Answer: A. Vertical angles
Step-by-step explanation:
They are vertically opposite angles.
Correct me if I am incorrect.
Answer:

Step-by-step explanation:
14 - 2xy when x = -6 and y = 2.

Question:
In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The longest and the shortest days of the year vary by 2 hours 53 minutes from the equinox. In this year, the equinox falls on March 21. In this task, you'll use a trigonometric function to model the hours of daylight hours on certain days of the year in New York City.
- Find amplitude and the period of the function
- Create a trigonometric function that describes the hours of sunlight for each day of the year
- Then use the function you built to find how fewer daylight hours February 10 will have then March 21
Answer:
(a)
--- Amplitude
---- Period
(b) Trigonometry function
![f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])](https://tex.z-dn.net/?f=f%28x%29%20%3D%2012.133%20%2B%202.883sin%28%5Cfrac%7B2%5Cpi%7D%7B365%7D%5Bx%20-%2080%5D%29)
(c) 
Step-by-step explanation:
Given


Solving (a): Amplitude (P) and Period (T)
The amplitude is the amount of time the longest and the shortest day vary.
So

Convert to hours


The period (T) is the duration i.e 1 year

Assume no leap year

Solving (b): Trigonometry function
The function follows a sinusoidal pattern and the general form is:

Where


Convert to hours



--- Amplitude
---- Period

So:

The function becomes:

![f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])](https://tex.z-dn.net/?f=f%28x%29%20%3D%2012.133%20%2B%202.883sin%28%5Cfrac%7B2%5Cpi%7D%7B365%7D%5Bx%20-%2080%5D%29)
Solving (c): Fewer daylight hours will Feb. 10 have.

So:
![f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])](https://tex.z-dn.net/?f=f%28x%29%20%3D%2012.133%20%2B%202.883sin%28%5Cfrac%7B2%5Cpi%7D%7B365%7D%5Bx%20-%2080%5D%29)
![f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[41 - 80])](https://tex.z-dn.net/?f=f%2841%29%20%3D%2012.133%20%2B%202.883sin%28%5Cfrac%7B2%5Cpi%7D%7B365%7D%5B41%20-%2080%5D%29)
![f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[-39])](https://tex.z-dn.net/?f=f%2841%29%20%3D%2012.133%20%2B%202.883sin%28%5Cfrac%7B2%5Cpi%7D%7B365%7D%5B-39%5D%29)

So:
![f(41) = 12.133 + 2.883sin(\frac{360}{365}[-39])](https://tex.z-dn.net/?f=f%2841%29%20%3D%2012.133%20%2B%202.883sin%28%5Cfrac%7B360%7D%7B365%7D%5B-39%5D%29)



The fewer daylight hours is the calculated as:



