Answer:
Option A.
Step-by-step explanation:
The given function is
[tex]f(x)=0.15x^2-6x+400[/tex]
Leading coefficient is positive, so it is an upward parabola. Vertex of an upward parabola is the point of minima.
Vertex of a parabola [tex]f(x)=ax^2+bx+c[/tex] is
[tex]Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)[/tex]
In the given function a=0.15, b=-6 and x=400.
[tex]-\dfrac{b}{2a}=-\dfrac{-6}{2(0.15)}=20[/tex]
At x=20,
[tex]f(20)=0.15(20)^2-6(20)+400=340[/tex]
So, minimum value of f(x) is 340 at x=20.
From the given table it is clear that the minimum value of g(x) is 55 at x=70.
Since 55 < 340, therefore, g(x) has a lower minimum values, i.e, (70,55).
Hence, option A is correct.
Answer:
A, I took the test and got it right
Step-by-step explanation:
