answer：Let's solve the problem step by step.**Step 1: Define Variables**Let the total distance from Rita's house to her sister's house be ( x ) miles.**Step 2: Calculate Rita's Usual Speed**Rita usually takes 200 minutes to drive the distance ( x ). Therefore, her usual speed is:[text{Usual Speed} = frac{x}{200} text{ miles per minute}]**Step 3: Distance Covered Before the Rainstorm**Rita drives ( frac{1}{4} ) of the way before encountering the rainstorm. The distance covered before the rain is:[text{Distance before rain} = frac{x}{4} text{ miles}]**Step 4: Speed During the Rainstorm**Rita reduces her speed by 15 miles per hour due to the rainstorm. Since speed is usually given in miles per hour, we need to convert her usual speed from miles per minute to miles per hour:[text{Usual Speed in mph} = frac{x}{200} times 60 = frac{3x}{10} text{ mph}]Therefore, her speed during the rainstorm is:[text{Speed during rain} = frac{3x}{10} - 15 text{ mph}]**Step 5: Time Taken Before and During the Rainstorm**The total time taken for the trip is 300 minutes. We can break this into two parts:1. Time taken to drive the first ( frac{x}{4} ) miles at her usual speed.2. Time taken to drive the remaining ( frac{3x}{4} ) miles at the reduced speed.First, convert the speeds to miles per minute for consistency:[text{Usual Speed} = frac{x}{200} text{ miles per minute}][text{Speed during rain} = left( frac{3x}{10} - 15 right) text{ mph} = left( frac{3x}{10} - 15 right) times frac{1}{60} text{ miles per minute} = frac{3x - 150}{600} text{ miles per minute}]Now, calculate the time taken for each part:1. Time before rain:[text{Time}_1 = frac{frac{x}{4}}{frac{x}{200}} = frac{x}{4} times frac{200}{x} = 50 text{ minutes}]2. Time during rain:[text{Time}_2 = frac{frac{3x}{4}}{frac{3x - 150}{600}} = frac{3x}{4} times frac{600}{3x - 150} = frac{450x}{3x - 150} text{ minutes}]**Step 6: Set Up the Equation for Total Time**The total time is the sum of the two times:[50 + frac{450x}{3x - 150} = 300]**Step 7: Solve for ( x )**Subtract 50 from both sides:[frac{450x}{3x - 150} = 250]Multiply both sides by ( 3x - 150 ):[450x = 250(3x - 150)][450x = 750x - 37500]Subtract ( 750x ) from both sides:[-300x = -37500]Divide both sides by -300:[x = frac{37500}{300} = 125]**Conclusion**The total distance from Rita's house to her sister's house is ( 125 ) miles.(boxed{C})

answer：First, I need to calculate the total length of lace Carolyn requires for each part of the dress. For the cuffs, there are two of them, each 50 cm long. So, the total length for the cuffs is 2 multiplied by 50 cm, which equals 100 cm.The hem is 300 cm long.The waist is a third of the length of the hem. Therefore, the waist length is 300 cm divided by 3, which equals 100 cm.For the neckline, there are 5 ruffles, each using 20 cm of lace. So, the total length for the neckline is 5 multiplied by 20 cm, which equals 100 cm.Next, I add up all these lengths to find the total length of lace Carolyn needs:- Cuffs: 100 cm- Hem: 300 cm- Waist: 100 cm- Neckline: 100 cmTotal lace length = 100 cm + 300 cm + 100 cm + 100 cm = 600 cmSince 1 meter is equal to 100 cm, I convert the total lace length from centimeters to meters:Total lace length in meters = 600 cm / 100 cm/meter = 6 metersCarolyn spent 36 on lace for a total of 6 meters. To find the cost per meter, I divide the total cost by the total length in meters:Cost per meter = Total cost / Total length in meters = 36 / 6 meters = 6 per meterTherefore, the lace costs 6 per meter.

answer：First, express the given ratio of ( a ) to ( b ) as ( frac{a}{b} = frac{4}{5} ).Next, calculate ( x ) as ( a ) increased by 25% of ( a ):[x = a + 0.25a = 1.25a]Let ( m ) be ( b ) decreased by ( p% ) of ( b ):[m = b - frac{p}{100}b = bleft(1 - frac{p}{100}right)]Given that ( frac{m}{x} = 0.8 ), substitute the expressions for ( m ) and ( x ):[frac{bleft(1 - frac{p}{100}right)}{1.25a} = 0.8]Using the ratio ( frac{a}{b} = frac{4}{5} ), express ( b ) in terms of ( a ):[b = frac{5a}{4}]Substitute ( b ) into the equation:[frac{frac{5a}{4}left(1 - frac{p}{100}right)}{1.25a} = 0.8]Simplify the equation:[frac{5}{4}left(1 - frac{p}{100}right) div 1.25 = 0.8]Since ( 1.25 = frac{5}{4} ), the equation simplifies to:[1 - frac{p}{100} = 0.8]Solve for ( p ):[frac{p}{100} = 1 - 0.8 = 0.2][p = 0.2 times 100 = 20]Therefore, the percentage decrease of ( b ) to get ( m ) is ( 20% ).

