💡Alright, so I have this problem to prove that the floor of the square roots of 4n+1, 4n+2, and 4n+3 are all equal for any natural number n. Hmm, okay. Let me try to understand what this means.First, the floor function, denoted by [x], gives the greatest integer less than or equal to x. So, [√(4n+1)] is the largest integer that is less than or equal to √(4n+1). Similarly for the others.I need to show that these three floor values are the same. That means that √(4n+1), √(4n+2), and √(4n+3) all lie within the same integer interval. In other words, there exists some integer k such that k ≤ √(4n+1) < k+1, and similarly for √(4n+2) and √(4n+3).So, maybe I can start by assuming that k is the floor of √(4n+1). That is, k = [√(4n+1)]. Then, by definition, k ≤ √(4n+1) < k+1. If I can show that this same k also satisfies k ≤ √(4n+2) < k+1 and k ≤ √(4n+3) < k+1, then I'm done.Let me write that down:Let k = [√(4n+1)]. Then,k ≤ √(4n+1) < k+1.I need to show that:k ≤ √(4n+2) < k+1,andk ≤ √(4n+3) < k+1.Okay, so starting with the first inequality, k ≤ √(4n+1) < k+1.If I square all parts of the inequality, since all terms are positive, the inequality directions will remain the same.So,k² ≤ 4n+1 < (k+1)².Now, let's see what this tells us about 4n+2 and 4n+3.Since 4n+2 is just 4n+1 +1, and 4n+3 is 4n+1 +2, we can write:4n+1 < 4n+2 < 4n+3.But from the earlier inequality, we have 4n+1 < (k+1)².So, 4n+2 and 4n+3 are both less than (k+1)² as well, right?Wait, let me check that.If 4n+1 < (k+1)², then adding 1 or 2 to 4n+1 will still keep it less than (k+1)², provided that (k+1)² is greater than 4n+3.Is that necessarily true?Hmm, maybe I need to find a relationship between k and n.Let's think about the value of k in terms of n.Since k = [√(4n+1)], we can write:k ≤ √(4n+1) < k+1.Squaring all parts,k² ≤ 4n+1 < (k+1)².So, 4n+1 is between k² and (k+1)².Therefore, 4n is between k² -1 and (k+1)² -1.So,k² -1 ≤ 4n < (k+1)² -1.Dividing by 4,(k² -1)/4 ≤ n < ((k+1)² -1)/4.Hmm, interesting. So n is in this interval defined by k.But I'm not sure if this helps directly. Maybe I need another approach.Let me consider the differences between 4n+1, 4n+2, and 4n+3.The differences are small: 1 and 2. So, their square roots shouldn't differ by much. Maybe the floor remains the same.But how can I formalize this?Suppose that √(4n+1) is just a little bit more than k. Then, √(4n+2) and √(4n+3) would be a bit more, but still less than k+1.But I need to make sure that adding 1 or 2 to 4n+1 doesn't push the square root over to the next integer.Is there a way to bound √(4n+3) in terms of k?Let me see.From k² ≤ 4n+1 < (k+1)²,I can write 4n+1 ≥ k²,so 4n ≥ k² -1,and 4n+3 ≥ k² +2.Similarly, 4n+3 < (k+1)² +2.Wait, that might not be helpful.Alternatively, maybe I can express 4n+2 and 4n+3 in terms of 4n+1.So, 4n+2 = 4n+1 +1,and 4n+3 = 4n+1 +2.So, √(4n+2) = √(4n+1 +1),and √(4n+3) = √(4n+1 +2).Now, since √(x + a) - √x is a decreasing function for x > 0 and a > 0, the increments in the square roots will be less than the increments in x.But I'm not sure if that helps.Alternatively, maybe I can use the mean value theorem or some approximation for the square root function.But perhaps that's overcomplicating.Wait, let's think about the maximum possible value of √(4n+3).From the initial inequality, 4n+1 < (k+1)².So, 4n+3 < (k+1)² +2.But that's not helpful because (k+1)² +2 is larger than (k+1)², so it doesn't give me an upper bound.Alternatively, maybe I can find an upper bound for √(4n+3).Since 4n+3 < 4n+4 = 4(n+1).So, √(4n+3) < √(4(n+1)) = 2√(n+1).But I don't know if that helps.Wait, perhaps I can relate this to k.From k² ≤ 4n+1 < (k+1)²,so 4n+1 is less than (k+1)²,so 4n+3 < (k+1)² +2.But again, not helpful.Alternatively, maybe I can consider the difference between (k+1)² and 4n+3.From k² ≤ 4n+1 < (k+1)²,so 4n+1 < (k+1)²,so 4n+3 < (k+1)² +2.But I need to show that √(4n+3) < k+1.Which would be true if 4n+3 < (k+1)².But from above, 4n+1 < (k+1)²,so 4n+3 < (k+1)² +2.But that doesn't necessarily mean 4n+3 < (k+1)².Wait, unless (k+1)² - (4n+1) > 2.Is that always true?From k² ≤ 4n+1 < (k+1)²,so (k+1)² - (4n+1) > 0.But how much is it?It depends on n.If 4n+1 is just slightly less than (k+1)², then (k+1)² - (4n+1) could be as small as just over 0.So, 4n+3 could be as large as (k+1)² -1 +2 = (k+1)² +1.Wait, that's not helpful.Hmm, maybe I need to approach this differently.Let me consider specific values of n and see what happens.For example, let n=1.Then, 4n+1=5, 4n+2=6, 4n+3=7.√5 ≈2.236, √6≈2.449, √7≈2.645.So, [√5]=[√6]=[√7]=2.Similarly, for n=2,4n+1=9, 4n+2=10, 4n+3=11.√9=3, √10≈3.162, √11≈3.316.So, [√9]=[√10]=[√11]=3.Wait, but 4n+1=9 is a perfect square, so [√9]=3, which is the same as [√10] and [√11].Similarly, for n=3,4n+1=13, 4n+2=14, 4n+3=15.√13≈3.606, √14≈3.741, √15≈3.872.So, [√13]=[√14]=[√15]=3.Okay, so in these examples, it's holding.But I need to prove it for any natural number n.Maybe I can consider the interval between k² and (k+1)².Suppose that 4n+1 is in [k², (k+1)²).Then, 4n+2 and 4n+3 are just 1 and 2 more than 4n+1.So, if 4n+1 is less than (k+1)², then 4n+2 and 4n+3 are also less than (k+1)² + something.But I need to ensure that 4n+3 is still less than (k+1)².Wait, no, because 4n+3 could potentially be equal to (k+1)².But since 4n+1 < (k+1)²,then 4n+3 < (k+1)² +2.But (k+1)² +2 is larger than (k+1)², so it doesn't help.Wait, maybe I need to find a lower bound for (k+1)² - (4n+1).From k² ≤ 4n+1 < (k+1)²,so (k+1)² - (4n+1) > 0.But how much is it?It depends on n.If 4n+1 is just above k², then (k+1)² - (4n+1) could be as large as (k+1)² - k² -1 = 2k.Wait, let's compute (k+1)² - k² = 2k +1.So, (k+1)² - (4n+1) = 2k +1 - (4n+1 -k²).But 4n+1 -k² is between 0 and 2k.Wait, maybe I'm overcomplicating.Let me think differently.Suppose that k = [√(4n+1)].Then, k ≤ √(4n+1) < k+1.Squaring,k² ≤ 4n+1 < (k+1)².So, 4n+1 is in [k², (k+1)²).Now, 4n+2 = 4n+1 +1,and 4n+3 = 4n+1 +2.So, 4n+2 is in [k² +1, (k+1)² +1),and 4n+3 is in [k² +2, (k+1)² +2).But I need to show that √(4n+2) and √(4n+3) are still less than k+1.Which would mean that 4n+2 < (k+1)²,and 4n+3 < (k+1)².But from the initial inequality, 4n+1 < (k+1)²,so 4n+2 < (k+1)² +1,and 4n+3 < (k+1)² +2.But that doesn't necessarily mean that 4n+2 or 4n+3 are less than (k+1)².Wait, unless (k+1)² - (4n+1) > 2.Is that always true?From k² ≤ 4n+1 < (k+1)²,so (k+1)² - (4n+1) > 0.But it could be as small as just over 0.For example, if 4n+1 is just less than (k+1)²,then (k+1)² - (4n+1) is just over 0.So, 4n+3 could be as large as (k+1)² -1 +2 = (k+1)² +1.Which is greater than (k+1)².So, √(4n+3) could be greater than k+1.Wait, but that contradicts the earlier examples.Wait, in the examples, when n=1, 4n+3=7, and √7≈2.645 <3.Similarly, when n=2, 4n+3=11, √11≈3.316 <4.Wait, but if