Question

fill in the blank 1 point
the average daily high temperature in seattle is modeled by the function t = -15 cos(\frac{\pi m}{6}) + 61, where t is the temperature in °f, and m is the number of months since the beginning of the year. what is the value of m when the average daily high temperature t is 76°f?
m= type your answer...

5 multiple choice 1 point
which of these is necessary to define the inverse cosine function?
the cosine function must be periodic.
the range of the cosine function must be y|y ≥ 0.
the domain of the cosine function must be restricted to 0 ≤ x ≤ π
the cosine function must be always increasing.

6 multiple answer 1 point
what is sin^{-1}(-\frac{\sqrt{2}}{2})? select all that apply.
-\frac{\pi}{4}
-45°
-\frac{\pi}{2}
-60°

Explanation:

Question 4

Step1: Substitute \(T = 76\) into the equation

Given \(T=-15\cos(\frac{\pi m}{6}) + 61\), substitute \(T = 76\):
\(76=-15\cos(\frac{\pi m}{6})+61\)

Step2: Solve for \(\cos(\frac{\pi m}{6})\)

Subtract \(61\) from both sides:
\(76 - 61=-15\cos(\frac{\pi m}{6})\)
\(15=-15\cos(\frac{\pi m}{6})\)
Divide both sides by \(- 15\):
\(\cos(\frac{\pi m}{6})=-1\)

Step3: Solve for \(m\)

We know that \(\cos\theta=-1\) when \(\theta=(2k + 1)\pi,k\in\mathbb{Z}\). So, \(\frac{\pi m}{6}=(2k + 1)\pi\).
Divide both sides by \(\pi\): \(\frac{m}{6}=2k + 1\).
Multiply both sides by \(6\): \(m = 12k+6\).
Since \(m\) represents the number of months (\(0\leq m\leq12\)), when \(k = 0\), \(m = 6\)

For a function \(y = f(x)\) to have an inverse, it must be one - to - one (pass the horizontal line test). The cosine function \(y=\cos(x)\) is periodic (\(y = \cos(x)=\cos(x + 2k\pi),k\in\mathbb{Z}\)) and not one - to - one on its natural domain \((-\infty,\infty)\).

• The cosine function is periodic, but this property alone does not help in defining an inverse. A periodic function is not one - to - one.
• The range of the cosine function is \(y\in[- 1,1]\), not \(y|y\geq0\)
• The cosine function \(y = \cos(x)\) is decreasing on \([0,\pi]\) and one - to - one on the restricted domain \(x\in[0,\pi]\). So, we can define the inverse cosine function \(y=\cos^{-1}(x)\) with domain \(x\in[-1,1]\) and range \(y\in[0,\pi]\)
• The cosine function is not always increasing. It is decreasing on \([0,\pi]\) and increasing on \([\pi,2\pi]\)

We know that the range of the inverse sine function \(y = \sin^{-1}(x)\) is \(y\in[-\frac{\pi}{2},\frac{\pi}{2}]\) (in radians) or \(y\in[-90^{\circ},90^{\circ}]\) (in degrees)

• For \(y=\sin^{-1}(x)\), if \(x=-\frac{\sqrt{2}}{2}\)
• In radians: \(\sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\) and \(-\frac{\pi}{4}\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
• In degrees: \(\sin(-45^{\circ})=-\frac{\sqrt{2}}{2}\) and \(-45^{\circ}\in[-90^{\circ},90^{\circ}]\)
• \(\sin(-\frac{\pi}{2})=-1\) and \(\sin(-60^{\circ})=-\frac{\sqrt{3}}{2}\)

Answer:

\(6\)

Question 5