answer：Okay, so I have this problem with two sets, A and B, and I need to find their intersection, A ∩ B. Let me try to understand what each set is first.Set A is defined as {x | -1 < x < 2}. That means A includes all real numbers x that are greater than -1 and less than 2. So, if I imagine a number line, A would be the segment from just above -1 to just below 2. It doesn't include the endpoints because it's an open interval.Now, set B is defined as {x | x² + 2x ≤ 0}. Hmm, this is a quadratic inequality. I think I need to solve this inequality to find out what values of x satisfy it. Let me try to factor the quadratic expression.x² + 2x can be factored as x(x + 2). So, the inequality becomes x(x + 2) ≤ 0. To solve this, I should find the critical points where the expression equals zero. Setting x(x + 2) = 0 gives x = 0 or x = -2. These are the points where the expression changes sign.Now, I need to determine the intervals where the product x(x + 2) is less than or equal to zero. The critical points divide the number line into three intervals:1. x < -22. -2 < x < 03. x > 0I'll test each interval to see where the product is negative or zero.For x < -2, let's pick x = -3:(-3)(-3 + 2) = (-3)(-1) = 3, which is positive. So, this interval doesn't satisfy the inequality.For -2 < x < 0, let's pick x = -1:(-1)(-1 + 2) = (-1)(1) = -1, which is negative. So, this interval satisfies the inequality.For x > 0, let's pick x = 1:(1)(1 + 2) = (1)(3) = 3, which is positive. So, this interval doesn't satisfy the inequality.Now, I should also check the critical points:At x = -2: (-2)(-2 + 2) = (-2)(0) = 0, which satisfies the inequality.At x = 0: (0)(0 + 2) = 0, which also satisfies the inequality.So, the solution to the inequality x² + 2x ≤ 0 is -2 ≤ x ≤ 0. Therefore, set B is {x | -2 ≤ x ≤ 0}.Now, I need to find the intersection of sets A and B, which is A ∩ B. The intersection includes all elements that are in both A and B.Set A is (-1, 2), and set B is [-2, 0]. So, the overlap between these two sets would be from the greater of the two lower bounds to the lesser of the two upper bounds.The lower bounds are -1 (from A) and -2 (from B). The greater of these is -1.The upper bounds are 2 (from A) and 0 (from B). The lesser of these is 0.Therefore, the intersection A ∩ B is the interval from -1 to 0. But I need to be careful about whether the endpoints are included or not.In set A, the lower bound is -1, but it's open, so -1 is not included. In set B, the lower bound is -2, which is included, but since we're taking the intersection, the lower bound for the intersection is -1, and since A doesn't include -1, the intersection doesn't include -1 either.For the upper bound, set A goes up to 2, but set B stops at 0, which is included in B. So, in the intersection, the upper bound is 0, and since B includes 0, the intersection includes 0.So, putting it all together, A ∩ B is the interval (-1, 0], which is all real numbers greater than -1 and less than or equal to 0.Looking back at the options provided:A: {x | 0 < x < 2} – This doesn't make sense because our intersection is between -1 and 0, not 0 and 2.B: {x | 0 ≤ x < 2} – Again, this is between 0 and 2, which isn't where our intersection lies.C: {x | -1 < x < 0} – This is close, but it doesn't include 0, whereas our intersection does include 0.D: {x | -1 < x ≤ 0} – This matches our conclusion exactly. It includes all numbers greater than -1 and up to and including 0.So, the correct answer should be D.

