Problem 80 - Intermediate Algebra

Question

The maximum distance \(D(h)\) in kilometers that a person can see from a height h kilometers above the ground is given by the func\(\operatorname{tion} D(h)=111.7 \sqrt{h} .\) Find the height that would allow a person to see 40 kilometers.

Step-by-Step Solution

Answer

A height of approximately 0.1283 kilometers allows seeing 40 kilometers.

Step 1: Understand the Problem

We are given a function for the maximum distance a person can see from a height, expressed as \(D(h)=111.7\sqrt{h}\). The task is to find the height \(h\) that allows seeing a distance of 40 kilometers.

Step 2: Set Up the Equation

We need to find the height \(h\) such that \(D(h) = 40\). Start by setting the equation \(111.7\sqrt{h} = 40\).

Step 3: Isolate the Square Root

Divide both sides of the equation by 111.7 to solve for the square root: \(\sqrt{h} = \frac{40}{111.7}\).

Step 4: Solve for h

Square both sides of the equation to get rid of the square root: \(h = \left(\frac{40}{111.7}\right)^2\).

Step 5: Calculate the Height

Calculate the value of \(h\) using a calculator: \(h \approx \left(\frac{40}{111.7}\right)^2 \approx 0.1283\). This means the height should be approximately 0.1283 kilometers.

Key Concepts

Understanding the Distance Formula

The distance formula is vital in many areas of mathematics and science. It allows us to calculate how far one point is from another. In this specific exercise, the formula is used to determine the distance visible from a certain height above the ground.

The given function, \(D(h) = 111.7 \sqrt{h}\), is applied here. It links the height \(h\) (in kilometers) to the maximum distance \(D\) (also in kilometers) a person can see. By understanding this relationship, you can find the height needed to see a specified distance by manipulating the formula.

In our given problem, you need to rearrange the formula to find the height \(h\) when the distance \(D\) is known. This exercise emphasizes understanding how to apply a given formula to compute unseen parameters.

Solving Square Root Equations

Square root equations are frequently encountered in algebra. The equation \(111.7\sqrt{h} = 40\) is a simple example. Here, we need to solve for \(h\), given a distance \(D\).

First, isolate the square root by dividing both sides by 111.7:

• \(\sqrt{h} = \frac{40}{111.7}\)

This eliminates the coefficient from the square root.

Next, eliminate the square root by squaring both sides:

• \(h = \left(\frac{40}{111.7}\right)^2\)

This operation is crucial as it converts the equation to not contain a square root, making it straightforward to solve for \(h\).

Calculate \(h\) to find the height. This involves basic arithmetic and using a calculator, ensuring precision in finding the height that allows visibility over a given distance.

Function Evaluation Essentials

Function evaluation involves plugging given values into a formula to find an unknown. In this exercise, the challenge is to find the height when the distance is known.

Think of the function \(D(h)=111.7\sqrt{h}\) as a machine. You input a height \(h\), and out comes \(D\), the distance.

If we know \(D\) and seek \(h\), we reverse the process: start from the known distance and work backwards. Follow these key steps:

• Set \(D(h) = 40\)
• Use algebraic manipulations to isolate and solve for \(h\)
• Double-check calculations for accuracy

Evaluation often involves such back-and-forth processes, testing your understanding of both direct and inverse operations on functions.