💡First, I need to determine the total area of the garden plot by calculating the area of the large rectangle.Next, I will subtract the area of the 4-foot by 4-foot square that has been cut out for the pond.Then, I will calculate the area of the 3-foot by 2-foot right triangle garden bed that has been added.Finally, I will add the area of the triangle to the remaining area of the garden plot to find the total remaining garden space.

💡Alright, so I have this logic problem here with two propositions, p and q, and I need to figure out which of the four given compound propositions are true. Let me take it step by step.First, let's understand what each proposition is saying.Proposition p: If x is greater than y, then -x is less than -y. Hmm, okay. So, if I have two numbers where x is bigger than y, then when I negate both, the order flips. That makes sense because multiplying both sides of an inequality by a negative number reverses the inequality sign. For example, if x = 5 and y = 3, then -x = -5 and -y = -3. Indeed, -5 is less than -3. So, p seems to be true.Proposition q: If x is less than y, then x squared is greater than y squared. Hmm, this one is a bit trickier. Let me test it with some numbers. Suppose x = 2 and y = 3. Then x < y is true because 2 < 3. Now, x squared is 4 and y squared is 9. Is 4 > 9? No, it's not. So, in this case, the conclusion is false even though the premise is true. That means proposition q is false.Okay, so now I know that p is true and q is false. Let's look at the four compound propositions:1. p ∧ q: This means both p and q have to be true for this to be true. But since q is false, p ∧ q is false.2. p ∨ q: This means either p or q has to be true. Since p is true, p ∨ q is true.3. p ∧ (¬q): This means p is true and q is false. Since p is true and q is indeed false, this is true.4. (¬p) ∨ q: This means either p is false or q is true. But p is true, so ¬p is false, and q is false, so the whole thing is false.So, out of the four, propositions 2 and 3 are true. Let me double-check to make sure I didn't make a mistake.For proposition q, I used x = 2 and y = 3, which showed that q is false. Maybe I should try another example to confirm. Let's say x = -1 and y = 0. Then x < y is true because -1 < 0. x squared is 1 and y squared is 0. Is 1 > 0? Yes, it is. So in this case, q is true. Wait, that contradicts my earlier conclusion. Hmm.But hold on, in logic, if there's even one case where the premise is true but the conclusion is false, the proposition is false. So even though q can be true in some cases, since there's at least one case where it's false, q is not universally true. Therefore, q is false.Okay, that clears up any confusion. So, sticking with my initial assessment, p is true and q is false. Therefore, among the four compound propositions, only 2 (p ∨ q) and 3 (p ∧ ¬q) are true.Looking back at the options given:A: (1) and (3) – We know (1) is false, so A is incorrect.B: (1) and (4) – Both are false, so B is incorrect.C: (2) and (3) – Both are true, so C is correct.D: (2) and (4) – (2) is true, but (4) is false, so D is incorrect.Therefore, the correct answer is C.**Final Answer**boxed{C}

💡First, I need to determine how much money Troy has saved initially and how much he plans to get from selling his old computer.Troy has saved 50 and plans to sell his old computer for 20. Adding these amounts together gives him a total of 70.Next, I need to find out how much more money Troy needs to afford the new computer, which costs 80.Subtracting the total amount Troy has (70) from the cost of the new computer (80) will give the additional amount he needs.So, 80 minus 70 equals 10. Therefore, Troy needs an additional 10 to buy the new computer.

💡First, I need to find the value of ( x ) given the equation (frac{x}{y} = frac{12}{5}) and the value of ( y = 25 ).I'll start by substituting ( y = 25 ) into the equation:[frac{x}{25} = frac{12}{5}]To isolate ( x ), I'll multiply both sides of the equation by 25:[x = frac{12}{5} times 25]Next, I'll simplify the multiplication:[x = frac{12 times 25}{5} = frac{300}{5} = 60]Therefore, the value of ( x ) is 60.

