To determine the dimensions of Ralph's monitor in terms of its width [tex]\( w \)[/tex], we start by setting up the relationship between the length [tex]\( l \)[/tex] and the width [tex]\( w \)[/tex] based on the given conditions.
1. Understanding the Length Expression:
The problem states that the length of the monitor is six times the quantity five less than half its width. Mathematically, this can be expressed as:
[tex]\[
l = 6 \left( \frac{w}{2} - 5 \right)
\][/tex]
2. Setting Up the Area Equation:
The area of the rectangle (the monitor) is given by the product of its length and width. So the area [tex]\( A \)[/tex] is:
[tex]\[
A = l \times w
\][/tex]
Given that the area is 384 square inches, we can substitute [tex]\( A \)[/tex] with 384:
[tex]\[
384 = l \times w
\][/tex]
3. Substituting the Length Expression:
Now we substitute the expression for [tex]\( l \)[/tex] into the area equation:
[tex]\[
384 = 6 \left( \frac{w}{2} - 5 \right) \times w
\][/tex]
4. Simplifying and Expanding the Equation:
First, simplify inside the parentheses:
[tex]\[
\frac{w}{2} - 5 = \frac{w - 10}{2}
\][/tex]
Then multiply by 6:
[tex]\[
6 \left( \frac{w - 10}{2} \right) = 3 (w - 10) = 3w - 30
\][/tex]
Now, our equation becomes:
[tex]\[
384 = (3w - 30) w
\][/tex]
5. Writing the Complete Equation:
Multiplying through by [tex]\( w \)[/tex], we get:
[tex]\[
384 = 3w^2 - 30w
\][/tex]
Rearranging this equation to bring all terms to one side yields:
[tex]\[
3w^2 - 30w - 384 = 0
\][/tex]
Thus, the complete equation that can be used to determine the dimensions of the monitor in terms of its width [tex]\( w \)[/tex] is:
[tex]\[
3w^2 - 30w - 384 = 0
\][/tex]
So, the correct sequence to complete the sentence is:
[tex]\[
3w^2 - 30w - 384 = 0
\][/tex]
