Question

a wing of a school is built around an outdoor courtyard.
drag the terms to show the combined area of the cafeteria and art room as a polynomial in standard form.
-5x -80x 16x³ 18x³ 16x² -40x

Explanation:

Step1: Find the height of the cafeteria

The height of the cafeteria is the sum of the heights of the art room, the middle section, and the band room. So, \(2x + (13x + 25)+ 2x=17x + 25\)? Wait, no, looking at the diagram, the left side of the cafeteria is labeled \(2x^{2}-5\), and the width of the cafeteria is \(8x\)? Wait, no, the art room has width \(x^{2}\) and height \(2x\), the cafeteria is a rectangle with width \(8x\) and height \(2x^{2}-5\)? Wait, no, let's re - examine.

The art room: width \(x^{2}\), height \(2x\), area \(A_{art}=x^{2}\times2x = 2x^{3}\)

The cafeteria: width \(8x\), height \(2x^{2}-5\), area \(A_{cafeteria}=8x\times(2x^{2}-5)=16x^{3}-40x\)

Wait, maybe I misread the diagram. Wait, the combined height of the art room, the middle (courtyard and hallway) and band room? No, the problem is to find the combined area of cafeteria and art room.

Wait, let's look at the dimensions again. The art room: length \(x^{2}\), height \(2x\), so area of art room is \(x^{2}\times2x = 2x^{3}\)

The cafeteria: length \(8x\), height \(2x^{2}-5\), so area of cafeteria is \(8x\times(2x^{2}-5)=16x^{3}-40x\)

Now, combined area of cafeteria and art room: \(A = 2x^{3}+16x^{3}-40x=18x^{3}-40x\)? Wait, no, maybe the height of the cafeteria is \(2x + (13x + 25)+2x\)? No, that seems wrong. Wait, maybe the vertical side of the cafeteria is \(2x^{2}-5\) and the horizontal side is \(8x\), and the art room has horizontal side \(x^{2}\) and vertical side \(2x\).

Wait, let's recalculate:

Area of Art Room: The art room is a rectangle with length \(x^{2}\) and width \(2x\), so area \(A_{art}=x^{2}\times2x = 2x^{3}\)

Area of Cafeteria: The cafeteria is a rectangle with length \(8x\) and width \(2x^{2}-5\), so area \(A_{cafeteria}=8x\times(2x^{2}-5)=16x^{3}-40x\)

Combined area \(A = A_{art}+A_{cafeteria}=2x^{3}+16x^{3}-40x = 18x^{3}-40x\)

Wait, but the given options are \(16x^{3},18x^{3}, - 5x,-80x,16x^{2},-40x\)

Wait, maybe I made a mistake in the art room's dimensions. Let's check the diagram again. Maybe the art room's height is \(2x\) and the length is \(x^{2}\), and the cafeteria's height is \(2x^{2}-5\) and length is \(8x\).

Wait, another approach: Maybe the art room has dimensions \(x^{2}\) (length) and \(2x\) (width), area \(2x^{3}\). The cafeteria has dimensions \(8x\) (length) and \(2x^{2}-5\) (width), area \(16x^{3}-40x\). Combining them: \(2x^{3}+16x^{3}-40x = 18x^{3}-40x\)

So the terms are \(18x^{3}\) and \(-40x\)

Step2: Identify the terms

From the given options, the terms that make up the combined area are \(18x^{3}\) and \(-40x\)

Answer:

The terms are \(18x^{3}\) and \(-40x\) (corresponding to the options \(18x^{3}\) and \(-40x\))