💡Okay, so I've got this problem here about 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}. Hmm, that means A includes all real numbers x such that x is greater than -1 and less than or equal to 2. So, in other words, A is all the numbers starting just above -1 up to and including 2. That would be like the interval (-1, 2]. I think that's how intervals work—parentheses mean the endpoint isn't included, and brackets mean it is included.Set B is given as {-1, 0, 1}. That's straightforward—it's just a set containing the integers -1, 0, and 1. So B has three elements: -1, 0, and 1.Now, I need to find the intersection of A and B, which is A ∩ B. The intersection of two sets is the set of elements that are common to both sets. So, I need to see which elements are in both A and B.Let's list out the elements of B: -1, 0, 1. Now, I need to check which of these are also in A.Starting with -1: Is -1 in A? Well, A is defined as x > -1. So, -1 is not greater than -1; it's equal to -1. Since the condition is x > -1, not x ≥ -1, -1 is not included in A. So, -1 is not in A ∩ B.Next, 0: Is 0 in A? Yes, because 0 is greater than -1 and less than 2. So, 0 is definitely in A. Therefore, 0 is in A ∩ B.Then, 1: Is 1 in A? Yes, because 1 is greater than -1 and less than 2. So, 1 is also in A. Therefore, 1 is in A ∩ B.So, from set B, the elements that are also in A are 0 and 1. Therefore, A ∩ B should be {0, 1}.Let me double-check to make sure I didn't miss anything. Set A doesn't include -1 because it's strictly greater than -1, so -1 is out. 0 and 1 are both within the range of A, so they stay. 2 is in A, but it's not in B, so it doesn't matter for the intersection.So, yeah, I think the intersection of A and B is {0, 1}. That should be option C.

💡First, I need to reverse the order of the numbers in the first and third rows of the given 3x3 matrix. Original matrix:[begin{array}{|c|c|c|}hline5 & 6 & 7 hline8 & 9 & 10 hline11 & 12 & 13 hlineend{array}]After reversing the first and third rows, the matrix becomes:[begin{array}{|c|c|c|}hline7 & 6 & 5 hline8 & 9 & 10 hline13 & 12 & 11 hlineend{array}]Next, I need to calculate the sum of the numbers on the main diagonal (from top left to bottom right) and the numbers occupying the edge positions of the matrix.- **Main diagonal elements**: 7, 9, 11- **Edge elements (excluding diagonal corners)**: 6, 5, 13, 12Now, I will add these numbers together:[7 + 9 + 11 + 6 + 5 + 13 + 12 = 63]Therefore, the required sum is 63.

💡Okay, so I have this problem about a geometric sequence. Let me try to understand what it's asking. It says that the terms of the geometric sequence {a_n} are positive, and the common ratio is q. It also tells me that q squared equals 4, and I need to find the value of (a₃ + a₄)/(a₄ + a₅). Hmm, okay.First, let me recall what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, which is q in this case. So, the nth term of a geometric sequence can be expressed as a_n = a₁ * q^(n-1), where a₁ is the first term.Given that q² = 4, I can solve for q. Since q squared is 4, q must be either 2 or -2. But the problem mentions that all the terms of the sequence are positive. If q were negative, then the terms would alternate between positive and negative, right? So, since all terms are positive, q must be positive. Therefore, q = 2.Now, I need to find (a₃ + a₄)/(a₄ + a₅). Let me write out expressions for each of these terms using the formula for the nth term.a₃ = a₁ * q^(3-1) = a₁ * q²a₄ = a₁ * q^(4-1) = a₁ * q³a₅ = a₁ * q^(5-1) = a₁ * q⁴So, substituting these into the expression:(a₃ + a₄)/(a₄ + a₅) = (a₁ * q² + a₁ * q³)/(a₁ * q³ + a₁ * q⁴)I notice that both the numerator and the denominator have a common factor of a₁ * q². Let me factor that out to simplify.Numerator: a₁ * q² (1 + q)Denominator: a₁ * q³ (1 + q)So, the expression becomes:(a₁ * q² (1 + q)) / (a₁ * q³ (1 + q))Now, I can cancel out the common terms in the numerator and the denominator. The a₁ terms cancel out, and the (1 + q) terms also cancel out. That leaves me with:q² / q³Which simplifies to 1/q.Since we already found that q = 2, substituting that in gives:1/2So, the value of (a₃ + a₄)/(a₄ + a₅) is 1/2.Wait, let me double-check to make sure I didn't make any mistakes. I started by finding q from q² = 4, which gave me q = 2 because all terms are positive. Then I expressed each term in the sequence using the formula for the nth term. I substituted these into the given expression and factored out the common terms. After simplifying, I ended up with 1/q, which is 1/2. That seems correct.Alternatively, I can think about the ratio of consecutive terms. Since each term is multiplied by q, the ratio of a₃ to a₄ is 1/q, a₄ to a₅ is also 1/q, and so on. So, when I take the sum of a₃ and a₄ in the numerator and the sum of a₄ and a₅ in the denominator, each pair is essentially a₄(1/q + 1) over a₅(1/q + 1). Then, since a₅ is a₄ * q, the ratio becomes (1/q + 1)/(q*(1/q + 1)) which simplifies to 1/q. So, that confirms my earlier result.Yeah, I think that's solid. The answer should be 1/2.**Final Answer**The value of frac{a_{3}+a_{4}}{a_{4}+a_{5}} is boxed{dfrac{1}{2}}.