💡Okay, so I have this problem where there's a rectangle ABCD with sides of lengths 12 and 5. I need to find the maximum possible area of an equilateral triangle that can fit entirely inside this rectangle. Hmm, interesting. Let me try to visualize this.First, I know that an equilateral triangle has all sides equal and all angles equal to 60 degrees. The area of an equilateral triangle with side length 's' is given by the formula (√3/4) * s². So, to maximize the area, I need to maximize the side length 's' such that the entire triangle fits inside the rectangle.Now, the rectangle has sides 12 and 5. So, the maximum possible side length of the triangle can't exceed the shorter side of the rectangle, right? Because if the triangle is placed with one side along the shorter side of the rectangle, the height of the triangle must also fit within the longer side of the rectangle.Wait, let me think about that again. If I place the triangle so that one of its sides is along the 12-unit side of the rectangle, then the height of the triangle must be less than or equal to 5 units. Similarly, if I place it along the 5-unit side, the height must be less than or equal to 12 units. But since the height of an equilateral triangle is (√3/2) * s, I need to make sure that this height doesn't exceed the other side of the rectangle.So, if I place the triangle along the 12-unit side, the height would be (√3/2) * s ≤ 5. Solving for 's', we get s ≤ (5 * 2)/√3 = 10/√3 ≈ 5.7735. But wait, the side of the triangle can't exceed 12 units because it's placed along that side. So, in this case, the limiting factor is the height, which restricts 's' to about 5.7735 units.On the other hand, if I place the triangle along the 5-unit side, the height would be (√3/2) * s ≤ 12. Solving for 's', we get s ≤ (12 * 2)/√3 = 24/√3 = 8√3 ≈ 13.8564. But wait, the side of the triangle can't exceed 5 units because it's placed along that side. So, in this case, the limiting factor is the side length itself, which is 5 units.Hmm, so placing the triangle along the longer side allows for a larger side length, but the height might be too big. Wait, no, actually, when placing along the longer side, the height is constrained by the shorter side, which gives a larger maximum side length than when placing along the shorter side.Wait, let me clarify. If I place the triangle with a side along the 12-unit side, the height of the triangle must be ≤ 5 units. So, s ≤ 10/√3 ≈ 5.7735. But if I place it along the 5-unit side, the height must be ≤ 12 units, which allows s ≤ 24/√3 ≈ 13.8564, but since the side can't exceed 5 units, that's not possible. So, actually, placing the triangle along the longer side allows for a larger side length than placing it along the shorter side.Wait, but 5.7735 is larger than 5, so if I place the triangle along the 12-unit side, I can have a side length of about 5.7735, which is larger than 5. But does that fit within the rectangle? Because the height would be 5 units, which is exactly the shorter side. So, yes, that should fit.But wait, if I place the triangle with a side along the 12-unit side, the height is 5 units, which is exactly the shorter side. So, the triangle would fit perfectly in terms of height, but what about the base? The base is 5.7735 units, which is less than 12 units, so that's fine.Alternatively, if I rotate the triangle so that it's not aligned with the sides of the rectangle, maybe I can fit a larger triangle inside the rectangle. Hmm, that might be possible. Because sometimes, rotating a shape can allow it to fit into a space more efficiently.Let me think about how to approach this. If I rotate the equilateral triangle inside the rectangle, the maximum side length might be larger than when it's aligned with the sides. But I need to ensure that all three vertices of the triangle stay within the rectangle.This seems more complicated. Maybe I can model the rectangle as a coordinate system and try to find the maximum 's' such that the triangle can fit inside.Let's set up a coordinate system where the rectangle has vertices at (0,0), (12,0), (12,5), and (0,5). Now, let's consider placing the triangle such that one vertex is at (0,0), another at (x,0), and the third somewhere inside the rectangle.Wait, but if I rotate the triangle, the coordinates of the vertices will change. Maybe I can use some trigonometry to find the maximum 's'.Alternatively, I can think about the maximum distance between two points inside the rectangle. The diagonal of the rectangle is √(12² + 5²) = √(144 + 25) = √169 = 13 units. So, the maximum possible distance between two points inside the rectangle is 13 units. But since an equilateral triangle has all sides equal, the side length 's' can't exceed 13 units. But wait, can we actually fit a triangle with side length 13 inside the rectangle?No, because the height of such a triangle would be (√3/2)*13 ≈ 11.258 units, which is larger than the shorter side of the rectangle (5 units). So, that won't fit.So, maybe the maximum side length is limited by the shorter side when placed along the longer side, as I initially thought. But earlier, I found that placing it along the longer side allows for a side length of about 5.7735 units, which is larger than 5 units.Wait, but 5.7735 is larger than 5, so if I place the triangle along the longer side, the height is 5 units, which is exactly the shorter side. So, that should fit.But let me double-check. If the side length is 5.7735, then the height is (√3/2)*5.7735 ≈ 5 units, which is exactly the shorter side. So, the triangle would fit perfectly in terms of height, and the base would be 5.7735 units, which is less than 12 units, so that's fine.Alternatively, if I place the triangle along the shorter side, the maximum side length is 5 units, which gives a height of (√3/2)*5 ≈ 4.330 units, which is less than 12 units, so that also fits.But since 5.7735 is larger than 5, placing the triangle along the longer side allows for a larger area. So, the maximum area would be (√3/4)*(10/√3)² = (√3/4)*(100/3) = (100√3)/12 = (25√3)/3 ≈ 14.433 units².Wait, but earlier I thought the area was (25√3)/4 ≈ 10.825 units² when placing along the shorter side. So, clearly, placing along the longer side gives a larger area.But wait, is there a way to place the triangle even larger by rotating it? Maybe not aligned with the sides.Let me think about this. If I rotate the triangle so that it's not aligned with the sides of the rectangle, perhaps I can fit a larger triangle.I recall that the maximum area of an equilateral triangle that can fit inside a rectangle can sometimes be found by considering the rectangle's diagonal, but in this case, the diagonal is 13 units, but as I saw earlier, a triangle with side length 13 won't fit because its height is too large.Alternatively, maybe the maximum side length is determined by the rectangle's shorter side when the triangle is placed at a certain angle.Wait, perhaps I can model this using some geometry. Let me consider placing the triangle such that one vertex is at a corner of the rectangle, and the other two vertices touch the opposite sides.Let me set up coordinates again. Let's say one vertex is at (0,0), another at (x,0), and the third at (a,b), where 0 ≤ x ≤ 12, 0 ≤ a ≤ 12, and 0 ≤ b ≤ 5.Since it's an equilateral triangle, the distance between (0,0) and (x,0) is x, the distance between (0,0) and (a,b) is √(a² + b²), and the distance between (x,0) and (a,b) is √((a - x)² + b²). All these distances must be equal to 's'.So, we have:1. x = s2. √(a² + b²) = s3. √((a - x)² + b²) = sFrom equation 1, x = s. So, equation 3 becomes √((a - s)² + b²) = s.Squaring both sides: (a - s)² + b² = s².Expanding: a² - 2as + s² + b² = s².Simplifying: a² - 2as + b² = 0.But from equation 2, a² + b² = s². So, substituting into the above equation:s² - 2as = 0 ⇒ s² = 2as ⇒ a = s/2.So, a = s/2.Now, from equation 2: (s/2)² + b² = s² ⇒ s²/4 + b² = s² ⇒ b² = (3/4)s² ⇒ b = (√3/2)s.But b must be ≤ 5, so (√3/2)s ≤ 5 ⇒ s ≤ (10)/√3 ≈ 5.7735.So, this confirms the earlier result. The maximum side length when placing the triangle with one side along the longer side is s = 10/√3 ≈ 5.7735, which gives a