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Sagot :
Answer:
[tex]\textsf{A)} \quad \ell = \dfrac{12x^2-15x}{3x}=4x-5[/tex]
B) Proof given below.
Step-by-step explanation:
Given values:
[tex]\textsf{Area}=12x^2-15x[/tex]
[tex]\textsf{Width} = 3x[/tex]
Area of a rectangle
[tex]A=w \cdot \ell[/tex]
where:
- [tex]w[/tex] = width
- [tex]\ell[/tex] = length
Part A
Substitute the given values into the area formula and solve for length:
[tex]\begin{aligned} A & = w \cdot \ell\\ \implies 12x^2-15x & = 3x \cdot \ell\\\ell & = \dfrac{12x^2-15x}{3x}\\\ell & = \dfrac{3x(4x-5)}{3x}\\\ell & = 4x-5\end{aligned}[/tex]
Part B
Prove by multiplying the given width by the found length:
[tex]\begin{aligned}A & = w \cdot \ell\\ \implies A & = 3x(4x-5)\\& = 3x \cdot 4x - 3x \cdot 5\\& = 12x^2-15x\end{aligned}[/tex]
Hence proving that the length of the rectangle is (4x - 5).
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