💡Okay, so I have this problem where I need to find p(6) for a monic quartic polynomial p(x). They give me the values of p at x=1,2,3,4: p(1)=3, p(2)=7, p(3)=13, p(4)=21. Hmm, quartic means degree 4, and monic means the leading coefficient is 1. First, I remember that if a polynomial passes through certain points, we can sometimes express it as another polynomial plus some function that fits those points. Maybe I can think of p(x) as a quartic polynomial plus a quadratic or something? Let me see.Looking at the given values: p(1)=3, p(2)=7, p(3)=13, p(4)=21. Let me compute the differences between consecutive p(x) values. From p(1)=3 to p(2)=7: difference is 4.From p(2)=7 to p(3)=13: difference is 6.From p(3)=13 to p(4)=21: difference is 8.Hmm, the differences are increasing by 2 each time. That suggests that the second differences are constant, which usually indicates a quadratic component. Wait, but p(x) is quartic, so maybe the quartic part is responsible for the higher-order differences?Alternatively, maybe p(x) can be expressed as a quadratic function plus a quartic polynomial that has roots at x=1,2,3,4. Let me try that.Let’s suppose that p(x) = q(x) + r(x), where q(x) is a quadratic polynomial and r(x) is a quartic polynomial that is zero at x=1,2,3,4. Since r(x) is quartic and has roots at 1,2,3,4, it can be written as r(x) = a(x-1)(x-2)(x-3)(x-4), where a is a constant.So, p(x) = a(x-1)(x-2)(x-3)(x-4) + q(x). Since p(x) is monic quartic, the leading term of p(x) must be x^4. The leading term of r(x) is a x^4, and q(x) is quadratic, so its leading term is at most x^2. Therefore, to make p(x) monic quartic, we must have a=1. So, r(x) = (x-1)(x-2)(x-3)(x-4).Therefore, p(x) = (x-1)(x-2)(x-3)(x-4) + q(x), where q(x) is a quadratic polynomial. Now, since p(x) is given at x=1,2,3,4, let's plug those into the equation to find q(x).At x=1: p(1) = 0 + q(1) = 3, so q(1)=3.At x=2: p(2) = 0 + q(2) =7, so q(2)=7.At x=3: p(3)=0 + q(3)=13, so q(3)=13.At x=4: p(4)=0 + q(4)=21, so q(4)=21.So, q(x) is a quadratic polynomial that passes through the points (1,3), (2,7), (3,13), (4,21). Wait, but a quadratic is determined by three points, so maybe I can find q(x) using three of these points and then verify with the fourth.Let me set up equations for q(x). Let’s assume q(x) = ax^2 + bx + c.Using x=1: a(1)^2 + b(1) + c = 3 => a + b + c = 3.Using x=2: a(4) + b(2) + c =7 => 4a + 2b + c =7.Using x=3: a(9) + b(3) + c =13 =>9a +3b +c=13.Now, we have a system of equations:1) a + b + c =32)4a +2b +c=73)9a +3b +c=13Let me subtract equation 1 from equation 2:(4a +2b +c) - (a + b + c)=7-33a + b =4. Let's call this equation 4.Similarly, subtract equation 2 from equation 3:(9a +3b +c) - (4a +2b +c)=13-75a + b=6. Let's call this equation 5.Now, subtract equation 4 from equation 5:(5a + b) - (3a + b)=6-42a=2 => a=1.Plugging a=1 into equation 4: 3(1) + b=4 => b=1.Then from equation 1:1 +1 +c=3 =>c=1.So, q(x)=x^2 +x +1.Wait, let's check if this works for x=4:q(4)=16 +4 +1=21, which matches p(4)=21. Perfect.So, p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +x +1.Now, I need to compute p(6). Let me compute each part.First, compute (6-1)(6-2)(6-3)(6-4)=5*4*3*2=120.Then, compute q(6)=6^2 +6 +1=36 +6 +1=43.So, p(6)=120 +43=163.Wait, but in the initial problem statement, the assistant had a different approach and got 158. Hmm, maybe I made a mistake.Wait, let me double-check my steps.First, I assumed p(x)=r(x)+q(x), with r(x)=(x-1)(x-2)(x-3)(x-4) and q(x) quadratic.Then, using the given points, I found q(x)=x^2 +x +1.Wait, let me check q(1)=1 +1 +1=3, correct.q(2)=4 +2 +1=7, correct.q(3)=9 +3 +1=13, correct.q(4)=16 +4 +1=21, correct.So, that seems right.Then, p(6)= (6-1)(6-2)(6-3)(6-4) +6^2 +6 +1=5*4*3*2 +36 +6 +1=120 +43=163.But the initial assistant answer was 158. Hmm, perhaps I made a miscalculation.Wait, let me recalculate (6-1)(6-2)(6-3)(6-4)=5*4*3*2=120, that's correct.q(6)=36 +6 +1=43, that's correct.120 +43=163. Hmm, so