💡Okay, so I need to find the smallest positive period of the function y = cos(2x + π/6). Hmm, I remember that for cosine functions, the period is usually 2π, but when there's a coefficient inside the cosine, it changes the period. Let me think about how that works.I recall that the general form of a cosine function is y = cos(Bx + C), and the period of this function is 2π divided by the absolute value of B. So, in this case, B is 2 because the function is cos(2x + π/6). That means the period should be 2π divided by 2, which is π. Wait, but let me make sure I'm not missing anything. The phase shift is π/6, but does that affect the period? I don't think so because the phase shift just moves the graph left or right, it doesn't change how often the function repeats. So, the period should still be determined by the coefficient in front of x, which is 2.Let me double-check by plugging in some values. If the period is π, then adding π to x should give the same value of the function. Let's take x = 0. Then y = cos(0 + π/6) = cos(π/6) = √3/2. Now, if I add π to x, I get x = π. Then y = cos(2π + π/6) = cos(13π/6). Cosine has a period of 2π, so cos(13π/6) is the same as cos(13π/6 - 2π) = cos(π/6) = √3/2. So, that checks out.What if I pick another point, like x = π/4. Then y = cos(2*(π/4) + π/6) = cos(π/2 + π/6) = cos(2π/3) = -1/2. Adding π to x gives x = 5π/4. Then y = cos(2*(5π/4) + π/6) = cos(5π/2 + π/6) = cos(16π/6) = cos(8π/3). Subtracting 2π (which is 6π/3) gives cos(2π/3) = -1/2. So, that also matches.I think I'm confident now that the period is indeed π. The phase shift doesn't affect the period, only the coefficient in front of x does. So, the smallest positive period is π.

💡Alright, so I've got this problem here about the golden section point on a line segment of length 10. I need to find the length of the shorter segment, and there are four options to choose from: A, B, C, and D. Let me try to figure this out step by step.First, I remember that the golden section, or golden ratio, is a special way of dividing a line segment into two parts such that the ratio of the whole segment to the longer part is the same as the ratio of the longer part to the shorter part. Mathematically, if we have a line segment of length ( L ) divided into two parts, ( a ) (the longer part) and ( b ) (the shorter part), then the golden ratio is defined as:[frac{L}{a} = frac{a}{b}]In this problem, the total length ( L ) is 10. So, substituting that in, we have:[frac{10}{a} = frac{a}{b}]Cross-multiplying gives:[a^2 = 10b]But I also know that the total length is the sum of the two parts:[a + b = 10]So now I have two equations:1. ( a^2 = 10b )2. ( a + b = 10 )I can use the second equation to express ( b ) in terms of ( a ):[b = 10 - a]Substituting this into the first equation:[a^2 = 10(10 - a)]Expanding the right side:[a^2 = 100 - 10a]Bringing all terms to one side to form a quadratic equation:[a^2 + 10a - 100 = 0]Now, I need to solve this quadratic equation for ( a ). Using the quadratic formula:[a = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Wait, that's confusing because the variable here is also ( a ). Let me rewrite the quadratic equation with different coefficients to avoid confusion.Let me denote the quadratic equation as:[x^2 + 10x - 100 = 0]Here, ( x ) represents the longer segment ( a ). Applying the quadratic formula:[x = frac{-10 pm sqrt{(10)^2 - 4(1)(-100)}}{2(1)}]Calculating the discriminant:[sqrt{100 + 400} = sqrt{500} = 10sqrt{5}]So, the solutions