answer：Okay, so I have this problem here where I need to figure out what happens when the graph of ( y = lg x ) is rotated 90 degrees counterclockwise about the origin. The options given are (A) through (E), and I need to pick the correct equation for the new graph ( G' ).First, let me recall what the graph of ( y = lg x ) looks like. Since ( lg x ) is the logarithm base 10, it's defined for ( x > 0 ). The graph passes through (1, 0) because ( lg 1 = 0 ), and it increases slowly as ( x ) increases. It also approaches the y-axis as ( x ) approaches 0 from the right, but never touches it.Now, the problem says we're rotating this graph 90 degrees counterclockwise about the origin. I remember that rotating a point (x, y) 90 degrees counterclockwise around the origin results in the point (-y, x). So, if I have a point (a, b) on the original graph ( G ), after rotation, it will be at (-b, a) on the new graph ( G' ).Let me try to apply this transformation to the equation ( y = lg x ). If I take a general point (x, y) on ( G ), then after rotation, it becomes (-y, x) on ( G' ). So, to find the equation of ( G' ), I need to express the relationship between the new x and y coordinates.Let me denote the new coordinates after rotation as (X, Y). So, according to the rotation, we have:[ X = -y ][ Y = x ]But on the original graph ( G ), we know that ( y = lg x ). So, substituting ( y ) from the original equation into the expression for ( X ), we get:[ X = -lg x ]But since ( Y = x ), we can substitute ( x ) with ( Y ):[ X = -lg Y ]Now, I need to solve this equation for Y in terms of X to get the equation of ( G' ). Let's rewrite the equation:[ X = -lg Y ]Multiplying both sides by -1:[ -X = lg Y ]To solve for Y, I can rewrite the logarithmic equation in its exponential form. Remember that ( lg Y = log_{10} Y ), so:[ Y = 10^{-X} ]So, the equation of the rotated graph ( G' ) is ( Y = 10^{-X} ). If I switch back to the standard variables x and y, the equation becomes:[ y = 10^{-x} ]Looking back at the options provided:(A) ( y = lg left( frac{x+90}{9} right) )(B) ( y = log_{x} 10 )(C) ( y = frac{1}{x+1} )(D) ( y = 10^{-x} )(E) ( y = 10^{x} )The equation I derived is ( y = 10^{-x} ), which corresponds to option (D). Let me just double-check to make sure I didn't make any mistakes in my reasoning.Starting from the original graph ( y = lg x ), I rotated it 90 degrees counterclockwise, which transformed each point (x, y) to (-y, x). Substituting ( y = lg x ) into this transformation gave me ( X = -lg Y ). Solving for Y, I got ( Y = 10^{-X} ), which is indeed ( y = 10^{-x} ). I think that makes sense because rotating the logarithmic graph should result in an exponential function, and since it's a counterclockwise rotation, the exponent should be negative. If it were a clockwise rotation, it might have been positive, but counterclockwise flips the sign.So, I feel confident that option (D) is the correct answer.**Final Answer**boxed{D}

answer：Okay, so I have this problem where the segments connecting the bases of the altitudes of an acute-angled triangle are 8, 15, and 17 units long. I need to find the radius of the circumcircle around the triangle. Hmm, let me try to figure this out step by step.First, I remember that in a triangle, the altitudes intersect at a point called the orthocenter. The bases of these altitudes form a smaller triangle called the orthic triangle. So, in this case, the orthic triangle has sides of lengths 8, 15, and 17. Wait, those numbers look familiar. 8, 15, 17 is a Pythagorean triple because 8² + 15² = 64 + 225 = 289 = 17². So, the orthic triangle is a right-angled triangle. That's interesting.Now, I need to relate this orthic triangle to the original triangle. I recall that the orthic triangle is similar to the original triangle but scaled down. But I'm not sure about the exact relationship. Maybe I should look into properties of orthic triangles and circumcircles.I remember that the circumradius of a triangle can be found using the formula R = a / (2 sin A), where a is the length of a side and A is the angle opposite that side. But I'm not sure how this applies directly here since I don't have the sides of the original triangle, only the sides of the orthic triangle.Wait, maybe there's a relationship between the orthic triangle and the original triangle's circumradius. I think the orthic triangle is related to the original triangle's circumcircle in some way. Let me think. If I extend the altitudes beyond the orthocenter, they meet