💡First, I need to determine the athlete's maximum heart rate by subtracting their age from 225. Given that the athlete is 30 years old, the calculation would be 225 minus 30, which equals 195 beats per minute.Next, I need to find 85% of this maximum heart rate to determine the target heart rate. To do this, I multiply 195 by 0.85, resulting in 165.75 beats per minute.Finally, I round 165.75 to the nearest whole number, which gives me 166 beats per minute. Therefore, the target heart rate of the athlete is 166 beats per minute.

💡First, I need to identify the elements of set A and set B.Set A is given as {1, 2, 3, 4, 5, 6, 7}.Set B is defined as {x ∈ N | 2 ≤ x < 6}. This means B includes all natural numbers x such that x is greater than or equal to 2 and less than 6. Therefore, B = {2, 3, 4, 5}.Next, to find the intersection of A and B (A ∩ B), I look for elements that are present in both sets.Comparing the elements:- Set A: 1, 2, 3, 4, 5, 6, 7- Set B: 2, 3, 4, 5The common elements are 2, 3, 4, and 5.Therefore, A ∩ B = {2, 3, 4, 5}, which corresponds to option B.

💡Okay, so I have this problem here: "If ( y ) varies inversely as ( x ), and if ( y = 6 ) when ( x = 3 ), find the value of ( y ) when ( x = 12 )." The options are given as (A) ( frac{2}{3} ), (B) ( frac{3}{2} ), (C) ( 6 ), and (D) ( 4 ).Alright, let's start by understanding what it means for ( y ) to vary inversely as ( x ). From what I remember, inverse variation means that as one quantity increases, the other decreases proportionally. The general form of an inverse variation is ( y = frac{k}{x} ), where ( k ) is a constant. So, in this case, ( y ) is equal to some constant divided by ( x ).Given that ( y = 6 ) when ( x = 3 ), I can plug these values into the equation to find ( k ). Let me write that out:( 6 = frac{k}{3} )To solve for ( k ), I can multiply both sides of the equation by 3:( 6 times 3 = k )( 18 = k )So, the constant ( k ) is 18. Now that I have ( k ), I can write the specific inverse variation equation for this problem:( y = frac{18}{x} )Now, the question asks for the value of ( y ) when ( x = 12 ). Let's substitute 12 into the equation:( y = frac{18}{12} )Simplifying that, I can divide both numerator and denominator by 6:( y = frac{3}{2} )So, ( y ) equals ( frac{3}{2} ) when ( x = 12 ).Let me double-check my steps to make sure I didn't make any mistakes. First, I identified the inverse variation equation correctly as ( y = frac{k}{x} ). Then, I substituted the given values ( y = 6 ) and ( x = 3 ) to find ( k ), which gave me 18. Plugging that back into the equation, I got ( y = frac{18}{x} ). Finally, substituting ( x = 12 ) gave me ( y = frac{3}{2} ). Looking at the answer choices, ( frac{3}{2} ) is option (B). So, I think that's the correct answer.Just to be thorough, let me consider if there's another way to approach this problem. Sometimes, with inverse variation, people use the concept that the product ( xy ) remains constant. So, if ( y ) varies inversely as ( x ), then ( xy = k ). Given ( y = 6 ) when ( x = 3 ), the product ( xy = 6 times 3 = 18 ). So, ( k = 18 ). When ( x = 12 ), then ( y = frac{18}{12} = frac{3}{2} ). Yep, that confirms the same result.I think I've got it right. The answer should be ( frac{3}{2} ), which is option (B).

💡Okay, so I have this sequence problem here. Let me try to figure it out step by step. The sequence is defined as {a_n}, where a₁ = 4, and for n ≥ 2, aₙ = aₙ₋₁ + 2^{n-1} + 3. Part (1) asks me to prove that the sequence {a_n - 2^n} is an arithmetic sequence and then find the general term formula for {a_n}. Hmm, okay. So, an arithmetic sequence is one where the difference between consecutive terms is constant. That means if I can show that (a_n - 2^n) - (a_{n-1} - 2^{n-1}) is a constant, then it's arithmetic.Let me write down the given recurrence relation again: aₙ = aₙ₋₁ + 2^{n-1} + 3. Maybe I can manipulate this to find a relation involving a_n - 2^n.So, let's compute a_n - 2^n:a_n - 2^n = (a_{n-1} + 2^{n-1} + 3) - 2^n.Simplify that:= a_{n-1} + 2^{n-1} + 3 - 2^n.Notice that 2^n = 2 * 2^{n-1}, so:= a_{n-1} + 2^{n-1} + 3 - 2 * 2^{n-1}.= a_{n-1} - 2^{n-1} + 3.So, a_n - 2^n = (a_{n-1} - 2^{n-1}) + 3.That looks like the difference between consecutive terms of the sequence {a_n - 2^n} is 3, which is a constant. Therefore, {a_n - 2^n} is indeed an arithmetic sequence with common difference 3.Now, to find the general term formula for {a_n}, I need to find the first term of the arithmetic sequence {a_n - 2^n}. Let's compute a₁ - 2¹:a₁ - 2¹ = 4 - 2 = 2.So, the first term is 2, and the common difference is 3. The general term of an arithmetic sequence is given by:b_n = b₁ + (n - 1)d.Here, b_n = a_n - 2^n, b₁ = 2, d = 3. So,a_n - 2^n = 2 + (n - 1)*3 = 2 + 3n - 3 = 3n - 1.Therefore, a_n = 2^n + 3n - 1.Okay, that seems to make sense. Let me verify it for n=1 and n=2.For n=1: a₁ = 2¹ + 3*1 -1 = 2 + 3 -1 = 4, which matches the given.For n=2: a₂ = a₁ + 2^{2-1} + 3 = 4 + 2 + 3 = 9. Using the formula: a₂ = 2² + 3*2 -1 = 4 + 6 -1 = 9. It matches.Good, so part (1) seems done.Now, part (2): Let b_n = a_n / 2^n. Find the sum of the first n terms, denoted as S_n.So, first, let's express b_n:b_n = a_n / 2^n = (2^n + 3n -1) / 2^n = 1 + (3n -1)/2^n.Therefore, S_n = sum_{k=1}^n b_k = sum_{k=1}^n [1 + (3k -1)/2^k] = sum_{k=1}^n 1 + sum_{k=1}^n (3k -1)/2^k.Compute the first sum: sum_{k=1}^n 1 = n.So, S_n = n + sum_{k=1}^n (3k -1)/2^k.Now, I need to compute the sum T_n = sum_{k=1}^n (3k -1)/2^k.Hmm, how do I compute this sum? Maybe I can split it into two separate sums:T_n = 3 sum_{k=1}^n k / 2^k - sum_{k=1}^n 1 / 2^k.So, T_n = 3A - B, where A = sum_{k=1}^n k / 2^k and B = sum_{k=1}^n 1 / 2^k.I know that sum_{k=1}^infty 1 / 2^k = 1, since it's a geometric series with ratio 1/2. But since we have finite n, B = sum_{k=1}^n 1 / 2^k = 1 - (1/2)^n.Similarly, for A, the sum sum_{k=1}^infty k / 2^k is known to be 2. But again, since we have finite n, we need a formula for A.I recall that for finite sums, sum_{k=1}^n k x^k = x(1 - (n+1)x^n + n x^{n+1}) / (1 - x)^2. Let me verify that.Yes, the formula for sum_{k=1}^n k x^{k} is x(1 - (n+1)x^n + n x^{n+1}) / (1 - x)^2.In our case, x = 1/2. So,A = sum_{k=1}^n k / 2^k = (1/2)(1 - (n+1)(1/2)^n + n (1/2)^{n+1}) / (1 - 1/2)^2.Simplify denominator: (1 - 1/2)^2 = (1/2)^2 = 1/4.So,A = (1/2)(1 - (n+1)/2^n + n / 2^{n+1}) / (1/4) = (1/2)(1 - (n+1)/2^n + n / 2^{n+1}) * 4.Simplify:= 2 [1 - (n+1)/2^n + n / 2^{n+1}].Let me compute the terms inside:1 is just 1.