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Destinyyyy

  • one year ago

Help with 5 problems...

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  1. Destinyyyy
    • one year ago
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    2 [7- 3/2x] +5=21 [] are absolute value lines

  2. Destinyyyy
    • one year ago
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    7- 3/2x =8

  3. Destinyyyy
    • one year ago
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    Not sure what to do next

  4. Destinyyyy
    • one year ago
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    @Nnesha @freckles @Vocaloid

  5. Vocaloid
    • one year ago
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    you're on the right track, but when we have absolute value signs, we must consider the positive and negative solutions to the equation so, going by what you have so far, we have: | 7 - 3/2x | = 8, where | | are the absolute value signs, with me so far?

  6. Destinyyyy
    • one year ago
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    Yes

  7. Vocaloid
    • one year ago
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    So, going from there, we need to write two different equations 7 - 3/2x = 8 and 7 - 3/2x = -8

  8. Vocaloid
    • one year ago
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    then we solve each one separately so, let's tackle the first one 7 - 3/2x = 8 first, subtract 7 from both sides and tell me what you get

  9. Destinyyyy
    • one year ago
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    - 3/2x= 1

  10. Vocaloid
    • one year ago
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    yup, now we divide both sides by (-3/2)

  11. Destinyyyy
    • one year ago
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    um -2/3

  12. Vocaloid
    • one year ago
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    yup, so x = -2/3 now, we're only halfway done, there's another value of x that works for the equation. now we solve 7 - (3/2)x = -8 using the same process as before. start by subtracting 7 from both sides

  13. Destinyyyy
    • one year ago
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    x=10

  14. Vocaloid
    • one year ago
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    yup, so x = 10 and x = -2/3 are your two solutions, good work ready for the next one?

  15. Destinyyyy
    • one year ago
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    Yes thank you

  16. Destinyyyy
    • one year ago
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    Solve. [x+8] +9=7 same as the one before

  17. Destinyyyy
    • one year ago
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    I got x=-10

  18. Vocaloid
    • one year ago
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    ah, not quite, there's an important rule to remember when it comes to absolute values. the absolute value of a mathematical expression cannot be negative as an example, if we have |x| = a, then a must be a positive number in order to have a solution so, looking at our problem, we first subtract 9 from each side to get |x + 8| = -2 the absolute value of any quantity must be positive, so this has no solution

  19. Destinyyyy
    • one year ago
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    Okay

  20. Vocaloid
    • one year ago
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    does that make sense? are you ready to move to the next one?

  21. Destinyyyy
    • one year ago
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    Yes it does. I thought it was but wanted to make sure.. Solve using appropriate substitution. (x^2 -6x)^2 -11(x^2 -6x) -30=0

  22. Destinyyyy
    • one year ago
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    I got 3+- square root 14 and 3+- square root 15 But its wrong...

  23. Vocaloid
    • one year ago
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    huh, this is a tricky one, hold on a moment

  24. Destinyyyy
    • one year ago
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    Okay

  25. Vocaloid
    • one year ago
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    well, I'm not sure if this is the right approach but...I'm guessing it wants us to substitute a variable for x^2 -6x so, if we let a = x^2 -6x, then our equation becomes a^2 - 11a - 30 after we apply the quadratic formula we get |dw:1440800955390:dw|

  26. Destinyyyy
    • one year ago
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    I factored (w-5)(w-6)

  27. Vocaloid
    • one year ago
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    I'm not really sure if my thinking is correct and that factoring doesn't work, that would only work for positive 30, not -30 (w-5)(w-6) = w^2 - 11w + 30

  28. Destinyyyy
    • one year ago
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    Oops

  29. Vocaloid
    • one year ago
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    ah, I wish I could be of more help on this one :( I feel like I'm missing an obvious trick here @welshfella @SithsAndGiggles

  30. Destinyyyy
    • one year ago
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    Its okay really. Your being super helpful :)

  31. anonymous
    • one year ago
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    Given that \(a^2 - 11a - 30=0\) yields roots \(a=\dfrac{11\pm\sqrt{241}}{2}\), you can find the roots to \(x^2-x-a=0\) by applying the QF again.

  32. anonymous
    • one year ago
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    typo: \(x^2-\color{red}6x-a=0\)

  33. Destinyyyy
    • one year ago
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    Um okay..

  34. dinamix
    • one year ago
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    w(1) = 5 or w(2) = 6

  35. Destinyyyy
    • one year ago
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    So... 1+- square root 5 over 2

  36. dinamix
    • one year ago
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    2 solution w(1) = 5 and w(2) = 6

  37. dinamix
    • one year ago
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    @Destinyyyy

  38. Destinyyyy
    • one year ago
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    Okay but how did you get to that?? Id like to know how to solve this

  39. dinamix
    • one year ago
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    \[\Delta = 1\]

  40. dinamix
    • one year ago
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    u should find that

  41. Destinyyyy
    • one year ago
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    The triangle is square root correct?

