Problem 26 - Algebra 2

Question

The optimal height \(h\) of the letters of a message printed on pavement is given by the formula \(h=\frac{0.0052 d^{227}}{e} .\) Here \(d\) is the distance of the driver from the letters and \(e\) is the height of the driver's eye above the pavement. All of the distances are in meters. Find \(h\) for the given values of \(d\) and \(e .\) $$d=100 \mathrm{m}, e=1.2 \mathrm{m}$$

Step-by-Step Solution

Answer

The optimal height \(h\) of the letters on pavement for a driver viewing from a distance of 100m and whose eyes are 1.2m above the ground is approximately 771.89 meters.

Step 1: Substitution of Values

Begin by substituting the given values into the formula. That is, \(d = 100\) meters and \(e = 1.2\) meters. Thus, the formula becomes: \(h = \frac{0.0052 \times 100^{2.27}}{1.2}\)

Step 2: Calculation of the Numerator

Next, calculate the value of the numerator, \(0.0052 \times 100^{2.27}\). Using a calculator or similar tool, this becomes approximately 926.27.

Step 3: Division

Now divide this result by the denominator to find the value of \(h\). This would be \(h = \frac{926.27}{1.2}\), which evaluates to approximately 771.89 m.

Key Concepts

Substitution

Substitution is a fundamental concept in algebra, where we replace variables with given numerical values. This allows us to simplify equations and solve them. In our exercise, we start with the formula for the height of letters:\[ h = \frac{0.0052 \times d^{2.27}}{e} \]We need to find the height, \(h\), when \(d = 100\) meters (distance from the driver to the letters) and \(e = 1.2\) meters (the driver's eye height above the pavement).
By substituting these values into the formula, we replace \(d\) with 100 and \(e\) with 1.2, turning the equation into:\[ h = \frac{0.0052 \times 100^{2.27}}{1.2} \]This simplification sets the stage for the next steps in solving the equation.

Exponents

Exponents signify repeated multiplication. When we see a number raised to a power, it means multiplying that number by itself as many times as the exponent indicates. For example, \(100^{2.27}\) involves raising the number 100 to the power of 2.27.
This notation indicates a not straightforward multiplication due to the decimal power.In our exercise, calculating \(100^{2.27}\) involves using a calculator or a software tool. This isn't a simple task by hand because it involves computing a non-integer power. However, understanding exponents will help you manage even complex equations by breaking them down into understandable parts.
This is essential for correctly applying the power to our distance value of 100 meters.Working through this calculation provides a portion of our numerator, which we will use in the division step.

Division

Division is another basic arithmetic operation where we determine how many times one number is contained within another. In algebraic formulas, it often complements multiplication. Here, we divide the calculated numerator by the denominator.After computing the numerator, \(0.0052 \times 100^{2.27} = 926.27\) approximately, the next task is to perform the division:
\[ h = \frac{926.27}{1.2} \]This requires us to divide 926.27 by 1.2. Helpful tools like calculators ensure high accuracy in such division operations. The division is essential to isolate the variable \(h\), giving us the answer in terms of meters.
It's important to take care to correctly implement each step and check calculations for precision.

Calculation

Calculation embodies the process of combining steps like substitution, exponentiation, and division into one cohesive process. It's the heart of solving mathematical problems and involves strategic sequencing of operations.By substituting values and using our understanding of exponents, we prepared the ground for accurate calculation. Staying organized and methodically applying each arithmetic operation ensures the reliability of our result.
The computed value of \(h\), approximately 771.89 meters, rounds the calculation task.
Tying all previous steps together correctly leads us to our final answer, highlighting the importance of each individual arithmetic operation.