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Answer :
We are given the function
[tex]$$
D(t) = 2\sin\left(\frac{\pi}{6} t\right) + 12,
$$[/tex]
where [tex]$D(t)$[/tex] represents the number of daylight hours and [tex]$t$[/tex] represents the time in months since January 1.
1. Notice that the sine function, [tex]$\sin(x)$[/tex], always produces a value between [tex]$-1$[/tex] and [tex]$1$[/tex]. That is,
[tex]$$
-1 \leq \sin\left(\frac{\pi}{6}t\right) \leq 1.
$$[/tex]
2. Multiply the inequality by [tex]$2$[/tex], the amplitude:
[tex]$$
-2 \leq 2\sin\left(\frac{\pi}{6}t\right) \leq 2.
$$[/tex]
3. Next, add [tex]$12$[/tex] (which is the vertical shift) to each part of the inequality:
[tex]$$
12 - 2 \leq 2\sin\left(\frac{\pi}{6}t\right) + 12 \leq 12 + 2.
$$[/tex]
This simplifies to:
[tex]$$
10 \leq D(t) \leq 14.
$$[/tex]
4. Therefore, the least number of daylight hours is [tex]$10$[/tex] and the greatest number is [tex]$14$[/tex].
The correct answer is:
Least: [tex]$10$[/tex] hours; greatest: [tex]$14$[/tex] hours.
[tex]$$
D(t) = 2\sin\left(\frac{\pi}{6} t\right) + 12,
$$[/tex]
where [tex]$D(t)$[/tex] represents the number of daylight hours and [tex]$t$[/tex] represents the time in months since January 1.
1. Notice that the sine function, [tex]$\sin(x)$[/tex], always produces a value between [tex]$-1$[/tex] and [tex]$1$[/tex]. That is,
[tex]$$
-1 \leq \sin\left(\frac{\pi}{6}t\right) \leq 1.
$$[/tex]
2. Multiply the inequality by [tex]$2$[/tex], the amplitude:
[tex]$$
-2 \leq 2\sin\left(\frac{\pi}{6}t\right) \leq 2.
$$[/tex]
3. Next, add [tex]$12$[/tex] (which is the vertical shift) to each part of the inequality:
[tex]$$
12 - 2 \leq 2\sin\left(\frac{\pi}{6}t\right) + 12 \leq 12 + 2.
$$[/tex]
This simplifies to:
[tex]$$
10 \leq D(t) \leq 14.
$$[/tex]
4. Therefore, the least number of daylight hours is [tex]$10$[/tex] and the greatest number is [tex]$14$[/tex].
The correct answer is:
Least: [tex]$10$[/tex] hours; greatest: [tex]$14$[/tex] hours.
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Rewritten by : Barada
