Solve the equation x² − 4x − 2 = 0, giving your answers correct to 2 decimal places.

Here are the solutions to the questions:

2. Solve the equation $x^2 - 4x - 2 = 0$, giving your answers correct to 2 decimal places.
This is a quadratic equation of the form $ax^2 + bx + c = 0$.
Here, $a=1$, $b=-4$, $c=-2$.
We use the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Substitute the values:

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)}$$ $$x = \frac{4 \pm \sqrt{16 + 8}}{2}$$ $$x = \frac{4 \pm \sqrt{24}}{2}$$

Now, calculate the two possible values for $x$:

$$x_1 = \frac{4 + \sqrt{24}}{2}$$ $$x_1 = \frac{4 + 4.898979...}{2}$$ $$x_1 = \frac{8.898979...}{2}$$ $$x_1 = 4.449489...$$

Rounding to 2 decimal places:

$$x_1 \approx 4.45$$ $$x_2 = \frac{4 - \sqrt{24}}{2}$$ $$x_2 = \frac{4 - 4.898979...}{2}$$ $$x_2 = \frac{-0.898979...}{2}$$ $$x_2 = -0.449489...$$

Rounding to 2 decimal places:

$$x_2 \approx -0.45$$

The solutions are:

$$x = \boxed{4.45 \text{ or } -0.45}$$

3. Simplify $\frac{7s^3t^2}{15u^2v^3} \times \frac{5u^3v}{28s^2t^2}$.
Multiply the numerators and the denominators:

$$\frac{7s^3t^2 \times 5u^3v}{15u^2v^3 \times 28s^2t^2}$$

Rearrange terms to group similar variables and constants:

$$\frac{(7 \times 5) \times (s^3 \times s^{-2}) \times (t^2 \times t^{-2}) \times (u^3 \times u^{-2}) \times (v \times v^{-3})}{15 \times 28}$$

Simplify the constants:

$$\frac{35}{420}$$

Simplify the variables using exponent rules ($a^m \times a^n = a^{m+n}$ and $a^m / a^n = a^{m-n}$):

$$s^{3-2} = s^1 = s$$ $$t^{2-2} = t^0 = 1$$ $$u^{3-2} = u^1 = u$$ $$v^{1-3} = v^{-2} = \frac{1}{v^2}$$

Combine these:

$$\frac{35}{420} \times s \times 1 \times u \times \frac{1}{v^2}$$

Simplify the fraction $\frac{35}{420}$. Both are divisible by 35 ($420 \div 35 = 12$):

$$\frac{1}{12}$$

So the simplified expression is:

$$\boxed{\frac{su}{12v^2}}$$

4. In a geometric progression, the third term is $\frac{2}{9}$ and the fourth term is $\frac{2}{27}$.
Let the first term be $a$ and the common ratio be $r$.
The $n$-th term of a geometric progression is $T_n = ar^{n-1}$.
Given:
$T_3 = ar^2 = \frac{2}{9}$ (Equation 1)
$T_4 = ar^3 = \frac{2}{27}$ (Equation 2)

i) Find the first term and the common ratio.
Divide Equation 2 by Equation 1 to find $r$:

$$\frac{ar^3}{ar^2} = \frac{\frac{2}{27}}{\frac{2}{9}}$$ $$r = \frac{2}{27} \times \frac{9}{2}$$ $$r = \frac{18}{54}$$ $$r = \frac{1}{3}$$

Now substitute $r = \frac{1}{3}$ into Equation 1 to find $a$:

$$a\left(\frac{1}{3}\right)^2 = \frac{2}{9}$$ $$a\left(\frac{1}{9}\right) = \frac{2}{9}$$ $$a = \frac{2}{9} \times 9$$ $$a = 2$$

The first term is $\boxed{2}$ and the common ratio is $\boxed*{\frac{1}{3}*}$.

