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anonymous

  • one year ago

4) What is the height of the apple when your friend catches it? (**HINT: The height would be the y-coordinate of the highest point.) Double click on the space below to show your work. h(t) = -16t^2 + 38.4t + 0.96 h= height t= time (seconds) I will fan and medal for work plus answers!

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  1. anonymous
    • one year ago
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    How do I find the height? How do I switch the equation to find the height instead of the time? I can solve it just not sure how to set it up.

  2. anonymous
    • one year ago
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    there is more to this question right?

  3. anonymous
    • one year ago
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    like more words

  4. anonymous
    • one year ago
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    Make t=0. :)

  5. anonymous
    • one year ago
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    The information is enough. :)

  6. anonymous
    • one year ago
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    naw dude its asking for the vertex

  7. anonymous
    • one year ago
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    Would I just substitute the time that we just figured out into the equation? The 2.4 seconds?

  8. anonymous
    • one year ago
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    |dw:1444445933109:dw|

  9. anonymous
    • one year ago
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    No you don't have to. Just make t=0. :) \(h(t) = -16(0)2 + 38.4(0) + 0.96\)

  10. anonymous
    • one year ago
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    why would you do that?? is the friend .96 ft off the ground

  11. anonymous
    • one year ago
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    The height we're finding is the same thing as the vertex. =_= Notice that the vertex in this case is the maximum because \(a\) is a negative.

  12. anonymous
    • one year ago
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    no not for t=0

  13. anonymous
    • one year ago
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    t=o is your initial starting point not the vertex

  14. anonymous
    • one year ago
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    or your y-intercept

  15. anonymous
    • one year ago
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    If (0,y) <--- this is the vertex. Please stop misguiding the one who asked.

  16. anonymous
    • one year ago
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    NO IT ISNT...

  17. anonymous
    • one year ago
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    If you solve the equation mathway gave me, I got 0.96.

  18. anonymous
    • one year ago
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    b/2a is the x coordinate and f(b/2a) is your Y

  19. anonymous
    • one year ago
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    "(**HINT: The height would be the y-coordinate of the highest point.)" Which means that you need to find the y-coordinate.

  20. anonymous
    • one year ago
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    Yes, it is \(\huge 0.96\). :)

  21. anonymous
    • one year ago
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    MAX HIGHT WHICH IS THE Y OF YOUR VERTEX

  22. anonymous
    • one year ago
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    t=0 is not a vertex

  23. anonymous
    • one year ago
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    that is 0 time e=meaning the apple hasnt been thrown yet...

  24. anonymous
    • one year ago
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    It is not a vertex because it is PART OF THE VERTEX. It is the x-intercept.

  25. anonymous
    • one year ago
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    how can your friend catch something that hasnt been thrown?

  26. anonymous
    • one year ago
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    @Nnesha Please confirm this.

  27. anonymous
    • one year ago
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    no x intercept is finding the root dude...

  28. anonymous
    • one year ago
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    you're really off

  29. anonymous
    • one year ago
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    @pooja195

  30. anonymous
    • one year ago
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    The question I posted before this was asking if a person threw an apple how many seconds would it take to reach the other person 3 stories up? We solved and got 2.4 seconds.

  31. anonymous
    • one year ago
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    yes

  32. anonymous
    • one year ago
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    Show me how you will find the height then. You keep on complaining, but you're not showing any work.

  33. anonymous
    • one year ago
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    Of course it is correct. I was the one who helped her. :P

  34. Nnesha
    • one year ago
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    `(**HINT: The height would be the y-coordinate of the highest point.)`

  35. anonymous
    • one year ago
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    but now plug that number into the function

  36. anonymous
    • one year ago
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    that is the hight

  37. anonymous
    • one year ago
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    not t=0...

  38. anonymous
    • one year ago
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    @mathway im not calling you dumb people make mistakes and this is one of yours

  39. anonymous
    • one year ago
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    You would still get the same answer either way, 0.96.

  40. anonymous
    • one year ago
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    yes buy he cant say thats what it is you have to explain it in the context of the problem

  41. anonymous
    • one year ago
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    since its a root ofcourse it would cancle out the other 2 but in a 4th power polynomial that would fail miserably

  42. anonymous
    • one year ago
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    so the fundamental way to explain the answer is V = {(b/2a),f(b/2a)} and you look for Y

  43. anonymous
    • one year ago
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    Okay guys last question! Using the same equation - h(t) = -16t^2 + 38.4t + 0.96 With h = height and t = time in seconds 5) Suppose your friend does not catch the apple and it falls back to the ground. How long did it take the apple to hit the ground? (**HINT: When the apple hits the ground is when the graph hits the x-axis.)

  44. anonymous
    • one year ago
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    find the root

  45. anonymous
    • one year ago
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    |dw:1444446732488:dw|

  46. Nnesha
    • one year ago
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    calm plz use the hint highest point = max so we need y -coordinate of the max point (max=vertex)

  47. Nnesha
    • one year ago
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    lol sorry for late reply >.< internet iz very slow -.-

  48. anonymous
    • one year ago
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    lol its ok

  49. anonymous
    • one year ago
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    ok hold up go back what did you get for your sec for the apple to get to him??????!!!!

