💡First, I need to determine the marks Eva scored in each subject for both semesters. I'll start by identifying the known marks from the second semester.Eva scored 80 marks in maths, 90 marks in arts, and 90 marks in science in the second semester.For the first semester:- In maths, she scored ten more marks than in the second semester, so her maths score is 80 + 10 = 90.- In science, she scored one-third less than in the second semester. One-third of 90 is 30, so her science score is 90 - 30 = 60.- In arts, let's denote the number of marks she scored less in the first semester than the second as ( x ). Therefore, her arts score in the first semester is ( 90 - x ).Next, I'll calculate the total marks for all semesters:- First semester: 90 (maths) + ( 90 - x ) (arts) + 60 (science) = ( 240 - x )- Second semester: 80 (maths) + 90 (arts) + 90 (science) = 260Adding both semesters together:( 240 - x + 260 = 485 )Simplifying the equation:( 500 - x = 485 )Solving for ( x ):( -x = 485 - 500 )( -x = -15 )( x = 15 )Therefore, Eva scored 15 less marks in arts in the first semester than the second.

💡First, let's define the number of children between Anna and Zara going anti-clockwise as ( n ). According to the problem, there are six times as many children between them going clockwise, which means there are ( 6n ) children in that direction.The total number of children in the class can be calculated by adding the children between Anna and Zara in both directions, plus Anna and Zara themselves:[text{Total children} = n + 6n + 2 = 7n + 2]We know the total number of children is between 20 and 30:[20 < 7n + 2 < 30]Subtracting 2 from all parts of the inequality:[18 < 7n < 28]Dividing by 7:[frac{18}{7} < n < 4]Since ( n ) must be an integer, the only valid value is ( n = 3 ).Substituting ( n = 3 ) back into the total number of children:[7(3) + 2 = 21 + 2 = 23]Therefore, there are 23 children in Miss Tree's class.

💡Okay, so I need to prove that if ( a > 1 ), then ( a^x ) can be made arbitrarily large by choosing ( x ) to be sufficiently large. Hmm, let me think about how to approach this.First, I know that exponential functions grow very quickly. Since ( a ) is greater than 1, as ( x ) increases, ( a^x ) should increase without bound. But I need to formalize this intuition into a mathematical proof.Maybe I can start by recalling some properties of exponential functions. I remember that for any ( a > 1 ), the function ( a^x ) is strictly increasing. That means as ( x ) gets larger, ( a^x ) will also get larger. But how do I show that it can be made as large as desired?Perhaps I can use the definition of limits at infinity. If I can show that the limit of ( a^x ) as ( x ) approaches infinity is infinity, then that would mean ( a^x ) can be made arbitrarily large by choosing a sufficiently large ( x ).Let me write that down: ( lim_{x to infty} a^x = infty ). To prove this, I might need to use the definition of limits. Specifically, for any large number ( M ), I should be able to find an ( x ) such that ( a^x > M ).So, let's formalize this. Given any ( M > 0 ), I need to find an ( x ) such that ( a^x > M ). How can I solve for ( x ) in terms of ( M )?Taking the natural logarithm of both sides might help. If ( a^x > M ), then ( ln(a^x) > ln(M) ). Using the logarithm power rule, this simplifies to ( x ln(a) > ln(M) ).Since ( a > 1 ), ( ln(a) ) is positive. Therefore, I can solve for ( x ) by dividing both sides by ( ln(a) ):( x > frac{ln(M)}{ln(a)} ).So, if I choose ( x ) to be greater than ( frac{ln(M)}{ln(a)} ), then ( a^x ) will be greater than ( M ). This shows that for any large ( M ), I can find an ( x ) such that ( a^x > M ), which means ( a^x ) can be made arbitrarily large by choosing a sufficiently large ( x ).Wait, let me double-check my steps. I started with ( a^x > M ), took the natural logarithm of both sides, used the power rule correctly, and solved for ( x ). Since ( ln(a) ) is positive, the inequality direction remains the same. So, yes, ( x > frac{ln(M)}{ln(a)} ) ensures ( a^x > M ).Is there another way to approach this? Maybe using sequences or induction? I think the logarithmic approach is straightforward and sufficient. It directly relates ( x ) to ( M ) in a clear manner.Alternatively, I could consider the definition of exponential growth. For ( a > 1 ), ( a^x ) grows exponentially, which means it increases faster than any linear or polynomial function. This inherent property