💡First, I need to recall the formula for the lateral surface area of a cone, which is ( pi r l ), where ( r ) is the radius of the base and ( l ) is the slant height.Given that the radius ( r ) is 10 cm and the slant height ( l ) is 50 cm, I can substitute these values into the formula.Calculating the lateral surface area:[pi times 10 , text{cm} times 50 , text{cm} = 500pi , text{cm}^2]Therefore, the lateral surface area of the conical paper hat is ( 500pi , text{cm}^2 ).

💡Okay, so I have this problem here: the function f(x) = 2 sin(2x + φ), where φ is between 0 and π/2. The graph of this function passes through the point (0, √3). I need to find the interval on [0, π] where the function is monotonically decreasing.Alright, let's break this down step by step. First, I know that the function is a sine function with some amplitude, frequency, and phase shift. The amplitude is 2, which means the maximum value is 2 and the minimum is -2. The frequency is 2, so the period is π, right? Because the period of sin(kx) is 2π/k, so here it's 2π/2 = π.Now, the phase shift is φ, which is given to be between 0 and π/2. So the graph is shifted to the left or right by φ/2, since the general form is sin(kx + φ) which is equivalent to sin(k(x + φ/k)). So in this case, it's shifted by φ/2.But before I get into the phase shift, the problem says the graph passes through the point (0, √3). That means when x = 0, f(x) = √3. Let me plug that into the function:f(0) = 2 sin(2*0 + φ) = 2 sin(φ) = √3.So, 2 sin(φ) = √3. Dividing both sides by 2, sin(φ) = √3/2. Hmm, okay, so sin(φ) is √3/2. I know that sin(π/3) = √3/2, and since φ is between 0 and π/2, which is 0 to 90 degrees, π/3 is 60 degrees, so that's within the given range. Therefore, φ must be π/3.So now, the function is f(x) = 2 sin(2x + π/3). Got that.Now, I need to find the interval where this function is monotonically decreasing on [0, π]. Monotonicity of a function is determined by its derivative. If the derivative is negative, the function is decreasing; if it's positive, the function is increasing.So, let's find the derivative of f(x):f'(x) = d/dx [2 sin(2x + π/3)] = 2 * cos(2x + π/3) * 2 = 4 cos(2x + π/3).Wait, hold on. Let me double-check that. The derivative of sin(u) is cos(u) * u', so here u = 2x + π/3, so u' = 2. Therefore, f'(x) = 2 * cos(2x + π/3) * 2 = 4 cos(2x + π/3). Yeah, that's correct.So, f'(x) = 4 cos(2x + π/3). We need to find where this derivative is negative because that's where the function is decreasing.So, 4 cos(2x + π/3) < 0. Since 4 is positive, this simplifies to cos(2x + π/3) < 0.Alright, so we need to solve the inequality cos(θ) < 0, where θ = 2x + π/3.I know that cosine is negative in the intervals (π/2 + 2πk, 3π/2 + 2πk) for any integer k. So, θ must lie in these intervals.Therefore, π/2 + 2πk < 2x + π/3 < 3π/2 + 2πk, for some integer k.Let me solve for x in this inequality.First, subtract π/3 from all parts:π/2 - π/3 + 2πk < 2x < 3π/2 - π/3 + 2πk.Calculating π/2 - π/3: π/2 is 3π/6, π/3 is 2π/6, so 3π/6 - 2π/6 = π/6.Similarly, 3π/2 - π/3: 3π/2 is 9π/6, π/3 is 2π/6, so 9π/6 - 2π/6 = 7π/6.So now, the inequality becomes:π/6 + 2πk < 2x < 7π/6 + 2πk.Divide all parts by 2:π/12 + πk < x < 7π/12 + πk.So, x lies