  50. anonymous
    • one year ago
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    the kid on the 3rd floor????

  51. Nnesha
    • one year ago
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    x coordinate of the vertex can be any number (0,y) = y-intercept :=)

  52. anonymous
    • one year ago
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    \[\frac{ -38.4 \pm \sqrt{38.4^{2}-4(-16)(0.96)} }{ 2(-16) }\]

  53. anonymous
    • one year ago
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    yea lol but only if it is centered on the orgin would that be a vertex

  54. anonymous
    • one year ago
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    yea @tbull1221 that works

  55. anonymous
    • one year ago
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    what does that give you?

  56. anonymous
    • one year ago
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    \[\frac{ -38.4 \pm \sqrt{1536} }{ -32 }\]

  57. anonymous
    • one year ago
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    Then you get: \[\frac{ -38.4 \pm 16\sqrt{6} }{ -32 }\]

  58. anonymous
    • one year ago
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    "x coordinate of the vertex can be any number (0,y) = y-intercept :=)" @Nnesha Thank you for backing up! At least I know that I'm not dumb and I don't necessarily have to say irrelevant things when I made a mistake. :)

  59. anonymous
    • one year ago
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    btw dude the other answer for the seconds it takes was wrong its 4.8sec

  60. anonymous
    • one year ago
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    plz read http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html

  61. anonymous
    • one year ago
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    im in 3rd year calc and 2 year stat i know how to find the vertex of a parabola...

  62. anonymous
    • one year ago
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    Can someone please just give me an answer for #5? My computer is about to die. I will fan and medal if i haven't already.

  63. anonymous
    • one year ago
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    I highly suggest that you don't call me "dude." We are not close.

  64. anonymous
    • one year ago
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    And is the seconds answer 2.4 or 4.8?

  65. anonymous
    • one year ago
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    sorry im a fairly welcoming person i dont show nor mean it in any form of disrespect

  66. anonymous
    • one year ago
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    the one that said "the amount of seconds for the apple to reach the friend" is 4.8 sec

  67. Nnesha
    • one year ago
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    people make mistakes i do all the time everyday more than 100 times it's okay we can correct it nicely :=) so please :=)

  68. anonymous
    • one year ago
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    because -b/2a =4.8

  69. anonymous
    • one year ago
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    so that is your seconds or X on your vertex

  70. anonymous
    • one year ago
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    then plug that into your equation and you will get the amount of feet it took for the friend to get the apple

  71. anonymous
    • one year ago
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    then for this what you did was fine you just plug the numbers into the quadratic equation and you will get the time it took to hit the ground

  72. anonymous
    • one year ago
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    he got offline...

  73. anonymous
    • one year ago
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    do you understand why you were wrong @mathway ??

  74. anonymous
    • one year ago
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    not trying to disrespect you. at this point im trying to help you

  75. anonymous
    • one year ago
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    @tbull1221 cool?

  76. anonymous
    • one year ago
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    go up like 5 comments and i did a step-by-step of the process

  77. anonymous
    • one year ago
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    So the seconds is 4.8? And What is the answer to #5?

  78. anonymous
    • one year ago
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    5 is the ft??

  79. anonymous
    • one year ago
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    5) Suppose your friend does not catch the apple and it falls back to the ground. How long did it take the apple to hit the ground? (**HINT: When the apple hits the ground is when the graph hits the x-axis.)

  80. anonymous
    • one year ago
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    but both of the ones before that were wrong because the sec were wrong so the FT to make to him wouldnt be .96

  81. anonymous
    • one year ago
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    so with the answer to -b/2a plug that number into the equation given and you will get the distance of the throw to the vertex (the friend)

  82. anonymous
    • one year ago
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    because mathway said it was 2.4 sec but it wasnt it was 4.8

  83. anonymous
    • one year ago
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    so then plug in 4.8 into the equation to get the answer to the Max height

  84. anonymous
    • one year ago
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    then what you did with the quadratic equation would give you #5

  85. anonymous
    • one year ago
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    Cool? @tbull1221

  86. anonymous
    • one year ago
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    @tbull1221 are you still there??

  87. anonymous
    • one year ago
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    aww tbull got off :(

  88. Jhannybean
    • one year ago
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    \[\begin{align} h(t) = -16t^2 + 38.4t + 0.96 &\implies h(t) = -16t^2+\frac{192}{5}t+\frac{24}{25} \\&\implies h(t) =t^2 -\frac{12}{5}t -\frac{3}{50} =0 \\&\implies h(t) = \left(t^2-\frac{12}{5}t \right) =\frac{3}{50} \\&\implies h(t) = \left(t^2 -\frac{12}{5}t +\frac{36}{25} \right)=\frac{3}{50}+\frac{36}{25} \\ &\implies h(t) = \left(t-\frac{6}{5}\right)^2=\frac{3}{2} \\&\implies h(t) : ~t=\frac{6}{5}\pm\sqrt{\frac{3}{2}} \end{align}\]

  89. anonymous
    • one year ago
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    yea she left:(

  90. Jhannybean
    • one year ago
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    time is always positive in this case, therefore \(t= \frac{6}{5} +\sqrt{\frac{3}{2}}\)

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