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anonymous
 one year ago
Find the vertices and foci of the hyperbola with equation quantity x minus five squared divided by eighty one minus the quantity of y minus one squared divided by one hundred and forty four = 1
anonymous
 one year ago
Find the vertices and foci of the hyperbola with equation quantity x minus five squared divided by eighty one minus the quantity of y minus one squared divided by one hundred and forty four = 1

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{(x5)^2}{81}\frac{(y1)^2}{4}=1\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is way easier than you think but you need to know two things first a) the center and b) which way it is oriented

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know that the hyperbolas are oriented left/right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i thought ya all that earlier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 you mean what i taught him?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so now what can I do?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{(x5)^2}{81}\frac{(y1)^2}{4}=1\]\[\frac{(xh)^2}{a^2}\frac{(yk)^2}{b^2}=1\] what is \(a\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats why i told ya to put em into ^2 form

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh sorry. a = 9, b = 12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, so c = 15 because c^2 = 225

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you know the hyperbola is transverse in xaxis so ( center +/ c, center) is your foci

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and same for the vertices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait. These are my choices: A. Vertices: (17, 1), (7, 1); Foci: (7, 1), (17, 1) B. Vertices: (14, 1), (4, 1); Foci: (10, 1), (20, 1) C. Vertices: (1, 14), (1, 4); Foci: (1, 10), (1, 20) D. Vertices: (1, 17), (1, 7); Foci: (1, 7), (1, 17) So I know it has to be between A and B.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find an equation in standard form for the hyperbola with vertices at (0, ±3) and foci at (0, ±7). If my answer choices are: A. y squared over nine minus x squared over forty = 1 B. y squared over forty minus x squared over nine = 1 C. y squared over forty nine minus x squared over nine = 1 D. y squared over nine minus x squared over forty nine = 1 I was thinking it was between B and D.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you see foci is c^2 = a^2+b^2 so we get 7^2  3^2 = 40

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok. So that would make it A.
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