the circumcircle again. So, each altitude is extended beyond the orthocenter to meet the circumcircle, effectively doubling the length from the orthocenter to the vertex.So, if the orthic triangle has sides 8, 15, 17, then the original triangle's sides might be related by some factor. Since the orthic triangle is formed by the feet of the altitudes, and the original triangle's sides are related through the altitudes, perhaps the sides of the original triangle are twice the lengths of the orthic triangle's sides?Wait, no, that might not be accurate. Let me think again. The orthic triangle's sides are the segments connecting the feet of the altitudes, so they are not directly proportional to the original triangle's sides. Instead, they are related through the original triangle's altitudes and angles.I remember that in an acute triangle, the orthocenter lies inside the triangle. Also, the orthic triangle is similar to the original triangle but scaled down by a factor. Maybe the ratio is related to the cosine of the angles or something like that.Alternatively, perhaps I can use the fact that the orthic triangle is similar to the original triangle. If the orthic triangle is similar, then the ratio of their sides would be the same. But I'm not sure about the exact ratio. Maybe I need another approach.Let me consider the nine-point circle. The nine-point circle passes through the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments from each vertex to the orthocenter. The radius of the nine-point circle is half the radius of the circumcircle. So, if I can find the radius of the nine-point circle, I can double it to get the circumradius.But how does the orthic triangle relate to the nine-point circle? The orthic triangle is inscribed in the nine-point circle. So, the circumradius of the orthic triangle is equal to the radius of the nine-point circle. Since the nine-point circle has a radius half of the circumradius, then the circumradius of the original triangle is twice the circumradius of the orthic triangle.Ah, that seems promising. So, if I can find the circumradius of the orthic triangle, I can double it to get the circumradius of the original triangle.Given that the orthic triangle has sides 8, 15, 17, which is a right-angled triangle, its circumradius can be easily found. In a right-angled triangle, the circumradius is half the hypotenuse. So, the hypotenuse here is 17, so the circumradius is 17 / 2 = 8.5.Therefore, the nine-point circle has a radius of 8.5, which means the circumradius of the original triangle is twice that, so 17.Wait, let me verify that again. The orthic triangle is inscribed in the nine-point circle, so its circumradius is equal to the nine-point circle's radius. Since the nine-point circle's radius is half the circumradius of the original triangle, then yes, the original triangle's circumradius is twice the orthic triangle's circumradius.So, since the orthic triangle's circumradius is 8.5, the original triangle's circumradius is 17.Hmm, that seems straightforward. Let me see if there's another way to think about it to confirm.Another approach: In an acute triangle, the orthocenter lies inside the triangle. The segments connecting the feet of the altitudes (the orthic triangle) have lengths 8, 15, 17. Since these form a right triangle, the original triangle must have some special properties.Wait, if the orthic triangle is right-angled, does that mean the original triangle is also right-angled? No, because the original triangle is acute-angled. So, the orthic triangle being right-angled doesn't imply the original triangle is right-angled.But perhaps the relationship between the sides still holds. Since the orthic triangle is similar to the original triangle scaled down by a factor, but I think the scaling factor is related to the cosine of the angles.Wait, maybe I should use trigonometric relationships. Let me denote the original triangle as ABC, with orthocenter H. The feet of the altitudes are D, E, F. So, DEF is the orthic triangle with sides 8, 15, 17.In triangle ABC, the sides can be related to the orthic triangle's sides. I remember that the length of the side opposite to vertex A in the orthic triangle is equal to BC cos A. Similarly for the other sides.So, if DE = 8, EF = 15, FD = 17, then:DE = BC cos A = 8EF = AC cos B = 15FD = AB cos C = 17So, BC = 8 / cos AAC = 15 / cos BAB = 17 / cos CBut in triangle ABC, by the Law of Cosines, we have:AB² = AC² + BC² - 2 AC BC cos ABut this might get complicated. Maybe