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:
<u>m= -19/1</u>
We need to use the slope equation
We are working with the points,
(17, 2) and (18, -17)
x1 y1 x2 y2
<u>m= -19/1</u>
Answer:
7/3
Step-by-step explanation:
i hope this works :)
C. Reflected over the y-axis and translated up 1 unit
Answer:
B
Step-by-step explanation:
In a triangle, the sum of any two sides must be bigger than the third.
For the first one, 10+20=30 is not greater than 30, so this is not correct.
For B, 122+137 = 259 > 257, 257+137>122 , and 257 + 122 > 137. This works
For C, 8.6 + 2.7 = 11.3 < 12.2, so this does not work
For D, 1/6 + 1/5 = 5/(6*5) + 6/(6*5) = 5/30 + 6/30 = 11/30 < 1/2 = 15/30, so this does not work
