly

  39. Astrophysics
    • one year ago
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    \[\frac{ f(5)-f(-2) }{ 5-(-2) }\]

  40. anonymous
    • one year ago
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    making my answer for number 2 , --1? lol

  41. Astrophysics
    • one year ago
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    Show your work

  42. Astrophysics
    • one year ago
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    If it makes more sense, y = 3x+2 just means f(x) = 3x+2

  43. Astrophysics
    • one year ago
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    You should not get -1

  44. anonymous
    • one year ago
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    I got 3

  45. Astrophysics
    • one year ago
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    That sounds good!

  46. anonymous
    • one year ago
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    quicl question. lol would this answer of mine be correct: what is the slope for the function y=-5x^2+2 at the point x=1? a- -5 ( my choice) b- -10 c--3 ( not the answer) d-slope no determined

  47. Astrophysics
    • one year ago
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    How did you get it

  48. anonymous
    • one year ago
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    I went with this: y=mx+b formula looked at the slope of my equation which is -5. is my reasoning wrong?

  49. Astrophysics
    • one year ago
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    No, all these problems require calculus

  50. Astrophysics
    • one year ago
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    You need to use \[\lim_{h \rightarrow 0} \frac{ f(a+h)-f(a) }{ h }\] use this definition to find it

  51. Astrophysics
    • one year ago
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    It's not a linear equation it's y = -5x^2+2, if it was y = -5x+2 we could say slope is -5.

  52. anonymous
    • one year ago
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    how would is et it up though?

  53. Astrophysics
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{ f(1+h)-f(1) }{ h }\]

  54. Astrophysics
    • one year ago
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    Can you do the rest?

  55. anonymous
    • one year ago
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    im assuming h is 0? making my anser =cannot be determined?

  56. Astrophysics
    • one year ago
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    No, limit h -> 0 does not mean h = 0, you need to simplify the problem, before you start taking the limit.

  57. Astrophysics
    • one year ago
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    Hey, why are they giving you these problems if you haven't learnt it yet..?

  58. anonymous
    • one year ago
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    I don't know im in a pickle?. lol ok then is it -10?

  59. Astrophysics
    • one year ago
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    If you show your work I will tell you whether you're right or not, use the draw tool if you want to, because otherwise to me it seems you're just guessing.

  60. anonymous
    • one year ago
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    in this question I am lol if h does not equal 0 its just approaching 0? how would I be able to find h before solving o.o

  61. Astrophysics
    • one year ago
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    Use your equation you are given, plug it into the "formula" I gave you \[\lim_{h \rightarrow 0} \frac{ -5(1+h)^2+2-(-5(1)^2+2 )}{ h }\]

  62. Astrophysics
    • one year ago
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    Now simplify

  63. anonymous
    • one year ago
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    -3? I put that previously a

  64. Astrophysics
    • one year ago
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    Oh, I don't know.

  65. Astrophysics
    • one year ago
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    Well since you're out of time I guess I will show you how to do it, the answer I got was -5. @magy33

  66. Astrophysics
    • one year ago
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    |dw:1435652296303:dw|

  67. Astrophysics
    • one year ago
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    Note that we keep the limit till we actually take the limit itself.

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