💡Okay, so I have this probability problem here: "Three points are chosen uniformly at random on a circle. What is the probability that no two of these points form an obtuse triangle with the circle's center?" Hmm, interesting. I need to figure this out step by step.First, let me visualize the problem. We have a circle, and we randomly pick three points on its circumference. The center of the circle is a fixed point, let's call it O. Now, for any two points, say A and B, the triangle formed by O, A, and B can be acute, right, or obtuse. I need to find the probability that none of these triangles are obtuse.So, the key here is to understand when triangle OAB is obtuse. Since OA and OB are radii of the circle, they are equal in length. So, triangle OAB is isosceles with OA = OB. In an isosceles triangle, the obtuse angle, if any, must be at the vertex opposite the base. In this case, the vertex is O.Wait, is that always true? Let me think. In an isosceles triangle, if the two equal sides are OA and OB, then the base is AB. So, the angles at A and B are equal. If the triangle is obtuse, the obtuse angle must be at O because the other two angles are equal and must be acute if the triangle is to have only one obtuse angle.So, triangle OAB is obtuse if and only if angle AOB is greater than 90 degrees. That makes sense. So, angle AOB is the central angle corresponding to arc AB. If arc AB is greater than a quarter of the circle (which is 90 degrees or π/2 radians), then angle AOB is obtuse.Therefore, for triangle OAB to not be obtuse, the arc AB must be less than or equal to π/2 radians. Similarly, this applies to arcs AC and BC as well, where C is the third point.So, the problem reduces to finding the probability that all three arcs between the points are less than or equal to π/2 radians. Wait, but hold on. If all three arcs are less than or equal to π/2, then the total circumference would be at most 3*(π/2) = 3π/2, but the circumference of the circle is 2π. So, 3π/2 is less than 2π, which is not possible because the sum of the arcs must equal the circumference.Hmm, that seems contradictory. Maybe I made a mistake here. Let me think again.If we fix one point, say A, then the other two points B and C must lie within a semicircle starting from A. Wait, no, because we need all arcs AB, BC, and CA to be less than or equal to π/2. But as I saw earlier, that's impossible because their sum would be less than 3π/2, which is less than 2π.So, perhaps my initial approach is wrong. Maybe instead of all arcs being less than π/2, we need to ensure that no arc is greater than π/2. But that also seems tricky because if one arc is greater than π/2, then the corresponding triangle would be obtuse.Wait, actually, the problem says "no two of these points form an obtuse triangle with the circle's center." So, for each pair of points, the triangle formed with the center should not be obtuse. That means, for every pair, the central angle should be less than or equal to π/2.But as I thought earlier, if all three central angles are less than or equal to π/2, their sum would be less than or equal to 3π/2, which is less than 2π. But the sum of all central angles around the circle must be 2π. Therefore, it's impossible for all three central angles to be less than or equal to π/2.Hmm, so does that mean the probability is zero? That can't be right because the problem is asking for a probability, not a certainty. Maybe I misunderstood the condition.Wait, perhaps the problem is not requiring all three arcs to be less than π/2, but rather that for each pair, the arc between them is less than π/2. But as I saw, that's impossible because the total circumference would be less than 3π/2, which is less than 2π.So, maybe the correct interpretation is that for each pair, the minor arc between them is less than π/2. The minor arc is the smaller of the two possible arcs between two points on a circle. So, if the minor arc between any two points is less than π/2, then the central angle is less than π/2, and the triangle is acute.But again, if all three minor arcs are less than π/2, their sum would be less than 3π/2, which is less than 2π. So, it's still impossible. Therefore, the probability should be zero? But that doesn't make sense because the problem is asking for a non-zero probability.Wait, maybe I'm overcomplicating this. Let's think differently. Instead of trying to fix all three arcs, perhaps we can fix one point and analyze the positions of the other two points relative to it.Let's fix point A at some position on the circle. Since the circle is uniform, we can fix A without loss of generality. Now, we need to place points B and C such that the arcs AB and AC are both less than π/2. Wait, but if both B and C are within π/2 of A, then the arc between B and C could be up to π/2 as well, but their positions relative to each other might still cause some arcs to exceed π/2.Wait, no. If both B and C are within π/2 of A, then the maximum arc between B and C would be π/2 as well, because they are both within a π/2 arc from A. So, in that case, all arcs AB, AC, and BC would be less than or equal to