💡Okay, so I have this problem here: If ( i ) is the imaginary unit, then ( left| frac{5 + 3i}{4 - i} right| ) equals what? Hmm, let me think about how to approach this.First, I remember that when dealing with complex numbers, especially when they're in the denominator, it's helpful to rationalize the denominator. That usually involves multiplying both the numerator and the denominator by the complex conjugate of the denominator. The complex conjugate of ( 4 - i ) is ( 4 + i ), right? So maybe I should multiply both the top and bottom by ( 4 + i ).Let me write that down:[frac{5 + 3i}{4 - i} times frac{4 + i}{4 + i}]Okay, so now I need to multiply out the numerators and the denominators. Let me start with the numerator:( (5 + 3i)(4 + i) ). I'll use the distributive property (FOIL) here.First, multiply 5 by 4: that's 20.Then, 5 times i: that's ( 5i ).Next, 3i times 4: that's ( 12i ).And finally, 3i times i: that's ( 3i^2 ).So putting it all together:20 + 5i + 12i + 3i².Wait, I know that ( i^2 = -1 ), so ( 3i^2 = 3(-1) = -3 ).So substituting that back in:20 + 5i + 12i - 3.Combine like terms:20 - 3 is 17, and 5i + 12i is 17i.So the numerator simplifies to ( 17 + 17i ).Now, let's work on the denominator:( (4 - i)(4 + i) ). This is a difference of squares, so it should be ( 4^2 - (i)^2 ).Calculating that:16 - ( i^2 ). Since ( i^2 = -1 ), this becomes 16 - (-1) = 16 + 1 = 17.So the denominator is 17.Putting it all together, the fraction becomes:[frac{17 + 17i}{17}]I can factor out 17 in the numerator:[frac{17(1 + i)}{17}]The 17s cancel out, leaving me with ( 1 + i ).So now, I have ( frac{5 + 3i}{4 - i} = 1 + i ).But the question asks for the modulus of this complex number. The modulus of a complex number ( a + bi ) is ( sqrt{a^2 + b^2} ).So for ( 1 + i ), ( a = 1 ) and ( b = 1 ).Calculating the modulus:[sqrt{1^2 + 1^2} = sqrt{1 + 1} = sqrt{2}]Therefore, the modulus of ( frac{5 + 3i}{4 - i} ) is ( sqrt{2} ).Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I multiplied numerator and denominator by the conjugate ( 4 + i ). That seems right.2. Expanded the numerator: ( (5 + 3i)(4 + i) ). I got 20 + 5i + 12i + 3i², which simplifies to 17 + 17i. That looks correct.3. Expanded the denominator: ( (4 - i)(4 + i) = 16 - i² = 17 ). That's correct.4. Simplified the fraction to ( 1 + i ). That makes sense.5. Calculated the modulus as ( sqrt{2} ). Yep, that's right.I think I did everything correctly. So the answer should be ( sqrt{2} ).**Final Answer**The modulus of ( frac{5 + 3i}{4 - i} ) is boxed{sqrt{2}}.

💡Alright, so I've got this math problem here with four options, and I need to figure out which one is correct. Let's take it step by step. First, let's look at Option A: (2a + 3b = 5ab). Hmm, okay, so this is saying that if you have two terms, (2a) and (3b), and you add them together, it equals (5ab). That doesn't seem right to me. I remember that when you add terms, they need to be like terms, meaning they have the same variables raised to the same powers. Here, (2a) and (3b) are not like terms because one has an (a) and the other has a (b). So, you can't combine them into a single term like (5ab). That must mean Option A is incorrect.Moving on to Option B: (2ab^{2} div b = 2b). Okay, so this is dividing (2ab^{2}) by (b). Let's break that down. When you divide terms with exponents, you subtract the exponents of the like bases. So, (b^{2} div b) would be (b^{2-1} = b). Then, the coefficient 2 remains the same. So, (2ab^{2} div b) should equal (2ab), not (2b). That means Option B is also incorrect.Next is Option C: (2a^{2} cdot 3a^{2} = 6a^{2}). Alright, multiplying two terms here. When you multiply terms with exponents, you add the exponents of the like bases. So, (a^{2} cdot a^{2} = a^{2+2} = a^{4}). Then, multiplying the coefficients: (2 cdot 3 = 6). So, the product should be (6a^{4}), not (6a^{2}). Therefore, Option C is incorrect as well.Finally, Option D: ((3ab)^{2} = 9a^{2}b^{2}). Let's see. When you raise a product to a power, you raise each factor to that power. So, (3^{2} = 9), (a^{2}) remains (a^{2}), and (b^{2}) remains (b^{2}). Multiplying those together gives (9a^{2}b^{2}), which matches the right side of the equation. So, Option D seems correct.Wait, let me double-check to make sure I didn't make any mistakes. For Option A, adding (2a) and (3b) definitely doesn't give (5ab) because they're not like terms. For Option B, dividing (2ab^{2}) by (b) gives (2ab), not (2b). For Option C, multiplying (2a^{2}) and (3a^{2}) gives (6a^{4}), not (6a^{2}). And for Option D, squaring (3ab) does indeed give (9a^{2}b^{2}). Yep, I think I got it right. So, the correct answer should be Option D.

