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anonymous

  • one year ago

What is the simplified form of ?

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. anonymous
    • one year ago
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    hold please<3

  3. Michele_Laino
    • one year ago
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    ok!

  4. anonymous
    • one year ago
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    |dw:1437504225310:dw|

  5. Michele_Laino
    • one year ago
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    the common denominator of these 2 fractions: 1/x and 1/y is xy

  6. anonymous
    • one year ago
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    i see

  7. Michele_Laino
    • one year ago
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    we can rewrite your numerator as follows: \[\frac{1}{y} - \frac{1}{x} = \frac{{x - y}}{{xy}}\]

  8. anonymous
    • one year ago
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    Okay:)

  9. Michele_Laino
    • one year ago
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    similarly, we have: \[\large \frac{1}{y} + \frac{1}{x} = \frac{{...}}{{xy}}\] please complete

  10. anonymous
    • one year ago
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    2 at the top

  11. Michele_Laino
    • one year ago
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    what is \[\frac{{xy}}{y}\]

  12. anonymous
    • one year ago
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    x

  13. Michele_Laino
    • one year ago
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    perfect, so we have: \[\frac{1}{y} + \frac{1}{x} = \frac{{x \cdot 1 + ...}}{{xy}}\]

  14. anonymous
    • one year ago
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    2?

  15. Michele_Laino
    • one year ago
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    now, what is: \[\frac{{xy}}{x}\]

  16. anonymous
    • one year ago
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    y

  17. Michele_Laino
    • one year ago
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    correct! so we have: \[\frac{1}{y} + \frac{1}{x} = \frac{{x \cdot 1 + y \cdot 1}}{{xy}} = \frac{{x + y}}{{xy}}\]

  18. Michele_Laino
    • one year ago
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    then your original expression can be rewritten as follows: \[\Large \frac{{\frac{1}{y} - \frac{1}{x}}}{{\frac{1}{y} + \frac{1}{x}}} = \frac{{\frac{{x - y}}{{xy}}}}{{\frac{{x + y}}{{xy}}}}\]

  19. anonymous
    • one year ago
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    can't we simplify that??

  20. Michele_Laino
    • one year ago
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    yes! sure we can write the next step as the multiplication of the first fraction, by the inverse of the second fraction, namely: \[\Large \frac{{\frac{1}{y} - \frac{1}{x}}}{{\frac{1}{y} + \frac{1}{x}}} = \frac{{\frac{{x - y}}{{xy}}}}{{\frac{{x + y}}{{xy}}}} = \frac{{x - y}}{{xy}} \cdot \frac{{xy}}{{x + y}} = ...\]

  21. anonymous
    • one year ago
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    x^2-y^2??

  22. Michele_Laino
    • one year ago
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    hint:|dw:1437504964076:dw|

  23. anonymous
    • one year ago
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    x-y/x+y

  24. Michele_Laino
    • one year ago
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    that's right!

  25. anonymous
    • one year ago
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    i need 1 more?

  26. Michele_Laino
    • one year ago
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    ok! :)

  27. anonymous
    • one year ago
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    Using a directrix of y = 5 and a focus of (4, 1), what quadratic function is created?

  28. anonymous
    • one year ago
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    thanks!

  29. Michele_Laino
    • one year ago
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    that is a parabola whose equation is like this: \[y = a{x^2} + bx + c\]

  30. Michele_Laino
    • one year ago
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    we have to determina the corfficients a, b, and c

  31. Michele_Laino
    • one year ago
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    determine*

  32. anonymous
    • one year ago
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    ok

  33. anonymous
    • one year ago
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    so then what would I do?? Plug in?

  34. Michele_Laino
    • one year ago
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    the coordinates of the focus, as a function of a, b, and c, are: \[F = \left( { - \frac{b}{{2a}},\frac{{1 - {b^2} + 4ac}}{{4a}}} \right)\]

  35. Michele_Laino
    • one year ago
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    so we can write: \[\begin{gathered} - \frac{b}{{2a}} = 4 \hfill \\ \hfill \\ \frac{{1 - {b^2} + 4ac}}{{4a}} = 1 \hfill \\ \end{gathered} \]

  36. anonymous
    • one year ago
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    Ok :)

  37. anonymous
    • one year ago
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    Where would the (4,1) go?