π/2.But as I thought earlier, the total circumference would be less than 3π/2, which is less than 2π. So, that's not possible. Therefore, it's impossible for all three arcs to be less than π/2. So, the probability is zero?But that contradicts the problem statement, which is asking for the probability, implying it's non-zero. Maybe I'm missing something here.Wait, perhaps the problem is not requiring all three arcs to be less than π/2, but rather that for each pair, the triangle formed with the center is not obtuse. So, for each pair, the central angle is less than or equal to π/2. But as we saw, that's impossible because the sum of the central angles would be less than 3π/2, which is less than 2π. Therefore, the probability is zero.But that seems counterintuitive. Let me check with an example. Suppose all three points are very close to each other. Then, the arcs between them are very small, so all central angles are small, less than π/2, so all triangles are acute. So, in that case, the probability is non-zero.Wait, but if all three points are close together, then the arcs between them are small, so the central angles are small, and the triangles are acute. So, that's possible. Therefore, the probability is not zero.So, maybe my earlier reasoning was flawed. Let me try again.Let me fix point A at angle 0. Then, points B and C are randomly placed on the circle. Let me represent their positions as angles θ and φ, uniformly distributed between 0 and 2π.But to simplify, since the circle is continuous and uniform, I can fix A at 0 and consider θ and φ relative to A. So, θ and φ are uniformly distributed between 0 and 2π, but we can also consider them as being placed on a circumference of length 2π, so we can fix A at 0 and consider θ and φ in [0, 2π).But to make it even simpler, we can fix A at 0 and consider the positions of B and C relative to A. Since the circle is rotationally symmetric, we can fix A at 0 without loss of generality.Now, let's define θ as the angle of point B relative to A, and φ as the angle of point C relative to A. Both θ and φ are uniformly distributed in [0, 2π).But since we're dealing with arcs, we can consider the positions of B and C in the interval [0, 2π) relative to A.Now, to ensure that the central angles for all pairs are less than or equal to π/2, we need:1. The arc between A and B is less than or equal to π/2.2. The arc between A and C is less than or equal to π/2.3. The arc between B and C is less than or equal to π/2.But as I thought earlier, if both B and C are within π/2 of A, then the arc between B and C is at most π/2 as well, because they are both within a π/2 arc from A.Wait, but if B is at θ and C is at φ, both within [0, π/2], then the arc between B and C is |θ - φ|, which is less than or equal to π/2. So, that works.But if B is at θ and C is at φ, both within [0, π/2], then the arc between B and C is |θ - φ|, which is less than or equal to π/2. So, all three arcs are less than or equal to π/2.But then, the total circumference covered by these three points is at most π/2 (from A to B) + π/2 (from A to C) + π/2 (from B to C) = 3π/2, which is less than 2π. So, that's possible because the three points are overlapping in their arcs.Wait, but actually, the three points are on the circumference, so the arcs between them are not overlapping. Wait, no, the arcs are just the distances between the points. So, if all three points are within a π/2 arc from A, then the total circumference covered is just π/2, not 3π/2.Wait, that makes more sense. So, if all three points are within a π/2 arc, then the maximum arc between any two points is π/2. So, that satisfies the condition that all central angles are less than or equal to π/2.Therefore, the probability we're looking for is the probability that all three points lie within a π/2 arc on the circle.But wait, the problem is not exactly that. The problem is that no two points form an obtuse triangle with the center. So, for each pair, the central angle is less than or equal to π/2. So, if all three points lie within a semicircle of π/2, then all central angles are less than or equal to π/2.But actually, the condition is slightly different. Because even if all three points lie within a semicircle, the central angles between pairs could still be up to π, which would make the triangle obtuse.Wait, no. If all three points lie within a π/2 arc, then the maximum central angle between any two points is π/2, so all triangles formed with the center are acute.But if the three points lie within a semicircle (π), but not necessarily within π/2, then some central angles could be up to π, making the triangle obtuse.So, to ensure that no two points form an obtuse triangle with the center, all three points must lie within a π/2 arc.Therefore, the problem reduces to finding the probability that all three points lie within a π/2 arc on the circle.Now, I need to calculate this probability.I remember that the probability that n points chosen uniformly at random on a circle all lie within some particular arc of angle θ is n*(θ/(2π))^{n-1}. But I'm not sure if that's correct.Wait, actually, for the