answer：Alright, so I've got these two geometry problems to solve. Let me take them one at a time and think through each step carefully.**Problem 1: Congruent Triangles**Okay, so we have two congruent triangles. Congruent means they're exactly the same in shape and size, right? So their corresponding sides must be equal. One triangle has sides of lengths 6, 8, and 10. The other triangle has sides of lengths 6, 2m - 2, and n + 1. I need to find the values of m and n.First, since the triangles are congruent, their corresponding sides must match. That means each side of the first triangle must equal a side of the second triangle. The first triangle has sides 6, 8, 10, and the second has sides 6, 2m - 2, n + 1.I notice that both triangles have a side of length 6. So that side corresponds to each other. Now, the other sides must correspond as well. So, the sides 8 and 10 in the first triangle must correspond to 2m - 2 and n + 1 in the second triangle. But I'm not sure which one corresponds to which. There are two possibilities:1. 2m - 2 = 8 and n + 1 = 102. 2m - 2 = 10 and n + 1 = 8Let me solve both sets of equations to see which ones make sense.**First Set:**2m - 2 = 8Add 2 to both sides: 2m = 10Divide by 2: m = 5n + 1 = 10Subtract 1: n = 9So, one possible solution is m = 5 and n = 9.**Second Set:**2m - 2 = 10Add 2 to both sides: 2m = 12Divide by 2: m = 6n + 1 = 8Subtract 1: n = 7Another possible solution is m = 6 and n = 7.Wait, so there are two possible solutions? That makes sense because the sides could correspond in two different ways. So, both sets of solutions are valid as long as the triangles are congruent.**Problem 2: Isosceles Triangle with Perimeter 16 and One Side 5**Alright, now this is an isosceles triangle, which means two sides are equal. The perimeter is 16, and one of the sides is 5. I need to find the lengths of the legs and the base.First, let's recall that in an isosceles triangle, the two equal sides are called the legs, and the third side is called the base. But sometimes, people might refer to the two equal sides as legs regardless of their position, so I need to clarify.Given that, there are two cases:1. The side of length 5 is one of the legs (equal sides).2. The side of length 5 is the base (the unequal side).Let me explore both cases.**Case 1: 5 is the length of one leg**If the two legs are each 5 units long, then the base can be found by subtracting the sum of the legs from the perimeter.Perimeter = sum of all sides = 5 + 5 + base = 16So, 10 + base = 16Subtract 10: base = 6So, in this case, the legs are 5 each, and the base is 6.**Case 2: 5 is the length of the base**If the base is 5, then the two legs are equal. Let's denote each leg as x.Perimeter = 5 + x + x = 16So, 5 + 2x = 16Subtract 5: 2x = 11Divide by 2: x = 5.5So, each leg is 5.5 units long, and the base is 5.Wait, but 5.5 seems a bit unusual. Is that okay? Well, triangles can have sides of any positive length as long as they satisfy the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the remaining side.Let me check both cases to make sure they satisfy the triangle inequality.**Checking Case 1: Sides 5, 5, 6**5 + 5 > 6: 10 > 6, which is true.5 + 6 > 5: 11 > 5, true.5 + 6 > 5: Same as above, true.So, this is a valid triangle.**Checking Case 2: Sides 5, 5.5, 5.5**5 + 5.5 > 5.5: 10.5 > 5.5, true.5.5 + 5.5 > 5: 11 > 5, true.5 + 5.5 > 5.5: Same as the first one, true.So, both cases are valid. Therefore, there are two possible sets of lengths for the triangle.But wait, the problem says "find the lengths of the legs and the base." So, depending on which side is 5, the legs and base change.In Case 1, legs are 5, base is 6.In Case 2, legs are 5.5, base is 5.So, both are correct, and the problem doesn't specify whether the given side is a leg or the base, so I think both solutions are acceptable.**Wait, but in the problem statement, it says "find the lengths of the legs and the base." So, it might be expecting both possibilities? Or maybe just one? Hmm.**But since both are possible, I think it's safe to present both solutions.**Summary of Thoughts:**For Problem 1, there are two possible solutions for m and n because the sides could correspond in two different ways.For Problem 2, there are two possible configurations depending on whether the given side is a leg or the base, both of which satisfy the triangle inequality.I think I've covered all the possibilities and checked my work to ensure the solutions make sense. I don't see any mistakes in my calculations or reasoning.**Final Answer**(1) The values of ( m ) and ( n ) are (boxed{m = 5}) and (boxed{n = 9}) or (boxed{m = 6}) and (boxed{n = 7}).(2) The lengths of the legs and the base are (boxed{5}) and (boxed{6}) or (boxed{5.5}) and (boxed{5}).

