PROBLEMAS QUE IMPLICAN UNA FUNCION DE 1ER GRADO
1 answer:
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Answer:
independent: day number; dependent: hours of daylight
d(t) = 12.133 +2.883sin(2π(t-80)/365.25)
1.79 fewer hours on Feb 10
Step-by-step explanation:
a) The independent variable is the day number of the year (t), and the dependent variable is daylight hours (d).
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b) The average value of the sinusoidal function for daylight hours is given as 12 hours, 8 minutes, about 12.133 hours. The amplitude of the function is given as 2 hours 53 minutes, about 2.883 hours. Without too much error, we can assume the year length is 365.25 days, so that is the period of the function,
March 21 is day 80 of the year, so that will be the horizontal offset of the function. Putting these values into the form ...
d(t) = (average value) +(amplitude)sin(2π/(period)·(t -offset days))
d(t) = 12.133 +2.883sin(2π(t-80)/365.25)
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c) d(41) = 10.34, so February 10 will have ...
12.13 -10.34 = 1.79
hours less daylight.
