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The altitude of the International Space Station 
t minutes after its perigee (closest point), in kilometers, is given by

A(t)=415-sin((2pi(t+23.2))/(92.8)).
The International Space Station reaches its perigee once in every orbit.
How long does the International Space Station take to orbit the earth? Give an exact answer.
minutes

The altitude of the International Space Station t t minutes after its perigee (closest point), in kilometers, is given by\newlineA(t)=415sin(2π(t+23.2)92.8). A(t)=415-\sin \left(\frac{2 \pi(t+23.2)}{92.8}\right) . \newlineThe International Space Station reaches its perigee once in every orbit.\newlineHow long does the International Space Station take to orbit the earth? Give an exact answer.\newline\square minutes

Full solution

Q. The altitude of the International Space Station t t minutes after its perigee (closest point), in kilometers, is given by\newlineA(t)=415sin(2π(t+23.2)92.8). A(t)=415-\sin \left(\frac{2 \pi(t+23.2)}{92.8}\right) . \newlineThe International Space Station reaches its perigee once in every orbit.\newlineHow long does the International Space Station take to orbit the earth? Give an exact answer.\newline\square minutes
  1. Altitude Function Period: The altitude function A(t)A(t) is periodic, with the period being the time it takes for the ISS to complete one orbit.
  2. Sine Function Period: The period of the sine function sin(Bt)\sin(Bt) is 2πB\frac{2\pi}{B}. Here, B=2π92.8B = \frac{2\pi}{92.8}.
  3. Calculation of Period: So, the period of A(t)A(t) is 92.892.8 minutes because that's the value that makes BtBt equal to 2π2\pi.
  4. Conclusion: Therefore, the International Space Station takes 92.892.8 minutes to orbit the Earth.

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