answer：First, I need to find the total marks of all 30 students by multiplying the average mark by the number of students.Next, I'll calculate the total marks of the 5 students who are to be excluded by multiplying their average mark by the number of these students.Then, I'll subtract the total marks of the excluded students from the total marks of all students to find the total marks of the remaining students.After that, I'll determine the number of remaining students by subtracting the number of excluded students from the total number of students.Finally, I'll find the average mark of the remaining students by dividing the total marks of the remaining students by the number of remaining students.

answer：First, I need to establish the relationship between A's and B's investments and the time periods for which they were invested. A's investment is three times that of B, and the time period for A's investment is twice that of B's.Next, I'll calculate the profit ratio by multiplying the investment amount by the time period for each partner. This gives me the proportion in which the profit should be divided between A and B.Once I have the profit ratio, I can determine the total number of parts in the ratio. Since B's share corresponds to a specific amount, I can use this to find the value of each part and subsequently calculate A's share.Finally, by adding both A's and B's shares, I'll obtain the total profit for the partnership.

answer：Alright, so I have this equation to solve: (5^{2x+1} - 5^{x+2} - 10 cdot 5^x + 50 = 0). Hmm, it looks a bit complicated with all these exponents. I remember that when dealing with exponential equations, a good strategy is to make a substitution to simplify the equation. Maybe I can let (a = 5^x) because that term appears multiple times.Let me try that substitution. If (a = 5^x), then (5^{2x}) would be ((5^x)^2 = a^2). But wait, the exponent in the first term is (2x + 1), so that would be (5^{2x} cdot 5^1 = 5a^2). Similarly, the second term is (5^{x+2}), which is (5^x cdot 5^2 = 25a). The third term is straightforward: (10 cdot 5^x = 10a). The last term is just 50.So substituting all of that into the equation, I get:[5a^2 - 25a - 10a + 50 = 0]Let me simplify that:[5a^2 - 35a + 50 = 0]Hmm, this is a quadratic equation in terms of (a). I can try to solve for (a) using the quadratic formula. But before that, maybe I can simplify the equation by dividing all terms by 5 to make the numbers smaller:[a^2 - 7a + 10 = 0]Okay, now it's easier. Applying the quadratic formula:[a = frac{7 pm sqrt{(-7)^2 - 4 cdot 1 cdot 10}}{2 cdot 1}][a = frac{7 pm sqrt{49 - 40}}{2}][a = frac{7 pm sqrt{9}}{2}][a = frac{7 pm 3}{2}]So, the solutions are:[a = frac{7 + 3}{2} = 5]and[a = frac{7 - 3}{2} = 2]Alright, so (a = 5) or (a = 2). But remember, (a = 5^x), so I need to solve for (x) in each case.First, if (5^x = 5), then (x = 1) because (5^1 = 5).Second, if (5^x = 2), then (x = log_5 2). I know that logarithms can give real numbers, so this is a valid solution as well.So, I have two solutions: (x = 1) and (x = log_5 2). Therefore, there are two real numbers (x) that satisfy the original equation.Let me just double-check to make sure I didn't make any mistakes. Starting from the substitution, everything seems correct. The quadratic equation was solved properly, and the solutions for (a) were correctly translated back to (x). I don't see any errors in my reasoning.Yeah, I think that's solid. The answer should be 2 real solutions.**Final Answer**boxed{C}

answer：Okay, so I have this problem where there are four projectors placed at four points on a plane. Each projector can illuminate a right angle, and the sides of these angles can be directed north, south, west, or east. I need to prove that these projectors can be directed in such a way that they illuminate the entire plane.Hmm, let me try to visualize this. Each projector is like a light that can shine in a 90-degree angle, and the sides of this angle can be aligned with the cardinal directions. So, for example, one projector could be pointed northeast and northwest, illuminating a right angle between those two directions. Another projector could be pointed southeast and southwest, and so on.Since there are four projectors, maybe each one can cover a quadrant of the plane? If I can direct each projector to cover a different quadrant, then together they could illuminate the entire plane. But wait, each projector only illuminates a right angle, which is 90 degrees. So each projector can cover a quadrant on its own? That seems too good to be true.Let me think again. If each projector is placed at a point, and it can illuminate a right angle, then depending on how we rotate the projector, it can cover different quadrants. So, if I have four projectors, each one can be rotated to cover one of the four quadrants: northeast, southeast, southwest, and northwest. If each projector is responsible for one quadrant, then together they can cover the entire plane.But wait, what if the projectors are not placed at the origin? The problem says they are placed at four given points on the plane. So, their positions might not be symmetric or at the origin. That complicates things because the coverage might overlap or leave