Find the vertices and foci of the hyperbola with equation quantity x plus 5 squared divided by 36 minus the quantity of y plus 1 squared divided by 64 equals 1.
vertices: (-1, 3), (-1, -13); foci: (-1, -13), (-1, 3)
vertices: (3, -1), (-13, -1); foci: (-13, -1), (3, -1)
vertices: (-1, 1), (-1, -11); foci: (-1, -15), (-1, 5)
vertices: (1, -1), (-11, -1); foci: (-15, -1), (5, -1)

Vertices: (1,-1), (-11, -1); Foci: (-15, -1), (5, -1)

Step-by-step explanation:

Center at (-5,-1) because of the plus 5 added to the x and the plus 1 added to the y.

a(squared)=36 which means a=6 and a=distance from center to vertices so add and subtract 6 from the x coordinate since this is a horizontal hyperbola, which is (1,-1), (-11,-1). From there you dont need to find the focus since there is only one option for this;

Vertices: (1,-1), (-11, -1); Foci: (-15, -1), (5, -1)

Vertices: (1, -1), (-11, -1); Foci: (-15, -1), (5, -1)

Step-by-step explanation:

Ok so we have 5x

As you know we have the equation of the hyperbola as (x-h)^2/a^2-(y-k)^2/b^2, so the formula of the foci is

then we have to calculate c using pythagoras theorem we have that

a=6 because is the root of 36

b=8 beacause is the root of 64

And then we have that

So the root of 100 is equal to 10

Hence c=10

Using the formula  given before and the equation we know that

h=-5

k=-1

And replacing those values on the equation we have that the foci are

And the vertices are:

So the correct answer is D

27/4x^6 y^8

step-by-step explanation:

4(3x^2y^4)^3 (2x^3y^5)^4

=

108x^6 y^12 /16x^12 y^20

=

108 /16x^6 y^8

=

27 /4x^6 y^8