Write the answer in QUESTION 1.5.1 as a binary number.

$$-3x(2x+1)+5=4$$ $$-6x^2 - 3x + 5 = 4$$

$$-6x^2 - 3x + 5 - 4 = 0$$ $$-6x^2 - 3x + 1 = 0$$ $$6x^2 + 3x - 1 = 0$$

Here, $a=6$, $b=3$, $c=-1$.

$$x = \frac{-3 \pm \sqrt{3^2 - 4(6)(-1)}}{2(6)}$$ $$x = \frac{-3 \pm \sqrt{9 + 24}}{12}$$ $$x = \frac{-3 \pm \sqrt{33}}{12}$$

$$x_1 = \frac{-3 + \sqrt{33}}{12} \approx \frac{-3 + 5.74456}{12} \approx 0.2287 \approx 0.23$$ $$x_2 = \frac{-3 - \sqrt{33}}{12} \approx \frac{-3 - 5.74456}{12} \approx -0.7287 \approx -0.73$$

The solutions are $\boxed{x \approx 0.23 \text{ or } x \approx -0.73}$.

1.2 Consider the number line given:
1.2.1 Write down the critical values.
The critical values are the points marked on the number line.
The critical values are $\boxed{-4 \text{ and } 2}$.

1.2.2 Write the solution in interval notation.
The number line shows a shaded region between -4 and 2, inclusive of these points (solid dots).
The solution in interval notation is $\boxed{[-4, 2]}$.

1.3 Given: $x+4y=0$ and $x^2+2xy+y=0$
1.3.1 Make x the subject of the equation $x+4y=0$.

$$x+4y=0$$ $$x = -4y$$

The subject of the equation is $\boxed{x = -4y}$.

1.3.2 Solve for x and y simultaneously.

$$(-4y)^2 + 2(-4y)y + y = 0$$ $$16y^2 - 8y^2 + y = 0$$ $$8y^2 + y = 0$$

$$y(8y + 1) = 0$$

This gives two possible values for y:

$$y_1 = 0$$ $$8y_2 + 1 = 0 \Rightarrow 8y_2 = -1 \Rightarrow y_2 = -\frac{1}{8}$$

For $y_1 = 0$:

$$x_1 = -4(0) = 0$$

For $y_2 = -\frac{1}{8}$:

$$x_2 = -4\left(-\frac{1}{8}\right) = \frac{4}{8} = \frac{1}{2}$$

The solutions are $\boxed{(x, y) = (0, 0) \text{ or } (x, y) = \left(\frac{1}{2}, -\frac{1}{8}\right)}$.

1.4 The formula used to determine the area (A) of a trapezium is given as: $A = \frac{1}{2}(a+b)h$
1.4.1 Make h the subject of the formula.

$$h = \frac{2A}{a+b}$$

The formula with h as the subject is $\boxed{h = \frac{2A}{a+b}}$.

1.4.2 Hence determine the height (h) of the trapezium if $A = 250 \text{ m}^2$; $a = 5 \text{ m}$ and $b = 15 \text{ m}$.

$$h = \frac{2 \times 250 \text{ m}^2}{5 \text{ m} + 15 \text{ m}}$$ $$h = \frac{500 \text{ m}^2}{20 \text{ m}}$$ $$h = 25 \text{ m}$$

The height of the trapezium is $\boxed{25 \text{ m}}$.

1.5 Given: $x = 50$ and $y = 2$
1.5.1 Evaluate $\frac{3x}{y}$.

$$\frac{150}{2} = 75$$

The value is $\boxed{75}$.

1.5.2 Write the answer in QUESTION 1.5.1 as a binary number.
The answer from 1.5.1 is 75. To convert 75 (base 10) to binary (base 2):
$75 \div 2 = 37$ remainder 1
$37 \div 2 = 18$ remainder 1
$18 \div 2 = 9$ remainder 0
$9 \div 2 = 4$ remainder 1
$4 \div 2 = 2$ remainder 0
$2 \div 2 = 1$ remainder 0
$1 \div 2 = 0$ remainder 1
Reading the remainders from bottom to top gives the binary number.
The binary number is $\boxed{1001011_2}$.