Calculate the probability of rolling a sum of 9 on the first roll and a sum of 6 on the second roll. Also, find the magnitude of vector r given 2p − 3q + r = 0, where p = 2i − 7j and q = -5i - 9j.

Here are the solutions to your questions:

Q5:
To find the probability, we first determine the number of favorable outcomes for each throw.

Step 1: Find the probability of obtaining a sum of nine in the first throw.
The possible outcomes when rolling two dice are $6 \times 6 = 36$.
The combinations that sum to nine are $(3,6), (4,5), (5,4), (6,3)$. There are 4 such outcomes.
The probability of a sum of nine is $P(S_1=9) = \frac{4}{36} = \frac{1}{9}$.

Step 2: Find the probability of obtaining a sum of six in the second throw.
The combinations that sum to six are $(1,5), (2,4), (3,3), (4,2), (5,1)$. There are 5 such outcomes.
The probability of a sum of six is $P(S_2=6) = \frac{5}{36}$.

Step 3: Calculate the combined probability.
Since the two throws are independent events, the total probability is the product of their individual probabilities.

$$P(\text{sum of 9 in 1st AND sum of 6 in 2nd}) = P(S_1=9) \times P(S_2=6)$$ $$P = \frac{1}{9} \times \frac{5}{36} = \frac{5}{324}$$

The probability is $\boxed{\frac{5}{324}}$.

Q6:
We are given vectors $p = 2i - 7j$ and $q = -5i - 9j$, and the equation $2p - 3q + r = 0$. We need to find $|r|$.

Step 1: Express vector $r$ in terms of $p$ and $q$.
From the given equation, $r = -2p + 3q$.

Step 2: Substitute the given vectors $p$ and $q$ into the expression for $r$.

$$r = -2(2i - 7j) + 3(-5i - 9j)$$ $$r = (-4i + 14j) + (-15i - 27j)$$

Step 3: Combine the $i$ and $j$ components to find vector $r$.

$$r = (-4 - 15)i + (14 - 27)j$$ $$r = -19i - 13j$$

Step 4: Calculate the magnitude of vector $r$.
The magnitude of a vector $r = xi + yj$ is given by $|r| = \sqrt{x^2 + y^2}$.

$$|r| = \sqrt{(-19)^2 + (-13)^2}$$ $$|r| = \sqrt{361 + 169}$$ $$|r| = \sqrt{530}$$

The magnitude of $r$ is $\boxed{\sqrt{530}}$.

Q7:
A body of mass $m = 2.5 \, \text{kg}$ is suspended by two strings inclined at $40^\circ$ and $50^\circ$ to the downward vertical. We need to find the tension in both strings, taking $g = 10 \, \text{ms}^{-2}$.

Step 1: Calculate the weight of the body.
The weight $W$ is the force due to gravity acting downwards.

$$W = mg = 2.5 \, \text{kg} \times 10 \, \text{ms}^{-2} = 25 \, \text{N}$$

Step 2: Resolve the forces horizontally and vertically.
Let $T_1$ be the tension in the string inclined at $40^\circ$ to the vertical, and $T_2$ be the tension in the string inclined at $50^\circ$ to the vertical.
For equilibrium, the sum of horizontal forces and the sum of vertical forces must be zero.
Horizontal forces: $T_1 \sin(40^\circ) - T_2 \sin(50^\circ) = 0$

$$T_1 \sin(40^\circ) = T_2 \sin(50^\circ) \quad (1)$$

Vertical forces: $T_1 \cos(40^\circ) + T_2 \cos(50^\circ) = W$

$$T_1 \cos(40^\circ) + T_2 \cos(50^\circ) = 25 \quad (2)$$

Step 3: Solve the system of equations for $T_1$ and $T_2$.
We know that $\sin(50^\circ) = \cos(40^\circ)$ and $\cos(50^\circ) = \sin(40^\circ)$.
Substitute these into the equations:
From (1): $T_1 \sin(40^\circ) = T_2 \cos(40^\circ) \implies T_1 = T_2 \frac{\cos(40^\circ)}{\sin(40^\circ)}$
Substitute this into (2):

$$T_2 \frac{\cos(40^\circ)}{\sin(40^\circ)} \cos(40^\circ) + T_2 \sin(40^\circ) = 25$$ $$T_2 \left( \frac{\cos^2(40^\circ)}{\sin(40^\circ)} + \sin(40^\circ) \right) = 25$$ $$T_2 \left( \frac{\cos^2(40^\circ) + \sin^2(40^\circ)}{\sin(40^\circ)} \right) = 25$$

Using the identity $\cos^2\theta + \sin^2\theta = 1$:

$$T_2 \left( \frac{1}{\sin(40^\circ)} \right) = 25$$ $$T_2 = 25 \sin(40^\circ)$$

Now find $T_1$:

$$T_1 = T_2 \frac{\cos(40^\circ)}{\sin(40^\circ)} = (25 \sin(40^\circ)) \frac{\cos(40^\circ)}{\sin(40^\circ)}$$ $$T_1 = 25 \cos(40^\circ)$$

Step 4: Calculate the numerical values for $T_1$ and $T_2$.
Using $\sin(40^\circ) \approx 0.6428$ and $\cos(40^\circ) \approx 0.7660$:

$$T_1 = 25 \times 0.7660 = 19.15 \, \text{N}$$ $$T_2 = 25 \times 0.6428 = 16.07 \, \text{N}$$

The tensions in the strings are $\boxed{T_1 \approx 19.15 \, \text{N}, T_2 \approx 16.07 \, \text{N}}$.

Q8:
We need to evaluate $\frac{3}{2-\sqrt{3}} - \frac{2}{3+\sqrt{3}}$ without a calculator and express the answer in the form $m+n\sqrt{3}$.

Step 1: Rationalize the denominator of the first term.
Multiply the numerator and denominator by the conjugate of the denominator, which is $2+\sqrt{3}$.

$$\frac{3}{2-\sqrt{3}} = \frac{3}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}$$ $$= \frac{3(2+\sqrt{3})}{2^2 - (\sqrt{3})^2} = \frac{6+3\sqrt{3}}{4-3} = \frac{6+3\sqrt{3}}{1} = 6+3\sqrt{3}$$

Step 2: Rationalize the denominator of the second term.
Multiply the numerator and denominator by the conjugate of the denominator, which is $3-\sqrt{3}$.

$$\frac{2}{3+\sqrt{3}} = \frac{2}{3+\sqrt{3}} \times \frac{3-\sqrt{3}}{3-\sqrt{3}}$$ $$$$