are:[x = frac{-10 pm 10sqrt{5}}{2}]Simplifying:[x = -5 pm 5sqrt{5}]Since length can't be negative, we take the positive solution:[x = -5 + 5sqrt{5}]So, the longer segment ( a ) is ( 5sqrt{5} - 5 ). Now, to find the shorter segment ( b ), we subtract ( a ) from the total length:[b = 10 - a = 10 - (5sqrt{5} - 5) = 10 - 5sqrt{5} + 5 = 15 - 5sqrt{5}]Looking at the options provided:A: ( 5sqrt{5} - 5 )B: ( 30 - 10sqrt{5} )C: ( 15 - 5sqrt{5} )D: ( 10sqrt{5} - 20 )The shorter segment ( b ) is ( 15 - 5sqrt{5} ), which corresponds to option C.Wait a second, let me double-check my calculations to make sure I didn't make a mistake. I started with the golden ratio definition, set up the equations correctly, solved the quadratic, and found the positive solution for ( a ). Then I subtracted ( a ) from 10 to get ( b ), which gave me ( 15 - 5sqrt{5} ). That seems right.Alternatively, I recall that the golden ratio ( phi ) is approximately 1.618, so the longer segment should be about 1.618 times the shorter segment. Let me check if ( 15 - 5sqrt{5} ) is approximately the shorter segment.Calculating ( 5sqrt{5} ):[5 times 2.236 approx 11.18]So, ( 15 - 11.18 approx 3.82 ).Now, the longer segment ( a ) is ( 5sqrt{5} - 5 approx 11.18 - 5 = 6.18 ).Adding them together: ( 3.82 + 6.18 = 10 ), which matches the total length. Also, checking the ratio:[frac{a}{b} = frac{6.18}{3.82} approx 1.618]Which is indeed the golden ratio. So, my calculations seem correct.Therefore, the length of the shorter segment is ( 15 - 5sqrt{5} ), which is option C.

💡Okay, so I have this problem about triangle ABC where a, b, and c are the lengths opposite angles A, B, and C respectively. They tell me that a is 2, and there's this equation involving sines: (sin A - sin B)/sin C equals (c - b)/(2 + b). I need to find the maximum area of triangle ABC.Hmm, let me start by writing down what I know. The Law of Sines says that in any triangle, a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe that can help me relate the sines of the angles to the side lengths.Given that a = 2, so 2/sin A = b/sin B = c/sin C. Let me denote this common ratio as 2R, so 2 = 2R sin A, which means R = 1/sin A. Similarly, b = 2R sin B and c = 2R sin C.Wait, maybe I can express sin A, sin B, and sin C in terms of a, b, c. Since a = 2, sin A = a/(2R) = 2/(2R) = 1/R. Similarly, sin B = b/(2R) and sin C = c/(2R).Let me substitute these into the given equation: (sin A - sin B)/sin C = (c - b)/(2 + b).Substituting the expressions in terms of a, b, c:( (1/R) - (b/(2R)) ) / (c/(2R)) = (c - b)/(2 + b)Simplify numerator: (1/R - b/(2R)) = (2 - b)/(2R)Denominator: c/(2R)So the left side becomes: (2 - b)/(2R) divided by c/(2R) = (2 - b)/cSo the equation simplifies to (2 - b)/c = (c - b)/(2 + b)Okay, so now I have (2 - b)/c = (c - b)/(2 + b). Let me cross-multiply to eliminate the fractions:(2 - b)(2 + b) = c(c - b)Left side: (2 - b)(2 + b) = 4 - b²Right side: c² - bcSo, 4 - b² = c² - bcLet me rearrange this equation: 4 = c² - bc + b²Hmm, that's 4 = b² + c² - bcWait, that looks similar to the Law of Cosines. The Law of Cosines says a² = b² + c² - 2bc cos A. Since a = 2, we have 4 = b² + c² - 2bc cos A.But in our equation, we have 4 = b² + c² - bc. So comparing the two, 2bc cos A = bc, which implies that cos A = 1/2.So, cos A = 1/2, which means angle A is 60 degrees or π/3 radians because that's where cosine is 1/2.Okay, so angle A is 60 degrees. That might be useful for calculating the area later.Now, I need to find the maximum area