answer：Okay, so I have this matrix, and I need to find the values of a, b, and c such that the matrix is its own inverse. Hmm, that means when I multiply the matrix by itself, I should get the identity matrix. Let me write down the matrix first:[begin{pmatrix} a & b & c 2 & -1 & 0 0 & 0 & 1 end{pmatrix}]Alright, so if this matrix is its own inverse, then multiplying it by itself should give me the identity matrix. Let me recall that the identity matrix is:[begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{pmatrix}]So, I need to compute the product of the given matrix with itself and set it equal to the identity matrix. Let me denote the given matrix as M. So, M squared should equal the identity matrix.Let me compute M squared step by step. To multiply two matrices, I need to take the dot product of the rows of the first matrix with the columns of the second matrix. Since both matrices are the same, it's just M multiplied by M.Let me write down the multiplication:[M times M = begin{pmatrix} a & b & c 2 & -1 & 0 0 & 0 & 1 end{pmatrix} times begin{pmatrix} a & b & c 2 & -1 & 0 0 & 0 & 1 end{pmatrix}]Let me compute each element of the resulting matrix one by one.First, the element in the first row and first column:- It's the dot product of the first row of the first matrix and the first column of the second matrix.- So, that's a*a + b*2 + c*0 = a² + 2b.Next, the element in the first row and second column:- It's the dot product of the first row of the first matrix and the second column of the second matrix.- So, that's a*b + b*(-1) + c*0 = ab - b.Then, the element in the first row and third column:- It's the dot product of the first row of the first matrix and the third column of the second matrix.- So, that's a*c + b*0 + c*1 = ac + c. Wait, hold on, is that right? Let me double-check. The third column of the second matrix is c, 0, 1. So, it's a*c + b*0 + c*1, which is ac + c. Hmm, okay.Moving on to the second row, first column:- It's the dot product of the second row of the first matrix and the first column of the second matrix.- So, that's 2*a + (-1)*2 + 0*0 = 2a - 2.Second row, second column:- It's the dot product of the second row of the first matrix and the second column of the second matrix.- So, that's 2*b + (-1)*(-1) + 0*0 = 2b + 1.Second row, third column:- It's the dot product of the second row of the first matrix and the third column of the second matrix.- So, that's 2*c + (-1)*0 + 0*1 = 2c.Third row, first column:- It's the dot product of the third row of the first matrix and the first column of the second matrix.- So, that's 0*a + 0*2 + 1*0 = 0.Third row, second column:- It's the dot product of the third row of the first matrix and the second column of the second matrix.- So, that's 0*b + 0*(-1) + 1*0 = 0.Third row, third column:- It's the dot product of the third row of the first matrix and the third column of the second matrix.- So, that's 0*c + 0*0 + 1*1 = 1.Putting all these together, the product matrix M squared is:[begin{pmatrix} a² + 2b & ab - b & ac + c 2a - 2 & 2b + 1 & 2c 0 & 0 & 1 end{pmatrix}]Now, since M squared should equal the identity matrix, we can set each corresponding element equal to the identity matrix's elements. Let's write down these equations:1. First row, first column: a² + 2b = 12. First row, second column: ab - b = 03. First row, third column: ac + c = 04. Second row, first column: 2a - 2 = 05. Second row, second column: 2b + 1 = 16. Second row, third column: 2c = 07. Third row, first column: 0 = 0 (which is always true)8. Third row, second column: 0 = 0 (which is always true)9. Third row, third column: 1 = 1 (which is always true)So, we have several equations here. Let's go through them one by one.Starting with equation 4: 2a - 2 = 0. Solving for a:2a - 2 = 0 2a = 2 a = 1Okay, so a is 1. That's straightforward.Moving to equation 6: 2c = 0. Solving for c:2c = 0 c = 0So, c is 0.Now, equation 5: 2b + 1 = 1. Solving for b:2b + 1 = 1 2b = 0 b = 0So, b is 0.Wait, so a is 1, b is 0, c is 0. Let me check if these values satisfy the other equations as well.Equation 1: a² + 2b = 1 Plugging in a=1 and b=0: 1² + 2*0 = 1 + 0 = 1 Which is correct.Equation 2: ab - b = 0 Plugging in a=1 and b=0: 1*0 - 0 = 0 - 0 = 0 Which is correct.Equation 3: ac + c = 0 Plugging in a=1, c=0: 1*0 + 0 = 0 + 0 = 0 Which is correct.So, all equations are satisfied with a=1, b=0, c=0.Wait, just to make sure, let me reconstruct the original matrix with these values and multiply it by itself to see if I get the identity