answer：Alright, so I came across this interesting geometry problem: "Show that there is no three-sided pyramid (tetrahedron) in which all four faces are right-angled triangles, and the right angles are each at a different vertex of the pyramid." At first glance, it seems pretty challenging, but I want to try to work through it step by step.First, let me make sure I understand the problem correctly. A tetrahedron has four triangular faces. The problem is asking if it's possible for all four of these faces to be right-angled triangles, with each right angle located at a different vertex. So, each of the four vertices of the tetrahedron would have a right angle on one of the faces meeting at that vertex.I know that in a right-angled triangle, the side opposite the right angle is the hypotenuse, which is the longest side. So, if all four faces are right-angled triangles, each face must have a hypotenuse. But in a tetrahedron, each edge is shared by two faces. That might complicate things because an edge that is a hypotenuse for one face would have to be a leg for another face.Let me try to visualize this. Imagine a tetrahedron with vertices A, B, C, and D. Suppose the right angles are at each of these vertices. So, the face ABC has a right angle at A, face ABD has a right angle at B, face ACD has a right angle at C, and face BCD has a right angle at D.Wait, but if each face has a right angle at a different vertex, then each vertex is the right angle for one face. That means each vertex is connected to two edges that form the legs of a right triangle, and one edge that is the hypotenuse.But in a tetrahedron, each vertex is connected to three edges. So, for each vertex, two of those edges would be the legs of a right triangle, and the third edge would be the hypotenuse for that face.Hmm, this seems tricky. Let me try to assign coordinates to the vertices to see if I can model this mathematically. Let's place vertex A at the origin (0,0,0). Since face ABC has a right angle at A, let's say point B is along the x-axis, point C is along the y-axis, and point D is somewhere in 3D space.So, let's assign coordinates:- A = (0, 0, 0)- B = (a, 0, 0) for some a > 0- C = (0, b, 0) for some b > 0- D = (c, d, e) for some c, d, eNow, since face ABD has a right angle at B, the vectors BA and BD should be perpendicular. Vector BA is A - B = (-a, 0, 0), and vector BD is D - B = (c - a, d, e). Their dot product should be zero:(-a)(c - a) + 0*d + 0*e = 0=> -a(c - a) = 0=> -ac + a² = 0=> a² = ac=> a = c (since a ≠ 0)So, c = a.Similarly, face ACD has a right angle at C. The vectors CA and CD should be perpendicular. Vector CA is A - C = (0, -b, 0), and vector CD is D - C = (c, d - b, e). Their dot product should be zero:0*c + (-b)(d - b) + 0*e = 0=> -b(d - b) = 0=> -bd + b² = 0=> b² = bd=> b = d (since b ≠ 0)So, d = b.Now, face BCD has a right angle at D. The vectors DB and DC should be perpendicular. Vector DB is B - D = (a - c, 0 - d, 0 - e) = (0, -b, -e) since c = a and d = b. Vector DC is C - D = (0 - c, b - d, 0 - e) = (-a, 0, -e). Their dot product should be zero:0*(-a) + (-b)*0 + (-e)*(-e) = 0=> e² = 0=> e = 0But if e = 0, then point D lies in the same plane as points A, B, and C, making the tetrahedron degenerate (i.e., it collapses into a flat triangle). That's not possible because a tetrahedron must be a three-dimensional figure.So, this leads to a contradiction. Therefore, our assumption that such a tetrahedron exists must be wrong.Wait, but let me double-check my steps to make sure I didn't make a mistake. I assigned coordinates to the vertices and used the condition that each face has a right angle at a different vertex. By setting up the dot products for perpendicularity, I derived that e must be zero, which collapses the tetrahedron. That seems correct.Another way to think about it is that if all four faces are right-angled triangles with right angles at different vertices, then each vertex is the right angle for one face. But in three-dimensional space, it's impossible to have all four vertices each being the right angle for their respective faces without causing some kind of overlap or degeneracy.I also recall that in a tetrahedron, the edges are shared between faces. If each face has a hypotenuse, those hypotenuses would have to correspond to edges that are shared between two faces. But if each face has a right angle at a different vertex, the hypotenuses would have to be arranged in a way that doesn't allow for a consistent three-dimensional structure.Maybe another approach is to consider the properties of right-angled tetrahedrons. I know that in a right-angled tetrahedron, three of the edges meeting at one vertex are mutually perpendicular. But in this problem, the right angles are at different vertices, so it's not just one vertex where all three edges are perpendicular.If I try to extend the concept of a right-angled tetrahedron to have right angles at all four vertices, it seems geometrically impossible because the mutual perpendicularity required for each vertex would conflict with each other in three-dimensional space.So, combining both the coordinate approach and the geometric intuition, it seems clear that such a tetrahedron cannot exist. The conditions imposed by having right angles at all four vertices lead to a contradiction, forcing the figure to collapse into a lower dimension or causing overlapping edges, which isn't possible for a valid tetrahedron.Therefore, after carefully working through the problem and checking my reasoning, I conclude that there is no such tetrahedron where all four faces are right-angled triangles with right angles at different vertices.

