Find the coordinates of points A and B, determine the equation of the straight line y = mx + c, and find the area between the curve y = (6−x)(x+2) and the line.

Here are the solutions to the problems from the image.

1. Curve and Line
The curve is given by $y = (6-x)(x+2)$.
Expanding the curve equation:

$$y = 6x + 12 - x^2 - 2x$$ $$y = -x^2 + 4x + 12$$

The line is given by $y = mx+c$.

a) Find the coordinates of the point A and B.
From the graph, points A and B are the intersection points of the curve and the line.
Point A is where $x=0$. Substitute $x=0$ into the curve equation:

$$y = (6-0)(0+2) = 6 \times 2 = 12$$

So, point A is $(0, 12)$.

Point B is where $x=4$. Substitute $x=4$ into the curve equation:

$$y = (6-4)(4+2) = 2 \times 6 = 12$$

So, point B is $(4, 12)$.
The coordinates are $\boxed{A(0, 12) \text{ and } B(4, 12)}$.

b) Determine the equation of the straight line $y = mx+c$.
The line passes through points A $(0, 12)$ and B $(4, 12)$.
The slope $m$ is given by:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 12}{4 - 0} = \frac{0}{4} = 0$$

The y-intercept $c$ is the y-coordinate when $x=0$, which is 12 (from point A).
So, the equation of the straight line is $y = 0x + 12$.
The equation of the straight line is $\boxed{y = 12}$.

c) Find the area between the curve $y = (6-x)(x+2)$ and the line $y = mx+c$.
The area between the curve $y = -x^2 + 4x + 12$ and the line $y = 12$ is given by the integral of the difference between the upper function and the lower function, from $x=0$ to $x=4$.
In the interval $[0, 4]$, the curve is above the line.
Area $= \int_{0}^{4} ((-x^2 + 4x + 12) - 12) \,dx$
Area $= \int_{0}^{4} (-x^2 + 4x) \,dx$
Integrate the expression:

$$\int (-x^2 + 4x) \,dx = -\frac{x^3}{3} + \frac{4x^2}{2} = -\frac{x^3}{3} + 2x^2$$

Now, evaluate the definite integral:

$$\text{Area} = \left[ -\frac{x^3}{3} + 2x^2 \right]_{0}^{4}$$ $$\text{Area} = \left( -\frac{4^3}{3} + 2(4^2) \right) - \left( -\frac{0^3}{3} + 2(0^2) \right)$$ $$\text{Area} = \left( -\frac{64}{3} + 2(16) \right) - (0)$$ $$\text{Area} = -\frac{64}{3} + 32$$ $$\text{Area} = \frac{-64 + 96}{3}$$ $$\text{Area} = \frac{32}{3}$$

The area between the curve and the line is $\boxed{\frac{32}{3} \text{ square units}}$.

2. Square-based pyramid ABCD V
Given: Square base ABCD with AD = DC = 6 cm.
Height of pyramid OV = 10 cm, where O is the center of the base.

a) State the projection of VA on the base ABCD.
The projection of the vertex V onto the base is the point O. The point A is already on the base.
Therefore, the projection of VA on the base ABCD is the line segment $\boxed{OA}$.

b) Find:

i) The length VA.
First, find the length of the diagonal AC of the square base. Using the Pythagorean theorem in triangle ADC:

$$AC^2 = AD^2 + DC^2$$ $$AC^2 = 6^2 + 6^2 = 36 + 36 = 72$$ $$AC = \sqrt{72} = 6\sqrt{2} \text{ cm}$$

The length OA is half of the diagonal AC:

$$OA = \frac{1}{2} AC = \frac{1}{2} (6\sqrt{2}) = 3\sqrt{2} \text{ cm}$$

Now, consider the right-angled triangle VOA. VO is the height of the pyramid (10 cm) and OA is the distance from the center to a vertex of the base.
Using the Pythagorean theorem in triangle VOA:

$$VA^2 = VO^2 + OA^2$$ $$VA^2 = 10^2 + (3\sqrt{2})^2$$ $$VA^2 = 100 + (9 \times 2)$$ $$VA^2 = 100 + 18 = 118$$ $$VA = \sqrt{118} \text{ cm}$$ $$$$