answer：Alright, so I've got this system of equations to solve:[begin{cases}9^{2 cdot tan x + cos y} = 3 9^{cos y} - 81^{tan x} = 2end{cases}]Hmm, okay. Let me try to figure this out step by step. First, I notice that both equations involve powers of 9 and 81, which are both powers of 3. Maybe I can rewrite everything in terms of base 3 to make it simpler.Starting with the first equation: (9^{2 cdot tan x + cos y} = 3). Since 9 is (3^2), I can rewrite this as:[(3^2)^{2 cdot tan x + cos y} = 3]Which simplifies to:[3^{4 cdot tan x + 2 cdot cos y} = 3^1]Since the bases are the same, the exponents must be equal. So:[4 cdot tan x + 2 cdot cos y = 1]Okay, that's one equation. Let's keep that in mind.Now, moving on to the second equation: (9^{cos y} - 81^{tan x} = 2). Again, 9 is (3^2) and 81 is (3^4), so let's rewrite this:[(3^2)^{cos y} - (3^4)^{tan x} = 2]Which simplifies to:[3^{2 cdot cos y} - 3^{4 cdot tan x} = 2]Hmm, so now I have two equations:1. (4 cdot tan x + 2 cdot cos y = 1)2. (3^{2 cdot cos y} - 3^{4 cdot tan x} = 2)This looks a bit complicated, but maybe I can introduce some substitutions to make it easier. Let me let (a = 3^{2 cdot tan x}) and (b = 3^{cos y}). Then, let's see what these substitutions do to the equations.Looking at the second equation:[b^2 - a^2 = 2]Because (3^{2 cdot cos y} = (3^{cos y})^2 = b^2) and (3^{4 cdot tan x} = (3^{2 cdot tan x})^2 = a^2).So, that's nice. The second equation becomes (b^2 - a^2 = 2).Now, looking back at the first equation:[4 cdot tan x + 2 cdot cos y = 1]But since (a = 3^{2 cdot tan x}), taking the logarithm base 3 of both sides gives:[log_3 a = 2 cdot tan x implies tan x = frac{log_3 a}{2}]Similarly, since (b = 3^{cos y}), taking the logarithm base 3 gives:[log_3 b = cos y]So, substituting these into the first equation:[4 cdot left( frac{log_3 a}{2} right) + 2 cdot (log_3 b) = 1]Simplifying:[2 cdot log_3 a + 2 cdot log_3 b = 1]Divide both sides by 2:[log_3 a + log_3 b = frac{1}{2}]Using logarithm properties, this becomes:[log_3 (a cdot b) = frac{1}{2}]Which means:[a cdot b = 3^{1/2} = sqrt{3}]So now I have two equations involving (a) and (b):1. (b^2 - a^2 = 2)2. (a cdot b = sqrt{3})This looks like a system of equations that I can solve for (a) and (b). Let me write them down again:1. (b^2 - a^2 = 2)2. (a cdot b = sqrt{3})Maybe I can express one variable in terms of the other from the second equation and substitute into the first. Let's solve the second equation for (b):[b = frac{sqrt{3}}{a}]Now, substitute this into the first equation:[left( frac{sqrt{3}}{a} right)^2 - a^2 = 2]Simplify:[frac{3}{a^2} - a^2 = 2]Multiply both sides by (a^2) to eliminate the denominator:[3 - a^4 = 2a^2]Bring all terms to one side:[-a^4 - 2a^2 + 3 = 0]Multiply both sides by -1 to make it a bit nicer:[a^4 + 2a^2 - 3 = 0]Hmm, this is a quartic equation, but it looks like a quadratic in terms of (a^2). Let me set (u = a^2), so the equation becomes:[u^2 + 2u - 3 = 0]Now, solve for (u):Using the quadratic formula:[u = frac{-2 pm sqrt{(2)^2 - 4 cdot 1 cdot (-3)}}{2 cdot 1} = frac{-2 pm sqrt{4 + 12}}{2} = frac{-2 pm sqrt{16}}{2} = frac{-2 pm 4}{2}]So, two solutions:1. (u = frac{-2 + 4}{2} = frac{2}{2} = 1)2. (u = frac{-2 - 4}{2} = frac{-6}{2} = -3)But (u = a^2), and (a^2) can't be negative, so we discard (u = -3). Thus, (u = 1), which means (a^2 = 1), so (a = pm 1).But (a = 3^{2 cdot tan x}), and since (3^{2 cdot tan x}) is always positive, (a) must be positive. Therefore, (a = 1).Now, substitute (a = 1) back into the second equation (a cdot b = sqrt{3}):[1 cdot b = sqrt{3} implies b = sqrt{3}]So, (a = 1) and (b = sqrt{3}).Now, let's recall what (a) and (b) represent:- (a = 3^{2 cdot tan x} = 1)- (b = 3^{cos y} = sqrt{3})Starting with (a = 1):[3^{2 cdot tan x} = 1]Since (3^0 = 1), this implies:[2 cdot tan x = 0 implies tan x = 0]The solutions for (tan x = 0) are:[x = kpi quad text{for integer } k]Now, for (b = sqrt{3}):[3^{cos y} = sqrt{3} = 3^{1/2}]Since the bases are the same, the exponents must be equal:[cos y = frac{1}{2}]The solutions for (cos y = frac{1}{2}) are:[y = pm frac{pi}{3} + 2pi n quad text{for integer } n]So, putting it all together, the solutions are:[x = kpi quad text{and} quad y = pm frac{pi}{3} + 2pi n quad text{for integers } k, n]Let me double-check to make sure these solutions satisfy the original equations.First, let's check the first equation:[9^{2 cdot tan x + cos y} = 3]Given (x = kpi), (tan x = 0), and (y = pm frac{pi}{3} + 2pi n), so (cos y = frac{1}{2}).Substituting:[9^{2 cdot 0 + frac{1}{2}} = 9^{frac{1}{2}} = 3]Which is correct.Now, the second equation:[9^{cos y} - 81^{tan x} = 2]Again, (cos y = frac{1}{2}) and (tan x = 0):[9^{frac{1}{2}} - 81^{0} = 3 - 1 = 2]Which is also correct.So, the solutions seem to work.I think that's it! The solutions are all real numbers (x) and (y) such that (x) is an integer multiple of (pi) and (y) is (pm frac{pi}{3}) plus any multiple of (2pi).**Final Answer**The solutions are (boxed{x = kpi}) and (boxed{y = pm frac{pi}{3} + 2npi}) for integers (k) and (n).

answer：Okay, so I have this problem where there are two functions, f(x) = 2(x + 1) and g(x) = x + ln x. Points A and B are on the graphs of f(x) and g(x) respectively, and their y-coordinates are always equal. I need to find the minimum distance between points A and B.First, I need to understand what it means for the y-coordinates of A and B to be equal. That means if point A is (m, f(m)) and point B is (n, g(n)), then f(m) must equal g(n). So, 2(m + 1) = n + ln n. From this equation, I can express m in terms of n. Let me rearrange the equation:2(m + 1) = n + ln n 2m + 2 = n + ln n 2m = n + ln n - 2 m = (n + ln n - 2)/2So, m is expressed in terms of n. Now, I need to find the distance between points A and B. Since both points are on their respective graphs, their coordinates are A(m, f(m)) and B(n, g(n)). But since their y-coordinates are equal, the vertical distance between them is zero. Therefore, the distance between A and B is purely horizontal, which is |m - n|.So, the distance |AB| is |m - n|. Substituting the expression for m from above:|m - n| = |(n + ln n - 2)/2 - n| = |(n + ln n - 2 - 2n)/2| = |(-n + ln n - 2)/2| = |(ln n - n - 2)/2|So, the distance is |(ln n - n - 2)/2|. To find the minimum distance, I need to minimize this expression with respect to n. Since the absolute value is always non-negative, I can instead minimize the function inside the absolute value, which is (ln n - n - 2)/2.Let me define a function h(n) = (ln n - n - 2)/2. To find its minimum, I can take its derivative and set it