there's a better way.Alternatively, since the orthic triangle is similar to the original triangle, but scaled by a factor. Wait, is that true? I think the orthic triangle is similar to the original triangle only in certain cases, like in equilateral triangles, but not necessarily in all acute triangles.Wait, no, actually, the orthic triangle is similar to the original triangle if and only if the original triangle is equilateral. So, that might not be the case here.Hmm, maybe I need to think about the Euler line. The Euler line connects the orthocenter, centroid, and circumcenter. The distance between the orthocenter and circumcenter is related to the circumradius. But I'm not sure how that helps here.Wait, another thought: The circumradius of the orthic triangle is equal to half the circumradius of the original triangle. Is that correct?Let me check. The nine-point circle has a radius equal to half the circumradius. Since the orthic triangle is inscribed in the nine-point circle, its circumradius is equal to the nine-point circle's radius, which is half the original circumradius. So yes, the orthic triangle's circumradius is R/2, where R is the original triangle's circumradius.Therefore, if I can find the orthic triangle's circumradius, I can double it to get R.Given that the orthic triangle has sides 8, 15, 17, which is a right-angled triangle, its circumradius is half the hypotenuse, which is 17/2 = 8.5. Therefore, the original triangle's circumradius is 8.5 * 2 = 17.That seems consistent with my earlier conclusion. So, I think the radius of the circumcircle around the triangle is 17.Just to make sure, let me recap:1. The orthic triangle has sides 8, 15, 17, which is a right-angled triangle.2. The circumradius of the orthic triangle is half the hypotenuse, so 17/2 = 8.5.3. The nine-point circle, which has a radius equal to the orthic triangle's circumradius, is half the original triangle's circumradius.4. Therefore, the original triangle's circumradius is 8.5 * 2 = 17.Yes, that all adds up. I think I'm confident with this answer.**Final Answer**The radius of the circumcircle around the triangle is boxed{17}.

answer：Okay, so I need to find the greatest common factor (GCF) of 36 and 45. Hmm, I remember that the GCF is the largest number that divides both numbers without leaving a remainder. But I'm not entirely sure how to find it step by step. Maybe I should start by listing out the factors of each number and then see which one is the largest common factor.First, let's list the factors of 36. To do this, I'll think of all the numbers that multiply together to give 36. Starting from 1, I know that 1 times 36 is 36, so 1 and 36 are factors. Then, 2 times 18 is 36, so 2 and 18 are factors. Next, 3 times 12 is 36, so 3 and 12 are factors. Continuing this way, 4 times 9 is 36, so 4 and 9 are factors. Finally, 6 times 6 is 36, so 6 is also a factor. So the factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36.Now, let's do the same for 45. Starting with 1 times 45, so 1 and 45 are factors. Then, 3 times 15 is 45, so 3 and 15 are factors. Next, 5 times 9 is 45, so 5 and 9 are factors. I don't think there are any more factors beyond that because 7 times something would be more than 45, and 8 times something doesn't divide evenly into 45. So the factors of 45 are: 1, 3, 5, 9, 15, 45.Now, I need to find the common factors between 36 and 45. Looking at the lists:Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.Factors of 45: 1, 3, 5, 9, 15, 45.The common factors are the numbers that appear in both lists. So, 1, 3, and 9 are common factors.Out of these, the greatest one is 9. So, the greatest common factor of 36 and 45 is 9.Wait, but I think there's another method using prime factorization. Maybe I should try that to double-check.For prime factorization, I break down each number into its prime factors.Starting with 36:36 divided by 2 is 18.18 divided by 2 is 9.9 divided by 3 is 3.3 divided by 3 is 1.So, the prime factors of 36 are 2 × 2 × 3 × 3, or (2^2 times 3^2).Now for 45:45 divided by 3 is 15.15 divided by 3 is 5.5 divided by 5 is 1.So, the prime factors of 45 are 3 × 3 × 5, or (3^2 times 5).To find the GCF using prime factors, I look for the common prime factors with the lowest exponents. Both numbers have (3^2) as a common factor. There are no other common prime factors.So, the GCF is (3^2 = 9).That matches what I found earlier by listing the factors. So, I'm confident that the greatest common factor of 36 and 45 is indeed 9.

