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Mathematics 18 Online
OpenStudy (anonymous):

What are the slopes of the asymptotes for the hyperbola with the equation (x - 2)2 /16 - (y + 3)2 / 25 = 1? A. +or- 16/25 B. +or- 25/16 C. +or- 5/4 D. +or- 4/5

OpenStudy (anonymous):

\[\left( x-2 \right)^{2}/16 - (y+3)^{2}/25 = 1\]

OpenStudy (anonymous):

equation of a hyperbola is x^2/a^2 - y^2/b^2 = 1 and the equation of the asymptotes will be y = -(b/a)x and y = (b/a)x

OpenStudy (anonymous):

That makes your two equations y = -(5/4)x and y = (5/4)x and as you know from y = mx + b the m = slope which is + or - (5/4) Make sense?

OpenStudy (anonymous):

Yep Thanks =)

OpenStudy (anonymous):

No prob

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