  38. Michele_Laino
    • one year ago
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    the directrix is: \[y = - \frac{{1 + {b^2} - 4ac}}{{4a}}\]

  39. Michele_Laino
    • one year ago
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    so we can write, by comparison: \[ - \frac{{1 + {b^2} - 4ac}}{{4a}} = 5\]

  40. anonymous
    • one year ago
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    ok :)

  41. anonymous
    • one year ago
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    now what would we do?

  42. Michele_Laino
    • one year ago
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    we have to solve this algebraic system: \[\left\{ \begin{gathered} - \frac{{1 + {b^2} - 4ac}}{{4a}} = 5 \hfill \\ - \frac{b}{{2a}} = 4 \hfill \\ \hfill \\ \frac{{1 - {b^2} + 4ac}}{{4a}} = 1 \hfill \\ \end{gathered} \right.\]

  43. anonymous
    • one year ago
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    Ohhh I understand!

  44. Michele_Laino
    • one year ago
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    I consider the first and the third equations

  45. anonymous
    • one year ago
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    would thr answer be 1/8(x+4)^2-3

  46. Michele_Laino
    • one year ago
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    I don't know

  47. anonymous
    • one year ago
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    ahh :( It's fine

  48. Michele_Laino
    • one year ago
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    If I multiply both sides of the first equation by -4a, I get: \[1 + {b^2} - 4ac = - 20a\]

  49. Michele_Laino
    • one year ago
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    whereas if I multiply both sides of the third equation, by 4a, I get: \[1 - {b^2} + 4ac = 4a\]

  50. Michele_Laino
    • one year ago
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    If I sum those 2 equations together, I get: \[2 = - 16a\]

  51. Michele_Laino
    • one year ago
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    am I right?

  52. anonymous
    • one year ago
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    yup! I believe in you :)

  53. Michele_Laino
    • one year ago
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    lol

  54. Michele_Laino
    • one year ago
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    ok! :)

  55. Michele_Laino
    • one year ago
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    what is a?

  56. anonymous
    • one year ago
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    -8

  57. Michele_Laino
    • one year ago
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    hint: |dw:1437506094548:dw|

  58. Michele_Laino
    • one year ago
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    so, a=...?

  59. anonymous
    • one year ago
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    1/-8

  60. Michele_Laino
    • one year ago
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    correct!

  61. Michele_Laino
    • one year ago
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    now, I take the second equation: \[ - \frac{b}{{2a}} = 4\]

  62. Michele_Laino
    • one year ago
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    Multiplying both sides by -2a, I can rewrite it as follows: \[b = - 8a\]

  63. Michele_Laino
    • one year ago
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    substituting a, we get: \[b = \left( { - 8} \right) \times \left( { - \frac{1}{8}} \right) = ...\]

  64. anonymous
    • one year ago
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    1

  65. Michele_Laino
    • one year ago
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    correct! so we have: a= -1/8, b=1

  66. Michele_Laino
    • one year ago
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    now I take the third equation: \[\frac{{1 - {b^2} + 4ac}}{{4a}} = 1\]

  67. anonymous
    • one year ago
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    |dw:1437506420592:dw|

  68. anonymous
    • one year ago
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    Is that the answer? :)

  69. Michele_Laino
    • one year ago
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    I don't know, since I solve the question now

  70. anonymous
    • one year ago
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    OK:) i'LL wait

  71. Michele_Laino
    • one year ago
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    I substitute into the third equation, these values: a= -1/8, and b=1, so I get: \[\frac{{1 - {1^2} + 4 \times \left( { - \frac{1}{8}} \right)c}}{{4 \times \left( { - \frac{1}{8}} \right)}} = 1\]

  72. Michele_Laino
    • one year ago
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    what is c?

  73. anonymous
    • one year ago
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    hmmm 4?

