💡Okay, so I've got this problem here about a half-sector of a circle being rolled up into a cone, and I need to find the volume of that cone. Hmm, let me try to visualize this. A half-sector means it's like half of a circle, right? So if the full circle has a radius of 6 inches, then the half-sector would have a radius of 6 inches as well, but only half the circumference.Alright, so when they roll this half-sector up into a cone, the two radii of the sector get taped together. That must form the lateral surface of the cone. I remember that the lateral surface area of a cone is given by πrl, where r is the radius of the base of the cone and l is the slant height. In this case, the slant height should be the same as the radius of the original sector, which is 6 inches. So, l = 6 inches.Now, the arc length of the half-sector becomes the circumference of the base of the cone. Let me calculate that. The circumference of the full circle would be 2π times the radius, so that's 2π*6 = 12π inches. Since it's a half-sector, the arc length is half of that, which is 6π inches. This arc length is equal to the circumference of the base of the cone, which is 2πr, where r is the radius of the base of the cone.So, setting up the equation: 2πr = 6π. If I divide both sides by π, I get 2r = 6, so r = 3 inches. Got it, the radius of the base of the cone is 3 inches.Next, I need to find the height of the cone to calculate the volume. I remember that in a right circular cone, the slant height (l), the radius (r), and the height (h) form a right triangle. So, using the Pythagorean theorem: l² = r² + h². Plugging in the known values: 6² = 3² + h². That simplifies to 36 = 9 + h². Subtracting 9 from both sides gives h² = 27. Taking the square root of both sides, h = √27. Simplifying that, √27 is equal to 3√3 inches.Now, with the radius and height known, I can find the volume of the cone. The formula for the volume of a cone is (1/3)πr²h. Plugging in the values: (1/3)π*(3)²*(3√3). Let's compute that step by step. First, 3 squared is 9. Then, multiplying by 3√3 gives 9*3√3 = 27√3. Now, multiplying by (1/3) gives (1/3)*27√3 = 9√3. So, the volume is 9√3π cubic inches.Let me double-check my steps to make sure I didn't make any mistakes. The arc length became the circumference, which gave me the radius of the base. Then, using the Pythagorean theorem with the slant height, I found the height. Plugging those into the volume formula gave me 9√3π. That seems right.I think I'm confident with this answer. It matches one of the options provided, which is option B: 9π√3.**Final Answer**boxed{B}

💡Alright, so I've got these four propositions to evaluate, and I need to figure out which ones are true and which ones aren't. Let me go through each one step by step.Starting with the first proposition: "If p and q are two propositions, and 'p OR q' is false, then 'NOT p AND NOT q' is true." Hmm, okay. So, in logic, 'p OR q' being false means that both p and q have to be false individually because for an OR statement to be false, both components need to be false. If p is false, then 'NOT p' is true, and similarly, if q is false, then 'NOT q' is true. So, 'NOT p AND NOT q' would be 'true AND true', which is true. That makes sense. So, proposition 1 seems correct.Moving on to proposition 2: "If the function y = f(x + 1) is an even function, then the graph of y = f(x) is symmetric about the line x = 1." Okay, let's unpack this. An even function satisfies f(-x) = f(x). So, if y = f(x + 1) is even, that means f(-(x) + 1) = f(x + 1). Wait, is that right? Let me think. If we have f(x + 1) being even, then f(-x + 1) should equal f(x + 1). So, that would imply that the function f is symmetric around x = 1. Because if you take a point x to the right of 1, it mirrors to a point x to the left of 1. Yeah, that seems correct. So, proposition 2 is also correct.Now, proposition 3: "The graph of the function y = f(x) has at most one intersection with the line x = a." Okay, so the line x = a is a vertical line. For a function, by definition, for each x-value, there's at most one y-value. So, when you draw a vertical line x = a, it can intersect the graph of y = f(x) at most once. That's the vertical line test for functions. So, this proposition is definitely true.Finally, proposition 4: "The contrapositive statement of 'If x ≠ y, then sin x ≠ sin y' is true." Alright, the contrapositive of "If P, then Q" is "If not Q, then not P." So, the contrapositive here would be "If sin x = sin y, then x = y." But wait, is that true? I know that sine is periodic, so sin x = sin y can happen even if x and y are different. For example, sin(0) = sin(π), but 0 ≠ π. So, the contrapositive isn't necessarily true because there are cases where sin x = sin y without x = y. Therefore, proposition 4 is incorrect.So, summarizing my thoughts: propositions 1, 2, and 3 are correct, while proposition 4 is incorrect.

