💡Okay, so I have this problem about the population of Town X and the employment statistics. Let me try to understand what it's asking. First, it says that 96% of the population are employed. That means out of every 100 people in Town X, 96 have jobs. Now, among these employed people, there's a breakdown by age groups: 40% are between 20-35 years old, 50% are between 36-50 years old, and 10% are between 51-65 years old. So, if I imagine the employed population, most of them are in the 36-50 age group, followed by the 20-35 group, and then the smallest portion is the 51-65 group.Next, it mentions that 24% of the population are employed males. So, out of the total population, 24% are males who have jobs. It also breaks down the educational levels of these employed males: 45% have a high school education, 35% have a college degree, and 20% have a postgraduate degree. The question is asking for the percentage of employed females in Town X across different age groups and educational levels. So, I need to figure out how many females are employed in each age group and each educational category.Let me start by figuring out the percentage of employed females. Since 96% of the population is employed, and 24% of the population are employed males, I can subtract the percentage of employed males from the total employed to get the percentage of employed females. So, 96% total employed minus 24% employed males equals 72% employed females. That seems straightforward.Now, I need to break this 72% down into the different age groups. The age groups for the employed population are 20-35, 36-50, and 51-65 years old, with percentages of 40%, 50%, and 10%, respectively. Assuming that the distribution of age groups is the same for both males and females, I can apply these percentages directly to the 72% employed females. For the 20-35 age group: 40% of 72% is 0.4 * 72 = 28.8%. For the 36-50 age group: 50% of 72% is 0.5 * 72 = 36%. For the 51-65 age group: 10% of 72% is 0.1 * 72 = 7.2%. So, the employed females are distributed as follows by age: 28.8% in the 20-35 group, 36% in the 36-50 group, and 7.2% in the 51-65 group.Next, I need to determine the educational levels of these employed females. The problem provides the educational distribution for employed males: 45% have a high school education, 35% have a college degree, and 20% have a postgraduate degree. Since the problem doesn't specify different educational distributions for females, I'll assume that the same percentages apply to employed females. So, for high school education: 45% of 72% is 0.45 * 72 = 32.4%. For college degree: 35% of 72% is 0.35 * 72 = 25.2%. For postgraduate degree: 20% of 72% is 0.2 * 72 = 14.4%. Therefore, the educational levels among employed females are: 32.4% have a high school education, 25.2% have a college degree, and 14.4% have a postgraduate degree.Let me double-check my calculations to make sure I didn't make any mistakes. Total employed: 96%Employed males: 24%Employed females: 96% - 24% = 72%Age groups:- 20-35: 40% of 72% = 28.8%- 36-50: 50% of 72% = 36%- 51-65: 10% of 72% = 7.2%Educational levels:- High school: 45% of 72% = 32.4%- College: 35% of 72% = 25.2%- Postgraduate: 20% of 72% = 14.4%Adding up the age groups: 28.8% + 36% + 7.2% = 72%, which matches the total employed females.Adding up the educational levels: 32.4% + 25.2% + 14.4% = 72%, which also matches.Everything seems to add up correctly. I think I've covered all the parts of the question.

💡First, I need to evaluate the expression ( (502)(502) - (501)(503) ).I notice that ( 501 ) and ( 503 ) are close to ( 502 ). Specifically, ( 501 = 502 - 1 ) and ( 503 = 502 + 1 ).This allows me to rewrite ( (501)(503) ) using the difference of squares formula:[(501)(503) = (502 - 1)(502 + 1) = 502^2 - 1^2]Now, I can substitute this back into the original expression:[(502)(502) - (501)(503) = 502^2 - (502^2 - 1)]Simplifying further:[502^2 - 502^2 + 1 = 1]Therefore, the final answer is ( 1 ).

