answer：First, I need to evaluate each option to determine which operation is correct.**Option A:**The expression is (-2x + 5x). Combining the like terms:[-2x + 5x = ( -2 + 5 )x = 3x]This does not match the given result of (-7x), so option A is incorrect.**Option B:**The expression is (yx - 3xy). Since (yx) and (xy) are like terms (commutative property of multiplication), we can combine them:[yx - 3xy = (1 - 3)xy = -2xy]This matches the given result, so option B is correct.**Option C:**The expression is (-x^2 - x^2). Combining the like terms:[-x^2 - x^2 = (-1 - 1)x^2 = -2x^2]This does not match the given result of (0), so option C is incorrect.**Option D:**The expression is (x^2 - x). Since (x^2) and (x) are not like terms, they cannot be combined directly:[x^2 - x neq x]This does not match the given result, so option D is incorrect.After evaluating all options, only option B is correct.

answer：First, I need to determine the charge for each additional 1/5 of a mile beyond the initial 1/5 mile charge of 3.00.The total fare for an 8-mile ride is 18.60. To find out how much of this fare is attributed to the additional distance beyond the first 1/5 mile, I subtract the initial charge from the total fare:18.60 - 3.00 = 15.60.Next, I need to calculate how many additional 1/5 mile segments are in the 8-mile ride. Since 1 mile is equivalent to 5 segments of 1/5 mile, 8 miles would be:8 miles × 5 = 40 segments of 1/5 mile.However, the first 1/5 mile is already covered by the initial 3.00 charge, so I subtract one segment:40 segments - 1 segment = 39 segments.Now, I can find the charge per additional 1/5 mile by dividing the additional fare by the number of additional segments:15.60 ÷ 39 segments = 0.40 per 1/5 mile.

answer：Alright, so I have this problem where I need to assign digits from 1 to 8 to Chinese characters in three different phrases: "尽心尽力", "力可拔山", and "山穷水尽". Each character represents a unique digit, and identical characters represent the same digit. The sum of the digits in each phrase must be 19, and there's an additional condition that "尽" > "山" > "力". I need to find the maximum possible value of "水".First, I'll break down each phrase into its constituent characters and set up equations based on the sums.1. "尽心尽力" translates to 尽 + 心 + 尽 + 力 = 192. "力可拔山" translates to 力 + 可 + 拔 + 山 = 193. "山穷水尽" translates to 山 + 穷 + 水 + 尽 = 19So, I have three equations:1. 尽 + 心 + 尽 + 力 = 192. 力 + 可 + 拔 + 山 = 193. 山 + 穷 + 水 + 尽 = 19I can simplify the first equation by combining like terms:2尽 + 心 + 力 = 19Similarly, the second equation remains as is:力 + 可 + 拔 + 山 = 19And the third equation:山 + 穷 + 水 + 尽 = 19Now, if I add all three equations together, I get:(2尽 + 心 + 力) + (力 + 可 + 拔 + 山) + (山 + 穷 + 水 + 尽) = 19 + 19 + 19 = 57Simplifying the left side:2尽 + 心 + 力 + 力 + 可 + 拔 + 山 + 山 + 穷 + 水 + 尽 = 57Combining like terms:(2尽 + 尽) + 心 + (力 + 力) + 可 + 拔 + (山 + 山) + 穷 + 水 = 57Which simplifies to:3尽 + 心 + 2力 + 可 + 拔 + 2山 + 穷 + 水 = 57Now, I know that each character represents a unique digit from 1 to 8, and the sum of all digits from 1 to 8 is 36. So, if I sum all the characters in the three phrases, I get:3尽 + 心 + 2力 + 可 + 拔 + 2山 + 穷 + 水 = 57But the total sum of all unique characters is 36, so the difference between 57 and 36 is 21. This means that the sum of the duplicated characters is 21. The duplicated characters are 尽, 力, and 山, each appearing twice or more.So, 21 = 2尽 + 2力 + 2山 + 心 + 可 + 拔 + 穷 + 水Wait, that doesn't seem right. Let me think again.Actually, when I added all three equations, I had:3尽 + 心 + 2力 + 可 + 拔 + 2山 + 穷 + 水 = 57But the sum of all unique characters is 36, so:3尽 + 心 + 2力 + 可 + 拔 + 2山 + 穷 + 水 = 57But the sum of all unique characters is 36, so:3尽 + 心 + 2力 + 可 + 拔 + 2山 + 穷 + 水 = 57But 36 is the sum of all unique characters, so:3尽 + 心 + 2力 + 可 + 拔 + 2山 + 穷 + 水 = 57This implies that:(3尽 + 2力 + 2山) + (心 + 可 + 拔 + 穷 + 水) = 57But the sum of all unique characters is 36, so:3尽 + 2力 + 2山 = 57 - (心 + 可 + 拔 + 穷 + 水)But I'm getting confused here. Maybe I should approach it differently.Let me list all the characters:From "尽心尽力": 尽, 心, 力From "力可拔山": 力, 可, 拔, 山From "山穷水尽": 山, 穷, 水, 尽So, the unique characters are: 尽, 心, 力, 可, 拔, 山, 穷, 水That's 8 characters, each representing a unique digit from 1 to 8.Now, the sum of all digits from 1 to 8 is 36. So, if I sum all the characters in the three phrases, I get:From "尽心尽力": 2尽 + 心 + 力From "力可拔山": 力 + 可 + 拔 + 山From "山穷水尽": 山 + 穷 + 水 + 尽Total sum: 2尽 + 心 + 力 + 力 + 可 + 拔 + 山 + 山 + 穷 + 水 + 尽Which simplifies to:3尽 + 心 + 2力 + 可 + 拔 + 2山 + 穷 + 水 = 57But since the sum of all unique characters is 36, we have:3尽 + 2力 + 2山 = 57 - (心 + 可 + 拔 + 穷 + 水)But I'm not sure if this is helpful. Maybe I should consider the individual equations.From the first phrase: 2尽 + 心 + 力 = 19From the second phrase: 力 + 可 + 拔 + 山 = 19From the third phrase: 山 + 穷 + 水 + 尽 = 19I need to find the maximum value of "水". To maximize "水", I need to minimize the other variables in the third equation: 山 + 穷 + 水 + 尽 = 19So, to maximize "水", I need to minimize 山, 穷, and 尽.But there's a condition: "尽" > "山" > "力"So, 尽 > 山 > 力Also, all characters are unique digits from 1 to 8.Let me try to assign values step by step.First, let's consider the first equation: 2尽 + 心 + 力 = 19Since 尽 is a digit from 1 to 8, and 2尽 must be less than or equal to 16 (since 2*8=16). So, 2尽 + 心 + 力 = 19Therefore, 心 + 力 = 19 - 2尽Since 心 and 力 are digits from 1 to 8, their sum must be at least 2 (1+1) and at most 16 (8+8). But since they are unique, their sum is at least 3 (1+2) and at most 15 (7+8).So, 19 - 2尽 must be between 3 and 15.Therefore, 2尽 must be between 4 and 16.But since 尽 is a digit from 1 to 8, 2尽 ranges from 2 to 16.So, 19 - 2尽 must be between 3 and 15.Thus, 2尽 must be between 4 and 16, which is always true since 2尽 is at least 2.But more specifically, since 19 - 2尽 must be at least 3, 2尽 must be at most 16, which is already covered.But let's think about possible values for 尽.Since 尽 > 山 > 力, and all are unique digits from 1 to 8.Let me assume that 尽, 山, and 力 are consecutive numbers.Let's say:力 = x山 = x + 1尽 = x + 2This way, 尽 > 山 > 力.Now, substitute into the first equation:2尽 + 心 + 力 = 192(x + 2) + 心 + x = 192x + 4 + 心 + x = 193x + 心 + 4 = 193x + 心 = 15So, 心 = 15 - 3xSince 心 must be a positive integer between 