💡First, I will distribute the coefficients inside each parenthesis to simplify the expression.For the term (-frac{1}{2}(4xy - 8x^{2}y^{2})), I will multiply (-frac{1}{2}) by each term inside the parentheses:[-frac{1}{2} times 4xy = -2xy][-frac{1}{2} times (-8x^{2}y^{2}) = 4x^{2}y^{2}]Next, for the term (2(3xy - 5x^{2}y^{2})), I will multiply (2) by each term inside the parentheses:[2 times 3xy = 6xy][2 times (-5x^{2}y^{2}) = -10x^{2}y^{2}]Now, I will combine all the simplified terms:[2xy - 2xy + 4x^{2}y^{2} + 6xy - 10x^{2}y^{2}]Combining like terms:[(2xy - 2xy + 6xy) + (4x^{2}y^{2} - 10x^{2}y^{2}) = 6xy - 6x^{2}y^{2}]Finally, I will substitute (x = 2) and (y = -1) into the simplified expression:[6xy - 6x^{2}y^{2} = 6 times 2 times (-1) - 6 times (2)^{2} times (-1)^{2}][= -12 - 24 = -36]

💡First, I need to determine the total area of Jesse's room. The room is 2 feet long and 12 feet wide, so the area is calculated by multiplying the length by the width.Next, I know that Jesse uses 6 tiles to cover the entire floor. To find the area of each tile, I divide the total area of the room by the number of tiles.Finally, by performing the division, I can find the size of each tile in square feet.

💡Okay, so I have this problem here: I need to find the derivative of the function y = x times e to the x. The options are A, B, C, D, with D being (1 + x)e^x. Hmm, let me think about how to approach this.First, I remember that when you have a function that's the product of two other functions, you need to use the product rule. The product rule says that if you have y = u*v, then the derivative y' is u'v + uv'. So in this case, my u is x and my v is e^x. That makes sense because y is x multiplied by e^x.Alright, so let me write that down. If u = x, then the derivative of u with respect to x, which is u', should be 1, right? Because the derivative of x is just 1. And then v is e^x, so the derivative of v with respect to x, which is v', is also e^x. I remember that the derivative of e^x is e^x, so that part is straightforward.Now, applying the product rule: y' = u'v + uv'. So plugging in the values I have, that would be y' = (1)(e^x) + (x)(e^x). Let me compute that. The first term is just e^x, and the second term is x times e^x. So combining those, I get y' = e^x + x e^x.Hmm, can I factor this expression further? Let me see. Both terms have an e^x in them, so I can factor that out. That would give me y' = e^x (1 + x). So that simplifies to y' = (1 + x)e^x.Looking back at the options, D is (1 + x)e^x, which matches what I just got. So that must be the correct answer. Let me double-check to make sure I didn't make any mistakes.Wait, let me go through the steps again. Starting with y = x e^x. Using the product rule, derivative of x is 1, derivative of e^x is e^x. So y' = 1*e^x + x*e^x. That's e^x + x e^x. Factor out e^x: e^x(1 + x). Yep, that's the same as (1 + x)e^x. So D is definitely the right choice.I don't think I made any errors here. The product rule was applied correctly, and the algebra seems fine. So I'm confident that D is the correct derivative of y = x e^x.**Final Answer**boxed{D}

💡First, I need to factor the quadratic expression ( x^2 + 23x + 132 ). To do this, I'll look for two numbers that multiply to 132 and add up to 23. I start by listing the factors of 132:- 1 and 132 (sum = 133)- 2 and 66 (sum = 68)- 3 and 44 (sum = 47)- 4 and 33 (sum = 37)- 6 and 22 (sum = 28)- 11 and 12 (sum = 23)The pair 11 and 12 meets the criteria because their product is 132 and their sum is 23. Therefore, the factored form of the first expression is:[x^2 + 23x + 132 = (x + 11)(x + 12)]So, ( a = 11 ) and ( b = 12 ).Next, I'll factor the second quadratic expression ( x^2 - 25x + 168 ). Here, I