(n+1)/2^n = (n+1)/2^n.n / 2^{n+1} = n / (2 * 2^n) = n / 2^{n+1}.So,A = 2 [1 - (n+1)/2^n + n / 2^{n+1}].Let me combine the terms:= 2 [1 - (n+1)/2^n + n / 2^{n+1}]= 2 [1 - (2(n+1))/2^{n+1} + n / 2^{n+1}]= 2 [1 - (2n + 2 - n)/2^{n+1}]= 2 [1 - (n + 2)/2^{n+1}]= 2 - (n + 2)/2^n.So, A = 2 - (n + 2)/2^n.Similarly, B = sum_{k=1}^n 1 / 2^k = 1 - 1/2^n.Therefore, T_n = 3A - B = 3[2 - (n + 2)/2^n] - [1 - 1/2^n].Compute this:= 6 - 3(n + 2)/2^n - 1 + 1/2^n= (6 - 1) + (-3(n + 2)/2^n + 1/2^n)= 5 - [3(n + 2) - 1]/2^n= 5 - [3n + 6 - 1]/2^n= 5 - (3n + 5)/2^n.Therefore, T_n = 5 - (3n + 5)/2^n.So, going back to S_n:S_n = n + T_n = n + 5 - (3n + 5)/2^n.Therefore, S_n = n + 5 - (3n + 5)/2^n.Let me check this for n=1 and n=2.For n=1:b₁ = a₁ / 2¹ = 4 / 2 = 2.S₁ = 2.Using the formula: S₁ = 1 + 5 - (3*1 +5)/2¹ = 6 - 8/2 = 6 - 4 = 2. Correct.For n=2:b₂ = a₂ / 2² = 9 / 4 = 2.25.S₂ = 2 + 2.25 = 4.25.Using the formula: S₂ = 2 + 5 - (6 +5)/4 = 7 - 11/4 = 7 - 2.75 = 4.25. Correct.Good, seems consistent.**Final Answer**(1) The general term is boxed{a_n = 2^n + 3n - 1}.(2) The sum of the first ( n ) terms is boxed{S_n = n + 5 - frac{3n + 5}{2^n}}.

💡Alright, so I've got this problem here: I need to find the values of ( m ) for which the equations ( m x - 1000 = 1001 ) and ( 1001 x = m - 1000 x ) have a common root. Hmm, okay, let's break this down step by step.First, I need to understand what it means for two equations to have a common root. I think it means that there's a specific value of ( x ) that satisfies both equations simultaneously. So, if I can find such an ( x ), and then find the corresponding ( m ), that should give me the answer.Let me start by looking at the first equation: ( m x - 1000 = 1001 ). I can try to solve for ( x ) in terms of ( m ). So, if I add 1000 to both sides, I get ( m x = 2001 ). Then, dividing both sides by ( m ), assuming ( m ) isn't zero, I get ( x = frac{2001}{m} ). Okay, so that's the expression for ( x ) from the first equation.Now, moving on to the second equation: ( 1001 x = m - 1000 x ). I can try to solve this for ( x ) as well. Let me bring all the terms involving ( x ) to one side. So, adding ( 1000 x ) to both sides gives me ( 1001 x + 1000 x = m ), which simplifies to ( 2001 x = m ). Then, dividing both sides by 2001, I get ( x = frac{m}{2001} ). Alright, so that's the expression for ( x ) from the second equation.Now, since both equations are supposed to have a common root, the ( x ) values I found should be equal. That means ( frac{2001}{m} = frac{m}{2001} ). Hmm, okay, so I can set these two expressions equal to each other and solve for ( m ).Let me write that equation down: ( frac{2001}{m} = frac{m}{2001} ). To solve for ( m ), I can cross-multiply. So, multiplying both sides by ( m ) and by 2001, I get ( 2001 times 2001 = m times m ). That simplifies to ( 2001^2 = m^2 ).Now, to find ( m ), I need to take the square root of both sides. Remembering that taking a square root can give both positive and negative results, I get ( m = pm 2001 ). So, ( m ) can be either 2001 or -2001.Wait, let me double-check that. If ( m = 2001 ), then from the first equation, ( x = frac{2001}{2001} = 1 ). Plugging ( x = 1 ) into the second equation: ( 1001 times 1 = 2001 - 1000 times 1 ), which simplifies to ( 1001 = 2001 - 1000 ), and that's ( 1001 = 1001 ). That