Question

a triangle’s area is given by the expression ( 3x^2 - x + 12 ). the height is ( 6x ). the base of the triangle is the quotient of the area and ( \frac{1}{2} ) of the height. which expression represents the base of the triangle? (\bigcirc \frac{x}{2} - \frac{1}{6} + \frac{4}{x}) (\bigcirc x - \frac{1}{3} + 4x) (\bigcirc \frac{x}{2} - \frac{1}{6} + 4x) (\bigcirc x - \frac{1}{3} + \frac{4}{x})

Explanation:

Step1: Find 1/2 of the height

The height is \( 6x \), so \( \frac{1}{2} \) of the height is \( \frac{1}{2} \times 6x = 3x \).

Step2: Set up the division for the base

The base is the area divided by \( \frac{1}{2} \) of the height. The area is \( 3x^2 - x + 12 \), and \( \frac{1}{2} \) of the height is \( 3x \). So we need to compute \( \frac{3x^2 - x + 12}{3x} \).

Step3: Divide each term in the numerator by the denominator

Divide \( 3x^2 \) by \( 3x \): \( \frac{3x^2}{3x} = x \).
Divide \( -x \) by \( 3x \): \( \frac{-x}{3x} = -\frac{1}{3} \).
Divide \( 12 \) by \( 3x \): \( \frac{12}{3x} = \frac{4}{x} \).
Combine these results: \( x - \frac{1}{3} + \frac{4}{x} \).

Answer:

\( x - \frac{1}{3} + \frac{4}{x} \) (corresponding to the last option: \( \boldsymbol{x - \frac{1}{3} + \frac{4}{x}} \))