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anonymous

  • one year ago

what are the equations to these diagrams? (attachment) please help will medal!!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    PLEASE HELP!

  4. welshfella
    • one year ago
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    the first one is a quadratic equation with a negative coefficient of x^2. It has zeros of 2 and -8. and a vertex at (-3,6) can you find the equation from this info?

  5. anonymous
    • one year ago
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    For the 1st diagram \[-(x+3)^2=4(y-6)\]

  6. anonymous
    • one year ago
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    is this right? −(x+3)2=4(y−6) @welshfella

  7. anonymous
    • one year ago
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    the ellipse equation is \[\frac{ (y-4)^2 }{ 6^2 }+\frac{ (x-3)^2 }{ 4^2 }=1\]

  8. anonymous
    • one year ago
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    for the first one? @saseal

  9. anonymous
    • one year ago
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    yes the first picture

  10. anonymous
    • one year ago
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    what about the second? @saseal

  11. welshfella
    • one year ago
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    yes that is correct for the first graph

  12. anonymous
    • one year ago
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    im doing the circle now, its \[(x+6)^2+(y+2)^2=5^2\]

  13. anonymous
    • one year ago
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    what about #13 and #14? then i will figure out the rest myself thank you for all ur help @saseal

  14. anonymous
    • one year ago
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    \[\frac{ (x-7)^2 }{ 2^2 }-\frac{ y-2)^2 }{ 2^2}=1\] for the hyperbola

  15. anonymous
    • one year ago
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    circle is #13 and hyperbola is #14

  16. anonymous
    • one year ago
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    what about the one on the top right?

  17. anonymous
    • one year ago
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    the oval @saseal

  18. anonymous
    • one year ago
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    2nd picture?

  19. anonymous
    • one year ago
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    yes

  20. anonymous
    • one year ago
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    gimme a few minutes

  21. anonymous
    • one year ago
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    thank you so much

  22. anonymous
    • one year ago
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    First oval is \[\frac{ (x-7)^2 }{ 6^2 }+\frac{ y^2 }{ 3^2 }=1\]

  23. anonymous
    • one year ago
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    if i tag you in some other ones on a different question could you help?

  24. welshfella
    • one year ago
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    plz dont just give answers saseal Code of Conduct asks us to Guide the user to the answers

  25. anonymous
    • one year ago
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    okies

  26. anonymous
    • one year ago
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    so I can't give ya answer now, work the last one out yourself; parabola equation for that curve looks something like this :) \[(y+1)^2=4p(x-4)^2\]

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