n is such that 4n+1 is just less than (k+1)²,then 4n+3 could be just less than (k+1)² +2.But √(4n+3) would still be less than k+1 + something.Wait, maybe I'm missing something.Let me think about the maximum possible value of √(4n+3).From k² ≤ 4n+1 < (k+1)²,so 4n+1 < (k+1)²,so 4n+3 < (k+1)² +2.But √(4n+3) < √((k+1)² +2).Now, √((k+1)² +2) = (k+1)√(1 + 2/(k+1)²).Using the approximation √(1 + x) ≈1 + x/2 for small x,so √(1 + 2/(k+1)²) ≈1 + 1/(k+1)².Thus, √((k+1)² +2) ≈ (k+1) + 1/(k+1).So, √(4n+3) < (k+1) + 1/(k+1).But since 1/(k+1) <1, this means that √(4n+3) < k+2.But we need to show that √(4n+3) < k+1.Wait, that's not helpful.Hmm, maybe I need to find a better bound.Alternatively, perhaps I can consider that between k² and (k+1)², the numbers 4n+1, 4n+2, 4n+3 are consecutive numbers, so their square roots are close enough that their floors are the same.But I need to formalize this.Maybe I can use the fact that the difference between consecutive squares is 2k+1.So, (k+1)² -k² = 2k+1.Now, 4n+1, 4n+2, 4n+3 are three consecutive numbers.So, the difference between 4n+3 and 4n+1 is 2.So, if 4n+1 is in [k², (k+1)²),then 4n+3 is in [k² +2, (k+1)² +2).But I need to ensure that 4n+3 is still less than (k+1)².Which would require that (k+1)² - (4n+1) >2.But from the initial inequality, (k+1)² - (4n+1) >0,but it could be less than or equal to 2.Wait, so if (k+1)² - (4n+1) >2,then 4n+3 < (k+1)²,so √(4n+3) <k+1.But if (k+1)² - (4n+1) ≤2,then 4n+3 ≥ (k+1)² -1,so √(4n+3) ≥ √((k+1)² -1).But √((k+1)² -1) = (k+1)√(1 -1/(k+1)²) ≈ (k+1) - 1/(2(k+1)).So, √(4n+3) is still less than k+1.Wait, but if 4n+3 is equal to (k+1)²,then √(4n+3)=k+1,but since 4n+1 < (k+1)²,4n+3 < (k+1)² +2,but 4n+3 could be equal to (k+1)²,which would make √(4n+3)=k+1,but in that case, [√(4n+3)]=k+1,which would contradict the initial assumption that [√(4n+1)]=k.Wait, but if 4n+1 < (k+1)²,then 4n+3 < (k+1)² +2,but 4n+3 could be equal to (k+1)²,which would mean that √(4n+3)=k+1,so [√(4n+3)]=k+1,which would not equal k.But in our earlier examples, when n=1,4n+3=7,and √7≈2.645,so [√7]=2,which is equal to k=2.Similarly, when n=2,4n+3=11,√11≈3.316,so [√11]=3,which is equal to k=3.Wait, but if 4n+3=(k+1)²,then √(4n+3)=k+1,so [√(4n+3)]=k+1,which is different from k.But in our initial assumption, k=[√(4n+1)].So, if 4n+3=(k+1)²,then 4n+1=(k+1)² -2.But from the initial inequality,k² ≤4n+1 < (k+1)²,so (k+1)² -2 < (k+1)²,which is true,but 4n+1=(k+1)² -2,so [√(4n+1)]=k,since √((k+1)² -2)=k+1 - ε,where ε is small.So, in this case, [√(4n+1)]=k,and [√(4n+3)]=k+1,which would contradict the statement.But in our earlier examples, this doesn't happen.Wait, maybe it's impossible for 4n+3 to be a perfect square.Let me check.Suppose that 4n+3=m²,for some integer m.Then, m² ≡3 mod4.But squares modulo 4 are 0 or 1.Because:0²=0,1²=1,2²=4≡0,3²=9≡1 mod4.So, m²≡0 or1 mod4.Thus, 4n+3≡3 mod4,which cannot be a square.Therefore, 4n+3 cannot be a perfect square.Therefore, √(4n+3) is never an integer,so [√(4n+3)] is always less than √(4n+3).Similarly, since 4n+3 is not a perfect square,√(4n+3) <k+1,because if it were equal to k+1,then 4n+3=(k+1)²,which is impossible.Therefore, [√(4n+3)]=k.Similarly, for 4n+2,since 4n+2 is not a perfect square either,because 4n+2≡2 mod4,and squares are 0 or1 mod4,so √(4n+2) is not an integer,so [√(4n+2)]=k.Therefore, [√(4n+1)]=[√(4n+2)]=[√(4n+3)]=k.Thus, the proof is complete.

