I need the answer to this plz

Question

The velocity of the satellite orbiting Earth is related to the distance the satellite is from the center of the Earth by the function, v=(sqrt)1.24•10^12/r, where the v is velocity in miles per hour and r is the distance from the center of the Earth in miles.

1)GPS satellites travel in mid Earth orbits. If mid Earth orbit is from 4077 to 26197 miles from the center of the Earth, What is the domain and range of the function?

2)What is the inverse of the function?

Answer 1

1) The problem states that the mid Earth orbit is between 4,077 to 26,197 miles from the center of the earth (r). Given the function v, we have r in the dominator – for the fraction to exis, r must not equal zero. Also, the square root requires that radicand (the part under the square root sign) be greater than zero. In the given range of values, r does not equal zero and the radicand is positive. Therefore the domain is all numbers between 4077 and 26197.

Domain: 4,077 <= r <= 26,197

To get the range, plug each of these values into the formula to find what values v can take.

v = sqrt(1.24 * 10^12)/4077)) = 17439.76

v = sqrt(1.24 * 10^12/26197) = 6879.95

The range is 6879.95 <= v <= 17439.76

2)     To get the inverse, replace r with v and v with r, solve for v.

v = sqrt(1.24 * 10^12/r)

making the change,

r = sqrt(1.24* 10^12/v)

r^2 = 1.24*10^12/v

r^2/1.24*10^12 = 1/v

v = 1.24*10^12/r^2