answer：Okay, so I need to prove that in a tetrahedron, the segments connecting each vertex to the centroid of the opposite face all meet at a single point called the centroid, and that this point divides each segment in a 3:1 ratio. Also, I need to show that the segments connecting the midpoints of opposite edges intersect at the same centroid and are bisected there. Hmm, this seems a bit complex, but I'll try to break it down step by step.First, let me recall what a tetrahedron is. It's a three-dimensional shape with four triangular faces, four vertices, and six edges. Each face is a triangle, and each vertex is connected to the other three vertices. Now, a median of a face would be a line segment from a vertex of that face to the centroid of the opposite face. Wait, no, actually, the median of a face is a line segment from a vertex to the centroid of the opposite face. But in a triangle, the centroid is the intersection point of the medians, right? So in each triangular face, the centroid is the point where the three medians intersect, and it's also the center of mass if the triangle is made of a uniform material.So, in the tetrahedron, each face has a centroid, and we can connect each vertex to the centroid of the opposite face. These connecting segments are called medians of the tetrahedron. The claim is that all these medians intersect at a single point, which is the centroid of the tetrahedron, and that this point divides each median in a 3:1 ratio, with the longer part being closer to the vertex.Alright, to prove this, maybe I can use coordinates. Let me assign coordinates to the vertices of the tetrahedron. Let's say the vertices are A, B, C, D, and I'll place them in a coordinate system. Maybe it's easiest to place one vertex at the origin, but I need to make sure the coordinates are general enough to represent any tetrahedron.Wait, actually, maybe it's better to use barycentric coordinates or some symmetric system. But I'm not sure. Maybe Cartesian coordinates would be more straightforward. Let me assign coordinates to the vertices as follows: Let A be at (0, 0, 0), B at (1, 0, 0), C at (0, 1, 0), and D at (0, 0, 1). This is a regular tetrahedron, but I think the proof should hold for any tetrahedron, not just regular ones. Hmm, maybe I should use a more general coordinate system.Alternatively, I can use vectors. Let me denote the position vectors of the vertices as vectors A, B, C, D. Then, the centroid of a face can be found by averaging the position vectors of its vertices. For example, the centroid of face BCD would be (B + C + D)/3. Similarly, the centroid of face ACD would be (A + C + D)/3, and so on.Now, the median from vertex A to the centroid of the opposite face BCD would be the line segment connecting A to (B + C + D)/3. Similarly, the median from vertex B would connect B to (A + C + D)/3, and so on for vertices C and D.I need to show that these four medians intersect at a single point, which is the centroid of the tetrahedron. The centroid of the tetrahedron should be the average of all four vertices, right? So, the centroid G would be (A + B + C + D)/4. Let me check that.If I consider the median from A to (B + C + D)/3, the parametric equation of this line can be written as A + t[(B + C + D)/3 - A], where t ranges from 0 to 1. Similarly, the median from B to (A + C + D)/3 can be written as B + s[(A + C + D)/3 - B], where s ranges from 0 to 1.If these two lines intersect at the centroid G, then there should be values of t and s such that:A + t[(B + C + D)/3 - A] = B + s[(A + C + D)/3 - B]Let me simplify this equation. First, expand the terms:A + t[(B + C + D)/3 - A] = A + t*(B + C + D)/3 - t*ASimilarly, the right side:B + s[(A + C + D)/3 - B] = B + s*(A + C + D)/3 - s*BSo, setting them equal:A + t*(B + C + D)/3 - t*A = B + s*(A + C + D)/3 - s*BLet me collect like terms:On the left side: (1 - t)A + t*(B + C + D)/3On the right side: (1 - s)B + s*(A + C + D)/3Now, let's rearrange the terms:(1 - t)A + t*(B + C + D)/3 - (1 - s)B - s*(A + C + D)/3 = 0Let me distribute the terms:(1 - t)A - (1 - s)B + [t/3 - s/3]B + [t/3 - s/3]C + [t/3 - s/3]D = 0Wait, this seems a bit messy. Maybe I should approach it differently. Let me consider the centroid G = (A + B + C + D)/4. Let's see if G lies on the median from A to (B + C + D)/3.The parametric equation of the median from A is A + t[(B + C + D)/3 - A]. Let's see if there exists a t such that A + t[(B + C + D)/3 - A] = (A + B + C + D)/4.Let me solve for t:A + t[(B + C + D)/3 - A] = (A + B + C + D)/4Subtract A from both sides:t[(B + C + D)/3 - A] = (A + B + C + D)/4 - ASimplify the right side:(A + B + C + D)/4 - A = (A + B + C + D - 4A)/4 = (-3A + B + C + D)/4So, we have:t[(B + C + D)/3 - A] = (-3A + B + C + D)/4Let me factor out the numerator on the right:(-3A + B + C + D) = -(3A) + (B + C + D)So, we can write:t[(B + C + D)/3 - A] = [-(3A) + (B + C + D)]/4Factor out (B + C + D):t[(B + C + D)/3 - A] = [ (B + C + D) - 3A ] /4Let me write the left side as:t[ (B + C + D)/3 - A ] = t[ (B + C + D - 3A)/3 ] = t*(B + C + D - 3A)/3So, we have:t*(B + C + D - 3A)/3 = (B + C + D - 3A)/4Assuming (B + C + D - 3A) is not zero, we can divide both sides by (B + C + D - 3A):t/3 = 1/4Thus, t = 3/4So, the point G lies on the median from A at t = 3/4, meaning it divides the median in the ratio t : (1 - t) = 3/4 : 1/4 = 3:1, counting from the vertex A. Similarly, the same would hold for the medians from B, C, and D.Therefore, all four medians intersect at the centroid G, which divides each median in the ratio 3:1.Now, for the second part, I need to show that the segments connecting the midpoints of opposite edges also intersect at G and are bisected there.First, let's define what opposite edges are. In a tetrahedron, each edge has an opposite edge. For example, edge AB is opposite to edge CD, edge AC is opposite to edge BD, and edge AD is opposite to edge BC.The midpoint of edge AB is (A + B)/2, and the midpoint of edge CD is (C + D)/2. Similarly, the midpoint of edge AC is (A + C)/2, and the midpoint of edge BD is (B + D)/2, and so on.So, the segment connecting the midpoints of opposite edges AB and CD is the line segment connecting (A + B)/2 and (C + D)/2. Similarly, the segment connecting the midpoints of AC and BD is the line segment connecting (A + C)/2 and (B + D)/2, and the segment connecting the midpoints of AD and BC is the line segment connecting (A + D)/2 and (B + C)/2.I need to show that these three segments intersect at G and that G divides each segment into two equal parts.Let's take one of these segments, say the one connecting (A + B)/2 and (C + D)/2. Let's find its midpoint. The midpoint would be [(A + B)/2 + (C + D)/2]/2 = (A + B + C + D)/4, which is exactly G. So, the midpoint of the segment connecting the midpoints of opposite edges AB and CD is G. Similarly, the same would hold for the other two segments.Therefore, all three segments connecting the midpoints of opposite edges intersect at G, and G is the midpoint of each segment, meaning each segment is bisected by G.So, to summarize, the medians from each vertex to the centroid of the opposite face intersect at G, the centroid of the tetrahedron, and G divides each median in a 3:1 ratio. Additionally, the segments connecting the midpoints of opposite edges also intersect at G, and G bisects each of these segments.I think this covers both parts of the proof. I used coordinate-free vector analysis to show the ratios and the intersection point. It might be helpful to visualize this with a specific example or to draw a diagram, but since I'm working this out in my mind, I'll rely on the algebraic proof.

