Answer:
The function for the outside temperature is represented by ![T(t) = 85\º + 15\º \cdot \sin \left[\frac{t-6\,h}{24\,h} \right]](https://tex.z-dn.net/?f=T%28t%29%20%3D%2085%5C%C2%BA%20%2B%2015%5C%C2%BA%20%5Ccdot%20%5Csin%20%5Cleft%5B%5Cfrac%7Bt-6%5C%2Ch%7D%7B24%5C%2Ch%7D%20%5Cright%5D)
, where t is measured in hours.
Step-by-step explanation:
Since outside temperature can be modelled as a sinusoidal function, the period is of 24 hours and amplitude of temperature and average temperature are, respectively:
Amplitude
Mean temperature
Given that average temperature occurs six hours after the lowest temperature is registered. The temperature function is expressed as:
![T(t) = \bar T + A \cdot \sin \left[2\pi\cdot\frac{t-6\,h}{\tau} \right]](https://tex.z-dn.net/?f=T%28t%29%20%3D%20%5Cbar%20T%20%2B%20A%20%5Ccdot%20%5Csin%20%5Cleft%5B2%5Cpi%5Ccdot%5Cfrac%7Bt-6%5C%2Ch%7D%7B%5Ctau%7D%20%5Cright%5D)
Where:
- Mean temperature, measured in degrees.
- Amplitude, measured in degrees.
- Daily period, measured in hours.
- Time, measured in hours. (where t = 0 corresponds with 5 AM).
If , and 
, the resulting function for the outside temperature is:
Answer:
true
Step-by-step explanation:
if its not a true or false than put the answer questions below
Answer:
In February, 423 daytime minutes is used
Step-by-step explanation:
Let the base plan charges be x
And cost per daytime minute be y
In December,
x + 510y = 92.25------------------(1)
In January,
x + 397y = 77.56---------------------(2)
Subtracting eq(2) from eq(1)
x + 510y = 92.25
x + 397y = 77.56
-------------------------------
0 + 113y = 14.69
-------------------------------
y = \frac{14.69}{113}
y = 0.13----------------------------------(3)
Substituting (3) in (1)
x + 510(0.13) = 92.25
x + 66.3 = 92.25
x = 92.25 - 66.3
x = 25.95
So In February
base plan + (daytime minute)(cost per daytime minute) = 80.9
25.95 + (daytime minute)(0.13) = 80.9
(daytime minute)(0.13) = 80.9 - 25.95
(daytime minute)(0.13) = 54.95
(daytime minute) =
daytime minutes = 422.69
daytime minute 
