answer：Okay, so I have this algebra problem here, and I need to figure it out step by step. Let me read it again to make sure I understand what's being asked.The problem says: For the algebraic expression in terms of ( x ), ( (ax - 3)(2x + 1) - 4x^2 + m ), after simplification, it does not contain the terms ( x^2 ) and constant term. Given that ( an + mn = 1 ), where ( an + mn = 1 ), find the value of ( 2n^3 - 9n^2 + 8n ).Alright, let's break this down. First, I need to simplify the given expression, and then use the conditions provided to find the values of ( a ), ( m ), and ( n ). Finally, I'll plug ( n ) into the polynomial ( 2n^3 - 9n^2 + 8n ) to get the answer.Starting with the expression ( (ax - 3)(2x + 1) - 4x^2 + m ). I think the first step is to expand this expression so I can see the coefficients of each term.Let me expand ( (ax - 3)(2x + 1) ). Using the distributive property (FOIL method):First, multiply ( ax ) by ( 2x ): ( ax * 2x = 2a x^2 ).Next, multiply ( ax ) by ( 1 ): ( ax * 1 = ax ).Then, multiply ( -3 ) by ( 2x ): ( -3 * 2x = -6x ).Finally, multiply ( -3 ) by ( 1 ): ( -3 * 1 = -3 ).So, putting it all together, the expanded form is:( 2a x^2 + ax - 6x - 3 ).Now, subtract ( 4x^2 ) and add ( m ):( 2a x^2 + ax - 6x - 3 - 4x^2 + m ).Let me combine like terms. The ( x^2 ) terms are ( 2a x^2 ) and ( -4x^2 ). The ( x ) terms are ( ax ) and ( -6x ). The constant terms are ( -3 ) and ( m ).So, combining the ( x^2 ) terms:( (2a - 4) x^2 ).Combining the ( x ) terms:( (a - 6) x ).Combining the constants:( (-3 + m) ).So, the simplified expression is:( (2a - 4)x^2 + (a - 6)x + (m - 3) ).Now, the problem states that after simplification, the expression does not contain the terms ( x^2 ) and the constant term. That means the coefficients of ( x^2 ) and the constant term must be zero.So, setting the coefficient of ( x^2 ) to zero:( 2a - 4 = 0 ).And setting the constant term to zero:( m - 3 = 0 ).Let me solve these equations to find ( a ) and ( m ).Starting with ( 2a - 4 = 0 ):Add 4 to both sides:( 2a = 4 ).Divide both sides by 2:( a = 2 ).Now, solving ( m - 3 = 0 ):Add 3 to both sides:( m = 3 ).Great, so ( a = 2 ) and ( m = 3 ).Now, moving on to the next part of the problem: ( an + mn = 1 ).Given that ( a = 2 ) and ( m = 3 ), let's substitute these values into the equation.So, substituting ( a = 2 ) and ( m = 3 ):( 2n + 3n = 1 ).Combine like terms:( (2 + 3)n = 1 ).( 5n = 1 ).Solving for ( n ):Divide both sides by 5:( n = frac{1}{5} ).Alright, so ( n = frac{1}{5} ).Now, the problem asks for the value of ( 2n^3 - 9n^2 + 8n ).Let me compute this step by step. First, I'll calculate each term separately.Compute ( n^3 ):( n = frac{1}{5} ), so ( n^3 = left(frac{1}{5}right)^3 = frac{1}{125} ).Multiply by 2:( 2n^3 = 2 * frac{1}{125} = frac{2}{125} ).Next, compute ( n^2 ):( n^2 = left(frac{1}{5}right)^2 = frac{1}{25} ).Multiply by 9:( 9n^2 = 9 * frac{1}{25} = frac{9}{25} ).But since the term is ( -9n^2 ), it becomes ( -frac{9}{25} ).Lastly, compute ( 8n ):( 8n = 8 * frac{1}{5} = frac{8}{5} ).Now, let's add all these terms together:( 2n^3 - 9n^2 + 8n = frac{2}{125} - frac{9}{25} + frac{8}{5} ).To add these fractions, they need a common denominator. The denominators are 125, 25, and 5. The least common denominator (LCD) is 125.Convert each term to have a denominator of 125:( frac{2}{125} ) remains the same.( -frac{9}{25} = -frac{9 * 5}{25 * 5} = -frac{45}{125} ).( frac{8}{5} = frac{8 * 25}{5 * 25} = frac{200}{125} ).Now, add them together:( frac{2}{125} - frac{45}{125} + frac{200}{125} ).Combine the numerators:( 2 - 45 + 200 = 157 ).So, the sum is ( frac{157}{125} ).Therefore, the value of ( 2n^3 - 9n^2 + 8n ) is ( frac{157}{125} ).Let me double-check my calculations to make sure I didn't make any mistakes.First, expanding ( (ax - 3)(2x + 1) ) gave me ( 2a x^2 + ax - 6x - 3 ). Subtracting ( 4x^2 ) and adding ( m ) gives ( (2a - 4)x^2 + (a - 6)x + (m - 3) ). Setting the coefficients of ( x^2 ) and the constant term to zero gives ( a = 2 ) and ( m = 3 ). Then, substituting into ( an + mn = 1 ) gives ( 5n = 1 ), so ( n = frac{1}{5} ). Plugging ( n = frac{1}{5} ) into ( 2n^3 - 9n^2 + 8n ) gives ( frac{2}{125} - frac{45}{125} + frac{200}{125} = frac{157}{125} ). That seems correct.I think I've covered all the steps and didn't skip anything important. So, I'm confident that the final answer is ( frac{157}{125} ).**Final Answer**The value of ( 2n^3 - 9n^2 + 8n ) is boxed{dfrac{157}{125}}.