answer：Okay, so I have this problem here where a snake has discovered 453 base 8 ways to shed its skin, and I need to figure out how many ways that is in base 10. Hmm, base conversions can be a bit tricky, but I think I remember that each digit in a base represents a power of that base. Let me try to recall how that works.First, I know that in base 8, each digit from right to left represents 8^0, 8^1, 8^2, and so on. So, for the number 453 in base 8, the rightmost digit is the ones place (8^0), the middle digit is the eights place (8^1), and the leftmost digit is the sixty-fours place (8^2).Let me write that out step by step. The number is 453_8. Breaking it down:- The rightmost digit is 3, which is in the 8^0 place. So, that's 3 * (8^0).- The middle digit is 5, which is in the 8^1 place. So, that's 5 * (8^1).- The leftmost digit is 4, which is in the 8^2 place. So, that's 4 * (8^2).Now, I need to calculate each of these components and then add them together to get the decimal (base 10) equivalent.Starting with the rightmost digit: 3 * (8^0). Since any number to the power of 0 is 1, this is just 3 * 1 = 3.Next, the middle digit: 5 * (8^1). 8^1 is 8, so this is 5 * 8 = 40.Finally, the leftmost digit: 4 * (8^2). 8^2 is 64, so this is 4 * 64 = 256.Now, I add all these results together: 3 + 40 + 256. Let's do that step by step to avoid mistakes.First, 3 + 40 is 43. Then, 43 + 256. Hmm, adding 43 and 256. Let's see, 43 + 200 is 243, and then adding 56 more gives 299. So, 43 + 256 = 299.Therefore, 453 in base 8 is equal to 299 in base 10.Wait, let me double-check my calculations to make sure I didn't make any errors. It's easy to mix up the exponents or multiply incorrectly.Starting again:- 4 * 8^2: 8^2 is 64, so 4 * 64. Let's calculate that. 64 * 4: 60*4=240 and 4*4=16, so 240 +16=256. That seems right.- 5 * 8^1: 8*5=40. That's straightforward.- 3 * 8^0: 3*1=3. Also straightforward.Adding them up: 256 + 40 = 296, and then 296 + 3 = 299. Yep, that checks out.I think I got it right. So, the snake has 299 ways to shed its skin in base 10.

answer：First, I need to determine the total number of students in Mr. Stewart's math class. I know there are 16 boys, and I need to find out how many girls there are.Next, I'll consider the number of students who brought calculators to class. Out of these 22 students, 13 are girls. This means the remaining students who brought calculators must be boys.By subtracting the number of girls who brought calculators from the total number of students who brought calculators, I can find out how many boys brought calculators.Finally, to find out how many boys didn't bring their calculators, I'll subtract the number of boys who brought calculators from the total number of boys in the class.

answer：First, I need to determine the new price of the suit after a 25% increase from the original price of 160.To calculate the increase:25% of 160 is (25/100) * 160 = 40.Adding this increase to the original price:160 + 40 = 200.Now, the store offers a 25% off coupon on the increased price of 200.To calculate the discount:25% of 200 is (25/100) * 200 = 50.Subtracting the discount from the increased price:200 - 50 = 150.Therefore, consumers who used the coupon paid 150 for the suit.

