Simplify: $\frac{\sqrt{16x^4 - 72x^2y^2 + 81y^4}}{\sqrt{4x^2 - 12xy + 9y^2}} - (2x - 3y)$, given that $2x > 3y$.
Let's simplify the given expression: \(\frac{\sqrt{16x^4 - 72x^2y^2 + 81y^4}}{\sqrt{4x^2 - 12xy + 9y^2}} - (2x - 3y)\).
First, consider the numerator: \(\sqrt{16x^4 - 72x^2y^2 + 81y^4}\).
This expression can be rewritten in the form of a square:
\(16x^4 - 72x^2y^2 + 81y^4 = (4x^2 - 9y^2)^2\)
Therefore, we have:
\(\sqrt{(4x^2 - 9y^2)^2} = |4x^2 - 9y^2|\)
Given that \(2x > 3y\), we know that \(4x^2 - 9y^2 \geq 0\), so \(|4x^2 - 9y^2| = 4x^2 - 9y^2\).
Now consider the denominator: \(\sqrt{4x^2 - 12xy + 9y^2}\). This expression can be simplified to:
\(4x^2 - 12xy + 9y^2 = (2x - 3y)^2\)
Hence:
\(\sqrt{(2x - 3y)^2} = |2x - 3y|\)
Given \(2x > 3y\), we have \(|2x - 3y| = 2x - 3y\).
Substituting back into the main expression, we have:
\(\frac{4x^2 - 9y^2}{2x - 3y} - (2x - 3y)\)
Which simplifies to:
\(\underset{}{} = \frac{(2x - 3y)(2x + 3y)}{2x - 3y} - (2x - 3y)\)\)
Since \(2x - 3y \neq 0\):
\(\underset{}{} = 2x + 3y - (2x - 3y) = 2x + 3y - 2x + 3y\)\)
This results in:
\(6y\)
Thus, the simplified form of the expression is \(6y\). The correct option is:
6y
Identify the number that will replace the question mark in the second equation based on the relationship represented in the first equation.

What value should come in the place of question mark (?) in the following equation?
$(0.008\div?) + (0.006 \div 0.03) + (0.008 \div 0.04) =0.5$
If 'I' stands for '+', 'J' stands for 'x', 'K' stands for '÷', and 'L' stands for '-', then what will come in place of the question mark (?) in the following equation?
6 K 2 J 4 L 3 I 9 = ?
If 'A' stands for '$\div $', 'B' stands for 'x', 'C' stands for '+' and 'D' stands for '-', then what will come in place of the question mark (?) in the following equation?
33 A 3 C 4 B 2 D 4 A 2 = ?