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BreezeFlow Group Title

I would really appreciate if someone could help me with this problem. It relates parabolas and archery. I've been trying to figure it out for about two hours now and I can't seem to get to the right answer, which I already know. Here's a link so you can see the picture with the problem: http://mrswhaples.cmswiki.wikispaces.net/file/view/Algebra+II+chapter+4+solutions.pdf The problem I need help with is on the second last page under Activity 2. Every time I attempt to get to the right answer, I am way off. D:

  • 2 years ago
  • 2 years ago

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  1. theEric Group Title
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    Hi! I want to help, but I'm not quite sure how this is working. So maybe we can work together to get to the right answer. You can tell me what you did and I'll try to understand it, or we can start from the beginning. Is it assumed that the archer shoots horizontally? And how high is the target? Are we able to know that? Sometimes I think we'd need to know things that we really don't. Variables cancel out, sometimes.. Also, I'm sorry you've spent so much time on this so far!

    • 2 years ago
  2. BreezeFlow Group Title
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    Thank you for being willing to help. So far I've been using three points to try to find an answer: (0, 1.39) (18,40) and (45,0). From top to bottom the whole target stand is supposed to be 1.5 meters. And the arrow is being shot at 1.39 meters.

    • 2 years ago
  3. Algebraic! Group Title
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    this question has been posted before. there's not enough info here to reach a conclusion without making an assumption...

    • 2 years ago
  4. BreezeFlow Group Title
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    Can you see the picture?

    • 2 years ago
  5. Algebraic! Group Title
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    "From top to bottom the whole target stand is supposed to be 1.5 meters." where does it say that??

    • 2 years ago
  6. BreezeFlow Group Title
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    On my homework sheet it says that. It's supposed to be a tip. Use the following piece of information to help you solve Activity 2: The distance from the top edge of the archery target to the ground is 1.5 meters

    • 2 years ago
  7. Algebraic! Group Title
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    then it's easy to solve.

    • 2 years ago
  8. theEric Group Title
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    Haha, that might be a necessary piece of information! It makes point A known! :)

    • 2 years ago
  9. BreezeFlow Group Title
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    Oh good it can be solved :D

    • 2 years ago
  10. theEric Group Title
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    And I was silly to think the shot was horizontal - the answer would be shoulder height :P Do you want to take it away, @Algebraic! ?

    • 2 years ago
  11. Algebraic! Group Title
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    not really... you have three points now. (0, 1.39) (18, 1.1) and (45,0) should not be a problem to find the vertex of that parabola...

    • 2 years ago
  12. BreezeFlow Group Title
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    IS 1.1 the height of A?

    • 2 years ago
  13. Algebraic! Group Title
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    OS crashed the pdf page... what were the measurements...

    • 2 years ago
  14. Algebraic! Group Title
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    1.5 - the distance from the top... whatever it was something like 4cm * 4 + 8cm?

    • 2 years ago
  15. theEric Group Title
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    The height was 1.5m, and the point was 24 cm down.

    • 2 years ago
  16. theEric Group Title
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    .24m down.

    • 2 years ago
  17. BreezeFlow Group Title
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    Here's the typed out problem An archer releases an arrow from a shoulder height of 1.39 m. When the arrow hits the target 18 m away, it hits point A. When the target is removed, the arrow lands 45 m away. Find the maximum height of the arrow along its parabolic path. 4 cm 4 cm 4 cm 4 cm 8 cm

    • 2 years ago
  18. Algebraic! Group Title
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    ah ok so (18, 1.26) is the second point...

    • 2 years ago
  19. BreezeFlow Group Title
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    Do I put the points into the quadratic equation form?

    • 2 years ago
  20. BreezeFlow Group Title
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    When I was working on this earlier I put c as being 1.39 for one of the equations.

    • 2 years ago
  21. Algebraic! Group Title
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    I'd use... y= a(x-h)^2 +k for ease... but yeah you can use y = ax^2 +bx +c with c=1.39m if you like... doesn't really matter...

    • 2 years ago
  22. Algebraic! Group Title
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    1.26 = a(18)^2 +b18 +1.39 0 = a(45)^2 +b(45) +1.39

    • 2 years ago
  23. Algebraic! Group Title
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    a is obviously going to be negative..

    • 2 years ago
  24. BreezeFlow Group Title
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    Do i solve it like linear systems?

    • 2 years ago
  25. Algebraic! Group Title
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    yep

    • 2 years ago
  26. BreezeFlow Group Title
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    Can I work this out and you guys tell me if i did it right?

    • 2 years ago
  27. Algebraic! Group Title
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    sure

    • 2 years ago
  28. BreezeFlow Group Title
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    Ok, thank you I'll get working on it.

