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anonymous

  • one year ago

y = 5 y = 3x - 4 PLEASE HELP ME WITH 3 QUESTIONS THAT'S ALL JUST 3!!

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  1. Michele_Laino
    • one year ago
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    here we can write this: \[\Large 3x - 4 = 5\]

  2. anonymous
    • one year ago
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    Okay

  3. Michele_Laino
    • one year ago
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    now I add 4 to both sides: \[\Large 3x - 4 + 4 = 5 + 4\] please continue

  4. anonymous
    • one year ago
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    3,9

  5. anonymous
    • one year ago
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    is 3,9 right or is it wrong what is it , I'm really frustrated so ....

  6. Michele_Laino
    • one year ago
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    after a simplification, we can write: \[\Large 3x = 9\]

  7. Michele_Laino
    • one year ago
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    now, please divide by 3, both sides

  8. anonymous
    • one year ago
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    what sides am I dividing it by

  9. anonymous
    • one year ago
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    I got 3

  10. Michele_Laino
    • one year ago
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    hint: |dw:1437154156003:dw|

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

  12. anonymous
    • one year ago
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    so is it 3,3

  13. Michele_Laino
    • one year ago
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    it is x=3

  14. anonymous
    • one year ago
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    I have to write it as an ordered pair

  15. Michele_Laino
    • one year ago
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    ok! we have: x=3 and y=5, so we can write: (3,5)

  16. anonymous
    • one year ago
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    Find the solution(s) to the system, if one or more exist. Write your answer as an ordered pair. Round to the nearest hundredth, if necessary. \begin{array}{l}y = {x^2} + 4x - 3\\x = 1\end{array} this is the other one, I have one more I need help on

  17. Michele_Laino
    • one year ago
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    we have to substitute x=1 into the first equation, so we can write: \[y = {1^2} + 4 \times 1 - 3 = ...?\]

  18. anonymous
    • one year ago
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    okay

  19. Michele_Laino
    • one year ago
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    please continue, what is y?

  20. anonymous
    • one year ago
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    42

  21. anonymous
    • one year ago
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    that;s what I got for y

  22. Michele_Laino
    • one year ago
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    hint: \[y = {1^2} + 4 \times 1 - 3 = 1 + 4 - 3 = ...?\]

  23. anonymous
    • one year ago
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    I got 3, 2

  24. Michele_Laino
    • one year ago
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    we have: y=2 and x=1 so we can write: (1,2)

  25. anonymous
    • one year ago
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    is that the answer?

  26. Michele_Laino
    • one year ago
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    yes!

  27. anonymous
    • one year ago
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    Explain how to find the solution(s) to the system. Give the solutions(s) as ordered pairs rounded to the nearest hundredth, if necessary. \begin{array}{l}y = - 0.5{x^2} + x + 6\\y = 3.25\end{array} this is the last question I need help on I need help finding the ordered pairs

  28. Michele_Laino
    • one year ago
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    we have to apply the transitive property, so we can write: \[\Large - 0.5{x^2} + x + 6 = 3.25\]

  29. anonymous
    • one year ago
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    ok

  30. Michele_Laino
    • one year ago
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    it is a quadratic equation, since we can rewrite it as follows: \[\Large 0.5{x^2} - x - 2.75 = 0\]

  31. anonymous
    • one year ago
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    ok

  32. Michele_Laino
    • one year ago
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    or using fraction instead decimal numbers, we get: \[\Large \frac{{{x^2}}}{2} - x - \frac{{11}}{4} = 0\]

  33. anonymous
    • one year ago
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    ok

  34. Michele_Laino
    • one year ago
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    now, I multiply both sides by 4, so I can write: \[\Large \begin{gathered} 4 \times \left( {\frac{{{x^2}}}{2} - x - \frac{{11}}{4}} \right) = 0 \times 4 \hfill \\ 4 \times \frac{{{x^2}}}{2} + 4 \times \left( { - x} \right) + 4 \times \left( { - \frac{{11}}{4}} \right) = 0 \hfill \\ 2{x^2} - 4x - 11 = 0 \hfill \\ \end{gathered} \]