💡Alright, so I have this problem here: I need to find the degree measure of angle LOQ in a regular hexagon LMNOPQ. There's also an Asymptote figure provided, which I can't see directly, but based on the labels, I can imagine the hexagon and the points mentioned.First, let me recall what a regular hexagon is. A regular hexagon is a six-sided polygon where all sides are equal in length and all interior angles are equal. I remember that the sum of the interior angles of an n-sided polygon is given by the formula 180(n-2) degrees. So, for a hexagon, n=6, which means the sum of the interior angles is 180*(6-2) = 720 degrees. Since it's regular, each interior angle is 720/6 = 120 degrees. So, each internal angle in the hexagon is 120 degrees.Now, the question is about angle LOQ. Let me try to visualize the hexagon. The points are labeled L, M, N, O, P, Q. So, starting from L, going to M, N, O, P, Q, and back to L. The Asymptote code draws the hexagon and some additional lines, specifically connecting L to O, O to Q, and Q back to L, forming triangle LOQ.Wait, actually, looking at the Asymptote code, it draws the hexagon and then draws lines from L to O, O to Q, and Q back to L, forming triangle LOQ. So, angle LOQ is the angle at point O, between points L, O, and Q.Hmm, okay. So, I need to find the measure of angle LOQ. Since it's a regular hexagon, all sides are equal, and all central angles are equal. Maybe I can use some properties of regular hexagons or triangles to find this angle.Let me think about the coordinates. The Asymptote code provides coordinates for each point:- L is at (-1, -1.73205081)- M is at (-2, 0)- N is at (-1, 1.73205081)- O is at (1, 1.73205081)- P is at (2, 0)- Q is at (1, -1.73205081)So, these coordinates are given, which might help me calculate the vectors or distances between points to find the angle.Alternatively, I can use the fact that in a regular hexagon, the distance between opposite vertices is twice the length of one side. Also, the hexagon can be inscribed in a circle, with all vertices lying on the circumference. So, the center of the circle is the center of the hexagon.Given that, the central angles between each vertex are 60 degrees because 360 degrees divided by 6 is 60 degrees. So, each vertex is 60 degrees apart from the next one around the center.Wait, but angle LOQ is not a central angle. It's an angle at point O, formed by points L, O, and Q. So, I need to figure out the positions of L, O, and Q relative to each other.Looking at the coordinates:- L is at (-1, -1.73205081)- O is at (1, 1.73205081)- Q is at (1, -1.73205081)So, point O is at (1, 1.73205081), which is the top right vertex, and point Q is at (1, -1.73205081), which is the bottom right vertex. Point L is at (-1, -1.73205081), which is the bottom left vertex.So, connecting L to O to Q forms a triangle. To find angle LOQ, which is at point O, I need to find the angle between vectors OL and OQ.Alternatively, since I have the coordinates, I can use coordinate geometry to find the angle.First, let's find the vectors OL and OQ.Vector OL goes from O to L: L - O = (-1 - 1, -1.73205081 - 1.73205081) = (-2, -3.46410162)Vector OQ goes from O to Q: Q - O = (1 - 1, -1.73205081 - 1.73205081) = (0, -3.46410162)Now, to find the angle between vectors OL and OQ, I can use the dot product formula:cos(theta) = (OL . OQ) / (|OL| |OQ|)First, compute the dot product OL . OQ:= (-2)(0) + (-3.46410162)(-3.46410162)= 0 + (3.46410162)^2= (3.46410162)^2Now, compute |OL| and |OQ|:|OL| = sqrt((-2)^2 + (-3.46410162)^2) = sqrt(4 + 12) = sqrt(16) = 4|OQ| = sqrt(0^2 + (-3.46410162)^2) = sqrt(12) ≈ 3.46410162So, cos(theta) = (12) / (4 * 3.46410162) = 12 / 13.85640648 ≈ 0.8660254Now, arccos(0.8660254) is approximately 30 degrees, since cos(30°) ≈ 0.8660254.Therefore, angle LOQ is 30 degrees.