💡Okay, so I need to figure out the percentage error when using the Egyptian method to calculate the area of an isosceles trapezoid. The example given has a lower base of 6 units, an upper base of 4 units, and a lateral side of 20 units. First, I remember that the standard formula for the area of a trapezoid is the average of the two bases multiplied by the height. So, the formula is:[ text{Area} = frac{(a + b)}{2} times h ]where ( a ) and ( b ) are the lengths of the two bases, and ( h ) is the height. But the Egyptian method uses the product of the half-sum of the bases and the lateral side. So, their formula would be:[ text{Area}_{text{Egyptian}} = frac{(a + b)}{2} times c ]where ( c ) is the lateral side. Given the values:- Lower base (( a )) = 6- Upper base (( b )) = 4- Lateral side (( c )) = 20So, plugging these into the Egyptian formula:[ text{Area}_{text{Egyptian}} = frac{(6 + 4)}{2} times 20 = frac{10}{2} times 20 = 5 times 20 = 100 ]Okay, so according to the Egyptian method, the area is 100 square units.Now, I need to find the actual area using the standard formula to compare and find the percentage error. For that, I need the height (( h )) of the trapezoid. Since it's an isosceles trapezoid, the legs are equal, and the height can be found using the Pythagorean theorem. The difference between the bases is ( 6 - 4 = 2 ) units. This difference is split equally on both sides, so each side extends beyond the upper base by ( 1 ) unit.So, each lateral side (which is 20 units) forms a right triangle with height ( h ) and base ( 1 ) unit. Therefore:[ h = sqrt{c^2 - left(frac{a - b}{2}right)^2} = sqrt{20^2 - 1^2} = sqrt{400 - 1} = sqrt{399} ]Calculating ( sqrt{399} ) gives approximately 19.975, but I'll keep it as ( sqrt{399} ) for exactness.Now, using the standard formula:[ text{Area}_{text{actual}} = frac{(6 + 4)}{2} times sqrt{399} = 5 times sqrt{399} ]So, the actual area is ( 5sqrt{399} ).To find the percentage error, I need to compare the Egyptian area to the actual area. The formula for percentage error is:[ text{Percentage Error} = left| frac{text{Approximate Value} - text{Exact Value}}{text{Exact Value}} right| times 100% ]Plugging in the values:[ text{Percentage Error} = left| frac{100 - 5sqrt{399}}{5sqrt{399}} right| times 100% ]Simplifying the numerator:[ 100 - 5sqrt{399} = 5 times (20 - sqrt{399}) ]So,[ text{Percentage Error} = left| frac{5 times (20 - sqrt{399})}{5sqrt{399}} right| times 100% = left| frac{20 - sqrt{399}}{sqrt{399}} right| times 100% ]This simplifies to:[ text{Percentage Error} = left| frac{20}{sqrt{399}} - 1 right| times 100% ]Calculating ( frac{20}{sqrt{399}} ):Since ( sqrt{399} approx 19.975 ), then:[ frac{20}{19.975} approx 1.00115 ]So,[ text{Percentage Error} approx left| 1.00115 - 1 right| times 100% = 0.115% ]Wait, that seems really small. Did I do that right? Let me double-check.I have:[ text{Percentage Error} = left| frac{20}{sqrt{399}} - 1 right| times 100% ]If ( sqrt{399} approx 19.975 ), then:[ frac{20}{19.975} approx 1.00115 ]So,[ 1.00115 - 1 = 0.00115 ]Then,[ 0.00115 times 100% = 0.115% ]Hmm, that does seem correct. The percentage error is approximately 0.115%.But wait, intuitively, using the lateral side instead of the height might lead to a larger error. Maybe I made a mistake in interpreting the Egyptian method.Let me revisit the Egyptian method. They used the lateral side as if it were the height. But in reality, the lateral side is the slant height, not the vertical height. So, by using the lateral side instead of the actual height, they are overestimating the area.But according to my calculations, the error is only about 0.115%. That seems surprisingly low. Maybe I should check the exact value without approximating.Let's compute ( frac{20}{sqrt{399}} ) more precisely.First, compute ( sqrt{399} ):[ sqrt{399} approx 19.974984355 ]So,[ frac{20}{19.974984355} approx 1.001152263 ]Thus,[ 1.001152263 - 1 = 0.001152263 ]Multiply by 100%:[ 0.001152263 times 100% approx 0.1152263% ]So, approximately 0.115%.Wait, that's less than 0.12%. That seems correct mathematically, but I thought the error would be larger. Maybe because the lateral side is much larger than the difference in the bases, the error is minimal.Alternatively, perhaps I should express the percentage error in terms of exact values without decimal approximation.So, going back:[ text{Percentage Error} = left| frac{20}{sqrt{399}} - 1 right| times 100% ]This can be written as:[ left( frac{20}{sqrt{399}} - 1 right) times 100% ]Since ( frac{20}{sqrt{399}} > 1 ), the absolute value is just the positive difference.Alternatively, rationalizing the denominator:[ frac{20}{sqrt{399}} = frac{20sqrt{399}}{399} ]So,[ text{Percentage Error} = left( frac{20sqrt{399}}{399} - 1 right) times 100% ]But this doesn't simplify much further. So, the exact percentage error is:[ left( frac{20sqrt{399} - 399}{399} right) times 100% ]But this is more complicated. I think the approximate decimal is more useful here.So, to summarize:- Egyptian area: 100- Actual area: ( 5sqrt{399} approx 5 times 19.975 approx 99.875 )- Difference: ( 100 - 99.875 = 0.125 )- Percentage error: ( frac{0.125}{99.875} times 100% approx 0.125% )Wait, that's slightly different from my earlier calculation. Maybe because I approximated ( sqrt{399} ) as 19.975, but actually, ( 5sqrt{399} ) is approximately 5 times that, which is 99.8749.So, the difference is ( 100 - 99.8749 = 0.1251 ).Thus,[ text{Percentage Error} = frac{0.1251}{99.8749} times 100% approx 0.1252% ]So, approximately 0.125%.Therefore, the percentage error is about 0.125%.But to express it more precisely, using exact values:[ text{Percentage Error} = left( frac{20}{sqrt{399}} - 1 right) times 100% ]Which is approximately 0.125%.I think that's the correct percentage error.

💡Okay, so I have this problem where I need to figure out the maximum number of dominoes I can place on a 2n x 2n grid. The condition is that for each square on the grid, there must be exactly one adjacent square covered by a domino. That sounds a bit tricky, but let me try to break it down.First, I know that a domino covers two squares, either horizontally or vertically. So, each domino will cover two adjacent squares. The grid is 2n x 2n, which means it's an even-sized grid, which is good because dominoes fit neatly on even-sized grids.Now, the key condition here is that for every square, there must be exactly one adjacent square covered by a domino. That means each square can't have zero or two adjacent dominoes; it has to have exactly one. So, if I imagine the grid, each square is like a node in a graph, and each domino is an edge connecting two nodes. The condition is essentially saying that each node must have exactly one edge connected to it. In graph theory terms, this is a 1-regular graph, which is a matching where every node has degree 1.Wait, so if each square must have exactly one adjacent domino, that means the dominoes must form a perfect matching on the grid. But a perfect matching would mean that every square is covered by a domino, right? But dominoes cover two squares, so a perfect matching would mean the entire grid is covered by dominoes without any overlaps or gaps. But the problem says "several dominoes," not necessarily covering the entire grid. Hmm, maybe I'm misunderstanding something.Let me read the problem again: "for each square in the grid, there is exactly one adjacent square covered by a domino." So, each square must have exactly one neighbor that's part of a domino. That doesn't necessarily mean the domino is covering that square, just that one of its neighbors is covered by a domino. So, it's not a perfect matching; it's more like each square has exactly one neighbor that's part of a domino.Okay, that makes more sense. So, each square must have exactly one adjacent square that's part of a domino. So, dominoes are placed such that every square has exactly one neighbor covered by a domino. That seems like a different condition. So, it's not about covering the entire grid, but about covering dominoes in such a way that every square has exactly one neighbor covered.So, how does that affect the number of dominoes? Each domino covers two squares, and each domino will affect the adjacent squares. Each domino placed will cover two squares, and each of those squares will have their adjacent squares affected.Let me think about a small grid first, like a 2x2 grid. If n=1, so 2x2 grid. How many dominoes can I place? Well, in a 2x2 grid, I can place two dominoes either horizontally or vertically. But let's check the condition. If I place two dominoes horizontally, covering the top two squares and the bottom two squares, then each square in the top row has a neighbor covered by a domino (the square below it), and each square in the bottom row has a neighbor covered by a domino (the square above it). Similarly, if I place them vertically, each square has a neighbor covered by a domino. So, in this case, the maximum number of dominoes is 2.Wait, but according to the initial thought, if it's a perfect matching, the entire grid is covered, which is 2 dominoes for 2x2. But the problem says "several dominoes," not necessarily covering the entire grid. But in this case, covering the entire grid satisfies the condition because each square has exactly one neighbor covered. So, maybe for n=1, the maximum is 2.But let's check n=2, which is a 4x4 grid. How many dominoes can I place? If I try to cover the entire grid, that would be 8 dominoes, but let me check the condition. Each square would have two neighbors covered by dominoes if it's in the middle, which violates the condition. So, covering the entire grid doesn't work because some squares would have more than one neighbor covered.So, I need to place dominoes in such a way that each square has exactly one neighbor covered. That sounds like a kind of independent set or something. Maybe it's similar to a checkerboard pattern where dominoes are placed in a way that they don't overlap and each square has only one neighbor covered.Let me try to visualize a 4x4 grid. If I place dominoes horizontally on the first two rows, covering rows 1 and 2, columns 1-2 and 3-4. Then, each square in rows 1 and 2 has a neighbor covered, but what about rows 3 and 4? They don't have any neighbors covered yet. So, I need to place dominoes in rows 3 and 4 as well. But if I place dominoes vertically in columns 1 and 2, covering rows 3-4, columns 1 and 2, then each square in columns 1 and 2 has a neighbor covered. Similarly, I can place dominoes vertically in columns 3 and 4 for rows 3-4.Wait, but then each square in rows 3 and 4 would have two neighbors covered if they are adjacent to both horizontal and vertical dominoes. Hmm, that's a problem. So, maybe I need a different arrangement.Alternatively, maybe I can place dominoes in a staggered pattern. For example, place a horizontal domino on the first row, columns 1-2, then a vertical domino on columns 3-4, rows 1-2. Then, on the second row, place a vertical domino on columns 1-2, rows 2-3, and a horizontal domino on columns 3-4, row 3. Wait, this might get complicated.Alternatively, maybe I can think of the grid as a graph where each square is a node, and edges connect adjacent squares. Then, the condition is that each node has exactly one neighbor that is part of a domino. So, each domino is an edge, and each node has exactly one edge incident to it. That sounds like a 1-factor or a perfect matching, but in this case, it's not a perfect matching because dominoes cover two squares, but each square only needs one neighbor covered.Wait, no, a perfect matching would mean every square is covered by a domino, but here, each square only needs one neighbor covered. So, it's a different kind of graph. It's like a directed graph where each node has out-degree 1, but in this case, it's undirected. So, each node has exactly one edge connected to it, which is a 1-regular graph, which is a matching where every node is matched.But in this case, the dominoes are edges, and each domino covers two nodes, so the total number of dominoes would be half the number of nodes, but each node only needs one edge. Wait, that doesn't make sense because each domino covers two nodes, so if each node has one edge, the total number of edges (dominoes) would be half the number of nodes.Wait, the grid has (2n)^2 = 4n^2 squares. If each square has exactly one neighbor covered by a domino, then the total number of "covered neighbors" is 4n^2. But each domino covers two squares, and each domino contributes two "covered neighbors" (one for each square it covers). So, the total number of dominoes would be 4n^2 / 2 = 2n^2. But that can't be right because the maximum number of dominoes on a 2n x 2n grid is (2n x 2n)/2 = 2n^2. So, is the maximum number of dominoes 2n^2?But wait, in the 2x2 grid, that would mean 2 dominoes, which is correct. In the 4x4 grid, that would mean 8 dominoes, but earlier I thought that covering the entire grid would cause some squares to have more than one neighbor covered. So, maybe I was wrong about that.Wait, if I cover the entire grid with dominoes, each square is covered by a domino, so each square has one neighbor covered (the one it's paired with). But actually, no, because each square is part of a domino, so it's covered, but the condition is about having exactly one adjacent square covered by a domino. So, if a square is covered by a domino, it's not necessarily that its neighbor is covered, but that one of its neighbors is covered.Wait, no, the condition is that for each square, there is exactly one adjacent square covered by a domino. So, if a square is covered by a domino, it's not necessarily that its neighbor is covered, but that one of its neighbors is covered. So, if I cover the entire grid, each square is part of a domino, so each square has one neighbor that is also part of a domino. So, in that case, each square would have exactly one neighbor covered by a domino, which satisfies the condition.Wait, but in that case, the maximum number of dominoes would be 2n^2, which is the entire grid covered. But earlier, I thought that in the 4x4 grid, covering the entire grid would cause some squares to have more than one neighbor covered. Maybe I was mistaken.Let me think again. If I cover the entire grid with dominoes, each square is part of a domino, so each square has exactly one neighbor that is part of the same domino. So, each square has exactly one adjacent square covered by a domino, which is the one it's paired with. So, in that case, the condition is satisfied.But wait, in the 4x4 grid, if I place dominoes horizontally on the first two rows, covering rows 1 and 2, columns 1-2 and 3-4, then each square in rows 1 and 2 has a neighbor covered (the one in the same domino). Similarly, if I place dominoes vertically on columns 1 and 2, rows 3-4, each square in columns 1 and 2 has a neighbor covered. But then, the squares in rows 3 and 4, columns 3 and 4, would not have any neighbors covered. So, that's a problem.Wait, so if I only cover part of the grid, some squares won't have any neighbors covered. So, to satisfy the condition, I need to cover the entire grid. But if I cover the entire grid, each square is part of a domino, so each square has exactly one neighbor covered. So, that seems to satisfy the condition.But earlier, I thought that in the 4x4 grid, covering the entire grid would cause some squares to have more than one neighbor covered. Maybe I was wrong. Let me visualize it.If I have a 4x4 grid and I place dominoes horizontally on the first two rows, covering rows 1 and 2, columns 1-2 and 3-4. Then, each square in rows 1 and 2 has a neighbor covered. Now, for rows 3 and 4, I need to place dominoes such that each square has exactly one neighbor covered. If I place dominoes vertically on columns 1 and 2, rows 3-4, then each square in columns 1 and 2 has a neighbor covered. Similarly, I can place dominoes vertically on columns 3 and 4, rows 3-4. So, in total, I have 8 dominoes, covering the entire grid.But wait, in this case, each square is part of a domino, so each square has exactly one neighbor covered. So, the condition is satisfied. So, maybe the maximum number of dominoes is indeed 2n^2.But wait, the problem says "several dominoes," which might imply that not necessarily the entire grid is covered. But in this case, covering the entire grid satisfies the condition, so it's allowed.Wait, but in the 2x2 grid, covering the entire grid with 2 dominoes satisfies the condition, and that's the maximum. For the 4x4 grid, covering the entire grid with 8 dominoes satisfies the condition, and that's the maximum. So, maybe for any 2n x 2n grid, the maximum number of dominoes is 2n^2.But wait, the problem says "for each square in the grid, there is exactly one adjacent square covered by a domino." So, if a square is covered by a domino, it's not necessarily that its neighbor is covered, but that one of its neighbors is covered. So, if I cover the entire grid, each square is part of a domino, so each square has exactly one neighbor that is part of the same domino. So, that satisfies the condition.But wait, in that case, the maximum number of dominoes is 2n^2, which is the entire grid. But the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But maybe it's allowed.Wait, but let me think again. If I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that satisfies the condition. So, the maximum number of dominoes is 2n^2.But wait, in the 2x2 grid, 2 dominoes cover the entire grid, and that's the maximum. For 4x4, 8 dominoes cover the entire grid, and that's the maximum. So, in general, for a 2n x 2n grid, the maximum number of dominoes is 2n^2.But wait, the problem says "for each square in the grid, there is exactly one adjacent square covered by a domino." So, if a square is covered by a domino, it's not necessarily that its neighbor is covered, but that one of its neighbors is covered. So, if I cover the entire grid, each square is part of a domino, so each square has exactly one neighbor that is part of the same domino. So, that satisfies the condition.But wait, in that case, the maximum number of dominoes is 2n^2, which is the entire grid. But the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But maybe it's allowed.Wait, but let me think again. If I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that satisfies the condition. So, the maximum number of dominoes is 2n^2.But wait, the problem might be interpreted differently. Maybe it's saying that for each square, there is exactly one adjacent square that is covered by a domino, regardless of whether the square itself is covered or not. So, if a square is not covered by a domino, it must have exactly one neighbor that is covered. If a square is covered by a domino, it's part of a domino, so it's covered, but the condition is about having exactly one neighbor covered, which would be the one it's paired with.Wait, no, if a square is covered by a domino, it's part of a domino, so it's covered, but the condition is about having exactly one neighbor covered. So, if a square is covered by a domino, it's part of a domino, so it's covered, but the condition is about having exactly one neighbor covered. So, if a square is covered by a domino, it's part of a domino, so it's covered, but the condition is about having exactly one neighbor covered.Wait, I'm getting confused. Let me rephrase. The condition is: for each square in the grid, there is exactly one adjacent square covered by a domino. So, regardless of whether the square itself is covered by a domino or not, it must have exactly one adjacent square that is covered by a domino.So, if a square is covered by a domino, it's part of a domino, so it's covered, but it still needs to have exactly one adjacent square covered by a domino. So, if a square is covered by a domino, it's part of a domino, so it's covered, but it still needs to have exactly one neighbor covered by a domino. So, the domino it's part of covers two squares, so each of those two squares has one neighbor covered by the domino. So, that satisfies the condition for those two squares.But for the other squares, which are not part of any domino, they must have exactly one neighbor covered by a domino. So, if I cover the entire grid, then all squares are part of dominoes, so each square has exactly one neighbor covered by a domino. So, that satisfies the condition.But if I don't cover the entire grid, then some squares are not part of any domino, and those squares must have exactly one neighbor covered by a domino. So, in that case, the number of dominoes would be less than 2n^2.But the problem is asking for the maximum number of dominoes that can be placed according to the requirement. So, if covering the entire grid satisfies the condition, then the maximum number of dominoes is 2n^2.But wait, in the 4x4 grid, if I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that works. So, maybe the maximum number of dominoes is 2n^2.But wait, let me think about a 6x6 grid. If I cover the entire grid with dominoes, each square has exactly one neighbor covered by a domino. So, that satisfies the condition. So, the maximum number of dominoes is 18.But wait, maybe there's a way to place more dominoes without covering the entire grid, but I don't think so because each domino covers two squares, and the grid has 36 squares, so 18 dominoes would cover the entire grid.Wait, but the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm not sure. Maybe I'm misunderstanding the condition. Let me read it again: "for each square in the grid, there is exactly one adjacent square covered by a domino." So, each square must have exactly one neighbor that is covered by a domino. So, if a square is covered by a domino, it's part of a domino, so it's covered, but it still needs to have exactly one neighbor covered by a domino. So, the domino it's part of covers two squares, so each of those two squares has one neighbor covered by the domino. So, that satisfies the condition for those two squares.But for the other squares, which are not part of any domino, they must have exactly one neighbor covered by a domino. So, if I cover the entire grid, then all squares are part of dominoes, so each square has exactly one neighbor covered by a domino. So, that satisfies the condition.But if I don't cover the entire grid, then some squares are not part of any domino, and those squares must have exactly one neighbor covered by a domino. So, in that case, the number of dominoes would be less than 2n^2.But the problem is asking for the maximum number of dominoes that can be placed according to the requirement. So, if covering the entire grid satisfies the condition, then the maximum number of dominoes is 2n^2.But wait, in the 4x4 grid, if I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that works. So, maybe the maximum number of dominoes is 2n^2.But wait, let me think about a 6x6 grid. If I cover the entire grid with dominoes, each square has exactly one neighbor covered by a domino. So, that satisfies the condition. So, the maximum number of dominoes is 18.But wait, maybe there's a way to place more dominoes without covering the entire grid, but I don't think so because each domino covers two squares, and the grid has 36 squares, so 18 dominoes would cover the entire grid.Wait, but the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm not sure. Maybe I'm misunderstanding the condition. Let me try to think differently.Suppose I have a grid where each square has exactly one neighbor covered by a domino. So, each square is either covered by a domino or has exactly one neighbor covered by a domino. So, the dominoes are placed in such a way that they form a kind of independent set where each domino is isolated in terms of coverage.Wait, no, because dominoes can be adjacent as long as each square only has one neighbor covered.Wait, maybe it's similar to a permutation matrix where each row and column has exactly one 1, but in this case, each square has exactly one neighbor covered.Wait, maybe it's a kind of graph where each node has exactly one edge, which is a 1-regular graph, which is a matching. So, the dominoes form a matching where each domino is an edge, and each node has exactly one edge.But in that case, the number of dominoes would be half the number of nodes, which is (2n x 2n)/2 = 2n^2. So, that would mean the maximum number of dominoes is 2n^2, which is the entire grid.But wait, in a 1-regular graph, each node has exactly one edge, so the number of edges is half the number of nodes. So, for a 2n x 2n grid, which has 4n^2 nodes, the number of edges (dominoes) would be 2n^2. So, that matches.But in that case, the entire grid is covered by dominoes, and each square has exactly one neighbor covered by a domino. So, that satisfies the condition.But wait, in the 4x4 grid, if I place dominoes in a way that each square has exactly one neighbor covered, but not necessarily covering the entire grid, can I place more dominoes than 8?Wait, no, because 8 dominoes cover the entire grid, and each domino covers two squares. So, 8 dominoes cover 16 squares, which is the entire 4x4 grid. So, you can't place more than 8 dominoes on a 4x4 grid without overlapping.So, in general, for a 2n x 2n grid, the maximum number of dominoes is 2n^2, which covers the entire grid, and each square has exactly one neighbor covered by a domino.But wait, the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm still a bit confused because in the 4x4 grid, if I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that works.Wait, but let me think about a different arrangement. Suppose I place dominoes in a checkerboard pattern, where each domino is placed on a black square and its adjacent white square. Then, each square would have exactly one neighbor covered by a domino. But in that case, the number of dominoes would be half the number of squares, which is 2n^2, same as covering the entire grid.So, maybe regardless of the arrangement, the maximum number of dominoes is 2n^2.But wait, in the 2x2 grid, 2 dominoes cover the entire grid, which is 2 dominoes. For 4x4, 8 dominoes cover the entire grid. So, in general, for 2n x 2n, it's 2n^2 dominoes.But wait, the problem says "for each square in the grid, there is exactly one adjacent square covered by a domino." So, if I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that satisfies the condition.But if I don't cover the entire grid, then some squares won't have any neighbors covered, which violates the condition. So, to satisfy the condition, I must cover the entire grid, which means the maximum number of dominoes is 2n^2.But wait, the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm still not sure. Maybe I'm missing something. Let me think about the domino placement.If I place a domino horizontally on the first row, covering squares (1,1) and (1,2). Then, square (1,1) has a neighbor covered by the domino, which is (1,2). Similarly, square (1,2) has a neighbor covered by the domino, which is (1,1). Now, square (2,1) is adjacent to (1,1), which is covered by a domino, so it has one neighbor covered. Similarly, square (2,2) is adjacent to (1,2), which is covered by a domino, so it has one neighbor covered.But wait, if I only place that one domino, then squares (1,3), (1,4), (2,3), (2,4), etc., don't have any neighbors covered. So, to satisfy the condition, I need to place dominoes such that every square has exactly one neighbor covered.So, in the 2x2 grid, placing two dominoes covers the entire grid, and each square has exactly one neighbor covered. So, that works.In the 4x4 grid, if I place dominoes in a way that each square has exactly one neighbor covered, I can do it by covering the entire grid with dominoes. So, 8 dominoes.But wait, maybe there's a way to place dominoes without covering the entire grid and still satisfy the condition. For example, placing dominoes in a pattern where each domino is isolated, but that might not cover all squares.Wait, no, because if I don't cover all squares, some squares won't have any neighbors covered, which violates the condition. So, to satisfy the condition, I must cover the entire grid, which means the maximum number of dominoes is 2n^2.But wait, the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm still not sure. Maybe I'm overcomplicating it. Let me think about the domino placement again.Each domino covers two squares, and each square must have exactly one neighbor covered by a domino. So, if I place a domino on two squares, those two squares each have one neighbor covered. For the other squares, they must have exactly one neighbor covered by a domino.So, if I place dominoes in such a way that each domino is isolated, meaning no two dominoes are adjacent, then each square covered by a domino has one neighbor covered, and each square adjacent to a domino has one neighbor covered. But in that case, the number of dominoes would be limited because each domino would require surrounding squares to not have any other dominoes adjacent.But in that case, the number of dominoes would be much less than 2n^2. So, maybe the maximum number of dominoes is actually less than 2n^2.Wait, but if I can cover the entire grid with dominoes, each square has exactly one neighbor covered, which satisfies the condition. So, why would the maximum number be less than 2n^2?I think I'm getting confused because I'm mixing two different interpretations of the condition. Let me try to clarify.The condition is: for each square in the grid, there is exactly one adjacent square covered by a domino.This means that for every square, regardless of whether it's covered by a domino or not, it must have exactly one neighbor that is covered by a domino.So, if a square is covered by a domino, it's part of a domino, so it's covered, but it still needs to have exactly one neighbor covered by a domino. So, the domino it's part of covers two squares, so each of those two squares has one neighbor covered by the domino. So, that satisfies the condition for those two squares.But for the other squares, which are not part of any domino, they must have exactly one neighbor covered by a domino. So, if I cover the entire grid, then all squares are part of dominoes, so each square has exactly one neighbor covered by a domino. So, that satisfies the condition.But if I don't cover the entire grid, then some squares are not part of any domino, and those squares must have exactly one neighbor covered by a domino. So, in that case, the number of dominoes would be less than 2n^2.But the problem is asking for the maximum number of dominoes that can be placed according to the requirement. So, if covering the entire grid satisfies the condition, then the maximum number of dominoes is 2n^2.But wait, in the 4x4 grid, if I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that works. So, maybe the maximum number of dominoes is 2n^2.But wait, let me think about a different arrangement. Suppose I place dominoes in a checkerboard pattern, where each domino is placed on a black square and its adjacent white square. Then, each square would have exactly one neighbor covered by a domino. But in that case, the number of dominoes would be half the number of squares, which is 2n^2, same as covering the entire grid.So, maybe regardless of the arrangement, the maximum number of dominoes is 2n^2.But wait, the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm still a bit confused because in the 4x4 grid, if I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that works.Wait, but let me think about a different arrangement. Suppose I place dominoes in a way that each domino is isolated, meaning no two dominoes are adjacent. Then, each domino covers two squares, and each of those squares has one neighbor covered by the domino. But for the other squares, they must have exactly one neighbor covered by a domino. So, in that case, the number of dominoes would be limited because each domino would require surrounding squares to not have any other dominoes adjacent.But in that case, the number of dominoes would be much less than 2n^2. So, maybe the maximum number of dominoes is actually less than 2n^2.Wait, but if I can cover the entire grid with dominoes, each square has exactly one neighbor covered, which satisfies the condition. So, why would the maximum number be less than 2n^2?I think I'm getting confused because I'm mixing two different interpretations of the condition. Let me try to clarify.The condition is: for each square in the grid, there is exactly one adjacent square covered by a domino.This means that for every square, regardless of whether it's covered by a domino or not, it must have exactly one neighbor that is covered by a domino.So, if a square is covered by a domino, it's part of a domino, so it's covered, but it still needs to have exactly one neighbor covered by a domino. So, the domino it's part of covers two squares, so each of those two squares has one neighbor covered by the domino. So, that satisfies the condition for those two squares.But for the other squares, which are not part of any domino, they must have exactly one neighbor covered by a domino. So, if I cover the entire grid, then all squares are part of dominoes, so each square has exactly one neighbor covered by a domino. So, that satisfies the condition.But if I don't cover the entire grid, then some squares are not part of any domino, and those squares must have exactly one neighbor covered by a domino. So, in that case, the number of dominoes would be less than 2n^2.But the problem is asking for the maximum number of dominoes that can be placed according to the requirement. So, if covering the entire grid satisfies the condition, then the maximum number of dominoes is 2n^2.But wait, in the 4x4 grid, if I cover the entire grid, each square has exactly one neighbor covered by a domino, which is the one it's paired with. So, that works. So, maybe the maximum number of dominoes is 2n^2.But wait, let me think about a different arrangement. Suppose I place dominoes in a checkerboard pattern, where each domino is placed on a black square and its adjacent white square. Then, each square would have exactly one neighbor covered by a domino. But in that case, the number of dominoes would be half the number of squares, which is 2n^2, same as covering the entire grid.So, maybe regardless of the arrangement, the maximum number of dominoes is 2n^2.But wait, the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm still not sure. Maybe I'm missing something. Let me think about the domino placement.If I place a domino horizontally on the first row, covering squares (1,1) and (1,2). Then, square (1,1) has a neighbor covered by the domino, which is (1,2). Similarly, square (1,2) has a neighbor covered by the domino, which is (1,1). Now, square (2,1) is adjacent to (1,1), which is covered by a domino, so it has one neighbor covered. Similarly, square (2,2) is adjacent to (1,2), which is covered by a domino, so it has one neighbor covered.But wait, if I only place that one domino, then squares (1,3), (1,4), (2,3), (2,4), etc., don't have any neighbors covered. So, to satisfy the condition, I need to place dominoes such that every square has exactly one neighbor covered.So, in the 2x2 grid, placing two dominoes covers the entire grid, and each square has exactly one neighbor covered. So, that works.In the 4x4 grid, if I place dominoes in a way that each square has exactly one neighbor covered, I can do it by covering the entire grid with dominoes. So, 8 dominoes.But wait, maybe there's a way to place dominoes without covering the entire grid and still satisfy the condition. For example, placing dominoes in a pattern where each domino is isolated, but that might not cover all squares.Wait, no, because if I don't cover all squares, some squares won't have any neighbors covered, which violates the condition. So, to satisfy the condition, I must cover the entire grid, which means the maximum number of dominoes is 2n^2.But wait, the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm still not sure. Maybe I'm overcomplicating it. Let me think about the domino placement again.Each domino covers two squares, and each square must have exactly one neighbor covered by a domino. So, if I place a domino on two squares, those two squares each have one neighbor covered. For the other squares, they must have exactly one neighbor covered by a domino.So, if I place dominoes in such a way that each domino is isolated, meaning no two dominoes are adjacent, then each domino covers two squares, and each of those squares has one neighbor covered by the domino. But for the other squares, they must have exactly one neighbor covered by a domino. So, in that case, the number of dominoes would be limited because each domino would require surrounding squares to not have any other dominoes adjacent.But in that case, the number of dominoes would be much less than 2n^2. So, maybe the maximum number of dominoes is actually less than 2n^2.Wait, but if I can cover the entire grid with dominoes, each square has exactly one neighbor covered, which satisfies the condition. So, why would the maximum number be less than 2n^2?I think I'm getting stuck in a loop here. Let me try to approach it differently.Let me consider the grid as a graph where each square is a vertex, and edges connect adjacent squares. The condition is that each vertex has exactly one neighbor that is part of a domino. So, each vertex has exactly one edge connected to it, which is a domino.In graph theory, this is called a 1-regular graph, which is a graph where each vertex has degree 1. A 1-regular graph is a matching, where each vertex is paired with exactly one other vertex.In a 2n x 2n grid, which has 4n^2 vertices, a 1-regular graph would consist of 2n^2 edges (dominoes), each connecting two vertices. So, the maximum number of dominoes is 2n^2.Therefore, the maximum number of dominoes that can be placed on a 2n x 2n grid, such that each square has exactly one adjacent square covered by a domino, is 2n^2.But wait, in the 2x2 grid, 2 dominoes cover the entire grid, which is 2 dominoes. For 4x4, 8 dominoes cover the entire grid. So, in general, for 2n x 2n, it's 2n^2 dominoes.But the problem says "several dominoes," which might imply that it's not necessarily the entire grid. But if covering the entire grid satisfies the condition, then it's allowed, and 2n^2 is the maximum.But I'm still a bit confused because I thought that covering the entire grid would cause some squares to have more than one neighbor covered, but now I realize that's not the case. Each square is part of a domino, so it has exactly one neighbor covered by the domino it's part of.So, in conclusion, the maximum number of dominoes that can be placed on a 2n x 2n grid, such that each square has exactly one adjacent square covered by a domino, is 2n^2.But wait, let me check for n=1 and n=2.For n=1, 2x2 grid: 2 dominoes, which is 2(1)^2 = 2. Correct.For n=2, 4x4 grid: 8 dominoes, which is 2(2)^2 = 8. Correct.For n=3, 6x6 grid: 18 dominoes, which is 2(3)^2 = 18. Correct.So, the formula seems to hold.Therefore, the maximum number of dominoes is 2n^2.

💡Alright, so I have this problem about a geometric sequence. Let me try to understand what it's asking for. First, part (1) says: Given the sequence {a_n} where n is a positive integer, it's a geometric sequence with a_n > 0, a_1 = 3, and a_3 = 27. I need to find the general term formula a_n of the sequence {a_n}. Okay, so a geometric sequence is one where each term is multiplied by a common ratio to get the next term. The general form of a geometric sequence is a_n = a_1 * r^(n-1), where r is the common ratio. Given that a_1 is 3, so the first term is 3. Then, a_3 is 27. Since it's a geometric sequence, a_3 should be a_1 multiplied by r squared, because it's the third term. So, a_3 = a_1 * r^2. Plugging in the values we have: 27 = 3 * r^2. To find r, I can divide both sides by 3, which gives me 9 = r^2. Taking the square root of both sides, r = 3 or r = -3. But since all terms a_n are positive, the common ratio must be positive. So, r = 3. Therefore, the general term formula should be a_n = 3 * 3^(n-1). Simplifying that, 3^(n-1) multiplied by 3 is 3^n. So, a_n = 3^n. That seems straightforward.Now, moving on to part (2): Let b_n = 2 log base 3 of a_n + 1. I need to find the sum of the first n terms, T_n, of the sequence {b_n}. First, let's write down what b_n is. Since a_n is 3^n, then log base 3 of a_n is log base 3 of 3^n, which is just n. So, b_n = 2n + 1. So, the sequence {b_n} is 2n + 1. Let me write out the first few terms to see the pattern. When n = 1: b_1 = 2*1 + 1 = 3 When n = 2: b_2 = 2*2 + 1 = 5 When n = 3: b_3 = 2*3 + 1 = 7 When n = 4: b_4 = 2*4 + 1 = 9 ... and so on.So, the sequence {b_n} is 3, 5, 7, 9, ... which looks like an arithmetic sequence. In an arithmetic sequence, each term increases by a constant difference. Let's check the difference between consecutive terms:5 - 3 = 2 7 - 5 = 2 9 - 7 = 2 ... etc.Yes, the common difference is 2. So, {b_n} is an arithmetic sequence with the first term b_1 = 3 and common difference d = 2. To find the sum of the first n terms of an arithmetic sequence, the formula is T_n = n/2 * (2a + (n - 1)d), where a is the first term. Alternatively, it's also written as T_n = n/2 * (first term + last term). Let me use the first formula: T_n = n/2 * [2*3 + (n - 1)*2]. Simplifying inside the brackets: 2*3 is 6, and (n - 1)*2 is 2n - 2. So, 6 + 2n - 2 = 2n + 4. Therefore, T_n = n/2 * (2n + 4). Simplifying further, n/2 * 2n is n^2, and n/2 * 4 is 2n. So, T_n = n^2 + 2n. Alternatively, using the second formula: T_n = n/2 * (first term + last term). The first term is 3, and the last term when n terms are considered is b_n = 2n + 1. So, T_n = n/2 * (3 + 2n + 1) = n/2 * (2n + 4) = n^2 + 2n. Either way, I get the same result. So, the sum of the first n terms of {b_n} is n squared plus 2n.Let me just recap to make sure I didn't make any mistakes. For part (1), I used the definition of a geometric sequence, found the common ratio by using the given terms a_1 and a_3, determined that the ratio is 3, and then wrote the general term as 3^n. That seems solid.For part (2), I substituted a_n into the expression for b_n, simplified the logarithm, recognized the resulting sequence as arithmetic, and then applied the arithmetic series sum formula. The steps all make sense, and I double-checked the formula by using two different forms, both leading to the same answer. I think that's it. I don't see any errors in my reasoning.**Final Answer**(1) The general term is boxed{a_n = 3^n}.(2) The sum of the first ( n ) terms is boxed{T_n = n^2 + 2n}.

