answer:Let's simplify the given complex number step by step. frac{1}{1+i} + i = frac{1}{1+i} + frac{i(1-i)}{1-i} = frac{1}{1+i} + frac{i - i^2}{1+i} = frac{1 + i + 1 - i^2}{1+i} Since i^2 = -1, we continue with frac{1 + i + 1 + 1}{1+i} = frac{3 + i}{1+i} = frac{3 + i}{1+i} cdot frac{1-i}{1-i} = frac{(3 + i)(1 - i)}{(1 + i)(1 - i)} = frac{3 - 3i + i - i^2}{1 - i^2} Now, distributing and applying i^2 = -1 again, we have frac{3 - 2i + 1}{2} = frac{4 - 2i}{2} = 2 - i This simplifies to frac{1}{2} + left(-frac{1}{2}right)i So the corresponding point in the complex plane is left(frac{1}{2}, -frac{1}{2}right). This point is located in the fourth quadrant. Therefore, the correct answer is: boxed{D: Fourth quadrant}

answer:Solution: (Ⅰ) According to the problem, since (x+1)(x-3) - (x+2)(x-4) = (x^2 - 2x - 3) - (x^2 - 2x - 8) = 5 > 0, thus, (x+1)(x-3) > (x+2)(x-4); (Ⅱ) Let the length of the rectangular vegetable garden be x meters and the width be y meters. Then 2(x+y) = 36, which means x+y = 18, and the area of the rectangular garden is xy square meters. According to sqrt{xy} leq frac{x+y}{2} = frac{18}{2} = 9, we get xy leq 81; Equality holds if and only if x = y, i.e., x = y = 9. Therefore, when the length and width of the rectangle are both 9m, the area of the garden is maximized, and the maximum area is boxed{81m^2}.

answer:To solve this problem, let's analyze each statement step by step: - **Statement ①**: An equilateral triangle, by definition, has all three sides equal. An isosceles triangle has at least two sides equal. Therefore, an equilateral triangle, having all sides equal, naturally satisfies the condition of having at least two sides equal, making it a specific type of isosceles triangle. Thus, statement ① is correct. - **Statement ②**: An isosceles triangle has two sides of equal length. It's possible for an isosceles triangle to also have a right angle, making it an isosceles right triangle. This means an isosceles triangle can indeed be a right triangle, so statement ② is correct. - **Statement ③**: Triangles can indeed be classified by their sides into different categories. However, the classification includes scalene triangles (no equal sides), isosceles triangles (at least two equal sides), and equilateral triangles (all sides equal). Since equilateral triangles are a subset of isosceles triangles (all sides equal is a special case of at least two sides being equal), it's incorrect to list equilateral triangles separately as if they are on the same level of classification as scalene and isosceles triangles. Therefore, statement ③ is incorrect. - **Statement ④**: Triangles can be classified by their angles into acute triangles (all angles less than 90 degrees), right triangles (one angle exactly 90 degrees), and obtuse triangles (one angle more than 90 degrees). This classification is based on the angles of the triangles, making statement ④ correct. Given the analysis, statements ①, ②, and ④ are correct, while statement ③ is incorrect. Therefore, the correct answer is that 3 statements are correct. boxed{C}

answer:Analysis: This problem primarily examines the dot product of vectors and operations related to the magnitude of a vector. The given vectors are mutually perpendicular unit vectors, which greatly facilitates calculations. When using the condition that the dot product equals zero, it's helpful to move terms around. Solution Enhancement: Given that |overset{to }{a}|=|overset{to }{b}|=1 and overset{to }{a}cdot overset{to }{b}=0, Since (overset{to }{a}-overset{to }{c})cdot (overset{to }{a}-overset{to }{c})=0, we can simplify it to |overset{to }{c}{|}^{2}=overset{to }{c}cdot (overset{to }{a}+overset{to }{b})=|overset{to }{c}|cdot |overset{to }{a}+overset{to }{b}|mathrm{cos}theta. Thus, |overset{to }{c}|=|overset{to }{a}+overset{to }{b}|mathrm{cos}theta =sqrt{2}mathrm{cos}theta. Given that mathrm{cos}theta in [-1,1], the maximum value of |overset{to }{c}| is sqrt{2}. Thus, the answer is boxed{C}.