Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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).
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.