Ben
contestada

Consider a rectangle that has a perimeter of [tex]80\ \text{cm}[/tex]. Write a function [tex]A(l)[/tex] that represents the area of the rectangle with length [tex]l[/tex].

Respuesta :

NoblePenguin

Answer:

[tex]\boxed{A(l)= 40l - l^2}[/tex]

Step-by-step explanation:

We know that a rectangle has four sides - two lengths and two widths. We also know that the lengths are equal and that the widths are equal. Therefore, to find the perimeter of a rectangle, we use:

[tex]P = 2L + 2W[/tex]

However, this problem asks to find the area of the rectangle. The area can be found by multiplying length and width (A = L × W).

Using our formula above, we can solve for the length.

[tex]80 = 2L + 2W[/tex] (divide everything by 2)

[tex]40 = L + W[/tex] (subtract W in order to solve for L)

[tex]40 - L = W[/tex] (rearrange W to the left - standard equation notation)

[tex]W = 40 - L[/tex]

The area of a rectangle is solved using the formula A = LW, so we just need to input our values and solve for A.

[tex]A(l) = (40 - l)(l)[/tex] (distribute the W throughout)

[tex]A(l)= 40l - l^2[/tex]

This is your final function. Hope this helps!

Supernova

Answer:

[tex]A(L) =40L-L^2[/tex]

Step-by-step explanation:

The perimeter of a rectangle is modeled by the function: [tex]P = 2L + 2W[/tex]

You are given that the perimeter is 80 cm, so substitute this value in for [tex]P[/tex]

  • [tex]80=2L+2W[/tex]

You want the function [tex]A(l)[/tex] to represent the area of the rectangle in terms of the length [tex]L[/tex], so, therefore, you want to solve for the width of the rectangle (so you can still have that [tex]L[/tex] variable).

Solve for [tex]W[/tex] by first subtracting [tex]2L[/tex] from both sides of the equation.

  • [tex]80-2L = 2W[/tex]

Then divide both sides of the equation by 2 to isolate the variable [tex]W[/tex].

  • [tex]\frac{80-2L}{2}=W[/tex]

This can be simplified even further:

  • [tex]W=40-L[/tex]

The area of a rectangle is modeled by the function: [tex]A=LW[/tex]

You have already solved for [tex]W[/tex], so substitute [tex]40-L[/tex] for it.

  • [tex]A=L(40-L)[/tex]

Use the distributive property to multiply everything inside the parentheses by [tex]L[/tex].

  • [tex]A=40L-L^2[/tex]

Since this function is in terms of [tex]L[/tex], you can write:

  • [tex]A(L) =40L-L^2[/tex]

This function represents the area of the rectangle with a perimeter of 80 cm and length [tex]L[/tex].

Otras preguntas