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, vectors:OL: from O(1, 1.73205081) to L(-1, -1.73205081) is (-2, -3.46410162)OQ: from O(1, 1.73205081) to Q(1, -1.73205081) is (0, -3.46410162)Dot product: (-2)(0) + (-3.46410162)(-3.46410162) = 0 + 12 = 12|OL|: sqrt((-2)^2 + (-3.46410162)^2) = sqrt(4 + 12) = sqrt(16) = 4|OQ|: sqrt(0^2 + (-3.46410162)^2) = sqrt(12) ≈ 3.46410162So, cos(theta) = 12 / (4 * 3.46410162) = 12 / 13.85640648 ≈ 0.8660254Yes, that's correct. And arccos(0.8660254) is indeed 30 degrees.Alternatively, I can think about the regular hexagon's properties without coordinates.In a regular hexagon, each side is equal, and the distance between opposite vertices is twice the side length. Also, the hexagon can be divided into six equilateral triangles by drawing lines from the center to each vertex.Given that, the central angles are 60 degrees each.Now, points L, O, and Q are vertices of the hexagon. Let me see their positions.Starting from L, going to M, N, O, P, Q, and back to L.So, from L to O, that's three sides apart: L to M to N to O.Similarly, from O to Q is two sides apart: O to P to Q.Wait, actually, in the hexagon, moving from L to O is three steps, which would make them opposite vertices, right? Because in a hexagon, opposite vertices are three apart.Similarly, from O to Q is two steps: O to P to Q.So, in terms of the central angles, the angle between L and O is 180 degrees because they are opposite vertices.Similarly, the angle between O and Q is 120 degrees because they are two steps apart, each step being 60 degrees.Wait, no. Actually, each side corresponds to a central angle of 60 degrees. So, moving from O to P is 60 degrees, and from P to Q is another 60 degrees, so O to Q is 120 degrees.Similarly, from L to M is 60 degrees, M to N is another 60 degrees, and N to O is another 60 degrees, so L to O is 180 degrees.So, in terms of vectors from the center, points L and O are directly opposite each other, 180 degrees apart, and points O and Q are 120 degrees apart.But angle LOQ is at point O, so it's not a central angle but rather an angle formed by two chords: OL and OQ.Hmm, perhaps I can use the law of cosines in triangle LOQ.Wait, but I already did that using coordinates, and it gave me 30 degrees. Let me see if I can confirm this another way.Alternatively, since the hexagon is regular, all sides are equal, and the triangles formed by the center are equilateral.So, the distance from the center to any vertex is the same, say, radius r.Then, the distance between two opposite vertices is 2r, which is twice the radius.In our case, the coordinates suggest that the center is at (0,0), because the hexagon is symmetric around the origin.So, point O is at (1, 1.73205081), which is approximately (1, sqrt(3)), and point Q is at (1, -sqrt(3)), and point L is at (-1, -sqrt(3)).So, the distance from the center to each vertex is sqrt(1^2 + (sqrt(3))^2) = sqrt(1 + 3) = sqrt(4) = 2. So, the radius r is 2.Therefore, the distance between opposite vertices, like L and O, is 2r = 4, which matches the distance between L(-1, -sqrt(3)) and O(1, sqrt(3)): sqrt[(1 - (-1))^2 + (sqrt(3) - (-sqrt(3)))^2] = sqrt[(2)^2 + (2sqrt(3))^2] = sqrt[4 + 12] = sqrt[16] = 4.Similarly, the distance between O and Q is sqrt[(1 - 1)^2 + (-sqrt(3) - sqrt(3))^2] = sqrt[0 + (-2sqrt(3))^2] = sqrt[12] = 2sqrt(3).So, in triangle LOQ, sides are:LO: 4OQ: 2sqrt(3)LQ: Let's compute that. From L(-1, -sqrt(3)) to Q(1, -sqrt(3)): sqrt[(1 - (-1))^2 + (-sqrt(3) - (-sqrt(3)))^2] = sqrt[(2)^2 + 0] = sqrt[4] = 2.So, triangle LOQ has sides of length 4, 2sqrt(3), and 2.Wait, that seems a bit off because in a triangle, the sum of any two sides must be greater than the third side. Let's check:4 + 2sqrt(3) > 2: Yes, because 4 + 3.464 > 2.4 + 2 > 2sqrt(3): 6 > 3.464: Yes.2sqrt(3) + 2 > 4: 3.464 + 2 = 5.464 > 4: Yes.So, it is a valid triangle.Now, to find angle LOQ, which is at point O, between sides LO and OQ.Using the law of cosines:c^2 = a^2 + b^2 - 2ab cos(theta)Where c is the side opposite angle theta.In this case, angle LOQ is at O, so the sides adjacent to it are OL and OQ, which