equal to zero.First, compute the derivative h'(n):h(n) = (ln n - n - 2)/2 h'(n) = (1/n - 1)/2Set h'(n) = 0:(1/n - 1)/2 = 0 1/n - 1 = 0 1/n = 1 n = 1So, the critical point is at n = 1. Now, I should check the second derivative to ensure it's a minimum.Compute the second derivative h''(n):h'(n) = (1/n - 1)/2 h''(n) = (-1/n²)/2At n = 1:h''(1) = (-1/1)/2 = -1/2Since h''(1) is negative, this critical point is actually a maximum. Hmm, that's not what I want. I need a minimum. Maybe I made a mistake in interpreting the function.Wait, h(n) is (ln n - n - 2)/2, and I found that the critical point at n=1 is a maximum. That means the function h(n) has a maximum at n=1, but I need to find the minimum of |h(n)|. Since h(n) can be negative or positive, the minimum distance would occur where |h(n)| is minimized.But let's analyze h(n):h(n) = (ln n - n - 2)/2Let me evaluate h(n) at n=1:h(1) = (0 - 1 - 2)/2 = (-3)/2 = -1.5So, at n=1, h(n) is -1.5. Let's see the behavior of h(n) as n approaches 0 and as n approaches infinity.As n approaches 0 from the right:ln n approaches negative infinity, and -n approaches 0. So, ln n - n - 2 approaches negative infinity. Therefore, h(n) approaches negative infinity.As n approaches infinity:ln n grows slower than n, so ln n - n - 2 approaches negative infinity. Therefore, h(n) approaches negative infinity.So, the function h(n) tends to negative infinity as n approaches both 0 and infinity, and it has a maximum at n=1 of -1.5. That means the maximum value of h(n) is -1.5, and it's the highest point on the curve.Since h(n) is always negative, the absolute value |h(n)| is always positive, and it's minimized when h(n) is closest to zero. But since h(n) is always negative, the minimum of |h(n)| occurs when h(n) is closest to zero from below.But wait, h(n) is (ln n - n - 2)/2. Let me see if h(n) can ever be zero.Set h(n) = 0:(ln n - n - 2)/2 = 0 ln n - n - 2 = 0 ln n = n + 2But ln n grows much slower than n, so ln n = n + 2 has no solution because for n > 0, ln n is always less than n + 2. Therefore, h(n) is always negative, and |h(n)| is always positive.So, the minimum of |h(n)| occurs at the point where h(n) is closest to zero. Since h(n) has a maximum at n=1 of -1.5, which is the highest point, and it goes to negative infinity on both sides, the closest h(n) gets to zero is at n=1, where it's -1.5. Therefore, the minimum |h(n)| is 1.5.Wait, but that seems contradictory. If h(n) is always negative and has a maximum at n=1, which is -1.5, then the minimum of |h(n)| would be 1.5, because that's the smallest absolute value it attains. Because as n moves away from 1, |h(n)| increases towards infinity.So, the minimum distance between points A and B is 1.5, which is 3/2.Let me double-check this reasoning.Given that the distance is |(ln n - n - 2)/2|, and we found that this expression reaches its maximum (most negative) at n=1, which is -1.5. Since the distance is the absolute value, the minimum distance is the smallest absolute value, which occurs at the point where the expression inside is closest to zero. But since the expression inside is always negative, the closest it gets to zero is at its maximum point, which is -1.5. Therefore, the minimum distance is 1.5.Yes, that makes sense. So, the minimum distance is 3/2.

