At an ocean depth of 8 meters, a buoy bobs up and then down 5 meters from the ocean's depth. Sixteen seconds pass from the time
the buoy is at its highest point to when it is at its lowest point. Assume at x = 0 the buoy is at normal ocean depth.
Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.
Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.
1 answer:
Answer:
See attached for a graph
Step-by-step explanation:
We're going to plot sea level as y=0 and a depth of 8 meters as y=-8.
The problem statement tells you the initial point (x=0) is at normal ocean depth (y=-8), so the first point you put into your sine tool is ...
(x, y) = (0, -8)
The buoy takes 16 seconds to go from a high point to a low point, so the time to the first high point is half that, or x=8 seconds. That high point is 5 meters above its average depth, so is at y=-3.
The second point you will put into your sine tool is ...
(x, y) = (8, -3)
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Hello here is a solution :
√(- 9 ) / (( 4 - 7 i ) - ( 6 - 6 i )) = 3 i / ( 4 - 7 i - 6 + 6 i ) =
= 3 i / ( - 2 - i ) = - 3 i / ( 2 + i ) =
= - 1 - 2 i
= 3 i / ( - 2 - i ) = - 3 i / ( 2 + i ) =
= - 1 - 2 i