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anonymous

  • one year ago

HELP!!! I WILL DO ANYTHING

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  1. anonymous
    • one year ago
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    The function H(t) = -16t2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second. Part A: The projectile was launched from a height of 82 feet with an initial velocity of 60 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. (2 points) Part B: What is the maximum height that the projectile will reach? Show your work. (2 points) Part C: Another object moves in the air along the path of g(t) = 10 + 63.8t where g(t) is the height, in feet, of the object from the ground at time t seconds. Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points) Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know? (2 points)

  2. anonymous
    • one year ago
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    @dan815

  3. anonymous
    • one year ago
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    @triciaal

  4. Plasmataco
    • one year ago
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    A:0=-16t2+60t+82

  5. anonymous
    • one year ago
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    YEAH, I have that part. but what do you mean by A;0 .. I have the numbers plugged in .. just solving is difficult

  6. anonymous
    • one year ago
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    @jim_thompson5910

  7. Plasmataco
    • one year ago
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    A: 0. Sry. Forgot the space

  8. Plasmataco
    • one year ago
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    Just sayin for problem a.

  9. anonymous
    • one year ago
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    either way what do you mean??

  10. anonymous
    • one year ago
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    ohh

  11. anonymous
    • one year ago
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    hello?

  12. Plasmataco
    • one year ago
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    The equation for problem a is 0=-16t2+60t+82

  13. anonymous
    • one year ago
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    I know that

  14. jim_thompson5910
    • one year ago
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    part A is definitely 0 = -16t^2+60t+82. So you have the correct answer there for that part. initial velocity = 60 ft/s v = 60 initial height = 82 ft s = 82

  15. jim_thompson5910
    • one year ago
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    for part b, consider the equation y = -16x^2+60x+82 it is in the form y = ax^2 + bx + c

  16. anonymous
    • one year ago
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    okay... now what do i do?

  17. anonymous
    • one year ago
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    yes

  18. jim_thompson5910
    • one year ago
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    notice how a = -16, b = 60, c = 82 plug the values of a and b into h = -b/(2a) and tell me what you get for h

  19. anonymous
    • one year ago
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    -1.875

  20. jim_thompson5910
    • one year ago
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    it should be positive

  21. jim_thompson5910
    • one year ago
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    you lost a sign somewhere

  22. anonymous
    • one year ago
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    yea. it is . i accidentally put it there

  23. jim_thompson5910
    • one year ago
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    now plug x = 1.875 into y = -16x^2+60x+82 to get the value of y what is the value of y?

  24. anonymous
    • one year ago
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    y=1094.5 ?

  25. jim_thompson5910
    • one year ago
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    that's too big

  26. anonymous
    • one year ago
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    ,,,,, but i plugged the number in and I'm sure i solved it correctly..

  27. jim_thompson5910
    • one year ago
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    -16x^2+60x+82 -16(1.875)^2+60(1.875)+82 ... replace every x with 1.875 now compute `-16(1.875)^2+60(1.875)+82`

  28. anonymous
    • one year ago
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    yeah.. thats what i had. .let me retry

  29. anonymous
    • one year ago
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    138.25??

  30. jim_thompson5910
    • one year ago
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    much better

  31. anonymous
    • one year ago
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    okay great.. next step?

  32. Plasmataco
    • one year ago
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    For c, u make the equation in a equal 10 +63.8t 10+63.8t=-16t2+60t+82 In which you solve for t, then plug it in in one of the equations

  33. anonymous
    • one year ago
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    HUH ^^^^

  34. jim_thompson5910
    • one year ago
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    So when the time is t = 1.875 seconds, the height of the object is 138.25 ft which is the peak height |dw:1439942710928:dw|

  35. jim_thompson5910
    • one year ago
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    |dw:1439942749753:dw|

  36. anonymous
    • one year ago
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    ohhhh okay

  37. Plasmataco
    • one year ago
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    I thought u were done with b, sry for interupting😰

  38. anonymous
    • one year ago
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    its ok...

  39. anonymous
    • one year ago
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    now what?

  40. jim_thompson5910
    • one year ago
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    what kind of calculator do you have?