answer：Okay, so I need to figure out how to derive the formula for the sum of the first n terms of an arithmetic sequence using the reverse order addition method. Hmm, I remember that an arithmetic sequence is one where each term increases by a constant difference. So, if the first term is a₁ and the common difference is d, then the second term is a₁ + d, the third is a₁ + 2d, and so on. Let me write down the sum of the first n terms, which is Sₙ. So, Sₙ = a₁ + a₂ + a₃ + ... + aₙ. Since it's an arithmetic sequence, each term can be expressed as a₁ + (k-1)d where k is the term number. So, a₂ = a₁ + d, a₃ = a₁ + 2d, etc., up to aₙ = a₁ + (n-1)d.Now, the problem mentions using the reverse order addition method. I think that means writing the sum forwards and then backwards and adding them together. Let me try that.First, write Sₙ forwards:Sₙ = a₁ + a₂ + a₃ + ... + aₙ₋₁ + aₙ.Now, write Sₙ backwards:Sₙ = aₙ + aₙ₋₁ + aₙ₋₂ + ... + a₂ + a₁.If I add these two equations together, term by term, what do I get? Let's see:(a₁ + aₙ) + (a₂ + aₙ₋₁) + (a₃ + aₙ₋₂) + ... + (aₙ₋₁ + a₂) + (aₙ + a₁).Wait, that seems a bit confusing. Let me think again. When I add the two equations, each pair of terms from the front and back should add up to the same value. For example, a₁ + aₙ, a₂ + aₙ₋₁, and so on. Since the sequence is arithmetic, each of these pairs should have the same sum.Let me verify that. Take a₁ + aₙ. Since aₙ = a₁ + (n-1)d, then a₁ + aₙ = a₁ + [a₁ + (n-1)d] = 2a₁ + (n-1)d.Similarly, a₂ + aₙ₋₁. a₂ = a₁ + d, and aₙ₋₁ = a₁ + (n-2)d. So, a₂ + aₙ₋₁ = [a₁ + d] + [a₁ + (n-2)d] = 2a₁ + (n-1)d. Oh, so each pair adds up to the same value, 2a₁ + (n-1)d. That makes sense because the sequence is arithmetic, so the terms are evenly spaced. Therefore, each pair from the start and end of the sequence will have the same sum.Since there are n terms in total, when I add the two equations, I have n pairs each adding up to 2a₁ + (n-1)d. So, the total sum of the two equations is n*(2a₁ + (n-1)d). But wait, that's actually 2Sₙ because I added Sₙ + Sₙ.So, 2Sₙ = n*(2a₁ + (n-1)d). Therefore, to find Sₙ, I can divide both sides by 2:Sₙ = [n*(2a₁ + (n-1)d)] / 2.Hmm, that seems correct, but I recall there's another way to write this formula using the first and last terms. Since aₙ = a₁ + (n-1)d, I can substitute that into the formula:Sₙ = [n*(a₁ + aₙ)] / 2.Yes, that looks familiar. So, the sum of the first n terms of an arithmetic sequence is equal to the number of terms multiplied by the average of the first and last terms.Let me double-check this with a simple example. Suppose I have the sequence 1, 3, 5, 7. Here, a₁ = 1, d = 2, and n = 4. The sum S₄ should be 1 + 3 + 5 + 7 = 16.Using the formula, S₄ = [4*(1 + 7)] / 2 = [4*8]/2 = 32/2 = 16. That matches. Another example: sequence 2, 4, 6, 8, 10. Here, a₁ = 2, d = 2, n = 5. Sum is 2 + 4 + 6 + 8 + 10 = 30.Using the formula: S₅ = [5*(2 + 10)] / 2 = [5*12]/2 = 60/2 = 30. Perfect, it works.So, the key idea is that by writing the sum forwards and backwards and adding them, each pair of terms adds up to the same value, allowing us to find a simple expression for the total sum.

