💡Carla starts her ride 30 minutes before Derek, so she has been biking for 3.5 hours when Derek has ridden for 3 hours. Carla's speed is 12 miles per hour, so she has ridden (12 times 3.5 = 42) miles. Derek's speed is 15 miles per hour, so he has ridden (15 times 3 = 45) miles. The difference in miles ridden is (45 - 42 = 3) miles.

💡First, I recognize that the problem involves an arithmetic sequence with the first term ( a_1 = 2 ) and a common difference ( d = 3 ).To find the general term formula for the sequence, I use the standard formula for the ( n )-th term of an arithmetic sequence:[a_n = a_1 + (n - 1)d]Substituting the given values into the formula:[a_n = 2 + 3(n - 1)]Next, I simplify the expression:[a_n = 2 + 3n - 3 = 3n - 1]Finally, I compare this result with the provided options and find that it matches option A.

💡Alright, so I have this trigonometric expression to simplify: sin(x - 2y)cos(3y) + cos(x - 2y)sin(3y). Hmm, okay. Let me think about how to approach this. I remember there are some trigonometric identities that might help here, especially the sine addition formula. The sine addition formula is sin(a + b) = sin(a)cos(b) + cos(a)sin(b). Wait, that looks a lot like the expression I have here. Let me write that down to compare:sin(a + b) = sin(a)cos(b) + cos(a)sin(b)Comparing this to my expression:sin(x - 2y)cos(3y) + cos(x - 2y)sin(3y)It seems like if I let a = x - 2y and b = 3y, then my expression would fit the sine addition formula. Let me try that substitution:sin((x - 2y) + 3y) = sin(x - 2y + 3y) = sin(x + y)So, does that mean my original expression simplifies to sin(x + y)? Let me double-check to make sure I didn't make a mistake. Starting with the original expression:sin(x - 2y)cos(3y) + cos(x - 2y)sin(3y)If I set a = x - 2y and b = 3y, then according to the sine addition formula, this should equal sin(a + b). Plugging in the values:sin((x - 2y) + 3y) = sin(x - 2y + 3y) = sin(x + y)Yes, that seems correct. So, the expression simplifies to sin(x + y). I don't think I made any mistakes here. The steps make sense, and the substitution fits perfectly with the sine addition formula. Just to be thorough, let me consider if there are any other identities that might apply here. For example, could I use the cosine addition formula or maybe some double-angle identities? Let me recall:The cosine addition formula is cos(a + b) = cos(a)cos(b) - sin(a)sin(b). That doesn't seem to fit here because my expression has a plus sign between the two terms, and the cosine addition formula has a minus sign. What about double-angle identities? Well, those typically involve expressions like sin(2θ) or cos(2θ), which doesn't directly apply here. Alternatively, could I expand sin(x - 2y) and cos(x - 2y) using the sine and cosine subtraction formulas? Let me try that just to see if it leads me somewhere.The sine subtraction formula is sin(a - b) = sin(a)cos(b) - cos(a)sin(b). So, sin(x - 2y) would be sin(x)cos(2y) - cos(x)sin(2y). Similarly, the cosine subtraction formula is cos(a - b) = cos(a)cos(b) + sin(a)sin(b). So, cos(x - 2y) would be cos(x)cos(2y) + sin(x)sin(2y).Let me substitute these back into the original expression:[sin(x)cos(2y) - cos(x)sin(2y)]cos(3y) + [cos(x)cos(2y) + sin(x)sin(2y)]sin(3y)Now, let's distribute the terms:sin(x)cos(2y)cos(3y) - cos(x)sin(2y)cos(3y) + cos(x)cos(2y)sin(3y) + sin(x)sin(2y)sin(3y)Hmm, that seems more complicated. Maybe I can factor out some terms or use other identities here. Let's see.Looking at the first term: sin(x)cos(2y)cos(3y). Is there a way to combine cos(2y)cos(3y)? I recall that there's a product-to-sum identity for cosine terms:cos(A)cos(B) = [cos(A + B) + cos(A - B)] / 2Similarly, for the term -cos(x)sin(2y)cos(3y), I can use the identity for sin(A)cos(B):sin(A)cos(B) = [sin(A + B) + sin(A - B)] / 2And for the term cos(x)cos(2y)sin(3y), again, it's similar to the previous one. And the last term is sin(x)sin(2y)sin(3y), which might be tricky. Maybe I can use the identity for sin(A)sin(B):sin(A)sin(B) = [cos(A - B) - cos(A + B)] / 2But this seems like it's getting too involved. Let me see if I can proceed step by step.First, let's handle the first term: sin(x)cos(2y)cos(3y). Using the product-to-sum identity for cos(2y)cos(3y):cos(2y)cos(3y) = [cos(5y) + cos(-y)] / 2 = [cos(5y) + cos(y)] / 2Since cosine is even, cos(-y) = cos(y). So, the first term becomes:sin(x) * [cos(5y) + cos(y)] / 2 = [sin(x)cos(5y) + sin(x)cos(y)] / 2Similarly, the second term is -cos(x)sin(2y)cos(3y). Let's apply the identity for sin(2y)cos(3y):sin(2y)cos(3y) = [sin(5y) + sin(-y)] / 2 = [sin(5y) - sin(y)] / 2So, the second term becomes:-cos(x) * [sin(5y) - sin(y)] / 2 = [-cos(x)sin(5y) + cos(x)sin(y)] / 2The third term is cos(x)cos(2y)sin(3y). Again, using the identity for cos(2y)sin(3y):cos(2y)sin(3y) = [sin(5y) + sin(y)] / 2So, the third term becomes:cos(x) * [sin(5y) + sin(y)] / 2 = [cos(x)sin(5y) + cos(x)sin(y)] / 2The fourth term is sin(x)sin(2y)sin(3y). Let's use the identity for sin(2y)sin(3y):sin(2y)sin(3y) = [cos(y) - cos(5y)] / 2So, the fourth term becomes:sin(x) * [cos(y) - cos(5y)] / 2 = [sin(x)cos(y) - sin(x)cos(5y)] / 2Now, let's put all four terms together:First term: [sin(x)cos(5y) + sin(x)cos(y)] / 2Second term: [-cos(x)sin(5y) + cos(x)sin(y)] / 2Third term: [cos(x)sin(5y) + cos(x)sin(y)] / 2Fourth term: [sin(x)cos(y) - sin(x)cos(5y)] / 2Let me write all of them out:1. [sin(x)cos(5y) + sin(x)cos(y)] / 22. [-cos(x)sin(5y) + cos(x)sin(y)] / 23. [cos(x)sin(5y) + cos(x)sin(y)] / 24. [sin(x)cos(y) - sin(x)cos(5y)] / 2Now, let's combine these four terms by adding them together:Term 1 + Term 2 + Term 3 + Term 4Let's handle each part step by step.First, let's look at the sin(x)cos(5y) terms:From Term 1: + sin(x)cos(5y)/2From Term 4: - sin(x)cos(5y)/2These cancel each other out: sin(x)cos(5y)/2 - sin(x)cos(5y)/2 = 0Next, the sin(x)cos(y) terms:From Term 1: + sin(x)cos(y)/2From Term 4: + sin(x)cos(y)/2Adding these together: sin(x)cos(y)/2 + sin(x)cos(y)/2 = sin(x)cos(y)Now, the -cos(x)sin(5y) terms:From Term 2: - cos(x)sin(5y)/2From Term 3: + cos(x)sin(5y)/2These also cancel each other out: -cos(x)sin(5y)/2 + cos(x)sin(5y)/2 = 0Lastly, the cos(x)sin(y) terms:From Term 2: + cos(x)sin(y)/2From Term 3: + cos(x)sin(y)/2Adding these together: cos(x)sin(y)/2 + cos(x)sin(y)/2 = cos(x)sin(y)So, after combining all the terms, we're left with:sin(x)cos(y) + cos(x)sin(y)Wait a minute, that's another sine addition formula! Because sin(a + b) = sin(a)cos(b) + cos(a)sin(b). So, if I let a = x and b = y, then:sin(x + y) = sin(x)cos(y) + cos(x)sin(y)Which is exactly what we have here. So, sin(x)cos(y) + cos(x)sin(y) simplifies to sin(x + y).Therefore, going all the way back, the original expression simplifies to sin(x + y). But wait, that's the same result I got earlier when I directly applied the sine addition formula without expanding everything out. So, both methods lead to the same conclusion, which is reassuring.I think I've verified it thoroughly. The expression simplifies neatly to sin(x + y) using the sine addition formula. I didn't need to go through all that expansion, but it was a good exercise to confirm that even if I break it down further, I still end up with the same simplified form.Just to recap, the key step was recognizing that the given expression matches the structure of the sine addition formula, which allows us to combine the angles. By setting a = x - 2y and b = 3y, we can directly apply the formula and simplify the expression without any complicated steps.I guess sometimes it's helpful to remember these fundamental identities because they can save a lot of time and effort. Instead of expanding everything out, which can get messy, recognizing the pattern early on can lead to a quicker solution.In summary, the expression sin(x - 2y)cos(3y) + cos(x - 2y)sin(3y) simplifies to sin(x + y) by applying the sine addition formula. It's always good to double-check using different methods to ensure the result is consistent, and in this case, both approaches confirmed the same answer.

