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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:

  • one year ago
  • one year 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!

    • one year 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.

    • one year 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...

    • one year ago
  4. BreezeFlow Group Title
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    Can you see the picture?

    • one year 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??

    • one year 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

    • one year ago
  7. Algebraic! Group Title
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    then it's easy to solve.

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

    • one year ago
  9. BreezeFlow Group Title
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    Oh good it can be solved :D

    • one year 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! ?

    • one year 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...

    • one year ago
  12. BreezeFlow Group Title
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    IS 1.1 the height of A?

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

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

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

    • one year ago
  16. theEric Group Title
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    .24m down.

    • one year 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

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

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

    • one year 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.

    • one year 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...

    • one year 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

    • one year ago
  23. Algebraic! Group Title
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    a is obviously going to be negative..

    • one year ago
  24. BreezeFlow Group Title
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    Do i solve it like linear systems?

    • one year ago
  25. Algebraic! Group Title
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    yep

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

    • one year ago
  27. Algebraic! Group Title
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    sure

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

    • one year 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?

    • one year 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?

    • one year ago
  31. Algebraic! Group Title
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    no.

    • one year ago
  32. BreezeFlow Group Title
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    What should I do instead?

    • one year 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

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

    • one year 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...

    • one year 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?

    • one year ago
  37. Algebraic! Group Title
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    10^-4

    • one year ago
  38. Algebraic! Group Title
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    we just found the equation for height...

    • one year ago
  39. BreezeFlow Group Title
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    We use the second equation?

    • one year 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....

    • one year 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

    • one year ago
  42. Algebraic! Group Title
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    the vertex is the location of the max height

    • one year ago
  43. Algebraic! Group Title
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    |dw:1350531626084:dw|

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

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

    • one year ago
  46. Algebraic! Group Title
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    yes

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

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

    • one year 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.

    • one year 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

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

    • one year 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

    • one year 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

    • one year 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.

    • one year ago
  55. Algebraic! Group Title
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    it's not.

    • one year 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.

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

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

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

    • one year ago
  60. BreezeFlow Group Title
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    Ok I'll try that

    • one year 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

    • one year ago
  62. Algebraic! Group Title
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    it's 1.41

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

    • one year ago
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