gaps depending on their positions.Hmm, so maybe I need a different approach. Perhaps I can think of the plane as being divided into regions, and each projector can cover a specific region. Since each projector can illuminate a right angle, maybe I can arrange their directions so that their illuminated areas overlap in such a way that every point on the plane is covered by at least one projector.Another thought: since each projector can be directed in any of the four cardinal directions, maybe I can use their orientations to cover all possible directions. For example, if one projector is pointing north and east, another is pointing north and west, another is pointing south and east, and another is pointing south and west, then together they might cover all directions.Wait, but each projector only illuminates a right angle, so it's not covering all directions, just a 90-degree sector. So, if I have four projectors each covering a different quadrant, their combined coverage would indeed cover the entire plane. But I need to make sure that regardless of where the projectors are placed, I can direct them in such a way that their illuminated areas overlap appropriately.Maybe I can use the fact that four right angles can cover a full 360 degrees. If each projector is responsible for a 90-degree sector, then four of them can cover the entire 360 degrees. But since the projectors are at different points, their coverage areas might not align perfectly, but perhaps by adjusting their directions, their illuminated areas can overlap sufficiently to cover the entire plane.Another idea: think of the plane as being covered by four overlapping right angles. Each projector can be directed to cover a specific area, and by adjusting their directions, their illuminated regions can overlap to ensure complete coverage.Wait, but how do I ensure that every point on the plane is illuminated by at least one projector? Maybe I can use the fact that any point on the plane can be reached by one of the four projectors if their directions are chosen appropriately.Let me try to formalize this. Suppose each projector can be assigned a direction, which is a pair of cardinal directions (north, south, east, west) that form a right angle. For example, a projector can be assigned to illuminate the northeast and northwest directions, forming a right angle. Similarly, another projector can be assigned to illuminate the southeast and southwest directions.If I have four projectors, each assigned to a different pair of cardinal directions, then their combined coverage would be the entire plane. But wait, each projector only illuminates a right angle, so it's not covering all directions, just a 90-degree sector. So, if I have four projectors each covering a different quadrant, their combined coverage would indeed cover the entire plane.But I need to make sure that regardless of where the projectors are placed, I can direct them in such a way that their illuminated areas overlap appropriately. Maybe I can use the fact that four right angles can cover a full 360 degrees. If each projector is responsible for a 90-degree sector, then four of them can cover the entire 360 degrees.Wait, but the projectors are at different points, so their coverage areas might not align perfectly. However, since they are projectors, their illuminated areas are infinite in extent, so even if they are at different points, as long as their directions are chosen such that every direction is covered by at least one projector, the entire plane will be illuminated.So, if I can assign each projector to cover a different quadrant, then together they can cover the entire plane. Since each projector can be directed to cover any right angle, I can choose their directions such that each one covers a different quadrant.Therefore, by directing each projector to cover a different quadrant, their combined coverage will illuminate the entire plane.Wait, but what if the projectors are placed in such a way that their coverage areas don't overlap properly? For example, if two projectors are placed very close together, their coverage areas might overlap too much, leaving gaps elsewhere. But since the projectors can be directed independently, regardless of their positions, I can adjust their directions to ensure that their coverage areas overlap appropriately.In other words, the key is that each projector can be independently directed to cover any right angle, regardless of its position. So, even if the projectors are placed at different points, I can still direct each one to cover a specific quadrant, ensuring that the entire plane is illuminated.Therefore, by assigning each projector to a different quadrant, their combined coverage will illuminate the entire plane.I think that makes sense. Each projector can be directed to cover a specific 90-degree sector, and with four projectors, each covering a different quadrant, their combined coverage will indeed illuminate the entire plane.So, in conclusion, by directing each of the four projectors to cover a different quadrant (northeast, southeast, southwest, and northwest), their combined illuminated areas will cover the entire plane, ensuring that every point is illuminated by at least one projector.