💡Alright, so I have this arithmetic sequence problem here. It says that in the arithmetic sequence {a_n}, the first term a₁ is 2, and the sum of the first three terms S₃ is 9. There are two parts to this problem: first, I need to find the general term formula a_n of the sequence, and second, I need to find the sum S_n of the first n terms of the sequence {2a_n}. Okay, let's start with part (1). I remember that in an arithmetic sequence, each term is obtained by adding a common difference d to the previous term. So, the general formula for the nth term of an arithmetic sequence is a_n = a₁ + (n-1)d. Here, a₁ is given as 2, so if I can find the common difference d, I can write the general term formula.Now, the sum of the first n terms of an arithmetic sequence is given by S_n = n/2 [2a₁ + (n-1)d]. In this case, S₃ is given as 9. So, plugging in n=3, a₁=2, and S₃=9 into the sum formula, I can solve for d.Let me write that out:S₃ = 3/2 [2*2 + (3-1)d] = 9Simplifying inside the brackets first: 2*2 is 4, and (3-1)d is 2d. So, the equation becomes:3/2 [4 + 2d] = 9Now, I can multiply both sides by 2 to eliminate the denominator:3*(4 + 2d) = 18Expanding the left side:12 + 6d = 18Subtracting 12 from both sides:6d = 6Dividing both sides by 6:d = 1Okay, so the common difference d is 1. Now, plugging this back into the general term formula:a_n = a₁ + (n-1)d = 2 + (n-1)*1 = 2 + n - 1 = n + 1So, the general term formula is a_n = n + 1. That seems straightforward.Now, moving on to part (2). I need to find the sum S_n of the first n terms of the sequence {2a_n}. So, first, let's understand what {2a_n} is. Since a_n = n + 1, then 2a_n = 2*(n + 1) = 2n + 2. So, the sequence {2a_n} is 2n + 2 for each term.Wait a second, is {2a_n} still an arithmetic sequence? Let me check. The original sequence {a_n} is arithmetic with common difference d=1. If I multiply each term by 2, does that preserve the arithmetic nature? Yes, because multiplying each term by a constant factor preserves the common difference. So, {2a_n} is also an arithmetic sequence with the first term 2a₁ = 2*2 = 4 and common difference 2d = 2*1 = 2.Therefore, the sequence {2a_n} is an arithmetic sequence with first term 4 and common difference 2. So, to find the sum S_n of the first n terms of this new sequence, I can use the sum formula for an arithmetic sequence again.The sum formula is S_n = n/2 [2a₁' + (n-1)d'], where a₁' is the first term of the new sequence and d' is its common difference.Plugging in the values:a₁' = 4, d' = 2So,S_n = n/2 [2*4 + (n-1)*2] = n/2 [8 + 2n - 2] = n/2 [2n + 6]Simplifying further:S_n = n/2 * 2(n + 3) = n(n + 3) = n² + 3nWait, that seems too simple. Let me double-check my steps.Starting from the sum formula:S_n = n/2 [2a₁' + (n-1)d']= n/2 [2*4 + (n-1)*2]= n/2 [8 + 2n - 2]= n/2 [2n + 6]= n/2 * 2(n + 3)= n(n + 3)= n² + 3nYes, that seems correct. Alternatively, I can think of {2a_n} as 2 times the original sequence {a_n}, so the sum S_n of {2a_n} would be 2 times the sum of the first n terms of {a_n}.The sum of the first n terms of {a_n} is S_n' = n/2 [2a₁ + (n-1)d] = n/2 [4 + (n-1)*1] = n/2 [n + 3]Therefore, the sum of {2a_n} would be 2*S_n' = 2*(n/2 [n + 3]) = n(n + 3) = n² + 3nSo, both methods give the same result, which is reassuring.Wait, but the problem says "the sum S_n of the first n terms of {2a_n}". So, is there another way to interpret this? Maybe {2a_n} is a geometric sequence instead? Let me see.Given that a_n = n + 1, then 2a_n = 2(n + 1). Is this a geometric sequence? A geometric sequence has a common ratio between consecutive terms, meaning each term is a multiple of the previous term. Let's check:First term: 2a₁ = 2*(1 + 1) = 4Second term: 2a₂ = 2*(2 + 1) = 6Third term: 2a₃ = 2*(3 + 1) = 8Fourth term: 2a₄ = 2*(4 + 1) = 10So, the terms are 4, 6, 8, 10,... which is an arithmetic sequence with common difference 2, not a geometric sequence. So, my initial thought was correct; {2a_n} is arithmetic, not geometric.Therefore, the sum S_n is n² + 3n.But just to be thorough, let me consider if the problem might have intended {2a_n} as a geometric sequence. If that were the case, then 2a_n would have to be a geometric sequence, which would require that each term is a multiple of the previous term by a common ratio r.But as we saw, the terms are 4, 6, 8, 10,... which does not have a common ratio. 6/4 = 1.5, 8/6 ≈ 1.333, which are not equal. So, it's definitely not a geometric sequence.Therefore, I can confidently say that {2a_n} is an arithmetic sequence, and its sum S_n is n² + 3n.So, summarizing my findings:1. The general term formula for {a_n} is a_n = n + 1.2. The sum of the first n terms of {2a_n} is S_n = n² + 3n.I think that's it. I don't see any mistakes in my reasoning, and both methods of calculating the sum gave me the same result, which adds to my confidence.**Final Answer**(1) The general term formula is boxed{a_n = n + 1}.(2) The sum of the first ( n ) terms of ({2a_n}) is boxed{n^2 + 3n}.

💡Okay, so I have this problem with two vectors, a and b. Vector a is given as (1, n, 1/2) and vector b is (-2, 1, -1). The problem says that 2a minus b is parallel to b, and I need to find the magnitude of vector a.First, I remember that if two vectors are parallel, one must be a scalar multiple of the other. So, if 2a - b is parallel to b, then 2a - b should equal some scalar k multiplied by b. That gives me an equation: 2a - b = k * b.Let me write that out in terms of components. Vector a is (1, n, 1/2), so 2a would be (2*1, 2*n, 2*(1/2)) which simplifies to (2, 2n, 1). Then, subtracting vector b, which is (-2, 1, -1), from 2a, I get:2a - b = (2 - (-2), 2n - 1, 1 - (-1)) = (4, 2n - 1, 2).So, 2a - b is (4, 2n - 1, 2). Since this is parallel to b, which is (-2, 1, -1), there must be a scalar k such that:(4, 2n - 1, 2) = k * (-2, 1, -1) = (-2k, k, -k).Now, I can set up equations by equating the corresponding components:1. 4 = -2k2. 2n - 1 = k3. 2 = -kLet me solve these equations one by one. Starting with equation 3: 2 = -k. Solving for k, I get k = -2.Now, plug k = -2 into equation 1: 4 = -2*(-2) = 4. That checks out, so no problem there.Next, plug k = -2 into equation 2: 2n - 1 = -2. Solving for n:2n - 1 = -2 2n = -2 + 1 2n = -1 n = -1/2.Okay, so n is -1/2. Now, I need to find the magnitude of vector a. Vector a is (1, n, 1/2), which with n = -1/2 becomes (1, -1/2, 1/2).The magnitude of a vector is calculated by the square root of the sum of the squares of its components. So:|a| = sqrt(1^2 + (-1/2)^2 + (1/2)^2) = sqrt(1 + 1/4 + 1/4) = sqrt(1 + 2/4) = sqrt(1 + 1/2) = sqrt(3/2) = sqrt(6)/2.Wait, let me double-check that. 1 squared is 1, (-1/2) squared is 1/4, and (1/2) squared is also 1/4. So adding them up: 1 + 1/4 + 1/4 = 1 + 2/4 = 1 + 1/2 = 3/2. The square root of 3/2 is sqrt(3)/sqrt(2), which can be rationalized as sqrt(6)/2. Yeah, that seems right.So, the magnitude of vector a is sqrt(6)/2.

💡Okay, so I have this problem here where an object is moving in a straight line according to the rule ( x = 4t^2 ) meters. The resistance of the medium is proportional to the velocity. When the velocity is ( 10 , text{m/s} ), the resistance is ( 2 , text{N} ). I need to find the integral expression for the work done by the resistance as the object moves from ( x = 0 ) to ( x = 2 ).Alright, let's break this down. First, I remember that work done by a force is given by the integral of the force over the distance it acts. The formula is ( W = int F , dx ). But in this case, the force is variable because it depends on the velocity, which in turn depends on time. So, I need to express the force in terms of ( x ) to set up the integral correctly.The problem says the resistance is proportional to the velocity. That means ( F = k v ), where ( k ) is the constant of proportionality. I need to find ( k ). They give me that when the velocity ( v = 10 , text{m/s} ), the resistance ( F = 2 , text{N} ). So, plugging these values into the equation, I get:( 2 = k times 10 )Solving for ( k ), I divide both sides by 10:( k = frac{2}{10} = frac{1}{5} )So, ( k = frac{1}{5} , text{N·s/m} ). That means the resistance force can be written as:( F = frac{1}{5} v )But I need ( F ) in terms of ( x ), not ( v ). To do that, I should express ( v ) in terms of ( x ). Since ( x = 4t^2 ), I can find ( v ) by differentiating ( x ) with respect to time ( t ):( v = frac{dx}{dt} = frac{d}{dt}(4t^2) = 8t )So, ( v = 8t ). Now, I need to express ( t ) in terms of ( x ) so that I can substitute it back into the expression for ( v ). From ( x = 4t^2 ), solving for ( t ):( t^2 = frac{x}{4} )Taking the square root of both sides:( t = sqrt{frac{x}{4}} = frac{sqrt{x}}{2} )Now, substitute this expression for ( t ) back into the equation for ( v ):( v = 8t = 8 times frac{sqrt{x}}{2} = 4sqrt{x} )So, ( v = 4sqrt{x} ). Now, substitute this back into the expression for ( F ):( F = frac{1}{5} v = frac{1}{5} times 4sqrt{x} = frac{4}{5}sqrt{x} )Therefore, the force as a function of ( x ) is ( F(x) = frac{4}{5}sqrt{x} ). But wait, the problem mentions that the resistance is in the opposite direction to the motion. That means the work done by the resistance will be negative because the force is opposing the displacement. So, the work done ( W ) is:( W = -int_{x_1}^{x_2} F(x) , dx )Here, ( x_1 = 0 ) and ( x_2 = 2 ). Plugging in the values, we get:( W = -int_{0}^{2} frac{4}{5}sqrt{x} , dx )I can factor out the constant ( frac{4}{5} ) from the integral:( W = -frac{4}{5} int_{0}^{2} sqrt{x} , dx )Now, I need to compute this integral. The integral of ( sqrt{x} ) is a standard one. Remember that ( sqrt{x} = x^{1/2} ), and the integral of ( x^n ) is ( frac{x^{n+1}}{n+1} ) for ( n neq -1 ). So, applying that here:( int sqrt{x} , dx = int x^{1/2} , dx = frac{x^{3/2}}{3/2} = frac{2}{3}x^{3/2} )So, putting it back into the expression for ( W ):( W = -frac{4}{5} left[ frac{2}{3}x^{3/2} right]_{0}^{2} )Now, evaluate this from 0 to 2:First, plug in ( x = 2 ):( frac{2}{3}(2)^{3/2} = frac{2}{3} times (2 times sqrt{2}) = frac{2}{3} times 2sqrt{2} = frac{4sqrt{2}}{3} )Then, plug in ( x = 0 ):( frac{2}{3}(0)^{3/2} = 0 )So, subtracting the lower limit from the upper limit:( frac{4sqrt{2}}{3} - 0 = frac{4sqrt{2}}{3} )Now, multiply by ( -frac{4}{5} ):( W = -frac{4}{5} times frac{4sqrt{2}}{3} = -frac{16sqrt{2}}{15} )Wait, hold on. I think I made a mistake here. Earlier, I expressed ( v = 4sqrt{x} ), but let me double-check that step.Starting from ( x = 4t^2 ), so ( t = sqrt{frac{x}{4}} = frac{sqrt{x}}{2} ). Then, ( v = 8t = 8 times frac{sqrt{x}}{2} = 4sqrt{x} ). That seems correct.Then, ( F = frac{1}{5} v = frac{4}{5}sqrt{x} ). That also seems correct.Then, setting up the integral ( W = -int_{0}^{2} frac{4}{5}sqrt{x} , dx ). That looks right.Computing the integral, ( int sqrt{x} , dx = frac{2}{3}x^{3/2} ). Correct.So, plugging in the limits, ( frac{2}{3}(2)^{3/2} = frac{2}{3} times 2sqrt{2} = frac{4sqrt{2}}{3} ). That's correct.Then, ( W = -frac{4}{5} times frac{4sqrt{2}}{3} = -frac{16sqrt{2}}{15} ). Hmm, but in the initial problem, the user had a different result. Wait, let me check my calculations again.Wait, in the initial problem, the user had ( W = -frac{16}{15} ). But in my calculation, I have ( -frac{16sqrt{2}}{15} ). So, which one is correct?Looking back, perhaps I made a mistake in evaluating ( (2)^{3/2} ). Let's compute ( (2)^{3/2} ). That is ( sqrt{2^3} = sqrt{8} = 2sqrt{2} ). So, that part is correct.So, ( frac{2}{3} times 2sqrt{2} = frac{4sqrt{2}}{3} ). Correct.Then, multiplying by ( -frac{4}{5} ):( -frac{4}{5} times frac{4sqrt{2}}{3} = -frac{16sqrt{2}}{15} ). So, that seems correct.But in the initial problem, the user had ( -frac{16}{15} ). So, perhaps they considered ( (2)^{3/2} ) as ( 2^{1.5} = 2.828 ), but in exact terms, it's ( 2sqrt{2} ). So, maybe the answer is supposed to be in terms of ( sqrt{2} ), or perhaps I made an error in expressing ( v ) in terms of ( x ).Wait, let's go back to the expression for ( F(x) ). The user had ( F(x) = frac{4}{5}sqrt{x} ). Then, the integral was set up as ( W = -int_{0}^{2} frac{4}{5}sqrt{x} , dx ), which evaluates to ( -frac{16}{15} ). But how?Wait, if ( int_{0}^{2} sqrt{x} , dx = frac{2}{3}x^{3/2} ) evaluated from 0 to 2 is ( frac{2}{3}(2)^{3/2} - 0 = frac{2}{3} times 2sqrt{2} = frac{4sqrt{2}}{3} ). So, multiplying by ( frac{4}{5} ), we get ( frac{16sqrt{2}}{15} ), and with the negative sign, ( -frac{16sqrt{2}}{15} ).But in the initial problem, the user had ( -frac{16}{15} ). So, perhaps they made a mistake in their calculation, or maybe I'm missing something.Wait, let me check the expression for ( F(x) ). The user had ( F(x) = frac{4}{5}sqrt{x} ). Let me verify that.Given ( F = kv ), and ( v = 8t ). When ( v = 10 ), ( F = 2 ). So, ( 2 = k times 10 ), so ( k = frac{1}{5} ). Then, ( F = frac{1}{5}v ). Since ( v = 8t ), and ( t = frac{sqrt{x}}{2} ), then ( v = 8 times frac{sqrt{x}}{2} = 4sqrt{x} ). Therefore, ( F = frac{1}{5} times 4sqrt{x} = frac{4}{5}sqrt{x} ). So, that's correct.Therefore, the integral should be ( -frac{4}{5} times frac{4sqrt{2}}{3} = -frac{16sqrt{2}}{15} ). So, perhaps the initial problem had a mistake, or maybe they considered ( x ) in a different way.Alternatively, maybe I should express the integral in terms of ( t ) instead of ( x ). Let's try that approach.Given ( x = 4t^2 ), so ( dx = 8t , dt ). Then, the work done is ( W = -int F , dx = -int F times dx ). Since ( F = frac{1}{5}v = frac{1}{5} times 8t = frac{8}{5}t ). So, ( W = -int frac{8}{5}t times 8t , dt ). Wait, that seems off.Wait, no. Let's clarify. The work done is ( W = int F cdot dx ). Since ( F ) is a function of ( t ), and ( dx ) is also a function of ( t ), we can express the integral in terms of ( t ).Given ( x = 4t^2 ), so ( dx = 8t , dt ). And ( F = frac{1}{5}v = frac{1}{5} times 8t = frac{8}{5}t ). Therefore, ( W = -int F , dx = -int frac{8}{5}t times 8t , dt = -int frac{64}{5}t^2 , dt ).Now, we need to find the limits of integration in terms of ( t ). When ( x = 0 ), ( t = 0 ). When ( x = 2 ), ( 2 = 4t^2 ), so ( t^2 = frac{1}{2} ), so ( t = frac{sqrt{2}}{2} ).Therefore, the integral becomes:( W = -frac{64}{5} int_{0}^{frac{sqrt{2}}{2}} t^2 , dt )Integrating ( t^2 ) gives ( frac{t^3}{3} ). So,( W = -frac{64}{5} left[ frac{t^3}{3} right]_{0}^{frac{sqrt{2}}{2}} = -frac{64}{5} times left( frac{(frac{sqrt{2}}{2})^3}{3} - 0 right) )Compute ( (frac{sqrt{2}}{2})^3 ):( (frac{sqrt{2}}{2})^3 = frac{(sqrt{2})^3}{2^3} = frac{2sqrt{2}}{8} = frac{sqrt{2}}{4} )So,( W = -frac{64}{5} times frac{sqrt{2}}{12} = -frac{64sqrt{2}}{60} = -frac{16sqrt{2}}{15} )So, same result as before. Therefore, the work done is ( -frac{16sqrt{2}}{15} , text{J} ).But in the initial problem, the user had ( -frac{16}{15} ). So, perhaps they made a mistake by not including the ( sqrt{2} ) term, or maybe they considered ( x ) differently.Alternatively, maybe I misread the problem. Let me check again.The problem states: "Find the integral expression for the work done by the resistance as the object moves from ( x = 0 ) to ( x = 2 )."Wait, maybe the user didn't compute the integral but just set up the expression. In their initial problem, they had:( W = -int_{0}^{2} frac{4}{5}sqrt{x} , dx )Which is correct, but when they computed it, they got ( -frac{16}{15} ), which is incorrect because ( int_{0}^{2} sqrt{x} , dx = frac{4sqrt{2}}{3} ), so multiplying by ( frac{4}{5} ) gives ( frac{16sqrt{2}}{15} ), and with the negative sign, ( -frac{16sqrt{2}}{15} ).Therefore, the correct integral expression is ( W = -frac{4}{5} int_{0}^{2} sqrt{x} , dx ), and the value is ( -frac{16sqrt{2}}{15} ).So, in conclusion, the integral expression is ( -frac{4}{5} int_{0}^{2} sqrt{x} , dx ), and the work done is ( -frac{16sqrt{2}}{15} , text{J} ).But wait, the user's initial problem had the integral expression as ( -frac{4}{5} int_{0}^{2} sqrt{x} , dx ), which is correct, and then they computed it as ( -frac{16}{15} ), which is incorrect because they forgot the ( sqrt{2} ) term.So, perhaps the user made a mistake in their calculation, but the integral expression itself is correct. Therefore, the answer should be the integral expression ( -frac{4}{5} int_{0}^{2} sqrt{x} , dx ), and the value is ( -frac{16sqrt{2}}{15} ).Alternatively, if the problem only asks for the integral expression, then the answer is ( -frac{4}{5} int_{0}^{2} sqrt{x} , dx ), without evaluating it.But in the initial problem, the user did compute it as ( -frac{16}{15} ), which is incorrect. So, to be precise, the integral expression is ( -frac{4}{5} int_{0}^{2} sqrt{x} , dx ), and the work done is ( -frac{16sqrt{2}}{15} , text{J} ).Therefore, the final answer is ( -frac{16sqrt{2}}{15} , text{J} ).

