answer：Alright, so I've got this geometry problem here with four propositions about lines and planes. I need to figure out how many of them are correct. Let me take it step by step.First, let me restate the propositions to make sure I understand them:1. If line a is perpendicular to line b, and line a is perpendicular to plane α, then line b is parallel to plane α.2. If line a is parallel to plane α, and plane α is perpendicular to plane β, then line a is perpendicular to plane β.3. If line a is perpendicular to plane β, and plane α is perpendicular to plane β, then line a is parallel to plane α.4. If line a is perpendicular to line b, line a is perpendicular to plane α, and line b is perpendicular to plane β, then plane α is perpendicular to plane β.Okay, so four statements. I need to evaluate each one.Starting with proposition ①: If a ⊥ b and a ⊥ α, then b // α.Hmm. So, if line a is perpendicular to both line b and plane α, does that mean line b is parallel to plane α? Let me visualize this. If a is perpendicular to α, that means a is a normal vector to the plane α. If a is also perpendicular to b, then b lies in a plane that's perpendicular to a. But does that necessarily mean b is parallel to α?Wait, actually, if a is perpendicular to α, then any line in α is perpendicular to a. So if b is perpendicular to a, b could either lie in α or be parallel to α. But since a and b are different lines, and a is perpendicular to α, b could be in α or parallel to it. So, is b necessarily parallel to α? Or could it lie in α?I think it's possible for b to lie in α. For example, if a is a vertical line, and α is a horizontal plane, then any horizontal line b in α would be perpendicular to a. So in that case, b is not parallel to α, it's lying in α. Therefore, proposition ① is not necessarily true. So ① is incorrect.Moving on to proposition ②: If a // α and α ⊥ β, then a ⊥ β.So, if line a is parallel to plane α, and plane α is perpendicular to plane β, does that mean line a is perpendicular to plane β?Hmm. Let me think. If a is parallel to α, then a doesn't intersect α and lies in a direction parallel to α. If α is perpendicular to β, then their dihedral angle is 90 degrees. But does that mean any line parallel to α is perpendicular to β?Wait, no. For example, imagine two perpendicular planes, like the floor and a wall. If I have a line parallel to the floor, it doesn't necessarily have to be perpendicular to the wall. It could be at some angle. For instance, a line running along the floor would be parallel to the floor but not necessarily perpendicular to the wall. It might intersect the wall at some angle, not necessarily 90 degrees. So proposition ② is incorrect.Next, proposition ③: If a ⊥ β and α ⊥ β, then a // α.So, if line a is perpendicular to plane β, and plane α is perpendicular to plane β, does that mean line a is parallel to plane α?Hmm. Let me visualize. If a is perpendicular to β, then a is a normal vector to β. If α is also perpendicular to β, then α and β intersect along a line, say l. Now, is a necessarily parallel to α?Wait, no. Because a is perpendicular to β, it could be that a is actually lying in α. For example, imagine two planes α and β intersecting at a line l. If a is perpendicular to β, it could be that a is in α and intersects l at a point. So a is not necessarily parallel to α; it could lie within α. Therefore, proposition ③ is incorrect.Finally, proposition ④: If a ⊥ b, a ⊥ α, and b ⊥ β, then α ⊥ β.So, if line a is perpendicular to line b, line a is perpendicular to plane α, and line b is perpendicular to plane β, does that mean plane α is perpendicular to plane β?Alright, let's break this down. If a is perpendicular to α, then a is a normal vector to α. Similarly, if b is perpendicular to β, then b is a normal vector to β. Now, if a is perpendicular to b, that means the normals of α and β are perpendicular. And if the normals of two planes are perpendicular, then the planes themselves are perpendicular. So, yes, this seems correct.Wait, let me make sure. If two planes have normals that are perpendicular, then the planes are indeed perpendicular. So, since a and b are normals to α and β respectively, and a is perpendicular to b, then α must be perpendicular to β. So proposition ④ is correct.So, in summary:①: Incorrect②: Incorrect③: Incorrect④: CorrectTherefore, only one proposition is correct. So the answer should be B: 1.**Final Answer**boxed{B}

answer：First, I need to determine the total amount Albert has and the additional amount he needs. He currently has 6.50 and needs 12 more, so the total amount he needs is 6.50 + 12 = 18.50.Next, I'll calculate the combined cost of the paintbrush and the set of paints. The paintbrush costs 1.50 and the paints cost 4.35, so together they cost 1.50 + 4.35 = 5.85.Finally, to find out how much the wooden easel costs, I'll subtract the combined cost of the paintbrush and paints from the total amount Albert needs. That is, 18.50 - 5.85 = 12.65.Therefore, the wooden easel costs 12.65.