💡Okay, so I need to find the coefficient of ( x^2 ) in the expansion of ( left(1 - frac{1}{2}xright)left(1 + 2sqrt{x}right)^5 ). Hmm, let me think about how to approach this step by step.First, I remember that when you have a product of two expressions, the coefficient of a specific term in the expansion can be found by considering how each term in the first expression multiplies with terms in the second expression to give the desired term. In this case, the first expression is ( 1 - frac{1}{2}x ) and the second expression is ( left(1 + 2sqrt{x}right)^5 ).So, I need to expand ( left(1 + 2sqrt{x}right)^5 ) first, or at least find the coefficients of the terms that, when multiplied by ( 1 ) and ( -frac{1}{2}x ) from the first expression, will result in an ( x^2 ) term.Let me recall the binomial theorem, which states that ( (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k ). Applying this to ( left(1 + 2sqrt{x}right)^5 ), I can write:[left(1 + 2sqrt{x}right)^5 = sum_{k=0}^{5} binom{5}{k} cdot 1^{5 - k} cdot left(2sqrt{x}right)^k]Simplifying this, since ( 1^{5 - k} = 1 ), it becomes:[sum_{k=0}^{5} binom{5}{k} cdot 2^k cdot x^{k/2}]So, each term in the expansion is ( binom{5}{k} cdot 2^k cdot x^{k/2} ). Now, I need to find the coefficients of ( x^1 ) and ( x^2 ) in this expansion because when I multiply by ( 1 ) and ( -frac{1}{2}x ) respectively, they will contribute to the ( x^2 ) term in the final product.Let me find the coefficient of ( x^1 ) first. For ( x^{k/2} = x^1 ), we have ( k/2 = 1 ), so ( k = 2 ). Therefore, the coefficient is:[binom{5}{2} cdot 2^2 = 10 cdot 4 = 40]Next, I need the coefficient of ( x^2 ). For ( x^{k/2} = x^2 ), we have ( k/2 = 2 ), so ( k = 4 ). The coefficient is:[binom{5}{4} cdot 2^4 = 5 cdot 16 = 80]Okay, so in the expansion of ( left(1 + 2sqrt{x}right)^5 ), the coefficient of ( x ) is 40 and the coefficient of ( x^2 ) is 80.Now, going back to the original expression ( left(1 - frac{1}{2}xright)left(1 + 2sqrt{x}right)^5 ), I can think of this as multiplying two polynomials. The ( x^2 ) term in the product will come from two sources:1. Multiplying the constant term ( 1 ) from the first expression by the ( x^2 ) term in the second expression.2. Multiplying the ( -frac{1}{2}x ) term from the first expression by the ( x ) term in the second expression.So, the coefficient of ( x^2 ) in the product will be:[1 times text{(coefficient of } x^2 text{ in the second expression)} + left(-frac{1}{2}right) times text{(coefficient of } x text{ in the second expression)}]Plugging in the values I found earlier:[1 times 80 + left(-frac{1}{2}right) times 40 = 80 - 20 = 60]Therefore, the coefficient of ( x^2 ) in the expansion is 60.Wait, let me double-check my calculations to make sure I didn't make a mistake. So, binomial coefficients for ( k = 2 ) and ( k = 4 ):For ( k = 2 ):[binom{5}{2} = 10, quad 2^2 = 4, quad 10 times 4 = 40]That seems correct.For ( k = 4 ):[binom{5}{4} = 5, quad 2^4 = 16, quad 5 times 16 = 80]Also correct.Then, multiplying by 1 and -1/2:[80 times 1 = 80, quad 40 times (-1/2) = -20]Adding them together: 80 - 20 = 60.Yes, that seems right. I don't think I made any errors in the calculations. So, I'm confident that the coefficient is 60.**Final Answer**The coefficient of ( x^2 ) is boxed{60}.