ensures that ( a^x ) will surpass any fixed bound ( M ) given enough time, i.e., a large enough ( x ).I should also consider if there are any edge cases or special conditions where this might not hold. Since ( a > 1 ) is given, and ( x ) is a real number, there doesn't seem to be any issue. If ( a ) were equal to 1, ( a^x ) would just be 1 for all ( x ), but since ( a > 1 ), we don't have that problem.Another thought: maybe using the definition of supremum or something related to the completeness of real numbers. But I think the logarithmic method is more straightforward.Let me summarize my thoughts:1. Start with the inequality ( a^x > M ).2. Take the natural logarithm of both sides: ( ln(a^x) > ln(M) ).3. Simplify using the logarithm power rule: ( x ln(a) > ln(M) ).4. Solve for ( x ): ( x > frac{ln(M)}{ln(a)} ).5. Conclude that for any ( M ), choosing ( x ) larger than ( frac{ln(M)}{ln(a)} ) ensures ( a^x > M ).This seems solid. I don't see any gaps or errors in this reasoning. It directly uses properties of logarithms and the fact that ( ln(a) ) is positive when ( a > 1 ).Just to make sure, let me test this with some numbers. Suppose ( a = 2 ) and ( M = 1000 ). Then ( ln(2) approx 0.6931 ), and ( ln(1000) approx 6.9078 ). So, ( x > frac{6.9078}{0.6931} approx 9.966 ). So, if I choose ( x = 10 ), then ( 2^{10} = 1024 ), which is indeed greater than 1000. That checks out.Another example: ( a = 3 ) and ( M = 1000000 ). ( ln(3) approx 1.0986 ), ( ln(1000000) approx 13.8155 ). So, ( x > frac{13.8155}{1.0986} approx 12.57 ). Choosing ( x = 13 ), ( 3^{13} = 1594323 ), which is greater than 1,000,000. Perfect.This method works consistently. Therefore, I'm confident in my proof.**Final Answer**boxed{a^x text{ can be arbitrarily large if } x text{ is sufficiently large.}}

💡Alright, so I have this problem about a regular dodecagon and a square. It's divided into two parts, a) and b). Let me try to understand what each part is asking and figure out how to approach them.Starting with part a): It says that four vertices of a regular dodecagon are located at the midpoints of the sides of a square. I need to prove that the area of the shaded part is 12 times smaller than the area of the dodecagon. Hmm, okay. So, first, I should probably visualize this setup. A regular dodecagon has 12 sides, all equal, and all internal angles equal. A square has four sides, and the midpoints of its sides would be the centers of each side.If four vertices of the dodecagon are at these midpoints, that means the dodecagon is somehow inscribed or related to the square. Maybe the square is inscribed in the dodecagon? Or perhaps the dodecagon is inscribed in the square? I need to clarify that.Wait, the problem says four vertices of the dodecagon are at the midpoints of the square's sides. So, the dodecagon is larger than the square, with four of its vertices touching the midpoints of the square. That makes sense. So, the square is kind of inscribed in the dodecagon, but only at the midpoints.Now, the shaded part is mentioned, but since I don't have the figure, I need to assume what the shaded part could be. Maybe it's the area inside the square but outside the dodecagon? Or perhaps it's the area inside the dodecagon but outside the square? The problem says the shaded part is 12 times smaller than the dodecagon's area. So, if the dodecagon's area is A, the shaded part is A/12.To approach this, I think I need to find the area of the dodecagon and then find the area of the shaded part, which is 1/12 of that. Alternatively, maybe the shaded part is a specific region related to the square and the dodecagon.Let me recall the formula for the area of a regular dodecagon. A regular dodecagon can be divided into 12 congruent isosceles triangles, each with a vertex angle of 30 degrees (since 360/12 = 30). The area of each triangle is (1/2)*r^2*sin(theta), where r is the radius and theta is the central angle.But wait, in this case, the dodecagon is related to a square. Maybe the radius of the dodecagon is related to the square's side length. If the midpoints of the square are vertices of the dodecagon, then the distance from the center of the square to the midpoint of its side is equal to the radius of the dodecagon.Let me denote the side length of the square as 's'. The distance from the center of the square to the midpoint of a side is s/2. So, the radius r of the dodecagon is s/2.Therefore, the area of the dodecagon would