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anonymous

  • 5 years ago

The vertices of the hyperbola 25^(2)-9x^(2)-200+36+139=0 are a) (-1,4) and (5,4) b) (2,9) and (2,-1) c) (2,7) and (2,-1) d) (2,9) and (2,-1)

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  1. anonymous
    • 5 years ago
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    25y^(2)-9x^(2)-200y+36x+139=0***

  2. anonymous
    • 5 years ago
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    hi chizzle, make them complete squares .This will give you (5y+20)^2 - (3x -6)^2 -100 +36 +139 = 0. This form should now look familiar.(standard hyperbola form).

  3. anonymous
    • 5 years ago
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    p.s. there might be a calculation error so check it once but I'm sure this is the general idea. you then can find out how much the graph has been shifted by looking a the modifiers for x and y .i.e x-> x-a; y->y-b

  4. anonymous
    • 5 years ago
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    hi chizzle, make them complete squares .This will give you (5y+20)^2 - (3x -6)^2 -100 +36 +139 = 0. This form should now look familiar.(standard hyperbola form).

  5. anonymous
    • 5 years ago
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    What I got before was \[2, 4\pm \sqrt{171/5}\] but that is not one of the answers. I need in the form of coordinates: (-1,4) and (5,4) (2,9) and (2,-1) (2,7) and (2,-1)

  6. anonymous
    • 5 years ago
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    hint: centre = (2,-5) ;a = 3 ;b = 5

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