are 4 and 2sqrt(3), and the side opposite is LQ, which is 2.So,2^2 = 4^2 + (2sqrt(3))^2 - 2*4*(2sqrt(3)) cos(theta)Simplify:4 = 16 + 12 - 16sqrt(3) cos(theta)4 = 28 - 16sqrt(3) cos(theta)Subtract 28 from both sides:4 - 28 = -16sqrt(3) cos(theta)-24 = -16sqrt(3) cos(theta)Divide both sides by -16sqrt(3):(-24)/(-16sqrt(3)) = cos(theta)Simplify:24/(16sqrt(3)) = cos(theta)Simplify numerator and denominator:Divide numerator and denominator by 8:3/(2sqrt(3)) = cos(theta)Rationalize the denominator:3/(2sqrt(3)) * (sqrt(3)/sqrt(3)) = (3sqrt(3))/(2*3) = sqrt(3)/2So, cos(theta) = sqrt(3)/2Therefore, theta = arccos(sqrt(3)/2) = 30 degrees.So, angle LOQ is 30 degrees.Wait, that's the same result as before. So, both methods confirm that angle LOQ is 30 degrees.Alternatively, I can think about the symmetry of the hexagon. Since it's regular, the triangle LOQ might have some special properties.Looking at the coordinates, point O is at (1, sqrt(3)), point Q is at (1, -sqrt(3)), and point L is at (-1, -sqrt(3)).So, triangle LOQ has vertices at (-1, -sqrt(3)), (1, sqrt(3)), and (1, -sqrt(3)).Plotting these points, I can see that point O is at the top right, Q is directly below O on the same vertical line, and L is diagonally opposite to O but on the bottom left.So, triangle LOQ is an isosceles triangle with sides LO = 4, OQ = 2sqrt(3), and LQ = 2.Wait, actually, LO is 4, OQ is 2sqrt(3), and LQ is 2. So, it's not isosceles in terms of sides, but maybe in terms of angles.But regardless, using the law of cosines, we found that angle LOQ is 30 degrees.Another way to think about it is by considering the vectors from point O to L and from point O to Q.Vector OL is from O(1, sqrt(3)) to L(-1, -sqrt(3)): (-2, -2sqrt(3))Vector OQ is from O(1, sqrt(3)) to Q(1, -sqrt(3)): (0, -2sqrt(3))So, vector OL is (-2, -2sqrt(3)) and vector OQ is (0, -2sqrt(3)).To find the angle between these two vectors, we can use the dot product formula again:cos(theta) = (OL . OQ) / (|OL| |OQ|)Compute the dot product:OL . OQ = (-2)(0) + (-2sqrt(3))(-2sqrt(3)) = 0 + (4*3) = 12Compute |OL|:|OL| = sqrt((-2)^2 + (-2sqrt(3))^2) = sqrt(4 + 12) = sqrt(16) = 4Compute |OQ|:|OQ| = sqrt(0^2 + (-2sqrt(3))^2) = sqrt(12) = 2sqrt(3)So, cos(theta) = 12 / (4 * 2sqrt(3)) = 12 / (8sqrt(3)) = (12)/(8sqrt(3)) = (3)/(2sqrt(3)) = sqrt(3)/2Therefore, theta = arccos(sqrt(3)/2) = 30 degrees.So, again, we get the same result.Alternatively, I can think about the triangle formed by points L, O, and Q in terms of the regular hexagon's structure.In a regular hexagon, each internal angle is 120 degrees. So, at point O, the internal angle is 120 degrees.But angle LOQ is not the internal angle; it's a different angle formed by connecting L to O to Q.Wait, perhaps I can consider the triangles within the hexagon.Looking at the Asymptote code, it draws the hexagon and then draws lines from L to O, O to Q, and Q back to L, forming triangle LOQ.Additionally, it draws lines from L to O, O to Q, and Q back to L, forming triangle LOQ.Wait, actually, the Asymptote code draws:draw((-2,0)--(-1,1.73205081)--(1,1.73205081)--(2,0)--(1,-1.73205081)--(-1,-1.73205081)--cycle);Then,draw((-1,-1.73205081)--(1,1.73205081)--(1,-1.73205081)--cycle);So, it's drawing the hexagon and then a triangle connecting L(-1,-1.73205081) to O(1,1.73205081) to Q(1,-1.73205081) and back to L.So, triangle LOQ is a triangle with vertices at L, O, and Q.Given that, and knowing the coordinates, we can use coordinate geometry to find the angle.But as I've already done that and confirmed it with the law of cosines and vector dot product, I'm confident that angle LOQ is 30 degrees.Another way to think about it is by considering the diagonals in the hexagon.In a regular hexagon, the diagonals can be of different lengths. The shortest diagonals connect vertices with one vertex in between, and the longest diagonals connect opposite vertices.In this case, LO is a long diagonal connecting L and O, which are opposite vertices, so its