¿Cómo puedes usar un punto en la gráfica de F –1(X) = 9 9X para determinar un punto en la gráfica de F(X) = Iniciar sesión9 9X?

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Ответ:

sanaiajohnson56

16.09.2021

Mathematics

Following are the responses to the given points:

For point a:

$Diameter\ (d)= 16\ m\\\\$

Calculating the 3 rotations for every minute:

Calculating time for completing 1 rotation:

$1\ rotation=\frac{60}{3}= 20\ second\\\\period=20 \ second\\\\$

The standard form of the equation of the sine and cosine function is:

$y=A \sin \{ B(x-c)\} +D\\\\y=A \cos \{ B(x-c)\} +D\\\\$

Calculating the Amplitue:

$A=\frac{max-min}{2}=\frac{17-1}{2}=\frac{16}{2}=8\\\\Period=\frac{2\pi}{B}\\\\20=\frac{2\pi}{B}\\\\B=\frac{2\pi}{20}\\\\B=\frac{\pi}{10}\\\\$

Calculating the phase shift:

for $\sin$ function: c=5

for $\cos$ function: c=10

Calculating the vertical shift:

$\to D=\frac{max+ min }{2}=\frac{17+ 1}{2}=\frac{18}{2}=9\\\\y=8 \sin \{ \frac{\pi}{10}(t-5)\} +9\\\\y=8 \cos \{ \frac{\pi}{10}(t-10)\} +9\\\\$

For point b:

$y 12\ m\\\\12=8 \sin \{ \frac{\pi}{10}(t-5)\} +9\\\\12-9=8 \sin \{ \frac{\pi}{10}(t-5)\} \\\\3=8 \sin \{ \frac{\pi}{10}(t-5)\} \\\\\frac{3}{8}=\sin \{ \frac{\pi}{10}(t-5)\} \\\\\sin^{-1}\frac{3}{8}=\frac{\pi}{10}(t-5) \\\\\frac{\pi}{10} (0.38439677)=(t-5) \\\\1.22357+5=t \\\\t=6.22357\ second\\\\t=6.22\ second\\\\\sin^{-1}\frac{3}{8}=\frac{\pi}{10}(t-5) \\\\\frac{\pi}{10} (2.7571961)=(t-5) \\\\t=8.7764+5\\\\t=13.78\ second\\\\t_2-t_1=13.7764-6.22357= 7.55283\approx 7.55\ second \\\\$

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