Select the correct answer from each drop-down menu. the equation (y-2)^2/3^2 - (x-2)^2/4^2=1 represents a hyperbola whose foci are blank and blank .

The foci are (2 , 7) and (2 , -3)

Step-by-step explanation:

* lets revise the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

- The coordinates of the vertices are ( h ± a , k )  

- The coordinates of the co-vertices are ( h , k ± b )

- The coordinates of the foci are (h , k ± c), where c² = a² + b²

* Now lets solve the problem

∵ The equation of the hyperbola of vertex (h , k) is

    (y - k)²/a² - (x - h)²/b² = 1

∵ The equation is (y - 2)²/3² - (x - 2)²/4² = 1

∴ k = 2 , h = 2 , a = 3 , b = 4

∵ The foci of it are (h , k + c) and (h , k - c)

- Lets find c from the equation c² = a² + b²

∵ a = 3

∴ a² = 3² = 9

∵ b = 4

∴ b² = 4² = 16

∴ c² = 9 + 16 = 25

∴ c = √25 =  5

- Lets find the foci

∵ The foci are (h , k + c) and (h , k - c)

∵ h = 2 , k = 2 , c = 5

∴ The foci are (2 , 2 + 5) and (2 , 2 - 5)

∴ The foci are (2 , 7) and (2 , -3)

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.