answer:1. **Calculate the initial sum of the 50 numbers:** We know from the problem that the average of the 50 numbers is 38. The sum of these 50 numbers can be calculated as follows: [ text{Sum} = text{Average} times text{Number of elements} = 38 times 50 = 1900 ] 2. **Determine the sum after removing two numbers (45 and 55):** Removing the numbers 45 and 55 from the set affects the sum: [ text{New Sum} = text{Initial Sum} - 45 - 55 = 1900 - 45 - 55 = 1800 ] 3. **Calculate the new number of elements:** Since we removed two numbers from the original 50 numbers, we have: [ text{New Number of elements} = 50 - 2 = 48 ] 4. **Calculate the new average:** The new average is given by the new sum divided by the new number of elements: [ text{New Average} = frac{text{New Sum}}{text{New Number of elements}} = frac{1800}{48} ] Performing the division: [ frac{1800}{48} = 37.5 ] # Conclusion: The arithmetic mean of the remaining 48 numbers is: [ boxed{B} ]

answer:First, rearrange the given circle equation by moving all terms to one side: [ x^2 - 4x + y^2 + 6y = 9 ] Next, complete the square for both the (x) and (y) terms: - For (x): Add and subtract (left(frac{-4}{2}right)^2 = 4), so: [ x^2 - 4x + 4 = (x-2)^2 ] - For (y): Add and subtract (left(frac{6}{2}right)^2 = 9), so: [ y^2 + 6y + 9 = (y+3)^2 ] Now, add these constants to the right side of the equation: [ (x-2)^2 + (y+3)^2 = 9 + 4 + 9 = 22 ] The equation of the circle in standard form is now: [ (x-2)^2 + (y+3)^2 = 22 ] Thus, the center of the circle is ((2, -3)). Therefore, [ x + y = 2 + (-3) = boxed{-1} ]

answer:(1) From (x+1)(2-x) geq 0, we get: -1 leq x leq 2, thus, when p is true: x in [-1, 2]; If the inequality x^2 + 2mx - m + 6 > 0 always holds, then Delta = 4m^2 - 4(-m+6) < 0, solving this gives: -3 < m < 2, so, when q is true, m in (-3, 2); (2) If p is a sufficient but not necessary condition for q, i.e., p subsetneq q, from p: [-1, 2] subsetneq (-3, 2], thus, m in (-3, 2]. Therefore, the solutions are: (1) For q to be true, m in boxed{(-3, 2)}; (2) For p to be a sufficient but not necessary condition for q, m in boxed{(-3, 2]}.

answer:1. **Express (a) in terms of (x):** Given (a equiv 3 pmod{8}), we can write (a) as: [ a = 8x + 3 ] for some integer (x). 2. **Lemma: (v_2(a^{2^{k-2}} - 1) = k):** We need to show that the 2-adic valuation of (a^{2^{k-2}} - 1) is (k). 3. **Proof using the Lifting The Exponent (LTE) lemma:** By the LTE lemma, for (a = 8x + 3) and (n = 2^{k-2}), we have: [ v_2(a^{2^{k-2}} - 1) = v_2(a - 1) + v_2(2^{k-2}) ] Since (a = 8x + 3), we have: [ a - 1 = 8x + 2 ] Therefore: [ v_2(a - 1) = v_2(8x + 2) = v_2(2(4x + 1)) = 1 + v_2(4x + 1) ] Since (4x + 1) is odd, (v_2(4x + 1) = 0). Thus: [ v_2(a - 1) = 1 ] Also: [ v_2(2^{k-2}) = k-2 ] Therefore: [ v_2(a^{2^{k-2}} - 1) = 1 + (k-2) = k-1 ] This contradicts the lemma, so we need to re-evaluate the LTE application. Instead, we use: [ v_2(a^2 - 1) = v_2((8x + 3)^2 - 1) = v_2(64x^2 + 48x + 8) = 3 + v_2(8x^2 + 6x + 1) ] Since (8x^2 + 6x + 1) is odd, (v_2(8x^2 + 6x + 1) = 0). Thus: [ v_2(a^2 - 1) = 3 ] Therefore: [ v_2(a^{2^{k-2}} - 1) = v_2(a^2 - 1) + v_2(2^{k-3}) = 3 + (k-3) = k ] 4. **Induction for (k = 1, 2, 3):** For (k = 1, 2, 3), we can take (m = 1). 5. **Assume (2^k mid a^m + a + 2) and (2^{k+1} nmid a^m + a + 2):** Suppose: [ a^m + a + 2 equiv 2^k pmod{2^{k+1}} ] 6. **Consider (a^{2^{k-2}} - 1 equiv 2^k pmod{2^{k+1}}):** We have: [ a^{2^{k-2}} - 1 equiv 2^k pmod{2^{k+1}} ] 7. **Evaluate (a^{m + 2^{k-2}} + a + 2):** [ a^{m + 2^{k-2}} + a + 2 = (a^m + a + 2) + a^m(a^{2^{k-2}} - 1) ] Since: [ a^{2^{k-2}} - 1 equiv 2^k pmod{2^{k+1}} ] We get: [ a^m(a^{2^{k-2}} - 1) equiv a^m cdot 2^k pmod{2^{k+1}} ] Therefore: [ a^{m + 2^{k-2}} + a + 2 equiv (2^k + 2^k) pmod{2^{k+1}} equiv 0 pmod{2^{k+1}} ] Hence: [ 2^{k+1} mid a^{m + 2^{k-2}} + a + 2 ] (blacksquare)