height of 5 units, fitting perfectly within the rectangle.Therefore, the maximum area is (√3/4)*(10/√3)² = (√3/4)*(100/3) = (100√3)/12 = (25√3)/3 ≈ 14.433 units².Wait, but earlier I thought the area was (25√3)/4 when placing along the shorter side, which is about 10.825 units². So, clearly, placing along the longer side gives a larger area.But wait, is there a way to place the triangle even larger by rotating it? Maybe not aligned with the sides.Wait, I just did a calculation where I considered placing the triangle with one vertex at a corner and the other two vertices on adjacent sides, which gave me s = 10/√3. But is this the maximum possible?Alternatively, maybe placing the triangle such that all three vertices touch the sides of the rectangle, not necessarily at the corners.This might allow for a larger side length. Let me try to model this.Suppose the triangle is placed inside the rectangle such that each vertex touches a different side. Let's denote the rectangle with vertices at (0,0), (12,0), (12,5), and (0,5). Let the triangle have vertices at (x,0), (12,y), and (z,5), where 0 < x < 12, 0 < y < 5, and 0 < z < 12.Since it's an equilateral triangle, all sides must be equal. So, the distance between (x,0) and (12,y) must equal the distance between (12,y) and (z,5), and also equal the distance between (z,5) and (x,0).This seems complicated, but maybe I can set up equations.Let me denote the three points as A(x,0), B(12,y), and C(z,5).Then, the distances are:AB: √[(12 - x)² + (y - 0)²] = √[(12 - x)² + y²]BC: √[(z - 12)² + (5 - y)²]CA: √[(x - z)² + (0 - 5)²] = √[(x - z)² + 25]Since AB = BC = CA = s.So, we have:√[(12 - x)² + y²] = √[(z - 12)² + (5 - y)²] = √[(x - z)² + 25]This is a system of equations with variables x, y, z, s. It seems quite involved, but maybe I can find a relationship between x, y, and z.Alternatively, maybe I can use some symmetry or geometric properties to simplify this.Wait, perhaps the triangle is placed such that it's symmetric with respect to the center of the rectangle. The center of the rectangle is at (6, 2.5). Maybe the triangle is rotated 30 degrees or something like that.Alternatively, maybe I can use some trigonometry. Let me consider the angles involved.Since it's an equilateral triangle, all angles are 60 degrees. If I place the triangle inside the rectangle, the orientation might affect how much it can be scaled.Wait, maybe I can model the triangle as being rotated by an angle θ from the horizontal axis. Then, the projections of the triangle's sides onto the rectangle's sides must fit within the rectangle's dimensions.Let me try this approach.Suppose the triangle is rotated by an angle θ. Then, the width it occupies along the x-axis would be s * cosθ + s * cos(60° - θ), and the height along the y-axis would be s * sinθ + s * sin(60° - θ). These must be less than or equal to 12 and 5, respectively.Wait, is that correct? Let me think.Actually, when you rotate an equilateral triangle, the bounding box (the minimum rectangle that can contain the triangle) has width and height determined by the projections of the triangle's sides.The width would be the maximum horizontal extent, which is s * cosθ + s * cos(60° - θ), and the height would be the maximum vertical extent, which is s * sinθ + s * sin(60° - θ).But I'm not sure if this is the right way to model it. Maybe I should consider the maximum and minimum x and y coordinates of the triangle's vertices after rotation.Alternatively, perhaps I can use the fact that the maximum width and height of the rotated triangle must fit within the rectangle's dimensions.Wait, maybe it's better to use some calculus here. Let me consider the area as a function of θ and find its maximum under the constraints that the triangle fits within the rectangle.But this might get complicated. Let me see if I can find a better approach.Wait, I remember that the maximum area of an equilateral triangle that can fit inside a rectangle can be found by considering the rectangle's shorter side as the limiting factor when the triangle is placed along the longer side, as I did earlier. But maybe there's a way to fit a larger triangle by rotating it.Wait, let me think about the case where the triangle is placed such that one vertex is at a corner, and the other two vertices are on adjacent sides. This is similar to what I did earlier, but maybe I can generalize it.Let me consider the triangle with vertices at (0,0), (a,0), and (b,c), where (b,c) is somewhere inside the rectangle. The distances between these points must all be equal.So, distance from (0,0) to (a,0) is 'a'.Distance from (0,0) to (b,c) is √(b² + c²).Distance from (a,0) to (b,c) is √((b - a)² + c²).Setting these equal:a = √(b² + c²) = √((b - a)² + c²)From the first equality: a² = b² + c².From the second equality: a² = (b - a)² + c².Expanding the second equation: a² = b² - 2ab + a² + c².Simplifying: 0 = -2ab + c².But from the first equation, c² = a² - b². Substituting into the second equation:0 = -2ab + (a² - b²)Rearranging: a² - b² - 2ab = 0 ⇒ a² - 2ab - b² = 0.This is a quadratic in terms of 'a'. Let me solve for 'a':a² - 2ab - b² = 0 ⇒ a = [2b ± √(4b² + 4b²)]/2 = [2b ± √(8b²)]/2 = [2b ± 2b√2]/2 = b(1 ± √2).Since 'a' must be positive, we take the positive root: a = b(1 + √2).Now, from the first equation, c² = a² - b² = [b²(1 + √2)²] - b² = b²[(1 + 2√2 + 2) - 1] = b²(2 + 2√2).So, c = b√(2 + 2√2).But we have constraints: a ≤ 12 and c ≤ 5.So, a = b(1 + √2) ≤ 12 ⇒ b ≤ 12 / (1 + √2).Similarly, c = b√(2 + 2√2) ≤ 5 ⇒ b ≤ 5 / √(2 + 2√2).Let me compute these values.First, compute 1 + √2 ≈ 1 + 1.4142 ≈ 2.4142.So, b ≤ 12 / 2.4142 ≈ 4.969.Next, compute √(2 + 2√2). Let's compute inside first: 2 + 2√2 ≈ 2 + 2.8284 ≈ 4.8284. Then, √4.8284 ≈ 2.197.So, b ≤ 5 / 2.197 ≈ 2.275.Therefore, the more restrictive constraint is b ≤ 2.275.So, the maximum 'b' is approximately 2.275, which gives a = 2.275 * (1 + √2) ≈ 2.275 * 2.4142 ≈ 5.5.And c = 2.275 * √(2 + 2√2) ≈ 2.275 * 2.197 ≈ 5.Wait, so c ≈ 5, which is exactly the shorter side of the rectangle. So, this configuration allows the triangle to have a side length of a ≈ 5.5, which is larger than the 5.7735 I found earlier? Wait, no, 5.5 is less than 5.7735.Wait, actually, 5.5 is less than 5.7735, so this configuration doesn't give a larger side length than the previous one.Hmm, so placing the triangle with one side along the longer side gives a larger side length than placing it with one vertex at a corner and the other two on adjacent sides.Therefore, the maximum side length is still 10/√3 ≈ 5.7735, giving an area of (25√3)/3 ≈ 14.433 units².But wait, earlier I thought that placing the triangle along the longer side gives s = 10/√3, which is approximately 5.7735, and that's the maximum.But let me confirm if this is indeed the maximum. Is there a way to place the triangle such that it's larger than this?Wait, another approach: The maximum area of an equilateral triangle that can fit inside a rectangle can be found by considering the rectangle's shorter side as the height of the triangle. So, if the height is 5, then the side length s = 2h / √3 = 10/√3 ≈ 5.7735, which is the same as before.So, this seems consistent.Alternatively, if we consider the rectangle's longer side as the base, then the height of the triangle would be 5, which is less than the longer side, so that's fine.Therefore, the maximum area is achieved when the triangle is placed with its base along the longer side of the rectangle, with side length 10/√3, giving an area of (25√3)/3.Wait, but earlier I thought the area was (25√3)/4 when placing along the shorter side, but that's smaller. So, the maximum area is indeed (25√3)/3.But let me double-check the calculations.Area = (√3/4) * s².If s = 10/√3, then s² = 100/3.So, Area = (√3/4) * (100/3) = (100√3)/12 = (25√3)/3 ≈ 14.433 units².Yes, that's correct.Alternatively, if I place the triangle along the shorter side, s = 5, then the area is (√3/4)*25 ≈ 10.825 units², which is smaller.Therefore, the maximum area is (25√3)/3.Wait, but I just realized something. When I place the triangle along the longer side, the height is exactly 5 units, which is the shorter side. So, the triangle is snug