in the intervals (π/12 + πk, 7π/12 + πk) for integer k.Now, since we're looking for x in [0, π], let's find the appropriate k values.For k = 0: π/12 ≈ 0.2618, 7π/12 ≈ 1.8326. So, the interval is (π/12, 7π/12).For k = 1: π/12 + π ≈ 0.2618 + 3.1416 ≈ 3.4034, and 7π/12 + π ≈ 1.8326 + 3.1416 ≈ 4.9742. But our domain is up to π ≈ 3.1416, so the next interval starts beyond π, so we don't need to consider k=1.For k = -1: π/12 - π ≈ negative, which is outside our domain, so we can ignore that.Therefore, the only interval within [0, π] where f'(x) < 0 is (π/12, 7π/12). But since the function is continuous, we can include the endpoints if the derivative is zero there, but let's check.At x = π/12: f'(π/12) = 4 cos(2*(π/12) + π/3) = 4 cos(π/6 + π/3) = 4 cos(π/2) = 0.Similarly, at x = 7π/12: f'(7π/12) = 4 cos(2*(7π/12) + π/3) = 4 cos(7π/6 + π/3) = 4 cos(7π/6 + 2π/6) = 4 cos(9π/6) = 4 cos(3π/2) = 0.So, at both endpoints, the derivative is zero, meaning the function has horizontal tangents there. Since we're looking for intervals where the function is decreasing, we can include the endpoints because the function stops increasing and starts decreasing at π/12, and stops decreasing and starts increasing at 7π/12.Therefore, the interval of monotonic decrease is [π/12, 7π/12].Let me just visualize this to make sure. The function is a sine wave with amplitude 2, period π, shifted left by π/6 (since phase shift is φ/2 = (π/3)/2 = π/6). So, the graph starts at x=0, f(0)=√3, which is somewhere between the maximum and the midpoint. Then it goes up to the maximum, comes back down, reaches a minimum, and then goes back up.Since the period is π, the function completes one full cycle from 0 to π. The critical points where the derivative is zero are at π/12 and 7π/12. So, between π/12 and 7π/12, the function is decreasing, and outside of that, it's increasing.Yes, that makes sense. So, I think I've got the right interval.**Final Answer**The interval of monotonic decrease of the function ( f(x) ) on ([0, pi]) is boxed{left[ dfrac{pi}{12}, dfrac{7pi}{12} right]}.

💡First, I need to identify the unknown number in the equation where a number plus x equals 13,800, and x is given as 3,927.To find the unknown number, I can set up the equation:number + 3,927 = 13,800Next, I will isolate the unknown number by subtracting 3,927 from both sides of the equation:number = 13,800 - 3,927Finally, I will perform the subtraction to determine the value of the unknown number:number = 9,873

💡Okay, so I have this problem here: If ( f(cos x) = cos 2x ), then what is ( fleft(sin frac{pi}{12}right) )? Hmm, let's see. I need to figure out what ( f ) does to its input and then apply that to ( sin frac{pi}{12} ).First, I know that ( f ) is defined in terms of cosine. Specifically, when the input is ( cos x ), the output is ( cos 2x ). So, ( f ) seems to be a function that takes a cosine value and gives back the cosine of twice the angle. That makes me think of the double-angle identity in trigonometry, which is ( cos 2x = 2cos^2 x - 1 ). So, maybe ( f(t) = 2t^2 - 1 ) where ( t = cos x ).Let me test that idea. If ( f(t) = 2t^2 - 1 ), then ( f(cos x) = 2cos^2 x - 1 = cos 2x ). Yeah, that works! So, the function ( f ) can be