Abig cruise ship dropped anchor off the caribbean island of antigua. the heavy anchor dropped into the water at a rate of 2.5 2.52, point, 5 meters per second. after 45 4545 seconds, the anchor was 40 4040 meters below the water's surface. from what height (above the water's surface) was the anchor released? meters how long did it take the anchor to reach the water's surface?

29 seconds

Explanation

anchor dropped in to water at the rate = 2.5 m/s

so for one second = 2.5 m

so for 45 sec= 2.5 x 45 = 112.5 m

but it  was already 40 m under water, so it have fallen from= 112.5-40= 72.5 m

so as we know the rate of falling was =2.5 m/s

so for 72.5 meter, it took = 72.5 /2.5= 29 sec

it took  for anchor to get down to the water surface after it started dropping.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°