answer：First, I need to determine how many pages Coral reads in the first week. Since she reads half of the 600-page book, I calculate:600 pages / 2 = 300 pagesThis means she has 300 pages remaining after the first week.Next, I calculate how many pages she reads in the second week, which is 30% of the remaining 300 pages:0.30 * 300 pages = 90 pagesAfter reading 90 pages in the second week, the number of pages she has left is:300 pages - 90 pages = 210 pagesTherefore, Coral must read 210 pages in the third week to finish the book.

answer：First, I need to determine the initial average daily production for the past 9 days. Let's denote this initial average as ( x ) units per day.The total production over these 9 days would be ( 9x ) units.Today, the company produced 90 units, so the new total production becomes ( 9x + 90 ) units.The new average production after including today's production is 54 units per day over a total of 10 days. Therefore, the total production over these 10 days is ( 54 times 10 = 540 ) units.Setting up the equation:[ 9x + 90 = 540 ]Solving for ( x ):[ 9x = 540 - 90 ][ 9x = 450 ][ x = frac{450}{9} ][ x = 50 ]Therefore, the initial average daily production was 50 units per day.

answer：Okay, so I need to find the equation of the tangent line to the curve ( y = -x^3 + 3x^2 ) at the point ( (1, 2) ). Hmm, I remember that to find the tangent line, I need two things: the slope of the tangent at that point and a point through which the tangent passes. Well, the point is already given as ( (1, 2) ), so I just need to find the slope.I think the slope of the tangent line at a particular point on a curve is given by the derivative of the function at that point. So, I need to find the derivative of ( y = -x^3 + 3x^2 ). Let me recall how to take derivatives. The derivative of ( x^n ) is ( nx^{n-1} ), right?So, applying that to each term:- The derivative of ( -x^3 ) is ( -3x^2 ).- The derivative of ( 3x^2 ) is ( 6x ).Putting those together, the derivative ( y' ) is ( -3x^2 + 6x ). Okay, so that's the general expression for the slope of the tangent at any point ( x ) on the curve.Now, I need the slope specifically at ( x = 1 ). So, I substitute ( x = 1 ) into the derivative:( y' = -3(1)^2 + 6(1) = -3(1) + 6 = -3 + 6 = 3 ).So, the slope of the tangent line at ( x = 1 ) is 3. Got that.Now, with the slope and a point, I can write the equation of the tangent line using the point-slope form. The point-slope form is ( y - y_1 = m(x - x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is the point.Substituting the known values:( y - 2 = 3(x - 1) ).Now, I can simplify this to get it into slope-intercept form ( y = mx + b ):( y - 2 = 3x - 3 ).Adding 2 to both sides:( y = 3x - 3 + 2 ).( y = 3x - 1 ).So, the equation of the tangent line is ( y = 3x - 1 ). Let me check the options given:A: ( y = 3x - 1 )B: ( y = -3x + 5 )C: ( y = 3x + 5 )D: ( y = 2x )Looks like option A matches my result. Just to make sure I didn't make any mistakes, let me go through the steps again.1. Find the derivative of ( y = -x^3 + 3x^2 ): ( y' = -3x^2 + 6x ). That seems correct.2. Plug in ( x = 1 ): ( y' = -3(1)^2 + 6(1) = -3 + 6 = 3 ). Yep, that's right.3. Use point-slope form: ( y - 2 = 3(x - 1) ). Simplify: ( y = 3x - 3 + 2 = 3x - 1 ). Perfect.So, I'm confident that option A is the correct answer.

answer：First, I need to understand the pseudocode provided. It initializes a variable ( S ) to 0 and then uses a loop to add values to ( S ). The loop starts at 1, increments by 3 each time, and continues until it reaches 28.I'll list out all the values of ( I ) that the loop will iterate through:[1, 4, 7, 10, 13, 16, 19, 22, 25, 28]Next, I'll calculate the sum of these values step by step:[1 + 4 = 5][5 + 7 = 12][12 + 10 = 22][22 + 13 = 35][35 + 16 = 51][51 + 19 = 70][70 + 22 = 92][92 + 25 = 117][117 + 28 = 145]After adding all the values, the final value of ( S ) is 145.