be 12*(1/2)*r^2*sin(30°) = 12*(1/2)*(s/2)^2*(1/2) = 12*(1/2)*(s^2/4)*(1/2) = 12*(s^2/16) = (12/16)*s^2 = (3/4)*s^2.Wait, that seems too small. Maybe I made a mistake in the formula. Let me double-check. The area of a regular polygon with n sides is (1/2)*n*r^2*sin(2π/n). For a dodecagon, n=12, so the area is (1/2)*12*r^2*sin(30°) = 6*r^2*(1/2) = 3*r^2.Ah, that's better. So, the area of the dodecagon is 3*r^2. Since r = s/2, then the area is 3*(s/2)^2 = 3*s^2/4.Okay, so the area of the dodecagon is (3/4)*s^2.Now, what is the shaded part? If the shaded part is 12 times smaller, then its area should be (3/4)*s^2 / 12 = (3/4)*(1/12)*s^2 = (1/16)*s^2.But wait, the area of the square is s^2. So, the shaded part is (1/16)*s^2, which is 1/16 of the square's area. That seems too small. Maybe I misunderstood the shaded part.Alternatively, perhaps the shaded part is the area inside the dodecagon but outside the square. If the square is inscribed in the dodecagon, then the area of the dodecagon is (3/4)*s^2, and the area of the square is s^2. But that would mean the shaded part (dodecagon minus square) is negative, which doesn't make sense.Wait, maybe the square is larger than the dodecagon? If four vertices of the dodecagon are at the midpoints of the square's sides, then the square must be larger. The distance from the center to the midpoint is s/2, which is the radius of the dodecagon. So, the square has side length s, and the dodecagon has a radius of s/2.But then, the area of the dodecagon is 3*(s/2)^2 = 3*s^2/4, and the area of the square is s^2. So, the shaded part could be the area of the square minus the area of the dodecagon, which is s^2 - 3*s^2/4 = s^2/4. But s^2/4 is 4 times smaller than the dodecagon's area (3*s^2/4). That doesn't match the 12 times smaller.Hmm, maybe the shaded part is something else. Perhaps it's one of the triangles formed by the dodecagon and the square. Since the dodecagon has 12 sides, and the square has 4 sides, maybe each side of the square corresponds to three sides of the dodecagon.Wait, if I imagine the square with midpoints connected to the dodecagon, each side of the square would have three sides of the dodecagon adjacent to it. So, maybe the shaded part is one of these small segments or triangles.Alternatively, perhaps the shaded part is the area of one of the triangles formed by the center of the square and two adjacent vertices of the dodecagon. Since the dodecagon has 12 vertices, each central angle is 30 degrees. So, each of these triangles would have an area of (1/2)*r^2*sin(30°) = (1/2)*(s/2)^2*(1/2) = (1/2)*(s^2/4)*(1/2) = s^2/16.So, if the shaded part is one of these triangles, its area is s^2/16, and the area of the dodecagon is 12 times that, which is 12*(s^2/16) = (12/16)*s^2 = (3/4)*s^2. That matches the area of the dodecagon we calculated earlier.Therefore, the shaded part is one of these small triangles, and its area is indeed 12 times smaller than the area of the dodecagon. So, part a) is proved.Moving on to part b): Prove that the area of a dodecagon inscribed in a circle of radius 1 is equal to 3.Okay, so a regular dodecagon inscribed in a circle of radius 1. The formula for the area of a regular polygon with n sides inscribed in a circle of radius r is (1/2)*n*r^2*sin(2π/n). For a dodecagon, n=12, so the area is (1/2)*12*1^2*sin(2π/12) = 6*sin(π/6).Wait, sin(π/6) is 1/2, so 6*(1/2) = 3. That's straightforward. So, the area is indeed 3.But let me verify this formula to be sure. The area of a regular polygon can be calculated by dividing it into n isosceles triangles, each with a central angle of 2π/n. The area of each triangle is (1/2)*r^2*sin(2π/n). So, total area is n*(1/2)*r^2*sin(2π/n) = (1/2)*n*r^2*sin(2π/n).For n=12 and r=1, it's (1/2)*12*1*sin(2π/12) = 6*sin(π/6) = 6*(1/2) = 3. Yep, that checks out.So, part b) is also proved.Wait, but in part a), I assumed the shaded part was one of the small triangles, but the problem didn't specify. Maybe I should double-check that assumption.If the shaded part is indeed one of those small triangles, then its area is 1/12 of the dodecagon's area. But if the shaded part is something else, like a different region, my conclusion might be wrong. However, given the problem states it's 12 times smaller, and the dodecagon is made up of 12 congruent triangles, it makes sense that each triangle is 1/12 of the total area.So, I think my reasoning holds. Therefore, both parts a) and b) are proved.