  74. Michele_Laino
    • one year ago
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    I got: c=1

  75. anonymous
    • one year ago
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    |dw:1437506785539:dw|

  76. Michele_Laino
    • one year ago
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    I don't understand the second fraction

  77. anonymous
    • one year ago
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    6Y^2/ 5X^4

  78. Michele_Laino
    • one year ago
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    we can make these simplifications: |dw:1437506923714:dw|

  79. anonymous
    • one year ago
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    so 3x^2/10y3 times 3/1

  80. Michele_Laino
    • one year ago
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    that's right!

  81. anonymous
    • one year ago
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    yay!

  82. anonymous
    • one year ago
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    would the answer be 9x^2/10y^3

  83. Michele_Laino
    • one year ago
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    yes! correct!

  84. anonymous
    • one year ago
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    last one?

  85. Michele_Laino
    • one year ago
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    ok! :)

  86. anonymous
    • one year ago
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    i got c

  87. Michele_Laino
    • one year ago
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    that's right, since we can write our polynomial as below: \[f\left( x \right) = \left( {x + 7} \right)q\left( x \right)\] where q(x) is a appropriate polynomial

  88. Michele_Laino
    • one year ago
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    we have to solve this equation: \[{x^2} + 2x - 15 = 0\]

  89. anonymous
    • one year ago
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    so then you get (x-5)(x+3)

  90. Michele_Laino
    • one year ago
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    the solutions are: \[\begin{gathered} x = \frac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times \left( { - 15} \right)} }}{{2 \times 1}} = \hfill \\ \hfill \\ = \frac{{ - 2 \pm \sqrt {4 + 60} }}{2} = \frac{{ - 2 \pm 8}}{2} = \begin{array}{*{20}{c}} {\frac{{ - 2 - 8}}{2} = - 5} \\ {\frac{{ - 2 + 8}}{2} = 3} \end{array} \hfill \\ \end{gathered} \]

  91. anonymous
    • one year ago
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    so would those be the discontuiny?

  92. Michele_Laino
    • one year ago
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    the factorization is: \[{x^2} + 2x - 15 = \left( {x + 5} \right)\left( {x - 3} \right)\]

  93. Michele_Laino
    • one year ago
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    yes! your discontinuities are located at x=-5 and x=3

  94. anonymous
    • one year ago
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    Which polynomial identity will prove that 16 = 25 − 9?

  95. anonymous
    • one year ago
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    Difference of Squares Difference of Cubes Sum of Cubes Square of a Binomial

  96. anonymous
    • one year ago
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    is it A?

  97. Michele_Laino
    • one year ago
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    I think difference of squares, since I can write: \[\left( {x - y} \right)\left( {x + y} \right) = {x^2} - {y^2},\quad x = 5,y = 3\]

  98. anonymous
    • one year ago
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    no. For the equation 16= 25-9

  99. Michele_Laino
    • one year ago
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    yes! please substitute x=5 and y=3, what do you get?

  100. anonymous
    • one year ago
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    sub. into what?

  101. Michele_Laino
    • one year ago
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    hint: we have the subsequent steps: \[\begin{gathered} \left( {5 - 3} \right)\left( {5 + 3} \right) = {5^2} - {3^2} \hfill \\ 2 \times 8 = 25 - 9 \hfill \\ 16 = 25 - 9 \hfill \\ \end{gathered} \]

  102. anonymous
    • one year ago
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    Oh! So it IS A

  103. Michele_Laino
    • one year ago
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    yes!

  104. anonymous
    • one year ago
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    yay! Solve 3x2 + 13x = 10.