case where we fix one point, the probability that the other n-1 points lie within an arc of length θ starting from the fixed point is (θ/(2π))^{n-1}.But since the arc can be anywhere on the circle, not necessarily starting from the fixed point, we need to adjust for that.I think the general formula for the probability that all n points lie within any arc of length θ is n*(θ/(2π))^{n-1}, provided that θ <= 2π/n.Wait, no, that doesn't seem right. Let me recall the correct formula.The probability that all n points lie within some arc of length θ is n*(θ/(2π))^{n-1}, but this is only valid when θ <= 2π/n.Wait, actually, I think the correct formula is n*(θ/(2π))^{n-1} when θ <= 2π/n, and 1 otherwise.But in our case, θ is π/2, and n is 3. So, 2π/n is 2π/3 ≈ 2.094 radians, and π/2 ≈ 1.571 radians, which is less than 2π/3. So, the formula applies.Therefore, the probability that all three points lie within some arc of length π/2 is 3*(π/2/(2π))^{2} = 3*(1/4)^{2} = 3*(1/16) = 3/16.Wait, but I'm not sure if that's correct. Let me think again.Actually, I think the formula is a bit different. The probability that all n points lie within any arc of length θ is n*(θ/(2π))^{n-1}, but only when θ <= 2π/n. Otherwise, it's 1.But in our case, θ is π/2, and n=3, so 2π/3 ≈ 2.094, and π/2 ≈ 1.571, which is less than 2π/3. So, the formula applies.Therefore, the probability is 3*(π/2/(2π))^{2} = 3*(1/4)^{2} = 3/16.But wait, let me verify this with a different approach.Another way to calculate this is to fix one point, say A, and then calculate the probability that the other two points, B and C, lie within a π/2 arc starting from A.Since the circle is continuous, we can fix A at angle 0 without loss of generality. Then, the positions of B and C are uniformly distributed in [0, 2π).The probability that both B and C lie within [0, π/2] is (π/2 / 2π)^2 = (1/4)^2 = 1/16.But since the arc can start anywhere on the circle, not just at A, we need to consider all possible positions of the arc.However, since we fixed A at 0, the arc can rotate around the circle, but we need to ensure that all three points lie within some π/2 arc.Wait, but if we fix A at 0, then the arc can be anywhere, but we need to find the probability that both B and C lie within a π/2 arc that also contains A.Wait, no. Actually, the arc can be anywhere on the circle, not necessarily containing A. So, fixing A at 0 might complicate things because we need to consider arcs that might not include A.Alternatively, perhaps it's better to not fix any points and consider the circle as a whole.Let me recall that the probability that all n points lie within some arc of length θ is n*(θ/(2π))^{n-1} when θ <= 2π/n.So, for n=3 and θ=π/2, which is less than 2π/3, the probability is 3*(π/2 / 2π)^{2} = 3*(1/4)^2 = 3/16.Therefore, the probability is 3/16.But let me think again. Suppose we fix point A at 0. Then, the probability that both B and C lie within [0, π/2] is (π/2 / 2π)^2 = 1/16. But since the arc can start anywhere, we need to consider all possible positions of the arc.However, because we fixed A at 0, we can consider the arc starting at A, but the arc could also be elsewhere. So, perhaps the total probability is the integral over all possible positions of the arc, but that seems complicated.Alternatively, I remember that the probability that all n points lie within some arc of length θ is n*(θ/(2π))^{n-1} when θ <= 2π/n.So, for n=3 and θ=π/2, which is less than 2π/3, the probability is 3*(π/2 / 2π)^{2} = 3*(1/4)^2 = 3/16.Therefore, the probability is 3/16.Wait, but I'm still a bit unsure. Let me try to calculate it differently.Suppose we fix point A at 0. Then, the positions of B and C are uniformly distributed in [0, 2π). Let me represent their positions as angles θ and φ, where θ and φ are in [0, 2π).We need to find the probability that there exists an arc of length π/2 that contains all three points A, B, and C.Since A is fixed at 0, we can consider the positions of B and C relative to A.Let me define the positions of B and C as θ and φ, both in [0, 2π).To have all three points within a π/2 arc, there must exist some angle α such that all three points lie within [α, α + π/2].But since A is at 0, we can consider α such that 0 is within [α, α + π/2]. So, α must be in [-π/2, 0], but since angles are modulo 2π, α can be in [2π - π/2, 2π], which is [3π/2, 2π].Wait, this is getting complicated. Maybe it's better to consider the positions of B and C relative to A.Let me sort the points in increasing order of their angles. Let's say θ ≤ φ.Then, the arcs between A and B is θ, between B and C is φ - θ, and between C and A is 2π - φ.We need all these arcs to be less than or equal to π/2.But wait, that's not possible because θ + (φ - θ) + (2π - φ) = 2π, which is greater than 3π/2.Wait, no, the arcs between the points are θ, φ - θ, and 2π - φ. We need all of these to be less than or equal to π/2.So, θ ≤ π/2, φ - θ ≤ π/2, and 2π - φ ≤ π/2.But 2π - φ ≤ π/2 implies that φ ≥ 2π - π/2 = 3π/2.But θ ≤ π/2 and φ ≥ 3π/2. Since θ ≤ π/2 and φ ≥ 3π/2, the arc between B and C is φ - θ ≥ 3π/2 - π/2 = π.But we need φ - θ ≤ π/2, which contradicts