6!+2!=[?] Evaluate.
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We presume the temperature is decreasing at midnight, so reaches a low at 6 a.m.. The amplitude of the variation is (87-63)/2 = 12 degrees. We want the period to be 24 hours, so the argument of the sine function will be 2π(t/24) = πt/12. Then we can write
... d = 75-12sin(π·t/12)
