in a given training week a swimmer completes 1/4 mile on day one 2/5 mile on day two and 3/7 mile on day three,,
2 answers:
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For the remaining integral, let
and
Now the integral is
The first integral is trivial, so we'll focus on the latter one. You have
Decompose the integrand into partial fractions:
so you have
which are all standard integrals. You end up with
To try to get the terms to match up with the available answers, let's add and subtract
which suggests A may be the answer. To make sure this is the case, show that
You have
So in the corresponding term of the antiderivative, you get
The
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.
