answer：Alright, so I have this problem here: I need to find the monotonic increasing interval of the function ( y = log_{0.2}(x^{2} - 6x + 8) ). Hmm, okay, let's break this down step by step.First, I remember that for logarithmic functions, the base is important. The base here is 0.2, which is between 0 and 1. I think that means the logarithmic function is decreasing because, as the input increases, the output decreases. Yeah, that sounds right. So, ( log_{0.2}(t) ) is a decreasing function in terms of ( t ).Now, the function inside the logarithm is a quadratic: ( x^{2} - 6x + 8 ). I should probably analyze this quadratic function first. Let me write it down: ( t = x^{2} - 6x + 8 ). To understand its behavior, I can find its vertex and determine where it's increasing or decreasing.The general form of a quadratic is ( ax^{2} + bx + c ), and in this case, ( a = 1 ), ( b = -6 ), and ( c = 8 ). The vertex of a parabola given by ( ax^{2} + bx + c ) is at ( x = -frac{b}{2a} ). Plugging in the values, that would be ( x = -frac{-6}{2*1} = frac{6}{2} = 3 ). So, the vertex is at ( x = 3 ).Since the coefficient of ( x^{2} ) is positive (1), the parabola opens upwards. That means the function ( t = x^{2} - 6x + 8 ) decreases to the left of the vertex and increases to the right of the vertex. So, it's decreasing on ( (-infty, 3) ) and increasing on ( (3, infty) ).But wait, I also need to consider the domain of the original logarithmic function. The argument of a logarithm must be positive, so ( x^{2} - 6x + 8 > 0 ). Let me solve this inequality.First, factor the quadratic: ( x^{2} - 6x + 8 ). Looking for two numbers that multiply to 8 and add up to -6. Hmm, -2 and -4. So, ( (x - 2)(x - 4) = 0 ). Therefore, the roots are at ( x = 2 ) and ( x = 4 ).Since the parabola opens upwards, the quadratic is positive outside the interval [2, 4]. So, the domain of the function ( y = log_{0.2}(x^{2} - 6x + 8) ) is ( (-infty, 2) cup (4, infty) ).Okay, so the function is defined on two intervals: ( (-infty, 2) ) and ( (4, infty) ). Now, I need to find where the function ( y ) is increasing.Remember, ( y = log_{0.2}(t) ) is a decreasing function in terms of ( t ). So, if ( t ) is increasing, ( y ) will be decreasing, and if ( t ) is decreasing, ( y ) will be increasing.From earlier, I know that ( t = x^{2} - 6x + 8 ) is decreasing on ( (-infty, 3) ) and increasing on ( (3, infty) ). But our domain is ( (-infty, 2) cup (4, infty) ). So, within the domain, ( t ) is decreasing on ( (-infty, 2) ) and increasing on ( (4, infty) ).Since ( y ) is a logarithm with base less than 1, it's decreasing in ( t ). Therefore, when ( t ) is decreasing, ( y ) is increasing. So, on the interval where ( t ) is decreasing, which is ( (-infty, 2) ), ( y ) is increasing.On the other interval ( (4, infty) ), ( t ) is increasing, so ( y ) would be decreasing there because the logarithm is decreasing in ( t ).Therefore, the function ( y = log_{0.2}(x^{2} - 6x + 8) ) is increasing on ( (-infty, 2) ).Wait, let me double-check. The quadratic ( t = x^{2} - 6x + 8 ) is decreasing on ( (-infty, 3) ), but our domain is ( (-infty, 2) cup (4, infty) ). So, on ( (-infty, 2) ), ( t ) is decreasing, and since ( y ) is decreasing in ( t ), ( y ) would be increasing. That makes sense.On ( (4, infty) ), ( t ) is increasing, so ( y ) is decreasing. So, the only interval where ( y ) is increasing is ( (-infty, 2) ).I think that's it. So, the monotonic increasing interval is ( (-infty, 2) ).**Final Answer**The monotonic increasing interval of the function is boxed{(-infty, 2)}.

