Solve the Problem $12.8$ -Hour Period $8 \mathrm {~m}$ And Ranges from 5 To

question 59

Multiple Choice

Solve the problem.
-Tides go up and down in a $12.8$ -hour period. The average depth of a certain river is $8 \mathrm {~m}$ and ranges from 5 to $11 \mathrm {~m}$ . The variation can be approximated by a sine curve. Write an equation that gives the approximate variation $y$ , if $x$ is the number of hours after midnight and high tide occurs at 10:00 am.

• A) \(y = 3 \sin \left( \frac { \pi x } { 6.4 } - \frac { 6.8 \pi } { 6.4 } \right) + 8\)
• B) \(\mathrm { y } = 8 \sin \left( \frac { \pi x } { 6.4 } - 10 \pi \right) + 3\)
• C) \(y = 3 \sin \left( \frac { \pi x } { 6.4 } - \frac { 10 \pi } { 6.4 } \right) + 8\)
• D) \(y = 8 \sin \left( \frac { \pi x } { 6.4 } - \frac { 10 \pi } { 6.4 } \right) + 3\)

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