1 and 8, and unique.Let's find possible x:x must be such that 15 - 3x ≥ 115 - 3x ≥ 1 → 3x ≤ 14 → x ≤ 4.666, so x ≤ 4Also, x must be at least 1.So, x can be 1, 2, 3, or 4.Let's test each:1. x = 1:力 = 1山 = 2尽 = 3心 = 15 - 3*1 = 12 → Not possible, since 心 must be ≤82. x = 2:力 = 2山 = 3尽 = 4心 = 15 - 6 = 9 → Not possible, 心 ≤83. x = 3:力 = 3山 = 4尽 = 5心 = 15 - 9 = 6Possible, 心 = 64. x = 4:力 = 4山 = 5尽 = 6心 = 15 - 12 = 3Possible, 心 = 3So, possible assignments:Either:力 = 3, 山 = 4, 尽 = 5, 心 = 6Or力 = 4, 山 = 5, 尽 = 6, 心 = 3Now, let's check which one allows us to maximize "水".First, let's try x = 4:力 = 4山 = 5尽 = 6心 = 3Now, let's look at the second equation: 力 + 可 + 拔 + 山 = 19Substitute 力 =4, 山=5:4 + 可 + 拔 +5 =19 → 可 + 拔 =10So, 可 + 拔 =10Possible pairs for 可 and 拔 (unique digits, not used yet: 1,2,7,8)Possible pairs:1+9=10 → 9 is not available2+8=107+3=10 → 3 is already used (心=3)So, only possible pair is 2 and8.Therefore, 可 and 拔 are 2 and8 in some order.Now, let's look at the third equation: 山 + 穷 + 水 + 尽 =19Substitute 山=5, 尽=6:5 + 穷 + 水 +6 =19 → 穷 + 水 =8We need to maximize "水", so we need to minimize "穷".The remaining digits are 1,7 (since used digits are 3,4,5,6, and 可 and 拔 are 2 and8)So, available digits: 1,7To minimize "穷", set 穷=1, then 水=7Thus, 水=7Now, let's check if all digits are unique:Used digits:力=4, 山=5, 尽=6, 心=3, 可=2, 拔=8, 穷=1, 水=7All unique, from 1 to8.So, this assignment works.Now, let's check the other possibility with x=3:力=3, 山=4, 尽=5, 心=6Second equation: 力 + 可 + 拔 + 山 =193 + 可 + 拔 +4=19 → 可 + 拔=12Possible pairs for 可 and 拔 (unique digits, not used yet:1,2,7,8)Possible pairs:4+8=12 → 4 is already used (山=4)5+7=12 →5 is used (尽=5)So, only possible pair is 5+7, but 5 is used.Wait, no, 5 is used, so 7+5 is not possible.Wait, 12 can be achieved by 4+8, but 4 is used.Alternatively, 3+9=12, but 9 is not available.So, no valid pairs for 可 + 拔=12 with available digits.Thus, this assignment is invalid.Therefore, the only valid assignment is when x=4:力=4, 山=5, 尽=6, 心=3, 可=2, 拔=8, 穷=1, 水=7Thus, the maximum value of "水" is7.

answer：Okay, so I have this problem where I need to find the number of ordered pairs (a, b) of integers such that |aω + b| = 2, where ω is a nonreal root of z³ = 1. Hmm, let me break this down step by step.First, I remember that the equation z³ = 1 has three roots: 1, ω, and ω², where ω is a primitive cube root of unity. Since ω is nonreal, it must satisfy ω² + ω + 1 = 0. That's a key equation I can use later.Now, the problem is about finding integer pairs (a, b) such that the modulus of aω + b is 2. To handle this, I think I should express ω in its standard form. I recall that ω can be written as (-1 + i√3)/2. So, ω = (-1 + i√3)/2.Let me substitute this into the expression aω + b. That would be a*(-1 + i√3)/2 + b. Let me simplify this:aω + b = (a*(-1)/2 + b) + i*(a√3)/2.So, the expression aω + b can be written as a complex number with real part (-a/2 + b) and imaginary part (a√3)/2.The modulus of this complex number is given by the square root of the sum of the squares of the real and imaginary parts. So,|aω + b| = sqrt[ (-a/2 + b)² + (a√3/2)² ].We are told that this modulus equals 