💡Alright, so I have this problem about a biased coin. The coin has a probability ( p ) of landing heads, and we're told that the probability of getting exactly 3 heads in 5 tosses is ( frac{144}{625} ). I need to figure out which of the options (A) through (E) is true.First, let me recall the formula for the probability of getting exactly ( k ) successes (in this case, heads) in ( n ) independent trials. That's the binomial probability formula:[P(X = k) = binom{n}{k} p^k (1 - p)^{n - k}]Here, ( n = 5 ) and ( k = 3 ). So plugging those values in, we get:[P(X = 3) = binom{5}{3} p^3 (1 - p)^{2}]I know that ( binom{5}{3} ) is the number of ways to choose 3 successes out of 5 trials. Calculating that:[binom{5}{3} = frac{5!}{3!(5 - 3)!} = frac{120}{6 times 2} = 10]So the equation becomes:[10 p^3 (1 - p)^2 = frac{144}{625}]Now, I need to solve for ( p ). Let me write that equation again:[10 p^3 (1 - p)^2 = frac{144}{625}]To simplify, I can divide both sides by 10:[p^3 (1 - p)^2 = frac{144}{6250}]Simplifying ( frac{144}{6250} ), I notice that both numerator and denominator are divisible by 2:[frac{144}{6250} = frac{72}{3125}]Hmm, 72 and 3125. 72 is ( 8 times 9 ) and 3125 is ( 5^5 ). So, I can write:[p^3 (1 - p)^2 = frac{8 times 9}{5^5}]Which simplifies to:[p^3 (1 - p)^2 = frac{2^3 times 3^2}{5^5}]Now, I need to find ( p ) such that this equation holds. Let me think about possible values of ( p ). The options given are fractions like ( frac{2}{5} ) and ( frac{3}{5} ). Maybe I can test these values to see if they satisfy the equation.First, let's try ( p = frac{2}{5} ):[left(frac{2}{5}right)^3 left(1 - frac{2}{5}right)^2 = left(frac{8}{125}right) left(frac{3}{5}right)^2 = left(frac{8}{125}right) left(frac{9}{25}right) = frac{72}{3125}]Which is exactly ( frac{72}{3125} ), so that works. So ( p = frac{2}{5} ) is a solution.Now, let's try ( p = frac{3}{5} ):[left(frac{3}{5}right)^3 left(1 - frac{3}{5}right)^2 = left(frac{27}{125}right) left(frac{2}{5}right)^2 = left(frac{27}{125}right) left(frac{4}{25}right) = frac{108}{3125}]Hmm, ( frac{108}{3125} ) is not equal to ( frac{72}{3125} ), so ( p = frac{3}{5} ) doesn't work.Wait, but the equation is a quintic equation, which might have multiple solutions. Maybe there's another value of ( p ) that also satisfies the equation.Let me consider the function ( f(p) = p^3 (1 - p)^2 ). I want to find all ( p ) such that ( f(p) = frac{72}{3125} ).I can analyze the behavior of ( f(p) ). Let's see:- When ( p = 0 ), ( f(0) = 0 ).- When ( p = 1 ), ( f(1) = 0 ).- At ( p = frac{1}{2} ), ( fleft(frac{1}{2}right) = left(frac{1}{2}right)^3 left(frac{1}{2}right)^2 = frac{1}{32} approx 0.03125 ).Calculating ( frac{72}{3125} approx 0.02304 ).So, ( fleft(frac{1}{2}right) approx 0.03125 ) is greater than ( frac{72}{3125} approx 0.02304 ). This suggests that the function ( f(p) ) crosses ( frac{72}{3125} ) twice: once between 0 and ( frac{1}{2} ), and once between ( frac{1}{2} ) and 1.Since ( p = frac{2}{5} = 0.4 ) is less than ( frac{1}{2} ), and we already saw that ( fleft(frac{2}{5}right) = frac{72}{3125} ), there must be another solution greater than ( frac{1}{2} ).Therefore, there are two values of ( p ) that satisfy the equation: one less than ( frac{1}{2} ) and one greater than ( frac{1}{2} ). Since ( p = frac{2}{5} ) is one solution, the other solution must be greater than ( frac{1}{2} ).Wait, but the options don't mention two solutions. Let me check the options again:(A) ( p ) must be ( frac{2}{5} ).(B) ( p ) must be ( frac{3}{5} ).(C) ( p ) must be greater than ( frac{3}{5} ).(D) ( p ) is not uniquely determined.(E) There is no value of ( p ) that makes ( w = frac{144}{625} ).So, since there are two possible values of ( p ), one is ( frac{2}{5} ) and the other is some value greater than ( frac{1}{2} ) but not necessarily greater than ( frac{3}{5} ). Therefore, ( p ) is not uniquely determined, so option (D) is correct.But wait, let me verify if the other solution is greater than ( frac{3}{5} ). Let me test ( p = frac{3}{5} ) again:We saw that ( fleft(frac{3}{5}right) = frac{108}{3125} approx 0.03456 ), which is greater than ( frac{72}{3125} approx 0.02304 ). So, since ( f(p) ) is decreasing from ( p = frac{1}{2} ) to ( p = 1 ), and ( fleft(frac{3}{5}right) > frac{72}{3125} ), the other solution must be greater than ( frac{3}{5} ).Wait, that contradicts my earlier thought. Let me think again.Wait, no. If ( fleft(frac{3}{5}right) = frac{108}{3125} approx 0.03456 ), which is greater than ( frac{72}{3125} approx 0.02304 ), and since ( f(p) ) is decreasing from ( p = frac{1}{2} ) to ( p = 1 ), then the solution greater than ( frac{1}{2} ) must be between ( frac{1}{2} ) and ( frac{3}{5} ), because ( fleft(frac{3}{5}right) ) is still greater than the desired value.Wait, no. Let me plot the function mentally. At ( p = frac{1}{2} ), ( f(p) approx 0.03125 ). At ( p = frac{3}{5} ), ( f(p) approx 0.03456 ). Wait, that can't be, because ( f(p) ) should be decreasing after the maximum point.Wait, actually, the function ( f(p) = p^3 (1 - p)^2 ) has a maximum somewhere. Let me find the maximum.To find the maximum, take the derivative of ( f(p) ) with respect to ( p ):[f'(p) = 3p^2 (1 - p)^2 + p^3 times 2(1 - p)(-1) = 3p^2 (1 - p)^2 - 2p^3 (1 - p)]Set ( f'(p) = 0 ):[3p^2 (1 - p)^2 - 2p^3 (1 - p) = 0]Factor out ( p^2 (1 - p) ):[p^2 (1 - p) [3(1 - p) - 2p] = 0]So, solutions are ( p = 0 ), ( p = 1 ), and:[3(1 - p) - 2p = 0 implies 3 - 3p - 2p = 0 implies 3 - 