answer：First, I need to determine the number of stripes each person has on their tennis shoes.Olga has 3 stripes on each shoe, so for two shoes, she has a total of 3 × 2 = 6 stripes.Rick has one less stripe per shoe than Olga, which means he has 3 - 1 = 2 stripes per shoe. For two shoes, Rick has 2 × 2 = 4 stripes.Hortense has double the number of stripes per shoe compared to Olga, so she has 3 × 2 = 6 stripes per shoe. For two shoes, Hortense has 6 × 2 = 12 stripes.Ethan has two more stripes per shoe than Hortense, which means he has 6 + 2 = 8 stripes per shoe. For two shoes, Ethan has 8 × 2 = 16 stripes.Next, I'll calculate the combined total number of stripes:Olga: 6 stripes Rick: 4 stripes Hortense: 12 stripes Ethan: 16 stripes Total stripes = 6 + 4 + 12 + 16 = 38 stripesNow, I'll divide the total number of stripes by 2:38 ÷ 2 = 19Since 19 is already a whole number, there's no need to round up.The final answer is 19 stripes.

answer：First, I'll define the variables:- Let ( r ) be the number of rabbits.- Let ( c ) be the number of chickens.From the problem, I know two things:1. The total number of heads is 51, so:[ r + c = 51 ]2. The total number of rabbit legs is 4 more than three times the total number of chicken legs. Since rabbits have 4 legs and chickens have 2 legs, this gives:[ 4r = 3(2c) + 4 ]Simplifying this equation:[ 4r = 6c + 4 ]Now, I'll solve for one variable in terms of the other. From the first equation:[ c = 51 - r ]Substituting this into the second equation:[ 4r = 6(51 - r) + 4 ][ 4r = 306 - 6r + 4 ][ 4r + 6r = 310 ][ 10r = 310 ][ r = 31 ]So, there are 31 rabbits in the cage.

answer：Okay, so I have this problem about an ellipse and two points on it. Let me try to figure out what's going on here.First, part (1) says that points A and B are both on the line y = 2x + 1, and the midpoint of segment AB has an x-coordinate of -1/3. I need to find the value of 'a'. Hmm, okay.I remember that for an ellipse, the standard equation is (x²/a²) + (y²/b²) = 1. But in the problem, it's given as (x²/a²) + y² = 1, so maybe b² is 1? Or maybe it's a typo? Wait, the problem says a > b and a ≠ 1, so probably b is another parameter. Maybe it's a typo and should be (x²/a²) + (y²/b²) = 1? Hmm, but the problem says (x²/a²) + y² = 1, so maybe b is 1? I'm not sure, but I'll proceed with the given equation.So, points A and B lie on both the ellipse and the line y = 2x + 1. Let me denote A as (x₁, y₁) and B as (x₂, y₂). Since they are on the line, y₁ = 2x₁ + 1 and y₂ = 2x₂ + 1.Also, the midpoint of AB has an x-coordinate of -1/3. The midpoint's x-coordinate is (x₁ + x₂)/2 = -1/3, so x₁ + x₂ = -2/3. Similarly, the y-coordinate of the midpoint would be (y₁ + y₂)/2. Since y = 2x + 1, substituting x = -1/3, the y-coordinate is 2*(-1/3) + 1 = -2/3 + 1 = 1/3. So, (y₁ + y₂)/2 = 1/3, which means y₁ + y₂ = 2/3.But since y₁ = 2x₁ + 1 and y₂ = 2x₂ + 1, adding them gives y₁ + y₂ = 2(x₁ + x₂) + 2 = 2*(-2/3) + 2 = -4/3 + 2 = 2/3. That checks out.Now, since both A and B lie on the ellipse, plugging their coordinates into the ellipse equation:For A: (x₁²/a²) + y₁² = 1For B: (x₂²/a²) + y₂² = 1Subtracting these two equations:(x₁² - x₂²)/a² + (y₁² - y₂²) = 0Factor the differences of squares:(x₁ - x₂)(x₁ + x₂)/a² + (y₁ - y₂)(y₁ + y₂) = 0We know x₁ + x₂ = -2/3 and y₁ + y₂ = 2/3. Also, since A and B lie