answer:Solution: z=2i(1+i)=2i+2i^2=-2+2i, the corresponding coordinates are (-2,2), which is located in the second quadrant. Therefore, the correct answer is boxed{text{B}}. This problem mainly tests the geometric meaning of complex numbers. Simplifying the expression using the basic operations of complex numbers is key to solving this problem.

answer:(Ⅰ) From the equation C: x^2 + y^2 - 4x + 2y + m = 0, we can complete the square to get (x-2)^2 + (y+1)^2 = 5 - m. Thus, the center of the circle, C, is at (2, -1), and the radius squared, r^2, is equal to 5 - m. As triangle ABC is an isosceles right triangle, let D be the midpoint of A and B. Drawing line CD, we find that triangle ACD is also an isosceles right triangle. Therefore, AC = sqrt{2}CD = 2sqrt{2}, so 5 - m = 8, which gives us m = -3. [boxed{m = -3}] (Ⅱ) Let's suppose the equation of the line is y = kx + 2. Then, the distance, d, from the center (2, -1) to the line y = kx + 2 is d = frac{|2k + 1 + 2|}{sqrt{1 + k^2}} = frac{|2k + 3|}{sqrt{1 + k^2}}. Since r = 2sqrt{2} and |MN| geq 4, we have frac{|2k + 3|}{sqrt{1 + k^2}} leq 2. Solving for k, we get k leq - frac{5}{12}. Thus, the range of k is (-infty, - frac{5}{12}]. [boxed{k in (-infty, - frac{5}{12}]}] (Ⅲ) Combine the equations of the line and the circle: [left{ begin{align*} x^2 + y^2 - 4x + 2y - 3 = 0 y = kx + 2 end{align*} right.] After eliminating the variable y, we get (1+k^2)x^2 + (6k-4)x + 5 = 0. Let M be at (x_1, y_1) and N at (x_2, y_2). By Vieta's formulas, we have x_1 + x_2 = frac{4 - 6k}{1 + k^2}. Since the line intersects circle C at two distinct points M and N, we have the discriminant Delta = (6k - 4)^2 - 20(1 + k^2) > 0. Simplifying, we get 4k^2 - 12k - 1 > 0. The solution to this inequality is k > frac{3 + sqrt{10}}{2} or k < frac{3 - sqrt{10}}{2}. y_1 + y_2 = k(x_1 + x_2) + 4 = frac{-2k^2 + 4k + 4}{1 + k^2}, thus overrightarrow{OM} + overrightarrow{ON} = left( frac{4 - 6k}{1 + k^2}, frac{-2k^2 + 4k + 4}{1 + k^2} right) and overrightarrow{OC} = (2, -1). If overrightarrow{OM} + overrightarrow{ON} is collinear with overrightarrow{OC}, then frac{6k - 4}{1 + k^2} = 2 times frac{-2k^2 + 4k + 4}{1 + k^2}, which simplifies to 2k^2 - k - 6 = 0. This gives us k = 2 or k = -frac{3}{2}. Upon verification, k=2 does not satisfy the earlier found conditions k > frac{3 + sqrt{10}}{2} or k < frac{3 - sqrt{10}}{2}. Therefore, a real number k = -frac{3}{2} satisfies the problem conditions. [boxed{k = -frac{3}{2}}]

answer:To find the percentage increase in state tax, we first need to determine the difference in cost between buying the bike this month and next month. The cost next month is 82,500, and the cost this month is 75,000. The difference in cost is 82,500 - 75,000 = 7,500. This 7,500 increase is due to the state tax increase alone. To find the percentage increase, we divide the increase by the original cost (this month's cost) and then multiply by 100 to get the percentage. Percentage increase = (7,500 / 75,000) * 100 Percentage increase = 0.1 * 100 Percentage increase = 10% So, the state tax will increase by boxed{10%} next month.

answer:1. **Position of the hands**: At 3:15 p.m., the minute hand points at the 3 (or 15 minutes), and the hour hand is a quarter of the way between 3 and 4. 2. **Calculate the hour hand's angular position**: The hour hand moves at a rate of 30^circ per hour (since 360^circ / 12 = 30^circ). At 3:00 it would be at 3 times 30^circ = 90^circ from the top (12:00). In 15 minutes, it moves an additional frac{15}{60} times 30^circ = 7.5^circ. Thus, at 3:15, the hour hand is at 90^circ + 7.5^circ = 97.5^circ. 3. **Calculate the minute hand's angular position**: The minute hand moves at a rate of 360^circ per hour, or 6^circ per minute (since 360^circ / 60 = 6^circ). At 15 minutes, it points at 90 degrees (as it points directly at the 3). 4. **Calculate the angle between the hands**: The difference between the hour and minute hand positions at 3:15 is |97.5^circ - 90^circ| = 7.5^circ. Conclusion: The smaller angle formed by the minute hand and the hour hand at 3:15 p.m. is boxed{7.5^circ}.