💡Okay, so I have this problem where I need to find the minimum value of the expression (x^2 + 4xy + 4y^2 + 2z^2) given that (xyz = 32) and (x), (y), and (z) are positive real numbers. Hmm, let me think about how to approach this.First, I notice that the expression (x^2 + 4xy + 4y^2) looks familiar. It seems like it could be a perfect square. Let me check that:(x^2 + 4xy + 4y^2 = (x + 2y)^2). Yeah, that works because ((x + 2y)^2 = x^2 + 4xy + 4y^2). So, the original expression simplifies to ((x + 2y)^2 + 2z^2).Now, I need to minimize ((x + 2y)^2 + 2z^2) with the constraint (xyz = 32). I remember that the AM-GM inequality is useful for finding minima or maxima of expressions involving products and sums. Maybe I can apply it here.Let me recall the AM-GM inequality: For non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, for positive numbers (a) and (b), (frac{a + b}{2} geq sqrt{ab}), with equality when (a = b).Looking at ((x + 2y)^2), I can think of (x + 2y) as a single term. Maybe I can apply AM-GM to (x) and (2y). Let's try that:(frac{x + 2y}{2} geq sqrt{x cdot 2y}). Multiplying both sides by 2, we get (x + 2y geq 2sqrt{2xy}). Squaring both sides, ((x + 2y)^2 geq 8xy).So, substituting back into the original expression, we have:((x + 2y)^2 + 2z^2 geq 8xy + 2z^2).Now, I need to minimize (8xy + 2z^2). Hmm, this still has three variables, but I have the constraint (xyz = 32). Maybe I can express one variable in terms of the others using the constraint.Let me solve for (z) from (xyz = 32): (z = frac{32}{xy}).Substituting this into (8xy + 2z^2), we get:(8xy + 2left(frac{32}{xy}right)^2 = 8xy + 2left(frac{1024}{x^2 y^2}right) = 8xy + frac{2048}{x^2 y^2}).Hmm, now I have an expression in terms of (xy). Let me set (t = xy) to simplify the expression:(8t + frac{2048}{t^2}).Now, I need to minimize (8t + frac{2048}{t^2}) with respect to (t). To do this, I can take the derivative with respect to (t) and set it equal to zero.Let me compute the derivative:(frac{d}{dt}left(8t + frac{2048}{t^2}right) = 8 - frac{4096}{t^3}).Setting this equal to zero:(8 - frac{4096}{t^3} = 0).Solving for (t):(8 = frac{4096}{t^3})Multiply both sides by (t^3):(8t^3 = 4096)Divide both sides by 8:(t^3 = 512)Take the cube root:(t = 8).So, the minimum occurs when (t = 8), which means (xy = 8).Now, recalling that (z = frac{32}{xy}), substituting (xy = 8):(z = frac{32}{8} = 4).So, (z = 4). Now, I need to find (x) and (y) such that (xy = 8). But I also have the earlier inequality where (x + 2y geq 2sqrt{2xy}). Since we're looking for the minimum, equality should hold, so (x + 2y = 2sqrt{2xy}).Given (xy = 8), let's compute (2sqrt{2xy}):(2sqrt{2 times 8} = 2sqrt{16} = 2 times 4 = 8).So, (x + 2y = 8). But we also have (xy = 8). Now, we can set up the equations:1. (x + 2y = 8)2. (xy = 8)Let me solve these equations. From equation 1, (x = 8 - 2y). Substitute into equation 2:((8 - 2y)y = 8)(8y - 2y^2 = 8)Divide both sides by 2:(4y - y^2 = 4)Rearrange:(-y^2 + 4y - 4 = 0)Multiply both sides by -1:(y^2 - 4y + 4 = 0)This factors as:((y - 2)^2 = 0)So, (y = 2). Then, from equation 1, (x = 8 - 2(2) = 8 - 4 = 4).So, (x = 4), (y = 2), and (z = 4).Now, let's compute the original expression with these values:(x^2 + 4xy + 4y^2 + 2z^2 = 4^2 + 4(4)(2) + 4(2)^2 + 2(4)^2)Calculate each term:- (4^2 = 16)- (4(4)(2) = 32)- (4(2)^2 = 16)- (2(4)^2 = 32)Add them up:(16 + 32 + 16 + 32 = 96).So, the minimum value is 96.Wait, let me double-check if I applied the AM-GM correctly. When I split (8xy + 2z^2) into three terms, I considered 4xy, 4xy, and 2z^2. Then, applying AM-GM:(frac{4xy + 4xy + 2z^2}{3} geq sqrt[3]{(4xy)(4xy)(2z^2)}).Calculating the geometric mean:(sqrt[3]{32x^2 y^2 z^2}).Given that (xyz = 32), so (x^2 y^2 z^2 = (xyz)^2 = 32^2 = 1024). Therefore, the geometric mean is (sqrt[3]{32 times 1024}).Wait, 32 times 1024 is 32768. The cube root of 32768 is 32, because (32^3 = 32768). So, the geometric mean is 32. Therefore, the AM-GM inequality gives:(frac{4xy + 4xy + 2z^2}{3} geq 32), so multiplying both sides by 3:(4xy + 4xy + 2z^2 geq 96).Which is exactly what we have. So, the minimum is indeed 96 when all the terms are equal, i.e., when (4xy = 2z^2). Since (4xy = 8) (because (xy = 8)), then (2z^2 = 8), so (z^2 = 4), which gives (z = 4), consistent with our earlier result.Therefore, everything checks out, and the minimum value is 96.