answer:To solve the equation 2(x-2)^2 = 6 - 3x, we proceed as follows: 1. Start with the given equation: 2(x-2)^2 = 6 - 3x 2. Notice that the right side can be rewritten as -3(x-2) by adding 3x to both sides and subtracting 6 from both sides: 2(x-2)^2 + 3x - 6 = 0 Which simplifies to: 2(x-2)^2 + 3(x-2) = 0 3. Factor out the common term (x-2): (x-2)(2(x-2) + 3) = 0 This simplifies to: (x-2)(2x - 4 + 3) = 0 Further simplifying gives: (x-2)(2x - 1) = 0 4. Set each factor equal to zero: - For x-2=0, solving for x gives x = 2. - For 2x-1=0, solving for x gives x = frac{1}{2}. Therefore, the solutions to the equation are x_1 = 2 and x_2 = frac{1}{2}. Encapsulating the final answers: - x_1 = boxed{2} - x_2 = boxed{frac{1}{2}}

answer:1. **Define Operations and Initial Condition**: The new operations are defined as [a diamondsuit b = a^{log_{10}(b)} quad text{and} quad a heartsuit b = a^{frac{1}{log_{10}(b)}}] and the given sequence starts with a_3 = 5^frac{1}{log_{10}(3)}. 2. **Recursive Formula Simplification**: The sequence updates with [ a_n = (n^frac{1}{log_{10}(n-1)})^{log_{10}(a_{n-1}+1)} ] Simplifying that, we have: [ a_n = n^{frac{log_{10}(a_{n-1}+1)}{log_{10}(n-1)}} ] 3. **Applying the Logarithm Function**: By applying the logarithmic base of 10, [ log_{10}(a_n) = frac{log_{10}(a_{n-1}+1)}{log_{10}(n-1)} cdot log_{10}(n) ] Since the function is recursive, the expression will follow [ log_{10}(a_n) = log_{10}(n) cdot log_{n-1}(10) cdot log_{10}(a_{n-1}+1) ] 4. **Recursive Expansion for Larger n**: Simplifying it recursively, as n approaches large numbers like 2020: [ log_{10}(a_{2020}) approx log_{10}(2020) cdot log_3(10) cdot log_{10}(5^frac{1}{log_{10}(3)}) ] 5. **Simplifying the Term**: Now calculating and simplifying leads to [ log_{10}(a_{2020}) approx log_{10}(2020) approx 3.3 ] Conclusion with boxed answer: [3] The final answer is boxed{B) 3}

answer:The product of two numbers is equal to the product of their Highest Common Factor (HCF) and their Least Common Multiple (LCM). This is known as the product of HCF and LCM. So, if the HCF of two numbers is 55 and their LCM is 1500, the product of the two numbers is: Product = HCF × LCM Product = 55 × 1500 Product = 82500 Therefore, the product of the two numbers is boxed{82,500} .

answer:# Part (a) 1. **Understanding the Problem:** We are given a family of sets mathcal{F} subseteq X^{(K)} where |X| = n. For any three distinct sets A, B, C in mathcal{F}, it holds that A cap B notsubset C. We need to show that the size of mathcal{F} is bounded by binom{k}{lfloor frac{k}{2} rfloor} + 1. 2. **Applying Sperner's Theorem:** Sperner's theorem states that the largest antichain in the power set mathcal{P}(X) of a set X of size k is binom{k}{lfloor frac{k}{2} rfloor}. An antichain is a family of sets where no set is a subset of another. 3. **Constructing the Antichain:** Consider the sets A, B in mathcal{F}. If A cap B subset C for some C in mathcal{F}, then A cap B would be a subset of C, violating the given condition. Hence, the sets A cap B must form an antichain in mathcal{P}(X). 4. **Bounding the Size of mathcal{F}:** By Sperner's theorem, the size of the largest antichain in mathcal{P}(X) is binom{k}{lfloor frac{k}{2} rfloor}. Since mathcal{F} must form an antichain, we have: [ |mathcal{F}| leq binom{k}{lfloor frac{k}{2} rfloor} + 1 ] Thus, we have shown the required inequality. # Part (b) 1. **Understanding the Problem:** Now, the elements of mathcal{F} do not necessarily have k elements, but the condition A cap B notsubset C for any three distinct sets A, B, C in mathcal{F} still holds. We need to show that |mathcal{F}| leq binom{n}{lceil frac{n-2}{3} rceil} + 2. 2. **Generalizing the Antichain Argument:** The condition A cap B notsubset C implies that the sets in mathcal{F} must still form an antichain in the power set mathcal{P}(X). However, since the sets do not necessarily have k elements, we need to consider the largest possible antichain in mathcal{P}(X). 3. **Applying Erdős–Ko–Rado Theorem:** The Erdős–Ko–Rado theorem provides a bound on the size of a family of sets where any two sets intersect. For a set X of size n, the largest such family has size binom{n-1}{lceil frac{n-2}{3} rceil}. 4. **Bounding the Size of mathcal{F}:** Given the condition A cap B notsubset C, the sets in mathcal{F} must form an antichain. By the Erdős–Ko–Rado theorem, the size of the largest such family is binom{n}{lceil frac{n-2}{3} rceil}. Adding the additional sets that do not necessarily intersect, we get: [ |mathcal{F}| leq binom{n}{lceil frac{n-2}{3} rceil} + 2 ] Thus, we have shown the required inequality. The final answer is boxed{binom{k}{lfloor frac{k}{2} rfloor} + 1} for part (a) and boxed{binom{n}{lceil frac{n-2}{3} rceil} + 2} for part (b).