💡Okay, so I have this problem here about a sequence {a_n} that satisfies the equation 2a₁ + 2²a₂ + 2³a₃ + … + 2ⁿaₙ = n, where n is a positive integer. Then, there's another sequence defined by the terms 1/(log₂aₙ · log₂a_{n+1}), and the sum of the first n terms of this sequence is denoted as Sₙ. The task is to find the product of S₁·S₂·S₃·…·S_{10}.Alright, let's break this down step by step. First, I need to figure out what the sequence {a_n} is. The given equation is 2a₁ + 2²a₂ + 2³a₃ + … + 2ⁿaₙ = n. Hmm, so for each n, the sum of these terms involving a₁ to aₙ equals n.I think I can find a recursive relation or maybe express aₙ in terms of the previous terms. Let me try subtracting the equation for n-1 from the equation for n. That might help me find a pattern or a formula for aₙ.So, for n, the equation is:2a₁ + 2²a₂ + 2³a₃ + … + 2ⁿaₙ = n.For n-1, it's:2a₁ + 2²a₂ + 2³a₃ + … + 2^{n-1}a_{n-1} = n - 1.If I subtract the second equation from the first, the left side becomes 2ⁿaₙ, and the right side becomes n - (n - 1) = 1. So, 2ⁿaₙ = 1, which implies that aₙ = 1/(2ⁿ).Wait, that seems straightforward. So, each term aₙ is 1 divided by 2 to the power of n. Let me check if this makes sense for small n.For n=1: 2a₁ = 1 ⇒ a₁ = 1/2. That works.For n=2: 2a₁ + 2²a₂ = 2 ⇒ 2*(1/2) + 4a₂ = 2 ⇒ 1 + 4a₂ = 2 ⇒ 4a₂ = 1 ⇒ a₂ = 1/4. That also works.Similarly, for n=3: 2a₁ + 2²a₂ + 2³a₃ = 3 ⇒ 2*(1/2) + 4*(1/4) + 8a₃ = 3 ⇒ 1 + 1 + 8a₃ = 3 ⇒ 8a₃ = 1 ⇒ a₃ = 1/8. Perfect, so the pattern holds.So, generalizing, aₙ = 1/(2ⁿ). Got it.Now, moving on to the next part. We have another sequence defined by the terms 1/(log₂aₙ · log₂a_{n+1}). Let's write that out:Termₙ = 1 / [log₂(aₙ) · log₂(a_{n+1})].Since we know aₙ = 1/(2ⁿ), let's compute log₂(aₙ). log₂(1/(2ⁿ)) is equal to log₂(2^{-n}) which is -n. Similarly, log₂(a_{n+1}) is log₂(1/(2^{n+1})) = -(n+1).So, substituting back into Termₙ:Termₙ = 1 / [(-n) · (-(n+1))] = 1 / [n(n+1)].That simplifies nicely. So, each term in the new sequence is 1/(n(n+1)). Hmm, that looks familiar. I think this can be expressed as a telescoping series.Indeed, 1/(n(n+1)) can be written as 1/n - 1/(n+1). Let me verify that:1/n - 1/(n+1) = (n+1 - n)/(n(n+1)) = 1/(n(n+1)). Yes, that's correct.So, each term Termₙ is equal to 1/n - 1/(n+1). Therefore, when we sum the first n terms of this sequence, Sₙ, we get:Sₙ = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + … + (1/n - 1/(n+1)).This is a telescoping series, meaning that most terms cancel out. Let's see:The first term is 1 - 1/2.The second term is 1/2 - 1/3.The third term is 1/3 - 1/4....The nth term is 1/n - 1/(n+1).When we add all these up, the -1/2 cancels with the +1/2, the -1/3 cancels with the +1/3, and so on, until the -1/n cancels with the +1/n. What's left is the first term of the first expression, which is 1, and the last term of the last expression, which is -1/(n+1).Therefore, Sₙ = 1 - 1/(n+1) = n/(n+1).So, Sₙ simplifies to n/(n+1). That's a neat result.Now, the problem asks for the product of S₁·S₂·S₃·…·S_{10}. Let's write out each Sₙ:S₁ = 1/2S₂ = 2/3S₃ = 3/4...S_{10} = 10/11So, the product is:(1/2) × (2/3) × (3/4) × … × (10/11)Looking at this product, I notice that each numerator cancels with the denominator of the next fraction. For example, the 2 in the numerator of S₂ cancels with the 2 in the denominator of S₁, the 3 in the numerator of S₃ cancels with the 3 in the denominator of S₂, and so on.This is another telescoping product. Let's see:(1/2) × (2/3) × (3/4) × … × (10/11) = (1 × 2 × 3 × … × 10) / (2 × 3 × 4 × … × 11)The numerator is 10! (10 factorial), and the denominator is (11!)/1, since 2×3×…×11 = 11!