answer：Okay, so I have this logarithmic equation to solve: log base 3 of y cubed plus log base 1/3 of y equals 7. Hmm, that sounds a bit tricky, but I think I can handle it. Let me write it down first to visualize it better.So, the equation is:[log_3 y^3 + log_{1/3} y = 7]Alright, I remember some logarithm rules. Maybe I can simplify each term separately before trying to combine them. Let's start with the first term: log base 3 of y cubed. I think there's a power rule that says log_b (a^n) equals n times log_b a. So, applying that here, I can bring the exponent 3 in front of the logarithm.So, that would make the first term:[log_3 y^3 = 3 log_3 y]Okay, that simplifies the first part. Now, the second term is log base 1/3 of y. Hmm, log base 1/3 is a bit confusing because the base is a fraction. I remember that log base 1/b of a is the same as negative log base b of a. Let me verify that.Yes, because log_{1/b} a = log_{b^{-1}} a = - log_b a. So, applying that here, log_{1/3} y is equal to negative log base 3 of y.So, the second term simplifies to:[log_{1/3} y = - log_3 y]Great, so now I can rewrite the original equation with these simplifications:[3 log_3 y - log_3 y = 7]Let me combine like terms. Both terms have log base 3 of y, so I can factor that out.[(3 - 1) log_3 y = 7][2 log_3 y = 7]Now, to solve for log base 3 of y, I can divide both sides by 2.[log_3 y = frac{7}{2}]Alright, so log base 3 of y equals 7/2. To find y, I need to rewrite this equation in exponential form. Remember that log_b a = c is equivalent to b^c = a.So, applying that here:[y = 3^{frac{7}{2}}]Hmm, 3 to the power of 7/2. That's the same as the square root of 3 to the 7th power, right? Because 3^{7/2} is equal to sqrt(3^7). Let me compute 3^7 first.3^1 is 3,3^2 is 9,3^3 is 27,3^4 is 81,3^5 is 243,3^6 is 729,3^7 is 2187.So, 3^{7/2} is sqrt(2187). Now, I can simplify sqrt(2187). Let me see if 2187 is a perfect square or if it can be factored into something with a square factor.Well, 2187 divided by 9 is 243, because 9 times 243 is 2187. And 243 is 9 times 27, which is 9 times 3^3. So, 2187 is 9 times 243, which is 9 times 9 times 27, which is 9 times 9 times 9 times 3, which is 9^3 times 3.Wait, that might not be the most straightforward way. Let me think differently. 2187 is 3^7, right? Because 3^7 is 2187. So, sqrt(3^7) is equal to 3^{7/2}, which is the same as 3^{3 + 1/2} = 3^3 times 3^{1/2} = 27 times sqrt(3).So, sqrt(2187) is 27 times sqrt(3). Therefore, y equals 27 times sqrt(3). Alternatively, I can write that as 9 times sqrt(27), but that might not be necessary. Let me check:Wait, sqrt(27) is 3 times sqrt(3), so 9 times sqrt(27) would be 9 times 3 times sqrt(3) which is 27 times sqrt(3). So, both expressions are equivalent. Maybe 27 sqrt(3) is simpler.But let me verify my steps again to make sure I didn't make any mistakes.Starting from the original equation:[log_3 y^3 + log_{1/3} y = 7]I used the power rule on the first term:[3 log_3 y]Then, for the second term, since the base is 1/3, I converted it to negative log base 3:[- log_3 y]So, combining the two:[3 log_3 y - log_3 y = 2 log_3 y = 7]Dividing both sides by 2:[log_3 y = frac{7}{2}]Converting to exponential form:[y = 3^{frac{7}{2}} = sqrt{3^7} = sqrt{2187} = 27 sqrt{3}]Yes, that seems correct. Alternatively, 3^{7/2} is also equal to (3^{1/2})^7, but that would be sqrt(3)^7, which is more complicated. So, expressing it as 27 sqrt(3) is better.Wait, just to make sure, let me compute 27 times sqrt(3):27 is 3^3, and sqrt(3) is 3^{1/2}, so multiplying them together gives 3^{3 + 1/2} = 3^{7/2}, which is consistent with our earlier result. So, that's correct.Alternatively, if I compute sqrt(2187), let's see:2187 divided by 9 is 243, as I thought earlier. So sqrt(2187) is sqrt(9 * 243) = sqrt(9) * sqrt(243) = 3 * sqrt(243). Then, sqrt(243) is sqrt(9 * 27) = 3 * sqrt(27). Then, sqrt(27) is 3 * sqrt(3). So, putting it all together:sqrt(2187) = 3 * 3 * 3 * sqrt(3) = 27 sqrt(3). Yep, that matches.So, I think I did everything correctly. Therefore, the value of y is 27 times the square root of 3.**Final Answer**The value of ( y ) is (boxed{27sqrt{3}}).