why does the assistant have 158?Wait, maybe I misread the problem. Let me check the original problem again.Wait, the problem says p(x) is a monic quartic polynomial such that p(1)=3, p(2)=7, p(3)=13, p(4)=21. Find p(6).Wait, in the initial solution, the assistant wrote p(x)=q(x)+x^2 +2, and then found q(x)= (x-1)(x-2)(x-3)(x-4). Then p(6)=5*4*3*2 +36 +2=120 +38=158.Wait, so in that solution, q(x)= (x-1)(x-2)(x-3)(x-4), and p(x)=q(x)+x^2 +2.But in my solution, I set p(x)=q(x)+r(x), where q(x)= (x-1)(x-2)(x-3)(x-4) and r(x)=x^2 +x +1.Wait, so why the difference? Because in the initial solution, the assistant assumed p(x)=q(x)+x^2 +2, but in my solution, I found that q(x)=x^2 +x +1.Wait, so perhaps the initial assumption was different. The assistant assumed p(x)=q(x)+x^2 +2, but in reality, the quadratic part is x^2 +x +1.So, which one is correct? Let's see.If p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +x +1, then p(1)=0 +1 +1 +1=3, correct.p(2)=0 +4 +2 +1=7, correct.p(3)=0 +9 +3 +1=13, correct.p(4)=0 +16 +4 +1=21, correct.So, that seems correct.But in the initial solution, the assistant had p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2, which would give p(1)=0 +1 +2=3, correct.p(2)=0 +4 +2=6, but p(2)=7, so that's incorrect. Wait, that can't be.Wait, no, the assistant said p(2)=7=4 +2 +q(2), so q(2)=1. But in their solution, they set q(x)=a(x-1)(x-2)(x-3)(x-4). So, q(2)=0, but p(2)=7=4 +2 +0=6, which is not correct. Wait, that can't be.Wait, no, in the initial solution, the assistant wrote p(x)=q(x)+x^2 +2, and then found q(x)= (x-1)(x-2)(x-3)(x-4). But then p(2)= q(2)+4 +2=0 +6=6, but p(2)=7, so that's a problem.Wait, so that initial solution must be wrong. Because if p(x)=q(x)+x^2 +2, then p(2)=q(2)+4 +2= q(2)+6. But p(2)=7, so q(2)=1. But q(x)=a(x-1)(x-2)(x-3)(x-4), so q(2)=0. Therefore, 0 +6=6≠7. So, that approach is flawed.Therefore, the correct approach is to set p(x)= (x-1)(x-2)(x-3)(x-4) + q(x), where q(x) is quadratic, and solve for q(x) as I did, getting q(x)=x^2 +x +1, leading to p(6)=163.But wait, the initial solution got 158, which is different. So, perhaps I made a mistake in my calculation.Wait, let me recalculate p(6). (6-1)(6-2)(6-3)(6-4)=5*4*3*2=120. Then q(6)=6^2 +6 +1=36 +6 +1=43. So, 120 +43=163.Alternatively, maybe the initial solution had a different approach, assuming p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2, but that doesn't fit p(2)=7, as we saw. So, perhaps the initial solution was incorrect.Alternatively, maybe I made a mistake in assuming q(x)=x^2 +x +1. Let me check again.From the system of equations:1) a + b + c =32)4a +2b +c=73)9a +3b +c=13We found a=1, b=1, c=1.So, q(x)=x^2 +x +1.Yes, that seems correct.Alternatively, maybe the problem is that p(x) is monic quartic, so when we add (x-1)(x-2)(x-3)(x-4) which is monic quartic, and q(x) is quadratic, so p(x) is indeed monic quartic. So, that's correct.Therefore, p(6)=163.But wait, in the initial problem, the assistant got 158, which is different. So, perhaps I made a mistake in my approach.Wait, let me think differently. Maybe instead of assuming p(x)= (x-1)(x-2)(x-3)(x-4) + q(x), I can consider the differences.Let me compute the values of p(x) and see if there's a pattern.Given p(1)=3, p(2)=7, p(3)=13, p(4)=21.Let me compute the first differences: 7-3=4, 13-7=6, 21-13=8.Second differences: 6-4=2, 8-6=2. So, second differences are constant at 2. That suggests that the polynomial is quadratic, but p(x) is quartic. So, perhaps the quartic part is responsible for higher-order differences.Wait, but if the second differences are constant, that would suggest that the quadratic part is x^2 + something, but since p(x) is quartic, the quartic part must be such that when combined with the quadratic, the second differences remain constant.Alternatively, maybe the quartic part has zero second differences, but that's not possible because quartic polynomials have non-zero third and fourth differences.Wait, perhaps