/1.So, the product becomes 10! / 11! = 1/11.Because 11! = 11 × 10!, so 10! / 11! = 1/11.Therefore, the product S₁·S₂·S₃·…·S_{10} is equal to 1/11.Let me just recap to make sure I didn't miss anything.1. Found aₙ by subtracting the equation for n-1 from n, leading to aₙ = 1/(2ⁿ).2. Expressed the term 1/(log₂aₙ · log₂a_{n+1}) as 1/(n(n+1)).3. Recognized that 1/(n(n+1)) can be written as 1/n - 1/(n+1), leading to a telescoping sum Sₙ = n/(n+1).4. Calculated the product of S₁ to S_{10}, noticing the telescoping nature, resulting in 1/11.Everything seems to check out. I don't see any mistakes in the reasoning or calculations.**Final Answer**The product of ( S_1 cdot S_2 cdot S_3 cdots S_{10} ) is boxed{dfrac{1}{11}}.

💡Okay, so I need to find the last two digits in the sum of the factorials of the first 50 positive integers. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I remember that the factorial of a number n, written as n!, is the product of all positive integers from 1 to n. So, 1! is 1, 2! is 2, 3! is 6, and so on. As the numbers get bigger, the factorials get really large, really fast.Now, the question is about the last two digits of the sum of these factorials from 1! up to 50!. That means I need to calculate 1! + 2! + 3! + ... + 50! and then find what the last two digits of that total are.But wait, calculating all these factorials up to 50! seems like a lot of work. There must be a smarter way to do this without computing each factorial individually. Maybe I can find a pattern or some property of factorials that can help simplify the problem.I recall that starting from a certain point, factorials have a lot of trailing zeros. For example, 10! is 3628800, which ends with two zeros. I think this is because factorials include multiples of 10, which are products of 2 and 5. Since there are more multiples of 2 than 5 in the factorial sequence, the number of trailing zeros is determined by the number of times 5 appears in the factors.So, for n ≥ 10, n! will have at least two trailing zeros. That means when we consider the last two digits of these factorials, they will be 00. Therefore, adding them to the sum won't affect the last two digits of the total. So, I only need to consider the factorials from 1! to 9! because starting from 10!, each term contributes 00 to the last two digits.Alright, that simplifies things a bit. Now, I need to calculate the sum of the factorials from 1! to 9! and then find the last two digits of that sum.Let me list out the factorials from 1! to 9!:1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 8! = 40320 9! = 362880Now, I need to add all these up. But since I'm only interested in the last two digits of the total sum, I can just consider the last two digits of each factorial before adding them together.Let's write down the last two digits of each factorial:1! = 01 (but since it's just 1, the last two digits are 01) 2! = 02 3! = 06 4! = 24 5! = 20 (since 120 ends with 20) 6! = 20 (since 720 ends with 20) 7! = 40 (since 5040 ends with 40) 8! = 20 (since 40320 ends with 20) 9! = 80 (since 362880 ends with 80)Wait, hold on. For 1!, it's just 1, so the last two digits are 01. But when adding, do I need to consider it as 01 or just 1? I think it's better to consider all of them as two-digit numbers for consistency, so 1! is 01, 2! is 02, and so on.So, now, let's list them again with two digits:1! = 01 2! = 02 3! = 06 4! = 24 5! = 20 6! = 20 7! = 40 8! = 20 9! = 80Now, I need to add all these two-digit numbers together. Let me write them down:01 02 06 24 20 20 40 20 80Let me add them step by step:Start with 01.01 + 02 = 0303 + 06 = 0909 + 24 = 3333 + 20 = 5353 + 20 = 7373 + 40 = 113113 + 20 = 133133 + 80 = 213So, the total sum of the last two digits is 213. But we're only interested in the last two digits of this sum, which is 13.Wait, let me