length is twice the side length.OQ is a shorter diagonal connecting O and Q, which are two vertices apart, so its length is sqrt(3) times the side length.Given that, and knowing the side length can be derived from the coordinates, we can use trigonometry to find the angle.But since I've already computed it using multiple methods, I think it's safe to conclude that angle LOQ is 30 degrees.Just to recap:1. Calculated using coordinates and vectors: 30 degrees.2. Used law of cosines in triangle LOQ: 30 degrees.3. Considered the properties of the regular hexagon and central angles: 30 degrees.All methods lead to the same conclusion. Therefore, I'm confident that the measure of angle LOQ is 30 degrees.

💡Alright, so I've got this problem here: I need to show that if a Borel function ( f = f(x) ) defined on ( mathbb{R}_{+} ) satisfies[int_{mathbb{R}_{+}} e^{-lambda x} f(x) , dx = 0 quad text{for all } lambda in mathbb{N},]then ( f ) must be zero almost everywhere with respect to the Lebesgue measure. Hmm, okay. Let me try to unpack this step by step.First, I know that ( mathbb{R}_{+} ) refers to the set of positive real numbers, so ( f ) is defined for all ( x > 0 ). The function ( f ) is a Borel function, which means it's measurable with respect to the Borel ( sigma )-algebra on ( mathbb{R}_{+} ). That's good because it ensures that the integral is well-defined.The integral given is similar to the Laplace transform of ( f ). In general, the Laplace transform of a function ( f ) is defined as[mathcal{L}{f}(lambda) = int_{0}^{infty} e^{-lambda x} f(x) , dx,]and it's a powerful tool in analysis. The problem states that this transform is zero for all natural numbers ( lambda ). So, ( mathcal{L}{f}(lambda) = 0 ) for ( lambda = 1, 2, 3, ldots ).Now, I remember that the Laplace transform is injective under certain conditions. Specifically, if two functions have the same Laplace transform, then they are equal almost everywhere. But in this case, the Laplace transform is zero for all ( lambda in mathbb{N} ). Does this imply that ( f ) is zero almost everywhere?I think so, but I need to be careful. The injectivity of the Laplace transform usually requires some conditions on the function ( f ), like being of exponential order or being in ( L^1 ) on ( mathbb{R}_{+} ). The problem doesn't specify these conditions, so I might need to assume that ( f ) is such that the integral converges for all ( lambda in mathbb{N} ).Alternatively, maybe I can approach this problem without directly invoking the injectivity of the Laplace transform. Let me think about other methods.Another idea is to use the concept of moments. In probability theory, if all the moments of a measure are zero, then the measure is zero. Maybe I can relate this integral to moments of some measure associated with ( f ).Let me try to make this precise. Suppose I consider the measure ( mu ) on ( mathbb{R}_{+} ) defined by[mu(A) = int_{A} f(x) , dx]for Borel sets ( A ). Then, the integral given is[int_{mathbb{R}_{+}} e^{-lambda x} , dmu(x) = 0 quad text{for all } lambda in mathbb{N}.]This looks like the Laplace transform of the measure ( mu ). If the Laplace transform of ( mu ) is zero for all ( lambda in mathbb{N} ), does that imply ( mu ) is the zero measure?I think yes, but I need to recall the exact theorem. I believe that if two measures have the same Laplace transform, they are equal. So, if the Laplace transform of ( mu ) is zero everywhere, then ( mu ) must be the zero measure.But wait, the problem only states that the Laplace transform is zero for ( lambda in mathbb{N} ), not for all ( lambda > 0 ). So, does knowing that the Laplace transform is zero on a countable set of points imply it's zero everywhere?I'm not entirely sure. Maybe I need a different approach.Let me try to use the Stone-Weierstrass theorem. This theorem states that if an algebra of functions separates points and contains the constant functions, then it's dense in the space of continuous functions with the uniform norm. Maybe I can use this to show that ( f ) must be zero.But how does this relate to the given integral condition?Alternatively, I can think about expanding ( e^{-lambda x} ) in a series. For example, ( e^{-lambda x} = sum_{n=0}^{infty} frac{(-lambda x)^n}{n!