I can model p(x) as a quadratic plus a quartic polynomial that has zero second differences. But that might not be straightforward.Alternatively, maybe I can use finite differences to find the coefficients of p(x).Since p(x) is quartic, the fourth differences should be constant.Let me set up a finite difference table.Given p(1)=3, p(2)=7, p(3)=13, p(4)=21.First differences: 4,6,8.Second differences: 2,2.Third differences: 0.Fourth differences: 0.Wait, but quartic polynomials have constant fourth differences, but here, the third differences are already zero, which would suggest that the polynomial is quadratic, which contradicts the fact that p(x) is quartic.Hmm, that's confusing. Maybe I need to extend the table.Wait, but I only have four points, so I can't compute higher differences beyond third.Wait, perhaps the initial assumption that p(x) is quadratic plus quartic is correct, but the finite difference approach is not sufficient here because we don't have enough points.Alternatively, maybe I can use the method of undetermined coefficients.Let me assume p(x)=x^4 + ax^3 +bx^2 +cx +d.Then, using the given points:p(1)=1 +a +b +c +d=3p(2)=16 +8a +4b +2c +d=7p(3)=81 +27a +9b +3c +d=13p(4)=256 +64a +16b +4c +d=21Now, we have a system of four equations:1) 1 +a +b +c +d=3 => a +b +c +d=22)16 +8a +4b +2c +d=7 =>8a +4b +2c +d= -93)81 +27a +9b +3c +d=13 =>27a +9b +3c +d= -684)256 +64a +16b +4c +d=21 =>64a +16b +4c +d= -235Now, let's write these equations:Equation 1: a + b + c + d = 2Equation 2:8a +4b +2c +d= -9Equation 3:27a +9b +3c +d= -68Equation 4:64a +16b +4c +d= -235Now, let's subtract equation 1 from equation 2:(8a +4b +2c +d) - (a +b +c +d)= -9 -27a +3b +c= -11. Let's call this equation 5.Similarly, subtract equation 2 from equation 3:(27a +9b +3c +d) - (8a +4b +2c +d)= -68 - (-9)19a +5b +c= -59. Let's call this equation 6.Subtract equation 3 from equation 4:(64a +16b +4c +d) - (27a +9b +3c +d)= -235 - (-68)37a +7b +c= -167. Let's call this equation 7.Now, we have:Equation 5:7a +3b +c= -11Equation 6:19a +5b +c= -59Equation 7:37a +7b +c= -167Now, subtract equation 5 from equation 6:(19a +5b +c) - (7a +3b +c)= -59 - (-11)12a +2b= -48 =>6a +b= -24. Let's call this equation 8.Similarly, subtract equation 6 from equation 7:(37a +7b +c) - (19a +5b +c)= -167 - (-59)18a +2b= -108 =>9a +b= -54. Let's call this equation 9.Now, subtract equation 8 from equation 9:(9a +b) - (6a +b)= -54 - (-24)3a= -30 =>a= -10.Now, plug a= -10 into equation 8:6*(-10) +b= -24 =>-60 +b= -24 =>b=36.Now, plug a= -10 and b=36 into equation 5:7*(-10) +3*36 +c= -11 =>-70 +108 +c= -11 =>38 +c= -11 =>c= -49.Now, plug a= -10, b=36, c= -49 into equation 1: -10 +36 -49 +d=2 =>(-10 +36)=26; 26 -49= -23; -23 +d=2 =>d=25.So, the polynomial is p(x)=x^4 -10x^3 +36x^2 -49x +25.Now, let's compute p(6):6^4=1296-10*6^3= -10*216= -216036*6^2=36*36=1296-49*6= -294+25.So, p(6)=1296 -2160 +1296 -294 +25.Let's compute step by step:1296 -2160= -864-864 +1296=432432 -294=138138 +25=163.So, p(6)=163.Wait, so this matches my earlier result of 163, but contradicts the initial assistant's answer of 158. Therefore, the correct answer should be 163.But why did the initial assistant get 158? Let me check their approach again.They assumed p(x)=q(x) +x^2 +2, where q(x)=a(x-1)(x-2)(x-3)(x-4). Then, they found q(1)=0, q(2)=1, q(3)=2, q(4)=3, leading to q(x)= (x-1)(x-2)(x-3)(x-4). Then p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2, and p(6)=5*4*3*2 +36 +2=120 +38=158.But wait, in this approach, p(2)= q(2)+4 +2=1 +6=7, which is correct. Similarly, p(3)= q(3)+9 +2=2 +11=13, correct. p(4)= q(4)+16 +2=3 +18=21, correct. p(1)=0 +1 +2=3, correct.Wait, so in this approach, p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2, which seems to fit all given points. Then, p(6)=120 +36 +2=158.But according to the method of undetermined coefficients, p(x)=x^4 -10x^3 +36x^2 -49x +25, which gives p(6)=163.So, which one is correct? There must be a mistake