answer：Okay, so I have this problem where I need to find the sum of the first 100 terms of a sequence defined by ( a_n = n cdot cos(npi) ). Hmm, let's see. I remember that ( cos(npi) ) alternates between -1 and 1 depending on whether ( n ) is odd or even. So, for each term in the sequence, if ( n ) is odd, ( cos(npi) ) will be -1, and if ( n ) is even, it will be 1. That means the sequence alternates between negative and positive terms, right?Let me write out the first few terms to get a better idea. For ( n = 1 ), ( a_1 = 1 cdot cos(pi) = -1 ). For ( n = 2 ), ( a_2 = 2 cdot cos(2pi) = 2 ). For ( n = 3 ), ( a_3 = 3 cdot cos(3pi) = -3 ). And for ( n = 4 ), ( a_4 = 4 cdot cos(4pi) = 4 ). So, the sequence goes -1, 2, -3, 4, -5, 6, and so on.I notice that the odd terms are negative and form an arithmetic sequence where each term increases by 2. Similarly, the even terms are positive and also form an arithmetic sequence with a common difference of 2. So, the odd terms are -1, -3, -5, ..., and the even terms are 2, 4, 6, ...Since we need the sum of the first 100 terms, and 100 is an even number, there will be 50 odd terms and 50 even terms. That makes sense because every pair of terms consists of one odd and one even.Now, I need to find the sum of these two separate arithmetic sequences and then add them together to get the total sum ( S_{100} ).Let's start with the odd terms. The first term ( a_1 ) is -1, and each subsequent odd term increases by 2. So, the sequence is -1, -3, -5, ..., up to the 50th odd term. Wait, actually, the 50th odd term would be ( -(2 times 50 - 1) = -99 ). Let me confirm that. The nth odd term can be represented as ( -(2n - 1) ), so for ( n = 50 ), it's ( -(2 times 50 - 1) = -99 ). Yes, that seems right.To find the sum of an arithmetic sequence, the formula is ( S = frac{k}{2} times (a_1 + a_k) ), where ( k ) is the number of terms, ( a_1 ) is the first term, and ( a_k ) is the last term. So, for the odd terms, ( k = 50 ), ( a_1 = -1 ), and ( a_k = -99 ). Plugging these into the formula: ( S_{odd} = frac{50}{2} times (-1 + (-99)) = 25 times (-100) = -2500 ).Wait, that doesn't seem right. Because when I add -1 and -99, I get -100, and multiplying by 25 gives -2500. But earlier, I thought the sum of the odd terms was -5000. Did I make a mistake?Let me double-check. The formula is correct: ( S = frac{k}{2} times (a_1 + a_k) ). So, 50 divided by 2 is 25, and -1 plus -99 is indeed -100. So, 25 times -100 is -2500. Hmm, maybe my initial thought was wrong. Let me see.Wait, actually, the odd terms are -1, -3, -5, ..., -99. So, how many terms are there? From 1 to 99, stepping by 2, that's 50 terms. So, the sum is ( frac{50}{2} times (-1 + (-99)) = 25 times (-100) = -2500 ). Okay, so the sum of the odd terms is -2500.Now, moving on to the even terms. The first even term is 2, and each subsequent even term increases by 2. So, the sequence is 2, 4, 6, ..., up to the 50th even term, which would be 100. Let me confirm that. The nth even term is ( 2n ), so for ( n = 50 ), it's ( 2 times 50 = 100 ). Correct.Using the same arithmetic series formula: ( S = frac{k}{2} times (a_1 + a_k) ). Here, ( k = 50 ), ( a_1 = 2 ), and ( a_k = 100 ). Plugging in: ( S_{even} = frac{50}{2} times (2 + 100) = 25 times 102 = 2550 ).Wait, 25 times 102 is 2550. But earlier, I thought the sum was 5100. Did I make a mistake here too?Hold on, let's recalculate. 2 plus 100 is 102, multiplied by 25 is indeed 2550. So, the sum of the even terms is 2550.Now, adding the two sums together: ( S_{100} = S_{odd} + S_{even} = -2500 + 2550 = 50 ). So, the total sum is 50.Wait, but in my initial thought process, I thought the sum of the odd terms was -5000 and the even terms was 5100, leading to a total of 50. But now, recalculating, I get -2500 and 2550, which also adds up to 50. So, both methods give the same result, but the intermediate steps were different.I think the confusion was in how I was counting the terms. Initially, I might have considered the odd terms as starting from -1 and going up to -99, which is 50 terms, and the even terms from 2 to 100, which is also 50 terms. But when I first calculated, I mistakenly thought the sum of the odd terms was -5000, which was incorrect. Upon recalculating, I realized it's -2500, and the even terms sum to 2550, leading to the correct total of 50.So, to summarize, the sum of the first 100 terms of the sequence ( a_n = n cdot cos(npi) ) is 50.