need two numbers that multiply to 168 and add up to 25 (since the middle term is negative, both numbers should be positive). Listing the factors of 168:- 1 and 168 (sum = 169)- 2 and 84 (sum = 86)- 3 and 56 (sum = 59)- 4 and 42 (sum = 46)- 6 and 28 (sum = 34)- 7 and 24 (sum = 31)- 8 and 21 (sum = 29)- 12 and 14 (sum = 26)The pair 7 and 24 meets the criteria because their product is 168 and their sum is 31, which is close to 25. However, since the middle term is negative, I'll use negative signs for both numbers:[x^2 - 25x + 168 = (x - 7)(x - 24)]So, ( c = 7 ) and ( d = 24 ).Finally, I'll calculate the sum ( a + c + d ):[a + c + d = 11 + 7 + 24 = 42]

💡Okay, so I need to find the intersection point of two lines. One of them is given by the equation y = 2x + 5, and the other line is perpendicular to this one and passes through the point (5, 5). Hmm, let me think about how to approach this step by step.First, I remember that if two lines are perpendicular, their slopes are negative reciprocals of each other. That means if one line has a slope of m, the other will have a slope of -1/m. So, the given line has a slope of 2. Therefore, the slope of the line perpendicular to it should be -1/2. Okay, that makes sense.Now, I need to find the equation of the line that is perpendicular to y = 2x + 5 and passes through the point (5, 5). To write the equation of a line, I can use the point-slope form, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Plugging in the values I have, m = -1/2 and the point (5, 5), the equation becomes:y - 5 = (-1/2)(x - 5)Let me simplify this equation to get it into slope-intercept form (y = mx + b) for easier comparison with the original line. Distribute the -1/2 on the right side:y - 5 = (-1/2)x + (5/2)Now, add 5 to both sides to solve for y:y = (-1/2)x + (5/2) + 5Hmm, I need to combine the constants. 5 is the same as 10/2, so:y = (-1/2)x + (5/2 + 10/2)y = (-1/2)x + (15/2)Okay, so the equation of the perpendicular line is y = (-1/2)x + 15/2.Now, to find the intersection point of the two lines y = 2x + 5 and y = (-1/2)x + 15/2, I need to set them equal to each other since at the intersection point, both y-values will be the same for the same x-value.So, set 2x + 5 equal to (-1/2)x + 15/2:2x + 5 = (-1/2)x + 15/2Let me solve for x. First, to eliminate the fractions, I can multiply every term by 2 to make the calculations easier:2*(2x) + 2*5 = 2*(-1/2)x + 2*(15/2)Simplifying each term:4x + 10 = -1x + 15Now, I'll bring all the x terms to one side and the constants to the other side. Let's add x to both sides:4x + x + 10 = 15Which simplifies to:5x + 10 = 15Next, subtract 10 from both sides:5x = 5Now, divide both sides by 5:x = 1Okay, so x is 1. Now, I need to find the corresponding y-value. I can plug x = 1 back into either of the original equations. Let's use y = 2x + 5 because it's simpler.y = 2*1 + 5y = 2 + 5y = 7So, the intersection point is (1, 7).Wait, let me double-check by plugging x = 1 into the other equation to make sure I didn't make a mistake.Using y = (-1/2)x + 15/2:y = (-1/2)*1 + 15/2y = -1/2 + 15/2y = ( -1 + 15 ) / 2y = 14/2y = 7Yes, that matches. So, both lines intersect at (1, 7). That seems correct.Just to recap, the steps I took were:1. Determine the slope of the given line.2. Find the slope of the perpendicular line.3. Use the point-slope form to write the equation of the perpendicular line.4. Simplify it to slope-intercept form.5. Set the two equations equal to each other to find the x-coordinate of the intersection.6. Solve for x.7. Plug x back into one of the original equations to find y.8. Verify by plugging x into the other equation.Everything checks out, so I'm confident that the intersection point is indeed (1, 7).

💡Okay, so I'm trying to solve this problem about pirates and their card game debts. There are 100 pirates, and after playing cards, each pirate has either won or lost some amount of gold sand. The key here is that the pirates can only exchange gold in specific ways: either one pirate pays an equal amount to every other pirate, or one pirate receives the same amount from every other pirate. I need to prove that after several such steps, it's possible for each winner to receive exactly what they've won and for each loser to pay exactly what they've lost.First, let me try to understand the problem better. There are 100 pirates, each with a certain amount they've won or lost. The total amount won by all pirates should equal the total amount lost because it's a zero-sum game. So, if I denote the amount each pirate has won or lost as ( w_1, w_2, ldots, w_{100} ), then the sum of all these ( w_i ) should be zero. That is, ( sum_{i=1}^{100} w_i = 0 ).Now, the pirates can only exchange gold in two ways: either one pirate pays an equal amount to every other pirate, or one pirate receives the same amount from every other pirate. Let me think about what these operations mean in terms of transferring gold.If a pirate pays an equal amount to every other pirate, that means they are distributing their gold equally among the other 99 pirates. Similarly, if a pirate receives the same amount from every other pirate, each of the other 99 pirates is giving them the same amount. So, these operations are somewhat symmetric but in opposite directions.I need to figure out how to use these operations to balance the debts so that each pirate ends up with exactly what they've won or lost. Let me consider a simpler case first, maybe with fewer pirates, to get an idea of how this might work.Suppose there are just two pirates. If one pirate has won some amount and the other has lost the same amount, then the winner can simply receive that amount from the loser. But in this case, since there are only two pirates, the operations would be straightforward. However, with 100 pirates, things are more complex.Let me think about the operations in terms of linear algebra. Each operation can be represented as a vector where one pirate is either paying or receiving a certain amount, and the rest are either receiving or paying the same amount. Since we have 100 pirates, each operation affects 99 pirates in some way.I recall that in problems involving balancing debts or resources, linear algebra and systems of equations can be useful. Maybe I can model the debts as a system where each operation corresponds to a vector, and I need to find a combination of these vectors that results in the desired debt distribution.But before diving into linear algebra, let me see if there's a more straightforward approach. Since the pirates can only pay or receive equal amounts from or to all others, perhaps there's a way to iteratively adjust the debts until they're balanced.Let me consider the total amount each pirate needs to pay or receive. Let's say pirate ( i ) needs to pay ( p_i ) or receive ( r_i ). The sum of all ( p_i ) should equal the sum of all ( r_i ) because the total amount lost equals the total amount won.Now, if I can find a sequence of operations where each pirate either pays or receives the exact amount they need, then the problem is solved. But how?Maybe I can use the fact that each operation affects all other pirates equally. So, if a pirate needs to pay a certain amount, they can do so by paying an equal amount to each of the other pirates. Similarly, if a pirate needs to receive a certain amount, they can receive an equal amount from each of the other pirates.Wait, but each operation affects all other pirates, so if I have multiple operations, the amounts can add up. Maybe I can decompose the required payments into a series of these equal payments or receipts.Let me try to formalize this. Suppose pirate ( A ) needs to pay a total of ( a ) gold. Pirate ( A ) can pay ( frac{a}{99} ) to each of the other 99 pirates. This way, pirate ( A ) pays a total of ( a ), and each of the other pirates receives ( frac{a}{99} ).Similarly, if pirate ( B ) needs to receive a total of ( b ) gold, pirate ( B ) can receive ( frac{b}{99} ) from each of the other 99 pirates. This way, pirate ( B ) receives a total of ( b ), and each of the other pirates pays ( frac{b}{99} ).But wait, if I do this for each pirate, won't the payments overlap and cause conflicts? For example, if pirate ( A ) pays ( frac{a}{99} ) to pirate ( B ), and pirate ( B ) also needs to pay ( frac{c}{99} ) to pirate ( A ), how does that balance out?Hmm, maybe I need to consider the net flow of gold between each pair of pirates. If I can ensure that the net amount each pirate pays or receives is exactly their debt, then the system will balance.Let me think about this in terms of linear equations. Each pirate has a net debt ( d_i ), which is positive if they need to pay and negative if they need to receive. The sum of all ( d_i ) is zero.I need to find a set of operations such that the sum of the operations equals the desired net debts. Each operation corresponds to a vector where one pirate pays or receives a certain amount, and the rest are adjusted accordingly.But this might get complicated with 100 pirates. Maybe there's a smarter way to approach this.I remember that in problems involving balancing resources, sometimes you can use the concept of linear combinations or spanning sets. If the set of possible operations can span the space of all possible debt distributions, then it's possible to reach any desired distribution.In this case, the operations are specific types of vectors where one pirate is either paying or receiving an equal amount from or to all others. I need to check if these operations can generate any possible debt distribution.Alternatively, maybe I can use induction. Start with a small number of pirates and see if the statement holds, then assume it holds for ( n ) pirates and prove it for ( n+1 ).But with 100 pirates, induction might not be the most straightforward approach. Maybe I can think about the problem in terms of linear algebra and matrices.Each operation can be represented as a matrix where one row (or column) has a certain value, and the others are adjusted accordingly. The key is to see if these matrices can generate any possible debt distribution.But I'm not sure if this is the right path. Maybe I should look for a constructive approach, where I can explicitly describe a sequence of operations that will balance the debts.Let me try to outline a possible sequence:1. Identify all pirates who need to pay (losers) and all pirates who need to receive (winners).2. For each loser, have them pay an equal amount to all other pirates until they've paid their total debt.3. For each winner, have them receive an equal amount from all other pirates until they've received their total winnings.But the problem is that these operations affect all pirates, so paying or receiving from all others will impact everyone's balance. I need to ensure that after all operations, each pirate's net change is exactly their debt.Maybe I can process one pirate at a time. For example, start with pirate 1. If pirate 1 needs to pay ( p_1 ), have them pay ( frac{p_1}{99} ) to each of the other pirates. This way, pirate 1's debt is settled, and each of the other pirates receives ( frac{p_1}{99} ).But now, the other pirates have their balances adjusted. If some of them were supposed to receive money, they've already started receiving it, but if they were supposed to pay, they've now received money, which might complicate things.Alternatively, maybe I can process the winners first. If pirate 1 needs to receive ( r_1 ), have them receive ( frac{r_1}{99} ) from each of the other pirates. This way, pirate 1's winnings are settled, and each of the other pirates pays ( frac{r_1}{99} ).But again, this affects the other pirates' balances. It seems like a delicate balance where each operation affects everyone else, making it tricky to isolate individual debts.Perhaps I need to consider the problem in terms of linear combinations. Each operation can be seen as adding a multiple of a specific vector to the current debt distribution. The question then becomes whether these vectors span the entire space of possible debt distributions.Given that we have 100 pirates, the space of debt distributions is 99-dimensional (since the total debt must sum to zero). Each operation corresponds to a vector where one pirate is either paying or receiving, and the others are adjusted accordingly. I need to check if these vectors form a spanning set for the 99-dimensional space.If they do, then any debt distribution can be expressed as a linear combination of these operations, meaning it's possible to balance the debts as required.But I'm not entirely sure how to formally prove that these operations span the space. Maybe I can consider the rank of the matrix formed by these operations. If the rank is 99, then they span the space.Alternatively, perhaps there's a more straightforward way. Since each operation allows a pirate to either pay or receive an equal amount from all others, maybe we can use these operations to zero out individual debts one by one.For example, suppose pirate 1 needs to pay ( p_1 ). Have pirate 1 pay ( frac{p_1}{99} ) to each of the other pirates. This settles pirate 1's debt, and each of the other pirates now has their balance adjusted by ( frac{p_1}{99} ).But now, the other pirates' balances are affected. If some of them were supposed to receive money, they've already started receiving it, but if they were supposed to pay, they've now received money, which might complicate things.Wait, maybe I can process the debts in a specific order. For example, process all the losers first, having