)find the value of \( A^{1} \cap B^{1} \cap C^{1} \) a). Make \( m \) the subject of the formula \( y=a \sqrt{\frac{m-3}{m-9}} \) ). find the value of \( m \) if \( y=4 \) and \( a=2 \)

Below is one acceptable solution. ────────────────────────────── Problem 1: Find the value of A¹ ∩ B¹ ∩ C¹ In many texts the notation A¹ is used to denote the complement of the set A (often written as Aᶜ). In that case the intersection   A¹ ∩ B¹ ∩ C¹ is the set of all elements which are not in A, not in B, and not in C. By one of De Morgan’s Laws we have   A¹ ∩ B¹ ∩ C¹ = (A ∪ B ∪ C)¹. That is, it is the complement of (A ∪ B ∪ C). (Without any further information about the sets A, B, and C, this is the most that can be said.) ────────────────────────────── Problem 2(a): Make m the subject of the formula   y = a √[(m – 3)/(m – 9)]. Step 1. Square both sides to eliminate the square root:   y² = a² · (m – 3)/(m – 9). Step 2. Multiply both sides by (m – 9) (assuming m ≠ 9):   y² (m – 9) = a² (m – 3). Step 3. Expand and bring like terms together:   y² m – 9y² = a² m – 3a². Step 4. Rearrange to group m terms on one side:   y² m – a² m = 9y² – 3a². Factor m out:   m (y² – a²) = 9y² – 3a². Step 5. Solve for m:   m = (9y² – 3a²)/(y² – a²). For a slightly neater form, you may factor 3 from the numerator:   m = [3(3y² – a²)]/(y² – a²), with the condition that y² ≠ a² (so that the denominator isn’t zero). ────────────────────────────── Problem 2(b): Find the value of m if y = 4 and a = 2. Substitute y = 4 (so that y² = 16) and a = 2 (so that a² = 4) into the expression we obtained:   m = (9y² – 3a²)/(y² – a²)     = (9·16 – 3·4)/(16 – 4)     = (144 – 12)/12     = 132/12     = 11. ────────────────────────────── Final Answers: 1. A¹ ∩ B¹ ∩ C¹ = (A ∪ B ∪ C)¹ (i.e. the complement of A ∪ B ∪ C). 2. (a) m = (9y² – 3a²)/(y² – a²)           (b) For y = 4 and a = 2, m = 11.