of triangle ABC. The area of a triangle can be given by (1/2)ab sin C, but in this case, since we have sides a, b, c and angle A, maybe it's better to use the formula (1/2)bc sin A.Since angle A is 60 degrees, sin A is √3/2. So the area is (1/2) * b * c * (√3/2) = (√3/4) * b * c.So, to maximize the area, I need to maximize the product bc. So, the problem reduces to finding the maximum value of bc given the constraint 4 = b² + c² - bc.Hmm, how can I maximize bc given that equation? Maybe I can use some algebraic manipulation or calculus.Let me consider the equation 4 = b² + c² - bc. I can think of this as a quadratic in terms of c: c² - bc + (b² - 4) = 0.Alternatively, maybe I can express c in terms of b or vice versa and then find the maximum.Alternatively, I can use the method of Lagrange multipliers, but that might be overkill. Maybe completing the square or using AM-GM inequality.Wait, let me think about AM-GM. The equation is 4 = b² + c² - bc. I need to find the maximum of bc.Let me denote bc as k. Then, 4 = b² + c² - k. But I don't know if that helps directly.Alternatively, maybe I can write b² + c² - bc = 4. Let me consider that as a quadratic in terms of b or c.Alternatively, I can use substitution. Let me set c = tb, where t is a positive real number. Then, substitute into the equation:4 = b² + (tb)² - b(tb) = b² + t²b² - tb² = b²(1 + t² - t)So, 4 = b²(1 + t² - t)Therefore, b² = 4 / (1 + t² - t)So, bc = b * tb = t b² = t * (4 / (1 + t² - t)) = 4t / (1 + t² - t)So, now, I need to maximize 4t / (1 + t² - t) with respect to t.Let me denote f(t) = 4t / (1 + t² - t). I need to find the maximum of f(t).To find the maximum, I can take the derivative of f(t) with respect to t and set it to zero.First, compute f'(t):f'(t) = [4(1 + t² - t) - 4t(2t - 1)] / (1 + t² - t)^2Simplify numerator:4(1 + t² - t) - 4t(2t - 1) = 4 + 4t² - 4t - 8t² + 4t = 4 + (4t² - 8t²) + (-4t + 4t) = 4 - 4t²So, f'(t) = (4 - 4t²) / (1 + t² - t)^2Set f'(t) = 0:4 - 4t² = 0 => 4t² = 4 => t² = 1 => t = 1 or t = -1But since t is positive (as c and b are lengths), t = 1.So, the maximum occurs at t = 1. Let me check the second derivative or test intervals to confirm it's a maximum.But intuitively, since the function f(t) tends to zero as t approaches 0 and as t approaches infinity, and there's only one critical point at t=1, it must be a maximum.So, at t=1, bc = 4*1 / (1 + 1 -1) = 4 /1 =4.So, the maximum value of bc is 4.Therefore, the maximum area is (√3/4)*4 = √3.Wait, but let me double-check. When t=1, c = b, so the triangle is isoceles with b=c.From the equation 4 = b² + c² - bc, if b=c, then 4 = 2b² - b² = b², so b²=4, so b=2, c=2.So, sides a=2, b=2, c=2. Wait, that's an equilateral triangle.But in that case, all angles are 60 degrees, which matches our earlier conclusion that angle A is 60 degrees.So, the maximum area occurs when the triangle is equilateral with all sides equal to 2, giving an area of (√3/4)*2² = √3.Wait, but hold on, the area formula I used was (√3/4)*b*c, but when b=c=2, that's (√3/4)*4 = √3, which is correct.Alternatively, since it's an equilateral triangle, area is (√3/4)*a² = (√3/4)*4 = √3.So, that seems consistent.But let me make sure that when t=1, we indeed get the maximum. Suppose t approaches 0, then bc approaches 0, so area approaches 0. If t approaches infinity, bc also approaches 0 because denominator grows faster. So, yes, t=1 gives the maximum.Therefore, the maximum area is √3.**Final Answer**The maximum area of triangle ABC is boxed{sqrt{3}}.