Arectangle 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?

Width: 3y²

Length: 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.

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

i recommend you do what your heart desires. if you're attracted to the same gender, then you're attracted to the same gender. you can't determine what your heart wants.