2, so:sqrt[ (-a/2 + b)² + (a√3/2)² ] = 2.If I square both sides to eliminate the square root, I get:[ (-a/2 + b)² + (a√3/2)² ] = 4.Let me compute each term inside the brackets:First, (-a/2 + b)² = (b - a/2)² = b² - a b + (a²)/4.Second, (a√3/2)² = (3 a²)/4.Adding these together:b² - a b + (a²)/4 + (3 a²)/4 = b² - a b + (a²)/4 + (3 a²)/4.Simplify the a² terms:(a²)/4 + (3 a²)/4 = (4 a²)/4 = a².So, the equation becomes:b² - a b + a² = 4.So, now I have the equation a² - a b + b² = 4, and I need to find integer solutions (a, b) for this.Hmm, okay, so I need to find all integer pairs (a, b) such that a² - a b + b² = 4.This looks like a quadratic in two variables. Maybe I can approach this by considering different cases for a and solving for b, or vice versa.Let me think about the possible values of a. Since a and b are integers, and the equation is symmetric in a and b (if we swap a and b, the equation remains the same), maybe I can consider a range of a values and find corresponding b values.But before that, let me note that a² - a b + b² is always non-negative because it's the sum of squares minus a product term. So, 4 is a positive number, so we're looking for integer pairs where this expression equals 4.Let me consider possible integer values for a. Since a and b are integers, the terms a² and b² can't be too large without making the whole expression exceed 4. So, let's see what possible a's we can have.Let me suppose that a is between, say, -3 and 3. Because if a is 4, then a² is 16, which is already larger than 4, so the equation would not hold unless b is also large, but let's check.Wait, actually, let's compute a² - a b + b² for a = 0, ±1, ±2, ±3, and see if we can find integer b's that satisfy the equation.Starting with a = 0:If a = 0, then the equation becomes 0 - 0 + b² = 4, so b² = 4. Therefore, b = ±2. So, we have two solutions: (0, 2) and (0, -2).Next, a = 1:Plugging a = 1 into the equation: 1 - b + b² = 4.So, b² - b + 1 = 4 => b² - b - 3 = 0.Let me solve this quadratic equation for b: b = [1 ± sqrt(1 + 12)] / 2 = [1 ± sqrt(13)] / 2.But sqrt(13) is irrational, so there are no integer solutions for b when a = 1.Similarly, a = -1:Plugging a = -1: (-1)^2 - (-1)b + b² = 1 + b + b² = 4.So, b² + b + 1 = 4 => b² + b - 3 = 0.Solutions: b = [-1 ± sqrt(1 + 12)] / 2 = [-1 ± sqrt(13)] / 2. Again, irrational, so no integer solutions.Moving on to a = 2:Plugging a = 2: 4 - 2b + b² = 4.So, b² - 2b + 4 = 4 => b² - 2b = 0 => b(b - 2) = 0.Thus, b = 0 or b = 2. So, two solutions: (2, 0) and (2, 2).Similarly, a = -2:Plugging a = -2: (-2)^2 - (-2)b + b² = 4 + 2b + b² = 4.So, b² + 2b + 4 = 4 => b² + 2b = 0 => b(b + 2) = 0.Thus, b = 0 or b = -2. So, two solutions: (-2, 0) and (-2, -2).Now, a = 3:Plugging a = 3: 9 - 3b + b² = 4.So, b² - 3b + 5 = 0.Discriminant: 9 - 20 = -11, which is negative. So, no real solutions, hence no integer solutions.Similarly, a = -3:Plugging a = -3: 9 - (-3)b + b² = 9 + 3b + b² = 4.So, b² + 3b + 5 = 0.Discriminant: 9 - 20 = -11, again negative. No real solutions.So, beyond a = ±3, the values of a² would be too large, making the equation impossible to satisfy with integer b.Wait, but