φ - θ ≥ π.Therefore, it's impossible for all three arcs to be less than or equal to π/2.Wait, so that means the probability is zero? But that contradicts my earlier conclusion.Hmm, I'm confused now. Let me try to clarify.If we fix A at 0, and require that all three points lie within a π/2 arc, then the arc must cover A, B, and C. So, the arc can be anywhere on the circle, not necessarily starting at A.Therefore, the arc can be placed such that it covers A, B, and C, even if A is not at the start of the arc.So, for example, the arc could start at some angle α, and cover α to α + π/2, and all three points must lie within this interval.Since the circle is continuous, we can fix A at 0, and then consider the positions of B and C relative to A.Let me define the positions of B and C as θ and φ, both in [0, 2π).We need to find the probability that there exists an α such that all three points lie within [α, α + π/2].Since A is at 0, we can consider α such that 0 is within [α, α + π/2]. So, α must be in [-π/2, 0], but since angles are modulo 2π, α can be in [2π - π/2, 2π], which is [3π/2, 2π].But this seems complicated. Maybe a better approach is to consider the circle as a line from 0 to 2π, and "wrap around" at 2π.Alternatively, we can "break" the circle at point A (0) and consider the circle as a line from 0 to 2π, with A at 0 and 2π.Then, the problem becomes finding the probability that both B and C lie within some interval of length π/2 on this line.But since the line is actually a circle, the interval can wrap around from 2π back to 0.This is similar to the problem of finding the probability that n points on a circle all lie within some arc of length θ.I think the general formula for this probability is n*(θ/(2π))^{n-1}, as I thought earlier, but only when θ <= 2π/n.In our case, n=3 and θ=π/2, which is less than 2π/3 ≈ 2.094.So, applying the formula, the probability is 3*(π/2 / 2π)^{2} = 3*(1/4)^2 = 3/16.Therefore, the probability is 3/16.But wait, let me verify this with a different approach.Suppose we fix point A at 0. Then, the positions of B and C are uniformly distributed in [0, 2π). Let me represent their positions as θ and φ, where θ and φ are in [0, 2π).We need to find the probability that there exists an arc of length π/2 that contains all three points A, B, and C.Since A is fixed at 0, we can consider the positions of B and C relative to A.Let me define the positions of B and C as θ and φ, both in [0, 2π).To have all three points within a π/2 arc, there must exist some angle α such that all three points lie within [α, α + π/2].But since A is at 0, we can consider α such that 0 is within [α, α + π/2]. So, α must be in [-π/2, 0], but since angles are modulo 2π, α can be in [2π - π/2, 2π], which is [3π/2, 2π].Wait, this is getting too convoluted. Maybe it's better to use the formula I mentioned earlier.The probability that all three points lie within some arc of length θ is 3*(θ/(2π))^{2} when θ <= 2π/3.So, for θ=π/2, the probability is 3*(π/2 / 2π)^{2} = 3*(1/4)^2 = 3/16.Therefore, the probability is 3/16.But I'm still a bit unsure. Let me try to calculate it using integration.Suppose we fix point A at 0. Then, the positions of B and C are θ and φ, uniformly distributed in [0, 2π).We need to find the probability that there exists an arc of length π/2 that contains all three points.This is equivalent to finding the probability that the maximum arc between any two consecutive points is less than or equal to π/2.Wait, no. Actually, the condition is that all three points lie within some π/2 arc, not that the arcs between them are all less than π/2.So, it's a different condition.To calculate this, we can use the formula for the probability that n points on a circle all lie within some arc of length θ, which is n*(θ/(2π))^{n-1} when θ <= 2π/n.In our case, n=3 and θ=π/2, which is less than 2π/3 ≈ 2.094.So, the probability is 3*(π/2 / 2π)^{2} = 3*(1/4)^2 = 3/16.Therefore, the probability is 3/16.I think this is the correct answer. So, the probability that no two of the three randomly chosen points form an obtuse triangle with the circle's center is 3/16.

💡To determine the area of triangle ( AMN ) in triangle ( ABC ) where ( M ) and ( N ) are midpoints of sides ( BC ) and ( AC ) respectively, and the area of triangle ( ABC ) is 180 square units, we can use the properties of midsegments in triangles.1. **Midsegments and Area Division**: - A midsegment in a triangle (a line segment connecting the midpoints of two sides) is parallel to the third side and half its length. - Connecting midpoints ( M ) and ( N ) forms the midsegment ( MN ), which is parallel to ( AB ) and half its length.2. **Area Ratio**: - The area of a triangle formed by a midsegment is one-fourth the area of the original triangle. This is because the midsegment divides the triangle into smaller triangles, each with one-fourth the area of the original.3. **Calculating the Area**: - Given the area of triangle ( ABC ) is 180 square units, the area of triangle ( AMN ) is: [ text{Area of } AMN = frac{1}{4} times 180 = 45 text{ square units} ]Thus, the area of triangle ( AMN ) is ( boxed{C} ).

