Amountain climber is at an altitude of 2.9 mi above the earth’s surface. from the climber’s viewpoint, what is the distance to the horizon? enter your answer as a decimal in the box. round only your final answer to the nearest tenth.

The distance to the horizon, from the climber’s viewpoint to the nearest tenth is, 151.6 mi

Step-by-step explanation:

Using Pythagoras theorem:

We know that:

Radius of the earth ≈ 3959 mi.

The radius of a circle meets a tangent at 90 degree.

As per the statement:

A mountain climber is at an altitude of 2.9 mi above the earth’s surface.

See the diagram as shown below in the attachment:

In triangle ABO

Hypotenuse side = AO = 3959 +2.9 = 3961.9 mi

Opposite side = AB = x mi and

Adjacent side = 3959 mi

Substitute these we have;

⇒

⇒

⇒

Simplify:

151.560219 mi = x

or

x =151.60219 mi

therefore, the distance to the horizon, from the climber’s viewpoint to the nearest tenth is, 151.6 mi

518,400

step-by-step explanation:

the area is multiplying all the given measurements together.

40 * 18 * 30 * 24 = 518,400