answer：Okay, so I need to figure out the total snowfall in Yardley. It snowed 0.125 inches in the morning and 0.5 inches in the afternoon. Hmm, I think I just need to add these two amounts together to get the total. But wait, let me make sure I'm doing this right.First, I'll write down the numbers: 0.125 and 0.5. Adding decimals can sometimes be tricky, especially when they have different numbers of decimal places. Maybe I should line them up to make it clearer. So, 0.125 has three decimal places, and 0.5 has one. To add them, I'll need to make sure they have the same number of decimal places. I can add a couple of zeros to 0.5 to make it 0.500. Now both numbers have three decimal places: 0.125 and 0.500.Alright, let's add them up. Starting from the rightmost digit:5 (from 0.125) plus 0 (from 0.500) equals 5.2 (from 0.125) plus 0 (from 0.500) equals 2.1 (from 0.125) plus 5 (from 0.500) equals 6.So, putting it all together, the sum is 0.625 inches. That seems right. But just to double-check, I'll convert the decimals to fractions to see if it makes sense.0.125 is the same as 1/8 of an inch, and 0.5 is the same as 1/2 of an inch. Adding 1/8 and 1/2, I need a common denominator. The least common denominator for 8 and 2 is 8. So, 1/2 is equivalent to 4/8. Now, adding 1/8 and 4/8 gives me 5/8. Wait, that's 0.625 inches. So, both methods give me the same result. That makes me more confident that the total snowfall is indeed 0.625 inches.I guess another way to think about it is in terms of money. If 0.125 inches is like 12.5 cents and 0.5 inches is like 50 cents, adding them together would be 62.5 cents, which is 0.625 dollars. That analogy helps me see that the addition is correct.Just to be thorough, I'll try adding them without converting to the same number of decimal places. So, 0.125 plus 0.5. Starting from the right:5 (hundredths place) plus nothing is 5.2 (tenths place) plus 5 (tenths place) is 7.0 (units place) plus 0 is 0.Wait, that gives me 0.75. That's different from before. Did I do something wrong? Oh, I see, I misaligned the decimal points. That's why it's important to line them up properly. So, I shouldn't have added the 2 and the 5 directly; instead, I should have added the 2 in the hundredths place to the 0 in the hundredths place of 0.5, which is actually 0.500. So, 2 plus 0 is 2, and then 1 plus 5 is 6, making it 0.625. That aligns with my previous calculations.Okay, I think I've confirmed it enough times. The total snowfall in Yardley was 0.625 inches.

answer：First, I need to evaluate the inner square root, which is √25. Since 5 multiplied by 5 equals 25, √25 is 5.Next, I substitute this value back into the original expression, resulting in √(49 × 5). Multiplying 49 by 5 gives me 245.Now, I need to find the square root of 245. I can simplify this by recognizing that 245 is equal to 49 multiplied by 5. Taking the square root of 49 gives me 7, and the square root of 5 remains as √5.Therefore, the square root of 245 simplifies to 7√5.