double-check my addition to make sure I didn't make a mistake.Starting over:01 (1!) +02 (2!) = 03 +06 (3!) = 09 +24 (4!) = 33 +20 (5!) = 53 +20 (6!) = 73 +40 (7!) = 113 +20 (8!) = 133 +80 (9!) = 213Yes, that seems correct. So, the sum is 213, and the last two digits are 13.But just to be thorough, let me verify each step:1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 8! = 40320 9! = 362880Adding them all up:1 + 2 = 3 3 + 6 = 9 9 + 24 = 33 33 + 120 = 153 153 + 720 = 873 873 + 5040 = 5913 5913 + 40320 = 46233 46233 + 362880 = 409113Wait a minute, that's different. The total sum is 409113, and the last two digits are 13. So, that matches my previous result.But hold on, in my initial step-by-step addition, I was only adding the last two digits of each factorial, which gave me 213, and the last two digits of that are 13. But when I added all the factorials normally, the total was 409113, which also ends with 13. So, both methods give the same result. That’s reassuring.Therefore, the last two digits of the sum of the factorials from 1! to 50! are 13.I think that makes sense because starting from 10!, all the factorials add numbers ending with 00, so they don't affect the last two digits. So, only the sum up to 9! matters, and that sum is 409113, which ends with 13.Just to make sure I didn't make any calculation errors, let me compute the sum of the factorials from 1! to 9! again:1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 8! = 40320 9! = 362880Adding them up:1 + 2 = 3 3 + 6 = 9 9 + 24 = 33 33 + 120 = 153 153 + 720 = 873 873 + 5040 = 5913 5913 + 40320 = 46233 46233 + 362880 = 409113Yes, that's correct. 409113 is the total sum, and the last two digits are indeed 13.So, even though the factorials get really large, the key insight is that beyond 9!, the factorials contribute 00 in the last two digits, so they don't affect the result. Therefore, only the sum up to 9! is necessary, and that sum ends with 13.I think I've got it. The last two digits are 13.

💡Alright, so I have this problem about two sequences, {a_n} and {b_n}, with some trigonometric relationships. It has two parts: first, I need to prove a certain identity involving sine and cosine squared terms, and second, I need to determine if there exist initial values a₁ and b₁ such that the sequence formed by the sum of sine squared and cosine squared terms is a geometric sequence. Hmm, okay, let me try to unpack this step by step.Starting with part (1): I need to prove that sin²a_{n+1} + cos²b_{n+1} equals twice the sum of sin²a_n + cos²b_n. The given relationships are sin a_{n+1} = sin a_n + cos b_n and cos b_{n+1} = cos b_n - sin a_n. So, maybe I can square both sides of these equations and then add them together? That seems like a common technique when dealing with trigonometric identities.Let me write down the given equations:1. sin a_{n+1} = sin a_n + cos b_n2. cos b_{n+1} = cos b_n - sin a_nIf I square both sides of equation 1, I get:sin²a_{n+1} = (sin a_n + cos b_n)² = sin²a_n + 2 sin a_n cos b_n + cos²b_nSimilarly, squaring both sides of equation 2 gives:cos²b_{n+1} = (cos b_n - sin a_n)² = cos²b_n - 2 sin a_n cos b_n + sin²a_nNow, if I add these two squared equations together:sin²a_{n+1} + cos²b_{n+1} = [sin²a_n + 2 sin a_n cos b_n + cos²b_n] + [cos²b_n - 2 sin a_n cos b_n + sin²a_n]Let me simplify the right-hand side:= sin²a_n + 2 sin a_n cos b_n + cos²b_n + cos²b_n - 2 sin a_n cos b_n + sin²a_nI notice that the 2 sin a_n cos b_n and -2 sin a_n cos b_n terms cancel each other out. So, we're left with:= sin²a_n + cos²b_n + cos²b_n + sin²a_n= 2 sin²a_n + 2 cos²b_n= 2 (sin²a_n + cos²b_n)So, that's exactly what we needed to prove for part (1)! That wasn't too bad. I just had to square both given equations, add