💡Okay, so I have this problem about the function f(x) = x ln x. It has two parts. Let me try to tackle them one by one.Starting with part (1): We need to find the intervals where the function g(x) = f(x) - a(x - 1) is monotonic. Monotonic means the function is either entirely non-increasing or non-decreasing over an interval. To find this, I remember that I should look at the derivative of g(x). If the derivative is always positive or always negative on an interval, then the function is monotonic there.So, first, let's write out g(x). Since f(x) is x ln x, then g(x) = x ln x - a(x - 1). Now, I need to find g'(x). The derivative of x ln x is ln x + 1, right? Because using the product rule: derivative of x is 1 times ln x plus x times derivative of ln x, which is 1/x. So that's ln x + 1. Then, the derivative of -a(x - 1) is just -a. So putting it all together, g'(x) = ln x + 1 - a.Now, to find where g(x) is increasing or decreasing, I need to analyze where g'(x) is positive or negative. So, let's set g'(x) equal to zero to find critical points. That gives ln x + 1 - a = 0. Solving for x, we get ln x = a - 1, so x = e^{a - 1}. This is the critical point.Now, to determine the intervals, I need to consider the behavior of g'(x) around x = e^{a - 1}. Since ln x is an increasing function, as x increases, ln x increases. So, for x < e^{a - 1}, ln x < a - 1, which means g'(x) = ln x + 1 - a < 0. Therefore, g(x) is decreasing on (0, e^{a - 1}).For x > e^{a - 1}, ln x > a - 1, so g'(x) = ln x + 1 - a > 0. Therefore, g(x) is increasing on (e^{a - 1}, ∞).So, summarizing, g(x) is decreasing on (0, e^{a - 1}) and increasing on (e^{a - 1}, ∞). That should answer part (1).Moving on to part (2): We need to find the equation of a line l that passes through the point (0, -1) and is tangent to the curve y = f(x) = x ln x. First, let's recall that the equation of a tangent line at a point (x0, y0) on the curve y = f(x) is given by y = f'(x0)(x - x0) + f(x0). So, we need to find the point (x0, y0) where the tangent line passes through (0, -1).Let me denote the point of tangency as (x0, y0). Then, y0 = f(x0) = x0 ln x0. The derivative f'(x) is ln x + 1, so the slope of the tangent line at x0 is m = f'(x0) = ln x0 + 1.So, the equation of the tangent line is y = (ln x0 + 1)(x - x0) + x0 ln x0. Simplifying this, it becomes y = (ln x0 + 1)x - x0(ln x0 + 1) + x0 ln x0. The last two terms simplify: -x0 ln x0 - x0 + x0 ln x0 = -x0. So, the equation is y = (ln x0 + 1)x - x0.Now, since this line passes through (0, -1), substituting x = 0 and y = -1 into the equation gives -1 = (ln x0 + 1)(0) - x0, which simplifies to -1 = -x0. Therefore, x0 = 1.So, substituting x0 = 1 back into y0, we get y0 = 1 * ln 1 = 0. So, the point of tangency is (1, 0). The slope m = ln 1 + 1 = 0 + 1 = 1. Therefore, the equation of the tangent line is y = 1*(x - 1) + 0, which simplifies to y = x - 1.Wait, let me double-check that. If x0 = 1, then the tangent line is y = (ln 1 + 1)(x - 1) + 1*ln 1. Since ln 1 is 0, this becomes y = (0 + 1)(x - 1) + 0, which is y = x - 1. And plugging in (0, -1) into this equation: y = 0 - 1 = -1, which works. So, that seems correct.So, part (2) gives the equation of the tangent line as y = x - 1.I think that covers both parts. Let me just recap:1. For g(x) = x ln x - a(x - 1), the derivative is ln x + 1 - a. Setting this equal to zero gives the critical point at x = e^{a - 1}. The function is decreasing before this point and increasing after, so the intervals are (0, e^{a - 1}) decreasing and (e^{a - 1}, ∞) increasing.2. For the tangent line passing through (0, -1), we found that the point of tangency is at x0 = 1, giving the tangent line equation y = x - 1.Everything seems to check out. I don't see any mistakes in my reasoning.**Final Answer**(1) The function ( g(x) ) is decreasing on the interval (boxed{(0, e^{a-1})}) and increasing on the interval (boxed{(e^{a-1}, +infty)}).(2) The equation of the tangent line ( l ) is (boxed{y = x - 1}).

💡Okay, so Xiaoming has some 2-cent and 5-cent coins in his piggy bank, and he thinks there's about five to six yuan in total. He divides these coins into two piles with equal amounts of money. In the first pile, the number of 2-cent coins is equal to the number of 5-cent coins. In the second pile, the total value of the 2-cent coins is equal to the total value of the 5-cent coins. I need to figure out how much money Xiaoming saved in total.First, let's understand the problem. Xiaoming has two types of coins: 2-cent and 5-cent. He estimates the total amount is between five and six yuan. Since one yuan is 100 cents, that means the total amount is between 500 and 600 cents.He divides these coins into two piles with equal amounts of money. So each pile has half of the total amount. That means each pile is between 250 and 300 cents.In the first pile, the number of 2-cent coins is equal to the number of 5-cent coins. Let's denote the number of 2-cent coins in the first pile as x. Therefore, the number of 5-cent coins in the first pile is also x.The total value of the first pile would be 2x + 5x = 7x cents.In the second pile, the total value of the 2-cent coins is equal to the total value of the 5-cent coins. Let's denote the number of 2-cent coins in the second pile as y and the number of 5-cent coins as z. The total value of the 2-cent coins in the second pile is 2y cents, and the total value of the 5-cent coins is 5z cents. According to the problem, 2y = 5z.Also, the total value of the second pile is 2y + 5z. Since 2y = 5z, we can substitute 5z for 2y, so the total value becomes 5z + 5z = 10z cents.But we also know that the total value of the second pile must be equal to the total value of the first pile because Xiaoming divided the coins into two piles with equal amounts of money. So, 7x = 10z.Now, let's summarize the equations we have:1. 2y = 5z2. 7x = 10zFrom equation 1, we can express y in terms of z: y = (5/2)z.From equation 2, we can express x in terms of z: x = (10/7)z.Since the number of coins must be whole numbers, z must be a multiple of 2 to make y an integer, and z must be a multiple of 7 to make x an integer. Therefore, z must be a multiple of 14.Let's denote z = 14k, where k is a positive integer.Then, y = (5/2)*14k = 35kAnd x = (10/7)*14k = 20kNow, let's calculate the total amount of money Xiaoming saved.Total number of 2-cent coins: x + y = 20k + 35k = 55kTotal number of 5-cent coins: x + z = 20k + 14k = 34kTotal value in cents: 2*(55k) + 5*(34k) = 110k + 170k = 280kWe know that the total amount is between 500 and 600 cents, so:500 ≤ 280k ≤ 600Divide all parts by 280:500/280 ≤ k ≤ 600/280Simplify:5/2.8 ≤ k ≤ 6/2.8Which is approximately:1.7857 ≤ k ≤ 2.1429Since k must be an integer, the only possible value is k = 2.Therefore, k = 2.Now, let's calculate the total value:280k = 280*2 = 560 cents, which is 5.6 yuan.So, Xiaoming saved a total of 5.6 yuan.

💡Okay, so I have this problem where I need to find all natural numbers ( n ) for which there exists a set of positive numbers ( x_1, x_2, ldots, x_n ) that satisfy the system of equations:[left{begin{array}{l}x_1 + x_2 + ldots + x_n = 9 frac{1}{x_1} + frac{1}{x_2} + ldots + frac{1}{x_n} = 1end{array}right.]And I also need to indicate all such sets for each found value of ( n ).Alright, let's start by understanding what this system is saying. We have a sum of numbers equal to 9 and the sum of their reciprocals equal to 1. So, it's like balancing the total sum and the total of their reciprocals.First, I think about inequalities that relate sums and reciprocals. The Cauchy-Schwarz inequality comes to mind, but maybe more directly, the AM-HM inequality. The Arithmetic Mean (AM) and Harmonic Mean (HM) inequality states that for positive real numbers,[frac{x_1 + x_2 + ldots + x_n}{n} geq frac{n}{frac{1}{x_1} + frac{1}{x_2} + ldots + frac{1}{x_n}}]with equality if and only if all ( x_i ) are equal.Let me plug in the given values into this inequality. The AM is ( frac{9}{n} ) and the HM is ( frac{n}{1} = n ). So,[frac{9}{n} geq n]Multiplying both sides by ( n ) (since ( n ) is positive),[9 geq n^2]Which implies that ( n leq 3 ). So, ( n ) can be 1, 2, or 3.But wait, let me check each case individually because sometimes inequalities can give a range, but specific cases might not hold.Starting with ( n = 1 ):If ( n = 1 ), then we have only one variable ( x_1 ). The system becomes:[x_1 = 9][frac{1}{x_1} = 1]But ( frac{1}{9} ) is not equal to 1, so ( n = 1 ) is not possible.Moving on to ( n = 2 ):We have two variables ( x_1 ) and ( x_2 ). The system is:[x_1 + x_2 = 9][frac{1}{x_1} + frac{1}{x_2} = 1]Let me denote ( x_1 = a ) and ( x_2 = b ) for simplicity.So,[a + b = 9][frac{1}{a} + frac{1}{b} = 1]I can express ( b ) in terms of ( a ): ( b = 9 - a ). Then substitute into the second equation:[frac{1}{a} + frac{1}{9 - a} = 1]To solve this, let's find a common denominator:[frac{9 - a + a}{a(9 - a)} = 1][frac{9}{a(9 - a)} = 1]Multiply both sides by ( a(9 - a) ):[9 = a(9 - a)][9 = 9a - a^2][a^2 - 9a + 9 = 0]This is a quadratic equation. Let's compute the discriminant:[D = 81 - 36 = 45]So,[a = frac{9 pm sqrt{45}}{2} = frac{9 pm 3sqrt{5}}{2}]Thus, the solutions are:[a = frac{9 + 3sqrt{5}}{2}, quad b = frac{9 - 3sqrt{5}}{2}]or[a = frac{9 - 3sqrt{5}}{2}, quad b = frac{9 + 3sqrt{5}}{2}]Since both ( a ) and ( b ) are positive (as ( sqrt{5} approx 2.236 ), so ( 3sqrt{5} approx 6.708 ), making ( 9 - 3sqrt{5} approx 2.292 ), which is positive), ( n = 2 ) is a valid solution.Now, for ( n = 3 ):We have three variables ( x_1, x_2, x_3 ). The system is:[x_1 + x_2 + x_3 = 9][frac{1}{x_1} + frac{1}{x_2} + frac{1}{x_3} = 1]I wonder if all three variables can be equal. Let's assume ( x_1 = x_2 = x_3 = x ). Then,[3x = 9 implies x = 3][frac{1}{3} + frac{1}{3} + frac{1}{3} = 1]Yes, that works! So, ( x_1 = x_2 = x_3 = 3 ) is a solution. But are there other solutions where the variables are not all equal?Hmm, that's a good question. Maybe, but the problem only asks for the existence of such sets, not necessarily all possible sets. Since we have at least one solution, ( n = 3 ) is valid.What about ( n = 4 )? Wait, earlier I concluded ( n leq 3 ) based on the inequality. Let me verify that.Using the AM-HM inequality again:[frac{9}{n} geq frac{n}{1} implies 9 geq n^2 implies n leq 3]So, ( n = 4 ) would give ( 9 geq 16 ), which is false. Therefore, ( n ) cannot be 4 or higher.Just to be thorough, let me see if ( n = 4 ) is possible. Suppose we have four positive numbers adding up to 9 and their reciprocals adding up to 1.Let me try setting three of them equal and the fourth different.Let’s say ( x_1 = x_2 = x_3 = a ) and ( x_4 = b ). Then,[3a + b = 9 implies b = 9 - 3a][frac{3}{a} + frac{1}{b} = 1]Substituting ( b ):[frac{3}{a} + frac{1}{9 - 3a} = 1]Let me solve this equation for ( a ):Multiply both sides by ( a(9 - 3a) ):[3(9 - 3a) + a = a(9 - 3a)][27 - 9a + a = 9a - 3a^2][27 - 8a = 9a - 3a^2][3a^2 - 17a + 27 = 0]Compute the discriminant:[D = 289 - 324 = -35]Negative discriminant, so no real solutions. Therefore, ( n = 4 ) is not possible.Similarly, for ( n > 4 ), the situation would be even worse because the sum of reciprocals would have to be even smaller, but the sum of numbers is fixed at 9. So, higher ( n ) would require smaller individual numbers, making their reciprocals larger, which conflicts with the sum of reciprocals being 1.Therefore, only ( n = 2 ) and ( n = 3 ) are possible.To summarize:- For ( n = 2 ), the solutions are ( left( frac{9 + 3sqrt{5}}{2}, frac{9 - 3sqrt{5}}{2} right) ) and vice versa.- For ( n = 3 ), the solution is ( (3, 3, 3) ).I think that's all. I don't see any other possibilities, and the inequalities confirm that higher ( n ) aren't feasible.

💡First, I need to find the total cost of the three items before tax by adding their individual prices.Next, I'll calculate the tax amount by applying the 5% tax rate to the total pre-tax cost.Finally, I'll add the tax amount to the total pre-tax cost to determine the final total cost after tax.