answer：To determine Xiao Ming's father's age, I start by noting that Xiao Ming is 9 years old this year.Since his father's age is 5 times Xiao Ming's age, I multiply 9 by 5, which equals 45. Therefore, Xiao Ming's father is 45 years old this year.Next, to find the total cost of buying 60 pens, each costing 2 yuan, I multiply the number of pens by the cost per pen. Multiplying 60 by 2 gives me 120 yuan. Thus, buying 60 pens costs 120 yuan.

answer：To solve the system of equations, I can use the elimination method.First, I'll multiply the first equation by 3 and the second equation by 2 to make the coefficients of y's equal.This gives me:9x + 6y = 64x + 6y = 16Next, I'll subtract the second equation from the first to eliminate y:5x = -10Solving for x, I find:x = -2Then, I'll substitute x = -2 back into the first original equation to find y:3(-2) + 2y = 2-6 + 2y = 22y = 8y = 4Finally, adding x and y together:x + y = -2 + 4 = 2

answer：To determine the value of ( f(4) ) for the given odd function, I start by recalling the definition of an odd function: ( f(-x) = -f(x) ).Given that ( f(x) = x(2 - x) ) when ( x < 0 ), I can find ( f(-4) ) by substituting ( x = -4 ) into this expression:[f(-4) = (-4)(2 - (-4)) = (-4)(6) = -24]Using the property of odd functions, I relate ( f(-4) ) to ( f(4) ):[f(-4) = -f(4) implies -24 = -f(4)]Solving for ( f(4) ), I find:[f(4) = 24]

answer：Okay, so I have this problem about the 12th National Games in Shenyang. They need to divide 6 volunteers into pairs and assign each pair to one of 3 different venues. But there's a condition: volunteers A and B must be in the same group. I need to figure out how many different allocation schemes there are under these conditions.First, let me try to understand the problem step by step. We have 6 volunteers, and we need to pair them up into 3 groups of 2. Then, each of these groups is assigned to a different venue. So, the problem involves two main parts: pairing the volunteers and assigning these pairs to venues.But there's a constraint: A and B must be in the same group. So, A and B are fixed as a pair. That simplifies things a bit because we don't have to consider all possible pairings; instead, we can fix A and B as one pair and then deal with the remaining volunteers.So, let's break this down:1. **Pairing Volunteers:** - Since A and B must be together, we can consider them as a single unit. That leaves us with 4 other volunteers: let's call them C, D, E, and F. - Now, we need to pair these 4 volunteers into two more groups. How many ways can we do this?2. **Assigning Pairs to Venues:** - Once we have the three pairs (including A and B), we need to assign each pair to one of the 3 different venues. Since the venues are different, the order matters here. So, this is a permutation problem.Let me tackle the first part: pairing the remaining 4 volunteers.When dealing with pairing, the number of ways to pair 2n people is given by the double factorial formula: (2n - 1)!!. For 4 people, that would be 3!! = 3 × 1 = 3. So, there are 3 ways to pair 4 people.Wait, let me verify that. If we have 4 people: C, D, E, F.- Pair 1: C-D and E-F- Pair 2: C-E and D-F- Pair 3: C-F and D-EYes, that's 3 different pairings. So, that's correct.So, the number of ways to pair the remaining 4 volunteers is 3.Now, moving on to the second part: assigning these pairs to the 3 different venues.We have 3 pairs: (A,B), (C,D), and (E,F) for example. Each pair needs to be assigned to a different venue. Since the venues are distinct, the order matters. So, this is a permutation of 3 items, which is 3! = 6.Therefore, for each pairing, there are 6 ways to assign them to the venues.So, putting it all together: the number of allocation schemes is the number of pairings multiplied by the number of assignments.That would be 3 (pairings) × 6 (assignments) = 18.Wait, hold on. Is that all? Let me think again.Wait, actually, maybe I oversimplified the pairing part. Because when we pair the remaining 4 volunteers, each pairing is unique, but when we assign them to venues, each pair is assigned to a specific venue, which is a different place.But let me think about the entire process step by step.1. First, fix A and B as a pair. So, we have one pair already: (A,B).2. Now, we have 4 volunteers left: C, D, E, F.3. We need to pair them into two more pairs. As I calculated earlier, there are 3 ways to do this. - Pair 1: (C,D) and (E,F) - Pair 2: (C,E) and (D,F) - Pair 3: (C,F) and (D,E)4. Now, for each of these pairings, we have 3 pairs in total: (A,B), (C,D), (E,F) for example.5. Now, we need to assign each of these 3 pairs to 3 different venues. Since the venues are different, the order matters. So, for each pairing, how many ways can we assign the pairs to the venues? - It's the number of permutations of 3 items, which is 3! = 6.So, for each of the 3 pairings, we have 6 assignments. So, total number of allocation schemes is 3 × 6 = 18.But wait, is that all? Or is there something else I'm missing?Let me think about whether the pairings themselves are considered different based on the order of selection. For example, when we pair C and D, is that different from pairing D and C? In this context, since we're just forming groups, the order within the pair doesn't matter. So, (C,D) is the same as (D,C). Similarly, (E,F) is the same as (F,E).Therefore, when we count the pairings, we are considering unordered pairs, so the count is correct as 3.Similarly, when assigning to venues, each pair is assigned as a whole, so the order of the pair doesn't matter, only which pair goes to which venue.Therefore, the calculation seems correct.But let me try another approach to confirm.Alternative approach:Total number of ways to divide 6 volunteers into 3 pairs without any restrictions is (6-1)!! = 5!! = 5 × 3 × 1 = 15.But since A and B must be in the same pair, we fix them as a pair, so we have 4 volunteers left, which can be paired in 3 ways, as before. Then, assigning to venues, 3! = 6. So, total is 3 × 6 = 18.Alternatively, if we didn't fix A and B first, the total number of ways without restrictions would be 15 pairings, and then assigning to 3 venues, which would be 15 × 6 = 90. But since A and B must be together, we have fewer pairings.Wait, but actually, the total number of pairings where A and B are together is 3, as we calculated earlier, so total allocation schemes would be 3 × 6 = 18.Therefore, the answer is 18.But wait, let me think again. Maybe I'm missing a step.Wait, the problem says "divided into pairs and assigned to 3 different venues." So, does that mean that the pairing is done first, and then assigned? Or is it that each pair is assigned to a venue, but the pairing is part of the assignment?Wait, no, it's two separate steps: first, divide into pairs, then assign each pair to a venue.So, if we fix A and B as a pair, then pair the remaining 4 into two more pairs, which is 3 ways. Then, assign these 3 pairs to 3 venues, which is 6 ways. So, 3 × 6 = 18.Alternatively, if we didn't fix A and B, the total number of ways would be (number of pairings) × (number of assignments). The number of pairings is 15, as I thought earlier, and assignments would be 6, so 15 × 6 = 90. But since A and B must be together, we have fewer pairings.But let me calculate the number of pairings where A and B are together.Total number of pairings: 15.Number of pairings where A and B are together: ?In general, the number of pairings where two specific elements are paired together is equal to the number of ways to pair the remaining elements. So, if we fix A and B as a pair, the remaining 4 can be paired in 3 ways, as we saw. So, the number of pairings where A and B are together is 3.Therefore, the number of allocation schemes where A and B are together is 3 × 6 = 18.Alternatively, another way to think about it is:First, pair A and B: 1 way.Then, pair the remaining 4: 3 ways.Then, assign the 3 pairs to 3 venues: 6 ways.Total: 1 × 3 × 6 = 18.Yes, that seems consistent.Wait, but let me think about another angle. Maybe considering permutations.Suppose we think of assigning each volunteer to a venue, but since they are in pairs, each venue gets exactly 2 volunteers.But since A and B must be together, they must be assigned to the same venue.So, first, choose a venue for A and B: there are 3 choices.Then, assign the remaining 4 volunteers into pairs and assign each pair to the remaining 2 venues.Wait, but this approach might complicate things, but let's try.1. Choose a venue for A and B: 3 choices.2. Now, we have 4 volunteers left: C, D, E, F.3. We need to divide them into two pairs and assign each pair to the remaining two venues. - First, pair the 4 volunteers: 3 ways. - Then, assign each pair to the remaining two venues: since there are two venues left, and two pairs, the number of