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bobobox

  • one year ago

I have seven math questions I need help answering please, It's my very very last assignement of the year and I need to pass.

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  1. bobobox
    • one year ago
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  2. anonymous
    • one year ago
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    Hit me with em (;

  3. anonymous
    • one year ago
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    I'm quite sure the answer is C on that one

  4. bobobox
    • one year ago
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    okay

  5. bobobox
    • one year ago
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  6. mathstudent55
    • one year ago
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    |dw:1435638657119:dw|

  7. mathstudent55
    • one year ago
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    1. \((3 + 2x)(5 + 2x) = 19.25\) \((2x + 3)(2x + 5) = 19.25\) \(4x^2 + 16x + 15 = 19.25\) \(4x^2 + 16x - 4.25 = 0\)

  8. bobobox
    • one year ago
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    Yay thanks so much for the explanation! It helps!

  9. bobobox
    • one year ago
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    Could you help me with the soccer one?

  10. bobobox
    • one year ago
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    @mathstudent55 @willb1220

  11. mathstudent55
    • one year ago
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    The acceleration due to gravity is g in the down direction, so let;'s call it -g. \(a = -g\) The velocity is \(v(t) = -gt + v_0\) where \(v_0\) is the initial velocity. The position (height) is \(h = -\dfrac{1}{2}gt^2 + v_0t + h_0\) where \(h_0\) is the initial height.

  12. bobobox
    • one year ago
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    okay

  13. mathstudent55
    • one year ago
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    We are told that the initial position is 0, so \(h_0 = 0\) We are also told the initial velocity is 42 ft/sec, so \(v_0 = 42\)

  14. bobobox
    • one year ago
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    yeah

  15. mathstudent55
    • one year ago
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    Also, g = 32 ft/sec^2

  16. bobobox
    • one year ago
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    yup

  17. mathstudent55
    • one year ago
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    \(h(t) = -16t^2 + 42t \) This is an inverted parabola. We can take the first derivative to find where the height is maximum: \(h'(t) = -32t + 42\) Now we set the derivative equal to zero and solve for t: \(-32t + 42 = 0\) \(-32t = -42\) \(t = 1.3125~sec\) At 1.3125 sec the ball reaches maximum height.

  18. bobobox
    • one year ago
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    So a since you round it

  19. bobobox
    • one year ago
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    Thank you for being awesome and walking me through it, do you think you could help me with the next?

  20. mathstudent55
    • one year ago
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    Now we plug in that time in the position (height) equation to find at what height the ball is at 1.3125 sec: \(h(t) = -16t^2 + 42t\) \(h(1.3125) = -16(1.3125)^2 + 42(1.3125) = 27.56\)

  21. bobobox
    • one year ago
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    oh wow

  22. mathstudent55
    • one year ago
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    Notice we first found the time was 1.3 sec. The question is what is the height, not the time, so we need to take the time in the position equation and find the position at that time. The answer is 27.6 ft, not 1.3 sec.

  23. bobobox
    • one year ago
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    Thank you

  24. bobobox
    • one year ago
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    COuld you help with another? It's simelar!

  25. mathstudent55
    • one year ago
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    It's so similar it's exactly the same. The answer is the same.

  26. bobobox
    • one year ago
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    it's 27.6? It's not the same question though

  27. mathstudent55
    • one year ago
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    Sorry. I read it too fast. The answer is the 1.3 sec we already got before.

  28. mathstudent55
    • one year ago
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    No. Wait.

  29. mathstudent55
    • one year ago
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    I finally read the whole question correctly.

  30. bobobox
    • one year ago
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    okay thanks! Can I pester you to help me with the last four questions? I really appreciate How hard you work to help others.

  31. mathstudent55
    • one year ago
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    Look at this graph: |dw:1435640162113:dw|

  32. mathstudent55
    • one year ago
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    |dw:1435640215914:dw|

  33. mathstudent55
    • one year ago
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    The ball is kicked at (0, 0), where time is zero, and the height of the ball is zero.

  34. bobobox
    • one year ago
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    okay

  35. mathstudent55
    • one year ago
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    As time goes by, the ball goes higher. It reaches a maximum height at 1.3 sec. |dw:1435640295295:dw|

  36. bobobox
    • one year ago
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    so is it 1.3?

  37. mathstudent55
    • one year ago
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    Notice that the shape of the curve is an inverted parabola. That means it is symmetric left to right. The amount of time it took for the ball to reach maximum height is the same amount of time it will take for the ball to hit the ground on the way down.

  38. bobobox
    • one year ago
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    okay so in total the ball was in the air for 2.6 seconds

  39. mathstudent55
    • one year ago
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    |dw:1435640433043:dw|

  40. mathstudent55
    • one year ago
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    Correct.

  41. bobobox
    • one year ago
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    thanks can I pester you to help with the last four?

  42. mathstudent55
    • one year ago
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    We can get this answer this way bec we already solved the part about maximum height. It's simply double the time it took to reach maximum height.

