answer:sure, i can provide an example of how to format a response with specific constraints. let's break down the requirements and see how they can be applied: * this is point 1: start by ensuring that the entire response is in lowercase letters, including the word 'example' which needs to appear at least three times. * this is point 2: include exactly three bullet points using markdown formatting. * this is point 3: organize the response into two paragraphs separated by a markdown divider. *** now, let's see how this example adheres to the constraints. the first paragraph introduced the points and explained how to follow the constraints, while this second paragraph continues the explanation and ensures that the word 'example' appears multiple times. this is an example of how to format a response with specific constraints. by following these steps, you can create a well-structured and compliant response.

answer:(1) According to the problem, we have: line l passes through the point (-2, 2) and has a slope of k= sqrt {3}, which means the angle of inclination is α= frac {π}{3}. Hence, the parametric equation of line l can be transformed into the standard form: begin{cases} x=-2+ frac {1}{2}t y=2+ frac { sqrt {3}}{2}tend{cases} where t is the parameter. (2) Substituting the parametric equations of line l into the equation of curve C, we get t^2+4t-10=0. Let t_1 and t_2 be the parameters corresponding to points A and B, respectively. Then, t_1+t_2=-4 and t_1t_2=-10. Thus, the length of |AB| is: |AB|=|t_{1}-t_{2}|=2 sqrt {14} (3) Using the conversion formulas between polar and rectangular coordinates, we find that the rectangular coordinates of point P are (-2, 2). Point P is on the line, and the parameter corresponding to the midpoint M is: t_{3}= frac {t_{1}+t_{2}}{2}=-2 Therefore, the distance from point P to the midpoint M is: |PM|=|t_{3}|=boxed{2}

answer:To begin, let's define the geometric mean (GM) and arithmetic mean (AM) of the given set {1, x, x^2}. The geometric mean is the nth root of the product of n numbers, so in this case, we have GM = (1 * x * x^2)^(1/3) = (x^3)^(1/3) = x. The arithmetic mean, on the other hand, is the sum of the numbers divided by the number of elements, so we have AM = (1 + x + x^2) / 3. Now, we want to compare GM and AM. Specifically, we want to show that GM ≤ AM for all positive x ≠ 1. To do this, we can use the AM-GM inequality, which states that the arithmetic mean of a set of non-negative numbers is greater than or equal to the geometric mean of the same set. Applying this inequality to our problem, we can write: (1 + x + x^2) / 3 ≥ (1 * x * x^2)^(1/3) = x. Simplifying this, we get (1 + x + x^2) / 3 ≥ x. To show that this inequality holds true for all positive x ≠ 1, we can subtract x from both sides and multiply by 3 to get: 1 + x + x^2 - 3x ≥ 0. Rearranging the terms, we get: x^2 - 2x + 1 ≥ 0, which can be factored into (x - 1)^2 ≥ 0. This inequality is true for all real numbers x, and since we know x is positive, the only value of x that would make the inequality an equality is x = 1. However, we are considering x ≠ 1, so the inequality is strict: x^2 - 2x + 1 > 0. Therefore, we have shown that (1 + x + x^2) / 3 > x for all positive x ≠ 1, which means that the geometric mean of {1, x, x^2} is less than the arithmetic mean of the same set.

answer:Yes, there is a straightforward way to show that the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM) for any set of non-negative numbers, with equality only when all the numbers in the set are equal. Let's consider a set of n non-negative numbers, denoted as {x1, x2, ..., xn}. The arithmetic mean (AM) is given by (x1 + x2 + ... + xn) / n, and the geometric mean (GM) is given by (x1 * x2 * ... * xn)^(1/n). We can start with the inequality that will lead us to the desired result. Using the fact that the exponential function is strictly increasing, we can write: exp((x1 + x2 + ... + xn) / n) ≥ (exp(x1) * exp(x2) * ... * exp(xn))^(1/n) Simplifying this inequality, we get: exp(AM) ≥ GM Now, we can take the logarithm of both sides (using any base, but we'll choose the natural logarithm for convenience): AM ≥ ln(GM) However, we want to show that AM ≥ GM, not just their logarithms. To get there, we can use Jensen's inequality, but a simpler approach involves using the concavity of the logarithm function. Let's define yi = ln(xi) for i = 1, 2, ..., n. Then, we can rewrite the geometric mean as: GM = exp((y1 + y2 + ... + yn) / n) Using the concavity of the logarithm function (or equivalently, the fact that the exponential function is convex), we can apply Jensen's inequality to get: GM ≤ (exp(y1) + exp(y2) + ... + exp(yn)) / n Now, recall that yi = ln(xi), so exp(yi) = xi. Therefore: GM ≤ (x1 + x2 + ... + xn) / n = AM This shows that the arithmetic mean is always greater than or equal to the geometric mean for any set of non-negative numbers. The equality case occurs when all the numbers in the set are equal, which can be seen by following the same steps with all xi being equal.