💡First, I need to determine the largest perfect square that is less than or equal to 92555. To do this, I'll start by finding the square root of 92555.Calculating the square root of 92555 gives approximately 304.22. Since this is not an integer, I'll consider the largest integer less than 304.22, which is 304.Next, I'll square 304 to find the perfect square:304 squared equals 92416.Finally, to find the number that should be subtracted from 92555 to make it a perfect square, I'll subtract 92416 from 92555:92555 minus 92416 equals 139.Therefore, the number that should be subtracted is 139.

💡Alright, so I'm trying to figure out this grid game problem. It seems like two players are drawing segments on a grid, and the rules are that each player has to continue the line from where the other left off, without retracing any segment. If a player can't move, they lose, and if the line reaches the edge, it's a draw. The goal is to prove that the second player can't lose, no matter how the game is played.Okay, let's break this down. First, the game starts with the first player drawing a segment. Then the second player continues from the end of that segment, and they alternate turns. The line can't go over the same side of a cell twice, and if it reaches the edge, it's a draw. If a player can't move on their turn, they lose.Hmm, so I need to show that the second player can never be the one who can't move. That is, the first player will always be the one to potentially lose or force a draw.Maybe I can think about this in terms of parity or something like that. Like, if the game can only end on an even number of moves or something. Wait, if the game ends when a player can't move, and the players alternate, then maybe the second player can always force the game to end on the first player's turn.Let me try to visualize this. Suppose the game starts at some internal point. The first player draws a segment, then the second player continues. Each move reduces the number of available segments from the current point. If the game is going to end, it's because all possible segments from the current point have been used.But how does that relate to who can't move? If the second player is always responding to the first player's move, maybe the second player can always mirror or respond in a way that forces the first player into a position where they can't move.Wait, maybe it's about the number of available moves. If the grid is finite, the number of segments is finite, so the game must end eventually. But who ends it? If the number of segments is even, then the second player makes the last move, right? If it's odd, the first player does.But the game doesn't necessarily have to use all segments. It can end earlier if someone can't move. So maybe the key is that the second player can always respond in a way that forces the first player to run out of moves first.Alternatively, maybe it's about the parity of the number of moves. If the game ends on an even move, the second player made the last move, so the first player loses. If it ends on an odd move, the first player made the last move, so the second player loses. But we need to show that the second player can't lose, so maybe the game can't end on an odd move.Wait, that might be the key. If the game must end on an even move, then the second player can't lose because they would have just made the last move, forcing the first player to lose.But why must the game end on an even move? Maybe because the game starts with the first player, and each pair of moves (first and second player) reduces the available segments in such a way that the game can only end after an even number of moves.Alternatively, maybe it's about the structure of the grid. If you think of the grid as a graph, each intersection is a vertex, and the segments are edges. The game is essentially building a path on this graph without repeating edges. If the path reaches the boundary, it's a draw. If it gets stuck, the player who can't move loses.In graph theory terms, this is similar to an Euler trail or Euler path, where you traverse each edge exactly once. But in this case, the game can end before all edges are used if a player gets stuck.But I'm not sure if that's directly applicable. Maybe I need to think about it differently. Let's consider small grids to see if I can spot a pattern.Suppose we have a 2x2 grid. The first player draws a segment, say horizontally. Then the second player has to continue from the end. Depending on the direction, the second player might have limited options. If the second player can always choose a direction that forces the first player to run out of moves first, then the second player can't lose.Wait, in a 2x2 grid, the maximum number of segments is 12 (each cell has 4 edges, but shared between cells). But the path can't use all of them because it would have to traverse each edge once, which might not be possible unless it's an Euler trail.But regardless, if the second player can always respond in a way that limits the first player's options, maybe the second player can always force the first player to lose.Another approach: think about the game as a series of moves where each player alternately extends the path. If the second player can always mirror the first player's moves in some way, they can ensure that the first player runs out of moves first.Wait, mirroring might not always be possible because the grid is finite and the path can get stuck in a corner. But maybe the second player can always choose a direction that doesn't lead to an immediate dead end, forcing the first player to eventually