Abig cruise ship dropped anchor off the caribbean island of antigua. the heavy anchor dropped into the water at a rate of 2.5 meters per second. after 45 seconds, the anchor was 40 meters below the water's surface.

1) from what height (above the water's surface) was the anchor released?
(meters)

2)how long did it take the anchor to reach the water's surface?
(seconds)

1) The anchor was  released from 72.5 meters above the water's surface.

2)The anchor took 29 seconds to reach the water's surface

Step-by-step explanation:

The rate of drop of anchor = 2.5 meters per second

So, total depth covered by anchor after 45 seconds

= 2.5 meters x 45 seconds

= 112.5 meters

But after 45 minutes, the anchor was 40 meters deep.

So, extra distance traveled by anchor = Total distance covered by anchor - depth of water

= 112.5 meters  - 40 meters  =  72.5 meters

Hence, the anchor was  released from 72.5 meters above the water's surface.

Now, speed of anchor = 2.5 m/sec

So, total time taken to cover extra 72. 5 meters = 72. 5 meters / 2.5 m/sec

= 29 seconds

Hence, the anchor took 29 seconds to reach the water's surface.

because they could drop even lower..

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!