answer:1. **Initial Price and First Discount**: Let the original price of the item be x dollars. The first discount is 15%. Therefore, the price after the first discount is: [ (1 - 0.15)x = 0.85x ] 2. **Second Discount**: The second discount is 10%. This discount is applied to the new price, 0.85x. Thus, the price after the second discount is: [ (1 - 0.10)(0.85x) = 0.90 times 0.85x = 0.765x ] 3. **Third Discount**: The third discount is 5%. This is applied to the price after the second discount, 0.765x. Therefore, the price after the third discount is: [ (1 - 0.05)(0.765x) = 0.95 times 0.765x = 0.72675x ] 4. **Equivalent Single Discount**: Let k be the fraction representing the equivalent single discount. The price after applying this single discount would also be 0.72675x. Therefore, we set up the equation: [ (1 - k)x = 0.72675x ] 5. **Solving for k**: To find k, we solve the equation: [ 1 - k = 0.72675 implies k = 1 - 0.72675 = 0.27325 ] Converting k to a percentage gives k = 27.325%, which is approximately 27.3%. Conclusion: The equivalent single discount is approximately 27.3%. Therefore, the final answer is: [ 27.3% ] The final answer is boxed{27.3%} (C)

answer:1. **Claim**: No matter the configuration of the labyrinth mathscr{L}, the maximum number k of knights that can be arranged such that no two knights meet is uniquely k=n+1. 2. To demonstrate this, we first use mathematical induction to prove that n walls divide the plane into binom{n+1}{2} + 1 regions. - **Base Case**: For n=0 (no walls), the whole plane is one region, thus binom{0+1}{2} + 1 = 1 region. The base case holds. - **Inductive Step**: Assume that the statement is true for n-1 walls, i.e., n-1 walls divide the plane into binom{n}{2} + 1 regions. - When we add the n-th wall, it intersects the other n-1 walls exactly once each (since no walls are parallel and no three walls are concurrent). Thus, the n-th wall is divided into n segments, each segment producing a new region by splitting an existing region into two. - Adding these n new regions to the binom{n}{2} + 1 regions from the assumption, we get: binom{n}{2} + 1 + n = binom{n+1}{2} + 1 Hence, the inductive step holds and by induction, the statement is proven. 3. Next, define G as a graph where each vertex corresponds to each of these binom{n+1}{2} + 1 regions. If two regions are connected by a door, there is an edge between the corresponding vertices in G. 4. **Graph Components**: - Each of the binom{n+1}{2} crossings corresponds to exactly one door since there is a unique edge in graph G for each door in the labyrinth. - When a new edge (door) is added to G, the number of connected components decreases by at most one, so the connected components (regions connected by doors) are at least binom{n+1}{2} + 1 - binom{n}{2} = n + 1. 5. **Coloring Argument**: - Color one side of each wall red and the other blue. - Any intersection of walls will have two opposing angles formed by different colors and one angle each formed by red and blue walls respectively, ensuring knights can move through doors without meeting. - Regardless of the coloring method, there are exactly binom{n}{2} crossings, and each crossing has a door, associating each door with one edge in G. 6. Using this setup, notice that knights can be placed in different regions corresponding to vertices in different connected components of G, ensuring they never meet. Since G has at least n+1 components, there are at least n+1 regions where knights can be placed. 7. Lastly, construct a scenario: - Place the coordinate system such that no wall is parallel to the coordinate axes. - Color each wall with its sides, say the west side red and the east side blue. - Label each region according to its position relative to the number of walls on its east side. - For each integer i in {0, 1, ldots, n}, there's a unique region touching the northern boundary, allowing knights to never meet if placed correspondingly. **Conclusion**: Hence, the maximum number of knights k(mathscr{D}) that can be placed such that no two knights meet is: boxed{n+1}