on the line y = 2x + 1, the slope between A and B is 2. So, (y₁ - y₂)/(x₁ - x₂) = 2, which means y₁ - y₂ = 2(x₁ - x₂).Let me denote x₁ - x₂ as Δx and y₁ - y₂ as Δy. Then Δy = 2Δx.Substituting back into the equation:(Δx)(-2/3)/a² + (Δy)(2/3) = 0But Δy = 2Δx, so:(Δx)(-2/3)/a² + (2Δx)(2/3) = 0Factor out Δx:Δx [ (-2)/(3a²) + 4/3 ] = 0Since x₁ ≠ x₂, Δx ≠ 0, so the term in brackets must be zero:(-2)/(3a²) + 4/3 = 0Multiply both sides by 3a² to eliminate denominators:-2 + 4a² = 0So, 4a² = 2 => a² = 1/2 => a = √(1/2) = √2/2.Alright, that seems to make sense. So, a is √2/2.Now, moving on to part (2). It says to let vectors m = (x₁/a, y₁) and n = (x₂/a, y₂). If m and n are perpendicular at the origin, we need to investigate whether the area of triangle OAB is a constant value.First, if m and n are perpendicular, their dot product is zero. So:m · n = (x₁/a)(x₂/a) + y₁ y₂ = 0Which simplifies to:(x₁ x₂)/a² + y₁ y₂ = 0So, that's the condition we have.We need to find the area of triangle OAB. The area can be calculated using the determinant formula:Area = (1/2) |x₁ y₂ - x₂ y₁|Alternatively, using vectors, the area is (1/2) |m × n|, where × denotes the cross product. Since m and n are in 2D, the magnitude of the cross product is |x₁ y₂ - x₂ y₁|.So, the area is (1/2)|x₁ y₂ - x₂ y₁|.But we have the condition that (x₁ x₂)/a² + y₁ y₂ = 0. Maybe we can relate this to the area.Alternatively, perhaps we can parameterize the points A and B on the ellipse and use the condition of perpendicularity to find a relation.Let me think about parametrizing the ellipse. For an ellipse (x²/a²) + y² = 1, a standard parametrization is x = a cosθ, y = sinθ. So, let me denote point A as (a cosθ, sinθ) and point B as (a cosφ, sinφ).Then, vectors m and n become:m = (cosθ, sinθ)n = (cosφ, sinφ)Their dot product is cosθ cosφ + sinθ sinφ = cos(θ - φ). Since they are perpendicular, cos(θ - φ) = 0, which implies θ - φ = ±π/2.So, φ = θ ± π/2.So, point B is (a cos(θ ± π/2), sin(θ ± π/2)).But cos(θ ± π/2) = ∓sinθ and sin(θ ± π/2) = ±cosθ.Therefore, point B can be written as (-a sinθ, cosθ) or (a sinθ, -cosθ). Let's take the first one: (-a sinθ, cosθ).So, points A and B are:A: (a cosθ, sinθ)B: (-a sinθ, cosθ)Now, let's compute the area of triangle OAB.Using the determinant formula:Area = (1/2) |x₁ y₂ - x₂ y₁|Substituting the coordinates:x₁ = a cosθ, y₁ = sinθx₂ = -a sinθ, y₂ = cosθSo,Area = (1/2) |(a cosθ)(cosθ) - (-a sinθ)(sinθ)|= (1/2) |a cos²θ + a sin²θ|= (1/2) |a (cos²θ + sin²θ)|= (1/2) |a (1)|= a/2Wow, so the area is a/2, which is a constant value independent of θ. So, regardless of the points A and B, as long as vectors m and n are perpendicular, the area of triangle OAB is always a/2.That's pretty neat! So, the area is indeed a constant value, and it's equal to a/2.I think that's the answer. Let me just double-check my steps.1. I parametrized the ellipse using standard parametric equations.2. Expressed vectors m and n in terms of θ and φ.3. Used the perpendicular condition to find that φ = θ ± π/2.4. Expressed point B in terms of θ.5. Calculated the area using the determinant formula.6. Simplified and found that the area is a/2.Everything seems to check out. So, yeah, the area is a constant value of a/2.**Final Answer**(1) boxed{dfrac{sqrt{2}}{2}}(2) The area of triangle OAB is a constant value of boxed{dfrac{a}{2}}.

