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anonymous

  • 5 years ago

ƒ(x) = √(x²+9) Is this a semi circle? :S

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  1. anonymous
    • 5 years ago
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    no it's not. you see y=√(x²+9)>0, also y^2/9-x^2/9=1, therefore the graph is half of the hyperbola whose focus is (0,3√2) and which is above the x-axis.

  2. anonymous
    • 5 years ago
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    have I made you understood??

  3. anonymous
    • 5 years ago
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    where you get y^2/9-x^2/9=1 from? is that equation for hyperbola?

  4. anonymous
    • 5 years ago
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    square y then move it over so that you have 9 = y^2-x^2 then divide by nine to get 1=y^2/9-x^2/9

  5. anonymous
    • 5 years ago
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    oh i see now, i was thinking the x was in the denominator for y² because of computer writing ;)

  6. anonymous
    • 5 years ago
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    haha no, ken is just showing you that the equation is actually a hyperbola, and he proved it by simplifying

  7. anonymous
    • 5 years ago
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    yes. it;s like that. a standard hyperbola equation is x^2/a^2-y^2/b^2=1 or y^2/a^2-x^2/b^2=1

  8. anonymous
    • 5 years ago
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    the former has its foci on the x-axis, and the latter has its foci on the y-axis.

  9. anonymous
    • 5 years ago
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    kk, all good now. cheers

  10. anonymous
    • 5 years ago
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    cheers(^^)

  11. anonymous
    • 5 years ago
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    www.aceyourcollegeclasses.com is a great place to post your questions

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