Looking out to the sea, the visibility in miles from a certain spot on a diff can be calculated using the function d(x) = V1.5%, where x is the height in
feet above sea level. Suppose Sasha's brother walks through elevations
ranging from 8 ft to 48 ft. What are the minimum and maximum distances
that he can see?

The exact coordinates of the point where the line through B and D intersects the y-axis is

Step-by-step explanation:

For the given question, we have the following parameters;

The shape of the quadrilateral ABCD in the question = A rhombus

The diagonals of the rhombus = AC and BD

The point of intersection of the diagonals, M = (6, -11)

The equation of the line points A and C lie = 2·y + 7·x = 20

A rhombus is an orthodiagonal quadrilateral (The diagonals of a rhombus are perpendicular)

Therefore;

AC ⊥ BD  and where the slope of the line with points A and C is 'm', the slope of the line with points B and D is (-1/m)

Rewriting the equation, 2·y + 7·x = 20, in slope and intercept form, we have;

2·y + 7·x = 20

y = (- 7·x + 20)/2 = -7·x/2 + 10

∴ y = -7·x/2 + 10

The slope of the line 'y = -7·x/2 + 10', m = -7/2

∴ The slope of the line having the points B and D is -1/m = -1/(-7/2) = 2/7

The equation of the line having the points B and D in point and slope form, using point (6, -11) is given as follows;

y - (-11) = 2/7 × (x - 6)

y + 11 = 2·x/7 - 12/7

∴ y = 2·x/7 - 12/7 - 11 = 2·x/7 - 89/7 = 2·x/7 -

The equation of the line having the points B and D in point is therefore;

y = 2·x/7 -

The coordinates of the point where the line having the points B and D intersects the y-axis is the y-intercept, of the equation,  y = 2·x/7 - , which is