Question

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 graphed below, where m = 0 represents January 1, m = 1 represents February 1, m = 2 represents March 1, and so on. If the equation is t = a cosine (StartFraction pi Over 6 EndFraction (m + 1)) + k, what are the values of a and k?

On a coordinate plane, a curve starts at (0, 42). It increases to (5, 80) and then decreases to (11, 40).

Answer 1

Given that the equation is t = a cosine (pi/6(m+1)) + k, we can use the given information to solve for a and k.

First, we can use the fact that the curve starts at (0,42) to solve for k:
t = acos(pi/6(0+1)) + k
42 = a*cos(pi/6) + k
k = 42 - a*cos(pi/6)

Next, we can use the fact that the curve increases to (5,80) to solve for a:
t = acos(pi/6(5+1)) + (42 - a*cos(pi/6))
80 = a*cos(pi/6*6) + (42 - a*cos(pi/6))
80 = a*cos(pi) + 42 - a*cos(pi/6)
80 = -a + 42 + a*cos(pi/6)
38 = a*cos(pi/6)
a = 38/cos(pi/6)

Finally, we can use the fact that the curve decreases to (11,40) to check our solution:
t = acos(pi/6(11+1)) + (42 - a*cos(pi/6))
40 = a*cos(2*pi) + (42 - a*cos(pi/6))
40 = a + 42 - a*cos(pi/6)
40 = 38/sqrt(3) + 42 - 38/2
40 = 67/sqrt(3)
40 = 38.7...

Therefore, our solution is:
a = 38/sqrt(3) or approximately 22
k = 42 - a*cos(pi/6) or approximately 33.5

Answer 2

I don’t really know sorry