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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)
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.025y^(2)9x^(2)200y+36x+139=0***

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hi 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0p.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> xa; y>yb

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hi 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hint: centre = (2,5) ;a = 3 ;b = 5
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