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ashontae19

  • one year ago

anybody in geomtry in ga virtual school?

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  1. k_lynn
    • one year ago
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    I'm not personally, but if you asked your question, I'm sure someone would be able to help you.

  2. ashontae19
    • one year ago
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    i need help with this

  3. k_lynn
    • one year ago
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    @freckles , @mathmate , or @Michele_Laino would probably be able to help you with that. I haven't gotten to that yet.

  4. ashontae19
    • one year ago
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    @Michele_Laino @mathmate @freckles could any one of yall help me

  5. ashontae19
    • one year ago
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    oh are you not in 10th grade?

  6. Michele_Laino
    • one year ago
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    question #1a the requested inequality is: \[ \Large - 12{x^2} + 960x \geqslant 18000\]

  7. ashontae19
    • one year ago
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    thank you but i need help pn 3a 3b

  8. ashontae19
    • one year ago
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    @Michele_Laino could you help me

  9. Michele_Laino
    • one year ago
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    question #3a the requested inequality is: \[\Large - 12{x^2} + 960x \leqslant 18432\]

  10. Michele_Laino
    • one year ago
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    dividing both sides by 12, we can rewrite that inequality as follows: \[\Large {x^2} - 80x + 1536 \geqslant 0\]

  11. ashontae19
    • one year ago
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    so thats the inequality?

  12. ashontae19
    • one year ago
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    would the answer be </ 32 and </ 48?

  13. Michele_Laino
    • one year ago
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    the correspnding equation is: \[\Large 12{x^2} - 960x + 18432 = 0\] whereas the critical values are: x=32, and x=48

  14. Michele_Laino
    • one year ago
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    finally the solution to that inequality is: \[\Large 0 \leqslant x \leqslant 32 \cup 48 \leqslant x\]

  15. ashontae19
    • one year ago
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    ok thank you so for 3a where as it says write an inequailty when she will not be taxes will it be >/ 50 because on the graph

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  16. Michele_Laino
    • one year ago
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    ok!

  17. ashontae19
    • one year ago
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    im asking is that correct?

  18. Michele_Laino
    • one year ago
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    the requested inequality is \[\Large x \geqslant 48\]

  19. ashontae19
    • one year ago
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    ok thank you. could you help me on more or are you stressed out?

  20. Michele_Laino
    • one year ago
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    I can help you! :)

  21. ashontae19
    • one year ago
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    ok thankyou here are the other ones

  22. Michele_Laino
    • one year ago
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    question #4a the requested inequality is: \[\Large - 12{x^2} + 960x \leqslant 14400\]

  23. ashontae19
    • one year ago
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    how would you do b?

  24. Michele_Laino
    • one year ago
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    here is the graph:

  25. ashontae19
    • one year ago
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    THANK YOU ! so i would shade below the graph correct?

  26. Michele_Laino
    • one year ago
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    yes! correct!

  27. Michele_Laino
    • one year ago
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  28. ashontae19
    • one year ago
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    ok for E it says are any of the values not in the orginal domain explain your answer and write your final solution as inequality.

  29. Michele_Laino
    • one year ago
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    yes all the values x, such that: \[\Large 0 \leqslant x < 30\] don't belong to the original domain of our function: \[\Large y=- 12{x^2} + 960x\]

  30. Michele_Laino
    • one year ago
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    so the solution is: \[\Large x \geqslant 60\]

  31. ashontae19
    • one year ago
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    wait so how would x> 60 belong in the equation ?

  32. Michele_Laino
    • one year ago
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    since the intersection between the graph: \[y = - 12{x^2} + 960x\] and the line: \[y = 14400\] occurs at x=20 and x=60. Now all x-values, such that 0</ x</20 don't belong to the domain of our function, so we have towrite x>/ 60

  33. ashontae19
    • one year ago
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    ok thank you what about these

  34. Michele_Laino
    • one year ago
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    question #5 as we can see from the drawing below, our solution is: \[x \leqslant 1 \cup x \geqslant 5\]

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  35. ashontae19
    • one year ago
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    what does u stand for?