somewhere.Wait, let me compute p(6) using both forms.First, using p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2.Compute (6-1)(6-2)(6-3)(6-4)=5*4*3*2=120.x^2 +2=36 +2=38.So, p(6)=120 +38=158.Now, using the expanded form p(x)=x^4 -10x^3 +36x^2 -49x +25.Compute p(6)=6^4 -10*6^3 +36*6^2 -49*6 +25=1296 -2160 +1296 -294 +25.Compute step by step:1296 -2160= -864-864 +1296=432432 -294=138138 +25=163.So, two different results: 158 vs 163.This is a problem. They can't both be correct.Wait, perhaps the initial assumption that p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2 is incorrect because when we expand it, it doesn't match the quartic polynomial we found.Let me expand p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2.First, compute (x-1)(x-2)(x-3)(x-4).Let me compute (x-1)(x-2)=x^2 -3x +2.Then, (x-3)(x-4)=x^2 -7x +12.Now, multiply these two quadratics: (x^2 -3x +2)(x^2 -7x +12).Multiply term by term:x^2*(x^2 -7x +12)=x^4 -7x^3 +12x^2-3x*(x^2 -7x +12)= -3x^3 +21x^2 -36x2*(x^2 -7x +12)=2x^2 -14x +24Now, add them up:x^4 -7x^3 +12x^2 -3x^3 +21x^2 -36x +2x^2 -14x +24Combine like terms:x^4 + (-7x^3 -3x^3)=x^4 -10x^3(12x^2 +21x^2 +2x^2)=35x^2(-36x -14x)= -50x+24.So, (x-1)(x-2)(x-3)(x-4)=x^4 -10x^3 +35x^2 -50x +24.Now, add x^2 +2: x^4 -10x^3 +35x^2 -50x +24 +x^2 +2= x^4 -10x^3 +36x^2 -50x +26.But according to the method of undetermined coefficients, p(x)=x^4 -10x^3 +36x^2 -49x +25.So, the two expressions are different. The difference is in the coefficients of x and the constant term.In the first approach, p(x)=x^4 -10x^3 +36x^2 -50x +26.In the second approach, p(x)=x^4 -10x^3 +36x^2 -49x +25.So, they differ by (-50x +26) vs (-49x +25), which is a difference of (-x +1).Therefore, one of the approaches is incorrect.Wait, but in the first approach, p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2, which when expanded gives x^4 -10x^3 +36x^2 -50x +26.But according to the method of undetermined coefficients, p(x)=x^4 -10x^3 +36x^2 -49x +25.So, which one is correct? Let's check p(1):First approach: (1-1)(1-2)(1-3)(1-4) +1 +2=0 +1 +2=3, correct.Second approach:1 -10 +36 -49 +25= (1 -10)= -9; (-9 +36)=27; (27 -49)= -22; (-22 +25)=3, correct.Similarly, p(2):First approach: (2-1)(2-2)(2-3)(2-4) +4 +2=0 +4 +2=6, but p(2)=7. Wait, that's a problem.Wait, no, in the first approach, p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2.At x=2, (2-1)(2-2)(2-3)(2-4)=1*0*(-1)*(-2)=0. So, p(2)=0 +4 +2=6, but p(2)=7. So, that's incorrect.Wait, but earlier, the assistant said p(2)=7=4 +2 +q(2)=6 +q(2)=7, so q(2)=1. But q(x)= (x-1)(x-2)(x-3)(x-4), so q(2)=0. Therefore, p(2)=0 +4 +2=6≠7. So, the initial approach is wrong.Therefore, the correct polynomial is the one found by the method of undetermined coefficients: p(x)=x^4 -10x^3 +36x^2 -49x +25, which gives p(6)=163.Therefore, the initial assistant's answer of 158 is incorrect because their assumption that p(x)= (x-1)(x-2)(x-3)(x-4) +x^2 +2 is flawed, as it doesn't satisfy p(2)=7.So, the correct answer is 163.

💡First, I need to determine the number of each type of chocolate in the box.Carlos bought a total of 50 chocolates.He has 3 caramels.The number of nougats is twice the number of caramels, so there are 2 * 3 = 6 nougats.The number of truffles is equal to the number of caramels plus 6, which means there are 3 + 6 = 9 truffles.Now, I'll add up the caramels, nougats, and truffles to find out how many chocolates are not peanut clusters:3 (caramels) + 6 (nougats) + 9 (truffles) = 18 chocolates.To find the number of peanut clusters, I'll subtract the number of non-peanut cluster chocolates from the total number of chocolates:50 (total chocolates) - 18 (caramels, nougats, and truffles) = 32 peanut clusters.Finally, to determine the percentage chance of picking a peanut cluster, I'll divide the number of peanut clusters by the total number of chocolates and multiply by 100:(32 / 50) * 100 = 64%.So, there is a 64% chance that Carlos will pick a peanut cluster.

