13
15
17
19
Those are the answers in order for the parrots age
Answer:
![5^8](https://tex.z-dn.net/?f=5%5E8)
Step-by-step explanation:
![5^{2^{3}}](https://tex.z-dn.net/?f=5%5E%7B2%5E%7B3%7D%7D)
![5^{2 \times 2 \times 2}](https://tex.z-dn.net/?f=5%5E%7B2%20%5Ctimes%202%20%5Ctimes%202%7D)
![5^8](https://tex.z-dn.net/?f=5%5E8)
![=390625](https://tex.z-dn.net/?f=%3D390625)
Answer:
13a+27
Step-by-step explanation:
8a+12+5a+15= 13a+27
Answer:
A. 8
Step-by-step explanation:
32 divided by 4 = 8
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) ![Hours= 1.794](https://tex.z-dn.net/?f=Hours%3D%201.794)
Step-by-step explanation:
Given
![Average\ Sunlight = 12hr\ 8 min](https://tex.z-dn.net/?f=Average%5C%20Sunlight%20%3D%2012hr%5C%208%20min)
![Variance = 2hr\ 53min](https://tex.z-dn.net/?f=Variance%20%3D%202hr%5C%2053min)
Solving (a): Amplitude (P) and Period (T)
The amplitude is the amount of time the longest and the shortest day vary.
So
![A = 2\ hr\ 53\ min](https://tex.z-dn.net/?f=A%20%3D%202%5C%20hr%5C%2053%5C%20min)
Convert to hours
![A = 2\ + \frac{53}{60}](https://tex.z-dn.net/?f=A%20%3D%202%5C%20%2B%20%5Cfrac%7B53%7D%7B60%7D)
![A = 2+0.883](https://tex.z-dn.net/?f=A%20%3D%202%2B0.883)
The period (T) is the duration i.e 1 year
![T = 1\ year](https://tex.z-dn.net/?f=T%20%3D%201%5C%20year)
Assume no leap year
![T = 365](https://tex.z-dn.net/?f=T%20%3D%20365)
Solving (b): Trigonometry function
The function follows a sinusoidal pattern and the general form is:
![f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cmu%2B%20Asin%28%5Cfrac%7B2%5Cpi%7D%7BT%7D%28x%20-n%29%29)
Where
![\mu = Average\ Value](https://tex.z-dn.net/?f=%5Cmu%20%3D%20Average%5C%20Value)
![\mu = 12\ hr 8\ min](https://tex.z-dn.net/?f=%5Cmu%20%3D%2012%5C%20hr%208%5C%20min)
Convert to hours
![\mu = 12 + \frac{8}{60}](https://tex.z-dn.net/?f=%5Cmu%20%3D%2012%20%2B%20%5Cfrac%7B8%7D%7B60%7D)
![\mu = 12 + 0.133](https://tex.z-dn.net/?f=%5Cmu%20%3D%2012%20%2B%200.133)
![\mu = 12.133](https://tex.z-dn.net/?f=%5Cmu%20%3D%2012.133)
--- Amplitude
---- Period
![n = Equinox](https://tex.z-dn.net/?f=n%20%3D%20Equinox)
So:
![n= 80](https://tex.z-dn.net/?f=n%3D%2080)
The function becomes:
![f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cmu%2B%20Asin%28%5Cfrac%7B2%5Cpi%7D%7BT%7D%28x%20-n%29%29)
![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.
![Feb\ 10 = 41st\ day](https://tex.z-dn.net/?f=Feb%5C%2010%20%3D%2041st%5C%20day)
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)
![2\pi = 360^\circ](https://tex.z-dn.net/?f=2%5Cpi%20%3D%20360%5E%5Ccirc)
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)
![f(41) = 12.133 + 2.883sin(-38.466)](https://tex.z-dn.net/?f=f%2841%29%20%3D%2012.133%20%2B%202.883sin%28-38.466%29)
![f(41) = 12.133 - 2.883*0.6221](https://tex.z-dn.net/?f=f%2841%29%20%3D%2012.133%20-%202.883%2A0.6221)
![f(41) = 10.339](https://tex.z-dn.net/?f=f%2841%29%20%3D%2010.339)
The fewer daylight hours is the calculated as:
![Hours= Average - f(41)](https://tex.z-dn.net/?f=Hours%3D%20Average%20-%20f%2841%29)
![Hours= \mu - f(41)](https://tex.z-dn.net/?f=Hours%3D%20%5Cmu%20-%20f%2841%29)
![Hours= 12.133 - 10.339](https://tex.z-dn.net/?f=Hours%3D%2012.133%20-%2010.339)
![Hours= 1.794](https://tex.z-dn.net/?f=Hours%3D%201.794)