A group of 4 frieds decide to attend a night of dinner and dancing on a cruise ship. The dinner and dance cruise will last 4 hours. The ship travels 18 km upstream and then travels back to the starting point. The river has a current of 2 km an hour. What is the cruise ship's speed and how long was the upstream journey

The speed of the cruise ship relative to the current is approximately 9.424 kilometers per hour.

The cruise ship's will need 2.425 hours for its upstream journey.

Step-by-step explanation:

The total time of the cruise (), in hours, is the sum of the times for the upstream () and downstream journeys (), in hours. Let suppose that both river and cruise ship travel at constant velocity, the cruise ship travels against the current in its upstream journey, whereas it is in favor of the current in the downstream journey, meaning that:

(1)

Where:

- Speed of the current, in kilometers per hour.

- Speed of the cruise ship relative to the current, in kilometers per hour.

If we know that , and , then we have the following expression:

The roots of the second order polynomial are, respectively:

, . Since the speed is the magnitude of the velocity, a realistic solution must be a positive quantity and the solution is:

The speed of the cruise ship relative to the current is approximately 9.424 kilometers per hour.

And the time needed for the upstream journey is:

(2)

If we know that , and , then time for the upstream journey is:

The cruise ship's will need 2.425 hours for its upstream journey.

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°

the anwers is yes.