💡First, I need to determine Matt's current age. I know that in 9 years, Jed will be 23.8 years old, so Jed's current age is 23.8 minus 9, which equals 14.8 years.Since Jed is 10.4 years older than Matt, I can set up the equation M + 10.4 = 14.8 to find Matt's age. Solving for M, I subtract 10.4 from both sides, resulting in M = 4.4 years.Next, I calculate Emily's age, who is 2.5 times older than Matt. Multiplying Matt's age by 2.5 gives Emily's age: 2.5 * 4.4 = 11 years.Finally, I sum up the current ages of Matt, Jed, and Emily: 4.4 + 14.8 + 11 = 30.2 years. Rounding this to the nearest whole number, the total is 30 years.

💡Okay, so I've got this problem here: Let ((x^2 + 1)(2x + 1)^9 = a + a_1(x + 2) + a_2(x + 2)^2 + ldots + a_n(x + 2)^n). Find the value of (a + a_1 + a_2 + ldots + a_n).Hmm, I need to find the sum of all the coefficients (a, a_1, a_2, ldots, a_n). I remember that in polynomial expressions, the sum of the coefficients can be found by plugging in (x = 1). But wait, in this case, the polynomial is expressed in terms of ((x + 2)) instead of just (x). So, does that change things?Let me think. If the polynomial is written as a sum of powers of ((x + 2)), then to find the sum of the coefficients (a, a_1, a_2, ldots, a_n), I should set (x + 2 = 1), right? Because when you substitute (x + 2 = 1), each term becomes just the coefficient times 1, so the sum of the coefficients is obtained. So, if (x + 2 = 1), then (x = -1). That makes sense. So, substituting (x = -1) into the original equation should give me the sum of the coefficients. Let me write that down:[(-1)^2 + 1 = 1 + 1 = 2]and[2(-1) + 1 = -2 + 1 = -1]So, ((2(-1) + 1)^9 = (-1)^9 = -1).Therefore, the left-hand side of the equation when (x = -1) is:[(2)(-1) = 2 times (-1) = -2]Wait, hold on. Let me recast that. The original expression is ((x^2 + 1)(2x + 1)^9). Plugging in (x = -1):First, compute (x^2 + 1):[(-1)^2 + 1 = 1 + 1 = 2]Then, compute (2x + 1):[2(-1) + 1 = -2 + 1 = -1]So, raising that to the 9th power:[(-1)^9 = -1]Therefore, multiplying these together:[2 times (-1) = -2]So, the left-hand side is (-2). The right-hand side, when (x = -1), becomes:[a + a_1(-1 + 2) + a_2(-1 + 2)^2 + ldots + a_n(-1 + 2)^n]Simplify each term:[a + a_1(1) + a_2(1)^2 + ldots + a_n(1)^n = a + a_1 + a_2 + ldots + a_n]So, the right-hand side is just the sum of the coefficients, which is exactly what we're looking for. Therefore, we have:[a + a_1 + a_2 + ldots + a_n = -2]Wait, that seems straightforward. But let me double-check if I did everything correctly. Maybe I made a mistake in substituting (x = -1). Let me go through it again.Compute (x^2 + 1) at (x = -1):[(-1)^2 = 1, so 1 + 1 = 2. That's correct.]Compute (2x + 1) at (x = -1):[2(-1) = -2, so -2 + 1 = -1. Correct.]Raise that to the 9th power:[(-1)^9 = -1. Correct.]Multiply the two results:[2 times (-1) = -2. Correct.]And on the right-hand side, substituting (x = -1) gives the sum of the coefficients. So, yes, that seems right.I think I might have been overcomplicating it initially, but it does make sense. When you have a polynomial expressed in terms of ((x + c)), substituting (x = -c) gives the sum of the coefficients because each ((x + c)) term becomes 1, leaving just the coefficients.So, in this case, substituting (x = -2) would give me the value of the polynomial at that point, but substituting (x = -1) gives me the sum of the coefficients because (-1 + 2 = 1). That's a neat trick.Just to make sure, let me think about another way to approach this. Maybe expanding the polynomial or something? But that seems too tedious, especially since ((2x + 1)^9) is a high power. It would take forever to expand that. So, substitution is definitely the smarter way.Alternatively, I could think about the fact that the sum of the coefficients is the value of the polynomial when (x + 2 = 1), which is when (x = -1). So, that's consistent with what I did earlier.Yeah, I think that's solid. So, the answer is (-2).**Final Answer**boxed{-2}

