in fairbanks alaska, the number of hours of daylight on day t (where t is the numbers of days after december…

Explanation:

Step1: Set up the inequality

We want to find when $L(t)\geq10$, so $10.1 + 0.2\sin\left(\frac{2\pi}{365}(t - 80)\right)\geq10$. Subtract 10.1 from both sides: $0.2\sin\left(\frac{2\pi}{365}(t - 80)\right)\geq10 - 10.1=- 0.1$. Then divide by 0.2: $\sin\left(\frac{2\pi}{365}(t - 80)\right)\geq-\frac{0.1}{0.2}=-0.5$.

Step2: Find the general solutions of the sine - inequality

The general solutions of $\sin x\geq - 0.5$ are $2k\pi-\frac{\pi}{6}\leq x\leq2k\pi+\frac{7\pi}{6},k\in\mathbb{Z}$. Let $x = \frac{2\pi}{365}(t - 80)$. Then $2k\pi-\frac{\pi}{6}\leq\frac{2\pi}{365}(t - 80)\leq2k\pi+\frac{7\pi}{6}$. First, divide each part of the compound - inequality by $2\pi$: $k-\frac{1}{12}\leq\frac{1}{365}(t - 80)\leq k+\frac{7}{12}$. Then multiply each part by 365: $365k-\frac{365}{12}\leq t - 80\leq365k+\frac{7\times365}{12}$. Add 80 to each part: $365k-\frac{365}{12}+80\leq t\leq365k+\frac{7\times365}{12}+80$. For $k = 0$: $365\times0-\frac{365}{12}+80\leq t\leq365\times0+\frac{7\times365}{12}+80$. $80-\frac{365}{12}\leq t\leq80+\frac{7\times365}{12}$. $80 - 30.42\leq t\leq80 + 212.92$. $49.58\leq t\leq292.92$. Rounding to the nearest whole number, $t = 50$ through $t = 293$.

Step3: Calculate the number of days

The number of days from $t = 50$ to $t = 293$ is $293 - 50+1=244$.

Answer:

(a) 50 through 293 (b) 244