assignments is 2! = 2.Therefore, for each choice of venue for A and B, we have 3 × 2 = 6 allocation schemes.Since there are 3 choices for the venue of A and B, total allocation schemes would be 3 × 6 = 18.Yes, same result.Therefore, regardless of the approach, the answer is 18.Wait, but let me make sure I didn't double-count or miss any constraints.Another way: think of the entire process as assigning each volunteer to a venue, but with the constraint that A and B are together.But since each venue must have exactly 2 volunteers, and A and B are together, we can model this as:1. Choose a venue for A and B: 3 choices.2. Assign the remaining 4 volunteers into the 3 venues, with each venue getting exactly 2 volunteers, but one venue already has A and B, so the remaining two venues need to have 2 volunteers each, and the third venue (the one with A and B) needs to have 0 more volunteers.Wait, no, actually, each venue must have exactly 2 volunteers. So, if A and B are assigned to a venue, that venue already has 2 volunteers, so the remaining 4 volunteers must be assigned to the other two venues, each getting 2 volunteers.So, step by step:1. Choose a venue for A and B: 3 choices.2. Now, we have 4 volunteers left: C, D, E, F.3. We need to divide them into two groups of 2 and assign each group to the remaining two venues. - The number of ways to divide 4 volunteers into two groups of 2 is 3 (as before). - Then, assign each group to a venue: since there are two venues left, it's 2! = 2 ways.Therefore, total number of ways: 3 (choices for A and B's venue) × 3 (ways to pair the remaining) × 2 (ways to assign pairs to venues) = 3 × 3 × 2 = 18.Same result.Alternatively, using multinomial coefficients:The number of ways to assign 6 volunteers into 3 groups of 2, with A and B in the same group.First, fix A and B in one group. Then, we need to partition the remaining 4 into two groups of 2.The number of ways to partition 4 into two groups of 2 is 3, as before.Then, assign these three groups to 3 venues: 3! = 6.So, total is 3 × 6 = 18.Alternatively, using the formula for the number of ways to partition into groups:The number of ways to partition 2n elements into n pairs is (2n)!)/(2^n n!).But in our case, since we have a constraint, it's better to fix A and B first.So, fixing A and B as a pair, the remaining 4 can be paired in 3 ways.Then, assigning to 3 venues: 6 ways.Total: 18.Therefore, I'm confident that the answer is 18.But just to make sure, let me think about it in terms of permutations.Suppose we think of assigning each volunteer to a venue, but with the constraint that A and B are together.Each venue must have exactly 2 volunteers.So, first, choose a venue for A and B: 3 choices.Then, assign the remaining 4 volunteers into the remaining two venues, each needing 2 volunteers.The number of ways to assign 4 volunteers into two groups of 2 is 3, as before.Then, assign these two groups to the two remaining venues: 2! = 2.So, total: 3 × 3 × 2 = 18.Same result.Alternatively, using multinomial coefficients:The number of ways to assign 6 volunteers into 3 groups of 2, with A and B in the same group, is:First, fix A and B in one group. Then, the number of ways to partition the remaining 4 into two groups of 2 is 3.Then, assign these three groups to 3 venues: 3! = 6.Total: 3 × 6 = 18.Yes, consistent.Alternatively, think of it as:Total number of ways without restrictions: (6)! / (2! 2! 2!) = 720 / 8 = 90.But since A and B must be together, we need to subtract the cases where A and B are not together.But that might be more complicated. Instead, it's easier to fix A and B together and calculate accordingly.But for thoroughness, let's see:Total number of ways without restrictions: 90.Number of ways where A and B are not together: ?To calculate this, we can use inclusion-exclusion.Total pairings: 15.Number of pairings where A and B are together: 3.Therefore, number of pairings where A and B are not together: 15 - 3 = 12.Then, number of assignments: 6.Therefore, total allocation schemes where A and B are not together: 12 × 6 = 72.Therefore, total allocation schemes where A and B are together: 90 - 72 = 18.Yes, same result.Therefore, regardless of the method, the answer is 18.I think I've considered multiple approaches and each time arrived at 18. So, I'm confident that the answer is 18.**Final Answer**The number of different allocation schemes is boxed{18}.