💡Alright, so I have this problem here about parametric equations and polar coordinates. It's divided into two parts, I and II. Let me try to work through each part step by step.Starting with part (I): I need to find the polar coordinates of the intersection point T between the curve C and the line l. First, let me write down the given parametric equations. For curve C, the parametric equations are:x = √6 cosθy = √2 sinθAnd for the line l, the parametric equations are:x = (√3 / 2) ty = 2 - (1/2) tSo, point T is where these two parametric equations meet, meaning their x and y coordinates are equal at some parameter values θ and t. I think the best approach is to substitute the parametric equations of line l into the equation of curve C. But wait, curve C is given in parametric form, but maybe it's easier if I convert it into a Cartesian equation first.For curve C, since x = √6 cosθ and y = √2 sinθ, I can express cosθ and sinθ in terms of x and y:cosθ = x / √6sinθ = y / √2Since cos²θ + sin²θ = 1, substituting these in gives:(x / √6)² + (y / √2)² = 1Which simplifies to:x² / 6 + y² / 2 = 1So that's the Cartesian equation of curve C. Now, I can substitute the parametric equations of line l into this equation to find the value of t where they intersect.Substituting x = (√3 / 2) t and y = 2 - (1/2) t into the equation:[( (√3 / 2) t )²] / 6 + [ (2 - (1/2) t )² ] / 2 = 1Let me compute each term step by step.First term: [( (√3 / 2) t )²] / 6= ( (3 / 4) t² ) / 6= (3 t²) / 24= t² / 8Second term: [ (2 - (1/2) t )² ] / 2First, expand the square:(2 - (1/2) t )² = 4 - 2*(2)*(1/2) t + (1/2 t)²= 4 - 2t + (1/4) t²So, dividing by 2:(4 - 2t + (1/4) t²) / 2= 2 - t + (1/8) t²Now, adding both terms together:t² / 8 + 2 - t + (1/8) t² = 1Combine like terms:(t² / 8 + 1/8 t²) + (-t) + 2 = 1(2/8 t²) - t + 2 = 1Simplify 2/8 to 1/4:(1/4) t² - t + 2 = 1Subtract 1 from both sides:(1/4) t² - t + 1 = 0Multiply all terms by 4 to eliminate the fraction:t² - 4t + 4 = 0This is a quadratic equation. Let's solve for t:t² - 4t + 4 = 0This factors as (t - 2)^2 = 0, so t = 2.So, t is 2. Now, plug t = 2 back into the parametric equations of line l to find the coordinates of T.x = (√3 / 2) * 2 = √3y = 2 - (1/2) * 2 = 2 - 1 = 1So, point T has coordinates (√3, 1) in Cartesian coordinates.Now, I need to convert this point into polar coordinates. Polar coordinates are given by (ρ, θ), where ρ is the distance from the origin and θ is the angle from the positive x-axis.First, calculate ρ:ρ = √(x² + y²) = √( (√3)^2 + 1^2 ) = √(3 + 1) = √4 = 2Next, find θ. Since the point is (√3, 1), it's in the first quadrant. The tangent of θ is y/x = 1/√3.So, tanθ = 1/√3. I remember that tan(π/6) = 1/√3, so θ = π/6.Therefore, the polar coordinates of point T are (2, π/6).Alright, that seems straightforward. Let me just double-check my steps to make sure I didn't make any mistakes.1. Converted parametric equations of curve C to Cartesian: x²/6 + y²/2 = 1. That looks correct.2. Substituted x and y from line l into the equation. Calculated each term step by step, which seems right.3. Solved the quadratic equation and got t = 2. Plugged back into line l to get (√3, 1). That seems correct.4. Converted to polar coordinates: ρ = 2, θ = π/6. Yep, that makes sense.So, part (I) is done. The polar coordinates of point T are (2, π/6).Moving on to part (II): We need to stretch the ordinate (y-coordinate) of all points on curve C by a factor of √3, keeping the abscissa (x-coordinate) unchanged. This will give us curve W. Then, we have to find the polar equation of line m that passes through point T and intercepts a segment of length 2√3 on curve W.First, let me understand what stretching the ordinate by √3 means. It means that for every point (x, y) on curve C, the corresponding point on curve W will be (x, √3 y). So, the transformation is:x' = xy' = √3 ySo, to find the equation of curve W, I can substitute y = y' / √3 into the equation of curve C.Original equation of curve C: x² / 6 + y² / 2 = 1Substitute y = y' / √3:x² / 6 + ( (y' / √3 )² ) / 2 = 1Simplify:x² / 6 + ( y'^2 / 3 ) / 2 = 1x² / 6 + y'^2 / 6 = 1Multiply both sides by 6:x² + y'^2 = 6So, the equation of curve W is x² + y² = 6. That's a circle centered at the origin with radius √6.Wait, hold on. If I stretch the y-coordinate by √3, the curve becomes a circle? Interesting.Now, we need to find the equation of line m that passes through point T and intercepts a segment of length 2√3 on curve W.First, let me recall that point T is (√3, 1) in Cartesian coordinates.So, line