Evaluate 9 + 12 / 4 - 2 x 5, express 125 as a product of its primes, find the cube root of 125, and find the L.C.M of 18 and 24.

Follow the order of operations (PEMDAS/BODMAS): Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).

First, perform division:

$$12 \div 4 = 3$$

Next, perform multiplication:

$$2 \times 5 = 10$$

Substitute these values back into the expression:

$$9 + 3 - 10$$

Now, perform addition and subtraction from left to right:

$$9 + 3 = 12$$ $$12 - 10 = 2$$

The evaluated expression is:

$$\boxed{2}$$

a) To express 125 as a product of its primes, we find the prime factors of 125.
125 is divisible by 5:

$$125 = 5 \times 25$$

25 is also divisible by 5:

$$25 = 5 \times 5$$

So, 125 can be written as:

$$125 = 5 \times 5 \times 5 = 5^3$$

The product of its primes is:

$$\boxed{5 \times 5 \times 5}$$

b) Since $125 = 5 \times 5 \times 5 = 5^3$, the cube root of 125 is the number that, when multiplied by itself three times, equals 125.

$$\sqrt[3]{125} = \sqrt[3]{5^3} = 5$$

The cube root of 125 is:

$$\boxed{5}$$

c) To find the Least Common Multiple (L.C.M) of 18 and 24, we first find their prime factorizations.
Prime factorization of 18:

$$18 = 2 \times 9 = 2 \times 3 \times 3 = 2 \times 3^2$$

Prime factorization of 24:

$$24 = 2 \times 12 = 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3$$

To find the L.C.M, we take the highest power of each prime factor present in either number:
The highest power of 2 is $2^3$.
The highest power of 3 is $3^2$.

$$L.C.M(18, 24) = 2^3 \times 3^2 = 8 \times 9 = 72$$

The L.C.M of 18 and 24 is:

$$\boxed{72}$$

Given two statements $p$ and $q$, we need to construct a truth table for the compound statement $\neg p \land q$.
$\neg p$ means "not p".
$\land$ means "and".

$p$	$q$	$\neg p$	$\neg p \land q$
T	T	F	F
T	F	F	F
F	T	T	T
F	F	T	F

a) R is 30km East of P, and Q is 40km South of R. This forms a right-angled triangle PQR, with the right angle at R.
The length of PR is 30 km.
The length of RQ is 40 km.
We can use the Pythagorean theorem to find the distance PQ (the hypotenuse):

$$PQ^2 = PR^2 + RQ^2$$ $$PQ^2 = (30 \text{ km})^2 + (40 \text{ km})^2$$ $$PQ^2 = 900 \text{ km}^2 + 1600 \text{ km}^2$$ $$PQ^2 = 2500 \text{ km}^2$$ $$PQ = \sqrt{2500 \text{ km}^2}$$ $$PQ = 50 \text{ km}$$

The distance PQ is:

$$\boxed{50 \text{ km}}$$

b) In the right-angled triangle PQR, we want to find the angle at P, denoted as $\angle QPR$.
We know the opposite side (RQ = 40 km) and the adjacent side (PR = 30 km) to $\angle QPR$.
We can use the tangent function:

$$\tan(\angle QPR) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{RQ}{PR}$$ $$\tan(\angle QPR) = \frac{40 \text{ km}}{30 \text{ km}}$$ $$\tan(\angle QPR) = \frac{4}{3}$$

To find the angle, we take the inverse tangent:

$$\angle QPR = \arctan\left(\frac{4}{3}\right)$$

Using a calculator:

$$\angle QPR \approx 53.13^\circ$$

Rounding to one decimal place:

$$\angle QPR \approx 53.1^\circ$$

The angle QPR is:

$$\boxed{53.1^\circ}$$