answer：Okay, so I have this problem: Given the expansion of (1 - 3x)^9, which is equal to a + a₁x + a₂x² + ... + a₉x⁹, I need to find the sum of the absolute values of the coefficients, that is |a| + |a₁| + |a₂| + ... + |a₉|. Hmm, that sounds a bit tricky, but let me see if I can figure it out.First, I remember that when you expand a binomial like (a + b)^n, you can use the binomial theorem. The binomial theorem says that (a + b)^n = Σ (from k=0 to n) [C(n, k) * a^(n-k) * b^k], where C(n, k) is the combination of n things taken k at a time. So in this case, our binomial is (1 - 3x)^9, so a is 1, b is -3x, and n is 9.So, expanding this, each term will be C(9, k) * 1^(9 - k) * (-3x)^k. Simplifying that, since 1 to any power is just 1, each term becomes C(9, k) * (-3)^k * x^k. Therefore, the coefficients a_k are C(9, k) * (-3)^k.Now, the problem is asking for the sum of the absolute values of these coefficients. So, I need to compute |a| + |a₁| + |a₂| + ... + |a₉|. Since each a_k is C(9, k) * (-3)^k, the absolute value of each a_k is C(9, k) * 3^k, because the absolute value of (-3)^k is 3^k regardless of whether k is even or odd.So, the sum we're looking for is Σ (from k=0 to 9) [C(9, k) * 3^k]. Wait a second, that looks familiar. Isn't that the expansion of (1 + 3)^9? Because according to the binomial theorem, (1 + 3)^9 = Σ (from k=0 to 9) [C(9, k) * 1^(9 - k) * 3^k], which is exactly the same as the sum I have here. So, that means Σ (from k=0 to 9) [C(9, k) * 3^k] = (1 + 3)^9 = 4^9.Therefore, the sum of the absolute values of the coefficients is 4^9. Let me just double-check that. If I set x = 1 in the original expression, I get (1 - 3*1)^9 = (-2)^9, which is -512. But that's not the sum of the absolute values; that's the sum of the coefficients without taking absolute values. So, setting x = 1 gives me the alternating sum, but I need the sum of absolute values.Wait, how do I get the sum of absolute values? Maybe I can substitute x with a value that makes all the coefficients positive. Since the coefficients alternate in sign because of the (-3x) term, if I substitute x with -1, then (-3*(-1))^k becomes (3)^k, which is positive for all k. So, substituting x = -1 into the original expression gives me (1 - 3*(-1))^9 = (1 + 3)^9 = 4^9, which is exactly the sum of the absolute values of the coefficients. That makes sense because when x = -1, all the terms become positive, so the result is the sum of the absolute values.So, putting it all together, the sum |a| + |a₁| + |a₂| + ... + |a₉| is equal to 4^9. Calculating 4^9, 4 squared is 16, 4 cubed is 64, 4^4 is 256, 4^5 is 1024, 4^6 is 4096, 4^7 is 16384, 4^8 is 65536, and 4^9 is 262144. So, the value is 262,144.I think that's the answer. Let me just recap to make sure I didn't make a mistake. The key idea was recognizing that substituting x = -1 into the binomial expression would yield the sum of the absolute values of the coefficients because it turns all the negative coefficients into positive ones. Then, using the binomial theorem, I saw that this substitution simplifies to (1 + 3)^9, which is 4^9. Calculating that gives me 262,144. Yeah, that seems right.**Final Answer**The value of ( |a| + |a_1| + |a_2| + ldots + |a_9| ) is boxed{262144}.

