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Answer :
One possible solution: dimensions [tex]\(6x^3y^5\) and \(4x^3y^{10}\)[/tex] create the area [tex]\(24x^6y^{15}\)[/tex] for the rectangle.
To find the dimensions of the rectangle given the area expression
[tex]\(24x^6 y^{15}\),[/tex] we must break it into two factors representing the length ((l)) and width (w)).
Start by prime factorizing the area expression: [tex]\(24x^6 y^{15}\).[/tex]
[tex]\[24 = 2 \times 2 \times 2 \times 3\][/tex]
So, we can represent (l) and (w) as the product of two factors of these prime factors and powers of (x) and (y) that sum up to (6) and (15) respectively.
One possible combination is [tex]\(l = 6x^3y^5\) and \(w = 4x^3y^{10}\):[/tex]
[tex]\[l \times w = (6x^3y^5) \times (4x^3y^{10}) = 24x^6y^{15}\][/tex]
Thus, [tex]\(6x^3y^5\) and \(4x^3y^{10}\)[/tex] are potential dimensions of the rectangle, resulting in the given area expression.
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