A rectangle has an area of 12y^2+21y^512 y 2 +21y 5 12, y, squared, plus, 21, y, start superscript, 5, end superscript.The width

Question

3 years ago

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A rectangle has an area of 12y^2+21y^512 y 2 +21y 5 12, y, squared, plus, 21, y, start superscript, 5, end superscript.The width

of the rectangle is equal to the greatest common monomial factor of 12y^212y 2 12, y, squared and 21y^521 y 5 21, y, start superscript, 5, end superscript.What is the length and width of the rectangle?

Mathematics

1 answer:

umka21 [38] · Answer 1 · 2022-07-30T11:16:28+03:00

Answer:

<em>Width</em>: 3y²

<em>Length</em>: 4 + 7y³

Step-by-step explanation:

The area of the rectangle is the length multiplied by the width. For the area given: 12y² + 21y⁵

12y² = 3y²*4

21y⁵ = 3y²*7y³

So, the greatest common monomial factor is 3y², which is the width.

The area is 3y²*(4 + 7y³), the first term is the width, the other must be the length: 4 + 7y³.