vera_ewing
  • vera_ewing
The temperature of a chemical reaction ranges between 30°C and 70°C. The temperature is at its lowest point when t = 0 and completes one cycle over a six-hour period. What is a sine function that would model this reaction?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
vera_ewing
  • vera_ewing
@Michele_Laino f(t) = 50 sin 10t + 20 ?
Michele_Laino
  • Michele_Laino
the amplitude of your function, is: \[\Large A = \frac{{\max value - \min value}}{2} = \frac{{70 - 30}}{2} = ...?\]
vera_ewing
  • vera_ewing
40/2 = 20

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

vera_ewing
  • vera_ewing
f(t) = 20 sin 10t + 50 is the answer?
Michele_Laino
  • Michele_Laino
we can write: \[\Large f\left( t \right) = 20\sin \left( {Bt} \right) + 50\] since we have these limit values: \[\Large 30 \leqslant 20\sin \left( {Bt} \right) + 50 \leqslant 70\]
Michele_Laino
  • Michele_Laino
now we have to determine the value of the constant B
vera_ewing
  • vera_ewing
So it's f(t) = 20 sin pi/5 t + 50 ?
Michele_Laino
  • Michele_Laino
I think that we have: \[\Large B = \frac{{2\pi }}{6}\] since our function has to assume the same values after 6 hours
Michele_Laino
  • Michele_Laino
therefore: \[\Large f\left( t \right) = 20\sin \left( {\frac{{2\pi }}{6}t} \right) + 50\] is the right function
vera_ewing
  • vera_ewing
vera_ewing
  • vera_ewing
Those are my answer choices ^
Michele_Laino
  • Michele_Laino
please wait a moment, I'm pondering...
Michele_Laino
  • Michele_Laino
please look at this drawing:
1 Attachment
Michele_Laino
  • Michele_Laino
maybe there is a typo since we can rewrite our function as below: \[\Large f\left( t \right) = 20\sin \left( {\frac{\pi }{3}t} \right) + 50\]
Michele_Laino
  • Michele_Laino
please wait a moment, I understand
Michele_Laino
  • Michele_Laino
I confirm my answer above
vera_ewing
  • vera_ewing
So which one should I choose?
Michele_Laino
  • Michele_Laino
try with the last option, please
Michele_Laino
  • Michele_Laino
is the period T=10 hours?
Michele_Laino
  • Michele_Laino
I think that there is a typo, namely: T=10 hours and not T=6 hours

Looking for something else?

Not the answer you are looking for? Search for more explanations.