answer：Okay, so I have this problem here where I need to figure out which of the four given expressions is the largest. The expressions are all combinations of tangent, cotangent, sine, and cosine of 48 degrees. Let me write them down again to make sure I have them right:(A) tan 48° + cot 48°(B) sin 48° + cos 48°(C) tan 48° + cos 48°(D) cot 48° + sin 48°Alright, so I need to compare these four expressions and determine which one is the biggest. Hmm, I remember that for angles between 0° and 90°, sine and cosine are positive, and tangent and cotangent are also positive. So all these expressions are positive numbers.Let me think about the values of these trigonometric functions at 48 degrees. I know that 48 degrees is less than 90 degrees, so all the functions are positive. I also recall that as the angle increases from 0° to 90°, sine increases, cosine decreases, tangent increases, and cotangent decreases.So, at 48 degrees, which is more than 45 degrees, I think that tan 48° is greater than 1 because tan 45° is 1, and it increases beyond that. Similarly, cot 48° would be less than 1 because cotangent decreases as the angle increases. So, tan 48° is greater than 1, and cot 48° is less than 1.Similarly, sin 48° is greater than sin 45°, which is √2/2 ≈ 0.707, so sin 48° is a bit higher, maybe around 0.74 or something. Cos 48° is less than cos 45°, so it's a bit less than 0.707, maybe around 0.66 or so.So, let me try to approximate these values to get a sense of which expressions might be larger.First, let's approximate tan 48° and cot 48°. I know that tan 45° is 1, and tan 60° is √3 ≈ 1.732. Since 48° is closer to 45°, tan 48° should be a bit more than 1. Maybe around 1.1 or 1.2? Let me check.Wait, actually, I think tan 48° is approximately 1.1106. Let me confirm that. Yes, using a calculator, tan 48° ≈ 1.1106. So, cot 48° is the reciprocal of tan 48°, which would be approximately 1 / 1.1106 ≈ 0.8999, roughly 0.9.So, tan 48° ≈ 1.1106 and cot 48° ≈ 0.9.Now, sin 48° is approximately 0.7431, and cos 48° is approximately 0.6691.So, let me write down these approximate values:tan 48° ≈ 1.1106cot 48° ≈ 0.9sin 48° ≈ 0.7431cos 48° ≈ 0.6691Now, let's compute each expression:(A) tan 48° + cot 48° ≈ 1.1106 + 0.9 ≈ 2.0106(B) sin 48° + cos 48° ≈ 0.7431 + 0.6691 ≈ 1.4122(C) tan 48° + cos 48° ≈ 1.1106 + 0.6691 ≈ 1.7797(D) cot 48° + sin 48° ≈ 0.9 + 0.7431 ≈ 1.6431So, from these approximations, it looks like expression (A) is the largest, followed by (C), then (B), and then (D). So, (A) is the largest.But wait, maybe I should think about this without relying solely on approximations. Maybe there's a trigonometric identity or inequality that can help me compare these expressions more precisely.I remember that for any angle θ, tan θ + cot θ can be expressed in terms of sine and cosine. Let me write that out:tan θ + cot θ = (sin θ / cos θ) + (cos θ / sin θ) = (sin² θ + cos² θ) / (sin θ cos θ)But sin² θ + cos² θ = 1, so this simplifies to 1 / (sin θ cos θ).So, tan θ + cot θ = 1 / (sin θ cos θ)Hmm, interesting. So, expression (A) is equal to 1 divided by (sin θ cos θ). Let me see if I can