ii) Find the sum of the first 5 terms of the geometric progression.
The sum of the first $n$ terms of a geometric progression is $S_n = \frac{a(1-r^n)}{1-r}$.
Here, $a=2$, $r=\frac{1}{3}$, and $n=5$.

$$S_5 = \frac{2\left(1 - \left(\frac{1}{3}\right)^5\right)}{1 - \frac{1}{3}}$$ $$S_5 = \frac{2\left(1 - \frac{1}{243}\right)}{\frac{2}{3}}$$ $$S_5 = \frac{2\left(\frac{243 - 1}{243}\right)}{\frac{2}{3}}$$ $$S_5 = \frac{2\left(\frac{242}{243}\right)}{\frac{2}{3}}$$ $$S_5 = 2 \times \frac{242}{243} \times \frac{3}{2}$$ $$S_5 = \frac{242 \times 3}{243}$$ $$S_5 = \frac{726}{243}$$

Both numerator and denominator are divisible by 3:

$$S_5 = \frac{242}{81}$$

The sum of the first 5 terms is $\boxed*{\frac{242}{81}*}$.

iii) Find the sum to infinity.
The sum to infinity of a geometric progression is $S_\infty = \frac{a}{1-r}$, provided $|r| < 1$.
Here, $a=2$ and $r=\frac{1}{3}$. Since $|\frac{1}{3}| < 1$, the sum to infinity exists.

$$S_\infty = \frac{2}{1 - \frac{1}{3}}$$ $$S_\infty = \frac{2}{\frac{2}{3}}$$ $$S_\infty = 2 \times \frac{3}{2}$$ $$S_\infty = 3$$

The sum to infinity is $\boxed{3}$.

5. The diagram below shows how learners in a Grade 12 class at Twaen travel to school. The learners use either buses (B), cars (C) or walk (W) to travel to school.
The numbers in the Venn diagram represent the number of learners in each category.
The regions are:
Only B: 14
Only C: 7
Only W: 7
B and C only: 2
B and W only: 4
C and W only: 3
B, C, and W: $x$

a) If 22 learners walk to school, find the value of x.
The total number of learners who walk to school is the sum of all regions within circle W.
From the diagram, these regions are $4, x, 3, 7$.
So, the total number of learners who walk is $4 + x + 3 + 7$.
We are given that this total is 22.

$$4 + x + 3 + 7 = 22$$ $$14 + x = 22$$ $$x = 22 - 14$$ $$x = \boxed{8}$$

b) How many learners use:

(a) only one mode of transport?
These are the learners in the regions that belong exclusively to one circle.
Only B: 14
Only C: 7
Only W: 7
Total learners using only one mode of transport = $14 + 7 + 7 = 28$.
Number of learners using only one mode of transport is $\boxed{28}$.

(b) two different modes of transport?
These are the learners in the regions that belong to the intersection of exactly two circles.
B and C only: 2
B and W only: 4
C and W only: 3
Total learners using two different modes of transport = $2 + 4 + 3 = 9$.
Number of learners using two different modes of transport is $\boxed{9}$.

6. Show that the points L(-2, -10), M(2, 2) and N(5, 11) are collinear.
To show that three points are collinear, we can demonstrate that the slope between any two pairs of points is the same.
Let's calculate the slope of segment LM ($m_{LM}$) and the slope of segment MN ($m_{MN}$).
The formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

For L(-2, -10) and M(2, 2):

$$m_{LM} = \frac{2 - (-10)}{2 - (-2)}$$ $$m_{LM} = \frac{2 + 10}{2 + 2}$$ $$m_{LM} = \frac{12}{4}$$ $$m_{LM} = 3$$

For M(2, 2) and N(5, 11):

$$m_{MN} = \frac{11 - 2}{5 - 2}$$ $$m_{MN} = \frac{9}{3}$$ $$m_{MN} = 3$$

Since $m_{LM} = m_{MN} = 3$, and both segments share the common point M, the points L, M, and N lie on the same straight line.
Therefore, the points L(-2, -10), M(2, 2) and N(5, 11) are indeed collinear.