**Final Answer**boxed{96}

💡Alright, so I've got this race problem between a tortoise and a hare. It's similar to the classic fable, but with some specific details. Let me try to break it down step by step.First, the hare is overconfident and dashes 50 meters quickly. Then it decides to rest. During this rest, the hare sleeps longer and only resumes the race after the tortoise passes it. Once the hare starts again, it races forward at a 45-degree angle on a distance-time (d-t) graph, which means it's moving at a constant speed equal to the tortoise's speed. The tortoise, on the other hand, moves at a constant speed throughout the entire race without stopping. The problem also mentions that the hare must get 50 meters ahead in 1/4 of the total time of the race. The tortoise's speed is given as 10 meters per minute, and the race ends at 100 meters for both participants. I need to find the total time for the race.Okay, let's start by understanding the hare's movement. The hare dashes 50 meters quickly and then rests until the tortoise passes it. After that, the hare resumes the race at the same speed as the tortoise. Since the hare must get 50 meters ahead in 1/4 of the total time, I can denote the total time as T. Therefore, the hare covers 50 meters in T/4 time.Given that the hare's speed during this initial dash is 50 meters divided by T/4, which simplifies to 200/T meters per minute. So, the hare's initial speed is 200/T m/min.Now, the tortoise is moving at a constant speed of 10 meters per minute. To cover 100 meters, the tortoise would take 100/10 = 10 minutes. So, the total time T for the race, from the tortoise's perspective, is 10 minutes.But wait, the hare has a different timeline because it rests for some time after the initial dash. Let me think about this. The hare dashes 50 meters in T/4 time, which is 10/4 = 2.5 minutes. So, the hare takes 2.5 minutes to cover the first 50 meters.After that, the hare rests until the tortoise passes it. The tortoise is moving at 10 meters per minute, so it takes the tortoise 50/10 = 5 minutes to reach the 50-meter mark. That means the hare rests for 5 - 2.5 = 2.5 minutes before resuming the race.Once the hare resumes, it moves at the same speed as the tortoise, which is 10 meters per minute. The hare now needs to cover the remaining 50 meters to reach the finish line. Since both the hare and the tortoise are moving at the same speed, they will finish the race together.Wait a minute, if the hare resumes at the same speed as the tortoise, and both have 50 meters left, they should finish at the same time. But the hare had already rested for 2.5 minutes, so does that affect the total time?Let me recast this. The total time T is the time it takes for the tortoise to finish the race, which is 10 minutes. The hare, on the other hand, spends 2.5 minutes dashing the first 50 meters, rests for 2.5 minutes, and then spends the remaining 5 minutes (from minute 5 to minute 10) moving at 10 meters per minute to finish the race.So, adding up the hare's time: 2.5 minutes (dashing) + 2.5 minutes (resting) + 5 minutes (moving at tortoise's speed) = 10 minutes, which matches the total time T.Therefore, the total time for the race is 10 minutes.