answer：To determine the price of the fourth variety of tea, we start by noting the given prices and the ratio in which they are mixed. The prices are Rs. 126, Rs. 135, Rs. 156, and Rs. x per kg, and the quantities are mixed in the ratio 2:3:4:5. The average price of the mixture is Rs. 165 per kg.First, we express the quantities in terms of a common variable, say 'a':- Quantity of the first variety, Q1 = 2a- Quantity of the second variety, Q2 = 3a- Quantity of the third variety, Q3 = 4a- Quantity of the fourth variety, Q4 = 5aNext, we calculate the total cost of the mixture by multiplying each price by its respective quantity and summing them up:- Total cost = 126*(2a) + 135*(3a) + 156*(4a) + x*(5a) = 252a + 405a + 624a + 5ax = 1281a + 5axThe total quantity of the mixture is:- Total quantity = 2a + 3a + 4a + 5a = 14aGiven the average price of the mixture is Rs. 165 per kg, we set up the equation:- 165 = (1281a + 5ax) / 14aMultiplying both sides by 14a to eliminate the denominator:- 165 * 14a = 1281a + 5ax- 2310a = 1281a + 5axSubtracting 1281a from both sides:- 1029a = 5axDividing both sides by 5a:- x = 1029a / 5a = 205.8Therefore, the price of the fourth variety is Rs. 205.8 per kg.

answer：Alright, so I have this geometry problem here. It says that points with masses equal to the lengths of the sides are placed at the midpoints of the sides of triangle ABC. I need to prove that the center of mass of this system of points is located at the center of the inscribed circle of the triangle formed by the midpoints of the sides of triangle ABC.Hmm, okay, let me try to unpack this step by step. First, let me visualize triangle ABC. Let's denote the midpoints of the sides BC, CA, and AB as A1, B1, and C1 respectively. So, A1 is the midpoint of BC, B1 is the midpoint of CA, and C1 is the midpoint of AB.Now, the problem states that at each of these midpoints, we place a point with a mass equal to the length of the side opposite to that midpoint. So, at A1, which is the midpoint of BC, the mass would be equal to the length of BC, which is typically denoted as 'a' in triangle notation. Similarly, at B1, the midpoint of CA, the mass would be equal to the length of CA, denoted as 'b', and at C1, the midpoint of AB, the mass would be equal to the length of AB, denoted as 'c'.So, we have masses m(A1) = a, m(B1) = b, and m(C1) = c. Now, the center of mass (or centroid) of this system of points is the point where the weighted average of the positions of these points is located. The formula for the center of mass in a system of particles is given by:G = (m1*r1 + m2*r2 + m3*r3) / (m1 + m2 + m3)Where m1, m2, m3 are the masses and r1, r2, r3 are the position vectors of the points.In this case, the masses are a, b, c, and the points are A1, B1, C1. So, the center of mass G would be:G = (a*A1 + b*B1 + c*C1) / (a + b + c)Now, the problem wants me to show that this center of mass G is located at the incenter of the triangle formed by the midpoints A1, B1, C1. The incenter is the point where the angle bisectors of the triangle meet and is equidistant from all sides of the triangle.So, I need to show that G is the incenter of triangle A1B1C1.First, let me recall that triangle A1B1C1 is called the medial triangle of triangle ABC. The medial triangle has some interesting properties. For example, it is similar to the original triangle ABC, scaled down by a factor of 1/2, and its sides are parallel to the sides of ABC.Now, the incenter of the medial triangle. Hmm, I wonder if there's a relationship between the incenter of the medial triangle and the centroid or some other center of the original triangle.Wait, but in this case, the center of mass G is defined with masses proportional to the lengths of the sides of ABC. So, it's not just the centroid of the medial triangle, which would be the average of the midpoints without any masses.So, perhaps I need to consider the properties of the center of mass when