Answer:
![\frac{dy}{dx}=-[(\frac{5x+24}{36x-6x^2})]](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D-%5B%28%5Cfrac%7B5x%2B24%7D%7B36x-6x%5E2%7D%29%5D)
Step-by-step explanation:
Given function:
y =
we know
= ln(A) - ln(B)
thus,
y = 
or
also,
ln(Aⁿ) = n × ln(A)
thus,
y = 
therefore,
![\frac{dy}{dx}=[(\frac{3}{2})\times\frac{1}{(6-x)}\times(0 - 1)] - [ (\frac{2}{3})\times\frac{1}{x}\times1]](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5B%28%5Cfrac%7B3%7D%7B2%7D%29%5Ctimes%5Cfrac%7B1%7D%7B%286-x%29%7D%5Ctimes%280%20-%201%29%5D%20-%20%5B%20%28%5Cfrac%7B2%7D%7B3%7D%29%5Ctimes%5Cfrac%7B1%7D%7Bx%7D%5Ctimes1%5D)
or

or
![\frac{dy}{dx}=-[(\frac{3(3x)+2\times2(6-x)}{2(6-x)\times(3x)})]](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D-%5B%28%5Cfrac%7B3%283x%29%2B2%5Ctimes2%286-x%29%7D%7B2%286-x%29%5Ctimes%283x%29%7D%29%5D)
or
![\frac{dy}{dx}=-[(\frac{5x+24}{36x-6x^2})]](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D-%5B%28%5Cfrac%7B5x%2B24%7D%7B36x-6x%5E2%7D%29%5D)
Answer:
-3 1/3
Step-by-step explanation:
The quadratic
... y = ax² +bx +c
has its extreme value at
... x = -b/(2a)
Since a = 3 is positive, we know the parabola opens upward and the extreme value is a minimum. (We also know that from the problem statement asking us to find the minimum value.) The value of x at the minimum is -(-4)/(2·3) = 2/3.
To find the minimum value, we need to evaluate the function for x=2/3.
The most straightforward way to do this is to substitue 2/3 for x.
... y = 3(2/3)² -4(2/3) -2 = 3(4/9) -8/3 -2
... y = (4 -8 -6)/3 = -10/3
... y = -3 1/3
_____
<em>Confirmation</em>
You can also use a graphing calculator to show you the minimum.
B. 0.6 seconds
Because they want you to round to the nearest 10th and 0 is the ground.