them, and simplify.Moving on to part (2): Does there exist a₁ and b₁ such that the sequence {sin²a_n + cos²b_n} forms a geometric sequence? Hmm, okay, so a geometric sequence is one where each term is a constant multiple of the previous term. That constant is called the common ratio.From part (1), we have that sin²a_{n+1} + cos²b_{n+1} = 2 (sin²a_n + cos²b_n). So, this suggests that each term is twice the previous term. That would mean that the sequence {sin²a_n + cos²b_n} is a geometric sequence with a common ratio of 2, provided that the initial term is non-zero.Wait, but the problem is asking if such a₁ and b₁ exist. So, does this mean that for any a₁ and b₁, the sequence will automatically be geometric with ratio 2? Or are there specific a₁ and b₁ that make it geometric?Let me think. If I choose a₁ and b₁ such that sin²a₁ + cos²b₁ is some non-zero value, say m, then according to the recurrence relation, sin²a₂ + cos²b₂ = 2m, sin²a₃ + cos²b₃ = 4m, and so on. So, the sequence would be m, 2m, 4m, 8m, ..., which is indeed a geometric sequence with ratio 2.But wait, is there any restriction on m? Since sin²a_n and cos²b_n are both non-negative, their sum must be non-negative as well. Moreover, sin²a_n ≤ 1 and cos²b_n ≤ 1, so their sum is at most 2. Therefore, m must satisfy 0 < m ≤ 2.But can m be any value in that interval? Let's see. If I choose a₁ and b₁ such that sin²a₁ + cos²b₁ = m, then the sequence will be m, 2m, 4m, 8m, etc. However, since each term is twice the previous one, unless m = 0, the terms will grow without bound. But wait, sin²a_n + cos²b_n cannot exceed 2, as each term is at most 1. So, if m > 0, then eventually, 2^{n}m will exceed 2, which is impossible because sin²a_n + cos²b_n can't be more than 2.This seems like a contradiction. So, does that mean that such a₁ and b₁ cannot exist? Because if we start with m > 0, the sequence would have to exceed 2 at some point, which is impossible. Therefore, the only way to avoid this is if m = 0, but sin²a₁ + cos²b₁ = 0 implies that both sin a₁ = 0 and cos b₁ = 0. But sin a₁ = 0 implies a₁ is an integer multiple of π, and cos b₁ = 0 implies b₁ is an odd multiple of π/2. However, in that case, sin a₁ = 0 and cos b₁ = 0, so the initial terms would be zero, but then the entire sequence would be zero, which is technically a geometric sequence with ratio 0, but that's trivial.But the problem probably expects a non-trivial geometric sequence, meaning a common ratio different from 0. Therefore, if we require the sequence to be non-trivial, such a₁ and b₁ do not exist because the terms would eventually exceed the maximum possible value of 2.Wait, but let me double-check. Suppose I choose a₁ and b₁ such that sin²a₁ + cos²b₁ = 1. Then the next term would be 2, which is still within the maximum limit. But the term after that would be 4, which is greater than 2, which is impossible. So, even starting with m = 1, the sequence would go 1, 2, 4, 8,... which is impossible because it exceeds 2 at the third term.Therefore, the only way to have a valid sequence is if m = 0, but that's trivial. Hence, there do not exist non-trivial initial values a₁ and b₁ such that {sin²a_n + cos²b_n} is a geometric sequence.So, summarizing my thoughts:1. For part (1), by squaring and adding the given recurrence relations, we can show that sin²a_{n+1} + cos²b_{n+1} = 2(sin²a_n + cos²b_n).2. For part (2), while the recurrence suggests a geometric sequence with ratio 2, the terms would exceed the maximum possible value of 2, making it impossible unless the initial term is zero, which is trivial. Therefore, no such non-trivial initial values exist.I think that covers both parts. I should probably write this up more formally to ensure all steps are clear.