answer：Okay, so I have this problem here about finding the range of a real number p. It involves two functions, f(x) and g(x), and some conditions about their domains. Let me try to unpack this step by step.First, the function f(x) is given as lg(x² - x - 2). I remember that the logarithm function, lg, is only defined when its argument is positive. So, for f(x) to be defined, x² - x - 2 must be greater than zero. That means I need to solve the inequality x² - x - 2 > 0.To solve this quadratic inequality, I should factor the quadratic expression. Let me see, x² - x - 2 can be factored as (x - 2)(x + 1). So, the inequality becomes (x - 2)(x + 1) > 0. To find where this product is positive, I can use a sign chart. The roots are x = 2 and x = -1, which divide the number line into three intervals: (-∞, -1), (-1, 2), and (2, ∞).Testing each interval:- For x < -1, say x = -2: (-2 - 2)(-2 + 1) = (-4)(-1) = 4 > 0.- For -1 < x < 2, say x = 0: (0 - 2)(0 + 1) = (-2)(1) = -2 < 0.- For x > 2, say x = 3: (3 - 2)(3 + 1) = (1)(4) = 4 > 0.So, the solution to x² - x - 2 > 0 is x ∈ (-∞, -1) ∪ (2, ∞). Therefore, the domain A of f(x) is (-∞, -1) ∪ (2, ∞).Next, the function g(x) is given as sqrt(3/x - 1). The square root function is defined when its argument is non-negative. So, 3/x - 1 ≥ 0. Let me solve this inequality.Starting with 3/x - 1 ≥ 0, I can rewrite it as 3/x ≥ 1. To solve for x, I can consider two cases because x can be positive or negative, and the inequality sign might change when multiplying both sides by x if x is negative.But first, let's note that x cannot be zero because we have 3/x. So, x ≠ 0.Case 1: x > 0.If x is positive, I can multiply both sides of the inequality 3/x ≥ 1 by x without changing the inequality sign. That gives 3 ≥ x, or x ≤ 3. Since we're in the case where x > 0, combining these gives 0 < x ≤ 3.Case 2: x < 0.If x is negative, multiplying both sides of 3/x ≥ 1 by x would reverse the inequality sign. So, 3 ≤ x. But wait, x is negative in this case, so 3 ≤ x would imply x ≥ 3, which contradicts x < 0. Therefore, there are no solutions in this case.So, the domain B of g(x) is (0, 3].Now, the problem mentions α: x ∈ A ∩ B, and β: x satisfies 2x + p < 0. It also says that α is a sufficient condition for β. I need to find the range of p.First, let me find A ∩ B.Set A is (-∞, -1) ∪ (2, ∞), and set B is (0, 3]. The intersection of these two sets would be the overlap between them.Looking at A: (-∞, -1) ∪ (2, ∞) and B: (0, 3]. The overlapping intervals are (2, ∞) intersected with (0, 3], which is (2, 3]. So, A ∩ B = (2, 3].So, α is x ∈ (2, 3]. Now, β is the condition 2x + p < 0, which can be rewritten as x < -p/2.The statement that α is a sufficient condition for β means that if x is in A ∩ B, then x must satisfy β. In logical terms, A ∩ B implies β. So, every x in (2, 3] must satisfy x < -p/2.In other words, the interval (2, 3] must be a subset of the interval (-∞, -p/2). For this to be true, the upper bound of (2, 3] must be less than or equal to -p/2.So, the maximum value in (2, 3] is 3. Therefore, 3 must be less than or equal to -p/2.Let me write that down: 3 ≤ -p/2.Solving