💡Okay, so I have this inequality: (m - 2)x^{|m - 1|} - 3 > 6. And I'm told that this is a linear inequality in x. Hmm, I need to find the value of m. Alright, let's break this down step by step.First, what does it mean for an inequality to be linear in x? Well, a linear inequality in x would mean that the highest power of x is 1. So, the exponent of x should be 1. In this case, the exponent is |m - 1|. So, for the inequality to be linear, |m - 1| must equal 1. That makes sense.So, let's set up the equation: |m - 1| = 1. Now, solving absolute value equations usually involves considering both the positive and negative scenarios. So, m - 1 could be equal to 1 or m - 1 could be equal to -1.Let's solve both cases:Case 1: m - 1 = 1If I add 1 to both sides, I get m = 2.Case 2: m - 1 = -1If I add 1 to both sides here, I get m = 0.So, the possible values for m are 2 and 0. But wait, I need to check if these values make sense in the original inequality.Looking back at the original inequality: (m - 2)x^{|m - 1|} - 3 > 6. If m = 2, then the coefficient of x becomes (2 - 2) which is 0. So, the inequality would simplify to 0*x^{|2 - 1|} - 3 > 6, which is just -3 > 6. That's not true, and also, if the coefficient of x is 0, then it's not really an inequality in x anymore—it's just a false statement. So, m = 2 doesn't work because it makes the coefficient of x zero, which isn't allowed for a linear inequality.Therefore, m = 2 is not a valid solution. That leaves us with m = 0. Let's check that.If m = 0, then the exponent |m - 1| becomes |0 - 1| = 1, which is good because we need the exponent to be 1 for it to be linear. The coefficient of x becomes (0 - 2) = -2. So, the inequality becomes -2x - 3 > 6. That's a valid linear inequality in x because we have x raised to the first power and the coefficient isn't zero.So, m = 0 is the valid solution. Therefore, the value of m is 0.Wait, let me just make sure I didn't miss anything. The key points were that the exponent has to be 1, which gave me m = 2 or m = 0, but m = 2 makes the coefficient zero, which isn't allowed. So, m = 0 is the only valid solution. Yeah, that seems right.I think I've covered all the steps and checked both possible solutions. So, I'm confident that m is 0.

💡Okay, so I have this problem here: If (x + y) squared equals 36 and xy equals 8, what is the value of x squared plus y squared? Hmm, let me think about how to approach this.First, I remember that when you square a binomial like (x + y), it expands to x squared plus 2xy plus y squared. So, (x + y)^2 is equal to x^2 + 2xy + y^2. That makes sense. So, if (x + y)^2 is 36, then x^2 + 2xy + y^2 must also be 36.But wait, the problem also gives me that xy is 8. Maybe I can use that to find x^2 + y^2. Let me write down what I know:1. (x + y)^2 = 362. xy = 8From the first equation, expanding it gives me x^2 + 2xy + y^2 = 36. Since I know that xy is 8, I can substitute that into the equation. So, replacing 2xy with 2 times 8, which is 16, I get:x^2 + 16 + y^2 = 36Now, I need to find x^2 + y^2. So, if I subtract 16 from both sides of the equation, I should be able to isolate x^2 + y^2. Let me do that:x^2 + y^2 = 36 - 16Calculating that, 36 minus 16 is 20. So, x^2 + y^2 equals 20. That seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.1. I started with the given equation (x + y)^2 = 36.2. Expanded it to x^2 + 2xy + y^2 = 36.3. Substituted xy = 8 into the equation, resulting in x^2 + 16 + y^2 = 36.4. Subtracted 16 from both sides to get x^2 + y^2 = 20.Everything seems to add up. I don't think I made any calculation errors. So, I'm pretty confident that the value of x^2 + y^2 is 20.Just to visualize it another way, maybe I can think of it as solving for x and y individually, but that might be more complicated. Since the problem only asks for x^2 + y^2, using the expansion of (x + y)^2 is the most efficient method. It avoids having to solve for x and y separately, which would involve more steps and potentially more complex calculations.Another way to look at it is by recognizing that x^2 + y^2 is a common expression in algebra, often related to the squares of sums and differences. Knowing that (x + y)^2 includes the cross term 2xy allows us to manipulate the equation to solve for x^2 + y^2 directly, which is exactly what I did.I think I've covered all the bases here. The key was remembering how to expand the square of a binomial and then using substitution to find the desired expression. It's a good reminder of how algebraic identities can simplify problems that might otherwise seem more difficult.Just to recap:- Start with (x + y)^2 = 36.- Expand to x^2 + 2xy + y^2 = 36.- Substitute xy = 8, so 2xy = 16.- Subtract 16 from both sides to get x^2 + y^2 = 20.Yep, that's solid. I don't see any issues with this approach. It's always reassuring when the steps are clear and lead directly to the answer without any unnecessary complexity. I think I'm good here!