💡First, calculate the credit Tom receives for each trade-in:1. **SNES Trade-In:** - Value: 150 - Credit: 80% of 150 = 0.80 × 150 = 1202. **Gameboy Trade-In:** - Value: 50 - Credit: 75% of 50 = 0.75 × 50 = 37.503. **PlayStation 2 Trade-In:** - Value: 100 - Credit: 60% of 100 = 0.60 × 100 = 60**Total Trade-In Credit:**- 120 (SNES) + 37.50 (Gameboy) + 60 (PlayStation 2) = 217.50Next, determine the total amount Tom pays:- Tom gives 80 and receives 10 back.- Net payment: 80 - 10 = 70**Total Amount Spent on NES:**- Trade-in credit + Net payment = 217.50 + 70 = 287.50Subtract the value of the free game:- NES cost before free game: 287.50 - 30 = 257.50Calculate the original price before the 15% discount:- Let X be the original price.- X - 15% of X = 257.50- 0.85X = 257.50- X = 257.50 / 0.85 ≈ 303.53Finally, find the sale price before the 5% sales tax:- Let Y be the sale price before tax.- Y + 5% of Y = 303.53- 1.05Y = 303.53- Y = 303.53 / 1.05 ≈ 289.08Therefore, the original NES was on sale for approximately 289.08 before the sales tax was added.

💡Alright, so I have this problem about a chessboard covered with 32 dominoes, each covering two adjacent squares. I need to show that the number of horizontal dominoes with a white square on the left equals the number with a white square on the right. Hmm, okay, let me try to wrap my head around this.First, I know a standard chessboard is 8x8, so it has 64 squares in total. Since each domino covers two squares, 32 dominoes will cover the entire board without overlapping or leaving any gaps. That makes sense. Now, dominoes can be placed either horizontally or vertically. The problem is specifically talking about horizontal dominoes, so I need to focus on those.Each horizontal domino covers two adjacent squares in a row. On a chessboard, adjacent squares alternate in color. So, if a horizontal domino is placed on two adjacent squares, one will be white and the other black. The question is about the orientation of these dominoes—specifically, whether the white square is on the left or the right.Let me visualize a chessboard. The top-left corner is typically white, and the colors alternate from there. So, in the first row, the squares go white, black, white, black, and so on. The second row starts with black, then white, and continues alternating. This pattern continues for all eight rows.Now, if I place a horizontal domino on the first row, starting from the left, it would cover the first two squares: white and black. So, in this case, the white square is on the left. If I place another horizontal domino starting from the second square, it would cover black and white, so the white square is on the right. Interesting.But wait, the dominoes can be placed anywhere on the board, not just in the first row. So, I need to consider all possible horizontal dominoes across all eight rows. Each horizontal domino will cover two squares of alternating colors, but depending on where it's placed, the white square could be on the left or the right.Let me think about the entire board. Since the board is symmetric, maybe there's a balance between the number of horizontal dominoes with white on the left and white on the right. But how can I formalize this?Perhaps I can consider the number of white squares on the left side of the board versus the right side. Wait, the chessboard is 8x8, so it's symmetric along the vertical axis. That means the number of white squares on the left half should be equal to the number of white squares on the right half. Is that true?Let me count. Each row has four white squares and four black squares. Since there are eight rows, each half of the board (left and right) will have four white squares per row. So, for eight rows, that's 4 x 8 = 32 white squares on each half. Wait, no, that can't be right because the entire board only has 32 white squares in total. So, each half must have 16 white squares.Yes, that makes sense. Each half of the board has 16 white squares and 16 black squares. So, the left half has 16 white squares, and the right half has 16 white squares.Now, each horizontal domino covers one square from the left half and one from the right half if it's placed across the middle. But actually, no, a horizontal domino placed entirely on the left half would cover two squares on the left, and similarly for the right half. Wait, no, a horizontal domino can only cover two adjacent squares in the same row, so if it's placed on the left half, it's entirely on the left, and if it's placed on the right half, it's entirely on the right. But dominoes can also be placed across the middle, covering one square on the left and one on the right.Wait, no, actually, a horizontal domino can only cover two squares in the same row. So, if it's placed near the middle, it might cover one square on the left half and one on the right half. But actually, the middle of the board is between the 4th and 5th columns. So, a horizontal domino starting in the 4th column would cover columns 4 and 5, which are on both halves.But wait, each domino is either entirely on the left half, entirely on the right half, or straddling the middle. Hmm, but since the dominoes are placed without overlapping, the ones straddling the middle would cover one square on each half.But I'm getting a bit confused. Maybe I should approach this differently.Let me think about the total number of horizontal dominoes. Each horizontal domino covers two squares in a row. Since there are 8 rows, and each row has 8 squares, the maximum number of horizontal dominoes per row is 4. So, in total, there can be up to 32 horizontal dominoes, but since we have 32 dominoes in total, some must be vertical.Wait, no, actually, the total number of dominoes is 32, regardless of their orientation. So, the number of horizontal dominoes plus the number of vertical dominoes equals 32.But I need to focus on the horizontal dominoes. Each horizontal domino covers two squares in a row, and since the board is 8x8, each row can have up to 4 horizontal dominoes. So, the total number of horizontal dominoes can vary, but their placement affects the count of white squares on the left and right.Wait, maybe I can model this as a graph or use some kind of parity argument. Let me think about the coloring.Each horizontal domino covers one white and one black square. If the domino is placed such that the white square is on the left, then it contributes to the count of white squares on the left. Similarly, if the white square is on the right, it contributes to the count on the right.But since the entire board has an equal number of white and black squares, and each domino covers one of each, the total number of white squares covered by horizontal dominoes must equal the number of black squares covered by horizontal dominoes.Wait, but the problem is specifically about the orientation—whether the white square is on the left or the right. So, maybe I need to consider the difference