matrix.Original matrix with a=1, b=0, c=0:[begin{pmatrix} 1 & 0 & 0 2 & -1 & 0 0 & 0 & 1 end{pmatrix}]Multiplying this by itself:First row, first column: 1*1 + 0*2 + 0*0 = 1 First row, second column: 1*0 + 0*(-1) + 0*0 = 0 First row, third column: 1*0 + 0*0 + 0*1 = 0 Second row, first column: 2*1 + (-1)*2 + 0*0 = 2 - 2 + 0 = 0 Second row, second column: 2*0 + (-1)*(-1) + 0*0 = 0 + 1 + 0 = 1 Second row, third column: 2*0 + (-1)*0 + 0*1 = 0 + 0 + 0 = 0 Third row, first column: 0*1 + 0*2 + 1*0 = 0 Third row, second column: 0*0 + 0*(-1) + 1*0 = 0 Third row, third column: 0*0 + 0*0 + 1*1 = 1So, the product matrix is:[begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{pmatrix}]Which is indeed the identity matrix. So, everything checks out.Therefore, the values are a=1, b=0, c=0.**Final Answer**The ordered triplet is boxed{(1, 0, 0)}.

answer：To determine the number of samples to be drawn from group C using stratified sampling, we first need to understand the ratio of employees in each group. The given ratio is 5:4:1 for groups A, B, and C respectively.1. **Calculate the total ratio:** [ 5 + 4 + 1 = 10 ] 2. **Determine the proportion of group C:** [ frac{1}{10} ] 3. **Apply this proportion to the total sample size of 20:** [ 20 times frac{1}{10} = 2 ] Therefore, the number of samples that should be drawn from group C is 2.

answer：Okay, so I have this problem about 20 pupils in a class, each sending 10 cards to 10 different classmates. The first part asks me to show that at least two pupils sent each other a card. Hmm, interesting. Let me try to wrap my head around this.First, I need to visualize the scenario. There are 20 pupils, and each one sends out 10 cards. Since you can't send a card to yourself, each pupil is sending cards to 10 others. So, in total, how many cards are being sent? Well, 20 pupils each sending 10 cards would result in 200 cards being sent in total. But since each card is sent from one pupil to another, the total number of directed edges in this graph would be 200.Wait, hold on. Maybe I should model this as a directed graph where each pupil is a node, and each card sent is a directed edge from one node to another. So, we have 20 nodes, each with an out-degree of 10. That makes sense.Now, the question is asking to show that at least two pupils sent each other a card. In graph terms, this means there's at least one pair of nodes where there's an edge going both ways between them, forming a mutual edge or a reciprocal connection.How can I approach this? Maybe using the Pigeonhole Principle? Let me think. Each pupil sends 10 cards, so each has 10 outgoing edges. Since there are 20 pupils, each pupil could potentially receive up to 19 cards, but in this case, each pupil is only sending 10, so the number of incoming edges per pupil could vary.Wait, but if each pupil sends 10 cards, the total number of incoming edges must also be 200, right? So, on average, each pupil receives 10 cards as well. So, the in-degree for each pupil is also 10 on average.Hmm, so we have a directed graph where each node has an in-degree and out-degree of 10. Now, I need to show that there's at least one mutual edge.Let me think about the total number of possible edges. In a directed graph with 20 nodes, there are 20*19 = 380 possible directed edges (since you can't have an edge from a node to itself). But in our case, we only have 200 edges.Wait, but the question is about mutual edges. So, how many mutual edges are possible? For each pair of nodes, there can be 0, 1, or 2 edges between them. If there are two edges, one in each direction, that's a mutual edge.So, the total number of mutual edges would be the number of pairs where both edges exist. Let me denote the number of mutual edges as M.Now, let's think about the total number of edges. Each mutual edge contributes 2 to the total count, and each non-mutual edge contributes 1. So, if we have M mutual edges, they contribute 2M edges, and the remaining edges are non-mutual, contributing (200 - 2M) edges.But wait, the total number of edges is 200, so:2M + (200 - 2M) = 200Hmm, that doesn't help much. Maybe I need a different approach.Let me consider the total number of ordered pairs (A, B) where A sends a card to B. That's 200. Now, if I consider unordered pairs {A, B}, each unordered pair can have 0, 1, or 2 ordered edges.If there are M mutual edges, that means there are M unordered pairs where both (A, B) and (B, A) are present. The remaining unordered pairs can have at most one