against the top and bottom of the rectangle.But what if I rotate the triangle slightly so that it's not aligned with the sides, but still fits within the rectangle? Could that allow for a larger side length?Wait, let me think about this. If I rotate the triangle, the projections along the x and y axes would change. The width and height required to contain the triangle would depend on the angle of rotation.The area of the triangle is maximized when the side length is as large as possible, but the side length is constrained by the rectangle's dimensions.I think the maximum side length is indeed 10/√3, as found earlier, because when the triangle is placed with its base along the longer side, the height is exactly the shorter side, and any rotation would require the height to be less than or equal to 5, which would not allow for a larger side length.Wait, actually, if I rotate the triangle, the height in the y-direction would be s * sinθ, and the width in the x-direction would be s * cosθ + s * cos(60° - θ). To fit within the rectangle, both of these must be ≤ 5 and 12, respectively.But I'm not sure if this approach would yield a larger side length than 10/√3.Alternatively, maybe I can set up the problem using calculus to maximize 's' subject to the constraints that the rotated triangle fits within the rectangle.Let me try that.Let me denote the angle of rotation as θ, measured from the x-axis. The triangle has vertices at (0,0), (s,0), and (s/2, (s√3)/2). When rotated by θ, the new coordinates become:A: (0,0)B: (s cosθ, s sinθ)C: (s/2 cosθ - (s√3)/2 sinθ, s/2 sinθ + (s√3)/2 cosθ)Now, the bounding box of the triangle must fit within the rectangle, so the maximum x-coordinate must be ≤ 12, and the maximum y-coordinate must be ≤ 5.The maximum x-coordinate is the maximum of 0, s cosθ, and s/2 cosθ - (s√3)/2 sinθ.Similarly, the maximum y-coordinate is the maximum of 0, s sinθ, and s/2 sinθ + (s√3)/2 cosθ.We need:max(s cosθ, s/2 cosθ - (s√3)/2 sinθ) ≤ 12andmax(s sinθ, s/2 sinθ + (s√3)/2 cosθ) ≤ 5This seems quite involved, but maybe I can find the maximum 's' that satisfies these inequalities.Alternatively, perhaps I can consider the case where the maximum x and y are achieved at different points.Wait, maybe the maximum x is s cosθ, and the maximum y is s sinθ + (s√3)/2 cosθ.Wait, no, let me compute the coordinates more carefully.Point A is (0,0).Point B is (s cosθ, s sinθ).Point C is (s/2 cosθ - (s√3)/2 sinθ, s/2 sinθ + (s√3)/2 cosθ).So, the x-coordinates are:0, s cosθ, s/2 cosθ - (s√3)/2 sinθ.The y-coordinates are:0, s sinθ, s/2 sinθ + (s√3)/2 cosθ.So, the maximum x is the maximum of s cosθ and s/2 cosθ - (s√3)/2 sinθ.Similarly, the maximum y is the maximum of s sinθ and s/2 sinθ + (s√3)/2 cosθ.We need both maximum x ≤ 12 and maximum y ≤ 5.Let me denote:MaxX = max(s cosθ, s/2 cosθ - (s√3)/2 sinθ)MaxY = max(s sinθ, s/2 sinθ + (s√3)/2 cosθ)We need MaxX ≤ 12 and MaxY ≤ 5.To find the maximum 's', we need to find the maximum 's' such that there exists a θ where both MaxX ≤ 12 and MaxY ≤ 5.This seems complex, but maybe I can find a relationship between MaxX and MaxY.Alternatively, perhaps I can set up the problem to maximize 's' subject to the constraints:s cosθ ≤ 12s/2 cosθ - (s√3)/2 sinθ ≤ 12s sinθ ≤ 5s/2 sinθ + (s√3)/2 cosθ ≤ 5But this is a system of inequalities, and it's not straightforward to solve.Alternatively, maybe I can consider the case where both MaxX and MaxY are achieved at their respective maximums, i.e.,s cosθ = 12s sinθ + (s√3)/2 cosθ = 5This would give us two equations to solve for 's' and θ.Let me write them down:1. s cosθ = 122. s sinθ + (s√3)/2 cosθ = 5From equation 1: cosθ = 12/s.From equation 2: s sinθ + (s√3)/2 * (12/s) = 5 ⇒ s sinθ + (12√3)/2 = 5 ⇒ s sinθ + 6√3 = 5 ⇒ s sinθ = 5 - 6√3.But sinθ must be between -1 and 1, and s is positive. However, 5 - 6√3 ≈ 5 - 10.392 ≈ -5.392, which is negative. But sinθ can't be negative if θ is between 0 and 60 degrees, as the triangle is above the x-axis.Wait, maybe θ is negative? If θ is negative, then sinθ is negative, which could make s sinθ negative, but then the