💡Okay, so I have this problem about average family income per capita in rural areas from 2018 to 2022. The data is given in a table with year codes and corresponding incomes. I need to do two things: first, determine the strength of the relationship between the year code (x) and the average income (y) using the correlation coefficient r. Second, find the linear regression equation to predict the average income for 2023.Alright, let's start with part (1). I remember that the correlation coefficient r measures how strong the linear relationship between two variables is. The formula for r is given, and I need to calculate it step by step. First, I need to find the means of x and y. The year codes are 1, 2, 3, 4, 5, so the mean of x, which I'll call x̄, is (1 + 2 + 3 + 4 + 5)/5. Let me compute that: 1+2 is 3, plus 3 is 6, plus 4 is 10, plus 5 is 15. So 15 divided by 5 is 3. So x̄ is 3.For y, the average incomes are 1.2, 1.4, 1.5, 1.6, 1.8. So ȳ is (1.2 + 1.4 + 1.5 + 1.6 + 1.8)/5. Let me add those up: 1.2 + 1.4 is 2.6, plus 1.5 is 4.1, plus 1.6 is 5.7, plus 1.8 is 7.5. So 7.5 divided by 5 is 1.5. So ȳ is 1.5.Next, I need to compute the numerator of r, which is the sum of (x_i * y_i) minus n times x̄ times ȳ. Let me compute the sum of x_i * y_i first.So, for each year:- 2018: x=1, y=1.2, so 1*1.2=1.2- 2019: x=2, y=1.4, so 2*1.4=2.8- 2020: x=3, y=1.5, so 3*1.5=4.5- 2021: x=4, y=1.6, so 4*1.6=6.4- 2022: x=5, y=1.8, so 5*1.8=9.0Adding these up: 1.2 + 2.8 is 4.0, plus 4.5 is 8.5, plus 6.4 is 14.9, plus 9.0 is 23.9. So the sum of x_i * y_i is 23.9.Now, n is 5, since there are 5 years. So n times x̄ times ȳ is 5 * 3 * 1.5. Let's compute that: 5*3 is 15, times 1.5 is 22.5.So the numerator is 23.9 - 22.5, which is 1.4.Now, the denominator of r is the square root of [sum of (x_i - x̄)^2] times [sum of (y_i - ȳ)^2]. Let's compute each part.First, sum of (x_i - x̄)^2. x̄ is 3, so:- For x=1: (1-3)^2 = (-2)^2 = 4- For x=2: (2-3)^2 = (-1)^2 = 1- For x=3: (3-3)^2 = 0- For x=4: (4-3)^2 = 1- For x=5: (5-3)^2 = 4Adding these up: 4 + 1 is 5, plus 0 is 5, plus 1 is 6, plus 4 is 10. So sum of (x_i - x̄)^2 is 10.Next, sum of (y_i - ȳ)^2. ȳ is 1.5, so:- For y=1.2: (1.2 - 1.5)^2 = (-0.3)^2 = 0.09- For y=1.4: (1.4 - 1.5)^2 = (-0.1)^2 = 0.01- For y=1.5: (1.5 - 1.5)^2 = 0- For y=1.6: (1.6 - 1.5)^2 = 0.1^2 = 0.01- For y=1.8: (1.8 - 1.5)^2 = 0.3^2 = 0.09Adding these up: 0.09 + 0.01 is 0.10, plus 0 is 0.10, plus 0.01 is 0.11, plus 0.09 is 0.20. So sum of (y_i - ȳ)^2 is 0.20.Now, the denominator is sqrt(10 * 0.20). Let's compute that. 10 * 0.20 is 2. So sqrt(2) is approximately 1.414.So putting it all together, r is numerator divided by denominator: 1.4 / 1.414. Let me compute that. 1.4 divided by 1.414 is approximately 0.99.So r is approximately 0.99. Since 0.99 is greater than 0.75, the relationship between y and x is strong.Okay, that was part (1). Now, part (2) is to find the linear regression equation y = b x + a, and predict the income for 2023.I remember the formulas for b and a. The slope b is [sum(x_i y_i) - n x̄ ȳ] divided by [sum(x_i^2) - n x̄^2]. And a is ȳ - b x̄.We already