edge.So, the total number of ordered edges can be expressed as:Total edges = 2M + SWhere S is the number of ordered edges in non-mutual pairs.But we know the total edges are 200, so:2M + S = 200Now, how many unordered pairs are there? There are C(20, 2) = 190 unordered pairs.Each unordered pair can contribute either 0, 1, or 2 to the total edges. So, the maximum number of edges without any mutual edges would be 190, since each pair can have at most one edge.But in our case, we have 200 edges, which is more than 190. Therefore, the excess edges must come from mutual edges.So, the number of mutual edges M can be calculated as:2M + S = 200But S ≤ 190, because S is the number of edges in non-mutual pairs, and there are only 190 unordered pairs.So, substituting S ≤ 190 into the equation:2M + 190 ≥ 200Therefore, 2M ≥ 10So, M ≥ 5Wait, that suggests there are at least 5 mutual edges. But the question only asks to show that at least two pupils sent each other a card, meaning at least one mutual edge. So, my calculation shows that there are actually at least 5 mutual edges, which is more than needed.But maybe I made a mistake somewhere. Let me double-check.Total edges: 200Total unordered pairs: 190If all 190 unordered pairs had one edge each, that would account for 190 edges. But we have 200 edges, which is 10 more.Each mutual edge adds an extra edge beyond the single edge in an unordered pair. So, each mutual edge contributes 2 edges instead of 1, which is an extra edge per mutual edge.Therefore, the number of extra edges is 200 - 190 = 10.Since each mutual edge contributes 1 extra edge, the number of mutual edges M must be at least 10.Wait, that contradicts my earlier conclusion. Hmm.Wait, no. Let me clarify.If we have M mutual edges, each contributes 2 edges, and the remaining (190 - M) unordered pairs contribute 1 edge each.So, total edges = 2M + (190 - M) = M + 190But we have 200 edges, so:M + 190 = 200Therefore, M = 10So, there must be at least 10 mutual edges.Wait, that makes more sense. So, actually, there are at least 10 mutual edges, meaning at least 10 pairs of pupils who sent each other cards.But the question only asks to show that at least two pupils sent each other a card, so my conclusion is stronger than needed, but it still answers the question.So, to summarize, in a directed graph with 20 nodes each having an out-degree of 10, the total number of edges is 200. Since the maximum number of edges without any mutual edges is 190, we must have at least 10 mutual edges. Therefore, at least two pupils sent each other a card.Now, moving on to part (b). It asks to generalize this result for n pupils each sending m cards. We need to find for which (m, n) the above statement is true, meaning that there must be at least one mutual edge.From part (a), we saw that when n=20 and m=10, we have at least 10 mutual edges. So, the condition is that m must be large enough such that the total number of edges exceeds the maximum number of edges possible without mutual edges.In general, for n pupils each sending m cards, the total number of edges is n*m. The maximum number of edges without any mutual edges is C(n, 2) = n(n-1)/2, since each unordered pair can have at most one edge.So, for there to be at least one mutual edge, we need:n*m > n(n-1)/2Simplifying this inequality:2m > n - 1So, m > (n - 1)/2Therefore, the minimal m(n) is m > (n - 1)/2, or equivalently, the maximal n(m) is n < 2m + 1.So, for any n and m where m > (n - 1)/2, we can guarantee that there is at least one mutual edge among the pupils.Let me verify this with the original problem. For n=20, m=10.(20 - 1)/2 = 19/2 = 9.5So, m=10 > 9.5, which satisfies the condition, and indeed, we have mutual edges.If m were 9, then 9 > (20 - 1)/2 = 9.5 is false, so we might not have any mutual edges. Let me check.If m=9, total edges = 20*9=180Maximum edges without mutual edges = 190Since 180 < 190, it's possible to have all edges be one-way without any mutual edges. So, indeed, when m ≤ (n - 1)/2, it's possible to have no mutual edges.Therefore, the condition m > (n - 1)/2 is both necessary and sufficient for guaranteeing at least one mutual edge.So, in part (b), the answer is that for n pupils each sending m cards, there must be at least one mutual edge if and only if m > (n - 1)/2.I think that's a solid conclusion. It makes sense because when m exceeds half of (n - 1), the total number of edges surpasses the maximum possible without mutual edges, forcing at least one mutual connection.