💡Okay, so I have this problem about an oblique parallelepiped. Hmm, I remember that a parallelepiped is like a 3D figure formed by six parallelograms. Since it's oblique, it means that the lateral edges aren't perpendicular to the base. The problem says that the projection of a lateral edge onto the base plane is 5, and the height is 12. I think the projection being 5 means that if I were to drop a perpendicular from the top vertex to the base, the length of that projection on the base plane is 5 units. The height is 12, which I believe is the actual length of the lateral edge from the top to the base, not just the projection. So, maybe I can use the Pythagorean theorem here? If the projection is 5 and the height is 12, then the actual length of the lateral edge should be the hypotenuse of a right triangle with legs 5 and 12. Let me calculate that: ( sqrt{5^2 + 12^2} = sqrt{25 + 144} = sqrt{169} = 13 ). Okay, so the lateral edge is 13 units long. That makes sense. Next, the problem mentions a cross-section perpendicular to the lateral edge, and it's a rhombus with an area of 24 and one diagonal of 8. I need to find the lateral surface area and the volume of the parallelepiped. Let me think about the cross-section first. A cross-section perpendicular to the lateral edge is a rhombus. The area of a rhombus is given by ( frac{d_1 times d_2}{2} ), where ( d_1 ) and ( d_2 ) are the lengths of the diagonals. We know the area is 24 and one diagonal is 8, so I can find the other diagonal. Let me set up the equation: ( frac{8 times d_2}{2} = 24 )Simplifying that:( 4 times d_2 = 24 )So, ( d_2 = 6 ). Alright, so the diagonals of the rhombus are 8 and 6. Now, in a rhombus, the diagonals bisect each other at right angles. So, each side of the rhombus can be found using the Pythagorean theorem as well. Each side is the hypotenuse of a right triangle with legs half of each diagonal. So, half of 8 is 4, and half of 6 is 3. Therefore, each side of the rhombus is:( sqrt{4^2 + 3^2} = sqrt{16 + 9} = sqrt{25} = 5 )So, each side of the rhombus is 5 units. Now, since the cross-section is a rhombus, the perimeter of this cross-section would be 4 times the side length, which is ( 4 times 5 = 20 ). But wait, how does this relate to the lateral surface area? The lateral surface area of a parallelepiped is the sum of the areas of all the lateral faces. In this case, since the cross-section is a rhombus, each lateral face is a parallelogram with one side being the lateral edge of the parallelepiped (which we found to be 13) and the other side being the side of the rhombus (which is 5). So, each lateral face has an area of base times height, but I need to be careful here. Actually, since the cross-section is perpendicular to the lateral edge, the area of each lateral face is the product of the side of the rhombus and the lateral edge. Wait, no, maybe not. Let me think again. The cross-section is a rhombus, so each lateral face is a parallelogram. The area of each parallelogram is the base times the height. Here, the base is the side of the rhombus, which is 5, and the height would be the height of the parallelepiped, which is 12? Or is it something else? Hmm, I'm getting confused. Maybe I should approach it differently. The lateral surface area can be calculated as the perimeter of the base times the height. But in this case, the base is not necessarily a rhombus. Wait, the cross-section is a rhombus, but the base is a different face. Wait, no, the cross-section is perpendicular to the lateral edge, so it's like slicing the parallelepiped with a plane perpendicular to the lateral edge, resulting in a rhombus. So, this cross-section is actually the shape that is extruded along the lateral edge to form the parallelepiped. Therefore, the lateral surface area would be the perimeter of this cross-section times the length of the lateral edge. So, the perimeter of the rhombus is 20, as I calculated earlier, and the lateral edge is 13. So, the lateral surface area should be ( 20 times 13 = 260 ). Okay, that seems reasonable. Now, for the