have some of these values from part (1). Let's recall:sum(x_i y_i) is 23.9, n is 5, x̄ is 3, ȳ is 1.5. So the numerator for b is 23.9 - 5*3*1.5, which we already calculated as 1.4.Now, we need sum(x_i^2). Let's compute that:- For x=1: 1^2 = 1- For x=2: 2^2 = 4- For x=3: 3^2 = 9- For x=4: 4^2 = 16- For x=5: 5^2 = 25Adding these up: 1 + 4 is 5, plus 9 is 14, plus 16 is 30, plus 25 is 55. So sum(x_i^2) is 55.Now, the denominator for b is sum(x_i^2) - n x̄^2. That's 55 - 5*(3)^2. 3 squared is 9, times 5 is 45. So 55 - 45 is 10.So b is 1.4 / 10, which is 0.14.Now, a is ȳ - b x̄. ȳ is 1.5, b is 0.14, x̄ is 3. So a is 1.5 - 0.14*3. Let's compute that: 0.14*3 is 0.42. So 1.5 - 0.42 is 1.08.So the regression equation is y = 0.14 x + 1.08.Now, to predict the income for 2023. The year code for 2023 would be 6, since 2018 is 1, so 2023 is 6.Plugging x=6 into the equation: y = 0.14*6 + 1.08. Let's compute that: 0.14*6 is 0.84. So 0.84 + 1.08 is 1.92.So the predicted average income for 2023 is 1.92 (in units of 10,000 yuan).Wait, let me double-check my calculations to make sure I didn't make any mistakes.For part (1), the correlation coefficient r was 0.99, which is strong. That seems correct because the data shows a clear increasing trend each year.For part (2), the regression equation: b was 0.14, a was 1.08, so y = 0.14x + 1.08. Plugging in x=6 gives 1.92. That seems reasonable given the trend.I think I did everything correctly. So, the final answers are that the relationship is strong, and the predicted income for 2023 is 1.92.

💡Alright, so I've got this problem here about fixed points and sequences. Let me try to unpack it step by step. First, the problem defines a fixed point of a function ( f(x) ) as a solution to the equation ( f(x) = x ). That makes sense; it's a point where the function's output equals the input. Cool.Next, it gives the function ( f(x) = frac{a x}{x + 1} ) and mentions that it has a unique fixed point. Hmm, okay. So, I need to find the value of ( a ) such that the equation ( f(x) = x ) has exactly one solution. Let me write that equation out:[frac{a x}{x + 1} = x]To solve for ( x ), I can multiply both sides by ( x + 1 ) to eliminate the denominator:[a x = x (x + 1)]Expanding the right side:[a x = x^2 + x]Let me bring all terms to one side to form a quadratic equation:[x^2 + x - a x = 0]Simplify the terms:[x^2 + (1 - a) x = 0]Factor out an ( x ):[x (x + (1 - a)) = 0]So, the solutions are ( x = 0 ) and ( x = a - 1 ). Wait, the problem says there's a unique fixed point. That means this quadratic equation should have only one solution. But quadratics usually have two solutions unless the discriminant is zero. Let me check the discriminant.The quadratic equation is ( x^2 + (1 - a) x = 0 ). The discriminant ( D ) is:[D = (1 - a)^2 - 4 times 1 times 0 = (1 - a)^2]For the quadratic to have a unique solution, the discriminant must be zero:[(1 - a)^2 = 0 implies 1 - a = 0 implies a = 1]So, ( a = 1 ). That means the function simplifies to:[f(x) = frac{x}{x + 1}]Alright, moving on to the sequence part. The sequence ( {a_n} ) is defined with ( a_1 = 1 ), and the recursive relation:[frac{1}{a_{n+1}} = fleft( frac{1}{a_n} right)]Let me substitute ( f(x) ) into this equation:[frac{1}{a_{n+1}} = frac{frac{1}{a_n}}{frac{1}{a_n} + 1}]Simplify the denominator:[frac{1}{a_{n+1}} = frac{frac{1}{a_n}}{frac{1 + a_n}{a_n}} = frac{1}{a_n} times frac{a_n}{1 + a_n} = frac{1}{1 + a_n}]So, we have:[frac{1}{a_{n+1}} = frac{1}{1 + a_n}]Taking reciprocals on both sides:[a_{n+1} = 1 + a_n]Oh, this is a simple recursive relation. It tells me that each term is just the previous term plus 1. That means the sequence is an arithmetic sequence with a common difference of 1.Given that ( a_1 = 1 ), the sequence would be:[a_1 = 1, quad a_2 = 2, quad a_3 = 3, quad dots]So, in general, the ( n )-th term is:[a_n = a_1 + (n - 1) times 1 = 1 + n - 1 = n]Therefore, ( a_n = n ) for all ( n ).To find ( a_{2019} ), I just substitute ( n = 2019 ):[a_{2019} = 2019]Let me double-check my steps to make sure I didn't make a mistake. Starting from the fixed point, solving for ( a ), getting ( a = 1 ), then substituting into the recursive relation, simplifying, and finding that each term increases by 1. It all seems to add up. So, yeah, I'm confident that ( a_{2019} ) is 2019.**Final Answer**boxed{2019}

💡Okay, so I have this equation to solve: (1/2)^(3x + 6) equals 8^(x + 2). Hmm, I remember that when dealing with exponential equations, it's helpful to express both sides with the same base if possible. Let me think about how to do that here.First, I know that 1/2 is the same as 2 to the power of -1, right? So, I can rewrite (1/2)^(3x + 6) as (2^-1)^(3x + 6). Using the exponent rule (a^m)^n = a^(m*n), that becomes 2^(-1*(3x + 6)). Let me calculate that exponent: -1 times 3x is -3x, and -1 times 6 is -6. So, the left side simplifies to 2^(-3x - 6).Now, looking at the right side, 8 is 2 cubed, so 8^(x + 2) can be written as (2^3)^(x + 2). Again, using the same exponent rule, that becomes 2^(3*(x + 2)). Calculating the exponent: 3 times x is 3x, and 3 times 2 is 6. So, the right side simplifies to 2^(3x + 6).Now the equation is 2^(-3x - 6) = 2^(3x + 6). Since the bases are the same, I can set the exponents equal to each other. That gives me -3x - 6 = 3x + 6.To solve for x, I'll move all the x terms to one side and the constants to the other. Let me add 3x to both sides to get rid of the -3x on the left. That gives me -6 = 6x + 6. Now, I'll subtract 6 from both sides to isolate the term with x. So, -6 - 6 equals 6x, which simplifies to -12 = 6x.Finally, I'll divide both sides by 6 to solve for x. That gives me x = -12/6, which simplifies to x = -2.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the beginning, expressing both sides with base 2 seems correct. Then, simplifying the exponents step by step, that looks good too. Setting the exponents equal because the bases are the same is a standard method, so that's right. Then solving the linear equation: -3x -6 = 3x +6. Adding 3x to both sides gives -6 = 6x +6. Subtracting 6 from both sides gives -12 = 6x, and dividing by 6 gives x = -2. Yeah, that all checks out. I think I got it right.

💡Okay, so I have this problem here: "For how many three-digit whole numbers does the sum of the digits equal 25?" The answer choices are A) 2, B) 4, C) 6, D) 8, E) 10. Hmm, I need to figure out how many three-digit numbers have digits that add up to 25. Let me think about how to approach this.First, let's understand what a three-digit number is. It has a hundreds place, a tens place, and a units place. So, if I denote the digits as ( a ), ( b ), and ( c ), the number is ( 100a + 10b + c ). The sum of the digits is ( a + b + c ), and we want this sum to be 25.Now, since it's a three-digit number, ( a ) can't be zero. It has to be at least 1 and at most 9. On the other hand, ( b ) and ( c ) can be from 0 to 9. So, ( a ) is in the range 1-9, and ( b ) and ( c ) are in the range 0-9.We need to find all possible combinations of ( a ), ( b ), and ( c ) such that ( a + b + c = 25 ). Let me think about how to count these combinations.One approach is to consider that since the maximum sum of three digits (each being 9) is 27, 25 is just 2 less than that. So, we can think of this as distributing the "deficit" of 2 across the three digits. However, since ( a ) has to be at least 1, we need to adjust for that.Alternatively, maybe it's easier to think about how many ways we can have ( a + b + c = 25 ) with the constraints on each digit. Let me try to fix one digit at a time and see how the others can vary.Let's start by considering the maximum possible value for each digit. Since each digit can be at most 9, let's see how many 9s we can have in the number.If all three digits are 9, the sum is 27, which is too high. So, we need to reduce the sum by 2. That means we need to decrease one or more digits by a total of 2.But we have to make sure that after decreasing, all digits are still within their valid ranges. For example, if we decrease ( a ) by 2, it would go from 9 to 7, which is still valid since ( a ) can be as low as 1. Similarly, decreasing ( b ) or ( c ) by 2 is fine as they can be 0.So, how many ways can we distribute this deficit of 2 across the three digits? This is similar to the stars and bars problem in combinatorics, where we want to distribute indistinguishable items (the deficit) into distinguishable bins (the digits).However, in this case, each digit has a maximum limit (9), so we have to ensure that we don't exceed that. But since we're only decreasing by 2, and starting from 9s, we don't have to worry about exceeding the maximum because we're only reducing.Wait, actually, since we're starting from 9s and reducing, the only constraint is that each digit must remain non-negative. But since we're only reducing by 2, and each digit is initially 9, reducing by 2 will still leave each digit at least 7, which is fine.So, the number of ways to distribute a deficit of 2 across three digits is the same as the number of non-negative integer solutions to ( x + y + z = 2 ), where ( x ), ( y ), and ( z ) are the amounts we decrease each digit by. The formula for this is ( binom{2 + 3 - 1}{3 - 1} = binom{4}{2} = 6 ).But wait, does this account for the fact that ( a ) must be at least 1? Because if we decrease ( a ) by 2, it becomes 7, which is still fine. If we decrease ( a ) by 1, it becomes 8, and if we don't decrease ( a ), it remains 9. So, all these cases are valid.Therefore, there are 6 ways to distribute the deficit of 2 across the three digits. Each of these corresponds to a unique three-digit number where the sum of the digits is 25.But let me double-check this because sometimes these combinatorial problems can have edge cases. Let me list out all possible combinations to make sure.If we have three digits ( a ), ( b ), and ( c ), each starting at 9. We need to reduce the total by 2. So, the possible reductions are:1. Reduce ( a ) by 2: ( a = 7 ), ( b = 9 ), ( c = 9 ). The number is 799.2. Reduce ( a ) by 1 and ( b ) by 1: ( a = 8 ), ( b = 8 ), ( c = 9 ). The number is 889.3. Reduce ( a ) by 1 and ( c ) by 1: ( a = 8 ), ( b = 9 ), ( c = 8 ). The number is 898.4. Reduce ( b ) by 2: ( a = 9 ), ( b = 7 ), ( c = 9 ). The number is 979.5. Reduce ( b ) by 1 and ( c ) by 1: ( a = 9 ), ( b = 8 ), ( c = 8 ). The number is 988.6. Reduce ( c ) by 2: ( a = 9 ), ( b = 9 ), ( c = 7 ). The number is 997.So, listing them out, we have 799, 889, 898, 979, 988, and 997. That's 6 numbers in total. Each of these has digits that add up to 25.Wait, but let me make sure that these are all unique numbers. 799, 889, 898, 979, 988, 997 – yes, each of these is distinct. So, that's 6 numbers.Is there any other way to get a sum of 25 without starting from three 9s? Let me think. Suppose we don't have all three digits as 9. For example, maybe two digits are 9 and the third is 7, but that's already covered in the cases above. Or maybe one digit is 9 and the other two add up to 16. Wait, that's another way to think about it.Let me try this approach. Let's fix one digit as 9 and see what the other two digits need to add up to.If ( a = 9 ), then ( b + c = 16 ). Since ( b ) and ( c ) are digits, they can be at most 9. So, what pairs of digits add up to 16?- ( b = 7 ), ( c = 9 )- ( b = 8 ), ( c = 8 )- ( b = 9 ), ( c = 7 )So, these are the possible pairs. Now, for each pair, how many three-digit numbers can we form?For ( b = 7 ), ( c = 9 ), the number is 979.For ( b = 8 ), ( c = 8 ), the number is 988.For ( b = 9 ), ( c = 7 ), the number is 997.So, that's 3 numbers where ( a = 9 ).Similarly, if ( b = 9 ), then ( a + c = 16 ). Let's see:- ( a = 7 ), ( c = 9 ) → number is 799- ( a = 8 ), ( c = 8 ) → number is 889- ( a = 9 ), ( c = 7 ) → number is 997 (but we already counted this when ( a = 9 ))Wait, so if ( b = 9 ), we get 799, 889, and 997. But 997 was already counted when ( a = 9 ). So, we have to be careful not to double-count.Similarly, if ( c = 9 ), then ( a + b = 16 ):- ( a = 7 ), ( b = 9 ) → number is 799- ( a = 8 ), ( b = 8 ) → number is 889- ( a = 9 ), ( b = 7 ) → number is 979Again, these are the same numbers as before, just approached from a different digit.So, in total, we have 6 unique numbers: 799, 889, 898, 979, 988, 997.Wait, but when I fixed ( a = 9 ), I got 979, 988, 997. When I fixed ( b = 9 ), I got 799, 889, 997. When I fixed ( c = 9 ), I got 799, 889, 979. So, combining all these, I have 799, 889, 898, 979, 988, 997. That's 6 numbers.So, both approaches give me 6 numbers. Therefore, the answer should be 6, which is option C.But just to be thorough, let me check if there are any other combinations where the digits don't include two 9s and a 7 or an 8.Suppose we have one 9 and the other two digits adding up to 16. Wait, that's what I just did. Alternatively, could we have digits like 8, 9, 8? That's 8 + 9 + 8 = 25, which is 889, which I already counted.Or 7, 9, 9, which is 799, also counted. Or 9, 7, 9, which is 979, and 9, 9, 7, which is 997. So, all these are covered.Is there a way to get 25 without having at least two 9s? Let's see. If we have only one 9, then the other two digits need to add up to 16. But as we saw, the only way to get 16 with two digits is 7 and 9, 8 and 8, or 9 and 7. So, that still requires at least one more 9 or an 8.Wait, if we have only one 9, then the other two digits must add up to 16, which as we saw requires at least one more 9 or an 8. So, actually, all numbers with digits summing to 25 must have at least two 9s or one 9 and two 8s.Wait, let me think. If I have two 8s and one 9, that's 8 + 8 + 9 = 25. So, that's another case. So, in addition to the numbers with two 9s and one 7, we also have numbers with one 9 and two 8s.So, let's see. For two 8s and one 9, the numbers would be:- 889- 898- 988These are the same numbers I got earlier when I fixed ( a = 9 ) and got 988, or fixed ( b = 9 ) and got 889, etc.So, in total, we have numbers with two 9s and one 7, and numbers with one 9 and two 8s. Each of these gives us 3 numbers, but some overlap when considering different positions.Wait, no, actually, when I fixed ( a = 9 ), I got 979, 988, 997. When I fixed ( b = 9 ), I got 799, 889, 997. When I fixed ( c = 9 ), I got 799, 889, 979. So, combining all these, I have 799, 889, 898, 979, 988, 997. That's 6 numbers in total.So, whether I think of it as two 9s and one 7 or one 9 and two 8s, I end up with the same set of numbers, just approached from different angles.Therefore, I'm confident that the total number of three-digit numbers where the sum of the digits is 25 is 6.**Final Answer**boxed{C}