} ). Then, interchanging the sum and the integral (if allowed), I get[sum_{n=0}^{infty} frac{(-lambda)^n}{n!} int_{mathbb{R}_{+}} x^n f(x) , dx = 0.]But this is zero for all ( lambda in mathbb{N} ). Hmm, not sure if this helps directly.Wait, maybe I can consider the function ( f ) in terms of its moments. The integral ( int_{mathbb{R}_{+}} x^n f(x) , dx ) is the ( n )-th moment of ( f ). If all these moments are zero, then under certain conditions, ( f ) must be zero almost everywhere.But again, I need to recall the exact conditions under which this holds. I think it's related to the uniqueness of moments. If two functions have the same moments, they are equal almost everywhere, provided they are in some suitable space, like ( L^1 ) or with some decay conditions.But in our case, the moments are all zero. So, if the moments uniquely determine the function, then ( f ) must be zero.Alternatively, maybe I can use the fact that the functions ( e^{-lambda x} ) for ( lambda in mathbb{N} ) form a total set in some function space. That is, their linear combinations are dense in that space.If that's the case, then if ( f ) is orthogonal to all these functions, it must be zero.But I need to make this precise.Let me think about the space ( L^2(mathbb{R}_{+}, e^{-lambda x} dx) ). Wait, no, that's not quite right. Maybe I need to consider a different space.Alternatively, perhaps I can use the fact that the functions ( e^{-lambda x} ) for ( lambda in mathbb{N} ) are linearly independent and form a basis for some space.But I'm not sure. Maybe another approach.Let me consider the Laplace transform again. If ( mathcal{L}{f}(lambda) = 0 ) for all ( lambda in mathbb{N} ), can I extend this to all ( lambda > 0 )?I know that analytic functions are determined by their values on a set with an accumulation point. The Laplace transform is analytic in the right half-plane where it converges. So, if ( mathcal{L}{f}(lambda) = 0 ) for ( lambda = 1, 2, 3, ldots ), which has an accumulation point at infinity, does that imply that ( mathcal{L}{f}(lambda) = 0 ) for all ( lambda ) in the domain of convergence?Hmm, I think that's the case. If an analytic function is zero on a set with an accumulation point in its domain, then it's zero everywhere in that domain. So, if the Laplace transform is zero for all ( lambda in mathbb{N} ), and since ( mathbb{N} ) has an accumulation point at infinity (though technically, infinity isn't in the complex plane), but in the sense that the function is zero on a sequence going to infinity, which might be enough to conclude that the function is identically zero.If that's the case, then ( mathcal{L}{f}(lambda) = 0 ) for all ( lambda ) in the domain of convergence, which would imply that ( f = 0 ) almost everywhere.But I'm not entirely sure about the accumulation point argument here. Maybe I need to think differently.Another idea: consider the function ( f ) multiplied by a smooth cutoff function to make it compactly supported, then use the fact that the Laplace transform of a compactly supported function is entire and can be extended to the whole complex plane. Then, if it's zero on a set with an accumulation point, it must be zero everywhere.But this seems a bit involved. Maybe there's a simpler way.Wait, let's think about the measure-theoretic perspective again. If ( int_{mathbb{R}_{+}} e^{-lambda x} f(x) , dx = 0 ) for all ( lambda in mathbb{N} ), then ( f ) is orthogonal to all functions of the form ( e^{-lambda x} ) in ( L^2(mathbb{R}_{+}) ).But are these functions dense in ( L^2(mathbb{R}_{+}) )? If they are, then ( f ) must be zero.I think that the set of functions ( { e^{-lambda x} }_{lambda in mathbb{N}} ) is