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bonufazy [111]
2 years ago
5

The average daily temperature, t, in degrees fahrenheit for a city as a function of the month of the year, m, can be modeled by

the equation t = 35 cosine (startfraction pi over 6 endfraction (m 3) 55, where m = 0 represents january 1, m = 1 represents february 1, m = 2 represents march 1, and so on. which equation also models this situation?
Mathematics
1 answer:
Vikentia [17]2 years ago
3 0

The answer will be t = sin((m.\frac{\pi}{6})+55) average daily temperature, t, in degrees Fahrenheit for a city as a function of the month of the year.

<h3>What is temperature?</h3>

Temperature is the degree or intensity of heat present in a substance or object, especially as expressed according to a comparative scale and shown by a thermometer or perceived by touch.

<h3>TO SOLVE:</h3>

35cos(\frac{\pi}{6(m+3)}+55)\\\\35 cos(\frac{\pi m}{6} + 55 + \frac{\pi}{2})\\

suppose x = m.\frac{\pi}{6}+55 and \frac{\pi}{2} = 90 degree

We know, cos(x+90°) = - sin(X)

⇒ cos((m.\frac{\pi}{6})+55+90degree)degree = -sin((m.\frac{\pi}{6})+55)

⇒ t = sin((m.\frac{\pi}{6})+55)

Hence the answer will be t = sin((m.\frac{\pi}{6})+55) average daily temperature, t, in degrees Fahrenheit for a city as a function of the month of the year.

To learn more visit:

brainly.com/question/27441712

#SPJ4

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Answer:

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Step-by-step explanation:

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qwelly [4]

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inessss [21]
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General Formulas and Concepts:
Order of Operations: BPEMDAS
- Substitution and Evaluation
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