answer：First, let's define the cost price of the badminton racket as ( C ) yuan.The marked price is 40% higher than the cost price, so the marked price ( M ) can be calculated as:[M = C + 0.40C = 1.40C]The selling price is 80% of the marked price, which means:[text{Selling Price} = 0.80M = 0.80 times 1.40C = 1.12C]The profit made from the sale is the difference between the selling price and the cost price:[text{Profit} = text{Selling Price} - C = 1.12C - C = 0.12C]According to the problem, the profit is 15 yuan:[0.12C = 15]To find the cost price ( C ), we solve the equation:[C = frac{15}{0.12} = 125]Therefore, the cost price of the badminton racket is 125 yuan.

answer：Okay, so I have this problem with four barrels. The capacities are 24, 13, 11, and 5 buckets. The first barrel is full with 24 buckets of water, and the others are empty. I need to divide this water into three equal parts. Hmm, that means each part should be 8 buckets because 24 divided by 3 is 8. So, I need to end up with three barrels each containing 8 buckets of water. Let me write down the capacities again to keep track: Barrel 1 is 24, Barrel 2 is 13, Barrel 3 is 11, and Barrel 4 is 5. All start empty except Barrel 1, which is full. I think the strategy here is to pour water from one barrel to another until either the source is empty or the destination is full. Maybe I can use the smaller barrels to measure out the water. Since I need to get 8 buckets into three barrels, perhaps I can use the 5 and 11 barrels to help with that. Let me start by trying to pour from Barrel 1 into Barrel 2. Barrel 2 can hold 13, so if I pour 13 buckets from Barrel 1 into Barrel 2, Barrel 1 will have 24 - 13 = 11 left. So, Barrel 1: 11, Barrel 2: 13, Barrel 3: 0, Barrel 4: 0. Now, maybe I can pour from Barrel 2 into Barrel 3. Barrel 3 can hold 11, so if I pour 11 buckets from Barrel 2 into Barrel 3, Barrel 2 will have 13 - 11 = 2 left. So, Barrel 1: 11, Barrel 2: 2, Barrel 3: 11, Barrel 4: 0. Hmm, now Barrel 3 is full. Maybe I can pour from Barrel 3 into Barrel 4. Barrel 4 can hold 5, so pouring 5 into it from Barrel 3 will leave Barrel 3 with 11 - 5 = 6. So, Barrel 1: 11, Barrel 2: 2, Barrel 3: 6, Barrel 4: 5. Now, Barrel 4 is full. Let me pour the 5 from Barrel 4 into Barrel 1. Barrel 1 currently has 11, so adding 5 will make it 16. So, Barrel 1: 16, Barrel 2: 2, Barrel 3: 6, Barrel 4: 0. Wait, that doesn't seem helpful. Maybe I should try a different approach. Let me go back a step. After pouring from Barrel 2 into Barrel 3, Barrel 2 had 2 left. Maybe instead of pouring into Barrel 3, I should pour into Barrel 4. So, starting over: Barrel 1: 24, Barrel 2: 0, Barrel 3: 0, Barrel 4: 0. Pour from Barrel 1 into Barrel 2 until Barrel 2 is full. So, Barrel 1: 24 - 13 = 11, Barrel 2: 13. Now, pour from Barrel 2 into Barrel 4. Barrel 4 can take 5, so Barrel 2 will have 13 - 5 = 8. So, Barrel 1: 11, Barrel 2: 8, Barrel 3: 0, Barrel 4: 5. Hmm, now Barrel 2 has 8, which is one of the parts I need. So, that's one barrel done. Now, I need to get two more barrels to 8. Let me pour the 5 from Barrel 4 into Barrel 1. Barrel 1 had 11, so now it's 16. Barrel 4 is empty. Now, pour from Barrel 1 into Barrel 4 again. Barrel 4 can take 5, so Barrel 1 will have 16 - 5 = 11, Barrel 4: 5. Hmm, this seems like I'm going in circles. Maybe I should try a different method. Let me try pouring from Barrel 1 into Barrel 3. Barrel 3 can hold 11, so pouring 11 into it from Barrel 1 leaves Barrel 1 with 24 - 11 = 13. So, Barrel 1: 13, Barrel 2: 0, Barrel 