Answer:
(f⋅g)(x)

Step-by-step explanation:
For the first part of the question we have two functions


If the expression refers to the multiplication of f and g, then:
(f⋅g)(x)
So we multiply the function f(x) with the function g(x)
(f⋅g)(x)
(f⋅g)(x)
For the second part we have the functions:

We wish to find (f - g) (x). We know that
![(f - g)(x) = f(x) - g(x)\\\\(f - g)(x) = x^3 - 2x^2 + 12x - 6 - [4x^2 - 6x + 4]\\\\(f - g)(x) = x^3 - 2x^2 + 12x - 6 -4x^2 + 6x - 4](https://tex.z-dn.net/?f=%28f%20-%20g%29%28x%29%20%3D%20f%28x%29%20-%20g%28x%29%5C%5C%5C%5C%28f%20-%20g%29%28x%29%20%3D%20x%5E3%20-%202x%5E2%20%2B%2012x%20-%206%20-%20%5B4x%5E2%20-%206x%20%2B%204%5D%5C%5C%5C%5C%28f%20-%20g%29%28x%29%20%3D%20x%5E3%20-%202x%5E2%20%2B%2012x%20-%206%20-4x%5E2%20%2B%206x%20-%204)

Answer:

Step-by-step explanation:
1) <u>The two points are</u>:
a) On the first swing she swings forward by 18 degrees: <em>(1, 18)</em>
b) On the second swing she only comes 13.5 degrees forward: <em>(2,13.5)</em>
2) <u>The general equation using the form given is</u>:

3) <u>Substitute the two points</u>:

4) <u>Divide the second equation by the first one</u>:
⇒ 13.5 / 18 = B
⇒ B = 0.75
5) <u>Substitue B = 0.75 into the first equation</u>:
18 = A (0.75) ⇒ A = 18 / 0.75 = 24
Hence, the equation is:

I'm pretty sure d sorry if not right
Answer:
The first graph. (an upside down U)
Step-by-step explanation:
The leading coefficient of a polynomial determines the direction of the graph's end behavior.
A positive leading coefficient has the end behavior point up when an even degree and point opposite directions when an odd degree with the left down and the right up.
A negative leading coefficient has the end behavior point down when an even degree and point opposite directions when an odd degree with the left up and the right down.
This graph has two evens. Because its negative, only one is possible - the first graph.
The other two graphs are odd with both starting down on the left and point up on the right which is a positive leading coefficient. These are not possible graphs.
The first graph is the solution.