masses are assigned to the midpoints.Let me think about barycentric coordinates. In barycentric coordinates, the centroid (center of mass) of a triangle is given by the average of the vertices. But here, the masses are not equal; they are proportional to the lengths of the sides.So, in barycentric coordinates, the center of mass G would have coordinates proportional to the masses. That is, if we consider triangle A1B1C1, the barycentric coordinates of G would be (a, b, c).But wait, in barycentric coordinates, the incenter of a triangle has coordinates proportional to the lengths of the sides. Specifically, the incenter has barycentric coordinates (a, b, c). So, if G has barycentric coordinates (a, b, c) with respect to triangle A1B1C1, then G must coincide with the incenter of A1B1C1.Is that correct? Let me verify.In barycentric coordinates, the incenter is indeed given by the normalized coordinates (a, b, c). That is, if we have a triangle with side lengths a, b, c, the incenter is at (a, b, c) in barycentric coordinates.But in our case, the masses are a, b, c, so the center of mass G would have barycentric coordinates (a, b, c) with respect to triangle A1B1C1. Therefore, G is the incenter of triangle A1B1C1.Wait, but triangle A1B1C1 is the medial triangle, so its side lengths are half of the original triangle's sides. So, the side lengths of A1B1C1 are a/2, b/2, c/2.Therefore, the incenter of A1B1C1 would have barycentric coordinates proportional to (a/2, b/2, c/2), which simplifies to (a, b, c). So, yes, the incenter of A1B1C1 has barycentric coordinates (a, b, c), which matches the center of mass G.Therefore, G is indeed the incenter of the medial triangle A1B1C1.Wait, but let me make sure I'm not confusing anything here. The incenter is defined as the intersection of the angle bisectors, and it's equidistant from all sides. The center of mass, on the other hand, is the weighted average of the points.But in this case, since the masses are proportional to the lengths of the sides, and the barycentric coordinates of the incenter are also proportional to the side lengths, they must coincide.So, putting it all together, the center of mass G of the system of points at the midpoints of ABC with masses equal to the side lengths is located at the incenter of the medial triangle A1B1C1.I think that makes sense. Let me try to write this more formally.First, define the midpoints A1, B1, C1 of sides BC, CA, AB respectively.Assign masses m(A1) = a, m(B1) = b, m(C1) = c.The center of mass G is given by:G = (a*A1 + b*B1 + c*C1) / (a + b + c)Now, consider triangle A1B1C1. The incenter of this triangle has barycentric coordinates proportional to the lengths of its sides. Since A1B1C1 is the medial triangle, its side lengths are a/2, b/2, c/2.Therefore, the incenter of A1B1C1 has barycentric coordinates (a/2, b/2, c/2), which simplifies to (a, b, c) when normalized.Thus, the center of mass G, which has barycentric coordinates (a, b, c) with respect to A1B1C1, coincides with the incenter of A1B1C1.Therefore, the center of mass of the system is located at the incenter of the medial triangle.I think that's a solid argument. I should probably check if there's any step where I might have made a mistake.Wait, when I say that the incenter has barycentric coordinates (a, b, c), is that correct? Let me recall. In barycentric coordinates, the incenter is given by (a : b : c), meaning the coordinates are proportional to the side lengths. So yes, if we consider the medial triangle, whose side lengths are half of the original, the incenter would still have coordinates proportional to (a, b, c), just scaled down.But since barycentric coordinates are homogeneous, scaling doesn't change the point. So, yes, the incenter of A1B1C1 is at (a, b, c) in barycentric coordinates, which matches the center of mass G.Okay, I think that's correct. So, the center of mass is indeed the incenter of the medial triangle.