volume. The volume of a parallelepiped is the area of the base times the height. But here, the cross-section perpendicular to the lateral edge is given, which is a rhombus with area 24. So, maybe the volume is the area of this cross-section times the length of the lateral edge? Let me check: Volume = area of cross-section × length along which it's extruded. So, yes, that would be ( 24 times 13 = 312 ). Wait, but the height of the parallelepiped is given as 12. Is that consistent? Because if the cross-section area is 24 and the height is 12, then the volume would be ( 24 times 12 = 288 ). Hmm, that contradicts my previous calculation. I must be making a mistake here. Let me clarify. In a parallelepiped, the volume can be calculated in two ways: either as the area of the base times the height (where height is the perpendicular distance from the base to the opposite face) or as the area of a cross-section times the length along which it's extruded (which is the lateral edge in this case). But in this problem, the height is given as 12, which is the perpendicular distance from the base to the top face. The cross-section is perpendicular to the lateral edge, which is 13. So, the volume should be the area of the cross-section times the length of the lateral edge, which is ( 24 times 13 = 312 ). But then why is the height given as 12? Maybe because the cross-section area is related to the height? Let me think. Wait, the cross-section is a rhombus with area 24, and one of its diagonals is 8. We found the other diagonal is 6, and the sides are 5. So, the area of the cross-section is 24, which is independent of the height of the parallelepiped. But the height of the parallelepiped is 12, which is the perpendicular distance between the bases. So, maybe the volume can also be calculated as the area of the base times the height. But we don't know the area of the base. Alternatively, since the cross-section is perpendicular to the lateral edge, the volume is indeed the area of the cross-section times the length of the lateral edge. So, 24 × 13 = 312. But then, how does the height of 12 factor into this? Maybe the height is related to the projection? Wait, earlier, we found that the lateral edge is 13, which is the hypotenuse of a right triangle with legs 5 and 12. So, 5 is the projection on the base plane, and 12 is the height. So, the height is 12, which is the perpendicular distance from the base to the top. So, if I were to calculate the volume as base area × height, I need to find the base area. But I don't have the base area directly. However, the cross-section area is 24, which is related to the lateral edge. Alternatively, since the cross-section is perpendicular to the lateral edge, the volume is indeed 24 × 13 = 312. I think that makes sense because the cross-section is like a slice that's being extended along the lateral edge to form the entire parallelepiped. So, the volume is the area of that slice times how far it's extended, which is the length of the lateral edge. Therefore, I think the volume is 312, and the lateral surface area is 260. Wait, just to double-check, the lateral surface area is the perimeter of the cross-section times the lateral edge. The cross-section is a rhombus with side 5, so perimeter is 20. 20 × 13 = 260. That seems correct. And the volume is 24 × 13 = 312. I think that's it. **Final Answer**The lateral surface area is boxed{260} and the volume is boxed{312}.

💡First, let's denote the factor by which Alice increases her money as ( x ). Therefore, Alice's final amount is ( 2000x ).Bob invests in real estate and makes five times more money than he invested. This means his final amount is ( 2000 + 5 times 2000 = 2000 times 6 = 12000 ).According to the problem, Bob has 8000 more than Alice now. We can set up the equation:[12000 = 2000x + 8000]Solving for ( x ):[12000 - 8000 = 2000x][4000 = 2000x][x = frac{4000}{2000} = 2]So, Alice's final amount is ( 2000 times 2 = 4000 ).The ratio of Alice's final amount to her initial investment is:[frac{4000}{2000} = 2]Therefore, the ratio is ( 2:1 ).