expressed as ( f(t) = 2t^2 - 1 ). That means for any input ( t ), ( f(t) ) is just twice the square of ( t ) minus one.Now, I need to find ( fleft(sin frac{pi}{12}right) ). Using the function I just found, that would be ( 2left(sin frac{pi}{12}right)^2 - 1 ). Let me write that down:[fleft(sin frac{pi}{12}right) = 2left(sin frac{pi}{12}right)^2 - 1]Hmm, okay. Now, I need to compute ( sin frac{pi}{12} ) first. I remember that ( frac{pi}{12} ) is 15 degrees. So, ( sin 15^circ ). I can use the sine of a difference formula here because 15 degrees is 45 degrees minus 30 degrees. The formula is:[sin(a - b) = sin a cos b - cos a sin b]Let me apply that:[sin 15^circ = sin(45^circ - 30^circ) = sin 45^circ cos 30^circ - cos 45^circ sin 30^circ]I know the exact values for these:- ( sin 45^circ = frac{sqrt{2}}{2} )- ( cos 30^circ = frac{sqrt{3}}{2} )- ( cos 45^circ = frac{sqrt{2}}{2} )- ( sin 30^circ = frac{1}{2} )Plugging these in:[sin 15^circ = left(frac{sqrt{2}}{2}right)left(frac{sqrt{3}}{2}right) - left(frac{sqrt{2}}{2}right)left(frac{1}{2}right)]Let me compute each term:First term: ( frac{sqrt{2}}{2} times frac{sqrt{3}}{2} = frac{sqrt{6}}{4} )Second term: ( frac{sqrt{2}}{2} times frac{1}{2} = frac{sqrt{2}}{4} )So, subtracting the second term from the first:[sin 15^circ = frac{sqrt{6}}{4} - frac{sqrt{2}}{4} = frac{sqrt{6} - sqrt{2}}{4}]Alright, so ( sin frac{pi}{12} = frac{sqrt{6} - sqrt{2}}{4} ). Now, let's square this value to plug it back into our function ( f ).Compute ( left(sin frac{pi}{12}right)^2 ):[left(frac{sqrt{6} - sqrt{2}}{4}right)^2 = frac{(sqrt{6} - sqrt{2})^2}{16}]Expanding the numerator:[(sqrt{6} - sqrt{2})^2 = (sqrt{6})^2 - 2 times sqrt{6} times sqrt{2} + (sqrt{2})^2 = 6 - 2sqrt{12} + 2 = 8 - 2sqrt{12}]Simplify ( sqrt{12} ):[sqrt{12} = sqrt{4 times 3} = 2sqrt{3}]So, substituting back:[8 - 2 times 2sqrt{3} = 8 - 4sqrt{3}]Therefore, the square of ( sin frac{pi}{12} ) is:[frac{8 - 4sqrt{3}}{16} = frac{8}{16} - frac{4sqrt{3}}{16} = frac{1}{2} - frac{sqrt{3}}{4}]Now, plug this back into the function ( f ):[fleft(sin frac{pi}{12}right) = 2left(frac{1}{2} - frac{sqrt{3}}{4}right) - 1]Let me compute each term step by step.First, multiply by 2:[2 times frac{1}{2} = 1][2 times left(-frac{sqrt{3}}{4}right) = -frac{sqrt{3}}{2}]So, putting it together:[1 - frac{sqrt{3}}{2} - 1]Wait, that simplifies to:[(1 - 1) - frac{sqrt{3}}{2} = 0 - frac{sqrt{3}}{2} = -frac{sqrt{3}}{2}]So, the value of ( fleft(sin frac{pi}{12}right) ) is ( -frac{sqrt{3}}{2} ).Let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning, I expressed ( f(t) ) as ( 2t^2 - 1 ), which seems correct because ( cos 2x = 2cos^2 x - 1 ). Then, I converted ( sin frac{pi}{12} ) into ( sin 15^circ ) and used the sine difference identity, which gave me ( frac{sqrt{6} - sqrt{2}}{4} ). Squaring that gave me ( frac{8 - 4sqrt{3}}{16} ), which simplifies to ( frac{1}{2} - frac{sqrt{3}}{4} ). Plugging that into ( f(t) ) gave me ( 2 times left(frac{1}{2} - frac{sqrt{3}}{4}right) - 1 ), which simplifies to ( -frac{sqrt{3}}{2} ). Hmm, that seems consistent. I don't see any errors in my calculations. So, I think the answer is correct.**Final Answer**The value of ( fleft(sin frac{pi}{12}right) ) is boxed{-dfrac{sqrt{3}}{2}}.

