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Solve the Problem 12.812.8 -Hour Period 8 m8 \mathrm {~m} And Ranges from 5 To

Question 59

Multiple Choice

Solve the problem.
-Tides go up and down in a 12.812.8 -hour period. The average depth of a certain river is 8 m8 \mathrm {~m} and ranges from 5 to 11 m11 \mathrm {~m} . The variation can be approximated by a sine curve. Write an equation that gives the approximate variation yy , if xx is the number of hours after midnight and high tide occurs at 10:00 am.


A) y=3sin(πx6.46.8π6.4) +8y = 3 \sin \left( \frac { \pi x } { 6.4 } - \frac { 6.8 \pi } { 6.4 } \right) + 8
B) y=8sin(πx6.410π) +3\mathrm { y } = 8 \sin \left( \frac { \pi x } { 6.4 } - 10 \pi \right) + 3
C) y=3sin(πx6.410π6.4) +8y = 3 \sin \left( \frac { \pi x } { 6.4 } - \frac { 10 \pi } { 6.4 } \right) + 8
D) y=8sin(πx6.410π6.4) +3y = 8 \sin \left( \frac { \pi x } { 6.4 } - \frac { 10 \pi } { 6.4 } \right) + 3

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