let me double-check for a = 3 and a = -3. Maybe I made a mistake in calculation.For a = 3:a² = 9, so 9 - 3b + b² = 4 => b² - 3b + 5 = 0. Discriminant is 9 - 20 = -11, correct. No solutions.Similarly, a = -3:a² = 9, so 9 + 3b + b² = 4 => b² + 3b + 5 = 0. Discriminant is 9 - 20 = -11. Correct, no solutions.So, the possible a's are -2, -1, 0, 1, 2. But from the above, only a = -2, 0, 2 give integer solutions for b.Wait, but when a = 1 and a = -1, we didn't get integer solutions, so those don't contribute.So, let me list all the solutions I found:For a = -2: b = 0, -2. So, (-2, 0) and (-2, -2).For a = 0: b = 2, -2. So, (0, 2) and (0, -2).For a = 2: b = 0, 2. So, (2, 0) and (2, 2).So, that's a total of 6 ordered pairs.Wait, but let me check if there are any other solutions I might have missed. Maybe for a = 1 or a = -1, even though the quadratic didn't yield integer solutions, perhaps I should check manually.For a = 1:We have the equation 1 - b + b² = 4 => b² - b - 3 = 0.Testing integer b's:b = 0: 0 - 0 - 3 = -3 ≠ 0.b = 1: 1 - 1 - 3 = -3 ≠ 0.b = 2: 4 - 2 - 3 = -1 ≠ 0.b = 3: 9 - 3 - 3 = 3 ≠ 0.b = -1: 1 + 1 - 3 = -1 ≠ 0.b = -2: 4 + 2 - 3 = 3 ≠ 0.So, no integer solutions for a = 1.Similarly, for a = -1:Equation: 1 + b + b² = 4 => b² + b - 3 = 0.Testing integer b's:b = 0: 0 + 0 - 3 = -3 ≠ 0.b = 1: 1 + 1 - 3 = -1 ≠ 0.b = 2: 4 + 2 - 3 = 3 ≠ 0.b = -1: 1 - 1 - 3 = -3 ≠ 0.b = -2: 4 - 2 - 3 = -1 ≠ 0.So, no solutions for a = -1.Therefore, the only solutions are the 6 ordered pairs I found earlier.Wait, but let me think again. The equation a² - a b + b² = 4 is symmetric in a and b, so for each solution (a, b), (b, a) is also a solution. But in our case, when a = 2, b = 0 or 2. So, (2, 0) and (0, 2) are both solutions, which we have included. Similarly, (2, 2) is symmetric, so it's the same when swapped.Similarly, (-2, 0) and (0, -2) are included, and (-2, -2) is symmetric.So, all symmetric solutions are accounted for.Wait, but let me check if there are any other solutions where a and b are both non-zero and different.For example, could a = 1 and b = 2 satisfy the equation?Let me compute a² - a b + b²: 1 - 2 + 4 = 3 ≠ 4.Similarly, a = 1, b = -1: 1 - (-1) + 1 = 1 + 1 + 1 = 3 ≠ 4.a = 2, b = 1: 4 - 2 + 1 = 3 ≠ 4.a = -1, b = 2: 1 - (-2) + 4 = 1 + 2 + 4 = 7 ≠ 4.a = -2, b = 1: 4 - (-2) + 1 = 4 + 2 + 1 = 7 ≠ 4.So, no, none of these give 4.Therefore, I think I have found all possible integer solutions.So, the ordered pairs are:(-2, 0), (-2, -2), (0, -2), (0, 2), (2, 0), (2, 2).That's six ordered pairs.Wait, but let me double-check each of these to ensure they satisfy the equation.For (-2, 0):a = -2, b = 0.Compute a² - a b + b²: (-2)^2 - (-2)(0) + 0^2 = 4 - 0 + 0 = 4. Correct.For (-2, -2):a = -2, b = -2.Compute: (-2)^2 - (-2)(-2) + (-2)^2 = 4 - 4 + 4 = 4. Correct.For (0, -2):a = 0, b = -2.Compute: 0 - 0 + (-2)^2 = 0 + 0 + 4 = 4. Correct.For (0, 2):a = 0, b = 2.Compute: 0 - 0 + 4 = 4. Correct.For (2, 0):a = 2, b = 0.Compute: 4 - 0 + 0 = 4. Correct.For (2, 2):a = 2, b = 2.Compute: 4 - 4 + 4 = 4. Correct.All six pairs satisfy the equation, so I think that's all of them.Therefore, the number of ordered pairs (a, b) is 6.

answer：Okay, so I have this problem where I need to find the minimum distance between a point P on the line l: √3 x - y + 2 = 0 and a point Q on the circle C: x² + y² + 2y = 0. Hmm, let me try to figure this out step by step.First, I remember that the distance between two points can be minimized by considering the distance from the line to the circle. Maybe I should start by understanding the given line and the circle better.Let me write down the equations again:Line l: √3 x - y + 2 = 0Circle C: x² + y² + 2y = 0I think it would help to rewrite the circle equation in standard form so I can easily identify its center and radius. To do that, I can complete the square for the y-terms.Starting with the circle equation:x² + y² + 2y = 0I can group the y terms:x² + (y² + 2y) = 0Now, to complete the square for y² + 2y, I take half of the coefficient of y, which is 1, square it to get 1, and add and subtract it inside the equation:x² + (y² + 2y + 1 - 1) = 0This simplifies to:x² + (y + 1)² - 1 = 0Adding 1 to both sides:x² + (y + 1)² = 1Okay, so now the circle is in standard form. That means the center of the circle is at (0, -1) and the radius is √1 = 1.Good, so the circle has center (0, -1) and radius 1.Now, I need to find the minimum distance between a point P on line l and a point Q on circle C. I think the minimum distance between the line and the circle would be the distance from the center of the circle to the line minus the radius of the circle.Wait, is that correct? Let me think. If the distance from the center to the line is greater than the radius, then the minimum distance between the line and the circle would be that distance minus the radius. If it's less than or equal to the radius, then the minimum distance would be zero because the line intersects or touches the circle.So, first, I need to find the distance from the center of the circle (0, -1) to the line l: √3 x - y + 2 = 0.I remember the formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is:Distance = |A x₀ + B y₀ + C| / √(A² + B²)Let me apply this formula.Given the line l: √3 x - y + 2 = 0, so A = √3, B = -1, C = 2.The center of the circle is (0, -1), so x₀ = 0, y₀ = -1.Plugging into the distance formula:Distance = |√3 * 0 + (-1) * (-1) + 2| / √( (√3)² + (-1)² )Calculating numerator:√3 * 0 = 0(-1) * (-1) = 1So, 0 + 1 + 2 = 3Absolute value of 3 is 3.Denominator:(√3)² = 3(-1)² = 1So, √(3 + 1) = √4 = 2Therefore, the distance from the center to the line is 3 / 2 = 1.5Hmm, so the distance from the center to the line is 1.5 units, and the radius of the circle is 1 unit.Since 1.5 > 1, the line does not intersect the circle, so the minimum distance between the line and the circle is 1.5 - 1 = 0.5 units.Therefore, the minimum distance between points P and Q is 0.5.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, the circle equation:x² + y² + 2y = 0Completing the square:x² + (y² + 2y + 1) = 1So, x² + (y + 1)² = 1Yes, center at (0, -1), radius 1.Distance from center (0, -1) to line √3 x - y + 2 = 0:|√3*0 - (-1) + 2| / √( (√3)^2 + (-1)^2 )Simplify numerator:0 + 1 + 2 = 3Denominator:√(3 + 1) = √4 = 2So, distance is 3/2 = 1.5Subtract radius 1: 1.5 - 1 = 0.5Yes, that seems correct.Alternatively, I can think of the minimum distance as the shortest distance between the line and any point on the circle. Since the circle is centered at (0, -1) with radius 1, the closest point on the circle to the line would be along the line connecting the center to the line l, and that distance would be 1.5 - 1 = 0.5.I think that makes sense. So, the minimum distance is 0.5 units.**Final Answer**The minimum distance between points ( P ) and ( Q ) is boxed{dfrac{1}{2}}.

answer：First, calculate the volume of one brick by multiplying its length, width, and height.Next, determine the volume of the wall by multiplying its length, width, and height. Since the height is given in meters, convert it to centimeters for consistency.Then, find the total volume of the bricks needed by multiplying the number of bricks by the volume of one brick.Finally, set the total volume of the bricks equal to the volume of the wall and solve for the unknown length of the wall.