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.

Welcome back Show — missed you this week. 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 × 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) = (4)/(36) = (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) = (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(sum of 9 in 1st AND sum of 6 in 2nd) = P(S_1=9) × P(S_2=6) P = (1)/(9) × (5)/(36) = (5)/(324) The probability is (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 sqrt(530). Q7: A body of mass m = 2.5 \, kg is suspended by two strings inclined at 40^ and 50^ to the downward vertical. We need to find the tension in both strings, taking g = 10 \, 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 \, kg × 10 \, ms^-2 = 25 \, N Step 2: Resolve the forces horizontally and vertically. Let T_1 be the tension in the string inclined at 40^ to the vertical, and T_2 be the tension in the string inclined at 50^ to the vertical. For equilibrium, the sum of horizontal forces and the sum of vertical forces must be zero. Horizontal forces: T_1 (40^) - T_2 (50^) = 0 T_1 (40^) = T_2 (50^) (1) Vertical forces: T_1 (40^) + T_2 (50^) = W T_1 (40^) + T_2 (50^) = 25 (2) Step 3: Solve the system of equations for T_1 and T_2. We know that (50^) = (40^) and (50^) = (40^). Substitute these into the equations: From (1): T_1 (40^) = T_2 (40^) T_1 = T_2 ((40^))/((40^)) Substitute this into (2): T_2 ((40^))/((40^)) (40^) + T_2 (40^) = 25 T_2 ( (^2(40^))/((40^)) + (40^) ) = 25 T_2 ( (^2(40^) + ^2(40^))/((40^)) ) = 25 Using the identity ^2 + ^2 = 1: T_2 ( (1)/((40^)) ) = 25 T_2 = 25 (40^) Now find T_1: T_1 = T_2 ((40^))/((40^)) = (25 (40^)) ((40^))/((40^)) T_1 = 25 (40^) Step 4: Calculate the numerical values for T_1 and T_2. Using (40^) ≈ 0.6428 and (40^) ≈ 0.7660: T_1 = 25 × 0.7660 = 19.15 \, N T_2 = 25 × 0.6428 = 16.07 \, N The tensions in the strings are T_1 ≈ 19.15 \, N, T_2 ≈ 16.07 \, N. Q8: We need to evaluate (3)/(2-sqrt(3)) - (2)/(3+sqrt(3)) without a calculator and express the answer in the form m+nsqrt(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). (3)/(2-sqrt(3)) = (3)/(2-sqrt(3)) × 2+sqrt(3)2+sqrt(3) = 3(2+sqrt(3))2^2 - (sqrt(3))^2 = 6+3sqrt(3)4-3 = 6+3sqrt(3)1 = 6+3sqrt(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). (2)/(3+sqrt(3)) = (2)/(3+sqrt(3)) × 3-sqrt(3)3-sqrt(3) = 2(3-sqrt(3))3^2 -