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Outside temperature over a day can be modeled as a sinusoidal function. Suppose

ID: 2920108 • Letter: O

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 87 degrees occurs at 4 PM and the average temperature for the day is 70 degrees. Find the temperature, to the nearest degree, at 5 AM.
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 87 degrees occurs at 4 PM and the average temperature for the day is 70 degrees. Find the temperature, to the nearest degree, at 5 AM.
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 87 degrees occurs at 4 PM and the average temperature for the day is 70 degrees. Find the temperature, to the nearest degree, at 5 AM.

Explanation / Answer

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

high temperature for the day is 87 degrees and the average temperature for the day is 70 degrees

k=70

A=87-70

A=17

period =24 hours

2/B =24

=>B= /12

D(t)= 17sin(( /12)(t+C)) +70

4 pm is 16 hours from midnight , high temperature of 87 degrees occurs at 4 PM

D(16)=87

=> 17sin(( /12)(16+C)) +70=87

=> 17sin(( /12)(16+C)) =17

=>sin(( /12)(16+C)) =1

=>(( /12)(16+C)) =/2

=>(16+C) =6

=>C=-10

so function is  D(t)= (17*sin(( /12)*(t-10))) +70 or you can use D(t)= 17sin(( /12)(t+14)) +70

5 pm is 17 hours from midnight

temperature at 5PM =D(17)

temperature at 5PM = (17*sin(( /12)*(17-10))) +70

temperature at 5PM = (17*sin(7 /12)) +70

temperature at 5PM = 86.42

temperature at 5PM = 86 degrees

please rate if helpful. please comment if you have any doubt

please rate if helpful. please comment if you have any doubt

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