💡Alright, so I have this problem about an arithmetic sequence. Let me try to understand what it's asking. We have an arithmetic sequence {a_n} with a non-zero common difference. S_n is the sum of the first n terms of this sequence. The problem gives two conditions: a_3 = S_5 and a_2 * a_4 = S_4. I need to find two things: (I) the general formula for the sequence {a_n}, and (II) the smallest value of n for which S_n > a_n holds.Okay, let's start with part (I). I remember that in an arithmetic sequence, the nth term is given by a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference. The sum of the first n terms, S_n, is given by S_n = n/2 * [2a_1 + (n-1)d].Given that a_3 = S_5, let's write that out. a_3 is the third term, so that's a_1 + 2d. S_5 is the sum of the first five terms, which is 5/2 * [2a_1 + 4d]. So, setting them equal:a_1 + 2d = (5/2) * [2a_1 + 4d]Let me simplify the right side. 2a_1 + 4d is just 2(a_1 + 2d), so:a_1 + 2d = (5/2) * 2(a_1 + 2d) = 5(a_1 + 2d)So, a_1 + 2d = 5a_1 + 10dHmm, subtracting a_1 + 2d from both sides:0 = 4a_1 + 8dDivide both sides by 4:0 = a_1 + 2dSo, a_1 + 2d = 0. That means a_1 = -2d.Alright, so the first term is -2d. Let's keep that in mind.Now, the second condition is a_2 * a_4 = S_4.Let's compute each part. a_2 is the second term, which is a_1 + d. a_4 is the fourth term, which is a_1 + 3d. So, their product is (a_1 + d)(a_1 + 3d).S_4 is the sum of the first four terms, which is 4/2 * [2a_1 + 3d] = 2*(2a_1 + 3d) = 4a_1 + 6d.So, setting them equal:(a_1 + d)(a_1 + 3d) = 4a_1 + 6dLet's expand the left side:a_1^2 + 4a_1 d + 3d^2 = 4a_1 + 6dNow, we already know that a_1 = -2d from the first condition. Let's substitute that into the equation.So, replace a_1 with -2d:(-2d)^2 + 4*(-2d)*d + 3d^2 = 4*(-2d) + 6dSimplify each term:First term: (-2d)^2 = 4d^2Second term: 4*(-2d)*d = -8d^2Third term: 3d^2So, left side: 4d^2 - 8d^2 + 3d^2 = (4 - 8 + 3)d^2 = (-1)d^2Right side: 4*(-2d) + 6d = -8d + 6d = -2dSo, the equation becomes:- d^2 = -2dLet's bring all terms to one side:- d^2 + 2d = 0Factor out a d:d(-d + 2) = 0So, d = 0 or d = 2But the problem states that the common difference is non-zero, so d = 2.Now that we have d = 2, we can find a_1. From earlier, a_1 = -2d = -4.So, the first term is -4, and the common difference is 2. Therefore, the general formula for the sequence is:a_n = a_1 + (n - 1)d = -4 + (n - 1)*2Simplify that:a_n = -4 + 2n - 2 = 2n - 6So, part (I) is done. The general term is a_n = 2n - 6.Now, moving on to part (II): Find the smallest value of n for which S_n > a_n.First, let's write expressions for S_n and a_n.We already have a_n = 2n - 6.For S_n, the sum of the first n terms of an arithmetic sequence is S_n = n/2 * [2a_1 + (n - 1)d]We know a_1 = -4 and d = 2, so plug those in:S_n = n/2 * [2*(-4) + (n - 1)*2] = n/2 * [-8 + 2n - 2] = n/2 * (2n - 10)Simplify:S_n = n/2 * 2(n - 5) = n(n - 5)So, S_n = n^2 - 5nNow, we need to find the smallest n such that S_n > a_n.So, set up the inequality:n^2 - 5n > 2n - 6Bring all terms to one side:n^2 - 5n - 2n + 6 > 0Simplify:n^2 - 7n + 6 > 0Now, let's solve this quadratic inequality.First, find the roots of the equation n^2 - 7n + 6 = 0.Using the quadratic formula: n = [7 ± sqrt(49 - 24)] / 2 = [7 ± sqrt(25)] / 2 = [7 ± 5]/2So, n = (7 + 5)/2 = 12/2 = 6, and n = (7 - 5)/2 = 2/2 = 1.So, the quadratic expression factors as (n - 1)(n - 6) > 0.Now, to solve the inequality (n - 1)(n - 6) > 0, we can consider the intervals determined by the roots: n < 1, 1 < n < 6, and n > 6.Since n is a positive integer (as it's the number of terms), we can ignore n < 1.Now, test the intervals:For 1 < n < 6, say n = 3: (3 - 1)(3 - 6) = 2*(-3) = -6 < 0For n > 6, say n = 7: (7 - 1)(7 - 6) = 6*1 = 6 > 0So, the inequality holds when n > 6.Since n must be an integer, the smallest n satisfying this is n = 7.Therefore, the smallest value of n for which S_n > a_n is 7.Let me double-check my calculations to make sure I didn't make any mistakes.For part (I):- a_3 = S_5 led to a_1 = -2d- Then, a_2 * a_4 = S_4 gave us d = 2, so a_1 = -4- Thus, a_n = 2n - 6For part (II):- S_n = n(n - 5) = n^2 - 5n- Set S_n > a_n: n^2 - 5n > 2n - 6- Simplify to n^2 - 7n + 6 > 0- Roots at n = 1 and n = 6- Test intervals: positive when n > 6- So, smallest n is 7Everything seems to check out. I think I'm confident with this solution.**Final Answer**Ⅰ. The general formula is boxed{a_n = 2n - 6}.Ⅱ. The smallest value of ( n ) is boxed{7}.

