Pluto's maximum distance from the sun (aphelion) is 7.47.47, point, 4 billion kilometers. Its minimum distance from the sun (perihelion) is 4.44.44, point, 4 billion kilometers. Pluto last reached its perihelion in the year 198919891989, and will next reach its perihelion in 22372237

Step-by-step explanation:

Assume a sine/cosine wave.

the period (min to max) is 2237–1989 = 248 years

so the first trial is D = M sin ((2πt/248) + k)

where k is the phase and M is the amplitude.

D = M sin (2π(t–1989)/248)

D = M sin ((2πt/248) – (2π1989/248))

at t = 1989, we have sin(0) = 0

at t = 2237, we have sin(56.67534–50.39216) = sin 2π = 0

but we want a min at those points, equivalent to sin 270º or 3π/2, or –π/2, so we have to add in additional phase shift.

D = M sin ((2πt/248) – (2π1989/248) – (π/2))

D = M sin (0.0253354t – 50.39216 – 1.570796)

D = M sin (0.0253354t – 51.96296)

at t = 1989,

D = M sin (50.39216 – 51.96296) = M sin 1.5708 = –1 ok

min to max distance is 7.4 – 4.4 = 3

so the equation is

D = 5.9 + 1.5sin (0.02534t – 51.96) (billions km)

in 2000

D = 5.9 + 1.5 sin (0.02534×2000 – 51.96) (billions km)

D = 5.9 + 1.5 sin (-1.28)

D = 5.9 – 1.5 × 0.02234

d = 5.87 billions km

1/5^2 = 7.56/45

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: