💡First, I need to determine the downstream speed of the man rowing the boat. I know the distance covered downstream is 80 meters and the time taken is 15.99872010239181 seconds. To find the downstream speed, I'll convert the distance from meters to kilometers and the time from seconds to hours.Converting 80 meters to kilometers:80 meters = 0.08 kilometersConverting 15.99872010239181 seconds to hours:15.99872010239181 seconds ÷ 3600 = 0.00444464447344217 hoursNow, calculate the downstream speed (Vd) in km/h:Vd = Distance / Time = 0.08 km / 0.00444464447344217 hours ≈ 18 km/hThe downstream speed is the sum of the man's rowing speed in still water (Vs) and the speed of the current (Vc):Vd = Vs + VcGiven that the man's rowing speed in still water (Vs) is 15 km/h, I can solve for the speed of the current (Vc):18 km/h = 15 km/h + VcVc = 18 km/h - 15 km/h = 3 km/hTherefore, the speed of the current is 3 km/h.

💡Alright, so I have this problem about subsets and sums. Let me try to wrap my head around it step by step. First, the problem says that for an integer ( n geq 3 ), we have a set ( M = {1, 2, 3, ldots, n} ). We need to consider all subsets of ( M ) that contain exactly 3 elements. These subsets are denoted as ( A_1, A_2, A_3, ldots, A_{C_n^3} ), where ( C_n^3 ) is the number of combinations of ( n ) elements taken 3 at a time. Then, ( S_n ) is defined as the sum of all elements in all these subsets. So, for each subset ( A_i ), we add up its three elements, and then we sum all those results together to get ( S_n ).The problem has two parts:Ⅰ. Find ( S_3, S_4, S_5 ), and derive a general formula for ( S_n ).Ⅱ. Prove that the sum ( S_3 + S_4 + ldots + S_n = 6C_{n+2}^5 ).Let me tackle part Ⅰ first.**Part Ⅰ: Finding ( S_3, S_4, S_5 ) and deriving ( S_n )**Starting with ( n = 3 ). The set ( M ) is ( {1, 2, 3} ). The only subset with 3 elements is the set itself. So, the sum ( S_3 ) is just ( 1 + 2 + 3 = 6 ). That seems straightforward.Moving on to ( n = 4 ). The set is ( {1, 2, 3, 4} ). The number of 3-element subsets is ( C_4^3 = 4 ). Let me list them:1. ( {1, 2, 3} ) with sum ( 6 )2. ( {1, 2, 4} ) with sum ( 7 )3. ( {1, 3, 4} ) with sum ( 8 )4. ( {2, 3, 4} ) with sum ( 9 )Adding these up: ( 6 + 7 + 8 + 9 = 30 ). So, ( S_4 = 30 ).Alternatively, instead of listing all subsets, maybe there's a smarter way. Each element in ( M ) appears in the same number of subsets. How many times does each element appear?For ( n = 4 ), each element is in ( C_3^2 = 3 ) subsets because once you fix an element, you need to choose 2 more from the remaining 3. So, each element appears 3 times. Therefore, the total sum ( S_4 ) is ( 3 times (1 + 2 + 3 + 4) = 3 times 10 = 30 ). That matches my earlier result.Okay, so that's a better approach. Instead of listing all subsets, figure out how many times each element appears and multiply by the sum of the elements.Now, for ( n = 5 ). The set is ( {1, 2, 3, 4, 5} ). The number of 3-element subsets is ( C_5^3 = 10 ). Instead of listing all 10 subsets, let's use the same logic.Each element appears in ( C_4^2 = 6 ) subsets, because fixing one element, we choose 2 from the remaining 4. So, each element appears 6 times. The sum of all elements is ( 1 + 2 + 3 + 4 + 5 = 15 ). Therefore, ( S_5 = 6 times 15 = 90 ).Let me verify this by another method. The sum of all elements across all subsets can also be calculated by considering each element's contribution. Since each element is in 6 subsets, each contributes 6 times its value. So, the total sum is ( 6 times (1 + 2 + 3 + 4 + 5) = 6 times 15 = 90 ). Yep, that works.So, generalizing this, for a set ( M ) with ( n ) elements, each element appears in ( C_{n-1}^2 ) subsets. Because, for each element, we need to choose 2 more elements from the remaining ( n - 1 ) elements. Therefore, the total sum ( S_n ) is ( C_{n-1}^2 times ) (sum of all elements in ( M )). The sum of all elements in ( M ) is ( frac{n(n + 1)}{2} ). So, ( S_n = C_{n-1}^2 times frac{n(n + 1)}{2} ).Let me compute ( C_{n-1}^2 ). That's ( frac{(n - 1)(n - 2)}{2} ). So, substituting back:( S_n = frac{(n - 1)(n - 2)}{2} times frac{n(n + 1)}{2} = frac{(n - 1)(n - 2)n(n + 1)}{4} ).Alternatively, this can be written as ( frac{n(n + 1)(n - 1)(n - 2)}{4} ).But the problem didn't specify the form, so either expression is acceptable. However, the original expression ( C_{n-1}^2 times frac{n(n + 1)}{2} ) is more concise.So, summarizing:- ( S_3 = 6 )- ( S_4 = 30 )- ( S_5 = 90 )- General formula: ( S_n = C_{n-1}^2 times frac{n(n + 1)}{2} )Moving on to part Ⅱ.**Part Ⅱ: Proving that ( S_3 + S_4 + ldots + S_n = 6C_{n+2}^5 )**Hmm, okay. So, we need to sum ( S_k ) from ( k = 3 ) to ( k = n ), and show that this sum equals ( 6C_{n+2}^5 ).First, let's recall that ( S_k = C_{k-1}^2 times frac{k(k + 1)}{2} ). So, ( S_k = frac{(k - 1)(k - 2)}{2} times frac{k(k + 1)}{2} = frac{(k - 1)(k - 2)k(k + 1)}{4} ).Therefore, the sum ( S_3 + S_4 + ldots + S_n ) is equal to the sum from ( k = 3 ) to ( k = n ) of ( frac{(k - 1)(k - 2)k(k + 1)}{4} ).Let me factor out the 1/4:( frac{1}{4} sum_{k=3}^n (k - 1)(k - 2)k(k + 1) ).So, I need to compute this sum and show it equals ( 6C_{n+2}^5 ).First, let's simplify the term inside the sum: ( (k - 1)(k - 2)k(k + 1) ).Let me note that ( (k - 2)(k - 1)k(k + 1) ) is equal to ( (k^2 - 3k + 2)(k^2 + k) ). But that might not be helpful.Alternatively, notice that ( (k - 2)(k - 1)k(k + 1) ) is equal to ( (k^2 - 3k + 2)(k^2 + k) ). Hmm, perhaps expanding it:( (k^2 - 3k + 2)(k^2 + k) = k^4 - 3k^3 + 2k^2 + k^3 - 3k^2 + 2k = k^4 - 2k^3 - k^2 + 2k ).But that seems messy. Maybe instead, think combinatorially.Wait, ( (k - 2)(k - 1)k(k + 1) ) is equal to ( k(k - 1)(k - 2)(k + 1) ). Hmm, that's four consecutive integers multiplied together, but not exactly in a straightforward way. Alternatively, perhaps express it as a combination.Wait, ( (k - 2)(k - 1)k(k + 1) ) can be written as ( (k + 1)k(k - 1)(k - 2) ), which is ( P(k + 1, 4) ), the number of permutations of 4 elements from ( k + 1 ). But I'm not sure if that helps.Alternatively, perhaps express it in terms of combinations. Remember that ( C(n, k) = frac{n!}{k!(n - k)!} ).Wait, ( (k - 2)(k - 1)k(k + 1) = (k + 1)k(k - 1)(k - 2) = 4! times C(k + 1, 4) ). Because ( C(k + 1, 4) = frac{(k + 1)k(k - 1)(k - 2)}{4!} ). So, multiplying by 4! gives ( (k + 1)k(k - 1)(k - 2) ).Therefore, ( (k - 2)(k - 1)k(k + 1) = 24 times C(k + 1, 4) ).So, substituting back into the sum:( frac{1}{4} sum_{k=3}^n 24 C(k + 1, 4) = 6 sum_{k=3}^n C(k + 1, 4) ).So, now the sum becomes ( 6 sum_{k=3}^n C(k + 1, 4) ).Let me change the index to make it easier. Let ( m = k + 1 ). When ( k = 3 ), ( m = 4 ); when ( k = n ), ( m = n + 1 ). So, the sum becomes:( 6 sum_{m=4}^{n + 1} C(m, 4) ).So, ( 6 sum_{m=4}^{n + 1} C(m, 4) ).Now, I recall that the sum of combinations ( sum_{m=r}^n C(m, r) = C(n + 1, r + 1) ). This is a combinatorial identity known as the hockey-stick identity or Christmas stocking theorem.In this case, ( r = 4 ), so ( sum_{m=4}^{n + 1} C(m, 4) = C(n + 2, 5) ).Therefore, substituting back:( 6 times C(n + 2, 5) ).Which is exactly what we needed to prove: ( S_3 + S_4 + ldots + S_n = 6C_{n + 2}^5 ).So, putting it all together:- For each ( k ), ( S_k = C_{k - 1}^2 times frac{k(k + 1)}{2} )- Summing ( S_k ) from ( k = 3 ) to ( n ) gives ( 6C_{n + 2}^5 )I think that makes sense. Let me double-check the combinatorial identity. The identity states that ( sum_{m=r}^n C(m, r) = C(n + 1, r + 1) ). So, in our case, ( r = 4 ), and the upper limit is ( n + 1 ), so the sum is ( C(n + 2, 5) ). Multiplying by 6 gives the desired result.Therefore, the proof holds.**Final Answer**Ⅰ. ( S_3 = boxed{6} ), ( S_4 = boxed{30} ), ( S_5 = boxed{90} ), and ( S_n = boxed{dfrac{(n - 1)(n - 2)n(n + 1)}{4}} ).Ⅱ. ( S_3 + S_4 + ldots + S_n = boxed{6C_{n+2}^5} ).

💡Okay, so I have this inequality to solve: 3 times the absolute value of x plus 8 is less than 29. Hmm, let me write that down to make it clearer. It's 3|x| + 8 < 29. Alright, my goal is to find the smallest integer value of x that satisfies this inequality. First, I think I need to isolate the absolute value term. To do that, I'll subtract 8 from both sides of the inequality. So, subtracting 8 from both sides gives me 3|x| < 21. Okay, that simplifies things a bit. Now, I need to get rid of that 3 that's multiplied by the absolute value of x. I can do that by dividing both sides of the inequality by 3. Dividing both sides by 3, I get |x| < 7. Alright, so the absolute value of x is less than 7. I remember that absolute value inequalities can be rewritten without the absolute value by considering two separate cases: one where the expression inside the absolute value is positive and one where it's negative. So, |x| < 7 means that x is between -7 and 7. Let me write that out: -7 < x < 7. So, x has to be greater than -7 and less than 7. Now, the question is asking for the smallest integer value of x that satisfies this inequality. Since x has to be greater than -7, the smallest integer greater than -7 would be -6. Wait, let me make sure I'm not making a mistake here. If x has to be greater than -7, then the integers that satisfy this are -6, -5, -4, and so on up to 6. So, the smallest integer in this range is indeed -6. But just to double-check, let me plug x = -6 back into the original inequality to see if it works. So, 3 times the absolute value of -6 is 3*6, which is 18. Adding 8 gives me 18 + 8 = 26. Is 26 less than 29? Yes, it is. So, x = -6 satisfies the inequality. What if I try x = -7? Let's see: 3 times the absolute value of -7 is 3*7 = 21. Adding 8 gives me 21 + 8 = 29. But the inequality is 3|x| + 8 < 29, which means it has to be strictly less than 29. Since 29 is not less than 29, x = -7 does not satisfy the inequality. Therefore, the smallest integer value of x that works is -6. I think that makes sense. I don't think I missed anything here. The key steps were isolating the absolute value, translating that into a compound inequality, and then identifying the smallest integer within that range. Just to recap: Starting with 3|x| + 8 < 29, subtract 8 to get 3|x| < 21, divide by 3 to get |x| < 7, which means x is between -7 and 7. The smallest integer in that range is -6. Plugging it back in confirms it works, and the next lower integer, -7, doesn't work because it makes the expression equal to 29, which doesn't satisfy the strict inequality. So, yeah, I'm pretty confident the answer is -6.

💡Okay, so I have this problem about a geometric sequence. Let me try to understand it step by step. The sequence is denoted as {a_n}, and all the terms are positive, which is good to know because it might help later on. The equation given is a_1a_5 + 2a_3a_5 + a_3a_7 = 25. I need to find the value of a_3 + a_5.First, I remember that in a geometric sequence, each term is a multiple of the previous term by a common ratio, usually denoted as r. So, the general formula for the nth term is a_n = a_1 * r^(n-1). That means I can express a_5, a_3, and a_7 in terms of a_1 and r.Let me write down the expressions for each term:- a_1 is just a_1.- a_3 = a_1 * r^(3-1) = a_1 * r^2.- a_5 = a_1 * r^(5-1) = a_1 * r^4.- a_7 = a_1 * r^(7-1) = a_1 * r^6.Okay, so now I can substitute these into the given equation:a_1a_5 + 2a_3a_5 + a_3a_7 = 25Substituting the expressions:a_1*(a_1*r^4) + 2*(a_1*r^2)*(a_1*r^4) + (a_1*r^2)*(a_1*r^6) = 25Let me simplify each term:First term: a_1 * a_1 * r^4 = a_1^2 * r^4Second term: 2 * (a_1*r^2) * (a_1*r^4) = 2 * a_1^2 * r^(2+4) = 2a_1^2 * r^6Third term: (a_1*r^2) * (a_1*r^6) = a_1^2 * r^(2+6) = a_1^2 * r^8So putting it all together:a_1^2 * r^4 + 2a_1^2 * r^6 + a_1^2 * r^8 = 25Hmm, this looks a bit complicated. Maybe I can factor out a common term. I notice that each term has a_1^2 * r^4. Let me factor that out:a_1^2 * r^4 * (1 + 2r^2 + r^4) = 25Now, let's look at the expression inside the parentheses: 1 + 2r^2 + r^4. Wait a minute, that looks familiar. It's a quadratic in terms of r^2. Let me set x = r^2 for a moment. Then the expression becomes 1 + 2x + x^2, which is (x + 1)^2. Substituting back, that's (r^2 + 1)^2.So now the equation becomes:a_1^2 * r^4 * (r^2 + 1)^2 = 25Hmm, maybe I can write this as (a_1 * r^2 * (r^2 + 1))^2 = 25. Let me check:(a_1 * r^2 * (r^2 + 1))^2 = a_1^2 * r^4 * (r^2 + 1)^2. Yes, that's correct.So, (a_1 * r^2 * (r^2 + 1))^2 = 25Taking the square root of both sides, since all terms are positive, we can ignore the negative root:a_1 * r^2 * (r^2 + 1) = 5Now, I need to find a_3 + a_5. Let me express a_3 and a_5 in terms of a_1 and r:a_3 = a_1 * r^2a_5 = a_1 * r^4So, a_3 + a_5 = a_1 * r^2 + a_1 * r^4 = a_1 * r^2 (1 + r^2)Wait, that's exactly the same as the expression I had earlier: a_1 * r^2 * (r^2 + 1). And from the equation above, that equals 5.So, a_3 + a_5 = 5.That seems straightforward, but let me double-check my steps to make sure I didn't make any mistakes.1. Expressed each term in the geometric sequence correctly in terms of a_1 and r.2. Substituted into the given equation and simplified each term.3. Factored out a common term, recognized the quadratic expression, and rewrote it as a square.4. Took the square root, considering only the positive root since all terms are positive.5. Noted that the expression a_1 * r^2 * (r^2 + 1) is equal to a_3 + a_5, which was the desired value.Everything seems to check out. I don't see any errors in my reasoning. So, the value of a_3 + a_5 is indeed 5.

💡To determine how many of the given propositions imply the truth of ((p wedge q) rightarrow r), we need to evaluate each proposition in the context of the implication.1. **Proposition 1: (p wedge q wedge sim r)** - Here, (p) and (q) are true, and (r) is false. - The implication ((p wedge q) rightarrow r) becomes (text{true} rightarrow text{false}), which is false. - Therefore, Proposition 1 does not imply the truth of the implication.2. **Proposition 2: (sim p wedge q wedge r)** - Here, (p) is false, (q) and (r) are true. - The implication ((p wedge q)) is false because (p) is false. - An implication with a false antecedent is always true, so ((p wedge q) rightarrow r) is true. - Therefore, Proposition 2 implies the truth of the implication.3. **Proposition 3: (p wedge sim q wedge r)** - Here, (p) is true, (q) is false, and (r) is true. - The implication ((p wedge q)) is false because (q) is false. - An implication with a false antecedent is always true, so ((p wedge q) rightarrow r) is true. - Therefore, Proposition 3 implies the truth of the implication.4. **Proposition 4: (sim p wedge sim q wedge sim r)** - Here, (p) and (q) are false, and (r) is false. - The implication ((p wedge q)) is false because both (p) and (q) are false. - An implication with a false antecedent is always true, so ((p wedge q) rightarrow r) is true. - Therefore, Proposition 4 implies the truth of the implication.In conclusion, Propositions 2, 3, and 4 imply the truth of ((p wedge q) rightarrow r). Thus, the number of propositions that imply the truth of the implication is 3.

💡First, I need to understand the problem. We have six integers: ( a, b, c, d, e, ) and ( f ). The condition given is that ( ab + cdef < 0 ). The goal is to determine the maximum number of these integers that can be negative while still satisfying the condition.To maximize the number of negative integers, I should consider how the signs of these integers affect the overall expression ( ab + cdef ).1. **Analyzing ( ab ):** - If both ( a ) and ( b ) are positive, their product ( ab ) is positive. - If one of ( a ) or ( b ) is negative and the other is positive, their product ( ab ) is negative. - If both ( a ) and ( b ) are negative, their product ( ab ) is positive. To make ( ab ) negative, one of ( a ) or ( b ) must be negative, and the other must be positive.2. **Analyzing ( cdef ):** - The product of an even number of negative integers is positive. - The product of an odd number of negative integers is negative. To make ( cdef ) negative, an odd number of ( c, d, e, ) and ( f ) must be negative.3. **Combining ( ab ) and ( cdef ):** - If ( ab ) is negative and ( cdef ) is also negative, their sum ( ab + cdef ) will be negative. - If ( ab ) is positive and ( cdef ) is negative, the sum could be negative depending on the magnitudes. - If ( ab ) is negative and ( cdef ) is positive, the sum could be negative depending on the magnitudes. - If both ( ab ) and ( cdef ) are positive, their sum will be positive. To ensure ( ab + cdef < 0 ), it's safest to have both ( ab ) and ( cdef ) negative.4. **Maximizing Negative Integers:** - From ( ab ), we can have 1 negative integer. - From ( cdef ), to make the product negative, we need an odd number of negative integers. The maximum odd number less than or equal to 4 is 3. Therefore, we can have 1 negative integer from ( ab ) and 3 negative integers from ( cdef ), totaling 4 negative integers.5. **Conclusion:** - The maximum number of negative integers possible while satisfying ( ab + cdef < 0 ) is 4.