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Let's analyze and solve the problem presented in the image.

We are given the equation for the shape of the arch:
\[
y = -\frac{1}{12}(x^2 - 36x)
\]
This equation represents a parabola where \(x\) is the width of the arch in inches, and \(y\) is the height in inches. The thickness of the arch is 0.3 inches, and we are asked to find the total height of the arch, excluding the thickness.

### Step 1: Simplify the equation
First, let's rewrite the equation to make it easier to analyze:
\[
y = -\frac{1}{12}(x^2 - 36x) = -\frac{1}{12}x^2 + 3x
\]
This is the equation of a downward-facing parabola.

### Step 2: Find the maximum height of the arch
The vertex of a parabola gives the maximum or minimum point. For a quadratic equation of the form:
\[
y = ax^2 + bx + c
\]
The x-coordinate of the vertex is given by:
\[
x = -\frac{b}{2a}
\]
Here, \(a = -\frac{1}{12}\) and \(b = 3\). Thus:
\[
x = -\frac{3}{2 \times -\frac{1}{12}} = \frac{3}{\frac{2}{12}} = \frac{3 \times 12}{2} = 18
\]
So, the maximum height occurs at \(x = 18\).

### Step 3: Calculate the height at \(x = 18\)
Substitute \(x = 18\) into the equation for \(y\):
\[
y = -\frac{1}{12}(18^2 - 36 \times 18)
\]
Simplify the expression inside the parentheses:
\[
18^2 = 324 \quad \text{and} \quad 36 \times 18 = 648
\]
Thus:
\[
y = -\frac{1}{12}(324 - 648) = -\frac{1}{12}(-324) = \frac{324}{12} = 27
\]
So, the height of the arch is 27 inches.

### Step 4: Exclude the thickness of the arch
The question asks for the height excluding the 0.3 inches of thickness. Since the thickness is negligible in terms of the height, the answer remains \(27\) inches.

### Final Answer:
The height of the arch, excluding the 0.3 inches thickness, is \(27\) inches, which corresponds to option C.

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Would you like a further breakdown of any part of the process, or do you have additional questions?

Here are 5 related questions to expand on this topic:
1. How would the shape of the arch change if the parabola was wider or narrower?
2. Can we derive the vertex form of a parabola from the given equation?
3. What happens to the maximum height if the parabola opens upwards instead of downwards?
4. How do the coefficients in the equation affect the overall geometry of the arch?
5. Could this equation be used to design other shapes besides an arch?

**Tip:** Always verify if the quadratic opens upward or downward by checking the sign of the leading coefficient \(a\). If \(a\) is negative, the parabola opens downward, giving a maximum point.

An architect designed a 0.3 inches thick arch-shaped monument for a park. The equation of the arch is given as y = -1/12(x^2 - 36x), where x is the width of the arch in inches, and y is the height in inches. Which of the following is the height of the arch, excluding the 0.3 inches thickness?

This problem involves calculating the height of an arch-shaped monument using the quadratic equation y = -1/12(x^2 - 36x). It covers key algebraic concepts like quadratic equations and vertex theorem. Suitable for students in Grades 9-12.

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