The side length of each square is 6 units. Find the areas of the inscribed shapes.

a) A₁ = 18 unit²

b) A₂ = 20 unit²

c) A₃ = 12 unit²

d) A₄ = 12 unit²

Step-by-step explanation:

a) Given that the side length of square is 6 units, we have;

The height of the square = The height of the triangle = 6 units

The base of the triangle = The side length of the square = 6 units

The area of a triangle A₁ = 1/2×base×height = 1/2×6×6 = 18 unit²

b) The side of the square A₂ forms an hypotenuse side to the side length 2 and 4 on sides of the circumscribing square

The length of the side = √(4^2 + 2^2) = 2·√5

A₂ = The area of a square =Side² = (2·√5)² = 20 unit²

c) The base length of the triangle, A₃ + 2 = The side length of the circumscribing square = 6 units

∴ The base length of the triangle, ₃₂ = 6 - 2 = 4 units

The height of the triangle, A₃ = The  side length of the circumscribing square = 6 units

The area of a triangle A₃ = 1/2×base×height = 1/2×4×6 = 12 unit²

d) Figure, A₄, is a parallelogram;

The area of a parallelogram = Base × Height

The base of the parallelogram, A₄ + 4 = 6 units

Therefore, the base of the parallelogram, A₄ = 6 - 4 = 2 units

The height of the parallelogram = The  side length of the circumscribing square = 6 units

The area of a parallelogram A₄ = 2× 6 = 12 unit².

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w