answer：First, I need to compute the value of the complex number ( Z = -1 + (1 - i)^2 ).I'll start by expanding the squared term:[(1 - i)^2 = 1^2 - 2 times 1 times i + i^2 = 1 - 2i + (-1) = -2i]Now, substitute this back into the original expression:[Z = -1 + (-2i) = -1 - 2i]The complex number ( Z = -1 - 2i ) corresponds to the point ((-1, -2)) in the complex plane.Next, I'll determine the quadrant of this point. In the complex plane:- The first quadrant has both coordinates positive.- The second quadrant has a negative x-coordinate and a positive y-coordinate.- The third quadrant has both coordinates negative.- The fourth quadrant has a positive x-coordinate and a negative y-coordinate.Since both coordinates of the point ((-1, -2)) are negative, it lies in the third quadrant.Therefore, the correct answer is (boxed{C}).

answer：Okay, so I have this problem here about a function f(x) that's differentiable everywhere on the real numbers. The problem states that the derivative of f(x), which is f'(x), is less than f(x) itself. Also, it says that f(x + 2) equals f(x - 2), which probably means the function is periodic with some period. And then, they give me that f(4) is equal to 1. I need to figure out where f(x) is less than e^x, and the options are different intervals.First, let me try to parse this information step by step. The function f is differentiable everywhere, so it's smooth, no sharp corners or anything. The derivative f'(x) is less than f(x). That seems like a differential inequality. Maybe I can relate this to some exponential function because the derivative of e^x is e^x, so maybe f(x) is somehow related to e^x but growing slower?Then, the function satisfies f(x + 2) = f(x - 2). Hmm, that seems like a periodicity condition. If I rearrange that, f(x + 2) = f(x - 2) implies that if I shift x by 4, the function repeats. Because if I let y = x + 2, then f(y) = f(y - 4). So, f(y) = f(y - 4), which means the function is periodic with period 4. So, f(x + 4) = f(x) for all x. That's a key point.Given that, and knowing that f(4) = 1, I can probably find other values of f(x) using the periodicity. For example, f(0) should be equal to f(4), right? Because f(0 + 4) = f(0). So, f(0) = f(4) = 1. Similarly, f(8) would also be 1, and so on. Also, f(-4) would be 1, because f(-4 + 4) = f(0) = 1.Now, the inequality I need to solve is f(x) < e^x. So, I need to find all real numbers x where f(x) is less than e^x. The options given are intervals starting from different points and going to infinity. So, maybe I need to figure out where f(x) crosses e^x or something.Given that f'(x) < f(x), maybe I can use this to analyze the behavior of f(x). Let me think about differential equations. If I have f'(x) = f(x), that's the exponential function. But here, f'(x) is less than f(x), so maybe f(x) grows slower than e^x? Or maybe decays faster?Wait, actually, if f'(x) < f(x), that could mean different things depending on the sign of f(x). If f(x) is positive, then f'(x) is less than f(x), so it's growing slower than e^x. If f(x) is negative, then f'(x) is less than f(x), which is negative, so it's decreasing faster than e^x. Hmm, but I don't know if f(x) is positive or negative.But given that f(4) = 1, which is positive, and the function is periodic with period 4, so f(0) = 1, which is also positive. So