relate this to the other expressions.Expression (B) is sin θ + cos θ. I know that sin θ + cos θ can be written as √2 sin(θ + 45°), which has a maximum value of √2 ≈ 1.4142. So, for θ = 48°, sin 48° + cos 48° is less than √2, which is consistent with my earlier approximation of approximately 1.4122.Expression (C) is tan θ + cos θ. Since tan θ is greater than 1 and cos θ is less than 1, this is going to be a bit more than 1.1106 + 0.6691 ≈ 1.7797.Expression (D) is cot θ + sin θ. Cot θ is less than 1, and sin θ is less than 1, so this is going to be a bit less than 0.9 + 0.7431 ≈ 1.6431.So, expression (A) is 1 / (sin θ cos θ). Let me compute sin θ cos θ for θ = 48°. I know that sin θ cos θ = (1/2) sin 2θ, so sin 48° cos 48° = (1/2) sin 96°. Since sin 96° is approximately sin 90° = 1, but slightly less. Let me compute sin 96°.Sin 96° is sin(90° + 6°) = cos 6° ≈ 0.9945. So, sin 96° ≈ 0.9945, so sin 48° cos 48° ≈ (1/2)(0.9945) ≈ 0.49725.Therefore, expression (A) is 1 / 0.49725 ≈ 2.0106, which matches my earlier approximation.So, expression (A) is approximately 2.0106, which is clearly larger than the other expressions, which are all less than 2.Wait, but just to make sure, let me think about whether there's a way to compare these expressions without calculating their approximate values.I know that tan θ + cot θ is always greater than or equal to 2 for any angle θ where tan θ and cot θ are defined, due to the AM-GM inequality. Because tan θ and cot θ are reciprocals, their arithmetic mean is greater than or equal to their geometric mean, which is 1. So, (tan θ + cot θ)/2 ≥ √(tan θ cot θ) = 1, so tan θ + cot θ ≥ 2.Since 48° is not 45°, tan 48° ≠ cot 48°, so the inequality is strict: tan 48° + cot 48° > 2.On the other hand, sin θ + cos θ has a maximum value of √2 ≈ 1.4142, which is less than 2. So, expression (B) is less than 2.Similarly, expressions (C) and (D) are sums of one function greater than 1 and one less than 1, but their sums are still less than 2. For example, tan 48° ≈ 1.11, and cos 48° ≈ 0.67, so their sum is about 1.78, which is less than 2. Similarly, cot 48° ≈ 0.9, and sin 48° ≈ 0.74, so their sum is about 1.64, which is also less than 2.Therefore, expression (A) is greater than 2, while the others are less than 2. Hence, expression (A) is the largest.Wait, but just to double-check, is there any chance that one of the other expressions could somehow be greater than 2? Let me think.Expression (C): tan θ + cos θ. Since tan θ is about 1.11 and cos θ is about 0.67, their sum is about 1.78, which is less than 2.Expression (D): cot θ + sin θ. Cot θ is about 0.9, and sin θ is about 0.74, so their sum is about 1.64, which is also less than 2.Expression (B): sin θ + cos θ is about 1.41, which is less than 2.So, none of the other expressions reach 2, while expression (A) is just over 2. Therefore, expression (A) is indeed the largest.I think that's solid reasoning. I used both approximate calculations and trigonometric identities to confirm that expression (A) is the largest. I don't see any flaws in this reasoning, so I'm confident that (A) is the correct answer.