answer:Let's call the certain number "X". We are looking for the smallest number that, when diminished by 12, is divisible by 12, 24, X, 48, and 56. We know this smallest number is 1020. So, when we diminish 1020 by 12, we get 1008. We need to find the value of X such that 1008 is divisible by X, along with 12, 24, 48, and 56. First, let's find the prime factorization of 1008 to see what numbers it is divisible by: 1008 = 2^4 * 3^2 * 7 Now, we know that 1008 is divisible by: - 12 (which is 2^2 * 3) - 24 (which is 2^3 * 3) - 48 (which is 2^4 * 3) - 56 (which is 2^3 * 7) Since 1008 must be divisible by X as well, X must be a factor of 1008. The prime factorization of 1008 includes the factors of 12, 24, 48, and 56. Therefore, X must be a divisor of 1008 that is not already listed (12, 24, 48, or 56). The prime factors of 1008 are 2, 3, and 7. Since 12, 24, 48, and 56 all include the prime factor 2, and 12 and 48 include the prime factor 3, and 56 includes the prime factor 7, X could be a product of these primes in a combination that is not already covered by 12, 24, 48, or 56. The only prime factor combination left that is not covered by 12, 24, 48, or 56 is 3^2 * 7, which is 63. Therefore, the number X that 1008 is also divisible by is boxed{63} .

answer:(I) Method 1: According to the geometric interpretation, "|x - 1| + |x + 2|" represents the sum of the distances from x to -2 and 1 on the number line. Therefore, the solution set of the inequality is {x|x ≤ -3 or x ≥ 2}. Method 2: By discussing the zero point intervals: ① When x ≤ -2, -x - 2 - x + 1 ≥ 5, which implies x ≤ -3. Hence, x ≤ -3. ② When -2 < x < 1, x + 2 - x + 1 ≥ 5, which implies 3 ≥ 5, contradicting the problem statement. ③ When x ≥ 1, x + 2 + x - 1 ≥ 5, which implies x ≥ 2. Hence, x ≥ 2. In summary, the solution set for the inequality is {x|x ≤ -3 or x ≥ 2}. Proof (II): Since f(x) = |x - 1| + |x + k| ≥ |(x - 1) - (x + k)| = |k + 1|, and the minimum value of function f(x) is 3 with k > 0, we have |k + 1| = 3, which yields k = 2. So, a + b + c = 2. By the Cauchy-Schwarz inequality, (a² + b² + c²)(1² + 1² + 1²) ≥ (a + b + c)² = 4. Therefore, a² + b² + c² ≥ boxed{frac{4}{3}}.

answer:First, let's calculate the total number of each baked good that the students will bring in: Brownies: 30 students * 12 brownies/student = 360 brownies Cookies: 20 students * 24 cookies/student = 480 cookies Donuts: 15 students * 12 donuts/student = 180 donuts Now, let's add up the total number of baked goods: Total baked goods = 360 brownies + 480 cookies + 180 donuts Total baked goods = 1020 items Since each item is sold for 2.00, we can calculate the total money raised by multiplying the total number of items by the price per item: Total money raised = 1020 items * 2.00/item = 2040 Therefore, they will raise boxed{2040} if they sell everything.

answer:1. First, we find ( g(2) ): [ g(2) = 3cdot2^2 - 3cdot2 - 4 = 3cdot4 - 6 - 4 = 12 - 6 - 4 = 2 ] 2. Now, substitute ( g(2) ) into ( f(x) ): [ f(g(2)) = f(2) = 3sqrt[3]{2} + frac{15}{sqrt[3]{2}} ] 3. Calculate separately: - ( sqrt[3]{2} ) remains as it is (since we are keeping exact values) - ( frac{15}{sqrt[3]{2}} ) is the reciprocal of the above times 15 Final expression: [ 3sqrt[3]{2} + frac{15}{sqrt[3]{2}} = 3sqrt[3]{2} + frac{15}{sqrt[3]{2}} = 3sqrt[3]{2} + 5sqrt[3]{2}^2 = 3sqrt[3]{2} + 5sqrt[3]{4} ] - Since ( sqrt[3]{4} ) is not further reducible in exact terms, the final answer can only be represented as: [ boxed{3sqrt[3]{2} + 5sqrt[3]{4}} ]