💡Okay, so I have this problem here where I need to find the intersection of two sets, A and B. The universal set is all real numbers, U = ℝ. Set A is defined as {x | x² + 2x - 3 < 0}, and set B is {x | x = 2k, k ∈ ℤ}. I need to find A ∩ B and choose the correct option from A to D.First, I think I should figure out what set A actually is. The inequality given is x² + 2x - 3 < 0. I remember that to solve quadratic inequalities, I should factor the quadratic if possible, find the roots, and then determine the intervals where the inequality holds.Let me try factoring x² + 2x - 3. Hmm, I need two numbers that multiply to -3 and add up to 2. That would be 3 and -1 because 3 * (-1) = -3 and 3 + (-1) = 2. So, the factored form is (x + 3)(x - 1) < 0.Now, the roots are x = -3 and x = 1. These roots divide the real number line into three intervals: (-∞, -3), (-3, 1), and (1, ∞). To determine where the inequality (x + 3)(x - 1) < 0 holds, I can pick test points from each interval.For the interval (-∞, -3), let's choose x = -4. Plugging into (x + 3)(x - 1): (-4 + 3)(-4 - 1) = (-1)(-5) = 5, which is positive. So, the inequality doesn't hold here.For the interval (-3, 1), let's choose x = 0. Plugging in: (0 + 3)(0 - 1) = 3*(-1) = -3, which is negative. So, the inequality holds in this interval.For the interval (1, ∞), let's choose x = 2. Plugging in: (2 + 3)(2 - 1) = 5*1 = 5, which is positive. So, the inequality doesn't hold here either.Therefore, set A is the open interval (-3, 1). So, A = {x | -3 < x < 1}.Next, set B is defined as {x | x = 2k, k ∈ ℤ}. That means B is the set of all even integers. So, B = {..., -4, -2, 0, 2, 4, ...}.Now, I need to find the intersection of A and B, which is A ∩ B. This means I need to find all elements that are both in A and in B. Since A is (-3, 1) and B is all even integers, I need to find all even integers that lie within the interval (-3, 1).Let me list the even integers around this interval. The even integers less than 1 and greater than -3 are -2, 0. Let's check:- -4 is less than -3, so it's not in A.- -2 is between -3 and 1, so it's in A.- 0 is also between -3 and 1, so it's in A.- 2 is greater than 1, so it's not in A.Therefore, the even integers in A are -2 and 0. So, A ∩ B = {-2, 0}.Looking at the options:A: {-2, 0}B: {0, 2}C: {-2, 0, 2}D: {-2, -1, 0}So, the correct answer is option A: {-2, 0}.Wait, let me double-check. Did I miss any even integers between -3 and 1? The even integers are spaced by 2, so starting from -4, -2, 0, 2, etc. Between -3 and 1, the even integers are indeed -2 and 0. So, yes, {-2, 0} is correct.I think that's it. I don't see any mistakes in my reasoning.