15 points and brainliest 1. a hyperbola has vertices (+-5,0) and one focus (6,0) what is the standard form equation of the hyperbola. 2. find an equation that models that path of a satellite if its path is a hyperbola, a=55,000 km, and c=81,000 km. assume that the center of the hyperbola is the origin and the transverse axis is horizontal.

1, The standard form equation of the hyperbola:

x² / a² - y² / b² = 1

a = 5,

a² = 25

c = 6

c² = a² + b²

36 = 25 + b²

b² = 36 - 25

b² = 11

x²/25  -  y²/11 = 1

2.

Given:

Hyperbola

a=55,000 km and c=81,000 km

hyperbola is the origin and the

transverse axis is horizontal

Required:

Equation of the path of a

satellite

Solution:

Formula for hyperbola, (x-h)2/a2

– (x-k)2/b2 = 1

At origin, (h, k) = (0, 0)

(x-0)2/(55000)2 – (x-0)2/(81000)2

= 1

X2/12100 – y2/26244 = 250000

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7