get stuck.Alternatively, maybe it's about the parity of the number of available moves from each point. If each internal point has an even number of edges, then the second player can always respond to the first player's move in such a way that the first player is left with no moves.Wait, in a grid, each internal point has four edges (up, down, left, right). So, if the second player can always respond to the first player's move by using another edge, they can ensure that the first player is left with no moves eventually.But how exactly? Let's say the first player moves right, the second player can move up, then the first player moves left, and the second player moves down, and so on. This way, the second player is always using a different direction, forcing the first player to eventually run out of moves.But I'm not sure if this strategy always works. What if the first player starts moving in a way that the second player can't mirror? Or what if the path starts to loop around, creating a closed loop?Wait, if the path forms a closed loop, then the number of segments used is even, right? Because to return to the starting point, you need an even number of moves. So, if the path forms a closed loop, the second player would have made the last move, forcing the first player to lose.But the problem says the broken line can intersect itself, but can't travel twice along the same side. So, forming a closed loop is allowed as long as you don't reuse the same edge.Hmm, so if the path forms a closed loop, it's a draw because it reaches the edge? Wait, no, forming a closed loop doesn't necessarily reach the edge. It just means the path has returned to a previous point.Wait, the problem says if the broken line reaches the edge, it's a draw. If a player can't continue, they lose. So, forming a closed loop doesn't end the game unless it reaches the edge.So, maybe the key is that the second player can always force the game to end on the first player's turn by making sure that the number of available moves is even, or something like that.Alternatively, maybe it's about the parity of the number of segments. If the total number of segments is even, the second player makes the last move. If it's odd, the first player does. But the game doesn't necessarily use all segments, so it's not directly about the total number.Wait, but the game can end before all segments are used if a player can't move. So, maybe the second player can always respond in a way that the first player is the one who can't move first.I think I'm getting closer. Maybe it's about the fact that the second player can always respond to the first player's move in such a way that the first player is left with no moves eventually.Alternatively, maybe it's about the fact that the second player can always mirror the first player's moves in a way that forces the first player into a dead end.Wait, let's think about it in terms of graph theory again. The grid is a graph where each internal node has degree 4. The game is building a path on this graph, and the players alternate extending the path.If the second player can always choose a direction that doesn't lead to an immediate dead end, they can force the first player to eventually get stuck.But how? Maybe the second player can always choose a direction that leaves an even number of available moves, ensuring that the first player is the one who runs out first.Alternatively, maybe it's about the fact that the second player can always respond to the first player's move in a way that the first player is left with an odd number of available moves, eventually leading to the first player being unable to move.Wait, I'm not sure. Maybe I need to think about it differently. Let's consider the game as a series of moves where each player alternately extends the path. The second player can always respond to the first player's move in a way that forces the first player to eventually run out of moves.Alternatively, maybe it's about the fact that the second player can always choose a direction that doesn't lead to the edge, forcing the first player to eventually reach the edge and cause a draw, but since the second player can't lose, that's okay.Wait, but the problem says if the line reaches the edge, it's a draw, not a win for either player. So, the second player can't lose, but the first player could lose or force a draw.Hmm, maybe the key is that the second player can always force the game to end on the first player's turn, either by making the first player lose or forcing a draw.But I'm not entirely sure. Maybe I need to think about it in terms of parity again. If the game can only end on an even number of moves, then the second player can't lose because they would have just made the last move, forcing the first player to lose.But why must the game end on an even number of moves? Maybe because the game starts with the first player, and each pair of moves (first and second player) reduces the available segments in such a way that the game can only end after an even number of moves.Alternatively, maybe it's about the fact that the second player can always respond to the first player's move in a way that the first player is left with no moves eventually.Wait, I think I'm going in circles here. Maybe I need to look for a more formal proof.Perhaps I can use induction. Suppose that for any grid of size n x n, the second player can't lose. Then, for a grid of size (n+1) x (n+1), the same applies.But I'm not sure how to set up the induction step. Maybe it's better to think about the game in terms of parity and available moves.Wait, another idea: if the second player can always mirror the first player's moves in such a way that the first player is forced to make the last move, then the second player can't lose.But mirroring might not always be possible because the grid is finite and the path can get stuck in a corner.Alternatively, maybe the second player can always choose a direction that doesn't lead to an immediate dead end, forcing the first player to eventually get stuck.Wait, I think I'm overcomplicating it. Maybe the key is that the second player can always respond to the first player's move in a way that the first player is left with no moves eventually.Alternatively, maybe it's about the fact that the second player can always choose a direction that doesn't lead to the edge, forcing the first player to eventually reach the edge and cause a draw, but since the second player can't lose, that's okay.Wait, but the problem says if the line reaches the edge, it's a draw, not a win for either player. So, the second player can't lose, but the first player could lose or force a draw.Hmm, maybe the key is that the second player can always force the game to end on the first player's turn, either by making the first player lose or forcing a draw.But I'm not entirely sure. Maybe I need to think about it in terms of parity again. If the game can only end on an even number of moves, then the second player can't lose because they would have just made the last move, forcing the first player to lose.But why must the game end on an even number of moves? Maybe because the game starts with the first player, and each pair of moves (first and second player) reduces the available segments in such a way that the game can only end after an even number of moves.Alternatively, maybe it's about the fact that the second player can always respond to the first player's move in a way that the first player is left with no moves eventually.Wait, I think I'm going in circles here. Maybe I need to look for a more formal proof.Perhaps I can use induction. Suppose that for any grid of size n x n, the second player can't lose. Then, for a grid of size (n+1) x (n+1), the same applies.But I'm not sure how to set up the induction step. Maybe it's better to think about the game in terms of parity and available moves.Wait, another idea: if the second player can always mirror the first player's moves in such a way that the first player is forced to make the last move, then the second player can't lose.But mirroring might not always be possible because the grid is finite and the path can get stuck in a corner.Alternatively, maybe the second player can always choose a direction that doesn't lead to an immediate dead end, forcing the first player to eventually get stuck.Hmm, I'm not making much progress here. Maybe I need to think about it differently. Let's consider the game as a series of moves where each player alternately extends the path. The second player can always respond to the first player's move in a way that forces the first player to eventually run out of moves.Wait, maybe it's about the parity of the number of available moves from each point. If each internal point has an even number of edges, then the second player can always respond to the first player's move by using another edge, ensuring that the first player is left with no moves eventually.But in a grid, each internal point has four edges, which is even. So, if the second player can always respond to the first player's move by using another edge, they can ensure that the first player is left with no moves.Wait, that makes sense. Since each internal point has an even number of edges, the second player can always respond to the first player's move by using another edge, leaving the first player with an even number of edges minus two, which is still even. Eventually, the first player will be left with no moves.But wait, what if the path reaches the edge before that? Then it's a draw, and the second player doesn't lose. So, either the first player loses or it's a draw, but the second player can't lose.Yes, that seems to be the key. Since each internal point has an even number of edges, the second player can always respond to the first player's move by using another edge, ensuring that the first player is the one who eventually can't move. If the path reaches the edge, it's a draw, but the second player still doesn't lose.So, in summary, because each internal point has an even number of edges, the second player can always respond to the first player's move in a way that forces the first player to eventually run out of moves. Therefore, the second player can't lose.

💡First, I need to convert the binary number (1011_2) to its decimal equivalent. I'll break it down by each digit's place value.Starting from the right, the first digit represents (2^0), the second digit represents (2^1), the third digit represents (2^2), and the fourth digit represents (2^3). Calculating each term:[1 times 2^3 = 8][0 times 2^2 = 0][1 times 2^1 = 2][1 times 2^0 = 1]Adding these together:[8 + 0 + 2 + 1 = 11]So, (1011_2) in decimal is (11).Next, I'll convert the ternary number (122_3) to decimal. Each digit represents a power of 3, starting from the right.The first digit from the right represents (3^0), the second digit represents (3^1), and the third digit represents (3^2). Calculating each term:[1 times 3^2 = 9][2 times 3^1 = 6][2 times 3^0 = 2]Adding these together:[9 + 6 + 2 = 17]So, (122_3) in decimal is (17).Finally, I'll multiply the two decimal numbers to find the product:[11 times 17 = 187]Therefore, the product of (1011_2) and (122_3) in base 10 is (187).