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ksaimouli

  • 3 years ago

find intersection in terms of y of y^2-4x=4and 4x-y=16

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  1. ksaimouli
    • 3 years ago
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    \[y^2-4x=4 and 4x-y=16\]

  2. ksaimouli
    • 3 years ago
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    @ZeHanz

  3. Hero
    • 3 years ago
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    Hint: y = 4x - 16

  4. ksaimouli
    • 3 years ago
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    i did but i dont know whre i was wrong

  5. ksaimouli
    • 3 years ago
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    i know dude to solve for x and set it eual and then solve for x

  6. ksaimouli
    • 3 years ago
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    |dw:1357491125167:dw|

  7. ksaimouli
    • 3 years ago
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    |dw:1357491191134:dw|

  8. Hero
    • 3 years ago
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    Don't make it complicated bro

  9. ksaimouli
    • 3 years ago
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    |dw:1357491240591:dw|

  10. ksaimouli
    • 3 years ago
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    solve for y am i right this is to find area between curve sooo i must to with respect y

  11. Hero
    • 3 years ago
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    Actually, that's a good approach after all.

  12. ksaimouli
    • 3 years ago
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    but something is wrong in solving hmm i am not getting the answer

  13. ksaimouli
    • 3 years ago
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    multiply both sides by 4

  14. Hero
    • 3 years ago
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    What did you get for the answer?

  15. ksaimouli
    • 3 years ago
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    .08 and -12.08

  16. Hero
    • 3 years ago
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    That's what you got for y values?

  17. ksaimouli
    • 3 years ago
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    nope

  18. ksaimouli
    • 3 years ago
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    i gote it 5 and -4

  19. Hero
    • 3 years ago
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    Okay, so what's the problem?

  20. ksaimouli
    • 3 years ago
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    arithmetic mistake lol

  21. ksaimouli
    • 3 years ago
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    now i have to integrate thats it

  22. ksaimouli
    • 3 years ago
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    |dw:1357491558149:dw|

  23. ksaimouli
    • 3 years ago
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    but not sure which one goes top and which goes bottom with respect to y

  24. Hero
    • 3 years ago
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    Here's another way to do it to avoid confusion: \[\frac{y^2 - 4}{4} - \frac{16 + y}{4} = 0\]

  25. ksaimouli
    • 3 years ago
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    ya i tried that worked well

  26. Hero
    • 3 years ago
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    Bro, a is always the lowest value. b is always the highest value.

  27. ksaimouli
    • 3 years ago
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    i mean the function not the points (curves)

  28. ksaimouli
    • 3 years ago
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    \[x=\frac{ y^2-4 }{ 4 }\]

  29. Hero
    • 3 years ago
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    Are you sure you're supposed to be using y values for a and b?

  30. ksaimouli
    • 3 years ago
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    or\[x=\frac{ 16+y }{ 4 }\]

  31. ksaimouli
    • 3 years ago
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    ya i am damn sure

  32. Hero
    • 3 years ago
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    I remember these. They can be tricky. What you do is just turn your graph sideways. And it will be the same as like when you're working with x-axis.

  33. ksaimouli
    • 3 years ago
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    if \[\sqrt{y}\]

  34. ksaimouli
    • 3 years ago
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    |dw:1357491948129:dw|

  35. ksaimouli
    • 3 years ago
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    ?

  36. ksaimouli
    • 3 years ago
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    alright thx

  37. Hero
    • 3 years ago
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    Bro, I graphed it as \[\frac{16 + x}{4}\] and \[\frac{x^2 - 4}{4}\] You get the same area when integrating from a = -4 to b = -5 The only difference is you're using variable y instead of variable x

  38. Hero
    • 3 years ago
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    @ksaimouli

  39. ksaimouli
    • 3 years ago
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    ya i noticed it thx

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