answer：Okay, so I need to figure out how many x-intercepts the graph of y = sin(1/x) has in the interval (0.001, 0.01). Hmm, x-intercepts occur where y = 0, which means sin(1/x) = 0. I remember that sin(θ) = 0 when θ is an integer multiple of π. So, 1/x should equal kπ, where k is an integer. That gives me x = 1/(kπ).Now, I need to find all such x that lie between 0.001 and 0.01. So, I can set up the inequality:0.001 < 1/(kπ) < 0.01I can rewrite this inequality to solve for k. Let me flip all parts of the inequality, remembering that flipping inequalities reverses them:1/0.01 > kπ > 1/0.001Calculating those values:1/0.01 = 1001/0.001 = 1000So, the inequality becomes:100 < kπ < 1000Now, divide all parts by π to solve for k:100/π < k < 1000/πI know π is approximately 3.14159, so let me compute 100/π and 1000/π.100/π ≈ 100 / 3.14159 ≈ 31.8309886181000/π ≈ 1000 / 3.14159 ≈ 318.30988618Since k has to be an integer, I need the smallest integer greater than 31.830988618 and the largest integer less than 318.30988618.The smallest integer greater than 31.83 is 32, and the largest integer less than 318.31 is 318.So, k can take integer values from 32 up to 318. To find how many integers are in this range, I subtract the lower bound from the upper bound and add 1 (since both endpoints are included).Number of integers = 318 - 32 + 1 = 287Wait, but in the initial solution, it was 318 - 32 = 286. Hmm, why is there a discrepancy? Let me double-check.If k starts at 32 and ends at 318, inclusive, then the count is 318 - 32 + 1. Let me compute that:318 - 32 = 286286 + 1 = 287So, there should be 287 integers. But the initial answer was 286. Did I make a mistake?Wait, maybe the initial calculation didn't include one of the endpoints. Let me see. If k starts at 32 and ends at 318, that's 318 - 32 + 1 = 287. But if k starts at 32 and ends at 318, but perhaps 318 is not included because 1/(318π) might be less than 0.01? Let me check.Compute 1/(318π):1/(318 * 3.14159) ≈ 1/(1000) = 0.001Wait, no, 318 * π is approximately 1000, so 1/(318π) ≈ 0.001, which is the lower bound. But the interval is (0.001, 0.01), so 0.001 is excluded. Therefore, x must be greater than 0.001, so 1/(kπ) > 0.001, which means k < 1000/π ≈ 318.309. So, k can be up to 318, but 1/(318π) ≈ 0.001, which is not included in the interval. Therefore, k must be less than 318.309, so the maximum integer k is 318, but 1/(318π) is exactly 0.001, which is excluded. Therefore, the maximum k is 317.Wait, that changes things. So, if k must satisfy 1/(kπ) > 0.001, then k < 1000/π ≈ 318.309, so the maximum integer k is 318, but 1/(318π) = 0.001, which is not in the interval. Therefore, the maximum k is 317.Similarly, the lower bound is 1/(kπ) < 0.01, so k > 100/π ≈ 31.83, so the minimum integer k is 32.Therefore, k ranges from 32 to 317 inclusive.So, the number of integers is 317 - 32 + 1 = 286.Ah, that's where the initial answer came from. So, I initially thought it was 287, but since 318 gives x = 0.001, which is excluded, we have to stop at k=317, giving 286 intercepts.So, the correct number is 286.