for p:Multiply both sides by 2: 6 ≤ -p.Multiply both sides by -1, remembering to reverse the inequality sign: -6 ≥ p, or p ≤ -6.So, p must be less than or equal to -6.But wait, let me double-check this. If p is less than or equal to -6, then -p/2 is greater than or equal to 3. So, the interval (-∞, -p/2) would include all numbers less than -p/2, which is at least 3. Therefore, the interval (2, 3] would indeed be a subset of (-∞, -p/2) because all x in (2, 3] are less than 3, which is less than or equal to -p/2.Wait, hold on. If p is less than or equal to -6, then -p is greater than or equal to 6, so -p/2 is greater than or equal to 3. So, the interval (-∞, -p/2) would start from negative infinity up to a point that is at least 3. Therefore, any x in (2, 3] is less than or equal to 3, which is less than or equal to -p/2. So, yes, (2, 3] is a subset of (-∞, -p/2).Therefore, p must satisfy p ≤ -6.But let me think again. If p is exactly -6, then -p/2 is 3. So, the interval (-∞, 3). But our set A ∩ B is (2, 3]. So, does (2, 3] being a subset of (-∞, 3) hold? Well, 3 is included in A ∩ B, but in the interval (-∞, 3), 3 is not included. So, actually, when p = -6, the condition 2x + p < 0 becomes 2x - 6 < 0, which is x < 3. But our set A ∩ B includes x = 3. So, x = 3 would not satisfy 2x - 6 < 0 because 2*3 -6 = 0, which is not less than 0. Therefore, p cannot be equal to -6 because then x = 3 would not satisfy β, even though x = 3 is in A ∩ B.Therefore, p must be strictly less than -6 to ensure that -p/2 is strictly greater than 3, so that the interval (-∞, -p/2) includes all of (2, 3], including x approaching 3 from the left, but not including 3 itself. Wait, but A ∩ B includes 3, so we need to make sure that even x = 3 satisfies β.But if p is less than -6, then -p/2 is greater than 3, so x = 3 would satisfy 2x + p < 0 because 2*3 + p = 6 + p. Since p < -6, 6 + p < 0. So, x = 3 would satisfy 2x + p < 0.Wait, let me test p = -7. Then, 2x + (-7) < 0 => 2x < 7 => x < 3.5. So, x = 3 is less than 3.5, so it satisfies. Similarly, for p = -6.5, 2x + (-6.5) < 0 => x < 3.25. So, x = 3 is less than 3.25, so it satisfies.But if p = -6, then 2x -6 < 0 => x < 3. So, x = 3 does not satisfy, which is a problem because x = 3 is in A ∩ B. Therefore, p must be strictly less than -6.Wait, but earlier I thought p ≤ -6, but now I see that p must be strictly less than -6 because at p = -6, x = 3 doesn't satisfy β. So, p must be less than -6.Therefore, the range of p is (-∞, -6).Let me confirm this with another example. Suppose p = -7. Then, β is 2x -7 < 0 => x < 3.5. Since A ∩ B is (2, 3], which is entirely less than 3.5, so yes, every x in (2, 3] satisfies x < 3.5. Similarly, if p = -10, β is x < 5, which also includes (2, 3].But if p = -5, which is greater than -6, then β is x < 2.5. But A ∩ B is (2, 3], which includes x = 2.5 and x = 3. So, x = 2.5 would satisfy 2x -5 = 0, which is not less than 0. So, p = -5 doesn't work.Therefore, p must be less than -6.So, the range of p is all real numbers less than -6, which is (-∞, -6).**Final Answer**The range of the real number ( p ) is boxed{(-infty, -6)}.