3: 11, Barrel 4: 0. Now, pour from Barrel 3 into Barrel 4. Barrel 4 can take 5, so Barrel 3 will have 11 - 5 = 6. So, Barrel 1: 13, Barrel 2: 0, Barrel 3: 6, Barrel 4: 5. Now, pour the 5 from Barrel 4 into Barrel 2. Barrel 2 can take 13, so it's now 5. So, Barrel 1: 13, Barrel 2: 5, Barrel 3: 6, Barrel 4: 0. Now, pour from Barrel 1 into Barrel 4. Barrel 4 can take 5, so Barrel 1 will have 13 - 5 = 8. So, Barrel 1: 8, Barrel 2: 5, Barrel 3: 6, Barrel 4: 5. Now, Barrel 1 has 8, which is another part. Now, I need to get the third part. Let me pour from Barrel 4 into Barrel 2. Barrel 2 can take 13 - 5 = 8 more. So, pouring 5 from Barrel 4 into Barrel 2 makes Barrel 2: 10, but wait, Barrel 2 can only hold 13, so 5 + 5 = 10, which is fine. So, Barrel 2: 10, Barrel 4: 0. Now, pour from Barrel 3 into Barrel 4. Barrel 3 has 6, Barrel 4 can take 5. So, Barrel 3: 1, Barrel 4: 5. Now, pour from Barrel 4 into Barrel 2. Barrel 2 has 10, can take 3 more. So, pour 3 from Barrel 4 into Barrel 2, making Barrel 2: 13, Barrel 4: 2. Now, Barrel 2 is full. Now, pour from Barrel 1 into Barrel 3. Barrel 1 has 8, Barrel 3 has 1. So, pour 7 into Barrel 3 to make it full? Wait, Barrel 3 can hold 11, so it can take 10 more. But Barrel 1 only has 8. So, pour all 8 from Barrel 1 into Barrel 3, making Barrel 3: 9, Barrel 1: 0. Hmm, that doesn't help. Maybe I should try a different approach again. Let me start over and try a different sequence. Barrel 1: 24, others: 0. Pour from Barrel 1 into Barrel 3 until Barrel 3 is full. So, Barrel 1: 24 - 11 = 13, Barrel 3: 11. Now, pour from Barrel 3 into Barrel 4. Barrel 4 can take 5, so Barrel 3: 6, Barrel 4: 5. Now, pour from Barrel 4 into Barrel 2. Barrel 2 can take 13, so Barrel 2: 5, Barrel 4: 0. Now, pour from Barrel 1 into Barrel 4. Barrel 1: 13 - 5 = 8, Barrel 4: 5. So now, Barrel 1: 8, Barrel 2: 5, Barrel 3: 6, Barrel 4: 5. Now, I need to get another 8. Let's pour from Barrel 3 into Barrel 2. Barrel 3 has 6, Barrel 2 has 5, can take 8 more. So, pour 6 into Barrel 2, making Barrel 2: 11, Barrel 3: 0. Now, pour from Barrel 1 into Barrel 3. Barrel 1 has 8, Barrel 3 can take 11. So, pour 8 into Barrel 3, making Barrel 3: 8, Barrel 1: 0. Now, we have Barrel 2: 11, Barrel 3: 8, Barrel 1: 0, Barrel 4: 5. Hmm, Barrel 2 has 11, which is more than 8. Maybe I can pour from Barrel 2 into Barrel 1. Barrel 1 can take 24, so pour 8 into it, making Barrel 2: 3, Barrel 1: 8. Now, we have Barrel 1: 8, Barrel 2: 3, Barrel 3: 8, Barrel 4: 5. Now, pour from Barrel 4 into Barrel 2. Barrel 2 can take 10 more (13 - 3 = 10), so pour 5 into it, making Barrel 2: 8, Barrel 4: 0. Now, we have Barrel 1: 8, Barrel 2: 8, Barrel 3: 8, Barrel 4: 0. Wait, that works! So, the steps are:1. Pour from Barrel 1 to Barrel 3 until Barrel 3 is full: Barrel 1:13, Barrel 3:11.2. Pour from Barrel 3 to Barrel 4 until Barrel 4 is full: Barrel 3:6, Barrel 4:5.3. Pour from Barrel 4 to Barrel 2: Barrel 2:5, Barrel 4:0.4. Pour from Barrel 1 to Barrel 4: Barrel 1:8, Barrel 4:5.5. Pour from Barrel 3 to Barrel 2 until Barrel 2 is full: Barrel 3:0, Barrel 2:11 (but wait, Barrel 2 can only hold 13, so after adding 5 from Barrel 4, it was at 5, then adding 6 from Barrel 3, it becomes 11, which is fine).6. Pour from Barrel 1 to Barrel 3: Barrel 1:0, Barrel 3:8.7. Pour from Barrel 2 to Barrel 1 until Barrel 1 is full: Barrel 1:8, Barrel 2:3.8. Pour from Barrel 4 to Barrel 2: Barrel 2:8, Barrel 4:0.Wait, that seems a bit convoluted, but it works. Let me recount the steps more clearly:1. Barrel 1 (24) → Barrel 