suppose the depth of the tide in a certain harbor can be modeled by y = 20 + 5 cos (pi/6 t), where y is the water depth in feet and t is the time in hours. Consider a day in which t = 0 represents 12:00 midnight. For that day, when are high tide and low tide and what is depth of each

- high tide occurs at 12 noon and 12 midnight

- Low tide occurs at 6 a.m and 6 p.m

- maximum depth value = 20 ft

- Minimum depth value = 15 ft

Step-by-step explanation:

The depth is modelled as;

y = 20 + 5 cos (πt/6)

We are told that t = 0 represents 12:00 midnight.

This is high tide because at t = 0, the cos function will be at it's maximum value of 1 since cos 0 = 1.

Max depth value is;

y = 20 + 5(0)

y = 20 ft

Minimum depth value will be the low tide and it will be when the cos function is equal to -1.

Thus;

y = 20 + 5(-1)

y = 15 ft

Since t represents number of hours and since at 12 midnight, t = 0, thus; high tide will occur again at;

12 noon

Also, let's check for low tide.

Let's try t = 6 which means 6 a.m

Thus;

y = 20 + 5 cos (π(6)/6)

y = 20 + 5 cos π

Cos π has a value of -1

Thus;

y = 20 + 5(-1)

y = 20 - 5

y = 15 ft

Thus;

Low tide occurs at 6 a.m and 6 p.m

180

step-by-step explanation:

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°