answer：First, I recognize that the problem involves an arithmetic sequence. In an arithmetic sequence, each term after the first is obtained by adding a constant difference to the previous term.Given that ( a_1 = 2 ) and ( a_5 = 6 ), I can use the formula for the ( n )-th term of an arithmetic sequence:[a_n = a_1 + (n - 1)d]where ( d ) is the common difference.Plugging in the known values for ( a_5 ):[6 = 2 + (5 - 1)d]Simplifying this equation:[6 = 2 + 4d]Subtracting 2 from both sides:[4 = 4d]Dividing both sides by 4:[d = 1]Now that I have determined the common difference ( d = 1 ), I can find ( a_{2015} ) using the same formula:[a_{2015} = 2 + (2015 - 1) times 1]Simplifying:[a_{2015} = 2 + 2014 = 2016]

answer：Okay, so I need to find the coordinates of point A, which is given as (0, y, 0), such that it's equidistant from points B and C. Points B and C are given as (-2, 4, -6) and (8, 5, 1) respectively. Hmm, let me think about how to approach this.First, I remember that if a point is equidistant from two other points, it lies on the perpendicular bisector of the segment connecting those two points. But since we're dealing with three-dimensional space, it's a bit more complex than just a line. Maybe I can use the distance formula in 3D to set up an equation.The distance formula in three dimensions between two points (x1, y1, z1) and (x2, y2, z2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]. Since point A is equidistant from B and C, the distance from A to B should equal the distance from A to C. So, I can set up the equation:Distance from A to B = Distance from A to CLet me write that out using the coordinates:sqrt[(-2 - 0)^2 + (4 - y)^2 + (-6 - 0)^2] = sqrt[(8 - 0)^2 + (5 - y)^2 + (1 - 0)^2]Simplifying inside the square roots:sqrt[(-2)^2 + (4 - y)^2 + (-6)^2] = sqrt[(8)^2 + (5 - y)^2 + (1)^2]Calculating the squares:sqrt[4 + (4 - y)^2 + 36] = sqrt[64 + (5 - y)^2 + 1]Simplify the constants:sqrt[40 + (4 - y)^2] = sqrt[65 + (5 - y)^2]Hmm, okay. Now, to get rid of the square roots, I can square both sides of the equation:40 + (4 - y)^2 = 65 + (5 - y)^2Let me expand both sides:Left side: 40 + (16 - 8y + y^2) = 40 + 16 - 8y + y^2 = 56 - 8y + y^2Right side: 65 + (25 - 10y + y^2) = 65 + 25 - 10y + y^2 = 90 - 10y + y^2So now the equation is:56 - 8y + y^2 = 90 - 10y + y^2Hmm, I can subtract y^2 from both sides to simplify:56 - 8y = 90 - 10yNow, let's get all the y terms on one side and constants on the other. I'll add 10y to both sides:56 + 2y = 90Then subtract 56 from both sides:2y = 34Divide both sides by 2:y = 17Wait, that seems straightforward. Let me check my steps to make sure I didn't make a mistake.Starting from the distance equation:sqrt[4 + (4 - y)^2 + 36] = sqrt[64 + (5 - y)^2 + 1]Which simplifies to sqrt[40 + (4 - y)^2] = sqrt[65 + (5 - y)^2]Squaring both sides gives 40 + (4 - y)^2 = 65 + (5 - y)^2Expanding both sides:Left: 40 + 16 - 8y + y^2 = 56 - 8y + y^2Right: 65 + 25 - 10y + y^2 = 90 - 10y + y^2Subtracting y^2 from both sides: 56 - 8y = 90 - 10yAdding 10y: 56 + 2y = 90Subtracting 56: 2y = 34Dividing by 2: y = 17Okay, that seems correct. So, point A has coordinates (0, 17, 0). Let me just verify this by plugging back into the distance formula.Distance from A to B:sqrt[(-2 - 0)^2 + (4 - 17)^2 + (-6 - 0)^2] = sqrt[4 + (-13)^2 + 36] = sqrt[4 + 169 + 36] = sqrt[209]Distance from A to C:sqrt[(8 - 0)^2 + (5 - 17)^2 + (1 - 0)^2] = sqrt[64 + (-12)^2 + 1] = sqrt[64 + 144 + 1] = sqrt[209]Yes, both distances are equal to sqrt(209), so that checks out. Therefore, y must be 17.**Final Answer**The coordinates of point ( A ) are boxed{(0, 17, 0)}.