  42. Destinyyyy
    • one year ago
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    What?? O.o

  43. dinamix
    • one year ago
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    \[w =\frac{ 11 \pm \sqrt{121-120} }{ 2} \]

  44. Destinyyyy
    • one year ago
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    Yes I have that

  45. Destinyyyy
    • one year ago
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    Except I have a positive 120

  46. Destinyyyy
    • one year ago
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    121+120=241

  47. Destinyyyy
    • one year ago
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    Now what?

  48. dinamix
    • one year ago
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    u have mistake

  49. Destinyyyy
    • one year ago
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    Where?

  50. dinamix
    • one year ago
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    b^2-4ac = \[(-11^2) - 4*1*30\]

  51. dinamix
    • one year ago
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    u looking for this w^2-11w+30 @Destinyyyy

  52. Destinyyyy
    • one year ago
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    its a negative 30 not positive...

  53. Destinyyyy
    • one year ago
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    The original problem contains a -30

  54. dinamix
    • one year ago
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    what is your problem could u tell me what is equation

  55. dinamix
    • one year ago
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    plz

  56. Destinyyyy
    • one year ago
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    Solve using appropriate substitution. (x^2 -6x)^2 -11(x^2 -6x) -30=0

  57. Destinyyyy
    • one year ago
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    w^2 -11w -30=0 w^2--> (x^2-6x)^2 w--> x^2 -6x

  58. dinamix
    • one year ago
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    are u sure its -30 ? cuz like that we dont have good numbers

  59. Destinyyyy
    • one year ago
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    Its actually -80 ... I miss read. So sorry

  60. Destinyyyy
    • one year ago
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    Everything else is correct

  61. dinamix
    • one year ago
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    so give me right equation without mistakes

  62. dinamix
    • one year ago
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    ok

  63. Destinyyyy
    • one year ago
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    Solve using appropriate substitution. (x^2 -6x)^2 -11(x^2 -6x) -80=0 I can screenshot it if you want

  64. dinamix
    • one year ago
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    ok so w(1) = 5 and w(2) = 16

  65. Destinyyyy
    • one year ago
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    ??

  66. dinamix
    • one year ago
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    i solve it when i do w^2-11w-80=0

  67. dinamix
    • one year ago
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    we have 2 solution w(1) = -5 and w(2) = 16

  68. dinamix
    • one year ago
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    u have too check it cuz i am sure about it

  69. dinamix
    • one year ago
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    @Destinyyyy did find like that or no ?

  70. dinamix
    • one year ago
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    \[\frac{ 11\pm \sqrt{121+320} }{ 2 }\]

  71. Destinyyyy
    • one year ago
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    I havent read a thing youve said... x= 5,1 x=-2,8

  72. Destinyyyy
    • one year ago
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    Correct?

  73. Destinyyyy
    • one year ago
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    I factored

  74. Destinyyyy
    • one year ago
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    ???

  75. dinamix
    • one year ago
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    u have 2 find 6 nomber to solve this problem 8;-2;5;1;16;-5

  76. Destinyyyy
    • one year ago
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    Why 6 numbers?

  77. Destinyyyy
    • one year ago
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    The last one I did like this was 2 numbers

  78. Destinyyyy
    • one year ago
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    I dont count the w only the x's

  79. Destinyyyy
    • one year ago
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    x= 5,-2,8 I should have one more step

  80. Destinyyyy
    • one year ago
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    @jim_thompson5910

  81. dinamix
    • one year ago
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    w^2-11w-80 = 0 i find 2 numbers is -5 and 16 this only solution of W w----> x^2-6x we have find x now w=16 and w=x^2-6x------> x^2-6x=16 and also for -5 we solve this equation x^2-6x-16=0 and x^2-6x+5=0 and we find x

  82. Destinyyyy
    • one year ago
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    Yeah I already found x

  83. Destinyyyy
    • one year ago
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    x= 5 and 1 x= -2 and 8

  84. Destinyyyy
    • one year ago
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    I should have another step where I do x^2

  85. dinamix
    • one year ago
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    yup this

  86. dinamix
    • one year ago
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    did u understand now

  87. Destinyyyy
    • one year ago
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    Are my answers correct? Are they the final answer?

  88. dinamix
    • one year ago
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    yup u right

  89. Destinyyyy
    • one year ago
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    And there are no more steps?

  90. dinamix
    • one year ago
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    yeah

  91. Destinyyyy
    • one year ago
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    They are right.. I have two more problems

  92. dinamix
    • one year ago
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    u still did understand or what ??o.O

  93. Destinyyyy
    • one year ago
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    I understand how to solve that last one.

  94. Destinyyyy
    • one year ago
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    Can you help me with two more?

  95. Destinyyyy
    • one year ago
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    Im going to post them in a new question and tag you.

  96. dinamix
    • one year ago
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    my time is 1 :00 Am

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