answer:Let z_A = 1-3i and z_B = (1+i)(2-i). First, we shall multiply the complex numbers to find z_B. begin{align*} z_B &= (1+i)(2-i) &= 1cdot2 + 1cdot(-i) + icdot2 + icdot(-i) &= 2 - i + 2i + 1 &= 3 + i. end{align*} Now we have z_A = 1 - 3i and z_B = 3 + i. Next, we identify the coordinates of points A and B. Point A corresponds to (1, -3) and point B to (3, 1) in the complex plane. To find the midpoint C of line segment AB, we use the midpoint formula: begin{align*} x_C &= frac{x_A + x_B}{2} = frac{1 + 3}{2} = 2, y_C &= frac{y_A + y_B}{2} = frac{-3 + 1}{2} = -1. end{align*} Thus, the coordinates of point C are (2, -1). The complex number corresponding to point C is: [ boxed{2 - i}. ] This matches option D.

answer:The derivative of the function f(x) = xe^x + 2 is given by f'(x) = e^x + xe^x. To find the slope of the tangent line at the point (0, f(0)), we need to evaluate the derivative at x = 0: f'(0) = e^0 + 0 cdot e^0 = 1 + 0 = 1. The slope of the tangent at that point is thus 1. The point of tangency is given by (0, f(0)), which we can calculate: f(0) = 0 cdot e^0 + 2 = 2. So the point of tangency is (0, 2). Using the point-slope form of a line, the equation of the tangent line can be written as: y - 2 = 1 cdot (x - 0), which simplifies to: y = x + 2. To express this in the general form, we rearrange the equation: x - y + 2 = 0. Therefore, the equation of the tangent line to the curve at the point (0, 2) is boxed{x - y + 2 = 0}.

answer:**Answer**: Let's assume the number is a two-digit number: Since (10a+b)^2 = 100a^2 + 20ab + b^2 = 100^2 + 10 times 2ab + b^2, the digit in the tens place is determined by 2ab and the tens digit of b^2, 2ab is definitely even. When b is an odd number less than 5, it does not affect the tens place; When b is equal to 5, 7, or 9, respectively, its tens place is even, thus, the tens digit of 10 times 2ab + b^2 is even. Therefore, an integer whose digits in both the units and tens places are odd cannot be a perfect square; Therefore, in decimal, perfect squares whose digits are all odd must be in the units place, Therefore, in decimal, there are only two perfect squares whose digits are all odd: 1 and 9, Hence, the correct choice is: boxed{B}.

answer:To negate the proposition P, which states that there exists at least one equilateral triangle, we need to construct a statement that indicates there is no such instance of an equilateral triangle. In other words, the negation must reject the possibility of any triangle being equilateral. Therefore, the negation of P must be a universally quantified statement. The correct negation would be: No triangles are equilateral triangles. Following this reasoning, we can eliminate options A and B, as they suggest the existence of some triangle(s) that do not match the criteria, but they do not negate the possibility of there being equilateral triangles. Option C asserts that all triangles are equilateral, which is not a negation of the original proposition but rather a different universal statement. Therefore, the answer must be: [boxed{D: text{¬P: No triangles are equilateral triangles}}]