answer：Okay, so I have this problem here with two functions, f(x) = x ln x and g(x) = -x² + a x - 3. There are two parts to the problem. The first part is to find the minimum value of f(x) on the interval (0, +∞). The second part is to find the range of the real number a such that for all x in (0, +∞), 2f(x) is greater than or equal to g(x). Hmm, let me try to work through each part step by step.Starting with part (1): Find the minimum value of f(x) on (0, +∞). So, f(x) = x ln x. I remember that to find the minimum or maximum of a function, we can use calculus, specifically finding the derivative and setting it equal to zero to find critical points.So, let's compute the derivative of f(x). The function is x multiplied by ln x. Using the product rule, the derivative of x is 1, and the derivative of ln x is 1/x. So, f'(x) = 1 * ln x + x * (1/x) = ln x + 1. That simplifies to f'(x) = ln x + 1.Now, to find critical points, set f'(x) = 0. So, ln x + 1 = 0. Solving for x, we get ln x = -1. Taking the exponential of both sides, x = e^(-1) = 1/e. So, x = 1/e is a critical point.Now, we need to determine if this critical point is a minimum or a maximum. We can use the second derivative test or analyze the sign changes of the first derivative.Let me compute the second derivative. The first derivative is f'(x) = ln x + 1, so the second derivative f''(x) is the derivative of ln x, which is 1/x. Since x is in (0, +∞), 1/x is always positive. Therefore, f''(x) > 0 for all x > 0, which means the function is concave upward everywhere, and the critical point at x = 1/e is a local minimum. Since it's the only critical point and the function tends to infinity as x approaches 0 and as x approaches infinity, this local minimum is actually the global minimum on the interval (0, +∞).So, the minimum value of f(x) is f(1/e). Let's compute that. f(1/e) = (1/e) * ln(1/e). I know that ln(1/e) is equal to -1 because ln(e^(-1)) = -1. So, f(1/e) = (1/e) * (-1) = -1/e.Alright, so that's part (1). The minimum value is -1/e.Moving on to part (2): Find the range of the real number a such that for all x in (0, +∞), 2f(x) ≥ g(x). So, substituting the given functions, 2f(x) = 2x ln x, and g(x) = -x² + a x - 3. So, the inequality is 2x ln x ≥ -x² + a x - 3.Let me rearrange this inequality to solve for a. Let's bring all terms to one side:2x ln x + x² - a x + 3 ≥ 0.Wait, actually, maybe it's better to isolate a. Let me see:Starting from 2x ln x ≥ -x² + a x - 3.Let's add x² to both sides:2x ln x + x² ≥ a x - 3.Then, add 3 to both sides:2x ln x + x² + 3 ≥ a x.Now, divide both sides by x (since x > 0, we can do this without changing the inequality direction):(2x ln x + x² + 3)/x ≥ a.Simplify the left side:2 ln x + x + 3/x ≥ a.So, the inequality becomes 2 ln x + x + 3/x ≥ a for all x > 0.Therefore, to satisfy this inequality for all x > 0, a must be less than or equal to the minimum value of the function h(x) = 2 ln x + x + 3/x on (0, +∞). So, we need to find the minimum of h(x) and set a to be less than or equal to that.Alright, so let's define h(x) = 2 ln x + x + 3/x. We need to find its minimum on (0, +∞). Again, we can use calculus for this.First, compute the derivative h'(x). Let's differentiate term by term:- The derivative of 2 ln x is 2*(1/x) = 2/x.- The derivative of x is 1.