not dense in ( L^2(mathbb{R}_{+}) ), but maybe their linear combinations are dense in some subspace.Alternatively, perhaps I can use the fact that the functions ( e^{-lambda x} ) can be used to approximate delta functions or something similar.Wait, another approach: consider the Fourier transform. If I can relate the Laplace transform to the Fourier transform, maybe I can use the uniqueness of the Fourier transform.Let me try a substitution. Let ( t = e^{-x} ), so ( x = -ln t ), and ( dx = -frac{1}{t} dt ). Then, the integral becomes[int_{0}^{1} t^{lambda} f(-ln t) frac{1}{t} dt = int_{0}^{1} t^{lambda - 1} f(-ln t) dt = 0]for all ( lambda in mathbb{N} ).So, defining ( g(t) = f(-ln t) ), we have[int_{0}^{1} t^{lambda - 1} g(t) dt = 0]for all ( lambda in mathbb{N} ).This looks like the integral of ( g(t) ) against the monomials ( t^{lambda - 1} ). So, ( g(t) ) is orthogonal to all monomials in ( L^2([0,1]) ).But the monomials ( { t^k }_{k=0}^{infty} ) form a basis for ( L^2([0,1]) ) under the inner product[langle f, g rangle = int_{0}^{1} f(t) g(t) dt.]Therefore, if ( g(t) ) is orthogonal to all basis elements, it must be the zero function almost everywhere.Hence, ( g(t) = 0 ) almost everywhere on ( [0,1] ), which implies that ( f(-ln t) = 0 ) almost everywhere. Since ( t ) ranges over ( (0,1) ), ( x = -ln t ) ranges over ( (0, infty) ). Therefore, ( f(x) = 0 ) almost everywhere on ( mathbb{R}_{+} ).Okay, that seems to make sense. Let me recap:1. I performed a substitution ( t = e^{-x} ) to transform the integral into one over ( [0,1] ).2. This led to defining a new function ( g(t) ) which is orthogonal to all monomials ( t^{lambda - 1} ).3. Since the monomials form a basis for ( L^2([0,1]) ), ( g(t) ) must be zero almost everywhere.4. Translating back to ( f(x) ), we conclude ( f(x) = 0 ) almost everywhere on ( mathbb{R}_{+} ).I think this is a solid argument. I used the substitution to make the problem fit into a space where I know the density of polynomials, which are the monomials here. Since ( g(t) ) is orthogonal to all polynomials, it must be zero, and hence ( f(x) ) is zero almost everywhere.I don't see any gaps in this reasoning. Maybe I should check if the substitution is valid. The substitution ( t = e^{-x} ) is smooth and invertible on ( mathbb{R}_{+} ), so it should preserve the measure class. Therefore, the integral transformation is valid, and the conclusion holds.Yeah, I think this works. So, the key steps were substitution to transform the integral into a more familiar space, recognizing the orthogonality to a dense set of functions, and then concluding that the function must be zero almost everywhere.**Final Answer**boxed{f = 0 text{ almost everywhere with respect to the Lebesgue measure.}}

💡First, I need to determine the number of schools in Pittsburgh based on the information provided about the new city's urban plan.1. **Identify the quantities for the new city:** - **Stores:** The new city should have half the number of stores as Pittsburgh. If Pittsburgh has 2000 stores, then the new city will have 2000 / 2 = 1000 stores. - **Hospitals:** The new city should have twice as many hospitals as Pittsburgh. With 500 hospitals in Pittsburgh, the new city will have 500 * 2 = 1000 hospitals. - **Schools:** The new city should have 50 fewer schools than Pittsburgh. If Pittsburgh has S schools, then the new city will have S - 50 schools. - **Police Stations:** The new city should have 5 more police stations than Pittsburgh. With 20 police stations in Pittsburgh, the new city will have 20 + 5 = 25 police stations.2. **Set up the equation for the total number of buildings in the new city:** [ 1000 text{ (stores)} + 1000 text{ (hospitals)} + (S - 50) text{ (schools)} + 25 text{ (police stations)} = 2175 ]3. **Simplify and solve for S:** [ 1000 + 1000 + S - 50 + 25 = 2175 ] [ 2000 + S - 25 = 2175 ] [ S = 2175 - 2000 + 25 ] [ S = 175 + 25 ] [ S = 200 ]Therefore, Pittsburgh has 200 schools.

