Question

The number of hours of​ daylight, H, on day t of any given year​ (on January​ 1, tequals​1) in a particular city can be modeled by the function ​H(t)equals12 plus 8.3 sine [StartFraction 2 pi Over 365 EndFraction (t minus 83 )]. a. March 24​, the 83rd day of the​ year, is the spring equinox. Find the number of hours of daylight in the city on this day. b. June 24​, the 175th day of the​ year, is the summer​ solstice, the day with the maximum number of hours of daylight. Find the number of hours of daylight in the city on this day. c. December 24​, the 358th day of the​ year, is the winter​ solstice, the day with the minimum number of hours of daylight. Find the number of hours of daylight in the city on this day.

Answer 1

Answer:

(a)12 Hours

(b)12.23 Hours

(c)12.68 Hours

Step-by-step explanation:

The number of hours of​ daylight, H, on day t of any given year​ (on January​ 1, t=​1) in a particular city can be modeled by the function:​$H(t)=12+8.3 sine \frac{2 \pi}{365}(t-83)$.

(a)On March 24​, the 83rd day

t=83

Therefore, the number of hours

$H(t)=12+8.3 sine [\frac{2 \pi}{365}(83-83)]$.

H(t)=12 Hours

(b)June 24​, the 175th day.

t=175

$H(175)=12+8.3 sine \frac{2 \pi}{365}(175-83)$.

$=12+8.3 sine [\frac{2 \pi}{365}(92)]$=12.23 Hours

(c)December 24​, the 358th day

t=358

$H(358)=12+8.3 sine [\frac{2 \pi}{365}(358-83)]$.

$=12+8.3 sine [\frac{2 \pi}{365}(275)]$=12.68 Hours