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To determine the months during which the temperature at the Willburn airport is [tex]\( 17^\circ C \)[/tex], we'll start by solving the given equation for [tex]\( x \)[/tex], the month of the year.
The given temperature model is:
[tex]\[ f(x) = 5 \cos \left(\frac{x}{12}\right) + 14.5 \][/tex]
We need to find the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 17 \)[/tex]. Thus, we set up the equation:
[tex]\[ 5 \cos \left(\frac{x}{12}\right) + 14.5 = 17 \][/tex]
Subtract 14.5 from both sides of the equation to isolate the cosine term:
[tex]\[ 5 \cos \left(\frac{x}{12}\right) = 2.5 \][/tex]
Divide both sides by 5:
[tex]\[ \cos \left(\frac{x}{12}\right) = \frac{2.5}{5} \][/tex]
[tex]\[ \cos \left(\frac{x}{12}\right) = \frac{1}{2} \][/tex]
The general solutions for the equation [tex]\( \cos y = \frac{1}{2} \)[/tex] are:
[tex]\[ y = \pm \frac{\pi}{3} + 2k\pi \][/tex]
(where [tex]\( k \)[/tex] is any integer)
Since [tex]\( y = \frac{x}{12} \)[/tex], substitute back:
[tex]\[ \frac{x}{12} = \pm \frac{\pi}{3} + 2k\pi \][/tex]
Multiply through by 12 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = 12 \left( \pm \frac{\pi}{3} + 2k\pi \right) \][/tex]
This gives:
[tex]\[ x = 12 \left( \frac{\pi}{3} + 2k\pi \right) \][/tex]
[tex]\[ x = 12 \cdot \frac{\pi}{3} + 24k\pi \][/tex]
[tex]\[ x = 4\pi + 24k\pi \][/tex]
And:
[tex]\[ x = 12 \left( -\frac{\pi}{3} + 2k\pi \right) \][/tex]
[tex]\[ x = 12 \cdot -\frac{\pi}{3} + 24k\pi \][/tex]
[tex]\[ x = -4\pi + 24k\pi \][/tex]
These values of [tex]\( x \)[/tex] represent months as a continuous variable. For simplicity, we then evaluate these solutions modulo [tex]\( 2\pi \)[/tex].
We note that since [tex]\( x \)[/tex] stands for months and should be within the range of a year, we simplify using modulo [tex]\( 24\pi \)[/tex] to find relevant months within the first 12 months (since [tex]\( 24\pi \)[/tex] corresponds to 12 months):
[tex]\[ 4\pi \mod 24\pi = 4\pi \][/tex]
For [tex]\( x = 4\pi \)[/tex], convert it to months:
[tex]\[ 4\pi \times \frac{12}{2\pi} = 24 \div 2 = 2 \][/tex]
Next,
[tex]\[ 20\pi \mod 24\pi = 20\pi \][/tex]
For [tex]\( x = 20\pi \)[/tex], convert it to months:
[tex]\[ 20\pi \times \frac{12}{2\pi} = 120 \div 2 = 10 \][/tex]
Thus, the months in which the temperature will be [tex]\( 17^\circ C \)[/tex] are:
[tex]\[ x = 4\pi + 24k\pi \][/tex]
[tex]\[ x = 20\pi + 24k\pi \][/tex]
These correspond to the months found through calculation: the 4th month (April) and the 10th month (October).
The correct answer is then the fourth given option:
[tex]\[ x = 4\pi + 2\pi n \][/tex]
[tex]\[ x = 20\pi + 2\pi n \][/tex]
The given temperature model is:
[tex]\[ f(x) = 5 \cos \left(\frac{x}{12}\right) + 14.5 \][/tex]
We need to find the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 17 \)[/tex]. Thus, we set up the equation:
[tex]\[ 5 \cos \left(\frac{x}{12}\right) + 14.5 = 17 \][/tex]
Subtract 14.5 from both sides of the equation to isolate the cosine term:
[tex]\[ 5 \cos \left(\frac{x}{12}\right) = 2.5 \][/tex]
Divide both sides by 5:
[tex]\[ \cos \left(\frac{x}{12}\right) = \frac{2.5}{5} \][/tex]
[tex]\[ \cos \left(\frac{x}{12}\right) = \frac{1}{2} \][/tex]
The general solutions for the equation [tex]\( \cos y = \frac{1}{2} \)[/tex] are:
[tex]\[ y = \pm \frac{\pi}{3} + 2k\pi \][/tex]
(where [tex]\( k \)[/tex] is any integer)
Since [tex]\( y = \frac{x}{12} \)[/tex], substitute back:
[tex]\[ \frac{x}{12} = \pm \frac{\pi}{3} + 2k\pi \][/tex]
Multiply through by 12 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = 12 \left( \pm \frac{\pi}{3} + 2k\pi \right) \][/tex]
This gives:
[tex]\[ x = 12 \left( \frac{\pi}{3} + 2k\pi \right) \][/tex]
[tex]\[ x = 12 \cdot \frac{\pi}{3} + 24k\pi \][/tex]
[tex]\[ x = 4\pi + 24k\pi \][/tex]
And:
[tex]\[ x = 12 \left( -\frac{\pi}{3} + 2k\pi \right) \][/tex]
[tex]\[ x = 12 \cdot -\frac{\pi}{3} + 24k\pi \][/tex]
[tex]\[ x = -4\pi + 24k\pi \][/tex]
These values of [tex]\( x \)[/tex] represent months as a continuous variable. For simplicity, we then evaluate these solutions modulo [tex]\( 2\pi \)[/tex].
We note that since [tex]\( x \)[/tex] stands for months and should be within the range of a year, we simplify using modulo [tex]\( 24\pi \)[/tex] to find relevant months within the first 12 months (since [tex]\( 24\pi \)[/tex] corresponds to 12 months):
[tex]\[ 4\pi \mod 24\pi = 4\pi \][/tex]
For [tex]\( x = 4\pi \)[/tex], convert it to months:
[tex]\[ 4\pi \times \frac{12}{2\pi} = 24 \div 2 = 2 \][/tex]
Next,
[tex]\[ 20\pi \mod 24\pi = 20\pi \][/tex]
For [tex]\( x = 20\pi \)[/tex], convert it to months:
[tex]\[ 20\pi \times \frac{12}{2\pi} = 120 \div 2 = 10 \][/tex]
Thus, the months in which the temperature will be [tex]\( 17^\circ C \)[/tex] are:
[tex]\[ x = 4\pi + 24k\pi \][/tex]
[tex]\[ x = 20\pi + 24k\pi \][/tex]
These correspond to the months found through calculation: the 4th month (April) and the 10th month (October).
The correct answer is then the fourth given option:
[tex]\[ x = 4\pi + 2\pi n \][/tex]
[tex]\[ x = 20\pi + 2\pi n \][/tex]
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