5p = 0 implies p = frac{3}{5}]So, the maximum occurs at ( p = frac{3}{5} ). Therefore, ( f(p) ) increases from ( p = 0 ) to ( p = frac{3}{5} ), and then decreases from ( p = frac{3}{5} ) to ( p = 1 ).Wait, that's different from what I thought earlier. So, at ( p = frac{3}{5} ), ( f(p) ) reaches its maximum. So, ( fleft(frac{3}{5}right) = frac{108}{3125} approx 0.03456 ).But our target is ( frac{72}{3125} approx 0.02304 ), which is less than the maximum. Therefore, the equation ( f(p) = frac{72}{3125} ) will have two solutions: one less than ( frac{3}{5} ) and one greater than ( frac{3}{5} ).Wait, but earlier I thought ( p = frac{2}{5} ) is a solution, which is less than ( frac{3}{5} ). So, there must be another solution greater than ( frac{3}{5} ).But wait, when ( p ) approaches 1, ( f(p) ) approaches 0. So, after ( p = frac{3}{5} ), ( f(p) ) decreases from ( frac{108}{3125} ) to 0. Therefore, since ( frac{72}{3125} ) is less than ( frac{108}{3125} ), there must be another solution greater than ( frac{3}{5} ).Therefore, there are two solutions: one at ( p = frac{2}{5} ) and another at some ( p > frac{3}{5} ). So, ( p ) is not uniquely determined, which makes option (D) correct.But wait, let me double-check if ( p = frac{2}{5} ) is indeed a solution:[fleft(frac{2}{5}right) = left(frac{2}{5}right)^3 left(frac{3}{5}right)^2 = frac{8}{125} times frac{9}{25} = frac{72}{3125}]Yes, that's correct. So, ( p = frac{2}{5} ) is a solution. And since the function ( f(p) ) has a maximum at ( p = frac{3}{5} ), and ( fleft(frac{3}{5}right) = frac{108}{3125} ), which is greater than ( frac{72}{3125} ), there must be another solution greater than ( frac{3}{5} ).Therefore, ( p ) is not uniquely determined, so the correct answer is (D).

💡First, I need to expand the product ((x + 3)(x^2 + x + 1)). To do this, I'll use the distributive property, which means I'll multiply each term in the first binomial by each term in the second trinomial.Starting with the first term in the binomial, (x), I'll distribute it across each term in the trinomial:[x cdot x^2 = x^3][x cdot x = x^2][x cdot 1 = x]Next, I'll distribute the second term in the binomial, (3), across each term in the trinomial:[3 cdot x^2 = 3x^2][3 cdot x = 3x][3 cdot 1 = 3]Now, I'll combine all these results:[x^3 + x^2 + x + 3x^2 + 3x + 3]Finally, I'll combine the like terms:[x^3 + (1x^2 + 3x^2) + (1x + 3x) + 3 = x^3 + 4x^2 + 4x + 3]

💡Okay, so I have this function g(x) defined piecewise. It's 3x + 6 when x is less than 10, and 5x - 5 when x is greater than or equal to 10. I need to find the inverse function values for 18 and 55, and then add them together. Hmm, inverses can be tricky, especially with piecewise functions, but I think I can handle it step by step.First, let me recall what an inverse function does. If g(x) maps x to some value, then g inverse would map that value back to the original x. So, g inverse of 18 is the x such that g(x) equals 18. Similarly, g inverse of 55 is the x such that g(x) equals 55. Got it.Alright, let's start with g inverse of 18. I need to find x where g(x) = 18. Since g(x) is piecewise, I have to check both cases: when x is less than 10 and when x is greater than or equal to 10.First case: x < 10. The function is 3x + 6. So, set that equal to 18:3x + 6 = 18Subtract 6 from both sides:3x = 12Divide both sides by 3:x = 4Okay, so x is 4. Now, I need to check if this x satisfies the condition x < 10. 4 is indeed less than 10, so this is a valid solution. Therefore, g inverse of 18 is 4.Now, let's check the second case just to be thorough. If x is greater than or equal to 10, the function is 5x - 5. So, set that equal to 18:5x - 5 = 18Add 5 to both sides:5x = 23Divide both sides by 5:x = 23/5, which is 4.6Wait, 4.6 is not greater than or equal to 10. So, this solution doesn't satisfy the condition for this piece of the function. Therefore, we can disregard this solution.So, for g inverse of 18, the only valid solution is 4.Alright, moving on to g inverse of 55. Again, I need to find x such that g(x) = 55. Let's check both cases.First case: x < 10. The function is 3x + 6. So, set that equal to 55:3x + 6 = 55Subtract 6 from both sides:3x = 49Divide both sides by 3:x = 49/3, which is approximately 16.333...Wait, but 16.333 is not less than 10. So, this solution doesn't satisfy the condition for this piece of the function. Therefore, this is not a valid solution.Now, let's check the second case where x is greater than or equal to 10. The function is 5x - 5. So, set that equal to 55:5x - 5 = 55Add 5 to both sides:5x = 60Divide both sides by 5:x = 1212 is indeed greater than or equal to 10, so this is a valid solution. Therefore, g inverse of 55 is 12.So, now I have both inverse values: 4 and 12. The problem asks for the sum of these two, so I just add them together:4 + 12 = 16Therefore, the final answer is 16.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For g inverse of 18:3x + 6 = 183x = 12x = 4. Correct.And 4 is less than 10, so that's good.For the other case:5x - 5 = 185x = 23x = 23/5 = 4.6, which is less than 10, so invalid. Correct.For g inverse of 55:3x + 6 = 553x = 49x = 49/3 ≈16.333, which is greater than 10, so invalid. Correct.Then, 5x - 5 = 555x = 60x = 12, which is greater than 10, so valid. Correct.Adding 4 and 12 gives 16. Yep, that seems right.I think I've covered all the steps and checked my work. I don't see any errors in my reasoning. So, I'm confident that the answer is 16.