  105. anonymous
    • one year ago
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    i got 2/3 and -5

  106. Michele_Laino
    • one year ago
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    I rewrite your equation as follows: \[3{x^2} + 13x - 10 = 0\]

  107. anonymous
    • one year ago
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    ok:)

  108. anonymous
    • one year ago
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    use the quad. formula

  109. Michele_Laino
    • one year ago
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    so I get: \[\begin{gathered} x = \frac{{ - 13 \pm \sqrt {169 + 4 \times 3 \times 10} }}{{2 \times 3}} = \hfill \\ \hfill \\ = \frac{{ - 13 \pm \sqrt {289} }}{6} = \begin{array}{*{20}{c}} {\frac{{ - 13 + 17}}{2} = 2} \\ {\frac{{ - 13 - 17}}{2} = - 15} \end{array} \hfill \\ \end{gathered} \]

  110. anonymous
    • one year ago
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    Thanks!!!!

  111. Michele_Laino
    • one year ago
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    oops.. I have made an error

  112. Michele_Laino
    • one year ago
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    \[\begin{gathered} x = \frac{{ - 13 \pm \sqrt {169 + 4 \times 3 \times 10} }}{{2 \times 3}} = \hfill \\ \hfill \\ = \frac{{ - 13 \pm \sqrt {289} }}{6} = \begin{array}{*{20}{c}} {\frac{{ - 13 + 17}}{6} = \frac{2}{3}} \\ {\frac{{ - 13 - 17}}{6} = - 5} \end{array} \hfill \\ \end{gathered} \]

  113. anonymous
    • one year ago
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    just factor instead of using the formula bc that diddn't work for me

  114. anonymous
    • one year ago
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    yayyy!!!!

  115. anonymous
    • one year ago
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    What are the zeros of the polynomial function f(x) = x3 + x2 − 20x?

  116. Michele_Laino
    • one year ago
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    we have to factor out x, so we can write: \[{x^3} + {x^2} - 20x = x\left( {{x^2} + x - 20} \right)\]

  117. Michele_Laino
    • one year ago
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    now I solve the quadratic equation with the standard formula: \[\begin{gathered} {x^2} + x - 20 = 0 \hfill \\ x = \frac{{ - 1 \pm \sqrt {1 + 4 \times 1 \times 20} }}{{2 \times 1}} = \frac{{ - 1 \pm 9}}{2} = \hfill \\ \hfill \\ = \begin{array}{*{20}{c}} {\frac{{ - 1 - 9}}{2} = - 5} \\ {\frac{{ - 1 + 9}}{2} = 4} \end{array} \hfill \\ \end{gathered} \]

  118. anonymous
    • one year ago
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    so 0,-5,4

  119. Michele_Laino
    • one year ago
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    that's right!

  120. anonymous
    • one year ago
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    |dw:1437508513154:dw|

  121. Michele_Laino
    • one year ago
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    the common denominator is: (x-5)(x+5)

  122. Michele_Laino
    • one year ago
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    so we can simplify your expression as follows: \[\frac{{\left( {4x + 2} \right)\left( {x + 5} \right) - \left( {3x - 1} \right)\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} = ...\] please complete my computation

  123. anonymous
    • one year ago
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    1x+1/(x+5)(X-5)

  124. Michele_Laino
    • one year ago
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    I got a different result

  125. anonymous
    • one year ago
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    expalin please

  126. Michele_Laino
    • one year ago
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    we can write this: \[\begin{gathered} \frac{{\left( {4x + 2} \right)\left( {x + 5} \right) - \left( {3x - 1} \right)\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} = \hfill \\ \hfill \\ = \frac{{4{x^2} + 22x + 10 - 3{x^2} + 16x - 5}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} = ... \hfill \\ \end{gathered} \]

  127. anonymous
    • one year ago
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    x^2+38x+5/(x-5)(x+5)

  128. Michele_Laino
    • one year ago
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    that's right!

  129. anonymous
    • one year ago
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    so now what do we do?

  130. Michele_Laino
    • one year ago
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    I think that we have completed our exercise

  131. anonymous
    • one year ago
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    KK :) tHANKS SO MUCH FOR ALL YOU HELP!!!!!!! <3 <3 <3

  132. Michele_Laino
    • one year ago
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    :)

  133. anonymous
    • one year ago
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  134. Michele_Laino
    • one year ago
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    thanks! :)

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spraguer (Moderator)
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