- The derivative of 3/x is -3/x².So, putting it all together, h'(x) = 2/x + 1 - 3/x².To find critical points, set h'(x) = 0:2/x + 1 - 3/x² = 0.Let me write this equation as:(2/x) + 1 - (3/x²) = 0.To solve this, let's multiply both sides by x² to eliminate denominators:2x + x² - 3 = 0.So, we have a quadratic equation: x² + 2x - 3 = 0.Let me solve this quadratic equation. The quadratic formula is x = [-b ± sqrt(b² - 4ac)]/(2a). Here, a = 1, b = 2, c = -3.So, discriminant D = b² - 4ac = 4 - 4*1*(-3) = 4 + 12 = 16.So, sqrt(D) = 4.Thus, solutions are x = [-2 ± 4]/2.So, x = (-2 + 4)/2 = 2/2 = 1, and x = (-2 - 4)/2 = -6/2 = -3.But since x > 0, we discard x = -3. So, the critical point is at x = 1.Now, we need to check if this critical point is a minimum or a maximum. We can use the second derivative test or analyze the sign changes of h'(x).Let me compute the second derivative h''(x). Starting from h'(x) = 2/x + 1 - 3/x².Differentiate term by term:- The derivative of 2/x is -2/x².- The derivative of 1 is 0.- The derivative of -3/x² is 6/x³.So, h''(x) = -2/x² + 6/x³.Simplify: h''(x) = (-2x + 6)/x³.Wait, let me compute it correctly.Wait, h'(x) = 2/x + 1 - 3/x².So, h''(x) is derivative of 2/x, which is -2/x², plus derivative of 1, which is 0, plus derivative of -3/x², which is 6/x³.So, h''(x) = -2/x² + 6/x³.We can factor this as h''(x) = (-2x + 6)/x³.Wait, let me see:-2/x² + 6/x³ = (-2x + 6)/x³. Yes, that's correct.So, h''(x) = (-2x + 6)/x³.Now, evaluate h''(x) at x = 1:h''(1) = (-2*1 + 6)/1³ = (4)/1 = 4.Since h''(1) = 4 > 0, the function is concave upward at x = 1, which means the critical point at x = 1 is a local minimum.Since h(x) approaches infinity as x approaches 0 and as x approaches infinity, this local minimum is actually the global minimum on (0, +∞).Therefore, the minimum value of h(x) is h(1). Let's compute h(1):h(1) = 2 ln 1 + 1 + 3/1.We know that ln 1 = 0, so h(1) = 0 + 1 + 3 = 4.So, the minimum value of h(x) is 4. Therefore, to satisfy the inequality 2 ln x + x + 3/x ≥ a for all x > 0, a must be less than or equal to 4.Hence, the range of a is (-∞, 4].Wait, let me double-check my steps to make sure I didn't make any mistakes.Starting from 2f(x) ≥ g(x):2x ln x ≥ -x² + a x - 3.Rearranged to 2x ln x + x² + 3 ≥ a x.Divided by x: 2 ln x + x + 3/x ≥ a.Defined h(x) = 2 ln x + x + 3/x.Computed h'(x) = 2/x + 1 - 3/x².Set h'(x) = 0, solved quadratic equation, got x = 1.Computed h''(x) at x = 1, found it positive, so minimum at x =1.h(1) = 4, so a ≤4.Seems correct.Just to be thorough, let me check the behavior of h(x) as x approaches 0 and infinity.As x approaches 0+, ln x approaches -infty, but 2 ln x approaches -infty, x approaches 0, and 3/x approaches +infty. So, h(x) = 2 ln x + x + 3/x. The dominant term as x approaches 0 is 3/x, which goes to +infty. So, h(x) approaches +infty.As x approaches +infty, ln x grows slower than x, so 2 ln x is negligible compared to x. Also, 3/x approaches 0. So, h(x) ≈ x, which approaches +infty. So, h(x) approaches +infty as x approaches both 0 and infinity, and has a minimum at x=1, which is 4. So, the minimum value is indeed 4.Therefore, the range of a is all real numbers less than or equal to 4.I think that's solid. I don't see any mistakes in the reasoning.**Final Answer**(1) The minimum value of ( f(x) ) is (boxed{-dfrac{1}{e}}).(2) The range of ( a ) is (boxed{(-infty, 4]}).