y-coordinate of point B would be negative, which is outside the rectangle since y must be ≥ 0.Therefore, this approach might not work because it leads to a negative y-coordinate.Alternatively, maybe the maximum y is achieved at point C, so:s/2 sinθ + (s√3)/2 cosθ = 5And the maximum x is achieved at point B: s cosθ = 12So, from equation 1: cosθ = 12/s.From equation 2: s/2 sinθ + (s√3)/2 * (12/s) = 5 ⇒ (s/2) sinθ + (12√3)/2 = 5 ⇒ (s/2) sinθ + 6√3 = 5 ⇒ (s/2) sinθ = 5 - 6√3 ≈ -5.392.Again, this leads to sinθ being negative, which would place point B below the x-axis, which is outside the rectangle.Therefore, this approach doesn't work.Alternatively, maybe the maximum x is achieved at point C: s/2 cosθ - (s√3)/2 sinθ = 12And the maximum y is achieved at point C: s/2 sinθ + (s√3)/2 cosθ = 5So, we have:1. s/2 cosθ - (s√3)/2 sinθ = 122. s/2 sinθ + (s√3)/2 cosθ = 5Let me write these as:(1) (s/2) cosθ - (s√3/2) sinθ = 12(2) (s/2) sinθ + (s√3/2) cosθ = 5Let me denote A = s/2 and B = s√3/2.Then, equations become:(1) A cosθ - B sinθ = 12(2) A sinθ + B cosθ = 5This is a system of linear equations in terms of cosθ and sinθ.Let me write it as:A cosθ - B sinθ = 12B cosθ + A sinθ = 5Let me solve for cosθ and sinθ.Let me denote x = cosθ and y = sinθ.Then, we have:A x - B y = 12B x + A y = 5This is a system of linear equations:A x - B y = 12B x + A y = 5We can solve this using substitution or matrix methods.Let me write it in matrix form:[ A -B ] [x] = [12][ B A ] [y] [5]The determinant of the coefficient matrix is A² + B².Since A = s/2 and B = s√3/2, A² + B² = (s²/4) + (3s²/4) = s².So, determinant = s².The solution is:x = (12*A + 5*B) / (A² + B²) = (12*(s/2) + 5*(s√3/2)) / s² = (6s + (5s√3)/2) / s² = (12s + 5s√3)/ (2s²) = (12 + 5√3)/(2s)Similarly,y = (5*A - 12*B) / (A² + B²) = (5*(s/2) - 12*(s√3/2)) / s² = (5s/2 - 6s√3) / s² = (5 - 12√3)/(2s)But we know that x² + y² = cos²θ + sin²θ = 1.So,[(12 + 5√3)/(2s)]² + [(5 - 12√3)/(2s)]² = 1Let me compute this:First term: [(12 + 5√3)/(2s)]² = (144 + 120√3 + 75)/(4s²) = (219 + 120√3)/(4s²)Second term: [(5 - 12√3)/(2s)]² = (25 - 120√3 + 432)/(4s²) = (457 - 120√3)/(4s²)Adding them together:(219 + 120√3 + 457 - 120√3)/(4s²) = (676)/(4s²) = 169/(s²)Set equal to 1:169/(s²) = 1 ⇒ s² = 169 ⇒ s = 13Wait, s = 13? But earlier, I saw that a triangle with side length 13 would have a height of (√3/2)*13 ≈ 11.258, which is larger than the rectangle's shorter side of 5. So, that can't be.But according to this calculation, s = 13. That seems contradictory.Wait, perhaps I made a mistake in the calculation.Let me recompute the squares.First term: [(12 + 5√3)/(2s)]²= (12 + 5√3)² / (4s²)= (144 + 120√3 + 75) / (4s²)= (219 + 120√3) / (4s²)Second term: [(5 - 12√3)/(2s)]²= (5 - 12√3)² / (4s²)= (25 - 120√3 + 432) / (4s²)= (457 - 120√3) / (4s²)Adding them:(219 + 120√3 + 457 - 120√3) / (4s²) = (676) / (4s²) = 169 / s²Set equal to 1:169 / s² = 1 ⇒ s² = 169 ⇒ s = 13.Hmm, so according to this, s = 13. But as I thought earlier, a triangle with side length 13 would have a height of (√3/2)*13 ≈ 11.258, which is larger than 5, so it can't fit inside the rectangle.This suggests that there's a mistake in my approach.Wait, maybe the assumption that both MaxX and MaxY are achieved at point C is incorrect. Because when I set up the equations, I assumed that both maximums are achieved at point C, but in reality, the maximum x and y might be achieved at different points.Alternatively, perhaps the triangle is placed such that point C is at (12,5), but that might not necessarily be the case.Wait, let me think differently. Maybe the maximum area occurs when the triangle is inscribed in the rectangle such that each vertex touches a different side.Let me consider the triangle with vertices on all four sides of the rectangle. Wait, but a triangle has only three vertices, so it can touch at most three sides.Wait, perhaps two vertices on one side and one on another, but that's not necessarily the case.Alternatively, maybe the triangle is placed such that each vertex touches a different side, but since the rectangle has four sides, one side will