3 (11): Barrel 1:13, Barrel 3:11.2. Barrel 3 (11) → Barrel 4 (5): Barrel 3:6, Barrel 4:5.3. Barrel 4 (5) → Barrel 2 (5): Barrel 2:5, Barrel 4:0.4. Barrel 1 (13) → Barrel 4 (5): Barrel 1:8, Barrel 4:5.5. Barrel 3 (6) → Barrel 2 (5 + 6=11): Barrel 3:0, Barrel 2:11.6. Barrel 1 (8) → Barrel 3 (8): Barrel 1:0, Barrel 3:8.7. Barrel 2 (11) → Barrel 1 (8): Barrel 2:3, Barrel 1:8.8. Barrel 4 (5) → Barrel 2 (3 +5=8): Barrel 4:0, Barrel 2:8.Yes, that works. So, the final state is Barrel 1:8, Barrel 2:8, Barrel 3:8, Barrel 4:0. So, the three equal parts are in Barrels 1, 2, and 3, each containing 8 buckets.

answer：Alright, so I have this problem here: I need to figure out if the product of two expressions, (7a + b - 2c + 1) and (3a - 5b + 4c + 10), is even or odd. The variables a, b, and c are integers. Hmm, okay. Let me think about how to approach this.First, I remember that when multiplying two numbers, the product is even if at least one of the numbers is even. If both numbers are odd, then the product is odd. So, maybe I can figure out whether each of these expressions is even or odd, and then see what their product would be.But wait, how do I determine if each expression is even or odd? I think I can analyze each expression modulo 2 because evenness and oddness are properties related to modulo 2. So, if I can find the parity (whether it's even or odd) of each expression, that should help.Let me write down each expression modulo 2.Starting with the first expression: 7a + b - 2c + 1.Let me compute each term modulo 2.7a modulo 2: 7 is odd, so 7a modulo 2 is the same as a modulo 2 because 7 is congruent to 1 mod 2. So, 7a ≡ a mod 2.b modulo 2: That's just b mod 2.-2c modulo 2: Well, 2c is obviously even, so -2c is also even. So, -2c ≡ 0 mod 2.1 modulo 2: That's 1.So, putting it all together: 7a + b - 2c + 1 ≡ a + b + 0 + 1 ≡ a + b + 1 mod 2.Okay, so the first expression is congruent to a + b + 1 mod 2. So, if a + b is even, then a + b + 1 is odd, and if a + b is odd, then a + b + 1 is even.So, the first expression is odd if a + b is even, and even if a + b is odd.Now, moving on to the second expression: 3a - 5b + 4c + 10.Again, let's compute each term modulo 2.3a modulo 2: 3 is odd, so 3a ≡ a mod 2.-5b modulo 2: 5 is odd, so -5b ≡ -b mod 2, which is the same as b mod 2 because -1 ≡ 1 mod 2.4c modulo 2: 4 is even, so 4c ≡ 0 mod 2.10 modulo 2: 10 is even, so 10 ≡ 0 mod 2.So, putting it all together: 3a - 5b + 4c + 10 ≡ a + b + 0 + 0 ≡ a + b mod 2.Therefore, the second expression is congruent to a + b mod 2. So, if a + b is even, the second expression is even, and if a + b is odd, the second expression is odd.Alright, so let me summarize:First expression: (7a + b - 2c + 1) ≡ a + b + 1 mod 2.Second expression: (3a - 5b + 4c + 10) ≡ a + b mod 2.So, depending on the parity of a + b, both expressions have specific parities.Case 1: a + b is even.Then, first expression: a + b + 1 ≡ 0 + 1 ≡ 1 mod 2, which is odd.Second expression: a + b ≡ 0 mod 2, which is even.So, in this case, we have an odd number multiplied by an even number. The product is even.Case 2: a + b is odd.Then, first expression: a + b + 1 ≡ 1 + 1 ≡ 0 mod 2, which is even.Second expression: a + b ≡ 1 mod 2, which is odd.So, in this case, we have an even number multiplied by an odd number. The product is even.Wait, hold on. In both cases, the product is even? So regardless of whether a + b is even or odd, the product is even?Let me double-check my reasoning.First expression: 7a + b - 2c + 1.Modulo 2: 7a ≡ a, b ≡ b, -2c ≡ 0, 