answer：Okay, so I'm trying to solve this math problem where the teacher starts a game by writing a number on the blackboard, and then students take turns writing new numbers based on specific rules. The goal is to reach the number 1. There are three parts to this problem: starting with 45, starting with 345, and explaining how to reach 1 from any non-zero natural number.First, let me make sure I understand the rules correctly. If the number written on the board has only one digit, the next number must be that number multiplied by 2. If the number has more than one digit, the student can choose to either erase the unit digit (which means removing the last digit) or multiply the number by 2. The game continues with the new number, following the same rules until we reach 1.Alright, starting with part (a): starting the game with the number 45 and showing a way to proceed until reaching 1.So, we begin with 45. Since 45 has two digits, we can either erase the unit digit or multiply by 2. Let's explore both options.Option 1: Erase the unit digit. That would take 45 to 4. Now, 4 is a single-digit number, so we have to multiply it by 2, resulting in 8. 8 is still a single-digit number, so we multiply by 2 again to get 16. Now, 16 has two digits, so we can either erase the unit digit or multiply by 2. If we erase the unit digit, we get 1, which is our target. So, this path works: 45 → 4 → 8 → 16 → 1.Option 2: Instead of erasing the unit digit from 45, we could multiply by 2, resulting in 90. Now, 90 has two digits, so we can either erase the unit digit or multiply by 2. If we erase the unit digit, we get 9, which is a single-digit number. Multiplying 9 by 2 gives 18. Then, erasing the unit digit of 18 gives us 1. So, this path also works: 45 → 90 → 9 → 18 → 1.Both options lead us to 1, so either path is valid. However, the first path seems shorter, so I'll go with that for part (a).Moving on to part (b): starting with the number 345 and showing a way to proceed until reaching 1.Starting with 345, which has three digits. We can either erase the unit digit or multiply by 2. Let's explore both options.Option 1: Erase the unit digit. That would take 345 to 34. Now, 34 has two digits, so we can erase the unit digit or multiply by 2. If we erase the unit digit, we get 3, which is a single-digit number. Multiplying 3 by 2 gives 6. Then, multiplying 6 by 2 gives 12. Erasing the unit digit of 12 gives us 1. So, this path works: 345 → 34 → 3 → 6 → 12 → 1.Option 2: Instead of erasing the unit digit from 345, we could multiply by 2, resulting in 690. Now, 690 has three digits, so we can erase the unit digit or multiply by 2. If we erase the unit digit, we get 69. Now, 69 has two digits, so we can erase the unit digit or multiply by 2. If we erase the unit digit, we get 6, which is a single-digit number. Multiplying 6 by 2 gives 12. Erasing the unit digit of 12 gives us 1. So, this path also works: 345 → 690 → 69 → 6 → 12 → 1.Again, both options lead us to 1, but the first path seems shorter, so I'll choose that for part (b).Now, part (c): explaining how to reach 1 starting the game with any non-zero natural number.To generalize the process, let's consider any non-zero natural number. The key is to reduce the number step by step until we reach 1. Here's a strategy:1. If the number has more than one digit, repeatedly erase the unit digit until you get a single-digit number. For example, starting with 3487, erase the unit digit to get 348, then erase again to get 34, and one more time to get 3.2. Once you have a single-digit number, you can multiply it by 2 until you reach a two-digit number, then erase the unit digit to get back to a single-digit number. Repeat this process until you reach 1. For example, starting with 3: - 3 × 2 = 6 - 6 × 2 = 12 - Erase the unit digit of 12 to get 1.This method works because multiplying by 2 eventually leads to a two-digit number, and erasing the unit digit reduces it back to a single-digit number, gradually decreasing the value until we reach 1.So, the strategy is to reduce the number of digits by erasing the unit digit when possible, and when you have a single-digit number, use multiplication by 2 to reach a two-digit number, then erase the unit digit again. This cycle continues until you reach 1.I think this approach covers all possible non-zero natural numbers, ensuring that we can always reach 1 by following the given rules.

