Given U = \1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\. Let A = : x is divisible by 3 B = : x is a factor of 18 and C = : x is even and not divisible by 5\. Find n[A (B C)] and solve 32y - 4 x 3^

Here are the solutions for both parts of question 4. a) Step 1: List the elements of sets A, B, and C from the universal set U = \1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\. • Set A = \x: x is divisible by 3\ The elements in U divisible by 3 are 3, 6, 9, 12. A = \3, 6, 9, 12\ • Set B = \x: x is a factor of 18\ The factors of 18 are 1, 2, 3, 6, 9, 18. From U, these are 1, 2, 3, 6, 9. B = \1, 2, 3, 6, 9\ • Set C = \x: x is even and not divisible by 5\ The even numbers in U are 2, 4, 6, 8, 10, 12. From these, the numbers not divisible by 5 are 2, 4, 6, 8, 12 (10 is divisible by 5). C = \2, 4, 6, 8, 12\ Step 2: Find the union of sets B and C, B C. B C = \1, 2, 3, 6, 9\ \2, 4, 6, 8, 12\ B C = \1, 2, 3, 4, 6, 8, 9, 12\ Step 3: Find the intersection of set A with (B C), A (B C). A (B C) = \3, 6, 9, 12\ \1, 2, 3, 4, 6, 8, 9, 12\ The common elements are 3, 6, 9, 12. A (B C) = \3, 6, 9, 12\ Step 4: Find the number of elements in A (B C), which is n[A (B C)]. The set A (B C) has 4 elements. n[A (B C)] = 4 The value of n[A (B C)] is 4. b) Solve the equation 3^2y - 4 × 3^y + 3 = 0. Step 1: Rewrite 3^2y as (3^y)^2. (3^y)^2 - 4 × 3^y + 3 = 0 Step 2: Let u = 3^y. Substitute u into the equation to form a quadratic equation. u^2 - 4u + 3 = 0 Step 3: Solve the quadratic equation for u. Factor the quadratic equation: We need two numbers that multiply to 3 and add to -4. These numbers are -1 and -3. (u-1)(u-3) = 0 Set each factor to zero to find the possible values for u: u-1 = 0 u = 1 u-3 = 0 u = 3 Step 4: Substitute back 3^y for u and solve for y. Case 1: u = 1 3^y = 1 Since any non-zero number raised to the power of 0 is 1, we have: 3^y = 3^0 y = 0 Case 2: u = 3 3^y = 3 Since 3 = 3^1, we have: 3^y = 3^1 y = 1 The solutions for y are y=0 or y=1. 3 done, 2 left today. You're making progress.