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anonymous

  • 5 years ago

im trying to do my math hw but i dont get it we are studying Quadratic functions abd the function is y=x^2-10x+2 help please :((

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    First of all: Don't call yourself a math retard

  3. myininaya
    • 5 years ago
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    lol

  4. anonymous
    • 5 years ago
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    haha (:

  5. myininaya
    • 5 years ago
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    the first step is do not be negative lol

  6. anonymous
    • 5 years ago
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    haha

  7. myininaya
    • 5 years ago
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    what would you like to do with the "quady"

  8. anonymous
    • 5 years ago
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    lol ok ok but can i get some gelp :/

  9. anonymous
    • 5 years ago
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    *help

  10. anonymous
    • 5 years ago
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    Second of all: Its just like solving an algebraic equation

  11. myininaya
    • 5 years ago
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    i like gelp better

  12. anonymous
    • 5 years ago
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    same lol

  13. anonymous
    • 5 years ago
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    lol i know its the same but i juss dnt get it

  14. myininaya
    • 5 years ago
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    so do you want to graph the "quady"?

  15. myininaya
    • 5 years ago
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    silence satellite!

  16. anonymous
    • 5 years ago
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    i dnt think im supposed to graph i think i juss needa find the vertex and coordinates :/ or maybe i am idk >.<

  17. anonymous
    • 5 years ago
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    myininaya izzat you?

  18. myininaya
    • 5 years ago
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    omg 465 so lame you are leaving me in the dust the more we find the better are graph will look lets start out with x-intercepts

  19. myininaya
    • 5 years ago
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    to find x-intercepts, set y=0 and solve for x

  20. myininaya
    • 5 years ago
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    x^2-10x+2=0

  21. anonymous
    • 5 years ago
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    okk then what :o

  22. myininaya
    • 5 years ago
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    now this is totally not factorble so you have to use the quady formula!

  23. myininaya
    • 5 years ago
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    \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  24. myininaya
    • 5 years ago
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    satellite i was kidding you can talk lol

  25. myininaya
    • 5 years ago
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    \[x=\frac{10 \pm \sqrt{10^2-4(1)(2)}}{2(1)}\]

  26. anonymous
    • 5 years ago
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    sooo after tht wat next o.o

  27. myininaya
    • 5 years ago
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    \[x=\frac{10 \pm \sqrt{100-8}}{2}=\frac{10 \pm \sqrt{92}}{2}=\frac{10 \pm \sqrt{4*23}}{2}\]

  28. myininaya
    • 5 years ago
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    \[x=\frac{10 \pm 2\sqrt{23}}{2}=5 \pm \sqrt{23}\]

  29. anonymous
    • 5 years ago
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    lordamercy \[x^2-10x+2=0\] \[x^2-20x=2\] \[(x-5)^2=-2+25=23\] \[x-5=\pm\sqrt{23}\] \[x=5\pm\sqrt{23}\]

  30. myininaya
    • 5 years ago
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    so these two are where your graph is going to the intercept the x axis

  31. myininaya
    • 5 years ago
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    satellite used completeing the square (which is used to proved the quady formula is the solution of the quady equation)

  32. anonymous
    • 5 years ago
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    oops typo. second line should have been \[x^2-10x=-2\]

  33. anonymous
    • 5 years ago
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    also it is nice to use when the middle coefficient is even so you do not have to simplify the radical, factor and cancel

  34. myininaya
    • 5 years ago
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    yes that too

  35. anonymous
    • 5 years ago
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    the quadratic formula forces a denominator on you but you see that there isn't one!

  36. anonymous
    • 5 years ago
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    i will be quiet now

  37. myininaya
    • 5 years ago
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    now lets take a break now from the hard stuff and see if we can find a y-intercept

  38. myininaya
    • 5 years ago
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    to find the y-intercept, set x=0 and solve for y

  39. myininaya
    • 5 years ago
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    y=0^2-10(0)+2=0-0+2=2

  40. anonymous
    • 5 years ago
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    smhh i know you guys are gonna get mad but...i still dnt get it :[

  41. myininaya
    • 5 years ago
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    i'm furious lol

  42. anonymous
    • 5 years ago
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    i'll just take my frends answers in school tommorow lol

  43. myininaya
    • 5 years ago
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    i was kidding

  44. myininaya
    • 5 years ago
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    what part is getting you down?

  45. anonymous
    • 5 years ago
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    i know lols

  46. anonymous
    • 5 years ago
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    the whooooole thing D:

  47. myininaya
    • 5 years ago
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    do you know the quadratic formula?

  48. anonymous
    • 5 years ago
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    ax^2=bx+c?

  49. anonymous
    • 5 years ago
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    *ax^2+bx+c

  50. myininaya
    • 5 years ago
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    \[ax^2+bx+c=0\] -> quad equation \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] -> quad formula (solution to the quad equation)

  51. anonymous
    • 5 years ago
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    mhmm :x

  52. anonymous
    • 5 years ago
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    i thought it was just -b/2a

  53. myininaya
    • 5 years ago
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    no

  54. anonymous
    • 5 years ago
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    wow ok :/

  55. anonymous
    • 5 years ago
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    bt anyway i go sleep now lols so ttyl

  56. myininaya
    • 5 years ago
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    ok gn

  57. anonymous
    • 5 years ago
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    gn ^_^

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