answer：Alright, so I have this problem here where a city is planning to select 2 projects each from 4 key projects and 6 general projects. I need to find the number of different ways to select these projects such that at least one of the key project A or the general project B is selected. The options given are A: 15, B: 45, C: 60, D: 75.Okay, let's break this down. First, there are two categories of projects: key projects and general projects. The city is selecting 2 from each category. So, in total, they're selecting 4 projects: 2 key and 2 general.Now, the condition is that at least one of project A (which is a key project) or project B (which is a general project) is selected. So, I need to count all the possible selections where either A is selected, B is selected, or both are selected.Hmm, when I see "at least one," it often makes me think of using the principle of inclusion-exclusion. That is, the total number of ways without any restrictions minus the number of ways where neither A nor B is selected. That should give me the number of ways where at least one of A or B is selected.Let me verify if that makes sense. If I calculate all possible selections and then subtract the ones that don't include A or B, I should be left with the selections that include at least one of them. Yeah, that sounds right.So, first, let's find the total number of ways to select 2 key projects out of 4 and 2 general projects out of 6 without any restrictions. For the key projects, the number of ways is the combination of 4 taken 2 at a time, which is denoted as C(4,2). Similarly, for the general projects, it's C(6,2).Calculating these:C(4,2) = 4! / (2! * (4-2)!) = (4*3)/(2*1) = 6C(6,2) = 6! / (2! * (6-2)!) = (6*5)/(2*1) = 15So, the total number of ways without restrictions is 6 * 15 = 90.Okay, now I need to find the number of ways where neither project A nor project B is selected. That means, for the key projects, we exclude A, so we're selecting 2 out of the remaining 3 key projects. Similarly, for the general projects, we exclude B, so we're selecting 2 out of the remaining 5 general projects.Calculating these:C(3,2) = 3! / (2! * (3-2)!) = (3*2)/(2*1) = 3C(5,2) = 5! / (2! * (5-2)!) = (5*4)/(2*1) = 10So, the number of ways where neither A nor B is selected is 3 * 10 = 30.Therefore, the number of ways where at least one of A or B is selected is the total number of ways minus the number of ways where neither is selected:90 - 30 = 60.So, the answer should be 60, which is option C.Wait, let me just make sure I didn't make any mistakes here. So, total ways: 6 * 15 = 90. Ways without A and without B: 3 * 10 = 30. So, 90 - 30 = 60. Yeah, that seems correct.Alternatively, I could think about it by considering the cases where A is selected, B is selected, and both are selected, and then use inclusion-exclusion to avoid double-counting. Let me try that approach to verify.First, the number of ways where A is selected. If A is selected, then we need to choose 1 more key project from the remaining 3, and 2 general projects from all 6.So, C(3,1) * C(6,2) = 3 * 15 = 45.Next, the number of ways where B is selected. If B is selected, then we need to choose 2 key projects from all 4, and 1 more general project from the remaining 5.So, C(4,2) * C(5,1) = 6 * 5 = 30.Now, the number of ways where both A and B are selected. If both are selected, then we choose 1 more key project from the remaining 3, and 1 more general project from the remaining 5.So, C(3,1) * C(5,1) = 3 * 5 = 15.Using inclusion-exclusion, the total number of ways where A or B is selected is:Number of ways with A + Number of ways with B - Number of ways with both A and B = 45 + 30 - 15 = 60.Okay, so that also gives me 60. So, both methods give the same answer, which is reassuring.Therefore, I'm confident that the correct answer is 60, which is option C.**Final Answer**boxed{C}

answer：Okay, so I have this problem where I need to find the smallest number of planes that can cover all the points in a set S, but none of the planes pass through the origin. The set S is defined as all points (x, y, z) where x, y, z are integers from 0 to n, and x + y + z is greater than 0. So, basically, S includes all points in a cube grid from (0,0,0) to (n,n,n) except the origin itself.First, I need to visualize this set S. It's like a 3D grid with points from 0 to n in each dimension, but excluding the origin. So, the total number of points in S is (n+1)^3 - 1. That makes sense because each dimension has n+1 points (including 0), so cubing that gives the total number of points, and subtracting 1 removes the origin.Now, the task is to cover all these points with the smallest number of planes, and none of these planes can pass through the origin. Hmm, okay. So, planes in 3D space can be defined by equations like ax + by + cz + d = 0. Since none of the planes can pass through the origin, d cannot be zero