between the number of horizontal dominoes with white on the left and white on the right.Let me denote:Let ( L ) be the number of horizontal dominoes with a white square on the left.Let ( R ) be the number of horizontal dominoes with a white square on the right.We need to show that ( L = R ).Each horizontal domino contributes either to ( L ) or ( R ). So, the total number of horizontal dominoes is ( L + R ).Now, each horizontal domino covers one white and one black square. So, the total number of white squares covered by horizontal dominoes is ( L + R ), and the same for black squares.But wait, the entire board has 32 white squares and 32 black squares. The horizontal dominoes cover some of these, and the vertical dominoes cover the rest.But vertical dominoes also cover one white and one black square each, just like horizontal ones. So, the total number of white squares covered by all dominoes is 32, and the same for black squares.Therefore, the number of white squares covered by horizontal dominoes plus the number covered by vertical dominoes equals 32.Similarly, the number of black squares covered by horizontal dominoes plus the number covered by vertical dominoes equals 32.But since each horizontal domino covers one white and one black, the number of white squares covered by horizontal dominoes is equal to the number of black squares covered by horizontal dominoes, which is ( L + R ).Similarly, the number of white squares covered by vertical dominoes is equal to the number of black squares covered by vertical dominoes, which is ( 32 - (L + R) ).But I'm not sure if this helps me directly. Maybe I need to think about the difference ( L - R ).Each horizontal domino with white on the left contributes +1 to ( L ), and each with white on the right contributes +1 to ( R ). So, the difference ( L - R ) represents the net number of white squares on the left minus white squares on the right among horizontal dominoes.But I need to show that this difference is zero, i.e., ( L = R ).How can I relate this to the overall structure of the board?Maybe I can consider the entire board and how the dominoes are arranged. Since the board is symmetric, perhaps the number of horizontal dominoes with white on the left equals those with white on the right.But I need a more rigorous argument.Wait, perhaps I can use the concept of a checkerboard and consider the number of white squares on the left and right halves.As I thought earlier, each half of the board has 16 white squares. So, the left half has 16 white squares, and the right half has 16 white squares.Now, each horizontal domino that is entirely on the left half covers one white and one black square on the left. Similarly, each horizontal domino entirely on the right half covers one white and one black square on the right.But what about horizontal dominoes that span the middle, covering one square on the left and one on the right? These dominoes would cover one white and one black square across the two halves.Wait, but if a horizontal domino spans the middle, it's actually covering two squares in the same row, one on the left half and one on the right half. So, depending on the row, the color of the squares it covers could vary.But since the board alternates colors, the parity of the row affects the color of the squares. For example, in even-numbered rows, the first square is black, while in odd-numbered rows, it's white.So, a horizontal domino spanning the middle in an odd-numbered row would cover a white square on the left and a black square on the right. Similarly, in an even-numbered row, it would cover a black square on the left and a white square on the right.Therefore, each horizontal domino spanning the middle contributes either a white on the left or a white on the right, depending on the row.But how does this affect the total count?Let me denote:Let ( S ) be the set of horizontal dominoes that span the middle, i.e., cover one square on the left half and one on the right half.Each domino in ( S ) will have either a white on the left or a white on the right, depending on the row.Now, the total number of white squares on the left half is 16, and the same for the right half.The white squares on the left half are covered by two types of dominoes:1. Horizontal dominoes entirely on the left half, which cover one white and one black square on the left.2. Horizontal dominoes in ( S ) that have a white square on the left.Similarly, the white squares on the right half are covered by:1. Horizontal dominoes entirely on the right half, which cover one white and one black square on the right.2. Horizontal dominoes in ( S ) that have a white square on the right.But the total number of white squares on the left is 16, and the same for the right.Let me denote:Let ( L_{text{left}} ) be the number of horizontal dominoes entirely on the left half with a white square on the left.Similarly, ( L_{text{right}} ) for the right half.But actually, each horizontal domino entirely on the left half will cover one white and one black square, but the orientation (which side is white) depends on the starting position.Wait, maybe I'm complicating things.Let me think differently. The total number of white squares on the left half is 16. These are covered by:- Horizontal dominoes entirely on the left half, each contributing one white square.- Horizontal dominoes in ( S ) that have a white square on the left.Similarly, the total number of white squares on the right half is 16, covered by:- Horizontal dominoes entirely on the right half, each contributing one white square.- Horizontal dominoes in ( S ) that have a white square on the right.But the number of horizontal dominoes entirely on the left half is equal to the number of horizontal dominoes entirely on the right half because the board is symmetric. Let's denote this number as ( H ).So, the total white squares on the left:( H + L_S = 16 )And on the right:( H + R_S = 16 )Where ( L_S ) is the number of dominoes in ( S ) with white on the left, and ( R_S ) is the number with white on the right.But since both equal 16, we have:( H + L_S = H + R_S )Subtracting ( H ) from both sides:( L_S = R_S )So, the number of horizontal dominoes spanning the middle with white on the left equals those with white on the right.But wait, the total number of horizontal dominoes is ( L + R ), where ( L = L_{text{left}} + L_S ) and ( R = R_{text{right}} + R_S ).But from the above, ( L_S = R_S ). Also, since the number of horizontal dominoes entirely on the left equals those on the right, ( L_{text{left}} = R_{text{right}} ).Therefore, ( L = L_{text{left}} + L_S ) and ( R = R_{text{right}} + R_S = L_{text{left}} + L_S ).Hence, ( L = R ).So, the number of horizontal dominoes with a white square on the left equals the number with a white square on the right.I think that makes sense. By considering the symmetry of the board and how the dominoes cover the squares, we can conclude that the counts must be equal.

💡First, I recognize that to convert radians to degrees, I need to use the conversion factor where (1 text{ radian} = left( frac{180}{pi} right)^circ).Next, I apply this conversion factor to the given angle of (frac{8pi}{5}) radians. This involves multiplying (frac{8pi}{5}) by (frac{180}{pi}) to convert it into degrees.As I perform the multiplication, I notice that the (pi) terms in the numerator and denominator cancel each other out, simplifying the calculation.Finally, I carry out the division to find the numerical value of the angle in degrees, which gives me (288^circ).

💡Alright, so I have this equation to solve: (frac{x-3}{x^2+3x+2} = frac{x-7}{x^2-7x+12}). Hmm, it's a rational equation, meaning it has fractions with variables in the denominators. I remember that to solve such equations, a common strategy is to eliminate the denominators by multiplying both sides by the least common denominator (LCD). First, I need to factor the denominators to see if they can be simplified or if there are any restrictions on the values of (x). Let me factor each denominator:The first denominator is (x^2 + 3x + 2). To factor this, I look for two numbers that multiply to 2 and add up to 3. That would be 1 and 2. So, (x^2 + 3x + 2 = (x + 1)(x + 2)).The second denominator is (x^2 - 7x + 12). I need two numbers that multiply to 12 and add up to -7. Those numbers are -3 and -4. So, (x^2 - 7x + 12 = (x - 3)(x - 4)).Now, I can rewrite the original equation with the factored denominators:[frac{x - 3}{(x + 1)(x + 2)} = frac{x - 7}{(x - 3)(x - 4)}]Before proceeding, I should note the restrictions on (x). The denominators cannot be zero, so:- (x + 1 neq 0 Rightarrow x neq -1)- (x + 2 neq 0 Rightarrow x neq -2)- (x - 3 neq 0 Rightarrow x neq 3)- (x - 4 neq 0 Rightarrow x neq 4)So, (x) cannot be -2, -1, 3, or 4. I'll keep that in mind in case any solutions come out to be these values, which would make them extraneous.Next, to eliminate the denominators, I'll multiply both sides of the equation by the LCD, which is ((x + 1)(x + 2)(x - 3)(x - 4)). This should clear the fractions:[(x - 3)(x - 3)(x - 4) = (x - 7)(x + 1)(x + 2)]Wait, hold on, let me make sure I did that correctly. When I multiply both sides by the LCD, each side will have the product of the denominators divided by their respective denominators, effectively canceling them. So, on the left side, multiplying by ((x + 1)(x + 2)(x - 3)(x - 4)) cancels out ((x + 1)(x + 2)), leaving ((x - 3)(x - 3)(x - 4)). Similarly, on the right side, it cancels out ((x - 3)(x - 4)), leaving ((x - 7)(x + 1)(x + 2)).Wait, actually, let me correct that. The left side after multiplying should be ((x - 3) times (x - 3)(x - 4)), and the right side should be ((x - 7) times (x + 1)(x + 2)). Hmm, no, that doesn't seem right. Let me think again.Actually, when I multiply both sides by the LCD, the left side becomes ((x - 3) times (x - 3)(x - 4)), and the right side becomes ((x - 7) times (x + 1)(x + 2)). Wait, no, that still doesn't seem correct. Let me try to write it step by step.Original equation:[frac{x - 3}{(x + 1)(x + 2)} = frac{x - 7}{(x - 3)(x - 4)}]Multiply both sides by ((x + 1)(x + 2)(x - 3)(x - 4)):Left side: ((x - 3) times (x - 3)(x - 4))Right side: ((x - 7) times (x + 1)(x + 2))Wait, no, that's not accurate. Let me clarify:When I multiply the left side by the LCD, the denominator ((x + 1)(x + 2)) cancels out, leaving ((x - 3) times (x - 3)(x - 4)). Similarly, on the right side, the denominator ((x - 3)(x - 4)) cancels out, leaving ((x - 7) times (x + 1)(x + 2)).So, the equation becomes:[(x - 3)(x - 3)(x - 4) = (x - 7)(x + 1)(x + 2)]Wait, that still seems a bit off. Let me think differently. Maybe I should cross-multiply instead, which is a simpler approach for proportions.Cross-multiplying gives:[(x - 3)(x - 3)(x - 4) = (x - 7)(x + 1)(x + 2)]Wait, no, cross-multiplying should be:[(x - 3)(x - 3)(x - 4) = (x - 7)(x + 1)(x + 2)]Wait, I'm getting confused here. Let me try again.Cross-multiplying the original equation:[(x - 3)(x - 3)(x - 4) = (x - 7)(x + 1)(x + 2)]Wait, that can't be right because the denominators are different. Let me correct this.Actually, cross-multiplying means multiplying the numerator of one side by the denominator of the other side. So, it should be:[(x - 3)(x - 3)(x - 4) = (x - 7)(x + 1)(x + 2)]Wait, no, that's not correct. Let me write it properly.Original equation:[frac{x - 3}{(x + 1)(x + 2)} = frac{x - 7}{(x - 3)(x - 4)}]Cross-multiplying gives:[(x - 3)(x - 3)(x - 4) = (x - 7)(x + 1)(x + 2)]Wait, that still seems incorrect. Let me think about it again.When cross-multiplying, it's numerator1 * denominator2 = numerator2 * denominator1.So, numerator1 is (x - 3), denominator2 is (x - 3)(x - 4).Numerator2 is (x - 7), denominator1 is (x + 1)(x + 2).So, cross-multiplying gives:[(x - 3)(x - 3)(x - 4) = (x - 7)(x + 1)(x + 2)]Wait, no, that's not right. It should be:[(x - 3) times (x - 3)(x - 4) = (x - 7) times (x + 1)(x + 2)]Wait, that's still the same as before. Maybe I'm overcomplicating it. Let me just proceed with expanding both sides.So, expanding the left side:[(x - 3)(x - 3)(x - 4)]First, multiply (x - 3)(x - 3):[(x - 3)^2 = x^2 - 6x + 9]Then multiply by (x - 4):[(x^2 - 6x + 9)(x - 4) = x^3 - 4x^2 - 6x^2 + 24x + 9x - 36]Combine like terms:[x^3 - 10x^2 + 33x - 36]Now, expanding the right side:[(x - 7)(x + 1)(x + 2)]First, multiply (x + 1)(x + 2):[x^2 + 3x + 2]Then multiply by (x - 7):[(x^2 + 3x + 2)(x - 7) = x^3 - 7x^2 + 3x^2 - 21x + 2x - 14]Combine like terms:[x^3 - 4x^2 - 19x - 14]Now, set both expanded sides equal to each other:[x^3 - 10x^2 + 33x - 36 = x^3 - 4x^2 - 19x - 14]Subtract (x^3) from both sides:[-10x^2 + 33x - 36 = -4x^2 - 19x - 14]Bring all terms to the left side:[-10x^2 + 33x - 36 + 4x^2 + 19x + 14 = 0]Combine like terms:[(-10x^2 + 4x^2) + (33x + 19x) + (-36 + 14) = 0][-6x^2 + 52x - 22 = 0]Multiply both sides by -1 to make the leading coefficient positive:[6x^2 - 52x + 22 = 0]Now, I have a quadratic equation: (6x^2 - 52x + 22 = 0). To find the sum of all real solutions, I can use Vieta's formulas, which state that for a quadratic equation (ax^2 + bx + c = 0), the sum of the roots is (-b/a).Here, (a = 6), (b = -52), so the sum of the roots is (-(-52)/6 = 52/6 = 26/3).But wait, I should check if both roots are real. The discriminant of the quadratic is (b^2 - 4ac = (-52)^2 - 4*6*22 = 2704 - 528 = 2176), which is positive, so there are two real roots. Therefore, the sum of all real solutions is indeed 26/3.I should also verify that these solutions don't make any denominators zero in the original equation. The solutions are from the quadratic, so they shouldn't be -2, -1, 3, or 4. Since the quadratic solutions are 26/3 divided by 2, which is 13/3, and the other solution is (26/3 - 13/3) = 13/3, wait, no, that's not correct. Wait, Vieta's formula gives the sum as 26/3, but the actual roots can be found using the quadratic formula.Let me find the roots:Quadratic equation: (6x^2 - 52x + 22 = 0)Using quadratic formula:[x = frac{52 pm sqrt{(-52)^2 - 4*6*22}}{2*6} = frac{52 pm sqrt{2704 - 528}}{12} = frac{52 pm sqrt{2176}}{12}]Simplify (sqrt{2176}):2176 ÷ 16 = 136, so (sqrt{2176} = 4sqrt{136})136 ÷ 4 = 34, so (sqrt{136} = 2sqrt{34})Thus, (sqrt{2176} = 4*2sqrt{34} = 8sqrt{34})So, the roots are:[x = frac{52 pm 8sqrt{34}}{12} = frac{13 pm 2sqrt{34}}{3}]These are two real numbers, and neither of them is equal to -2, -1, 3, or 4, so they are valid solutions.Therefore, the sum of all real solutions is indeed 26/3.