4. Consider the figure below. (5x - 14) ft
(2x) - 5x2 - 12x + 14) ft
What is the total area, in square feet, of the figure

(10x⁴ - 52x³ + 5x² + 246x - 196) ft²

Step-by-step explanation:

The figure can be divided into two rectangles: the small rectangle and the large rectangle.

Dimensions of the small rectangle:

Length (l) = (x) ft

Width (w) = (x² - 6) - (5x - 14) = x² - 6 - 5x + 14 = (x² - 5x + 8) ft

Area of small rectangle = l*w = (x)(x² - 5x + 8)

Area = x(x²) - x(5x) + x(8) (using distributive property)

Area of small rectangle = (x³ - 5x² + 8x) ft²

Dimensions of large rectangle:

Length (l) = (2x³ - 5x² - 12x + 14) ft

Width (w) = (5x - 14) ft

Area of large rectangle = l*w = (2x³ - 5x² - 12x + 14)(5x - 14)

2x³(5x - 14) - 5x²(5x - 14) - 12x(5x - 14) + 14(5x - 14) (distributive property)

10x⁴ - 28x³ - 25x³ + 70x² - 60x² + 168x + 70x - 196

Combine like terms

Area of large rectangle = (10x⁴ - 53x³ + 10x² + 238x - 196) ft²

Area of figure = area of small rectangle + area of large rectangle

= (x³ - 5x² + 8x) + (10x⁴ - 53x³ + 10x² + 238x - 196)

Open parentheses

= x³ - 5x² + 8x + 10x⁴ - 53x³ + 10x² + 238x - 196

Combine like terms and arrange in standard form (from the greatest degree to the least)

= 10x⁴ + x³ - 53x³ - 5x² + 10x² + 8x + 238x - 196

= 10x⁴ - 52x³ + 5x² + 246x - 196

Area of the figure = (10x⁴ - 52x³ + 5x² + 246x - 196) ft²

water molecules will move from the side of higher water concentration to the side of lower concentration until both solutions are isotonic. at this point, equilibrium is reached. in an isotonic solution, the flow of water in and out of the cell is happening at the same rate.

hope this < 3