each pay their debt in equal installments to all others. Then, process the winners, having each receive their winnings in equal installments from all others.But I'm not sure if this will work because the payments from the losers will affect the winners' balances, and vice versa.Alternatively, maybe I can use a combination of both types of operations. For example, have some pirates pay while others receive, in a way that the net effect is the desired debt distribution.I think I need to formalize this a bit more. Let me denote the debt of each pirate as ( d_1, d_2, ldots, d_{100} ), where ( d_i ) is positive if pirate ( i ) needs to pay and negative if pirate ( i ) needs to receive.The goal is to find a sequence of operations such that the sum of the operations equals the debt vector ( (d_1, d_2, ldots, d_{100}) ).Each operation can be represented as a vector where one pirate pays or receives a certain amount, and the rest are adjusted accordingly. Specifically, if pirate ( k ) pays ( x ) to each of the other pirates, the debt vector changes by ( (-x, x, x, ldots, x) ) (with the first component being pirate ( k ) paying ( x ), and the rest receiving ( x )).Similarly, if pirate ( k ) receives ( x ) from each of the other pirates, the debt vector changes by ( (x, -x, -x, ldots, -x) ).So, each operation corresponds to adding a vector of the form ( (x, -x, -x, ldots, -x) ) or ( (-x, x, x, ldots, x) ) to the current debt vector.Now, the question is whether these operations can generate any possible debt vector ( (d_1, d_2, ldots, d_{100}) ) with ( sum d_i = 0 ).I think the key is to note that these operations can be combined to create any desired change in the debt vector. Specifically, by choosing appropriate values of ( x ) for each operation, we can adjust the debts to match the desired distribution.Let me try to construct such a sequence. Suppose I want to adjust pirate ( k )'s debt by ( Delta_k ). I can perform an operation where pirate ( k ) pays ( frac{Delta_k}{99} ) to each of the other pirates. This will decrease pirate ( k )'s debt by ( Delta_k ) and increase each of the other pirates' debts by ( frac{Delta_k}{99} ).Similarly, if I want to increase pirate ( k )'s debt by ( Delta_k ), I can have pirate ( k ) receive ( frac{Delta_k}{99} ) from each of the other pirates.By carefully choosing the order and amounts of these operations, I can adjust each pirate's debt to match their desired amount.But I need to ensure that after all operations, the total debt remains zero and that each pirate's net change is exactly their desired debt.Let me consider an example with three pirates to see how this might work.Suppose we have three pirates: A, B, and C. Pirate A needs to pay 3 units, pirate B needs to pay 2 units, and pirate C needs to receive 5 units. The total debt is zero.To balance this, I can have pirate A pay 1 unit to each of the other pirates (totaling 2 units paid by A), and pirate B pay 1 unit to each of the other pirates (totaling 2 units paid by B). However, pirate C needs to receive 5 units, so this approach doesn't quite work.Alternatively, I can have pirate C receive 1 unit from each of the other pirates. Since there are two other pirates, pirate C would receive 2 units, which is less than the required 5 units. So, I need to perform multiple operations.Maybe I can have pirate C receive 2 units from each of the other pirates, which would give pirate C 4 units, and then have pirate C receive 1 more unit from each pirate, totaling 5 units. However, this would require pirate A and pirate B to each pay 3 units in total, which matches their debts.So, in this case, it's possible to balance the debts by having pirate C receive 2 units from each pirate in one operation and then 1 more unit from each pirate in another operation. This way, pirate A pays 3 units, pirate B pays 3 units, and pirate C receives 5 units.Extending this idea to 100 pirates, it seems plausible that we can decompose the required payments into a series of equal payments or receipts from all other pirates.Another approach is to consider that each pirate's debt can be represented as a linear combination of the operations. Since the operations correspond to vectors that can be combined to reach any point in the space, it should be possible to find such a combination.I think the key insight is that the operations allow us to adjust each pirate's debt by a multiple of 1/99 of the total debt, which, when combined appropriately, can sum up to the desired individual debts.Therefore, by carefully choosing the sequence and amounts of these operations, we can ensure that each pirate ends up paying or receiving exactly their share of the total debt.In summary, the problem can be approached by recognizing that each operation allows us to adjust the debts in a controlled manner, and by combining these operations, we can achieve the desired distribution of debts among the pirates.