  36. Michele_Laino
    • one year ago
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    it is the union of sets

  37. ashontae19
    • one year ago
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    do i put and or or ? or u

  38. Michele_Laino
    • one year ago
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    that writing is equivalent to this one: \[( - \infty ,1] \cup [5, + \infty )\] you can put "or"

  39. Michele_Laino
    • one year ago
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    Question #6 as we can see from the drawing below, our solution is: \[ - 1 \leqslant x \leqslant 5\]

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  40. Michele_Laino
    • one year ago
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    Question #7 here you have to solve the quadratic equation associated to your inequality, first. For example, if we consider case a), we get: \[{x^2} - 4x + 3 < 0\] so you have to solve this equation: \[{x^2} - 4x + 3 = 0\] what do you get?

  41. ashontae19
    • one year ago
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    x=3,1?

  42. Michele_Laino
    • one year ago
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    ok!

  43. Michele_Laino
    • one year ago
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    so we can write the solution as an inequality below: \[1 < x < 3\]

  44. ashontae19
    • one year ago
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    shaded to the left?

  45. Michele_Laino
    • one year ago
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    it is: |dw:1434396507609:dw| and x=1 , x=3 don't belong to the solution interval

  46. Michele_Laino
    • one year ago
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    case b) here after a simplification, we get: \[3{x^2} - 5x - 12 > 0\] so we have to solve this equation: \[3{x^2} - 5x - 12 = 0\] what do you get?

  47. ashontae19
    • one year ago
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    1.93?

  48. Michele_Laino
    • one year ago
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    are you sure?

  49. Michele_Laino
    • one year ago
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    please wait a moment

  50. ashontae19
    • one year ago
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    ya...

  51. Michele_Laino
    • one year ago
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    I got: x=3, and x=-4/3 so our solution is: \[x < - 4/3,\;x > 3\] |dw:1434399244525:dw|

  52. Michele_Laino
    • one year ago
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    x=3, and x=-4/3 don't belong to the solution interval

  53. Michele_Laino
    • one year ago
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    case c) we can write: \[{x^2} + 6x + 9 \geqslant 0\] now we have to nothe that: \[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2}\]

  54. Michele_Laino
    • one year ago
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    so our inequality is checked for all real values of x: |dw:1434399797411:dw| in x=-3 we have: \[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2} = 0\]

  55. ashontae19
    • one year ago
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    i got x>/ -3?

  56. ashontae19
    • one year ago
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    ok thank you for 1d on paula peaches i got 30 </ x</ 50 is that correct?

  57. Michele_Laino
    • one year ago
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    the inequality: \[{x^2} + 6x + 9 \geqslant 0\] is always true, and the equal sign holds, when x=-3

  58. Michele_Laino
    • one year ago
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    30</x</50 is not correct, since the production of peaches can not be higher than 14400, so the solution is x>/60

  59. Michele_Laino
    • one year ago
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    more precisely: 60</x</80

  60. ashontae19
    • one year ago
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    for 2b i got x would equal 50, 30 and c im a little lost on thats the last two

  61. Michele_Laino
    • one year ago
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    there we have to solve this inequality: \[ - 12{x^2} + 960x \geqslant 18000\]

  62. Michele_Laino
    • one year ago
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    which, after a simplification, becomes: \[{x^2} - 80x + 1500 \leqslant 0\]

  63. Michele_Laino
    • one year ago
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    now the roots of the associated quadratic equation: \[{x^2} - 80x + 1500 = 0\] are: x=50 and x=30

  64. Michele_Laino
    • one year ago
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    so the requested solution is: \[30 \leqslant x \leqslant 50\]

  65. Michele_Laino
    • one year ago
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    so we have the subsequent graph: |dw:1434401181442:dw|

  66. ashontae19
    • one year ago
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    thank you so much for the help you answered my questions fully

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