Find the value of p from 24 = 6 + 2p and the value of y from y2/8 = 8. Describe the shape formed by coordinates A(2,4), B(4,1), C(2,-2), D(-4,1).

The value of $p$ is $\boxed{9}$.

c) Find the value of $y$ from equation $\frac{y^2}{8} = 8$

Remember to include both positive and negative roots.

The value of $y$ is $\boxed{\pm 8}$.

SECTION C: 10 MARKS

7.

a) Describe the shape that is formed by the coordinates A(2,4), B(4,1), C(2,-2), D(-4,1).

Length AB: $\sqrt{(4-2)^2 + (1-4)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$
Length BC: $\sqrt{(2-4)^2 + (-2-1)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$
Length CD: $\sqrt{(-4-2)^2 + (1-(-2))^2} = \sqrt{(-6)^2 + (3)^2} = \sqrt{36+9} = \sqrt{45}$
Length DA: $\sqrt{(2-(-4))^2 + (4-1)^2} = \sqrt{6^2 + 3^2} = \sqrt{36+9} = \sqrt{45}$

We have two pairs of equal-length adjacent sides: AB = BC and CD = DA. This is the definition of a kite.
We can also check the diagonals:
Diagonal AC connects (2,4) and (2,-2). This is a vertical line segment.
Diagonal BD connects (4,1) and (-4,1). This is a horizontal line segment.
Vertical and horizontal lines are perpendicular. In a kite, the diagonals are perpendicular.
The shape formed by the coordinates is a $\boxed{\text{kite}}$.

b) Find the average of $20 \text{ kg}, 35 \text{ kg}, 15 \text{ kg}, 10 \text{ kg},$ and $20 \text{ kg}$.

There are 5 values.

The average is $\boxed{20 \text{ kg}}$.

8.

a) Find the size of angle EBD.

The angles $\angle CBE$, $\angle EBD$, and $\angle DBA$ lie on the straight line CA.
So, $\angle CBE + \angle EBD + \angle DBA = 180^\circ$.

The size of angle EBD is $\boxed{\frac{50}{3}^\circ}$ or approximately $16.67^\circ$.

b) Find the area of the shaded part.

The shaded part is a right-angled triangle. The sides are $(6x-2)$ m, $(20-5x)$ m, and the hypotenuse is 26 m.

This quadratic equation is difficult to factor directly. Let's check for integer solutions by testing small integer values for $x$.
If $x=4$:
$61(4^2) - 224(4) - 272 = 61(16) - 896 - 272 = 976 - 896 - 272 = 80 - 272 = -192 \neq 0$.
If $x=5$:
$61(5^2) - 224(5) - 272 = 61(25) - 1120 - 272 = 1525 - 1120 - 272 = 405 - 272 = 133 \neq 0$.

Let's re-examine the problem. It's possible there's a simpler way or an integer value for $x$ that makes the sides work out.
If the sides are integers, we could look for Pythagorean triples. 26 is $2 \times 13$. Common triples are $(5,12,13)$, so $(10,24,26)$ is a possibility.
If the legs are 10 and 24:
Case 1: $6x-2 = 10 \implies 6x = 12 \implies x=2$.
Then $20-5x = 20-5(2) = 20-10 = 10$. This doesn't match 24.
Case 2: $6x-2 = 24 \implies