💡Okay, so I have this problem here: "How many different positive integers can be represented as a difference of two distinct members of the set {1, 2, 3, ..., 19, 20}?" Hmm, let me try to figure this out step by step.First, I need to understand what the question is asking. It wants to know how many unique positive integers you can get by subtracting two different numbers from this set. So, if I take any two numbers from 1 to 20, subtract the smaller one from the larger one, and collect all those results, how many different positive numbers will I have?Alright, let me think about how differences work. If I have two numbers, say a and b, where a > b, then the difference is a - b. Since all the numbers are positive integers, the differences will also be positive integers. The smallest possible difference would be 1 (like 2 - 1), and the largest possible difference would be 19 (like 20 - 1).So, does that mean all the integers from 1 to 19 are possible? Let me check. If I take 1 and 2, the difference is 1. If I take 1 and 3, the difference is 2. Continuing this way, 1 and 4 give 3, and so on up to 1 and 20, which gives 19. So, it seems like I can get every integer from 1 to 19 by subtracting 1 from each number up to 20.But wait, is that the only way? What if I don't use 1? For example, if I take 2 and 3, the difference is 1. If I take 3 and 4, the difference is also 1. So, 1 can be obtained in multiple ways. Similarly, 2 can be obtained by 3 - 1, 4 - 2, 5 - 3, etc. So, even if I don't use 1, I can still get all the differences.But does that mean there are more differences beyond 19? Well, the maximum difference is 19, as I thought earlier, because 20 - 1 is 19. So, I don't think there are any differences larger than 19 in this set.Now, let me think if there are any gaps. For example, can I get every integer between 1 and 19? Let's see:- 1: Yes, as 2 - 1, 3 - 2, etc.- 2: Yes, as 3 - 1, 4 - 2, etc.- 3: Yes, as 4 - 1, 5 - 2, etc.- ...- 19: Yes, as 20 - 1.It seems like every integer from 1 to 19 is achievable. Therefore, the number of different positive integers that can be represented as a difference is 19.But just to be thorough, let me consider another approach. Maybe I can think about how many possible differences there are in total and then see if there are any duplicates or missing numbers.The set has 20 numbers. The number of ways to choose two distinct numbers is C(20, 2), which is 190. However, many of these differences will repeat. For example, as I mentioned earlier, the difference of 1 can be obtained in 19 different ways (2-1, 3-2, ..., 20-19). Similarly, the difference of 2 can be obtained in 18 ways (3-1, 4-2, ..., 20-18), and so on.So, the number of unique differences is not 190 but much less. From my earlier reasoning, it seems like the unique differences are exactly the integers from 1 to 19. Therefore, there are 19 unique positive integers that can be represented as differences.Wait a second, let me make sure I'm not missing something. Is there any difference that I can't get? For example, can I get 0? Well, the problem specifies "positive integers," so 0 is excluded. What about negative numbers? Again, since we're only considering positive differences, negatives are out of the question.Also, I should confirm that I can indeed get every integer from 1 to 19. Let's list them out:1: 2-1, 3-2, ..., 20-192: 3-1, 4-2, ..., 20-183: 4-1, 5-2, ..., 20-17...19: 20-1Yes, each integer from 1 to 19 is covered. There's no integer in that range that I can't get by subtracting two distinct numbers from the set.Therefore, the answer should be 19 different positive integers.