1st of all, this makes my brain hurt
2nd of all, which one was the first marble?
A quadrilateral I believe.
Answer:
Domain
Step-by-step explanation:
Input Values Form the Domain of the Function
Mathematicians call the set of all input values for a function its domain.
Answer:
d) The limit does not exist
General Formulas and Concepts:
<u>Calculus</u>
Limits
- Right-Side Limit:

- Left-Side Limit:

Limit Rule [Variable Direct Substitution]: 
Limit Property [Addition/Subtraction]: ![\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20c%7D%20%5Bf%28x%29%20%5Cpm%20g%28x%29%5D%20%3D%20%20%5Clim_%7Bx%20%5Cto%20c%7D%20f%28x%29%20%5Cpm%20%5Clim_%7Bx%20%5Cto%20c%7D%20g%28x%29)
Step-by-step explanation:
*Note:
In order for a limit to exist, the right-side and left-side limits must equal each other.
<u>Step 1: Define</u>
<em>Identify</em>

<u>Step 2: Find Right-Side Limit</u>
- Substitute in function [Limit]:

- Evaluate limit [Limit Rule - Variable Direct Substitution]:

<u>Step 3: Find Left-Side Limit</u>
- Substitute in function [Limit]:

- Evaluate limit [Limit Rule - Variable Direct Substitution]:

∴ Since
, then 
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
Let point (x, y) be any point on the graph, than the distance between (x, y) and the focus (3, 6) is sqrt((x - 3)^2 + (y - 6)^2) and the distance between (x, y) and the directrix, y = 4 is |y - 4|
Thus sqrt((x - 3)^2 + (y - 6)^2) = |y - 4|
(x - 3)^2 + (y - 6)^2 = (y - 4)^2
x^2 - 6x + 9 + y^2 - 12y + 36 = y^2 - 8y + 16
x^2 - 6x + 29 = -8y + 12y = 4y
(x - 3)^2 + 20 = 4y
y = 1/4(x - 3)^2 + 5
Required answer is f(x) = one fourth (x - 3)^2 + 5