  43. bobobox
    • one year ago
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    Cool ill post the next

  44. bobobox
    • one year ago
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    A golf ball is hit from the ground with an initial velocity of 208 feet per second. Assume the starting height of the ball is 0 feet. How long will it take the golf ball to hit the ground? (1 point) 13 secs 21 secs 42 secs 6 secs

  45. mathstudent55
    • one year ago
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    If we had not solved that part, then we could solve this problem on its own. The equation of the position is \(h(t) = -16t^2 + 42t\) The height is 0 at the beginning (when the ball is kicked) and at the end (when the ball falls on the ground) \(0 = -16t^2 + 42t\) \(t(-16t + 42) = 0\) \(t = 0\) or \(-16t + 42 = 0\) \(t = 0\) or \(-16t = -42\) \(t = 0\) or \(t = 2.6\) The ball is at zero height at 0 sec and at 2.6 sec. That means it was in the air for 2.6 sec.

  46. bobobox
    • one year ago
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    yup :)

  47. mathstudent55
    • one year ago
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    Use the general equation of height we have: \(h(t) = -16t^2 + v_0t + h_0\)

  48. mathstudent55
    • one year ago
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    \(h_0 = 0\) \(v_0 = 208\) Plug these values in and solve the equation for t.

  49. bobobox
    • one year ago
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    okay one sec

  50. bobobox
    • one year ago
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    I tried I got a fraction so I am not even sure I did it right

  51. mathstudent55
    • one year ago
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    \(h(t) = -16t^2 + v_0t + h_0\) \(h(t) = -16t^2 + 208t = 0\) \(-16t^2 + 208t = 0\) \(-2t^2 + 26t = 0\) \(t(-2t + 26) = 0\) \(t = 0\) or \(-2t + 26 = 0\) \(~~~~~~~~~~~~~~~~~~-2t = -26\) \(~~~~~~~~~~~~~~~~~~~~~~~~~t = 13\)

  52. bobobox
    • one year ago
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    oohhhhh

  53. bobobox
    • one year ago
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    so is 13 our answer?

  54. mathstudent55
    • one year ago
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    The ball is at height zero at 0 seconds, when it is hit, and at 13 seconds when if falls on the ground.

  55. bobobox
    • one year ago
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    cool thanks! only three more left!

  56. bobobox
    • one year ago
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    A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second. What is the maximum height of the particle? (1 point) 128 feet 224 feet 272 feet 484 feet

  57. bobobox
    • one year ago
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    @mathstudent55 please help me I need you without you I might fail please please please

  58. bobobox
    • one year ago
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    @mathstudent55 I will fan write a testimony and I will give you a medal please

  59. bobobox
    • one year ago
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    @willb1220 help please help

  60. mathstudent55
    • one year ago
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    We use the same height equation again, and we plug in the info of this problem. Here we have an initial position.

  61. bobobox
    • one year ago
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    is the answer A? because I tried figuring it out

  62. bobobox
    • one year ago
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    crap no I meant d

  63. mathstudent55
    • one year ago
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    \(h(t) = -16t^2 + v_0t + h_0\) \(h(t) = -16t^2 + 144t + 160\) \(h'(t) = -32t + 144\) \(-32t + 144 = 0\) \(-32t = -144\) \(t = 4.5~sec\) Maximum height is achieved at 4.5 sec. Now we plug in 4.5 sec in the height equation to find the actual height.

  64. mathstudent55
    • one year ago
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    \(h(t) = -16(4.5)^2 + 144(4.5) + 160\)

  65. mathstudent55
    • one year ago
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    What DDDDDDDdo you get?

  66. bobobox
    • one year ago
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    let me see

  67. bobobox
    • one year ago
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    484

  68. mathstudent55
    • one year ago
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    You DDDDDDid get the correct answer.

  69. bobobox
    • one year ago
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    yay! Okay just two more questions okay?

  70. mathstudent55
    • one year ago
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    ok

  71. bobobox
    • one year ago
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    I think it's the one that I choose

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  72. bobobox
    • one year ago
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    @mathstudent55

  73. mathstudent55
    • one year ago
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    correct

  74. bobobox
    • one year ago
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    Yay the last one I am not to sure about

  75. mathstudent55
    • one year ago
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    |dw:1435642461182:dw|

  76. bobobox
    • one year ago
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    Help

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  77. bobobox
    • one year ago
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    @mathstudent55

  78. mathstudent55
    • one year ago
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    Can you solve it for c^2? Since c^2 is being multiplied by m, divide both sides by m. |dw:1435642705027:dw|

  79. mathstudent55
    • one year ago
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    |dw:1435642763475:dw|

  80. mathstudent55
    • one year ago
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    ok so far?

  81. bobobox
    • one year ago
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    omg yes

  82. mathstudent55
    • one year ago
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    Now you want c but you have c squared. The opposite of squaring is taking the square root, so you take the square root of both sides: |dw:1435642840101:dw|

  83. bobobox
    • one year ago
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    I just want to say thank you, I don't think I could have said it enough, really

  84. mathstudent55
    • one year ago
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    You're welcome. I hope you pass & graduate.

  85. bobobox
    • one year ago
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    I got a 100%!!!!!!!!!!!!!!!!!!!!!!!!! @mathstudent55

  86. mathstudent55
    • one year ago
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    Great!!!

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