  35. anonymous
    • one year ago
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    ok

  36. Michele_Laino
    • one year ago
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    note that equation is an equation like this: \[\Large a{x^2} + bx + c = 0\] where: \[\Large \begin{gathered} a = 2 \hfill \\ b = - 4 \hfill \\ c = - 11 \hfill \\ \end{gathered} \]

  37. anonymous
    • one year ago
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    okay

  38. Michele_Laino
    • one year ago
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    so we can apply the standard formula: \[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}\]

  39. Michele_Laino
    • one year ago
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    in order to find the values of x

  40. anonymous
    • one year ago
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    okay

  41. Michele_Laino
    • one year ago
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    after a simple substitution, we get: \[\large \begin{gathered} x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}} = \frac{{4 \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \times 2 \times \left( { - 11} \right)} }}{{2 \times 2}} = \hfill \\ \hfill \\ = \frac{{4 \pm \sqrt {16 + 88} }}{4} \hfill \\ \end{gathered} \] please continue

  42. anonymous
    • one year ago
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    I got 7

  43. Michele_Laino
    • one year ago
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    hint: we have 2 values for x: \[\Large \begin{gathered} {x_1} = \frac{{4 + \sqrt {16 + 88} }}{4} = ...? \hfill \\ \hfill \\ {x_2} = \frac{{4 - \sqrt {16 + 88} }}{4} = ...? \hfill \\ \end{gathered} \]

  44. anonymous
    • one year ago
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    42

  45. Michele_Laino
    • one year ago
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    hint: \[\begin{gathered} {x_1} = \frac{{4 + \sqrt {16 + 88} }}{4} = \frac{{4 + \sqrt {104} }}{4} = \frac{{4 + 10.2}}{4} = \frac{{14.2}}{4} = ...? \hfill \\ \hfill \\ {x_2} = \frac{{4 - \sqrt {16 + 88} }}{4} = \frac{{4 - \sqrt {104} }}{4} = \frac{{4 - 10.2}}{4} = \frac{{ - 6.2}}{4} = ...? \hfill \\ \end{gathered} \]

  46. anonymous
    • one year ago
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    14.24 -6.24

  47. Michele_Laino
    • one year ago
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    you have to divide by 4, namely 14.20/4=...? -6.20/4=...?

  48. anonymous
    • one year ago
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    3.55 , 1.55

  49. anonymous
    • one year ago
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    -1.55

  50. Michele_Laino
    • one year ago
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    we have: x1= 3.55 x2=-1.55

  51. Michele_Laino
    • one year ago
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    ok!

  52. anonymous
    • one year ago
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    ok

  53. Michele_Laino
    • one year ago
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    so the requested pairs, are: (3.55, 3.25) (-1.55, 3.25)

  54. anonymous
    • one year ago
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    Thank you

  55. anonymous
    • one year ago
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    i need two that I wanted you to check

  56. Michele_Laino
    • one year ago
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    ok!

  57. anonymous
    • one year ago
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    Find the solution(s) to the system, if one or more exist. Write your answer as an ordered pair. \begin{array}{l}y = 3{x^2} + 2x + 4\\x = 4\end{array} I have 156,4 as my answer

  58. Michele_Laino
    • one year ago
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    substituting x=4 into the first equatiuon, we get: \[y = 3 \times {4^2} + 2 \times 4 + 4 = 3 \times 16 + 8 + 4 = 48 + 12 = 60\]

  59. Michele_Laino
    • one year ago
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    so the requested pair is: (4, 60)

  60. anonymous
    • one year ago
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    one of the answer they have is (4,60) and (60 ,4)

  61. anonymous
    • one year ago
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    thank you

  62. Michele_Laino
    • one year ago
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    :)

  63. anonymous
    • one year ago
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    the one with 3,5 wasn't right

  64. anonymous
    • one year ago
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    right

  65. Michele_Laino
    • one year ago
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    please what question do you refer to?

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