1 ≡1. So, a + b + 1.Second expression: 3a -5b +4c +10.Modulo 2: 3a ≡ a, -5b ≡ b, 4c ≡0, 10≡0. So, a + b.So, first expression is a + b +1, second is a + b.So, if a + b is even:First expression: even +1 = oddSecond expression: evenProduct: odd * even = evenIf a + b is odd:First expression: odd +1 = evenSecond expression: oddProduct: even * odd = evenSo, in both cases, the product is even.Therefore, regardless of the values of a, b, c (as integers), the product is always even.Wait, so is the product always even? Hmm, that seems to be the case.But let me test with some specific numbers to make sure.Let's pick a = 0, b = 0, c = 0.First expression: 7*0 + 0 - 2*0 +1 = 1, which is odd.Second expression: 3*0 -5*0 +4*0 +10 = 10, which is even.Product: 1*10 = 10, which is even.Another example: a =1, b=1, c=1.First expression: 7*1 +1 -2*1 +1 =7 +1 -2 +1=7, which is odd.Second expression:3*1 -5*1 +4*1 +10=3 -5 +4 +10=12, which is even.Product:7*12=84, which is even.Another example: a=1, b=0, c=1.First expression:7*1 +0 -2*1 +1=7 -2 +1=6, which is even.Second expression:3*1 -5*0 +4*1 +10=3 +0 +4 +10=17, which is odd.Product:6*17=102, which is even.Another example: a=2, b=3, c=4.First expression:7*2 +3 -2*4 +1=14 +3 -8 +1=10, which is even.Second expression:3*2 -5*3 +4*4 +10=6 -15 +16 +10=17, which is odd.Product:10*17=170, which is even.Wait, so in all these examples, the product is even. So, seems consistent.Therefore, regardless of the integers a, b, c, the product is always even.Hence, the answer is that the product is even.**Final Answer**The product is always even, so the answer is boxed{even}.

answer：First, I recognize that the problem involves distributing identical coins into distinct boxes. This is a classic combinatorics problem that can be solved using the stars and bars theorem.The formula for the number of ways to distribute ( n ) identical items into ( k ) distinct groups is given by:[binom{n + k - 1}{k - 1}]In this case, there are ( n = 5 ) identical coins and ( k = 3 ) different boxes. Plugging these values into the formula:[binom{5 + 3 - 1}{3 - 1} = binom{7}{2}]Calculating the combination:[binom{7}{2} = frac{7!}{2!(7-2)!} = frac{7 times 6}{2 times 1} = 21]Therefore, there are 21 different ways to distribute the 5 identical coins into the 3 different boxes.

answer：To find the value of ( x ) such that the vectors ( overrightarrow{a} = (2, -1, x) ) and ( overrightarrow{b} = (3, 2, -1) ) are perpendicular, I need to use the property that the dot product of two perpendicular vectors is zero.First, I'll compute the dot product of ( overrightarrow{a} ) and ( overrightarrow{b} ):[overrightarrow{a} cdot overrightarrow{b} = (2)(3) + (-1)(2) + (x)(-1) = 6 - 2 - x = 4 - x]Since the vectors are perpendicular, their dot product must equal zero:[4 - x = 0]Solving for ( x ):[x = 4]

answer：First, I need to calculate the sums of the numerators and denominators in both fractions.For the first fraction, the numerator is (2 + 4 + 6 = 12) and the denominator is (1 + 3 + 5 = 9). So, the first fraction simplifies to (frac{12}{9}), which can be reduced to (frac{4}{3}).For the second fraction, the numerator is (1 + 3 + 5 = 9) and the denominator is (2 + 4 + 6 = 12). This fraction simplifies to (frac{9}{12}), which reduces to (frac{3}{4}).Next, I subtract the second fraction from the first: (frac{4}{3} - frac{3}{4}). To do this, I find a common denominator, which is 12. Converting both fractions, I get (frac{16}{12} - frac{9}{12} = frac{7}{12}).Therefore, the final answer is (frac{7}{12}).