An algebra tile configuration. There are 3 large tiles, 5 tiles each half the size of a large tile, and 8 tiles each one-quarter the size of a large tile. Two of the large tiles are labeled plus x squared and 1 is labeled negative x square. Two smaller tiles are labeled plus x and 3 are labeled negative x. Six of the smallest tiles are labeled + and 2 are labeled minus. Which polynomial is represented by the algebra tiles? x2 – x – 4 x2 – x + 4 3x2 – 5x + 8 3x2 – 5x – 8

x² - x + 4

Step-by-step explanation:

Algebra tile configuration involve the use of tiles to represent algebra problem in polynomial form.

Given that:

There are 3 large tiles. Two of the large tiles are labeled plus x squared and 1 is labeled negative x square. Therefore the large tiles is given by:

2(x²) + (-x²) = 2x² - x² = x²

5 tiles are each half the size of a large tile with Two labeled + x and 3 are labeled - x. The algebra of the small tiles is:

2(+x) + 3(-x) = 2x - 3x = -x

8 tiles are each one-quarter the size of a large tile, Six of the smallest tiles are labeled +1 and 2 are labeled -1. The algebra of the smallest tiles is:

6(+1) + 2(-1) = 6 - 2 = 4

Therefore the polynomial is given by:

x² + (-x) + (4) = x² - x + 4

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: