💡Okay, so I have this problem here: The sum of two numbers is 40, and their difference is 6. I need to find the larger of the two numbers. Alright, let's see how to approach this.First, I think I should define the two numbers. Let me call the larger number something like "x" and the smaller one "y." That makes sense because "x" is often used for variables, and "y" is just another variable. So, I'll say:Let x be the larger number.Let y be the smaller number.Now, the problem says their sum is 40. So, I can write that as an equation:x + y = 40And their difference is 6. Since x is the larger number, subtracting y from x should give me 6. So, that equation would be:x - y = 6Alright, now I have two equations:1. x + y = 402. x - y = 6I think I can solve these two equations simultaneously to find the values of x and y. I remember from algebra that if I have two equations with two variables, I can use either substitution or elimination. Let me try elimination because the equations are set up nicely for that.If I add the two equations together, the y terms will cancel out. Let's see:(x + y) + (x - y) = 40 + 6Simplifying the left side:x + y + x - y = 2xAnd the right side:40 + 6 = 46So now I have:2x = 46To solve for x, I just need to divide both sides by 2:2x / 2 = 46 / 2Which simplifies to:x = 23Okay, so the larger number is 23. But wait, let me double-check to make sure I didn't make a mistake. If x is 23, then y should be 40 - x, right?So, y = 40 - 23 = 17Let me check if the difference is indeed 6:23 - 17 = 6Yes, that works out. So, x is 23 and y is 17. Therefore, the larger number is 23.I think that's it. It makes sense because when you add 23 and 17, you get 40, and their difference is 6. So, everything checks out.

💡Alright, so I've got this problem about a sequence defined by a starting positive integer ( a_1 ) and each subsequent term being the previous term plus the radical of the previous term. The radical of a number, ( text{rad}(n) ), is the product of its distinct prime factors. For example, ( text{rad}(12) = 2 times 3 = 6 ) because 12 factors into ( 2^2 times 3 ).The problem asks me to prove that for any positive integer ( N ), there exist ( N ) consecutive terms in this sequence that form an arithmetic progression. An arithmetic progression is a sequence of numbers where the difference between consecutive terms is constant. So, for example, 2, 4, 6, 8 is an arithmetic progression with a common difference of 2.First, I need to understand how the sequence behaves. Let's start with an example. Suppose ( a_1 = 1 ). Then:- ( a_2 = a_1 + text{rad}(a_1) = 1 + 1 = 2 )- ( a_3 = a_2 + text{rad}(a_2) = 2 + 2 = 4 )- ( a_4 = 4 + 2 = 6 )- ( a_5 = 6 + 6 = 12 )- ( a_6 = 12 + 6 = 18 )- ( a_7 = 18 + 6 = 24 )- ( a_8 = 24 + 6 = 30 )- ( a_9 = 30 + 30 = 60 )- ( a_{10} = 60 + 30 = 90 )- ( a_{11} = 90 + 30 = 120 )- ( a_{12} = 120 + 30 = 150 )- ( a_{13} = 150 + 30 = 180 )- ( a_{14} = 180 + 180 = 360 )Looking at this, I can see that sometimes the radical increases, but often it stays the same. For instance, from ( a_2 ) to ( a_4 ), the radical is 2, then it jumps to 6 at ( a_5 ), and then stays at 6 for several terms. Then it jumps again to 30 at ( a_9 ), and then stays at 30 for several terms.This makes me think that the radical can stay the same for multiple steps, which would mean that the sequence is increasing by the same amount repeatedly. If the radical remains constant for ( N ) consecutive steps, then those ( N ) terms would form an arithmetic progression with a common difference equal to that radical.So, perhaps the key is to show that for any ( N ), there will eventually be a point in the sequence where the radical doesn't change for ( N ) consecutive terms. That would mean those ( N ) terms are in arithmetic progression.But how can I be sure that such a point exists for any ( N )? Maybe by analyzing the properties of the radical function and how it affects the sequence.I know that the radical of a number is always less than or equal to the number itself, and it's equal to the number only when the number is square-free (i.e., it has no repeated prime factors). So, ( text{rad}(n) leq n ), with equality if and only if ( n ) is square-free.Given that ( a_{n+1} = a_n + text{rad}(a_n) ), each term is at least ( a_n + 1 ) and at most ( 2a_n ). So the sequence grows at least linearly and at most exponentially.But if the radical doesn't change for several steps, the growth becomes linear for those steps, forming an arithmetic progression.To formalize this, maybe I can consider the behavior of the radical function over the sequence. If the radical remains constant for ( N ) steps, then those ( N ) terms are in arithmetic progression. So, I need to show that for any ( N ), such a stretch exists.Perhaps I can use the fact that the radical can only take on so many values before it must repeat or stay constant. Since the radical is multiplicative and depends on the prime factors, maybe after a certain point, the prime factors stabilize or repeat in a way that causes the radical to stay constant.Alternatively, maybe I can argue by contradiction. Suppose that for some ( N ), there are no ( N ) consecutive terms in arithmetic progression. Then, the radical must change frequently enough to prevent such a progression. But if the radical changes too often, does that lead to some contradiction with the growth rate of the sequence?Let me think about the growth rate. If the radical is changing every few steps, the sequence's growth is more erratic. But if the radical remains constant for ( N ) steps, the growth is linear for those steps. So, if the radical changes too often, the sequence might grow faster, but I'm not sure how that leads to a contradiction.Wait, maybe I can bound the sequence. Since each term is at most double the previous term, the sequence grows at most exponentially. On the other hand, if the radical remains constant for ( N ) steps, the sequence grows linearly for those steps, which is much slower.But how does that help? Maybe if the radical changes too often, the sequence would have to grow faster, but it's already bounded by exponential growth. So, perhaps that doesn't lead to a contradiction.Alternatively, maybe I can consider the prime factors of the terms in the sequence. Each time the radical changes, it's because the prime factors have changed. If the radical stays the same, it means the prime factors haven't changed.So, if the radical stays the same for ( N ) steps, the prime factors of those terms are the same, meaning the terms are multiples of the same radical. So, they form an arithmetic progression with common difference equal to the radical.Therefore, to have ( N ) consecutive terms in arithmetic progression, we need ( N ) consecutive terms with the same radical. So, I need to show that for any ( N ), there exists a point in the sequence where the radical remains the same for ( N ) consecutive terms.How can I ensure that? Maybe by considering that the radical can only take on a finite number of values before it must repeat or stay constant.But the radical can take on infinitely many values as the sequence progresses, since the terms can have more prime factors. However, for any finite ( N ), perhaps after some point, the radical must stay the same for ( N ) steps.Alternatively, maybe I can use the fact that the radical divides the term. So, ( text{rad}(a_n) ) divides ( a_n ), which means that ( a_{n+1} = a_n + text{rad}(a_n) ) is divisible by ( text{rad}(a_n) ). Therefore, ( text{rad}(a_{n+1}) ) must divide ( text{rad}(a_n) ), because ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ).Wait, that's interesting. So, ( text{rad}(a_{n+1}) ) divides ( text{rad}(a_n) ). That means the radical can only stay the same or decrease in prime factors as the sequence progresses.But the radical is a product of primes, so if it can only stay the same or lose prime factors, it can't gain new prime factors. Wait, that seems contradictory because earlier in my example, the radical increased from 2 to 6 when moving from 4 to 6. So, maybe my reasoning is flawed.Wait, no. Let me check. ( a_4 = 6 ), so ( text{rad}(6) = 2 times 3 = 6 ). Then ( a_5 = 6 + 6 = 12 ), and ( text{rad}(12) = 2 times 3 = 6 ). So, actually, the radical stayed the same from ( a_4 ) to ( a_5 ). Then ( a_6 = 12 + 6 = 18 ), ( text{rad}(18) = 2 times 3 = 6 ). So, it stayed the same again.Wait, so in this case, the radical didn't increase; it stayed the same. So, perhaps my earlier thought that the radical can only stay the same or decrease is correct, but in the sense that it can't introduce new prime factors beyond what it already has.Wait, but in the example, when ( a_1 = 1 ), ( a_2 = 2 ), ( a_3 = 4 ), ( a_4 = 6 ). So, from ( a_3 = 4 ) to ( a_4 = 6 ), the radical went from 2 to 6, which is an increase. So, that contradicts the idea that the radical can't increase.Hmm, so maybe my earlier reasoning was wrong. Let me think again.Given ( a_{n+1} = a_n + text{rad}(a_n) ), and ( text{rad}(a_{n+1}) ) is the product of the distinct prime factors of ( a_{n+1} ).Since ( a_{n+1} = a_n + text{rad}(a_n) ), and ( text{rad}(a_n) ) divides ( a_n ) only if ( a_n ) is square-free. Wait, no, ( text{rad}(a_n) ) is always a divisor of ( a_n ) only if ( a_n ) is square-free. Otherwise, ( text{rad}(a_n) ) is less than ( a_n ).Wait, actually, ( text{rad}(a_n) ) is always a divisor of ( a_n ) only if ( a_n ) is square-free. If ( a_n ) has square factors, then ( text{rad}(a_n) ) is less than ( a_n ), but not necessarily a divisor.Wait, no, actually, ( text{rad}(a_n) ) is the product of the distinct prime factors, so it's always a divisor of ( a_n ) only if ( a_n ) is square-free. If ( a_n ) has square factors, then ( text{rad}(a_n) ) is not a divisor of ( a_n ).Wait, let me clarify. For example, ( a_n = 4 ). ( text{rad}(4) = 2 ), which divides 4. ( a_n = 12 ), ( text{rad}(12) = 6 ), which divides 12. ( a_n = 18 ), ( text{rad}(18) = 6 ), which divides 18. ( a_n = 24 ), ( text{rad}(24) = 6 ), which divides 24. ( a_n = 30 ), ( text{rad}(30) = 30 ), which divides 30. ( a_n = 60 ), ( text{rad}(60) = 30 ), which divides 60. ( a_n = 90 ), ( text{rad}(90) = 30 ), which divides 90. ( a_n = 120 ), ( text{rad}(120) = 30 ), which divides 120. ( a_n = 150 ), ( text{rad}(150) = 30 ), which divides 150. ( a_n = 180 ), ( text{rad}(180) = 30 ), which divides 180. ( a_n = 360 ), ( text{rad}(360) = 30 ), which divides 360.Wait, so in all these cases, ( text{rad}(a_n) ) divides ( a_n ). So, perhaps my earlier thought was incorrect. Maybe ( text{rad}(a_n) ) always divides ( a_n ), regardless of whether ( a_n ) is square-free or not.Wait, let's test with ( a_n = 8 ). ( text{rad}(8) = 2 ), which divides 8. ( a_n = 9 ), ( text{rad}(9) = 3 ), which divides 9. ( a_n = 16 ), ( text{rad}(16) = 2 ), which divides 16. So, yes, it seems that ( text{rad}(a_n) ) always divides ( a_n ), because ( a_n ) is a multiple of each of its prime factors, so their product (the radical) must also divide ( a_n ).Therefore, ( a_{n+1} = a_n + text{rad}(a_n) ) is a multiple of ( text{rad}(a_n) ). Therefore, ( text{rad}(a_{n+1}) ) must divide ( text{rad}(a_n) ), because ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ), and the radical of a multiple of a number can't have more prime factors than the original number.Wait, that makes sense. So, ( text{rad}(a_{n+1}) ) divides ( text{rad}(a_n) ). Therefore, the radical can only stay the same or decrease in prime factors as the sequence progresses.But in my earlier example, when ( a_n = 4 ), ( text{rad}(4) = 2 ), then ( a_{n+1} = 6 ), ( text{rad}(6) = 6 ), which is larger than 2. So, that contradicts the idea that the radical can only stay the same or decrease.Wait, so my reasoning must be wrong. Let me think again.If ( a_{n+1} = a_n + text{rad}(a_n) ), and ( text{rad}(a_n) ) divides ( a_n ), then ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ). Therefore, ( text{rad}(a_{n+1}) ) must divide ( text{rad}(a_n) ), because ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ), and the radical of a multiple can't have more prime factors than the original.But in the example, ( a_n = 4 ), ( text{rad}(4) = 2 ), then ( a_{n+1} = 6 ), ( text{rad}(6) = 6 ), which is not a divisor of 2. So, this seems contradictory.Wait, perhaps I made a mistake in assuming that ( text{rad}(a_{n+1}) ) divides ( text{rad}(a_n) ). Let's see.If ( a_{n+1} = a_n + text{rad}(a_n) ), and ( text{rad}(a_n) ) divides ( a_n ), then ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ). Therefore, ( text{rad}(a_{n+1}) ) must divide ( text{rad}(a_n) ), because the radical of a multiple can't have more prime factors than the original.But in the example, ( a_n = 4 ), ( text{rad}(4) = 2 ), then ( a_{n+1} = 6 ), ( text{rad}(6) = 6 ), which is not a divisor of 2. So, this seems to be a contradiction.Wait, perhaps I'm misunderstanding the relationship. Let me think again.If ( a_{n+1} = a_n + text{rad}(a_n) ), and ( text{rad}(a_n) ) divides ( a_n ), then ( a_{n+1} ) is congruent to 0 modulo ( text{rad}(a_n) ). Therefore, ( text{rad}(a_{n+1}) ) must divide ( text{rad}(a_n) ), because ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ), and the radical of a multiple can't have more prime factors than the original.But in the example, ( a_n = 4 ), ( text{rad}(4) = 2 ), then ( a_{n+1} = 6 ), ( text{rad}(6) = 6 ), which is not a divisor of 2. So, this seems to be a contradiction.Wait, perhaps the key is that ( text{rad}(a_{n+1}) ) divides ( text{rad}(a_n) ) only if ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ). But in the example, ( a_{n+1} = 6 ) is a multiple of ( text{rad}(a_n) = 2 ), so ( text{rad}(6) = 6 ) must divide ( text{rad}(a_n) = 2 ), which it doesn't. Therefore, my earlier conclusion must be wrong.Wait, perhaps the correct statement is that ( text{rad}(a_n) ) divides ( text{rad}(a_{n+1}) ). Let me check.In the example, ( text{rad}(a_n) = 2 ), ( text{rad}(a_{n+1}) = 6 ). So, 2 divides 6, which is true. So, perhaps it's the other way around: ( text{rad}(a_n) ) divides ( text{rad}(a_{n+1}) ).Wait, but in another example, ( a_n = 6 ), ( text{rad}(6) = 6 ), then ( a_{n+1} = 12 ), ( text{rad}(12) = 6 ). So, ( text{rad}(a_n) = 6 ) divides ( text{rad}(a_{n+1}) = 6 ).Another example: ( a_n = 12 ), ( text{rad}(12) = 6 ), ( a_{n+1} = 18 ), ( text{rad}(18) = 6 ). So, again, ( text{rad}(a_n) = 6 ) divides ( text{rad}(a_{n+1}) = 6 ).Another example: ( a_n = 18 ), ( text{rad}(18) = 6 ), ( a_{n+1} = 24 ), ( text{rad}(24) = 6 ). So, same as before.Then, ( a_n = 24 ), ( text{rad}(24) = 6 ), ( a_{n+1} = 30 ), ( text{rad}(30) = 30 ). So, ( text{rad}(a_n) = 6 ) divides ( text{rad}(a_{n+1}) = 30 ), which is true because 6 divides 30.So, it seems that ( text{rad}(a_n) ) divides ( text{rad}(a_{n+1}) ). Therefore, the radical can only stay the same or increase in prime factors as the sequence progresses.Wait, that's the opposite of my earlier conclusion. So, perhaps I was wrong before. Let me try to formalize this.Given ( a_{n+1} = a_n + text{rad}(a_n) ), and since ( text{rad}(a_n) ) divides ( a_n ), we have ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ). Therefore, ( text{rad}(a_{n+1}) ) must be a multiple of ( text{rad}(a_n) ), because ( a_{n+1} ) is a multiple of ( text{rad}(a_n) ), and the radical of a multiple can't have fewer prime factors than the original.Wait, that makes sense. So, ( text{rad}(a_{n+1}) ) is a multiple of ( text{rad}(a_n) ). Therefore, the radical can only stay the same or increase in prime factors as the sequence progresses.This is important because it means that once a prime factor is introduced into the radical, it remains in all subsequent radicals. So, the radical is a non-decreasing function in terms of its prime factors.This has implications for the sequence. If the radical increases, it means that a new prime factor has been introduced. If it stays the same, no new prime factors are introduced.Now, going back to the problem, we need to show that for any ( N ), there are ( N ) consecutive terms in arithmetic progression. As I thought earlier, this would happen if the radical remains constant for ( N ) consecutive steps, because then each term would be the previous term plus the same radical, forming an arithmetic progression.So, to prove this, I need to show that for any ( N ), there exists a point in the sequence where the radical remains constant for ( N ) consecutive terms.Given that the radical is non-decreasing in its prime factors, once a new prime is added, it can't be removed. Therefore, the radical can only stay the same or increase. So, if the radical increases, it's because a new prime has been added.But how does this help me? Well, if the radical can only increase or stay the same, and each time it increases, it's adding a new prime factor, then the number of distinct prime factors in the radical is non-decreasing.However, the number of distinct prime factors is bounded by the number of primes less than or equal to ( a_n ), which is infinite, but for any finite ( N ), we can only have a finite number of new primes added before the radical must stay the same for ( N ) steps.Wait, that might not be the right way to think about it. Let me try a different approach.Suppose that for some ( N ), there are no ( N ) consecutive terms in arithmetic progression. That would mean that the radical changes (i.e., increases) at least every ( N-1 ) steps. So, in other words, every ( N-1 ) steps, a new prime factor is added to the radical.But if that's the case, then the number of distinct prime factors in the radical would grow without bound, because each time we add a new prime, the radical increases, and this would continue indefinitely.However, the sequence ( a_n ) grows at most exponentially, as each term is at most double the previous term. On the other hand, the product of the first ( k ) primes grows faster than exponentially. For example, the product of the first ( k ) primes is known to grow faster than ( 2^{2^k} ), which is much faster than exponential.Therefore, if the radical were to keep adding new primes indefinitely, the radical itself would grow faster than exponentially, which would cause ( a_n ) to grow faster than exponentially, contradicting the fact that ( a_n ) grows at most exponentially.Therefore, our assumption that the radical changes (i.e., increases) at least every ( N-1 ) steps must be false. Hence, for any ( N ), there must exist a point in the sequence where the radical remains constant for ( N ) consecutive steps, leading to ( N ) consecutive terms in arithmetic progression.So, to summarize, if the radical were to change too often, it would cause the sequence to grow too fast, which contradicts the known upper bound on the sequence's growth. Therefore, the radical must remain constant for arbitrarily long stretches, ensuring that for any ( N ), there are ( N ) consecutive terms in arithmetic progression.I think this line of reasoning makes sense. By assuming the opposite and showing that it leads to a contradiction with the growth rate of the sequence, we can conclude that the original statement must be true.

💡Okay, so I have this differential equation:[(x - x^3) y' + (2x^2 - 1) y - x^3 = 0]And I need to check if the function ( y = x sqrt{1 - x^2} + x ) is a solution. Hmm, let me think about how to approach this.First, I remember that to verify if a function is a solution to a differential equation, I need to substitute the function and its derivative into the equation and see if the equation holds true.Alright, so let's start by finding the derivative of ( y ). The function is ( y = x sqrt{1 - x^2} + x ). I'll need to differentiate this with respect to ( x ).Breaking it down, the derivative of ( x sqrt{1 - x^2} ) can be found using the product rule. The product rule says that ( frac{d}{dx}(u cdot v) = u'v + uv' ). Here, ( u = x ) and ( v = sqrt{1 - x^2} ).So, let's compute ( u' ) and ( v' ):- ( u = x ) implies ( u' = 1 ).- ( v = sqrt{1 - x^2} ) implies ( v' = frac{1}{2}(1 - x^2)^{-1/2} cdot (-2x) = -frac{x}{sqrt{1 - x^2}} ).Putting it all together:[frac{d}{dx}(x sqrt{1 - x^2}) = (1) cdot sqrt{1 - x^2} + x cdot left(-frac{x}{sqrt{1 - x^2}}right) = sqrt{1 - x^2} - frac{x^2}{sqrt{1 - x^2}}]Now, the derivative of the entire function ( y ) is:[y' = sqrt{1 - x^2} - frac{x^2}{sqrt{1 - x^2}} + 1]Okay, so that's ( y' ). Now, I need to substitute both ( y ) and ( y' ) into the differential equation and see if it simplifies to zero.Let me write down the differential equation again:[(x - x^3) y' + (2x^2 - 1) y - x^3 = 0]Substituting ( y ) and ( y' ):[(x - x^3) left( sqrt{1 - x^2} - frac{x^2}{sqrt{1 - x^2}} + 1 right) + (2x^2 - 1) left( x sqrt{1 - x^2} + x right) - x^3 = 0]Hmm, this looks a bit complicated, but let's try to expand and simplify term by term.First, let's expand ( (x - x^3) y' ):[(x - x^3) left( sqrt{1 - x^2} - frac{x^2}{sqrt{1 - x^2}} + 1 right)]Multiplying each term inside the parentheses by ( (x - x^3) ):1. ( (x - x^3) sqrt{1 - x^2} )2. ( (x - x^3) left( -frac{x^2}{sqrt{1 - x^2}} right) )3. ( (x - x^3) cdot 1 )So, that gives us:[(x - x^3) sqrt{1 - x^2} - frac{x^2 (x - x^3)}{sqrt{1 - x^2}} + (x - x^3)]Simplify each term:1. ( (x - x^3) sqrt{1 - x^2} ) can be written as ( x(1 - x^2) sqrt{1 - x^2} )2. ( - frac{x^2 (x - x^3)}{sqrt{1 - x^2}} = - frac{x^3 - x^5}{sqrt{1 - x^2}} )3. ( (x - x^3) ) remains as is.Now, let's look at the second part of the equation: ( (2x^2 - 1) y ):[(2x^2 - 1) left( x sqrt{1 - x^2} + x right)]Again, multiply each term inside the parentheses by ( (2x^2 - 1) ):1. ( (2x^2 - 1) x sqrt{1 - x^2} )2. ( (2x^2 - 1) x )Simplify each term:1. ( (2x^2 - 1) x sqrt{1 - x^2} = x(2x^2 - 1) sqrt{1 - x^2} )2. ( (2x^2 - 1) x = 2x^3 - x )Putting it all together, the entire expression becomes:[x(1 - x^2) sqrt{1 - x^2} - frac{x^3 - x^5}{sqrt{1 - x^2}} + (x - x^3) + x(2x^2 - 1) sqrt{1 - x^2} + (2x^3 - x) - x^3 = 0]Now, let's try to combine like terms.First, let's look at the terms involving ( sqrt{1 - x^2} ):1. ( x(1 - x^2) sqrt{1 - x^2} )2. ( x(2x^2 - 1) sqrt{1 - x^2} )Combine these:[x(1 - x^2 + 2x^2 - 1) sqrt{1 - x^2} = x(x^2) sqrt{1 - x^2} = x^3 sqrt{1 - x^2}]Next, the terms involving ( frac{1}{sqrt{1 - x^2}} ):1. ( - frac{x^3 - x^5}{sqrt{1 - x^2}} )That's just:[- frac{x^3(1 - x^2)}{sqrt{1 - x^2}} = -x^3 sqrt{1 - x^2}]Wait, that's interesting. So, the ( sqrt{1 - x^2} ) terms and the ( frac{1}{sqrt{1 - x^2}} ) terms seem to cancel each other out:[x^3 sqrt{1 - x^2} - x^3 sqrt{1 - x^2} = 0]Now, let's look at the remaining terms:1. ( (x - x^3) )2. ( (2x^3 - x) )3. ( -x^3 )Combine these:[(x - x^3) + (2x^3 - x) - x^3 = x - x^3 + 2x^3 - x - x^3 = 0]So, all the terms cancel out, and we're left with zero.Therefore, substituting ( y = x sqrt{1 - x^2} + x ) into the differential equation satisfies it, meaning it is indeed a solution.Now, the second part of the question is whether this is a general solution. A general solution to a first-order differential equation typically contains an arbitrary constant. However, the given function ( y = x sqrt{1 - x^2} + x ) doesn't have any arbitrary constants; it's a specific solution. Therefore, it's not a general solution but a particular solution.So, in conclusion, the function ( y = x sqrt{1 - x^2} + x ) is a solution to the differential equation, but it's not the general solution because it lacks an arbitrary constant.**Final Answer**The function is a solution, but not a general solution. boxed{y = x sqrt{1 - x^2} + x}

💡First, I need to determine how many hours Jill studies each day.On the first day, she studies for 2 hours.On the second day, she doubles the amount from the first day, so she studies for 4 hours.On the third day, she studies one hour less than the previous day, which means she studies for 3 hours.On the fourth day, she studies half the time she studied on the third day, so she studies for 1.5 hours.Next, I need to convert the total hours studied over the four days into minutes. Since 1 hour equals 60 minutes, I can calculate the minutes for each day and then sum them up.First day: 2 hours × 60 minutes/hour = 120 minutesSecond day: 4 hours × 60 minutes/hour = 240 minutesThird day: 3 hours × 60 minutes/hour = 180 minutesFourth day: 1.5 hours × 60 minutes/hour = 90 minutesFinally, adding up all the minutes: 120 + 240 + 180 + 90 = 630 minutes.

💡First, I recognize that a standard chess game begins with a total of 32 pieces, evenly divided between two players, meaning each player starts with 16 pieces.Thomas has lost 5 pieces, so the number of pieces Thomas has remaining is:16 (initial pieces) - 5 (lost pieces) = 11 pieces.The total number of pieces on the chessboard is given as 21. If Thomas has 11 pieces, then Audrey must have:21 (total pieces) - 11 (Thomas's pieces) = 10 pieces.Audrey started with 16 pieces and now has 10 pieces remaining, which means she has lost:16 (initial pieces) - 10 (remaining pieces) = 6 pieces.

💡Okay, so I have this problem where a triangle has altitudes of lengths 9, 12, and 18, and I need to determine the largest angle in the triangle. The options are 90°, 100°, 104.5°, 110°, and 120°. Hmm, let me think about how to approach this.First, I remember that the area of a triangle can be calculated using the formula: Area = (1/2) * base * height. Since the triangle has three different altitudes, each corresponding to a different side, I can express the area in terms of each altitude and its corresponding base.Let me denote the sides opposite to the altitudes of 9, 12, and 18 as a, b, and c respectively. Then, the area of the triangle can be written as:Area = (1/2) * a * 9 = (1/2) * b * 12 = (1/2) * c * 18Since all these expressions equal the same area, I can set them equal to each other:(1/2) * a * 9 = (1/2) * b * 12 = (1/2) * c * 18Simplifying each equation by multiplying both sides by 2, I get:9a = 12b = 18cLet me denote this common value as k. So, 9a = 12b = 18c = k.From this, I can express each side in terms of k:a = k / 9b = k / 12c = k / 18Now, I need to find the actual lengths of the sides. To do this, I can choose a value for k that makes all sides integers. The least common multiple (LCM) of 9, 12, and 18 is 36. So, let me set k = 36.Then, the sides become:a = 36 / 9 = 4b = 36 / 12 = 3c = 36 / 18 = 2So, the sides of the triangle are 4, 3, and 2 units long.Now, I need to determine the largest angle in this triangle. The largest angle is opposite the longest side, which is 4. Let's denote the angle opposite side a (which is 4) as θ.To find θ, I can use the Law of Cosines, which states:c² = a² + b² - 2ab cosθWait, actually, in this case, since θ is opposite side a, the formula should be:a² = b² + c² - 2bc cosθPlugging in the values:4² = 3² + 2² - 2 * 3 * 2 * cosθCalculating each term:16 = 9 + 4 - 12 cosθ16 = 13 - 12 cosθSubtracting 13 from both sides:3 = -12 cosθDividing both sides by -12:cosθ = -3 / 12 = -1/4So, cosθ = -1/4. Now, I need to find the angle θ whose cosine is -1/4. Since cosine is negative, the angle must be greater than 90°, which makes sense because it's the largest angle.Using a calculator, I can find θ:θ = arccos(-1/4) ≈ 104.4775°Rounding this to one decimal place, it's approximately 104.5°, which matches one of the answer choices.Let me double-check my steps to make sure I didn't make a mistake. I started by relating the altitudes to the sides using the area, which seems correct. Then, I found the sides by choosing k as the LCM, which gave me integer lengths. I used the Law of Cosines correctly to find the angle opposite the longest side, and my calculations seem accurate. The result of approximately 104.5° is one of the options, so I think that's the correct answer.I don't see any errors in my reasoning, so I feel confident that the largest angle in the triangle is 104.5°.**Final Answer**The largest angle in the triangle is boxed{C}.