💡Okay, so I need to find the limit as n approaches infinity of the expression (n + 5)(1 - 3n) divided by (2n + 1) squared. Hmm, let me write that down to visualize it better:Limit = limₙ→∞ [(n + 5)(1 - 3n) / (2n + 1)²]Alright, so first, I remember that when dealing with limits involving polynomials, especially as n approaches infinity, the highest degree terms dominate the behavior of the expression. So, maybe I can simplify this by looking at the leading terms of the numerator and the denominator.Let me start by expanding the numerator. The numerator is (n + 5)(1 - 3n). Let me multiply these two binomials:First, multiply n by 1, which is n. Then, n multiplied by -3n is -3n². Next, 5 multiplied by 1 is 5, and 5 multiplied by -3n is -15n. So, putting it all together:Numerator = n * 1 + n * (-3n) + 5 * 1 + 5 * (-3n)= n - 3n² + 5 - 15nNow, let's combine like terms. The n terms are n and -15n, which combine to -14n. So, the numerator simplifies to:Numerator = -3n² - 14n + 5Okay, so the numerator is a quadratic polynomial with leading term -3n².Now, the denominator is (2n + 1) squared. Let me expand that as well:Denominator = (2n + 1)² = (2n)² + 2*(2n)*(1) + (1)²= 4n² + 4n + 1So, the denominator is also a quadratic polynomial with leading term 4n².Now, the expression becomes:Limit = limₙ→∞ [(-3n² - 14n + 5) / (4n² + 4n + 1)]Since both the numerator and the denominator are quadratic polynomials, their degrees are the same. I remember that when the degrees of the numerator and denominator are equal, the limit as n approaches infinity is the ratio of their leading coefficients.In this case, the leading coefficient of the numerator is -3, and the leading coefficient of the denominator is 4. So, the limit should be -3/4.But wait, let me make sure I'm not missing anything. Sometimes, even if the degrees are the same, there might be some cancellation or something else, but in this case, since both are quadratics, and the leading terms are the dominant ones as n grows large, I think it's safe to say that the limit is just the ratio of the leading coefficients.Just to double-check, maybe I can divide both the numerator and the denominator by n², the highest power of n in the denominator. Let's try that:Numerator: (-3n² - 14n + 5) / n² = -3 - 14/n + 5/n²Denominator: (4n² + 4n + 1) / n² = 4 + 4/n + 1/n²So, the expression becomes:Limit = limₙ→∞ [(-3 - 14/n + 5/n²) / (4 + 4/n + 1/n²)]Now, as n approaches infinity, the terms with 1/n and 1/n² will approach zero. So, the expression simplifies to:Limit = (-3 - 0 + 0) / (4 + 0 + 0) = -3/4Yep, that confirms it. So, the limit is indeed -3/4.I think that's solid. I don't see any mistakes in my reasoning. I expanded both the numerator and the denominator, simplified them, and then applied the rule for limits of rational functions where the degrees of the numerator and denominator are equal. Dividing by n² also helped to see that the other terms become negligible as n grows large.Just to recap:1. Expand the numerator: (n + 5)(1 - 3n) = -3n² - 14n + 52. Expand the denominator: (2n + 1)² = 4n² + 4n + 13. Observe that both are quadratic, so the limit is the ratio of leading coefficients: -3/44. Verified by dividing numerator and denominator by n², leading to the same result.So, I'm confident that the limit is -3/4.**Final Answer**The limit is boxed{-dfrac{3}{4}}.