Here are the solutions to the mathematics problems: SECTION A (35 MARKS) 1. (6 Marks) a) Simplify: rac34 imes 3-232 b) Express 0.00032 in standard form. c) Rationalize and simplify: rac5sqrt(3)

$$
3^4 \times 3^{-2} = 3^{4 + (-2)} = 3^{4-2} = 3^2
$$

$$
\frac{3^2}{3^2}
$$

$$
3^{2-2} = 3^0
$$

$$
3^0 = 1
$$
The simplified expression is $\boxed{1}$.

$$
0.00032 \to 3.2
$$

$$
3.2 \times 10^{-4}
$$
In standard form, it is $\boxed{3.2 \times 10^{-4}}$.

$$
\frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}
$$

$$
\frac{5\sqrt{3}}{(\sqrt{3})^2}
$$

$$
\frac{5\sqrt{3}}{3}
$$
The rationalized and simplified expression is $\boxed{\frac{5\sqrt{3}}{3}}$.

$$
\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}
$$

$$
\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}
$$

$$
5\sqrt{2} + 3(2\sqrt{2})
$$

$$
5\sqrt{2} + 6\sqrt{2}
$$

$$
(5+6)\sqrt{2} = 11\sqrt{2}
$$
The evaluated expression is $\boxed{11\sqrt{2}}$.

$$
(2x)(x) + (2x)(5) + (-3)(x) + (-3)(5)
$$
$$
2x^2 + 10x - 3x - 15
$$
$$
2x^2 + 7x - 15
$$

$$
(x)^2 - 2(x)(4) + (4)^2
$$
$$
x^2 - 8x + 16
$$

$$
(2x^2 - x^2) + (7x + 8x) + (-15 - 16)
$$
$$
x^2 + 15x - 31
$$
The expanded and simplified expression is $\boxed{x^2 + 15x - 31}$.

$$
3x^2 - 12x + x - 4
$$

$$
3x(x - 4) + 1(x - 4)
$$

$$
(x-4)(3x+1)
$$
The completely factorized expression is $\boxed{(x-4)(3x+1)}$.

$$
4x^2 = 9
$$

$$
x^2 = \frac{9}{4}
$$

$$
x = \pm\sqrt{\frac{9}{4}}
$$
$$
x = \pm\frac{\sqrt{9}}{\sqrt{4}}
$$
$$
x = \pm\frac{3}{2}
$$
The solutions are $\boxed{x = \frac{3}{2} \text{ or } x = -\frac{3}{2}}$.

$$
(3x + 2y) + (4x - 2y) = 17 + 8
$$
$$
7x = 25
$$

$$
x = \frac{25}{7}
$$

$$
2\left(\frac{25}{7}\right) - y = 4
$$
$$
\frac{50}{7} - y = 4
$$

$$
-y = 4 - \frac{50}{7}
$$
$$
-y = \frac{28}{7} - \frac{50}{7}
$$
$$
-y = -\frac{22}{7}
$$
$$
y = \frac{22}{7}
$$
The solution is $\boxed{x = \frac{25}{7}, y = \frac{22}{7}}$.

$$
a = -2, \quad b = 12
$$

$$
t = \frac{-12}{2(-2)} = \frac{-12}{-4} = 3
$$

$$
h = -2(3)^2 + 12(3) + 5
$$
$$
h = -2(9) + 36 + 5
$$
$$
h = -18 + 36 + 5
$$
$$
h = 18 + 5
$$
$$
h = 23
$$
The maximum height is $\boxed{23 \text{ units}}$.

$$
-2t^2 + 12t + 5 = 0
$$

$$
t_1 = \frac{-12 + 13.56}{-4} = \frac{1.56}{-4} = -0.39
$$
$$
t_2 = \frac{-12 - 13.56}{-4} = \frac{-25.56}{-4} = 6.39
$$
Since time cannot be negative, we take the positive value.  
The projectile hits the ground at approximately $\boxed{6.39 \text{ units of time}}$.

$$
h(1) = -2(1)^2 + 12(1) + 5
$$
$$
h(1) = -2 + 12 + 5 = 15
$$

$$
h(3) = -2(3)^2 + 12(3) + 5
$$
$$
h(3) = -2(9) + 36 + 5
$$
$$
h(3) = -18 + 36 + 5 = 23
$$

$$
\text{Average rate of change} = \frac{h(3) - h(1)}{3 - 1}
$$
$$
= \frac{23 - 15}{2}
$$
$$
= \frac{8}{2}
$$
$$
= 4
$$
The average rate of change between $t=1$ and $t=3$ is $\boxed{4 \text{ units of height per unit of time}}$.