    • 2 years ago
  29. BreezeFlow Group Title
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    Ok so far I simplified the expressions to 324a + 18b +1.39= 1.26 and 2025a+45b+1.39=0 Do I then subtract 1.39 from both sides and use the method of multiplying on equation by a number so that the coefficients of one variable will be opposites?

    • 2 years ago
  30. BreezeFlow Group Title
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    From subtracting 1.39 from both sides I got 324a + 18b = -0.13 and 2025a+45b=-1.39 Now, should I multiply one equation like 324a + 18b = -0.13 by -2.5 so the coefficient of b will be -45?

    • 2 years ago
  31. Algebraic! Group Title
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    no.

    • 2 years ago
  32. BreezeFlow Group Title
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    What should I do instead?

    • 2 years ago
  33. Algebraic! Group Title
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    I solved -.13 = 18^2a +18b for b and subsed in to the other equation a~ -8.765E-4 b~ .008555

    • 2 years ago
  34. BreezeFlow Group Title
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    To find the maximum height do I use the formula -b/2a?

    • 2 years ago
  35. Algebraic! Group Title
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    that's the x location of the max height... you plug that into the equation to find the actual height...

    • 2 years ago
  36. BreezeFlow Group Title
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    What's the equation for height. Oh and with the number for a should i type e-4 in my calculator?

    • 2 years ago
  37. Algebraic! Group Title
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    10^-4

    • 2 years ago
  38. Algebraic! Group Title
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    we just found the equation for height...

    • 2 years ago
  39. BreezeFlow Group Title
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    We use the second equation?

    • 2 years ago
  40. Algebraic! Group Title
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    the equation of the parabola... we were using the points... to find the equation of the parabola to find its vertex....

    • 2 years ago
  41. BreezeFlow Group Title
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    I have to find the vertex before I find the height? Sorry, if I sound stupid, I've been really confused with this problem for awhile. XP

    • 2 years ago
  42. Algebraic! Group Title
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    the vertex is the location of the max height

    • 2 years ago
  43. Algebraic! Group Title
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    |dw:1350531626084:dw|

    • 2 years ago
  44. BreezeFlow Group Title
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    Does that mean the y-coordinate of the vertex is the max height?

    • 2 years ago
  45. Algebraic! Group Title
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    (x,y) = (x location of max height, max height)

    • 2 years ago
  46. Algebraic! Group Title
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    yes

    • 2 years ago
  47. BreezeFlow Group Title
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    Do i already know the vertex? Was it one of the points from earlier?

    • 2 years ago
  48. Algebraic! Group Title
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    yeah it was (45,0)

    • 2 years ago
  49. Algebraic! Group Title
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    you already knew the answer before you did the problem, so this was all a huge waste of time.

    • 2 years ago
  50. BreezeFlow Group Title
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    I still don't get how to arrive to the answer and I need to show the work

    • 2 years ago
  51. Algebraic! Group Title
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    find a and b y= ax^2 +bx +c find the vertex

    • 2 years ago
  52. BreezeFlow Group Title
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    If the vertex is supposed to be (45,0) wouldn't that make the maximum height 0? o_o

    • 2 years ago
  53. Algebraic! Group Title
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    (0, 1.39) was a point so y= a(0)^2 +b(0) +c 1.39 =c (18, 1.26) was a point so 1.26 = a(18^2) + b(18) +1.39 (45,0) was a point so 0 = a(45)^2 +b(45)+ 1.39

    • 2 years ago
  54. BreezeFlow Group Title
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    When I try to find the height I get -4.88025002715, but I know that can't be write.

    • 2 years ago
  55. Algebraic! Group Title
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    it's not.

    • 2 years ago
  56. BreezeFlow Group Title
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    I don't know how the answer is supposed to be 1.41 and I don't know how to find the vertex that has 1.41 as the y-coordinate.

    • 2 years ago
  57. Algebraic! Group Title
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    plug x =-b/2a into the equation of the parabola

    • 2 years ago
  58. BreezeFlow Group Title
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    The ax^2+bx+c type equation?

    • 2 years ago
  59. Algebraic! Group Title
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    alternately, calculate (4ac -b^2) / 4a your choice.

    • 2 years ago
  60. BreezeFlow Group Title
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    Ok I'll try that

    • 2 years ago
  61. BreezeFlow Group Title
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    I got 1.3691, but I probably miscalculated somewhere unless I'm supposed to round it to 1.4

    • 2 years ago
  62. Algebraic! Group Title
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    it's 1.41

    • 2 years ago
  63. Algebraic! Group Title
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    1.410876 using the rough values for a and b that I gave

    • 2 years ago
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