works.Now, if ( m = -2001 ), then from the first equation, ( x = frac{2001}{-2001} = -1 ). Plugging ( x = -1 ) into the second equation: ( 1001 times (-1) = -2001 - 1000 times (-1) ), which simplifies to ( -1001 = -2001 + 1000 ), and that's ( -1001 = -1001 ). That also works.So, both values of ( m ) satisfy the condition that the two equations have a common root. Therefore, the values of ( m ) are 2001 and -2001.I think that's it. I don't see any mistakes in my calculations, and both values check out when I plug them back into the original equations. So, I'm confident that ( m = pm 2001 ) are the correct answers.

💡Alright, so I've got this problem here: 48 blacksmiths need to shoe 60 horses. Each blacksmith takes 5 minutes per horseshoe. I need to find the minimum time they'll spend on the job. Hmm, okay, let's break this down step by step.First off, I know that each horse has 4 hooves, right? So, if each horse needs 4 horseshoes, then for 60 horses, the total number of horseshoes needed is 60 multiplied by 4. Let me write that down:Total horseshoes = 60 horses * 4 hooves/horse = 240 horseshoes.Okay, so there are 240 horseshoes that need to be made in total. Now, each blacksmith can make one horseshoe in 5 minutes. So, if I have 48 blacksmiths working simultaneously, how many horseshoes can they make in a given time?Well, if each blacksmith makes one horseshoe in 5 minutes, then in one minute, a blacksmith can make 1/5 of a horseshoe. So, 48 blacksmiths working together can make 48*(1/5) = 9.6 horseshoes per minute.Wait, that doesn't sound quite right. Let me think again. If each blacksmith takes 5 minutes per horseshoe, then in 5 minutes, one blacksmith can make one horseshoe. So, in 5 minutes, 48 blacksmiths can make 48 horseshoes.Ah, that makes more sense. So, in 5 minutes, they can produce 48 horseshoes. Since we need 240 horseshoes, how many 5-minute intervals do we need?Number of intervals = Total horseshoes / Horseshoes per interval = 240 / 48 = 5 intervals.Each interval is 5 minutes, so total time = 5 intervals * 5 minutes/interval = 25 minutes.Wait, but is that the minimum time? Let me double-check. If they work in parallel, with each blacksmith making one horseshoe at a time, then yes, 48 horseshoes every 5 minutes. So, 5 intervals of 5 minutes each would give us 240 horseshoes.But hold on, is there a way to overlap the work or make it more efficient? For example, can blacksmiths work on different parts of the same horse simultaneously? But I think each horseshoe is an independent task, so overlapping might not reduce the total time.Alternatively, maybe some blacksmiths can work on multiple horses at the same time, but since each horseshoe takes 5 minutes, and each blacksmith can only work on one horseshoe at a time, I don't think that would help.So, it seems like 25 minutes is indeed the minimum time required for 48 blacksmiths to shoe 60 horses, given that each blacksmith takes 5 minutes per horseshoe.Just to recap:- Total horseshoes needed: 240- Horseshoes produced per 5 minutes by 48 blacksmiths: 48- Number of 5-minute intervals needed: 5- Total time: 25 minutesYeah, that seems to add up. I don't see any shortcuts or more efficient ways to distribute the work, so I think 25 minutes is the answer.