answer:**Analysis** Utilize the relationship between the original function and its derivative. For f(x) to be monotonically decreasing in the interval (2m, m + 1), the derivative f'(x) = 3x^2 - 12 leqslant 0 must hold true in the interval (2m, m + 1). This leads to the following system of inequalities: begin{cases} f'(2m) leqslant 0 f'(m+1) leqslant 0 2m < m+1 end{cases} Solve for the range of possible values of m. **Step-by-step solution** 1. Find the derivative of f(x): f'(x) = frac{d}{dx}(x^3 - 12x) = 3x^2 - 12 2. Set up the system of inequalities based on the given conditions: begin{cases} f'(2m) leqslant 0 f'(m+1) leqslant 0 2m < m+1 end{cases} 3. Substitute f'(x) into the inequalities: begin{cases} 3(2m)^2 - 12 leqslant 0 3(m+1)^2 - 12 leqslant 0 2m < m+1 end{cases} 4. Simplify and solve the inequalities: begin{cases} 12m^2 - 12 leqslant 0 3m^2 + 6m + 3 - 12 leqslant 0 m < 1 end{cases} begin{cases} m^2 - 1 leqslant 0 m^2 + 2m - 3 leqslant 0 m < 1 end{cases} begin{cases} (m - 1)(m + 1) leqslant 0 (m + 3)(m - 1) leqslant 0 m < 1 end{cases} 5. Solve for m: From (m - 1)(m + 1) leqslant 0, we find that -1 leqslant m leqslant 1. From (m + 3)(m - 1) leqslant 0, we find that -3 leqslant m leqslant 1. However, since m < 1, the final range of values for m is -1 leqslant m < 1. Thus, the correct answer is: boxed{text{D: } -1 leqslant m < 1}.

answer:Given the quadratic equation in two variables: [ x^2 + 2xy + 3y^2 - 2x + y + 1 = 0 ] 1. **Rewrite the given quadratic equation as a quadratic equation in terms of ( x ):** [ x^2 + 2(y-1)x + (3y^2 + y + 1) = 0 ] 2. **For the equation to have integer solutions ((x, y)), the discriminant must be a non-negative perfect square. The discriminant (Delta) of the quadratic equation ( ax^2 + bx + c = 0 ) is given by:** [ Delta = b^2 - 4ac ] 3. **In this case, the discriminant is:** [ b = 2(y-1), quad a = 1, quad c = 3y^2 + y + 1 ] [ Delta = [2(y-1)]^2 - 4 cdot 1 cdot (3y^2 + y + 1) ] [ Delta = 4(y-1)^2 - 4(3y^2 + y + 1) ] [ Delta = 4[(y-1)^2 - (3y^2 + y + 1)] ] [ Delta = 4[y^2 - 2y + 1 - 3y^2 - y - 1] ] [ Delta = 4[-2y^2 - 3y] ] [ Delta = -4(2y^2 + 3y) ] 4. **For the discriminant to be non-negative:** [ -4(2y^2 + 3y) geq 0 ] [ 2y^2 + 3y leq 0 ] 5. **Solve the inequality (2y^2 + 3y leq 0):** [ y(2y + 3) leq 0 ] 6. **Find the roots of the quadratic equation:** [ y = 0 ] [ 2y + 3 = 0 ] [ y = -frac{3}{2} ] 7. **The solutions to the inequality fall within the interval:** [ -frac{3}{2} leq y leq 0 ] 8. **Since ( y ) must be an integer, possible values of ( y ) are:** [ y = 0, -1 ] 9. **Substitute back ( y = 0 ) into the quadratic equation and solve for ( x ):** [ x^2 - 2x + 1 = 0 ] [ (x-1)^2 = 0 ] [ x = 1 ] 10. **Substitute ( y = -1 ) into the quadratic equation and solve for ( x ):** [ x^2 - 4x + 3 = 0 ] [ (x-1)(x-3) = 0 ] [ x = 1, 3 ] 11. **Therefore, the integer solutions are:** [ (x, y) = (1, 0), (1, -1), (3, -1) ] Conclusion: [ boxed{(x, y) = (1, 0), (1, -1), (3, -1)} ]