💡Alright, so I've got these two equations to solve. Let's take them one at a time. The first one is 3x + 6 = 31 - 2x. Hmm, okay, so I need to find the value of x that makes this equation true. First, I remember that to solve for x, I need to get all the x terms on one side and the constants on the other side. So, maybe I can start by adding 2x to both sides to get rid of the -2x on the right. Let's see:3x + 6 + 2x = 31 - 2x + 2xThat simplifies to:5x + 6 = 31Okay, now I have 5x + 6 = 31. Next, I need to get rid of that +6 on the left side. I can do that by subtracting 6 from both sides:5x + 6 - 6 = 31 - 6Which simplifies to:5x = 25Now, to solve for x, I just need to divide both sides by 5:5x / 5 = 25 / 5So, x = 5. That seems straightforward enough. Let me double-check by plugging it back into the original equation:3(5) + 6 = 31 - 2(5)15 + 6 = 31 - 1021 = 21Yep, that works out. So, x = 5 is the solution for the first equation.Now, onto the second equation: 1 - 8(1/4 + 0.5x) = 3(1 - 2x). This one looks a bit more complicated because of the fractions and decimals, but I think I can handle it.First, I should probably simplify both sides of the equation. Let's start by distributing the -8 on the left side:1 - 8*(1/4) - 8*(0.5x) = 3*(1) - 3*(2x)Calculating each term:-8*(1/4) is -2, and -8*(0.5x) is -4x. On the right side, 3*(1) is 3, and -3*(2x) is -6x.So, the equation becomes:1 - 2 - 4x = 3 - 6xSimplifying the left side:1 - 2 is -1, so:-1 - 4x = 3 - 6xNow, I need to get all the x terms on one side and the constants on the other. Let's add 6x to both sides to move the x terms to the left:-1 - 4x + 6x = 3 - 6x + 6xSimplifying:-1 + 2x = 3Next, I'll add 1 to both sides to isolate the x term:-1 + 1 + 2x = 3 + 1Which simplifies to:2x = 4Finally, divide both sides by 2 to solve for x:2x / 2 = 4 / 2So, x = 2. Let me check this by plugging it back into the original equation:1 - 8*(1/4 + 0.5*2) = 3*(1 - 2*2)First, calculate inside the parentheses:1/4 + 0.5*2 = 0.25 + 1 = 1.25Then, multiply by -8:1 - 8*1.25 = 1 - 10 = -9On the right side:3*(1 - 4) = 3*(-3) = -9Both sides equal -9, so x = 2 is correct.Overall, solving these equations wasn't too bad once I broke them down step by step. It's important to carefully distribute and combine like terms to avoid mistakes.

💡First, I need to determine the length of the diagonal that the lady crosses. I know her speed is 2.4 km/hr and the time taken is 3.0004166666666667 hours. To find the distance, which is the diagonal of the square field, I'll use the formula:[ text{Distance} = text{Speed} times text{Time} ]Since the area is required in square meters, I'll convert the speed from km/hr to m/s:[ 1 , text{km/hr} = frac{1000 , text{m}}{3600 , text{s}} = frac{5}{18} , text{m/s} ][ 2.4 , text{km/hr} = 2.4 times frac{5}{18} , text{m/s} = frac{12}{18} , text{m/s} = frac{2}{3} , text{m/s} ]Next, I'll calculate the distance:[ text{Distance} = frac{2}{3} , text{m/s} times 3.0004166666666667 , text{hours} times 3600 , text{seconds/hour} ][ text{Distance} = frac{2}{3} times 10800.15 , text{meters} ][ text{Distance} = 7200.1 , text{meters} ]Now that I have the diagonal, I'll find the side of the square field using the Pythagorean theorem:[ d = ssqrt{2} ][ s = frac{d}{sqrt{2}} ][ s = frac{7200.1}{sqrt{2}} ][ s approx 5092.9 , text{meters} ]Finally, I'll calculate the area of the square field:[ A = s^2 ][ A = (5092.9)^2 ][ A approx 25939764.41 , text{square meters} ]