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.

  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)

chase is right because he already been to the swings and he know how far it will be to get there

step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."