maybe f(x) is always positive? Or at least, it's positive at integer multiples of 4.Wait, but without knowing more about f(x), it's hard to say. Maybe I can define a new function to make this easier. Let me try defining g(x) = f(x)/e^x. Then, maybe I can analyze g(x) instead.So, if g(x) = f(x)/e^x, then the derivative of g(x) would be g'(x) = [f'(x)e^x - f(x)e^x]/(e^{2x}) = [f'(x) - f(x)]/e^x. Since f'(x) < f(x), that means f'(x) - f(x) is negative. So, g'(x) is negative, which means g(x) is decreasing everywhere.Okay, so g(x) is decreasing on the real line. That's a useful property. So, if I can find some value of g(x), I can figure out where it's less than 1 or something.Given that f(4) = 1, so g(4) = f(4)/e^4 = 1/e^4. Similarly, f(0) = 1, so g(0) = 1/e^0 = 1. So, at x = 0, g(x) is 1, and at x = 4, it's 1/e^4, which is a much smaller number.Since g(x) is decreasing, that means as x increases, g(x) decreases. So, for x > 0, g(x) is less than g(0) = 1. So, f(x)/e^x < 1, which implies f(x) < e^x for x > 0.Wait, but hold on. The function is periodic with period 4, right? So, does that affect the behavior of g(x)? Because g(x) is f(x)/e^x, and f(x) repeats every 4, but e^x doesn't repeat. So, g(x) isn't periodic, but it's decreasing everywhere.So, for x > 0, g(x) is less than 1, meaning f(x) < e^x. But what about for x < 0? Since g(x) is decreasing, for x < 0, g(x) would be greater than g(0) = 1, meaning f(x)/e^x > 1, so f(x) > e^x for x < 0.But wait, let me check. If g(x) is decreasing, then as x increases, g(x) decreases. So, for x > 0, moving to the right, g(x) goes down, so it's less than 1. For x < 0, moving to the left, g(x) increases, so it's greater than 1. Therefore, f(x) < e^x only when x > 0.But hold on, the function f(x) is periodic with period 4, so f(x + 4) = f(x). But e^x is not periodic, it's increasing everywhere. So, does that mean that f(x) is always less than e^x for x > 0? Because for x > 0, e^x is increasing, and f(x) is periodic, so it's oscillating between some values, but since f(x)/e^x is decreasing, f(x) must be less than e^x for x > 0.But let me think about specific points. At x = 0, f(0) = 1, and e^0 = 1, so f(0) = e^0. So, at x = 0, they are equal. Then, for x > 0, since g(x) is decreasing, f(x)/e^x < 1, so f(x) < e^x. For x < 0, g(x) is increasing, so f(x)/e^x > 1, meaning f(x) > e^x.Therefore, the inequality f(x) < e^x holds for x > 0. So, the solution set is (0, +∞), which is option A.Wait, but let me double-check. The function f(x) is periodic with period 4, so f(4) = f(0) = 1. So, at x = 4, f(4) = 1, and e^4 is a much larger number, so f(4) < e^4. Similarly, at x = 8, f(8) = 1, and e^8 is even larger. So, as x increases, e^x grows exponentially, while f(x) just oscillates between some fixed values because it's periodic. So, after x = 0, f(x) is always less than e^x.But what about between x = 0 and x = 4? Since f(x) is periodic, it's going to repeat its values every 4 units. So, between 0 and 4, f(x) goes from 1 back to 1, but depending on its behavior, it might dip below or go above e^x. But wait, since g(x) is decreasing, f(x)/e^x is decreasing. So, at x = 0, it's 1, and as x increases, it decreases. So, for x > 0, it's always less than 1, meaning f(x) < e^x.Therefore, the solution set is all x > 0, which is (0, +∞). So, the answer should be option A.But wait, let me think again. The function f(x) is periodic with period 4, so it's going to have the same behavior every 4 units. So, for example, at x = -4, f(-4) = 1, and e^{-4} is a small number, so f(-4) = 1 > e^{-4}. Similarly, at x = -2, f(-2) = f(2), which is some value, but e^{-2} is still less than 1, so f(-2) is likely greater than e^{-2}.But the key point is that for x > 0, f(x) is less than e^x, and for x < 0, f(x) is greater than e^x. So, the inequality f(x) < e^x holds for x > 0.Therefore, the solution set is (0, +∞), which is option A.

answer：First, I need to determine the ratio of A's investment to the total investment. A invested Rs. 6300, and the total investment is the sum of A's, B's, and C's investments.Next, I'll set up a proportion based on the profit shares. A's share of the profit is Rs. 3900 out of the total profit of Rs. 13000. This proportion should be equal to the ratio of A's investment to the total investment.By setting up the equation ( frac{3900}{13000} = frac{6300}{6300 + 4200 + x} ), I can solve for ( x ), which represents C's investment.Cross-multiplying and simplifying the equation will allow me to isolate ( x ) and find its value. After performing the calculations, I find that C invested Rs. 10500 in the business.

answer：First, I need to calculate the annual return on investment for both Emma and Briana based on their respective investment amounts and interest rates.For Emma:- Investment Amount: 300- Annual Interest Rate: 15%- Annual ROI = 15% of 300 = 0.15 * 300 = 45For Briana:- Investment Amount: 500- Annual Interest Rate: 10%- Annual ROI = 10% of 500 = 0.10 * 500 = 50Next, I calculate the total return on investment after 2 years by multiplying the annual ROI by 2.Emma's ROI after 2 years = 45 * 2 = 90Briana's ROI after 2 years = 50 * 2 = 100Finally, I find the difference between their returns after 2 years.Difference = Briana's ROI - Emma's ROI = 100 - 90 = 10

answer：Okay, so I have this algebra problem here: (3x + 2) - 2(2x - 1). And I need to remove the parentheses and simplify it. The options are A, B, C, and D. Let me try to work through this step by step.First, I remember that when you have parentheses like this, you need to distribute any numbers outside the parentheses to the terms inside. So, looking at the first part, (3x + 2), there's nothing outside it, so I can just keep that as it is. But then there's a minus sign followed by 2 times (2x - 1). That means I need to distribute that -2 to both terms inside the second parentheses.Alright, so let's break it down. The original expression is (3x + 2) - 2(2x - 1). I'll start by distributing the -2 to both 2x and -1 inside the second parentheses. So, -2 times 2x would be -4x, right? Because -2 multiplied by 2 is -4, and then times x is -4x. Then, -2 times -1 is... hmm, negative times negative is positive, so that should be +2. Wait, let me double-check that. If I have -2 times 2x, that's -4x, and -2 times -1 is +2. Yeah, that seems correct.So now, after distributing the -2, the expression becomes (3x + 2) - 4x + 2. But I still have the first parentheses, (3x + 2), which I can just remove since there's no operation outside of it that affects it. So, removing the parentheses, the expression is 3x + 2 - 4x + 2.Now, I can combine like terms. Let's see, the terms with x are 3x and -4x. Combining those gives me -x. Then, the constant terms are 2 and 2, which add up to 4. So, putting it all together, the simplified expression is -x + 4.But wait, the question only asks for removing the parentheses, not necessarily combining like terms. Let me check the options again to see which one matches before combining like terms.Looking at the options:A: 3x + 2 - 2x + 1B: 3x + 2 - 4x + 1C: 3x + 2 - 4x - 2D: 3x + 2 - 4x + 2So, after distributing, I had 3x + 2 - 4x + 2, which matches option D. Just to make sure I didn't make any mistakes, let me go through it again. Starting with (3x + 2) - 2(2x - 1). Distribute the -2: -2 * 2x is -4x, and -2 * -1 is +2. So, it becomes 3x + 2 - 4x + 2. Yep, that's definitely option D.I think I got it right. So, the correct answer should be D.