  41. Plasmataco
    • one year ago
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    So basically, 138.25 meters is the answer to b

  42. anonymous
    • one year ago
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    Texas Instruments TI 30X

  43. Plasmataco
    • one year ago
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    Right?

  44. jim_thompson5910
    • one year ago
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    have you used a website called desmos before? It's a free graphing calculator

  45. jim_thompson5910
    • one year ago
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    yes @Plasmataco

  46. anonymous
    • one year ago
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    yes!!

  47. jim_thompson5910
    • one year ago
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    ok we're going to use that to answer part c

  48. Plasmataco
    • one year ago
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    Oh ok

  49. anonymous
    • one year ago
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    ok what do i type in?

  50. Plasmataco
    • one year ago
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    Scroll upsidoodles

  51. jim_thompson5910
    • one year ago
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    type in f(x) = -16x^2 + 60x + 82 into the first box and g(x) = 10 + 63.8x into the second box let me know when you have that in and I'll move onto the next step

  52. Plasmataco
    • one year ago
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    You want go and fx to be the same

  53. Plasmataco
    • one year ago
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    Gx dang it autocorrect

  54. anonymous
    • one year ago
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    okay done.. but i don't know how to put the f(x) and g(x) into the equation.. I don't see those as options.. like i can type the like equation after the = sign but nothing before

  55. jim_thompson5910
    • one year ago
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    you should have this typed in so far

  56. anonymous
    • one year ago
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    how did you get the f(x) there... just type?

  57. jim_thompson5910
    • one year ago
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    yes

  58. anonymous
    • one year ago
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    ok.. but no lines appear

  59. Plasmataco
    • one year ago
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    If Gx=Fx, the -16t2+60t+82= 10+63.8t

  60. jim_thompson5910
    • one year ago
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    your graph window may be off

  61. anonymous
    • one year ago
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    i have a mac.. nothing is wrong with it.. so how is that possible?

  62. Plasmataco
    • one year ago
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    Des ps is a lot quicker :3

  63. jim_thompson5910
    • one year ago
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    try setting the y min to -20 and the y max to 150

  64. jim_thompson5910
    • one year ago
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    click the wrench on the right side to adjust the graph window

  65. anonymous
    • one year ago
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    i have lines now..

  66. jim_thompson5910
    • one year ago
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    you should see something like this

  67. anonymous
    • one year ago
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    yeah

  68. jim_thompson5910
    • one year ago
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    now onto the table part select the box under the g(x). So select the third box then move up to the + button and add a table change the \(\Large x_1\) to just x. In the second column, replace \(\Large y_1\) with f(x). You'll see a list of numbers pop up in the third column, type g(x) at the top of this column. More numbers will pop up

  69. jim_thompson5910
    • one year ago
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    the goal is to have the numbers be equal. Which row does this? Or do we get pretty close at all?

  70. anonymous
    • one year ago
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    they aren't appearing with numbers. i have the table and stuff.. but there aren't any numbers

  71. jim_thompson5910
    • one year ago
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    you should see this

  72. anonymous
    • one year ago
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    okay I got it now

  73. jim_thompson5910
    • one year ago
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    do you see any rows where f(x) = g(x) or pretty close to it?

  74. anonymous
    • one year ago
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    yea. row 2

  75. jim_thompson5910
    • one year ago
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    yep, so at about t = 2 seconds, the two objects will be at the same height (approximately)

  76. anonymous
    • one year ago
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    awesome! So is that part c??

  77. jim_thompson5910
    • one year ago
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    yes

  78. anonymous
    • one year ago
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    okay and Finally, Part D/

  79. jim_thompson5910
    • one year ago
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    and it might be hard to see, but the projectile is falling down when they finally met up here is a zoomed in look |dw:1439944300088:dw|

  80. anonymous
    • one year ago
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    ohhh. okay

  81. jim_thompson5910
    • one year ago
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    if you click the intersection point between the two graphs, you'll see the point (2.006, 137.98) pop up. This is the more accurate approximate intersection point So at approx t = 2.006 seconds, the two objects are at an approx height of 137.98 ft

  82. anonymous
    • one year ago
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    Great :)

  83. anonymous
    • one year ago
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    @jim_thompson5910 so if that is all.. I know your probably exhausted ...but could you help me with another problem I only need help on one part and its fairly simple???

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