Part 2)- Trigonometric Equations and Inequalitites Due Mon 12/11/20177: Show Int

Question

Part 2)- Trigonometric Equations and Inequalitites Due Mon 12/11/20177: Show Intro/Instru Score on last attempt: 0 out of 1 (parts: O) score in gradebook: L-] 0 out of 1 (parts: O) Try another similar question, or select another question This question, with your last answer and correct answer, is displayed below Suppose outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature is at a low of 57 degrees at midnight and a high of 83 degrees at noon. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t. Use -cosine with the smallest horizontal shift possible. D(t) Preview What is the smallest positive number of hours since midnight when the temperature reaches 78 degrees? (round to 4 decimal places) hours Answer: 70 13 cost Answer: 8.5319914963235 12 When you are done, click here to see a summary of your scores.

Explanation / Answer

comparing with D(t)=Acos(B(t+C))+ k

minimum = 57

maximum =83

amplitude ,A= (83-57)/2

amplitude ,A=13

average ,k=(83+57)/2

average ,k=70

period , 2/B = 24 hours

=>B=/12

horizontal shift=12 hours to right

=>C=-12

equation is D(t)=13cos((/12)(t-12))+ 70 or D(t)=13cos((t/12) -)+ 70

temperature =78 degrees

=>13cos((/12)t -)+ 70 =78

=>-13cos((/12)t) +70=78

=>13cos((/12)t) =70-78

=>cos((/12)t) =-8/13

=>((/12)t) =-cos-1(8/13)

=>t =12-((12/)cos-1(8/13))

=>t =12-((12/)*0.90792250314476086555134325338939)

=>t =12-3.4680085

=>t =8.5320 hours

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