answer：Okay, so I have these two problems to solve related to powers. Let me take them one at a time and try to figure them out step by step. Starting with the first problem: If (9 times 27^{x} = 3^{17}), find the value of (x). Hmm, okay. I remember that when dealing with equations involving exponents, it's helpful to express all terms with the same base if possible. Looking at the numbers 9 and 27, I know they are both powers of 3. Specifically, 9 is (3^2) and 27 is (3^3). So maybe I can rewrite the equation using base 3. Let me try that.So, replacing 9 with (3^2) and 27 with (3^3), the equation becomes:[3^2 times (3^3)^x = 3^{17}]Now, I need to simplify the left side of the equation. I remember that when you raise a power to another power, you multiply the exponents. So, ((3^3)^x) becomes (3^{3x}). So now the equation is:[3^2 times 3^{3x} = 3^{17}]Next, I recall that when you multiply exponents with the same base, you add the exponents. So, (3^2 times 3^{3x}) becomes (3^{2 + 3x}). Now the equation is:[3^{2 + 3x} = 3^{17}]Since the bases are the same (both are base 3), I can set the exponents equal to each other:[2 + 3x = 17]Now, I just need to solve for (x). Subtract 2 from both sides:[3x = 17 - 2][3x = 15]Then, divide both sides by 3:[x = frac{15}{3}][x = 5]Okay, that seems straightforward. So, (x = 5). Moving on to the second problem: Given (a^{x} = -2), (a^{y} = 3), find the value of (a^{3x - 2y}). Hmm, this one looks a bit trickier. I need to find (a^{3x - 2y}). I remember that exponents can be manipulated using properties like (a^{m + n} = a^m times a^n) and (a^{m - n} = frac{a^m}{a^n}). Also, (a^{mn} = (a^m)^n). So, let's break down (a^{3x - 2y}). I can write this as:[a^{3x - 2y} = frac{a^{3x}}{a^{2y}}]Now, I can express (a^{3x}) and (a^{2y}) in terms of (a^x) and (a^y), which are given.Starting with (a^{3x}), this is the same as ((a^x)^3). Since (a^x = -2), this becomes:[(a^x)^3 = (-2)^3 = -8]Similarly, (a^{2y}) is ((a^y)^2). Given that (a^y = 3), this becomes:[(a^y)^2 = 3^2 = 9]Now, substituting these back into the expression for (a^{3x - 2y}):[a^{3x - 2y} = frac{-8}{9}]So, the value of (a^{3x - 2y}) is (-frac{8}{9}).Wait, let me double-check that. I know that when raising a negative number to an odd power, the result remains negative, which is why ((-2)^3 = -8). And squaring 3 gives 9, which is correct. So, dividing -8 by 9 gives (-frac{8}{9}). That seems right.I also need to make sure that the properties of exponents I used are applicable here. Since (a^{x}) and (a^{y}) are given, and assuming (a) is a real number (though (a^x = -2) suggests that (a) might be negative or a complex number, but I think for the purposes of this problem, we can treat it as a real number with the given exponents). So, I think my steps are correct. I expressed (a^{3x - 2y}) in terms of (a^x) and (a^y), substituted the given values, and simplified. Just to recap:1. (a^{3x} = (a^x)^3 = (-2)^3 = -8)2. (a^{2y} = (a^y)^2 = 3^2 = 9)3. Therefore, (a^{3x - 2y} = frac{-8}{9})Yes, that all checks out. I don't see any mistakes in my reasoning.So, to summarize:1. For the first problem, (x = 5).2. For the second problem, (a^{3x - 2y} = -frac{8}{9}).I think I've solved both problems correctly. It's always good to go through the steps again to make sure I didn't skip anything or make any calculation errors, but everything seems consistent.

answer：Okay, so I have this system of equations to solve:[begin{cases}3x + 4y = 16 5x - 6y = 33end{cases}]Hmm, I remember that to solve a system like this, I can use either substitution or elimination. I think elimination might be easier here because the coefficients are manageable. Let me try that.First, I need to make the coefficients of one of the variables the same in both equations so that when I subtract or add the equations, that variable will cancel out. Looking at the coefficients of x and y:In the first equation, the coefficients are 3 and 4 for x and y respectively. In the second equation, they are 5 and -6. I'm thinking maybe I can eliminate y because 4 and 6 have a common multiple of 12. Alternatively, I could also eliminate x since 3 and 5 have a common multiple of 15. Let me try eliminating x first.To eliminate x, I need to make the coefficients of x in both equations equal. The least common multiple of 3 and 5 is 15. So, I'll multiply the first equation by 5 and the second equation by 3.Multiplying the first equation by 5:[5 times (3x + 4y) = 5 times 16 15x + 20y = 80]Multiplying the second equation by 3:[3 times (5x - 6y) = 3 times 33 15x - 18y = 99]Now I have the two new equations:1. (15x + 20y = 80)2. (15x - 18y = 99)Next, I'll subtract the second equation from the first to eliminate x:[(15x + 20y) - (15x - 18y) = 80 - 99 15x + 20y - 15x + 18y = -19 38y = -19]Now, solving for y:[y = frac{-19}{38} y = -frac{1}{2}]Okay, so y is -1/2. Now I need to find x. I'll substitute y back into one of the original equations. Let me use the first equation:[3x + 4y = 16 3x + 4left(-frac{1}{2}right) = 16 3x - 2 = 16 3x = 18 x = 6]So, x is 6. Let me check if these values satisfy both original equations.First equation:[3(6) + 4left(-frac{1}{2}right) = 18 - 2 = 16]Yes, that works.Second equation:[5(6) - 6left(-frac{1}{2}right) = 30 + 3 = 33]That also works. So, the solution is x = 6 and y = -1/2.Wait, let me think if there's another way to solve this, just to verify. Maybe using substitution instead of elimination.Starting again with the original equations:1. (3x + 4y = 16)2. (5x - 6y = 33)Let me solve the first equation for x:[3x = 16 - 4y x = frac{16 - 4y}{3}]Now, substitute this expression for x into the second equation:[5left(frac{16 - 4y}{3}right) - 6y = 33 frac{80 - 20y}{3} - 6y = 33]Multiply both sides by 3 to eliminate the denominator:[80 - 20y - 18y = 99 80 - 38y = 99 -38y = 19 y = -frac{1}{2}]Same result for y. Then, substituting back into x:[x = frac{16 - 4left(-frac{1}{2}right)}{3} x = frac{16 + 2}{3} x = frac{18}{3} x = 6]Consistent with the previous method. So, definitely, x is 6 and y is -1/2.I think that's solid. Both methods give the same answer, and plugging back into the original equations works. So, I'm confident that's the correct solution.

answer：Okay, so I have this problem about two circles with their centers 50 inches apart. The smaller circle has a radius of 7 inches, and the larger one has a radius of 8 inches. I need to find the length of the common internal tangent between them. Hmm, let me think about how to approach this.First, I remember that a common internal tangent is a line that touches both circles and lies between them, meaning it doesn't cross the line segment connecting their centers. I think the key here is to visualize the two circles and the tangent line. Maybe drawing a diagram would help, but since I can't draw right now, I'll try to imagine it.Let me denote the centers of the two circles as points A and B. So, point A is the center of the larger circle with radius 8 inches, and point B is the center of the smaller circle with radius 7 inches. The distance between A and B is 50 inches.Now, the common internal tangent will touch the larger circle at some point, say point C, and the smaller circle at another point, say point D. The tangent line CD is what I need to find the length of.I recall that for two circles, the length of the common internal tangent can be found using the formula:Length = sqrt(d^2 - (r1 + r2)^2)where d is the distance between the centers, and r1 and r2 are the radii of the two circles.Wait, let me make sure I have the right formula. I think for the internal tangent, it's sqrt(d^2 - (r1 + r2)^2), whereas for the external tangent, it's sqrt(d^2 - (r1 - r2)^2). Yeah, that sounds right because for internal tangent, the radii effectively add up when considering the triangle formed by the centers and the tangent points.So, plugging in the values:d = 50 inchesr1 = 8 inchesr2 = 7 inchesSo, the length should be sqrt(50^2 - (8 + 7)^2) = sqrt(2500 - 225) = sqrt(2275)Hmm, sqrt(2275) is not a perfect square. Let me calculate that. 2275 divided by 25 is 91, so sqrt(2275) = sqrt(25 * 91) = 5 * sqrt(91). Now, sqrt(91) is approximately 9.539, so 5 * 9.539 is approximately 47.695 inches.Looking at the answer choices, 47.7 inches is option B. So, that must be the answer.Wait, let me double-check my formula. Sometimes I mix up internal and external tangents. For internal tangent, the formula is indeed sqrt(d^2 - (r1 + r2)^2), because the tangent line is between the circles, so the effective distance to consider is the sum of the radii. For external tangent, it's the difference of the radii because the tangent lines are outside the space between the circles.Yes, that makes sense. So, I think my calculation is correct. The length of the common internal tangent is approximately 47.7 inches.Just to be thorough, let me think about another approach. Maybe using similar triangles or coordinate geometry.If I place the two circles on a coordinate plane, with center A at (0,0) and center B at (50,0). The larger circle has radius 8, so its equation is x^2 + y^2 = 64. The smaller circle has radius 7, so its equation is (x - 50)^2 + y^2 = 49.The common internal tangent will have some slope m, and it will touch both circles. The tangent line can be written as y = mx + c. The condition for tangency is that the distance from the center to the line equals the radius.So, for the larger circle, the distance from (0,0) to the line y = mx + c is |0 - (mx + c)| / sqrt(m^2 + 1) = |c| / sqrt(m^2 + 1) = 8.Similarly, for the smaller circle, the distance from (50,0) to the line y = mx + c is |0 - (m*50 + c)| / sqrt(m^2 + 1) = | -50m - c | / sqrt(m^2 + 1) = 7.So, we have two equations:1) |c| / sqrt(m^2 + 1) = 82) | -50m - c | / sqrt(m^2 + 1) = 7Assuming the tangent is below the line connecting the centers, the signs might be consistent. Let me assume c is negative, so |c| = -c. Similarly, | -50m - c | would be | -50m - c |. Let me proceed without worrying about the signs for now.From equation 1: c = ±8 sqrt(m^2 + 1)From equation 2: | -50m - c | = 7 sqrt(m^2 + 1)Let me substitute c from equation 1 into equation 2.Case 1: c = 8 sqrt(m^2 + 1)Then, | -50m - 8 sqrt(m^2 + 1) | = 7 sqrt(m^2 + 1)This gives two possibilities:a) -50m - 8 sqrt(m^2 + 1) = 7 sqrt(m^2 + 1)Which simplifies to -50m = 15 sqrt(m^2 + 1)But sqrt(m^2 + 1) is positive, so -50m must be positive, which implies m is negative.Square both sides: (50m)^2 = (15)^2 (m^2 + 1)2500 m^2 = 225 m^2 + 2252500 m^2 - 225 m^2 = 2252275 m^2 = 225m^2 = 225 / 2275 = 9/91m = -3/sqrt(91) ≈ -0.314Then, c = 8 sqrt(m^2 + 1) = 8 sqrt(9/91 + 1) = 8 sqrt(100/91) = 8 * (10)/sqrt(91) = 80 / sqrt(91) ≈ 80 / 9.539 ≈ 8.38But wait, if c is positive, and we assumed c = 8 sqrt(m^2 + 1), but earlier I thought c might be negative. Maybe I need to check the other case.Case 2: c = -8 sqrt(m^2 + 1)Then, | -50m - (-8 sqrt(m^2 + 1)) | = | -50m + 8 sqrt(m^2 + 1) | = 7 sqrt(m^2 + 1)Again, two possibilities:a) -50m + 8 sqrt(m^2 + 1) = 7 sqrt(m^2 + 1)Which simplifies to -50m = -1 sqrt(m^2 + 1)So, 50m = sqrt(m^2 + 1)Square both sides: 2500 m^2 = m^2 + 12499 m^2 = 1m^2 = 1/2499m ≈ ±0.020But 50m = sqrt(m^2 + 1) implies m is positive, so m ≈ 0.020Then, c = -8 sqrt(m^2 + 1) ≈ -8 sqrt(1/2499 + 1) ≈ -8 sqrt(2500/2499) ≈ -8 * (50)/sqrt(2499) ≈ -8 * 50 / 49.99 ≈ -8.0016Wait, that seems off because if m is very small, the tangent line is almost horizontal, but the internal tangent should be more steep. Maybe I made a mistake in the assumption.Alternatively, perhaps the other possibility in case 2:b) -(-50m + 8 sqrt(m^2 + 1)) = 7 sqrt(m^2 + 1)Which is 50m - 8 sqrt(m^2 + 1) = 7 sqrt(m^2 + 1)So, 50m = 15 sqrt(m^2 + 1)Again, sqrt(m^2 + 1) is positive, so m is positive.Square both sides: 2500 m^2 = 225 (m^2 + 1)2500 m^2 = 225 m^2 + 2252275 m^2 = 225m^2 = 225 / 2275 = 9/91m = 3/sqrt(91) ≈ 0.314Then, c = -8 sqrt(m^2 + 1) = -8 sqrt(9/91 + 1) = -8 sqrt(100/91) = -80 / sqrt(91) ≈ -8.38So, the tangent line is y = (3/sqrt(91))x - 80/sqrt(91)Now, to find the length of the tangent, which is the distance between points C and D on the two circles.Alternatively, since I have the slope and the intercept, I can find the points of tangency and then compute the distance.But maybe there's a simpler way. I remember that the length of the tangent can be found using the formula I used earlier, which gave me approximately 47.7 inches. Since this coordinate geometry approach is getting complicated, and I already have a formula that gives me the answer, I think I'll stick with that.Just to confirm, the formula for the length of the common internal tangent is sqrt(d^2 - (r1 + r2)^2). Plugging in the numbers:sqrt(50^2 - (8 + 7)^2) = sqrt(2500 - 225) = sqrt(2275) ≈ 47.7 inches.Yes, that seems consistent. So, I think the answer is 47.7 inches, which is option B.

answer：First, I recognize that Jason has some violet balloons and 4 red balloons. The key information is that he lost 3 violet balloons and now has 4 violet balloons left.To find out how many violet balloons Jason had initially, I need to consider the number of violet balloons he lost and the number he has remaining.By adding the number of violet balloons he lost (3) to the number of violet balloons he has now (4), I can determine the initial number of violet balloons.So, the initial number of violet balloons is 3 plus 4, which equals 7.Therefore, Jason initially had 7 violet balloons.

answer：Alright, so I have this problem here about sets. The universal set U is {1,2,3,4,5,6}, and there are two subsets, A and B. A is {1,2,3,5} and B is {3,5,6}. I need to find two things: first, the intersection of A and B, which is part (I), and second, the union of the complement of A with respect to U and B, which is part (II). Okay, starting with part (I): A ∩ B. I remember that the intersection of two sets is the set of elements that are common to both sets. So, I need to look at the elements in A and see which ones are also in B.Let me list out the elements of A: 1, 2, 3, 5. And the elements of B: 3, 5, 6. Now, comparing these, I see that 3 is in both A and B, and 5 is also in both A and B. The other elements, 1 and 2, are only in A, and 6 is only in B. So, the common elements are 3 and 5. Therefore, A ∩ B should be {3,5}. That seems straightforward.Moving on to part (II): (complement of A with respect to U) union B. I need to recall what the complement of a set is. The complement of A, denoted as ∁ₐ U or sometimes A', is the set of all elements in the universal set U that are not in A. So, first, I need to find ∁ₐ U.Given that U is {1,2,3,4,5,6} and A is {1,2,3,5}, the complement of A would be the elements in U that are not in A. Let's list them out: 4 and 6. So, ∁ₐ U is {4,6}.Now, I need to find the union of this complement set with B. The union of two sets is the set of all elements that are in either of the sets. So, ∁ₐ U is {4,6} and B is {3,5,6}. Combining these, we get 3, 4, 5, and 6. Wait, let me make sure I'm not missing anything. So, starting with ∁ₐ U: 4 and 6. Then, adding the elements from B: 3 and 5. So, putting them all together, we have 3, 4, 5, 6. Is there any overlap? Yes, 6 is in both ∁ₐ U and B, but in the union, we only list each element once. So, the union should be {3,4,5,6}.Let me double-check my work. For part (I), A ∩ B: A has 1,2,3,5 and B has 3,5,6. The common elements are indeed 3 and 5. So, that's correct.For part (II), the complement of A is everything in U not in A. U is {1,2,3,4,5,6}, so removing A's elements {1,2,3,5} leaves us with 4 and 6. Then, taking the union with B, which is {3,5,6}, we add 3 and 5 to the complement set. So, combining {4,6} and {3,5,6} gives us {3,4,5,6}. That seems right.I think I've got both parts correct. The intersection is straightforward, and the union after taking the complement also makes sense. I don't see any mistakes in my reasoning.**Final Answer**(I) The intersection of A and B is boxed{{3, 5}}.(II) The union of the complement of A with B is boxed{{3, 4, 5, 6}}.