answer：Alright, so I've got this math problem here, and I need to figure out what (35)^{mn} equals in terms of P and Q, where P is 5^m and Q is 7^n. The options are given, but I should try to solve it step by step without just guessing.First, let me write down what I know:- P = 5^m- Q = 7^n- I need to find (35)^{mn}Okay, so 35 is a number that I recognize as 5 multiplied by 7. So, 35 = 5 * 7. That might be useful because I see both 5 and 7 in the definitions of P and Q.So, if I rewrite 35 as 5 * 7, then (35)^{mn} becomes (5 * 7)^{mn}. Now, I remember that when you have an exponent applied to a product, you can distribute the exponent to each factor. So, (a * b)^c = a^c * b^c. Applying that here, I get:(5 * 7)^{mn} = 5^{mn} * 7^{mn}Alright, so now I have 5^{mn} times 7^{mn}. But I need to express this in terms of P and Q. Let's recall that P is 5^m and Q is 7^n. So, can I express 5^{mn} and 7^{mn} in terms of P and Q?Let me think about 5^{mn}. Since P = 5^m, if I raise P to the power of n, I get P^n = (5^m)^n = 5^{mn}. That works! Similarly, for 7^{mn}, since Q = 7^n, if I raise Q to the power of m, I get Q^m = (7^n)^m = 7^{mn}. Perfect!So, substituting back, 5^{mn} * 7^{mn} is equal to P^n * Q^m. That seems to match one of the answer choices. Let me check the options again:- A: P^2Q- B: P^nQ^m- C: P^nQ^{2m}- D: P^{2m}Q^n- E: P^{2n}Q^mYes, option B is P^nQ^m, which is exactly what I got. So, I think that's the correct answer.But wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from (35)^{mn}:1. 35 = 5 * 7, so (35)^{mn} = (5 * 7)^{mn}2. Applying the exponent rule: (5 * 7)^{mn} = 5^{mn} * 7^{mn}3. Expressing in terms of P and Q: - 5^{mn} = (5^m)^n = P^n - 7^{mn} = (7^n)^m = Q^m4. Combining them: 5^{mn} * 7^{mn} = P^n * Q^mEverything seems to check out. I don't see any errors in my reasoning. So, yeah, I'm confident that option B is the right answer.Just to be thorough, let me consider if any other options could make sense or if I might have misread something. For example, option A is P^2Q, which would mean (5^m)^2 * 7^n = 5^{2m} * 7^n. That's not the same as 5^{mn} * 7^{mn} unless m = 2 and n = 1, which isn't generally true for all integers m and n. So, A doesn't work.Option C is P^nQ^{2m}, which would be (5^m)^n * (7^n)^{2m} = 5^{mn} * 7^{2mn}. That's not the same as 5^{mn} * 7^{mn} unless m = 0 or n = 0, which isn't the case here. So, C is out.Option D is P^{2m}Q^n, which is (5^m)^{2m} * (7^n)^n = 5^{2m^2} * 7^{n^2}. That's definitely different from 5^{mn} * 7^{mn} unless m and n are specific values, which they aren't. So, D isn't correct.Option E is P^{2n}Q^m, which is (5^m)^{2n} * (7^n)^m = 5^{2mn} * 7^{mn}. Again, that's not the same as 5^{mn} * 7^{mn} unless m or n is 1, which isn't specified. So, E is also incorrect.Therefore, after eliminating the other options, B is the only one that fits for all integers m and n. I think I've covered all bases here, so I'm pretty sure B is the right choice.**Final Answer**boxed{B}

answer：Okay, so I need to figure out how many numbers between 1 and 2005 are multiples of 3 or 4 but not 12. Hmm, let me break this down step by step. First, I remember that when we're dealing with "or" in counting problems, it usually involves using the principle of inclusion-exclusion. That means I need to count the multiples of 3, count the multiples of 4, and then subtract the ones that are multiples of both 3 and 4 because those would have been counted twice. But wait, the problem also says "but not 12." So, after finding the total number of multiples of 3 or 4, I need to exclude those that are multiples of 12. That makes sense because 12 is the least common multiple of 3 and 4, so any multiple of 12 is already a multiple of both 3 and 4. Let me start by calculating the number of multiples of 3 between 1 and 2005. To do this, I can divide 2005 by 3 and take the floor of that value. So, 2005 divided by 3 is approximately 668.333. Taking the floor gives me 668. That means there are 668 multiples of 3 between 1 and 2005. Next, I'll do the same for multiples of 4. Dividing 2005 by 4 gives me 501.25. Taking the floor of that gives me 501. So, there are 501 multiples of 4 between 1 and 2005. Now, I need to find how many numbers are multiples of both 3 and 4, which are the multiples