💡First, determine the time Lauren took to run 5 miles. Since she ran it in one-third of the time it took Alex to run 6 miles, and Alex took 36 minutes, Lauren's time is ( frac{1}{3} times 36 = 12 ) minutes.Next, calculate Lauren's running rate. She ran 5 miles in 12 minutes, so her rate is ( frac{12}{5} = 2.4 ) minutes per mile.Finally, to find out how long it would take Lauren to run 8 miles, multiply her rate by the distance: ( 8 times 2.4 = 19.2 ) minutes.

💡Alright, let's try to figure out this stock price problem. So, we have a stock that goes through four weeks of price changes, and we need to find out what percentage increase or decrease happens in the fourth week to bring the stock price back to its original value. First, let me break down the problem week by week. **Week 1:** The stock price rises by 30%. Okay, so if I assume the initial price is, say, 100 (just to make the math easier), then after a 30% increase, the price becomes 130. That makes sense because 30% of 100 is 30, so 100 + 30 = 130.**Week 2:** The stock price falls by 25%. Now, this is a decrease from the new price of 130. So, 25% of 130 is... let me calculate that. 25% is the same as a quarter, so a quarter of 130 is 32.5. So, subtracting that from 130 gives us 130 - 32.5 = 97.5. So, the price at the end of the second week is 97.50.**Week 3:** The stock price rises again by 20%. This time, the increase is based on the price at the end of week two, which is 97.50. So, 20% of 97.50 is... let's see, 10% of 97.50 is 9.75, so 20% is double that, which is 19.50. Adding that to 97.50 gives us 97.50 + 19.50 = 117. So, the price at the end of the third week is 117.**Week 4:** Now, we need to find out what percentage change is needed to bring the price back to its original value, which was 100. So, starting from 117, we need to get back to 100. To find the percentage change, I can use the formula:[text{Percentage Change} = left( frac{text{Final Value} - text{Initial Value}}{text{Initial Value}} right) times 100%]In this case, the final value is 100, and the initial value is 117. Plugging in the numbers:[text{Percentage Change} = left( frac{100 - 117}{117} right) times 100% = left( frac{-17}{117} right) times 100%]Calculating that gives:[text{Percentage Change} approx -14.53%]Hmm, that's approximately a 14.53% decrease. But looking at the answer choices, we have -15% as one of the options. Since -14.53% is very close to -15%, I think the answer is -15%.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Starting with 100, after a 30% increase, it's 130. Then a 25% decrease brings it down to 97.50. A 20% increase takes it to 117. To get back to 100 from 117, the percentage decrease needed is indeed around 14.53%, which rounds to -15%. So, I think I'm confident that -15% is the correct answer.Just to be thorough, let me verify the overall effect of all four weeks. Starting with 100:- After Week 1: 100 * 1.30 = 130- After Week 2: 130 * 0.75 = 97.50- After Week 3: 97.50 * 1.20 = 117- After Week 4: 117 * (1 + y/100) = 100Solving for y:[117 * (1 + y/100) = 100 1 + y/100 = 100/117 y/100 = (100/117) - 1 y/100 = -17/117 y = (-17/117) * 100 y approx -14.53]Yes, that confirms it. So, the percentage change needed in the fourth week is approximately -14.53%, which is closest to -15%. Therefore, the answer is -15%.

💡Alright, let me try to figure out which of these statements are correct. I'm a bit new to this, so I'll take it step by step.First, statement ① says: "The coefficient and degree of the monomial x are both 0." Hmm, okay. So, a monomial is like a single term, right? And x is just x, which is the same as 1x. So, the coefficient should be 1, not 0. And the degree is the exponent on x, which is 1. So, both the coefficient and degree are 1, not 0. So, statement ① is incorrect.Next, statement ②: "The degree of 3x⁴ - 5x²y² - 6y³ + 2 is 11." Okay, the degree of a polynomial is the highest degree of its terms. Let's look at each term:- 3x⁴: degree is 4- -5x²y²: degree is 2 + 2 = 4- -6y³: degree is 3- 2: degree is 0 (constant term)So, the highest degree is 4, not 11. So, statement ② is incorrect.Moving on to statement ③: "The polynomial 1 - 2x + (1/2)x² is composed of three terms: 1, -2x, and (1/2)x²." Okay, so terms are separated by addition or subtraction. So, yes, 1 is the constant term, -2x is the linear term, and (1/2)x² is the quadratic term. So, it's composed of three terms. Therefore, statement ③ is correct.Lastly, statement ④: "Among (1/3)a², (x - y)/π, (5y)/(4x), there are 2 polynomials." Let's see. A polynomial is an expression consisting of variables and coefficients, constructed using only addition, subtraction, multiplication, and non-negative integer exponents of variables.- (1/3)a²: This is a polynomial because it's a single term with a variable raised to a non-negative integer exponent.- (x - y)/π: This can be rewritten as (1/π)x - (1/π)y, which is still a polynomial because π is just a constant.- (5y)/(4x): This is not a polynomial because it involves division by a variable, which is not allowed in polynomials.So, out of the three, two are polynomials: (1/3)a² and (x - y)/π. Therefore, statement ④ is correct.So, summarizing:- ①: Incorrect- ②: Incorrect- ③: Correct- ④: CorrectThat means two statements are correct: ③ and ④. So, the answer should be B: 2.

💡Okay, so I have this equation: 2a² = 3b³, and I need to find all pairs of positive integers (a, b) that satisfy it. Hmm, let me think about how to approach this.First, I know that both a and b have to be positive integers, so I can't have fractions or anything like that. The equation relates a squared term to a cubed term, which might mean that a and b have to be multiples of certain numbers to make the equation balance out.Let me write down the equation again: 2a² = 3b³. Maybe I can rearrange it to see if that helps. If I divide both sides by 2, I get a² = (3/2)b³. Hmm, but a² has to be an integer, so (3/2)b³ must also be an integer. That means that b³ must be divisible by 2, so b has to be even. Let me note that down: b is even.Since b is even, let me write b as 2k, where k is a positive integer. Substituting that into the equation: 2a² = 3(2k)³. Let's compute that: (2k)³ is 8k³, so 3*8k³ is 24k³. So now the equation becomes 2a² = 24k³. If I divide both sides by 2, I get a² = 12k³.Alright, so a² = 12k³. Now, a² has to be an integer, so 12k³ must be a perfect square. Let me think about what that means. For 12k³ to be a perfect square, all the exponents in its prime factorization must be even.Let me factorize 12: 12 = 2² * 3¹. So, 12k³ = 2² * 3¹ * k³. For this to be a perfect square, the exponents of all prime factors must be even. So, let's look at the exponents:- The exponent of 2 is 2, which is already even.- The exponent of 3 is 1, which is odd. So, we need another factor of 3 to make it even.- The exponent of any other prime in k³ must also be even.Therefore, k must be a multiple of 3. Let me write k as 3m, where m is a positive integer. Substituting back, k = 3m, so k³ = 27m³. Then, 12k³ = 12*27m³ = 324m³. So, a² = 324m³.Now, 324 is 18², right? Because 18*18 = 324. So, a² = (18)² * m³. Therefore, a must be a multiple of 18. Let me write a as 18n, where n is a positive integer. Then, a² = (18n)² = 324n². So, substituting back into a² = 324m³, we get 324n² = 324m³. Dividing both sides by 324, we get n² = m³.So now, we have n² = m³. This is another equation where n and m are positive integers. Let me think about this. For n² to equal m³, both sides must be perfect squares and perfect cubes. The smallest such number is 1, then 64 (since 8²=64 and 4³=64). Wait, actually, 64 is 8² and 4³. So, in general, n² = m³ implies that n is a cube and m is a square. Let me explain.If n² = m³, then n must be a perfect cube, and m must be a perfect square. Let me denote n as p³ and m as p², where p is a positive integer. Then, n² = (p³)² = p⁶ and m³ = (p²)³ = p⁶, so they are equal. That works.So, substituting back, n = p³ and m = p². Then, a = 18n = 18p³, and k = 3m = 3p². Since b = 2k, b = 2*3p² = 6p².Therefore, the solutions are a = 18p³ and b = 6p², where p is any positive integer.Let me check with p=1: a=18, b=6. Plugging into the original equation: 2*(18)² = 2*324 = 648, and 3*(6)³ = 3*216 = 648. Yes, that works.How about p=2: a=18*8=144, b=6*4=24. Then, 2*(144)² = 2*20736 = 41472, and 3*(24)³ = 3*13824 = 41472. That works too.So, it seems like the general solution is a=18p³ and b=6p² for positive integers p.I think that's all the solutions because we've broken down the equation step by step, ensuring that each substitution maintains the equality and that all variables are positive integers. There doesn't seem to be any other way to satisfy the equation without following this structure.

💡Okay, so I need to find the range of set A, which consists of all the prime numbers between 15 and 36. Hmm, let's start by understanding what a prime number is. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So, I need to list all the prime numbers in that range first.Starting from 15, the next number is 16, but 16 is not a prime because it's divisible by 2. Then 17—let's see, 17 is only divisible by 1 and 17, so that's a prime. Next is 18, which is even, so not prime. 19 is next, and that's a prime too because it doesn't have any divisors other than 1 and 19.Moving on, 20 is even, so not prime. 21 is divisible by 3 and 7, so not prime. 22 is even, not prime. 23 is a prime because it's only divisible by 1 and 23. Then 24 is even, not prime. 25 is divisible by 5, so not prime. 26 is even, not prime. 27 is divisible by 3, not prime. 28 is even, not prime. 29 is a prime because it's only divisible by 1 and 29.Next, 30 is even, not prime. 31 is a prime because it's only divisible by 1 and 31. 32 is even, not prime. 33 is divisible by 3 and 11, not prime. 34 is even, not prime. 35 is divisible by 5 and 7, not prime. 36 is even, not prime.So, the prime numbers between 15 and 36 are: 17, 19, 23, 29, and 31.Now, to find the range of set A, I need to subtract the smallest number in the set from the largest number. The smallest prime number in the set is 17, and the largest is 31. So, the range would be 31 minus 17.Let me do that calculation: 31 minus 17 equals 14.Wait, is that correct? Let me double-check. 17 is the first prime after 15, and 31 is the last prime before 36. Subtracting 17 from 31 gives me 14. That seems right.I think I've got it. The range of set A is 14.

💡Okay, so I have this problem where I need to find all functions ( f: mathbb{N} rightarrow mathbb{N} ) that satisfy the equation ( f(x + f(y)) = f(x) + y ) for all ( x, y in mathbb{N} ). Hmm, functional equations can be tricky, but let me try to break it down step by step.First, I should recall that ( mathbb{N} ) typically refers to the set of positive integers, so ( x ) and ( y ) are positive integers here. The function ( f ) takes a positive integer and maps it to another positive integer. The equation given is a bit complex, so maybe I can start by plugging in some specific values for ( x ) and ( y ) to see if I can find a pattern or figure out some properties of ( f ).Let me try setting ( x = 1 ) and see what happens. Then the equation becomes ( f(1 + f(y)) = f(1) + y ). Hmm, interesting. So for any ( y ), ( f ) evaluated at ( 1 + f(y) ) is equal to ( f(1) + y ). That seems like it could be useful. Maybe I can denote ( f(1) ) as some constant, say ( c ). Then the equation becomes ( f(1 + f(y)) = c + y ). So ( f ) evaluated at ( 1 + f(y) ) is linear in ( y ). That suggests that ( f ) might have some linear properties.What if I set ( y = 1 )? Then the original equation becomes ( f(x + f(1)) = f(x) + 1 ). Since ( f(1) = c ), this simplifies to ( f(x + c) = f(x) + 1 ). So, adding ( c ) to the input of ( f ) increases the output by 1. That seems like a sort of periodicity or linearity. Maybe ( f ) is linear? Let me test that idea.Suppose ( f(x) = kx + b ) for some constants ( k ) and ( b ). Let's plug this into the original equation and see if it works. So, ( f(x + f(y)) = k(x + f(y)) + b = kx + k f(y) + b ). On the other hand, ( f(x) + y = kx + b + y ). So, setting these equal:( kx + k f(y) + b = kx + b + y )Simplify both sides by subtracting ( kx + b ):( k f(y) = y )So, ( f(y) = frac{y}{k} ). But ( f(y) ) must be a natural number, so ( k ) must divide ( y ) for all ( y in mathbb{N} ). The only way this can happen is if ( k = 1 ), because 1 is the only positive integer that divides all natural numbers. So, ( f(y) = y ). Let me check if this works in the original equation.If ( f(y) = y ), then ( f(x + f(y)) = f(x + y) = x + y ). On the other hand, ( f(x) + y = x + y ). So, both sides are equal. Therefore, ( f(x) = x ) is indeed a solution.But wait, is this the only solution? I need to make sure there aren't any other functions that satisfy the equation. Let me think about the properties I found earlier.From setting ( y = 1 ), I found that ( f(x + c) = f(x) + 1 ), where ( c = f(1) ). If ( f ) is linear, then ( c = 1 ), which would make ( f(x) = x ). But what if ( f ) isn't linear? Could there be another function that satisfies ( f(x + c) = f(x) + 1 )?Suppose ( c = 2 ). Then ( f(x + 2) = f(x) + 1 ). Let's see what this implies. Starting from ( x = 1 ), ( f(3) = f(1) + 1 = c + 1 ). Then ( f(5) = f(3) + 1 = c + 2 ), and so on. Similarly, for even numbers, starting from ( x = 2 ), ( f(4) = f(2) + 1 ). But I don't know what ( f(2) ) is yet. Hmm, this seems a bit messy. Maybe I can find another property to pin down ( f ).Earlier, I set ( x = 1 ) and got ( f(1 + f(y)) = c + y ). If I assume ( f ) is injective or surjective, that might help. Let me check if ( f ) is injective. Suppose ( f(a) = f(b) ). Then, for any ( x ), ( f(x + f(a)) = f(x) + a ) and ( f(x + f(b)) = f(x) + b ). Since ( f(a) = f(b) ), the left sides are equal, so ( f(x) + a = f(x) + b ), which implies ( a = b ). Therefore, ( f ) is injective.Now, is ( f ) surjective? Let's see. For any ( z in mathbb{N} ), can I find a ( y ) such that ( f(y) = z )? From the equation ( f(x + f(y)) = f(x) + y ), if I fix ( x ) and vary ( y ), the right side can take any value greater than ( f(x) ), which suggests that ( f ) is surjective.Since ( f ) is both injective and surjective, it's bijective. That means ( f ) has an inverse function. Let me denote the inverse function as ( f^{-1} ). So, ( f^{-1}(f(y)) = y ) and ( f(f^{-1}(z)) = z ).Going back to the equation ( f(x + f(y)) = f(x) + y ), I can apply ( f^{-1} ) to both sides:( x + f(y) = f^{-1}(f(x) + y) )Hmm, not sure if that helps directly. Maybe I can find another substitution. Let me try setting ( x = f(z) ) for some ( z ). Then the equation becomes:( f(f(z) + f(y)) = f(f(z)) + y )But from earlier, when I set ( x = 0 ) (wait, hold on, the original problem says ( mathbb{N} ), which usually starts at 1, but sometimes people include 0. Hmm, the problem statement doesn't specify, but in the initial problem, the user wrote ( mathbb{N} ), so maybe it's safer to assume it's positive integers starting at 1. So, setting ( x = 1 ) earlier was fine, but maybe I can't set ( x = 0 ). Hmm, that complicates things a bit.Wait, in the initial problem, the user wrote ( f: mathbb{N} rightarrow mathbb{N} ), and in the equation, ( x ) and ( y ) are in ( mathbb{N} ). So, if ( mathbb{N} ) is positive integers starting at 1, then I can't set ( x = 0 ). That might limit some substitutions. Let me think.Alternatively, maybe I can set ( x = 1 ) and ( y = 1 ). Then, ( f(1 + f(1)) = f(1) + 1 ). Let me denote ( f(1) = c ), so ( f(1 + c) = c + 1 ). From earlier, when I set ( y = 1 ), I had ( f(x + c) = f(x) + 1 ). So, if I set ( x = 1 ), then ( f(1 + c) = f(1) + 1 = c + 1 ), which is consistent.What if I set ( x = 2 ) and ( y = 1 )? Then, ( f(2 + c) = f(2) + 1 ). Similarly, ( x = 3 ), ( y = 1 ): ( f(3 + c) = f(3) + 1 ). It seems like adding ( c ) to the argument of ( f ) increases the output by 1. This suggests that ( f ) is linear with a slope of ( 1/c ). But since ( f ) maps naturals to naturals, ( c ) must be 1, because otherwise, ( f ) wouldn't map naturals to naturals if ( c > 1 ). For example, if ( c = 2 ), then ( f(2) = 2 ), ( f(3) = f(2 + 1) = f(2) + 1 = 3 ), ( f(4) = f(3 + 1) = f(3) + 1 = 4 ), and so on. Wait, actually, even if ( c = 2 ), ( f ) could still be linear. Hmm, maybe my earlier reasoning was off.Wait, if ( c = 2 ), then ( f(x + 2) = f(x) + 1 ). So, starting from ( f(1) = 2 ), ( f(3) = f(1) + 1 = 3 ), ( f(5) = f(3) + 1 = 4 ), and so on. Similarly, ( f(2) ) would need to satisfy ( f(2 + 2) = f(2) + 1 ), so ( f(4) = f(2) + 1 ). But from the other side, ( f(4) = f(3 + 1) = f(3) + 1 = 4 ). So, ( f(2) + 1 = 4 ), which means ( f(2) = 3 ). Then, ( f(4) = 4 ), ( f(6) = f(4) + 1 = 5 ), etc. So, it seems like ( f ) could be defined as ( f(x) = lceil frac{x + 1}{2} rceil + 1 ) or something like that. Wait, let me check:If ( f(1) = 2 ), ( f(2) = 3 ), ( f(3) = 3 ), ( f(4) = 4 ), ( f(5) = 4 ), ( f(6) = 5 ), etc. Hmm, but then ( f ) wouldn't be injective because ( f(3) = 3 ) and ( f(4) = 4 ), but ( f(5) = 4 ), which would mean ( f(4) = f(5) ), contradicting injectivity. Wait, but earlier I concluded that ( f ) is injective. So, if ( f ) is injective, then ( f(4) ) can't equal ( f(5) ). Therefore, my assumption that ( c = 2 ) might not hold because it leads to a contradiction with injectivity.Wait, let's see. If ( c = 2 ), then ( f(1) = 2 ), ( f(3) = 3 ), ( f(5) = 4 ), etc., and ( f(2) = 3 ), ( f(4) = 4 ), ( f(6) = 5 ), etc. So, ( f(2) = 3 ), ( f(3) = 3 ). Oh, that's a problem because ( f(2) = f(3) ), which would mean ( f ) is not injective. But earlier, I proved that ( f ) is injective. Therefore, ( c ) cannot be 2. So, ( c ) must be 1.Therefore, ( f(1) = 1 ). Then, from ( f(x + 1) = f(x) + 1 ), which is the case when ( c = 1 ). So, ( f(x + 1) = f(x) + 1 ). This suggests that ( f ) is a linear function with slope 1. Therefore, ( f(x) = x + k ) for some constant ( k ). Let me test this.Suppose ( f(x) = x + k ). Then, ( f(x + f(y)) = f(x + y + k) = (x + y + k) + k = x + y + 2k ). On the other hand, ( f(x) + y = (x + k) + y = x + y + k ). For these to be equal, ( x + y + 2k = x + y + k ), which implies ( k = 0 ). But ( f: mathbb{N} rightarrow mathbb{N} ), and if ( k = 0 ), then ( f(x) = x ), which is valid because ( f ) maps naturals to naturals.Wait, but earlier I thought ( f(x) = x + k ) might work, but it only works if ( k = 0 ). So, ( f(x) = x ) is the only linear solution. But is it the only solution overall?Let me think again. From the equation ( f(x + f(y)) = f(x) + y ), and knowing that ( f ) is bijective, maybe I can find an expression for ( f ) in terms of itself. Let me try to express ( f(y) ) in terms of ( f^{-1} ).From the equation, ( f(x + f(y)) = f(x) + y ). Let me apply ( f^{-1} ) to both sides:( x + f(y) = f^{-1}(f(x) + y) )Hmm, not sure if that helps directly. Maybe I can set ( x = f(z) ) for some ( z ). Then, ( f(f(z) + f(y)) = f(f(z)) + y ). But since ( f ) is bijective, ( f(f(z)) = z + k ) for some ( k ). Wait, no, that might not necessarily be the case. Alternatively, since ( f ) is bijective, there exists a ( z ) such that ( f(z) = w ) for any ( w ).Wait, maybe I can use the fact that ( f ) is linear. Since ( f(x + 1) = f(x) + 1 ), and ( f(1) = 1 ), then by induction, ( f(x) = x ) for all ( x ). Let me check that.Base case: ( x = 1 ), ( f(1) = 1 ).Inductive step: Assume ( f(k) = k ) for some ( k geq 1 ). Then, ( f(k + 1) = f(k) + 1 = k + 1 ). Therefore, by induction, ( f(x) = x ) for all ( x in mathbb{N} ).So, it seems that ( f(x) = x ) is indeed the only solution. Let me verify this once more in the original equation.If ( f(x) = x ), then ( f(x + f(y)) = f(x + y) = x + y ). On the other hand, ( f(x) + y = x + y ). So, both sides are equal, confirming that ( f(x) = x ) satisfies the equation.I think I've covered all the bases here. I started by testing specific values, found that ( f ) must be injective and surjective, leading to it being bijective. Then, by considering the functional equation and the properties derived, I concluded that ( f(x) = x ) is the only function that satisfies the given condition.