m passes through (√3, 1) and intersects curve W (the circle x² + y² = 6) such that the length of the intercepted segment is 2√3.I need to find the equation of such a line m, and then express it in polar coordinates.Let me think about how to approach this.First, the line passes through (√3, 1). Let me denote the equation of line m in Cartesian coordinates. Since it's a line, it can be written as y = k(x - √3) + 1, where k is the slope. Alternatively, in standard form: kx - y - k√3 + 1 = 0.But maybe it's easier to use the general line equation: Ax + By + C = 0, passing through (√3, 1). So, A√3 + B*1 + C = 0.But perhaps another approach is better. Since the line intersects the circle x² + y² = 6, and the length of the chord intercepted is 2√3, we can use the formula for the length of a chord in a circle.The formula for the length of a chord is 2√(r² - d²), where r is the radius and d is the distance from the center to the line.Given that the length is 2√3, and the radius r is √6, we can set up the equation:2√( (√6)^2 - d^2 ) = 2√3Divide both sides by 2:√(6 - d²) = √3Square both sides:6 - d² = 3So, d² = 3, which means d = √3 or d = -√3, but since distance is non-negative, d = √3.Therefore, the distance from the center (0,0) to the line m is √3.So, now, we need to find the equation of line m passing through (√3, 1) such that its distance from (0,0) is √3.Let me denote the equation of line m as ax + by + c = 0. Since it passes through (√3, 1), we have:a√3 + b*1 + c = 0 --> equation (1)Also, the distance from (0,0) to the line is |a*0 + b*0 + c| / √(a² + b²) = |c| / √(a² + b²) = √3 --> equation (2)But we have two equations and three variables (a, b, c), so we need another condition. However, since the equation of a line is determined up to a scalar multiple, we can set one of the variables to 1 or some other convenient value.Alternatively, let me express the line in terms of its slope. Let me assume the line has a slope k, so its equation can be written as y = kx + c. But since it passes through (√3, 1), we have:1 = k√3 + c --> c = 1 - k√3So, the equation is y = kx + (1 - k√3). Rewriting in standard form:kx - y + (1 - k√3) = 0So, a = k, b = -1, c = 1 - k√3Now, the distance from (0,0) to this line is |c| / √(a² + b²) = |1 - k√3| / √(k² + 1) = √3So, set up the equation:|1 - k√3| / √(k² + 1) = √3Since distance is non-negative, we can drop the absolute value by considering both cases:Case 1: 1 - k√3 = √3 * √(k² + 1)Case 2: -(1 - k√3) = √3 * √(k² + 1) --> k√3 - 1 = √3 * √(k² + 1)Let me solve Case 1 first:1 - k√3 = √3 * √(k² + 1)Square both sides:(1 - k√3)^2 = 3(k² + 1)Expand left side:1 - 2k√3 + 3k² = 3k² + 3Simplify:1 - 2k√3 + 3k² - 3k² - 3 = 01 - 3 - 2k√3 = 0-2 - 2k√3 = 0-2k√3 = 2k√3 = -1k = -1 / √3Case 2:k√3 - 1 = √3 * √(k² + 1)Square both sides:(k√3 - 1)^2 = 3(k² + 1)Expand left side:3k² - 2k√3 + 1 = 3k² + 3Simplify:3k² - 2k√3 + 1 - 3k² - 3 = 0-2k√3 - 2 = 0-2k√3 = 2k√3 = -1k = -1 / √3Wait, both cases led to the same solution: k = -1 / √3So, the slope k is -1/√3.Therefore, the equation of line m is:y = (-1/√3)x + (1 - (-1/√3)*√3)Simplify the constant term:1 - (-1/√3)*√3 = 1 + 1 = 2So, the equation is y = (-1/√3)x + 2Alternatively, in standard form:(1/√3)x + y - 2 = 0Multiply both sides by √3 to rationalize:x + √3 y - 2√3 = 0But let me keep it as y = (-1/√3)x + 2 for now.Now, I need to express this line in polar coordinates.Recall that in polar coordinates, x = ρ cosθ and y = ρ sinθ.So, substitute these into the equation:ρ sinθ = (-1/√3) ρ cosθ + 2Bring all terms to one side:ρ sinθ + (1/√3) ρ cosθ - 2 = 0Factor out ρ:ρ (sinθ + (1/√3) cosθ) = 2So, the polar equation is:ρ (sinθ + (1/√3) cosθ) = 2Alternatively, we can write it as:ρ = 2 / (sinθ + (1/√3) cosθ)But usually, polar equations are expressed in terms of ρ and θ without fractions, so the first form is acceptable.Alternatively, we can write it as:ρ sinθ + (1/√3) ρ cosθ = 2Which is the same as:ρ (sinθ + (1/√3) cosθ) = 2Alternatively, we can factor out 2/√3:Let me see, sinθ + (1/√3) cosθ can be expressed as 2/√3 sin(θ + φ), where φ is some angle, but I don't think that's necessary here.Alternatively, another approach is to write the line in polar form using the standard formula.The general equation of a line in polar coordinates is ρ = e / (sinθ + f cosθ), but in our case, we have:ρ (sinθ + (1/√3) cosθ) = 2So, that's already a valid polar equation.Alternatively, we can write it as:ρ = 2 / (sinθ + (1/√3) cosθ)But perhaps it's better to rationalize or make it look neater.Alternatively, multiply numerator and denominator by √3:ρ = 2√3 / (√3 sinθ + cosθ)So, that might be another way to write it.But both forms are correct. So, depending on how the answer is expected, either form is acceptable.Alternatively, another way to express the line is to use the normal form.The normal form of a line in polar coordinates is ρ cos(θ - α) = p, where p is the distance from the origin and α is the angle of the normal.But in our case, the distance from the origin is √3, and the line is at a distance √3 from the origin. Wait, earlier we found that the distance from the origin to the line is √3.So, using the normal form, we can write:ρ cos(θ - α) = √3But we need to find α such that the line passes through (√3, 1). Hmm, maybe that's more complicated.Alternatively, since we already have the equation in terms of ρ and θ, perhaps it's better to leave it as:ρ sinθ + (1/√3) ρ cosθ = 2Or, as I did earlier, ρ = 2 / (sinθ + (1/√3) cosθ)Alternatively, multiplying numerator and denominator by √3:ρ = 2√3 / (√3 sinθ + cosθ)This might be a more standard form.Alternatively, we can write it as:√3 sinθ + cosθ = 2 / ρBut I think the most straightforward form is:ρ sinθ + (1/√3) ρ cosθ = 2Alternatively, factor out ρ:ρ (sinθ + (1/√3) cosθ) = 2So, that's the polar equation of line m.Wait, but earlier, when I considered the slope, I found that the line could also be vertical, x = √3. Because when I considered the chord length, I might have missed that case.Wait, let me think again. When I used the chord length formula, I assumed that the line is not vertical, because if the line is vertical, the formula still applies, but the distance from the center is just the x-coordinate.Wait, in this case, if the line is vertical, x = √3, then the distance from the origin is |√3| = √3, which matches our earlier distance. So, that's another possible line.So, line m could be x = √3, which is a vertical line passing through (√3, 1), and it's at a distance √3 from the origin.Therefore, there are two possible lines: one with slope -1/√3 and another vertical line x = √3.So, in polar coordinates, the vertical line x = √3 can be written as ρ cosθ = √3.Because x = ρ cosθ, so ρ cosθ = √3.So, that's another polar equation.Therefore, the polar equations of line m are:1. ρ cosθ = √32. ρ sinθ + (1/√3) ρ cosθ = 2So, both are valid.Therefore, the answer for part (II) includes both equations.Let me just verify that both lines indeed pass through point T (√3, 1).For the vertical line x = √3: yes, because x = √3, and y can be anything, so (√3, 1) is on this line.For the other line: plug x = √3, y = 1 into the equation y = (-1/√3)x + 2:1 = (-1/√3)(√3) + 21 = -1 + 21 = 1, which is correct.So, both lines pass through T and are at a distance √3 from the origin, intercepting a chord of length 2√3 on curve W.Therefore, both are valid solutions.So, summarizing:(I) The polar coordinates of point T are (2, π/6).(II) The polar equations of line m are ρ cosθ = √3 and ρ sinθ + (1/√3) ρ cosθ = 2.I think that's it. Let me just make sure I didn't miss any steps or make any calculation errors.For part (I):- Converted parametric to Cartesian: correct.- Substituted line into curve: correct.- Solved quadratic: correct, got t=2.- Found coordinates (√3,1): correct.- Converted to polar: ρ=2, θ=π/6: correct.For part (II):- Stretching y by √3: correct, got x² + y² =6.- Chord length formula: correct, found distance d=√3.- Found line equation with slope -1/√3: correct.- Also considered vertical line x=√3: correct.- Converted both lines to polar: correct.Yes, everything seems to check out.

💡First, determine Kim's age in two years by adding 2 to her current age of 10, resulting in 12 years.Next, since Sandy will be three times as old as Kim in two years, multiply Kim's future age by 3 to find Sandy's age in two years: 3 * 12 = 36 years.Subtract 2 years from Sandy's future age to find her current age: 36 - 2 = 34 years.Finally, calculate Sandy's monthly phone bill expense by multiplying her current age by 10: 10 * 34 = 340.