of 12. Dividing 2005 by 12 gives me approximately 167.083. Taking the floor gives me 167. So, there are 167 multiples of 12 between 1 and 2005. Using the principle of inclusion-exclusion, the total number of multiples of 3 or 4 is the sum of the multiples of 3 and 4 minus the multiples of 12. That would be 668 + 501 - 167. Let me calculate that: 668 + 501 is 1169, and 1169 - 167 is 1002. But wait, the problem says "but not 12," so I think I need to subtract the multiples of 12 again because they were included in the initial count. Hmm, no, actually, in the inclusion-exclusion principle, we subtract the multiples of 12 once to correct for double-counting. So, the number 1002 already excludes the multiples of 12. Wait, no, that doesn't seem right. Let me think again. The inclusion-exclusion gives me the count of numbers that are multiples of 3 or 4, including those that are multiples of both. But since the problem says "but not 12," I need to exclude those multiples of 12 from the total. So, if I have 1002 numbers that are multiples of 3 or 4, and 167 of those are multiples of 12, then the number of multiples of 3 or 4 but not 12 would be 1002 - 167. Let me calculate that: 1002 - 167 is 835. Wait, but that contradicts my earlier thought. I'm confused now. Let me verify. When I used inclusion-exclusion, I added the multiples of 3 and 4, which included the multiples of 12 twice, so I subtracted them once to get the correct count of multiples of 3 or 4. So, the 1002 includes all multiples of 3 or 4, including those that are multiples of 12. Therefore, to exclude the multiples of 12, I need to subtract them again. So, 1002 - 167 = 835. That would be the number of multiples of 3 or 4 but not 12. But looking back at the answer choices, 835 is one of them, option C. However, I remember that sometimes when we use inclusion-exclusion, we might have to adjust differently. Let me double-check my calculations. Multiples of 3: floor(2005/3) = 668. Multiples of 4: floor(2005/4) = 501. Multiples of 12: floor(2005/12) = 167. Inclusion-exclusion: 668 + 501 - 167 = 1002. But since we need to exclude multiples of 12, we subtract 167 again: 1002 - 167 = 835. Wait, but another way to think about it is that the multiples of 12 are the intersection of multiples of 3 and 4. So, when we do inclusion-exclusion, we already accounted for them once. So, if we want to exclude them entirely, we need to subtract them from the total. Alternatively, maybe I should calculate it differently. Instead of inclusion-exclusion, I can calculate the number of multiples of 3 not divisible by 12 plus the number of multiples of 4 not divisible by 12. Let me try that approach. Multiples of 3: 668. Multiples of 3 that are also multiples of 12: 167. So, multiples of 3 not divisible by 12: 668 - 167 = 501. Similarly, multiples of 4: 501. Multiples of 4 that are also multiples of 12: 167. So, multiples of 4 not divisible by 12: 501 - 167 = 334. Now, adding these two together: 501 + 334 = 835. Okay, that matches my previous result. So, the number of multiples of 3 or 4 but not 12 is 835. But wait, looking at the answer choices, option C is 835, but I recall that sometimes in these problems, the wording can be tricky. Let me make sure I interpreted it correctly. The problem says "multiples of 3 or 4 but not 12." So, it's the union of multiples of 3 and multiples of 4, excluding those that are multiples of 12. Yes, that's exactly what I calculated. So, 835 should be the correct answer. But hold on, when I first did inclusion-exclusion, I got 1002, which is the total number of multiples of 3 or 4, including those divisible by 12. Then, subtracting the 167 multiples of 12 gives me 835. Alternatively, if I think about it as (multiples of 3 + multiples of 4 - 2*multiples of 12), that would be 668 + 501 - 2*167 = 668 + 501 - 334 = 835. Yes, that also gives me 835. So, I think 835 is the correct answer, which is option C. But wait, let me check the answer choices again. The options are: (A) 501 (B) 668 (C) 835 (D) 1002 (E) 1169 So, 835 is option C. But I'm a bit confused because sometimes in these problems, people might interpret "but not 12" differently. Maybe they mean numbers that are multiples of 3 or 4 but not both, which would be different. But in this case, the problem says "but not 12," which specifically refers to excluding multiples of 12. So, I think my approach is correct. Just to be thorough, let me consider another angle. Total numbers from 1 to 2005: 2005. Numbers that are multiples of 3 or 4: 1002. Numbers that are multiples of 12: 167. So, numbers that are multiples of 3 or 4 but not 12: 1002 - 167 = 835. Yes, that makes sense. Alternatively, if I think about it as the union of multiples of 3 and multiples of 4, minus the intersection (which is multiples of 12), that's exactly what inclusion-exclusion does. So, the count is 1002, and since we want to exclude the intersection, we subtract 167, giving 835. I think I'm confident now that 835 is the correct answer. But wait, let me check if my initial counts are correct. Multiples of 3: floor(2005/3) = 668. 3*668 = 2004, which is less than 2005, so that's correct. Multiples of 4: floor(2005/4) = 501. 4*501 = 2004, which is also less than 2005, so that's correct. Multiples of 12: floor(2005/12) = 167. 12*167 = 2004, which is less than 2005, so that's correct. Therefore, my calculations are accurate. So, the final answer is 835, which is option C. Wait, but hold on a second. I just realized that when I subtracted the multiples of 12 from the inclusion-exclusion result, I got 835. But another way to think about it is that the multiples of 12 are already included in both the multiples of 3 and 4 counts, so when I subtract them once in inclusion-exclusion, I'm left with the count of numbers that are multiples of 3 or 4 but not both. But in this problem, we want to exclude multiples of 12 regardless of whether they are multiples of both or not. Wait, no, actually, multiples of 12 are exactly the numbers that are multiples of both 3 and 4. So, when we do inclusion-exclusion, we subtract the multiples of 12 once to avoid double-counting. But the problem wants to exclude all multiples of 12, regardless of whether they were counted once or twice. So, in the inclusion-exclusion count of 1002, the multiples of 12 are counted once (since we subtracted them once). Therefore, to exclude them entirely, we need to subtract them again. So, 1002 - 167 = 835. Yes, that makes sense. Alternatively, if I think about it as the union of multiples of 3 and multiples of 4, minus the intersection (which is multiples of 12), that's exactly what inclusion-exclusion does. So, the count is 1002, and since we want to exclude the intersection, we subtract 167, giving 835. I think I'm confident now that 835 is the correct answer. But just to be absolutely sure, let me consider a smaller example to test my logic. Suppose we have numbers from 1 to 12. Multiples of 3: 3,6,9,12 → 4 numbers. Multiples of 4: 4,8,12 → 3 numbers. Multiples of 12: 12 → 1 number. Inclusion-exclusion: 4 + 3 - 1 = 6 numbers. But we want multiples of 3 or 4 but not 12. So, from the 6 numbers, we need to exclude 12. So, 6 - 1 = 5 numbers. Let's list them: 3,4,6,8,9. That's 5 numbers. Alternatively, using my method: Multiples of 3 not divisible by 12: 3,6,9 → 3 numbers. Multiples of 4 not divisible by 12: 4,8 → 2 numbers. Total: 3 + 2 = 5 numbers. So, that works. In this case, inclusion-exclusion gave me 6, and subtracting the multiples of 12 gave me 5, which is correct. Similarly, in the original problem, inclusion-exclusion gave me 1002, and subtracting multiples of 12 gave me 835, which should be correct. Therefore, I'm confident that the answer is 835, which is option C. But wait, looking back at the problem, the options are: (A) 501 (B) 668 (C) 835 (D) 1002 (E) 1169 So, 835 is option C. But just to make sure, let me check if I made any calculation errors. Multiples of 3: floor(2005/3) = 668. Yes, because 3*668 = 2004. Multiples of 4: floor(2005/4) = 501. Because 4*501 = 2004. Multiples of 12: floor(2005/12) = 167. Because 12*167 = 2004. Inclusion-exclusion: 668 + 501 - 167 = 1002. Subtracting multiples of 12: 1002 - 167 = 835. Yes, all calculations are correct. Therefore, the answer is 835, which is option C. But wait, I just noticed that in my initial thought process, I considered subtracting multiples of 12 again, but in the smaller example, that worked. So, I think my approach is correct. Yes, I think I've thought it through thoroughly and my answer is correct.