for any of these planes because if d were zero, plugging in (0,0,0) would satisfy the equation, meaning the plane passes through the origin.So, each plane must have a non-zero constant term. That's an important point. Now, how can I cover all the points in S with such planes?Let me think about the structure of S. Since S includes all points except the origin, it's like a cube with a missing corner. Each face of the cube is a plane, but those planes do pass through the origin, so they're not allowed. So, I need other planes that can cover the points without passing through (0,0,0).Maybe I can think about the coordinate planes shifted away from the origin. For example, the plane x = 1, y = 1, z = 1, etc. These planes would cover points where x, y, or z is 1, but they wouldn't cover all points. Wait, actually, if I take all planes x = i for i from 1 to n, y = i for i from 1 to n, and z = i for i from 1 to n, that would give me 3n planes. Each of these planes is parallel to one of the coordinate planes but shifted by 1 unit along the respective axis.Let me check if these planes cover all points in S. For any point (x, y, z) in S, at least one of x, y, or z is greater than 0. So, if x is at least 1, then the plane x = x will cover it. Similarly for y and z. So, yes, these 3n planes would cover all points in S. Also, none of these planes pass through the origin because, for example, x = 1 doesn't include (0,0,0), and similarly for the others.But is 3n the minimal number? Maybe there's a way to cover all points with fewer planes. Let me think about that.Another approach could be to use planes that are not axis-aligned. For example, planes like x + y + z = k for some constant k. These planes are diagonal and might cover multiple points with a single plane. Let's see how many such planes we would need.The smallest value of x + y + z in S is 1 (when one coordinate is 1 and the others are 0). The largest value is 3n (when all coordinates are n). So, if we take all planes x + y + z = k for k from 1 to 3n, that would cover all points in S. However, that's 3n planes as well. So, same number as before.But wait, maybe some of these diagonal planes can cover multiple points more efficiently? Hmm, but each plane x + y + z = k can only cover points where the sum is exactly k. So, for each k, we need a separate plane. Therefore, it's still 3n planes.Is there a smarter way to arrange the planes so that each plane covers more points, thus reducing the total number needed? Maybe using a combination of axis-aligned and diagonal planes?Let me think about the properties of planes in 3D space. A plane can be defined by three non-collinear points. So, if I can find a set of planes where each plane covers as many points as possible, that might help reduce the total number.But considering the grid structure of S, it's not obvious how to arrange planes to cover multiple points efficiently without overlapping too much or missing some points.Wait, maybe I can use the concept of hyperplanes. In 3D, a hyperplane is just a plane. Each hyperplane can cover an entire face of the cube, but as I thought earlier, those pass through the origin, which is not allowed.Alternatively, maybe using planes that are offset from the origin but still cover multiple layers of the cube. For example, a plane that covers all points where x = 1, 2, ..., n, but that's just the same as the axis-aligned planes I considered earlier.Hmm, perhaps another approach is needed. Maybe using the concept of linear algebra or combinatorics.Let me think about the dual problem. Instead of trying to cover all points, maybe think about how many points each plane can cover and then calculate the minimal number needed.Each plane can cover at most how many points? In the case of axis-aligned planes, x = i covers (n+1)^2 points, but since we're excluding the origin, it's actually n^2 + 2n points? Wait, no. If x = i, then y and z can range from 0 to n, so that's (n+1)^2 points. But since we're excluding the origin, which is only one point, it's still (n+1)^2 points. But actually, when x = i, the origin isn't on that plane unless i = 0, which we aren't considering.Wait, actually, for x = i where i >= 1, the plane x = i doesn't include the origin, so it's fine. Each such plane covers (n+1)^2 points. Similarly for y = i and z = i.So, each axis-aligned plane covers (n+1)^2 points. How many such planes do we have? 3n, as I thought earlier.Now, the total number of points to cover is (n+1)^3 - 1. Let's compute that:(n+1)^3 - 1 = n^3 + 3n^2 + 3n + 1 - 1 = n^3 + 3n^2 + 3n.Each plane covers (n+1)^2 points, which is n^2 + 2n + 1.If we have 3n planes, the total number of points covered would be 3n*(n^2 + 2n + 1). But wait, that's way more than the total number of points we need to cover. So, clearly, there's a lot of overlap.But since we're looking for the minimal number, maybe we can do better by