Answer: a. Here is a sketch of the situation described above:
80 + . July (x = 7)
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60 + . .
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40 + . . .
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0 +_______________________________________________
1 2 3 4 5 6 7 8 9 10 11 12
January December
b. One possible trigonometric equation that models the temperature throughout the year is:
T(x) = (36cos((2π/12)(x-7))) + 41
where T(x) represents the average temperature in degrees Fahrenheit for month x (with January corresponding to x=1), and the constant term of 41 is added to shift the curve up to match the lowest average temperature recorded.
c. To find the average monthly temperature for the month of March, we simply plug in x=3 into the equation above:
T(3) = (36cos((2π/12)(3-7))) + 41
= (36*cos(-π/3)) + 41
≈ 51.4°F
So the average monthly temperature for the month of March is approximately 51.4 degrees Fahrenheit.
d. To find the period of time during which the average temperature is less than 41°F, we need to solve the inequality:
T(x) < 41
Substituting the equation for T(x) from part b, we get:
(36cos((2π/12)(x-7))) + 41 < 41
Simplifying this inequality, we get:
cos((2π/12)*(x-7)) < 0
We can solve this inequality by finding the values of x for which the cosine function is negative. The cosine function is negative in the second and third quadrants of the unit circle, so we have:
(2π/12)*(x-7) ∈ (π, 2π) ∪ (3π, 4π)
Simplifying this expression, we get:
π/6 < x-7 < π/2 or 5π/6 < x-7 < 2π/3
Adding 7 to both sides of each inequality, we get:
7 + π/6 < x < 7 + π/2 or 7 + 5π/6 < x < 7 + 2π/3
Simplifying these expressions, we get:
7.524 < x < 8.571 or 11.286 < x < 11.857
Therefore, the average temperature is less than 41°F during the period of time from approximately November 24th to December 19th, and from approximately February 15th to March 20th.
Explanation: