answer：I need to find a positive number such that two-thirds of it is equal to sixteen over two hundred sixteen of its reciprocal. Let me denote this number as ( x ).First, I'll set up the equation based on the problem statement:[frac{2}{3}x = frac{16}{216} times frac{1}{x}]To eliminate the fractions, I'll cross-multiply:[frac{2}{3}x times x = frac{16}{216}][frac{2}{3}x^2 = frac{16}{216}]Next, I'll simplify the equation by multiplying both sides by 216 to eliminate the denominator:[216 times frac{2}{3}x^2 = 16][144x^2 = 16]Now, I'll solve for ( x^2 ) by dividing both sides by 144:[x^2 = frac{16}{144}][x^2 = frac{1}{9}]Taking the square root of both sides gives me two possible solutions:[x = frac{1}{3} quad text{or} quad x = -frac{1}{3}]Since we're looking for a positive number, I'll discard the negative solution:[x = frac{1}{3}]Therefore, the positive number is ( frac{1}{3} ).

answer：Okay, so I have this geometry problem here about two intersecting circles, Γ₁ and Γ₂, which intersect at points P and Q. There are these variable points A₁ and B₁ on Γ₁, and then A₂ and B₂ are defined as the second intersections of lines A₁P and B₁P with Γ₂. Then, C is the intersection of lines A₁B₁ and A₂B₂. I need to prove that the center O of the circumcircle of triangle CA₁A₂ always lies on a fixed circle as A₁ and A₂ vary.Hmm, let me try to visualize this. There are two circles intersecting at P and Q. Points A₁ and B₁ are on Γ₁, and when I draw lines from A₁ and B₁ through P, they intersect Γ₂ again at A₂ and B₂. Then, connecting A₁B₁ and A₂B₂ gives me point C. So, triangle CA₁A₂ is formed, and I need to find the center of its circumcircle and show that this center moves along a fixed circle as A₁ and A₂ change.I remember something about Miquel's theorem, which involves cyclic quadrilaterals and circles. Maybe that's applicable here? Let me recall: Miquel's theorem states that for any quadrilateral, if you draw a circle through three of its vertices, the fourth vertex lies on a circle determined by the other three. But I'm not sure how that directly applies here.Alternatively, maybe I should consider the properties of cyclic quadrilaterals. Since A₁ and A₂ are on Γ₁ and Γ₂ respectively, and lines A₁P and A₂P intersect at P, which is a common point. Similarly, B₁ and B₂ are connected through P.Wait, maybe I can use the concept of power of a point. The power of point C with respect to both Γ₁ and Γ₂ might be useful. Since C lies on both A₁B₁ and A₂B₂, the power of C with respect to Γ₁ is CA₁ * CB₁, and with respect to Γ₂ is CA₂ * CB₂. But I'm not sure how this helps with finding the circumcircle of CA₁A₂.Let me think about the circumcircle of CA₁A₂. The center O of this circle must satisfy OA₁ = OA₂ = OC. So, O is equidistant from A₁, A₂, and C. Maybe I can find some relationship between O and the centers of Γ₁ and Γ₂.Let me denote the centers of Γ₁ and Γ₂ as O₁ and O₂, respectively. Perhaps O lies on some circle related to O₁ and O₂? If I can show that O moves on a circle defined by O₁, O₂, and some other fixed points, that would solve the problem.Another thought: since A₂ is the second intersection of A₁P with Γ₂, maybe there's a spiral similarity or some inversion that maps A₁ to A₂. Similarly for B₁ and B₂. Inversion might be a powerful tool here, but I'm not too familiar with it. Maybe I can avoid inversion and stick to more basic concepts.Wait, maybe I can use the concept of radical axes. The radical axis of Γ₁ and Γ₂ is the line PQ. Since C is the intersection of A₁B₁ and A₂B₂, and both A₁B₁ and A₂B₂ are chords of Γ₁ and Γ₂ respectively, perhaps C lies on the radical axis? But I don't think that's necessarily true unless certain conditions are met.Alternatively, maybe I can consider the circumcircle of CA₁A₂ and see how it relates to Γ₁ and Γ₂. Since A₁ and A₂ are on Γ₁ and Γ₂, and C is their intersection, perhaps there's a way to relate the angles in triangle CA₁A₂ to the angles in Γ₁ and Γ₂.Let me try to draw the diagram step by step. I have Γ₁ and Γ₂ intersecting at P and Q. Let me pick A₁ on Γ₁, draw line A₁P, which intersects Γ₂ again at A₂. Similarly, pick B₁ on Γ₁, draw line B₁P, intersecting Γ₂ again at B₂. Then, lines A₁B₁ and A₂B₂ intersect at C. So, C is the intersection point of these two lines.Now, considering triangle CA₁A₂, its circumcircle must pass through C, A₁, and A₂. The center O is the intersection of the perpendicular bisectors of CA₁ and A₁A₂. Maybe I can find some relationship between O and the centers of Γ₁ and Γ₂.Wait, if I can show that O lies on the perpendicular bisector of O₁O₂, then it would lie on a fixed circle. Or perhaps O lies on some other fixed circle related to O₁ and O₂.Alternatively, maybe I can use the fact that O is equidistant from A₁, A₂, and C. Since A₁ and A₂ are on Γ₁ and Γ₂, their distances from O₁ and O₂ are fixed. Maybe I can express OA₁ and OA₂ in terms of O₁ and O₂.This is getting a bit abstract. Maybe I should consider specific cases. For example, what happens when A₁ approaches P? Then A₂ would also approach P, since A₁P intersects Γ₂ at P and A₂. Similarly, if A₁ approaches Q, then A₂ would approach Q as well. So, in these cases, O would approach the center of Γ₂ or some other fixed point.Wait, if A₁ approaches P, then A₂ approaches P, and C would approach the intersection of A₁B₁ and A₂B₂. If A₁ and A₂ are both near P, then C might approach P as well. So, the circumcircle of CA₁A₂ would collapse to a point near P, and its center O would approach P. But P is fixed, so O approaches P.Similarly, if A₁ approaches Q, then A₂ approaches Q, and C approaches Q as well. So, O approaches Q. So, in these extreme cases, O approaches P or Q.What if A₁ is somewhere else? Maybe O moves along a circle passing through P and Q. That could be the fixed circle.Wait, if O always lies on the circle passing through P and Q, and maybe another fixed point, then that would be the fixed circle. But I need to verify this.Alternatively, maybe O lies on the circumcircle of triangle O₁O₂Q or something like that. Since O₁ and O₂ are fixed, their circumcircle is fixed, so if O lies on that, then it's fixed.Wait, let me think about the circumcircle of CA₁A₂. Since A₁ and A₂ are on Γ₁ and Γ₂, and C is their intersection, maybe there's a way to relate the angles in triangle CA₁A₂ to the angles in Γ₁ and Γ₂.Alternatively, maybe I can use the fact that the circumcircle of CA₁A₂ passes through Q. Is that true? Let me see.If I can show that Q lies on the circumcircle of CA₁A₂, then the center O would lie on the perpendicular bisector of CQ and A₁A₂. But I'm not sure if Q lies on that circumcircle.Wait, maybe using Miquel's theorem here. Miquel's theorem says that for a quadrilateral, the circumcircles of the triangles formed by three of its vertices concur at a point. In this case, maybe the quadrilateral is A₁B₁A₂B₂, and the circumcircle of CA₁A₂ is one of those circles.Wait, actually, Miquel's theorem might say that the circumcircle of CA₁A₂ passes through Q. Let me check.If I consider the complete quadrilateral formed by lines A₁B₁, A₂B₂, A₁A₂, and B₁B₂, then Miquel's theorem says that the circumcircles of the four triangles formed by omitting each line concur at a point. In this case, the circumcircle of CA₁A₂ would pass through Q, which is the Miquel point.So, if that's the case, then Q lies on the circumcircle of CA₁A₂. Therefore, the center O must lie on the perpendicular bisector of CQ and A₁A₂.But since Q is fixed, the perpendicular bisector of CQ would vary as C varies. Hmm, not sure if that helps.Wait, but if Q is on the circumcircle of CA₁A₂, then the center O is equidistant from Q, A₁, and A₂. So, O lies on the perpendicular bisectors of QA₁ and QA₂.But QA₁ and QA₂ are chords of Γ₁ and Γ₂, respectively. The perpendicular bisectors of QA₁ and QA₂ would pass through O₁ and O₂, the centers of Γ₁ and Γ₂.Wait, so the perpendicular bisector of QA₁ passes through O₁, and the perpendicular bisector of QA₂ passes through O₂. Therefore, the center O lies on both these perpendicular bisectors. So, O is the intersection point of the perpendicular bisectors of QA₁ and QA₂, which pass through O₁ and O₂.But O is the center of the circumcircle of CA₁A₂, which passes through Q. So, O is the intersection of the perpendicular bisectors of QA₁ and QA₂, which are lines through O₁ and O₂.Therefore, O lies on the perpendicular bisector of QA₁ (which passes through O₁) and on the perpendicular bisector of QA₂ (which passes through O₂). So, O is the intersection of these two lines.But as A₁ and A₂ vary, these perpendicular bisectors change, but O must lie on some fixed circle. Wait, but how?Wait, maybe O lies on the circumcircle of triangle O₁O₂Q. Because O is the intersection of two lines, each passing through O₁ and O₂, and O is equidistant from Q, A₁, and A₂. So, perhaps O lies on the circumcircle of O₁O₂Q.Let me see. If O lies on the circumcircle of O₁O₂Q, then that would mean that O is moving on a fixed circle as A₁ and A₂ vary, since O₁, O₂, and Q are fixed.Yes, that makes sense. Because O is determined by the intersection of two lines through O₁ and O₂, and since O must lie on the circumcircle of O₁O₂Q, it follows that O moves on a fixed circle.Therefore, the center O of the circumcircle of triangle CA₁A₂ always lies on the circumcircle of triangle O₁O₂Q, which is fixed.So, the fixed circle is the circumcircle of the centers of Γ₁ and Γ₂ and the intersection point Q.**Final Answer**The center ( O ) always lies on a fixed circle, specifically the circumcircle of triangle ( O_1O_2Q ). Thus, the final answer is boxed{O}.

answer：First, I identify the unknown number as ( x ).Next, I set up the equation based on the given expression:[0.8x + 0.1 times 0.5 = 0.29]I simplify the multiplication:[0.8x + 0.05 = 0.29]Then, I isolate the term with ( x ) by subtracting 0.05 from both sides:[0.8x = 0.24]Finally, I solve for ( x ) by dividing both sides by 0.8:[x = frac{0.24}{0.8} = 0.3]

answer：Okay, so I have this problem about a children's clothing factory that has some fabric and wants to make clothes to maximize profit. Let me try to understand it step by step.First, the factory has 38 meters of Type A fabric and 26 meters of Type B fabric. They plan to make a total of 50 sets of clothing, which can be either size L or size M. Each size L clothing set requires 0.5 meters of Type A fabric and 1 meter of Type B fabric, and it gives a profit of 45 yuan. On the other hand, each size M clothing set needs 0.9 meters of Type A fabric and 0.2 meters of Type B fabric, with a profit of 30 yuan.They want me to write a function for the total profit, y, in terms of the number of size L sets produced, x. Also, I need to find the range of possible values for x.Alright, let's break this down.First, let's define x as the number of size L sets. Since they are making a total of 50 sets, the number of size M sets must be 50 - x. That makes sense.Now, the total profit y would be the profit from size L sets plus the profit from size M sets. So, y = 45x + 30*(50 - x). Let me compute that:y = 45x + 30*(50 - x)y = 45x + 1500 - 30xy = (45x - 30x) + 1500y = 15x + 1500Okay, so that gives me the profit function: y = 15x + 1500. That seems straightforward.But now, I need to make sure that the number of sets they produce doesn't exceed the fabric they have. So, I need to set up some inequalities based on the fabric constraints.First, let's consider Type A fabric. Each size L set uses 0.5 meters, and each size M set uses 0.9 meters. The total Type A fabric used should be less than or equal to 38 meters.So, the inequality for Type A fabric is:0.5x + 0.9*(50 - x) ≤ 38Let me compute that:0.5x + 0.9*50 - 0.9x ≤ 380.5x + 45 - 0.9x ≤ 38(0.5x - 0.9x) + 45 ≤ 38-0.4x + 45 ≤ 38Now, subtract 45 from both sides:-0.4x ≤ -7Multiply both sides by -1 (remembering to reverse the inequality sign):0.4x ≥ 7Divide both sides by 0.4:x ≥ 7 / 0.4x ≥ 17.5Okay, so x has to be at least 17.5. But since x represents the number of sets, which has to be a whole number, x has to be at least 18. Hmm, but the problem might be okay with x being a decimal for the sake of the model, so maybe we can keep it as 17.5 for now.Now, let's consider Type B fabric. Each size L set uses 1 meter, and each size M set uses 0.2 meters. The total Type B fabric used should be less than or equal to 26 meters.So, the inequality for Type B fabric is:1*x + 0.2*(50 - x) ≤ 26Let me compute that:x + 0.2*50 - 0.2x ≤ 26x + 10 - 0.2x ≤ 26(1x - 0.2x) + 10 ≤ 260.8x + 10 ≤ 26Subtract 10 from both sides:0.8x ≤ 16Divide both sides by 0.8:x ≤ 16 / 0.8x ≤ 20So, x has to be less than or equal to 20.Putting it all together, from the Type A fabric constraint, x ≥ 17.5, and from Type B fabric constraint, x ≤ 20. So, x has to be between 17.5 and 20.But wait, x has to be an integer because you can't produce half a set of clothing. So, x can be 18, 19, or 20. But the problem didn't specify that x has to be an integer, so maybe it's okay to have x as a real number between 17.5 and 20.So, summarizing:The profit function is y = 15x + 1500, and the range of x is 17.5 ≤ x ≤ 20.Let me double-check my calculations to make sure I didn't make any mistakes.For Type A fabric:0.5x + 0.9*(50 - x) ≤ 380.5x + 45 - 0.9x ≤ 38-0.4x + 45 ≤ 38-0.4x ≤ -7x ≥ 17.5That looks correct.For Type B fabric:x + 0.2*(50 - x) ≤ 26x + 10 - 0.2x ≤ 260.8x + 10 ≤ 260.8x ≤ 16x ≤ 20That also looks correct.So, I think my answer is right. The profit function is y = 15x + 1500, and x can range from 17.5 to 20.

answer：Okay, so I have this problem about a parallelepiped with some midpoints and perpendicular lines. I need to find its volume given the lengths of these lines. Hmm, let me try to visualize this first.A parallelepiped is like a 3D figure formed by six parallelograms. So, it's similar to a cube but with all sides being parallelograms instead of squares. The vertices are labeled ABCD A1B1C1D1, which means A is connected to A1, B to B1, and so on.Points M and N are midpoints of edges AA1 and CC1, respectively. So, M is halfway up the edge from A to A1, and N is halfway up the edge from C to C1. Got it.The lines A1C, B1M, and BN are mutually perpendicular. That means each of these lines forms a right angle with the other two. Interesting. So, A1C is perpendicular to B1M, which is perpendicular to BN, and BN is also perpendicular to A1C. That's a key piece of information.Given lengths: A1C = a, B1M = b, BN = c. I need to find the volume of the parallelepiped.I remember that the volume of a parallelepiped can be found using the scalar triple product of its edge vectors. If I can express the vectors A1C, B1M, and BN in terms of the edges of the parallelepiped, maybe I can use their perpendicularity to find the necessary components.Let me denote the vectors along the edges as follows: Let’s say vector AB is **u**, vector AD is **v**, and vector AA1 is **w**. So, the parallelepiped is defined by these three vectors.Then, the coordinates of the points can be expressed in terms of **u**, **v**, and **w**. For example, point A is at the origin, B is at **u**, D is at **v**, and A1 is at **w**. Similarly, C would be at **u** + **v**, and C1 would be at **u** + **v** + **w**.Given that, point M is the midpoint of AA1, so its position vector is (0 + **w**)/2 = **w**/2. Similarly, point N is the midpoint of CC1, so its position vector is (**u** + **v** + (**u** + **v** + **w**))/2 = (**u** + **v** + **u** + **v** + **w**)/2 = (2**u** + 2**v** + **w**)/2 = **u** + **v** + **w**/2.Now, let's find the vectors A1C, B1M, and BN.Vector A1C goes from A1 to C. Since A1 is at **w** and C is at **u** + **v**, the vector A1C is (**u** + **v**) - **w** = **u** + **v** - **w**.Vector B1M goes from B1 to M. B1 is at **u** + **w**, and M is at **w**/2. So, vector B1M is (**w**/2) - (**u** + **w**) = -**u** - **w**/2.Vector BN goes from B to N. B is at **u**, and N is at **u** + **v** + **w**/2. So, vector BN is (**u** + **v** + **w**/2) - **u** = **v** + **w**/2.Now, these three vectors A1C, B1M, and BN are mutually perpendicular. So, their dot products should be zero.Let me write down the vectors again:- A1C = **u** + **v** - **w**- B1M = -**u** - (1/2)**w**- BN = **v** + (1/2)**w**First, let's compute the dot product of A1C and B1M:(**u** + **v** - **w**) · (-**u** - (1/2)**w**) = 0Expanding this:(**u** · -**u**) + (**u** · - (1/2)**w**) + (**v** · -**u**) + (**v** · - (1/2)**w**) + (-**w** · -**u**) + (-**w** · - (1/2)**w**) = 0Simplify each term:- **u** · -**u** = -|**u**|²- **u** · - (1/2)**w** = - (1/2)(**u** · **w**)- **v** · -**u** = -(**v** · **u**)- **v** · - (1/2)**w** = - (1/2)(**v** · **w**)- -**w** · -**u** = (**w** · **u**)- -**w** · - (1/2)**w** = (1/2)|**w**|²Combine like terms:-|**u**|² - (1/2)(**u** · **w**) - (**v** · **u**) - (1/2)(**v** · **w**) + (**w** · **u**) + (1/2)|**w**|² = 0Notice that (**u** · **w**) and (**w** · **u**) are the same, so they combine:-|**u**|² - (**v** · **u**) - (1/2)(**u** · **w**) - (1/2)(**v** · **w**) + (**u** · **w**) + (1/2)|**w**|² = 0Simplify:-|**u**|² - (**v** · **u**) + (1/2)(**u** · **w**) - (1/2)(**v** · **w**) + (1/2)|**w**|² = 0Hmm, this is getting a bit complicated. Maybe I should consider another approach.Alternatively, since the vectors are mutually perpendicular, the volume of the parallelepiped formed by these three vectors would be the product of their lengths, because the scalar triple product is equal to the product of their magnitudes when they are orthogonal.Wait, but the volume of the original parallelepiped is given by the scalar triple product of **u**, **v**, and **w**. So, Volume = |(**u** × **v**) · **w**|.But I have these three vectors A1C, B1M, and BN, which are perpendicular. Maybe I can relate their lengths and the scalar triple product.Given that A1C, B1M, and BN are mutually perpendicular, the volume of the parallelepiped formed by them is |A1C · (B1M × BN)|. But since they are mutually perpendicular, this volume is simply |A1C| |B1M| |BN|.But wait, the original parallelepiped's volume is |(**u** × **v**) · **w**|. How can I relate this to the other volume?Alternatively, maybe I can express the vectors A1C, B1M, and BN in terms of **u**, **v**, and **w**, and then use their perpendicularity to set up equations.Let me denote:A1C = **u** + **v** - **w** = a (given length a)B1M = -**u** - (1/2)**w** = b (given length b)BN = **v** + (1/2)**w** = c (given length c)Since these vectors are mutually perpendicular, their dot products are zero.So, first, A1C · B1M = 0:(**u** + **v** - **w**) · (-**u** - (1/2)**w**) = 0As I computed earlier, this gives:-|**u**|² - (**v** · **u**) + (1/2)(**u** · **w**) - (1/2)(**v** · **w**) + (1/2)|**w**|² = 0Let me denote |**u**|² = u², |**v**|² = v², |**w**|² = w², and the dot products as **u** · **v** = uv, **u** · **w** = uw, **v** · **w** = vw.So, substituting:- u² - uv + (1/2)uw - (1/2)vw + (1/2)w² = 0Similarly, let's compute the dot product of A1C and BN:(**u** + **v** - **w**) · (**v** + (1/2)**w**) = 0Expanding:**u** · **v** + (1/2)**u** · **w** + **v** · **v** + (1/2)**v** · **w** - **w** · **v** - (1/2)|**w**|² = 0Simplify:uv + (1/2)uw + v² + (1/2)vw - vw - (1/2)w² = 0Combine like terms:uv + (1/2)uw + v² - (1/2)vw - (1/2)w² = 0Third, the dot product of B1M and BN:(-**u** - (1/2)**w**) · (**v** + (1/2)**w**) = 0Expanding:-**u** · **v** - (1/2)**u** · **w** - (1/2)**w** · **v** - (1/4)|**w**|² = 0Simplify:- uv - (1/2)uw - (1/2)vw - (1/4)w² = 0So now I have three equations:1. - u² - uv + (1/2)uw - (1/2)vw + (1/2)w² = 02. uv + (1/2)uw + v² - (1/2)vw - (1/2)w² = 03. - uv - (1/2)uw - (1/2)vw - (1/4)w² = 0This is a system of three equations with variables u², v², w², uv, uw, vw. It might be a bit complex, but perhaps I can solve it step by step.Let me write the equations again for clarity:Equation 1: -u² - uv + (1/2)uw - (1/2)vw + (1/2)w² = 0Equation 2: uv + (1/2)uw + v² - (1/2)vw - (1/2)w² = 0Equation 3: -uv - (1/2)uw - (1/2)vw - (1/4)w² = 0Let me try to add Equations 1 and 2 to see if something cancels:Equation 1 + Equation 2:(-u² - uv + (1/2)uw - (1/2)vw + (1/2)w²) + (uv + (1/2)uw + v² - (1/2)vw - (1/2)w²) = 0 + 0Simplify:- u² + (1/2)uw + (1/2)uw + v² - (1/2)vw - (1/2)vw + (1/2)w² - (1/2)w² = 0Which simplifies to:- u² + uw + v² - vw = 0So, Equation 4: -u² + uw + v² - vw = 0Similarly, let's look at Equation 3:-uv - (1/2)uw - (1/2)vw - (1/4)w² = 0Let me multiply Equation 3 by 2 to eliminate fractions:-2uv - uw - vw - (1/2)w² = 0So, Equation 3a: -2uv - uw - vw - (1/2)w² = 0Now, let's see if we can express some variables in terms of others.From Equation 4: -u² + uw + v² - vw = 0Let me rearrange:u² = uw + v² - vwSimilarly, from Equation 2: uv + (1/2)uw + v² - (1/2)vw - (1/2)w² = 0Let me express this as:uv + (1/2)uw + v² - (1/2)vw = (1/2)w²Hmm, maybe I can express v² from Equation 4 and substitute into Equation 2.From Equation 4: v² = u² - uw + vwSubstitute into Equation 2:uv + (1/2)uw + (u² - uw + vw) - (1/2)vw = (1/2)w²Simplify:uv + (1/2)uw + u² - uw + vw - (1/2)vw = (1/2)w²Combine like terms:u² + uv + (1/2)uw - uw + vw - (1/2)vw = (1/2)w²Simplify:u² + uv - (1/2)uw + (1/2)vw = (1/2)w²Hmm, not sure if this helps directly. Maybe I should consider another approach.Alternatively, since the vectors A1C, B1M, and BN are mutually perpendicular, the volume of the parallelepiped they form is |A1C · (B1M × BN)|. But since they are mutually perpendicular, this volume is simply |A1C| |B1M| |BN| = a b c.But the original parallelepiped's volume is |(**u** × **v**) · **w**|. How are these related?Wait, maybe the volume of the original parallelepiped can be related to the volume formed by A1C, B1M, and BN. Since these vectors are combinations of **u**, **v**, and **w**, perhaps there's a scaling factor involved.Let me express the vectors A1C, B1M, and BN in terms of **u**, **v**, and **w**:A1C = **u** + **v** - **w**B1M = -**u** - (1/2)**w**BN = **v** + (1/2)**w**So, the matrix formed by these vectors as columns (or rows) would be:| 1 -1 0 || 1 0 1 ||-1 -1/2 1/2|Wait, actually, each vector is a combination of **u**, **v**, **w**, so the transformation matrix from (**u**, **v**, **w**) to (A1C, B1M, BN) is:[1, -1, 0;1, 0, 1;-1, -1/2, 1/2]The determinant of this matrix would give the scaling factor between the volumes.So, the volume of the parallelepiped formed by A1C, B1M, BN is |det(T)| times the volume of the original parallelepiped.But since A1C, B1M, BN are mutually perpendicular, their parallelepiped volume is a b c. So,|det(T)| * V = a b cTherefore, V = (a b c) / |det(T)|So, I need to compute the determinant of the transformation matrix T.Let me write the matrix T:T = [1, -1, 0;1, 0, 1;-1, -1/2, 1/2]Compute det(T):det(T) = 1*(0*(1/2) - 1*(-1/2)) - (-1)*(1*(1/2) - 1*(-1)) + 0*(1*(-1/2) - 0*(-1))Simplify term by term:First term: 1*(0 - (-1/2)) = 1*(1/2) = 1/2Second term: -(-1)*(1/2 - (-1)) = 1*(1/2 + 1) = 1*(3/2) = 3/2Third term: 0*(something) = 0So, det(T) = 1/2 + 3/2 + 0 = 2Therefore, |det(T)| = 2Thus, V = (a b c) / 2So, the volume of the original parallelepiped is (a b c)/2.Wait, but let me double-check the determinant calculation because it's crucial.Matrix T:Row 1: 1, -1, 0Row 2: 1, 0, 1Row 3: -1, -1/2, 1/2Compute det(T):Using the rule of Sarrus or cofactor expansion. Let's use cofactor expansion along the first row.det(T) = 1 * det([0, 1; -1/2, 1/2]) - (-1) * det([1, 1; -1, 1/2]) + 0 * det(...)Compute each minor:First minor: det([0, 1; -1/2, 1/2]) = 0*(1/2) - 1*(-1/2) = 0 + 1/2 = 1/2Second minor: det([1, 1; -1, 1/2]) = 1*(1/2) - 1*(-1) = 1/2 + 1 = 3/2Third term is 0, so it doesn't contribute.Thus, det(T) = 1*(1/2) - (-1)*(3/2) + 0 = 1/2 + 3/2 = 2Yes, that's correct. So, det(T) = 2.Therefore, the volume V = (a b c)/2.So, the volume of the parallelepiped is (a b c)/2.**Final Answer**The volume of the parallelepiped is boxed{dfrac{abc}{2}}.

answer：Okay, so I need to figure out what two digits I can add to the right of the number 2013 to make the resulting six-digit number divisible by 101. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I know that when we add two digits to the right of 2013, we're essentially creating a new number that's 2013 followed by those two digits. Let's call those two digits 'x' and 'y', so the new number would be 2013xy. Since we're adding two digits, this will make it a six-digit number.Now, the goal is for this six-digit number, 2013xy, to be divisible by 101. Divisibility by 101 means that when we divide 2013xy by 101, there should be no remainder. So, mathematically, we can write this as:2013xy ≡ 0 mod 101But how do I translate this into something I can work with? Well, I think I can express 2013xy in terms of 2013 and the two digits x and y. Since x is the tens place and y is the ones place, the number 2013xy can be written as:2013 * 100 + 10x + ySo, 201300 + 10x + yNow, I need this entire expression to be divisible by 101. That means:201300 + 10x + y ≡ 0 mod 101Okay, so I need to find x and y such that when I add 10x + y to 201300, the result is divisible by 101. To do this, I think I should first find out what 201300 is modulo 101. That way, I can figure out what 10x + y needs to be to make the whole thing divisible by 101.Let me calculate 201300 divided by 101 and find the remainder. Hmm, 101 times 2000 is 202000, which is more than 201300. So, let's try 101 times 1990. 101*1990 = 200,990. Wait, that's still more than 201,300. Hmm, maybe I should approach this differently.Alternatively, I can use the property that if a number is divisible by 101, then the alternating sum of its blocks of two digits is also divisible by 101. But I'm not sure if that applies here directly. Maybe it's better to just compute 201300 mod 101.Let me try to compute 201300 divided by 101:First, 101 * 2000 = 202000, which is larger than 201300. So, let's try 101 * 1990 = 200,990. Now, 201300 - 200,990 = 310. So, 201300 = 101*1990 + 310.But 310 is still larger than 101, so let's divide 310 by 101. 101*3 = 303, which leaves a remainder of 7. So, 310 = 101*3 + 7.Therefore, 201300 = 101*1990 + 101*3 + 7 = 101*(1990 + 3) + 7 = 101*1993 + 7.So, 201300 mod 101 is 7. That means:201300 ≡ 7 mod 101So, going back to our original equation:201300 + 10x + y ≡ 0 mod 101We can substitute 201300 with 7:7 + 10x + y ≡ 0 mod 101Which simplifies to:10x + y ≡ -7 mod 101But since we're dealing with mod 101, -7 is equivalent to 94 (because 101 - 7 = 94). So:10x + y ≡ 94 mod 101Now, x and y are single digits, meaning they can only be from 0 to 9. So, 10x + y can range from 0 (if x=0 and y=0) to 99 (if x=9 and y=9). Therefore, 10x + y must be equal to 94 because 94 is within the range of 0 to 99, and it's the only value that satisfies the congruence.So, 10x + y = 94Now, we need to find digits x and y such that 10x + y = 94. Let's solve for x and y.Divide 94 by 10 to get x:x = 94 // 10 = 9Then, y = 94 % 10 = 4So, x = 9 and y = 4.Therefore, the two digits that need to be added to the right of 2013 are 9 and 4, making the number 201394.Let me double-check to make sure this is correct. Let's divide 201394 by 101 and see if there's no remainder.201394 ÷ 101First, 101 * 2000 = 202000, which is larger than 201394. So, let's try 101 * 1990 = 200,990.201394 - 200,990 = 404Now, 404 ÷ 101 = 4, because 101*4 = 404.So, 201394 = 101*(1990 + 4) = 101*1994Yes, that works out perfectly. So, 201394 is indeed divisible by 101.Wait a minute, but the problem says "find all possible answers." Did I miss something? I thought there was only one solution because 10x + y had to equal 94, and since x and y are single digits, there's only one combination: x=9 and y=4.But let me think again. The equation was 10x + y ≡ 94 mod 101. Since 10x + y can only be between 0 and 99, the only possible value is 94. So, there's only one solution.Therefore, the only two digits that can be added to 2013 to make the resulting number divisible by 101 are 9 and 4.I think that's it. I don't see any other possibilities because 10x + y can't be 94 + 101 = 195, which is way beyond the two-digit range. So, yeah, 94 is the only solution.