have no vertex.Wait, perhaps the triangle is placed such that one vertex is on the bottom side, one on the right side, and one on the top side.Let me try to model this.Let me denote the rectangle with vertices at (0,0), (12,0), (12,5), (0,5).Let the triangle have vertices at (a,0), (12,b), and (c,5), where 0 < a < 12, 0 < b < 5, and 0 < c < 12.Since it's an equilateral triangle, all sides must be equal.So, the distance between (a,0) and (12,b) must equal the distance between (12,b) and (c,5), and also equal the distance between (c,5) and (a,0).This gives us the following equations:√[(12 - a)² + b²] = √[(c - 12)² + (5 - b)²] = √[(a - c)² + 25]This is a system of equations with variables a, b, c.It's quite complex, but maybe I can find a relationship between them.Let me denote the common side length as s.So,(12 - a)² + b² = s² ...(1)(c - 12)² + (5 - b)² = s² ...(2)(a - c)² + 25 = s² ...(3)From equations (1) and (2):(12 - a)² + b² = (c - 12)² + (5 - b)²Expanding both sides:(144 - 24a + a²) + b² = (c² - 24c + 144) + (25 - 10b + b²)Simplify:144 - 24a + a² + b² = c² - 24c + 144 + 25 - 10b + b²Cancel b² and 144 from both sides:-24a + a² = c² - 24c + 25 - 10bRearranging:a² - 24a = c² - 24c + 25 - 10b ...(4)From equations (1) and (3):(12 - a)² + b² = (a - c)² + 25Expanding:144 - 24a + a² + b² = a² - 2ac + c² + 25Simplify:144 - 24a + b² = -2ac + c² + 25Rearranging:-24a + b² = -2ac + c² - 119 ...(5)Now, we have equations (4) and (5):From (4): a² - 24a = c² - 24c + 25 - 10bFrom (5): -24a + b² = -2ac + c² - 119This is getting quite involved. Maybe I can express 'b' from equation (4) and substitute into equation (5).From equation (4):a² - 24a - c² + 24c - 25 = -10bSo,b = [ -a² + 24a + c² - 24c + 25 ] / 10Now, substitute this into equation (5):-24a + [ (-a² + 24a + c² - 24c + 25 ) / 10 ]² = -2ac + c² - 119This is a very complicated equation, and I don't think it's practical to solve this without more advanced methods or computational tools.Alternatively, maybe I can assume some symmetry or make an educated guess about the values of 'a', 'b', and 'c'.Wait, perhaps the triangle is placed such that the points (a,0), (12,b), and (c,5) form an equilateral triangle with certain symmetries.Alternatively, maybe I can consider that the triangle is placed such that the point (c,5) is the reflection of (a,0) over the center of the rectangle, but I'm not sure.Alternatively, maybe I can use some geometric transformations or complex numbers to model this, but that might be beyond my current knowledge.Given the complexity of this approach, I think it's safe to stick with the earlier result where placing the triangle with its base along the longer side of the rectangle gives the maximum area of (25√3)/3.Therefore, the maximum possible area of such a triangle is (25√3)/3.

💡First, I need to determine the amount of yellow paint in each of the two paints being mixed. The light green paint is 20% yellow, and I have 5 gallons of it. So, the amount of yellow paint in the light green paint is 0.20 multiplied by 5 gallons.Next, the darker green paint has an unknown percentage of yellow paint, which I'll denote as ( x ). I have 1.66666666667 gallons of this darker green paint, so the amount of yellow paint in it is ( x ) multiplied by 1.66666666667 gallons.When these two paints are mixed together, the total volume of the mixture is the sum of the two volumes, which is ( 5 + 1.66666666667 ) gallons. The desired concentration of yellow paint in the mixture is 25%, so the total amount of yellow paint in the mixture should be 0.25 multiplied by the total volume.Setting up the equation, the sum of the yellow paint from both paints equals the total yellow paint in the mixture:[0.20 times 5 + x times 1.66666666667 = 0.25 times (5 + 1.66666666667)]Simplifying the equation:[1 + x times 1.66666666667 = 1.66666666667]Subtracting 1 from both sides:[x times 1.66666666667 = 0.66666666667]Finally, solving for ( x ) by dividing both sides by 1.66666666667:[x = frac{0.66666666667}{1.66666666667} approx 0.4]Converting this decimal back to a percentage gives approximately 40%.