The function r(x) = –0.000345(x – 50)2 + 2.1 can be used to model the curve of a road designed to allow for water to run off in rainstorms depending on the distance from the shoulder of the road, x, in feet. A European firm studying these curves is going to convert feet to meters using the function m(x) = 0.3048x. Which composition can be used to model the curve of the road in terms of meters? m(r(x)) = –0.000105156(x – 50)2 + 2.1 m(r(x)) = –0.000105156(x – 50)2 + 0.64008 m(r(x)) = –0.000345(0.3048x – 50)2 + 2.1 m(r(x)) = –0.000345(0.3048x – 50)2 + 0.64008

Respuesta :

Answer:

[tex]m(r(x)) = -0.000105156(x-50)^2 + 0.64008[/tex]

Step-by-step explanation:

Composite Function

The composition of functions f(x) and g(x) is denoted as [tex](f\circ g)(x)[/tex] and can be obtained by the expression

[tex]f\circ g(x)=f(g(x))[/tex]

In practice, it means to replace into each x of f, the whole expression for g.

We know the function

[tex]r(x) = -0.000345(x-50)^2 + 2.1[/tex]

can be used to model some reality, being x in feet. We also know that another function

[tex]m(x) = 0.3048x[/tex]

is going to be used to convert feet to meters. We are required to find

[tex]m\circ r(x)=m(r(x))[/tex]

We'll replace the expression for r into m as:

[tex]m(x) = 0.3048\left[-0.000345(x-50)^2 + 2.1\right][/tex]

Operating

[tex]\boxed{m(r(x)) = -0.000105156(x-50)^2 + 0.64008}[/tex]

Answer:

B: (m(r(x))=-0.000105156(x-50)^2+0.64008

Step-by-step explanation:

On Edge