considering that some points are covered by multiple planes.Wait, but each point in S must be covered by at least one plane. So, even though there's overlap, we need to ensure that every point is covered.But if I try to minimize the number of planes, I need to maximize the coverage per plane without missing any points.Alternatively, maybe I can think about the problem in terms of linear algebra. Each plane can be represented as a linear equation, and the set of planes must cover all points except the origin.But I'm not sure how to translate that into a minimal set.Wait, another idea: maybe use the concept of hyperplane arrangements. In 3D, the minimal number of hyperplanes needed to cover all points except the origin.But I'm not sure about the exact theory here.Wait, perhaps I can think about the dual problem: instead of covering points, think about the dual space where points become planes and planes become points. But that might complicate things.Alternatively, think about the problem in terms of linear algebra. Each plane equation is a linear equation, and we need a set of linear equations such that every point in S satisfies at least one equation, but the origin doesn't satisfy any.But how does that help me find the minimal number?Wait, maybe consider that each plane can be represented as a linear form, and the product of all these linear forms would be a polynomial that vanishes on all points in S but not at the origin.Then, the degree of this polynomial would be equal to the number of planes, and by some theorem, maybe we can bound the degree.Wait, that sounds promising. Let me think about that.Suppose I have N planes, each defined by a linear equation L_i(x, y, z) = 0. Then, the product P(x, y, z) = L_1 L_2 ... L_N is a polynomial of degree N that vanishes on all points in S, but P(0,0,0) ≠ 0 because none of the planes pass through the origin.Now, by some theorem, maybe we can say that the degree of P must be at least something, which would give us a lower bound on N.I recall that in algebraic geometry, there's a result called the Combinatorial Nullstellensatz which might be relevant here. It states that if a polynomial vanishes on all points of a grid except one, then its degree must be at least the size of the grid in some way.Wait, let me recall the exact statement. The Combinatorial Nullstellensatz says that if a polynomial P in k variables vanishes on all points of {0, 1, ..., n}^k except one, then the degree of P is at least kn.But in our case, the polynomial P vanishes on all points of S, which is {0, 1, ..., n}^3 except the origin. So, maybe we can apply a similar idea.If we consider P as a polynomial in three variables, then the degree of P must be at least 3n. Because each variable can take n+1 values, and the polynomial must vanish on all combinations except one.Wait, but in our case, the polynomial doesn't vanish on all combinations except one, but rather on all points except the origin. So, it's similar but not exactly the same.But perhaps we can still use a similar argument. If P vanishes on all points (x, y, z) where x + y + z > 0, then P must have a certain degree.In fact, I think there's a lemma that states that if a polynomial vanishes on all points of {0, 1, ..., n}^k except the origin, then its degree is at least kn. So, in our case, k = 3, so the degree must be at least 3n.Therefore, since P is the product of N linear forms, each of degree 1, the total degree is N. So, N must be at least 3n.Thus, the minimal number of planes needed is 3n.Wait, but earlier I thought of 3n planes as axis-aligned planes, which also gives 3n. So, this seems to confirm that 3n is indeed the minimal number.But let me double-check. Suppose n = 2. Then, the minimal number of planes should be 6.Let's see: for n = 2, S consists of all points (x, y, z) where x, y, z ∈ {0,1,2} and x + y + z > 0. So, total points are 3^3 - 1 = 26.If I take the planes x = 1, x = 2, y = 1, y = 2, z = 1, z = 2, that's 6 planes. Each plane x = i covers 3x3 = 9 points, similarly for y and z. But since some points are covered by multiple planes, the total coverage is more than 26, but we only need to cover 26 points.Alternatively, using the planes x + y + z = k for k from 1 to 6, that's also 6 planes. Each plane x + y + z = k covers a certain number of points, and together they cover all 26 points.So, for n = 2, 6 planes are sufficient and necessary.Similarly, for n = 3, it would be 9 planes, and so on.Therefore, it seems that the minimal number of planes required is indeed 3n.I think I've convinced myself that 3n is the minimal number. The key was realizing that the polynomial constructed from the product of the linear forms of the planes must have a degree of at least 3n, which implies that the number of planes must be at least 3n. And since we can construct such a set of planes with exactly 3n planes, that must be the minimal number.