💡Alright, so I came across this interesting problem that says there doesn't exist a polynomial ( P(x) ) such that all the values ( P(0), P(1), P(2), ldots ) are prime numbers. At first glance, it seems a bit tricky, but I think I can break it down.Let me start by recalling what a polynomial is. A polynomial is an expression consisting of variables and coefficients, constructed using only addition, subtraction, multiplication, and non-negative integer exponents of variables. In this case, the polynomial is given as:[P(x) = a_0 x^n + a_1 x^{n-1} + ldots + a_{n-1} x + a_n]where ( a_0, a_1, ldots, a_n ) are coefficients, and ( n ) is the degree of the polynomial.The problem states that if such a polynomial existed, then every value ( P(0), P(1), P(2), ldots ) would have to be a prime number. My goal is to show that this is impossible.First, let me consider what happens when we plug in different integer values into the polynomial. For ( x = 0 ), the polynomial simplifies to ( P(0) = a_n ). So, ( a_n ) must be a prime number. Similarly, ( P(1) = a_0 + a_1 + ldots + a_n ) must also be prime, and so on.Now, if I think about polynomials, they can grow quite rapidly depending on their degree. For example, a linear polynomial grows linearly, a quadratic polynomial grows quadratically, and so on. Since primes become less frequent as numbers get larger, it's unlikely that a polynomial of higher degree could consistently produce primes for all integer inputs.But this is just an intuition. I need a more rigorous argument.I remember something about polynomials and modular arithmetic. Maybe I can use that. Let's suppose that such a polynomial ( P(x) ) exists where ( P(k) ) is prime for all integers ( k geq 0 ).Let me pick a specific value, say ( x = N ), where ( N ) is some integer. Then ( P(N) = M ), which is a prime number. Now, consider ( P(N + kM) ) for some integer ( k ). I want to see what happens when I plug in ( N + kM ) into the polynomial.Expanding ( P(N + kM) ), we get:[P(N + kM) = a_0 (N + kM)^n + a_1 (N + kM)^{n-1} + ldots + a_{n-1} (N + kM) + a_n]Now, subtracting ( P(N) ) from both sides:[P(N + kM) - P(N) = a_0 left[(N + kM)^n - N^nright] + a_1 left[(N + kM)^{n-1} - N^{n-1}right] + ldots + a_{n-1} (kM)]Looking at each term, like ( (N + kM)^l - N^l ), I can factor out ( kM ). For example, using the binomial theorem, ( (N + kM)^l = N^l + binom{l}{1} N^{l-1} (kM) + ldots + (kM)^l ). Subtracting ( N^l ) leaves terms that are all multiples of ( kM ). So each term in the expression ( P(N + kM) - P(N) ) is divisible by ( kM ).Therefore, ( P(N + kM) - P(N) ) is divisible by ( M ), which means:[P(N + kM) equiv P(N) pmod{M}]Since ( P(N) = M ), this simplifies to:[P(N + kM) equiv 0 pmod{M}]This tells me that ( P(N + kM) ) is divisible by ( M ). But ( M ) is a prime number, so ( P(N + kM) ) must be either ( M ) or ( -M ) or a multiple of ( M ). However, since we're dealing with positive integers (as primes are positive), ( P(N + kM) ) must be either ( M ) or a multiple of ( M ).But wait, ( P(N + kM) ) is supposed to be prime for all ( k ). If ( P(N + kM) ) is a multiple of ( M ), the only way it can be prime is if ( P(N + kM) = M ). So, for all ( k ), ( P(N + kM) = M ).This would mean that the polynomial ( P(x) ) takes the value ( M ) infinitely often, specifically at ( x = N, N + M, N + 2M, ldots ). But a non-constant polynomial can only take a particular value finitely many times. The only way this could happen is if the polynomial is constant, but a constant polynomial would have to be equal to ( M ) everywhere, which isn't possible because ( P(0) = a_n ) must also be prime, and unless ( a_n = M ), which would make the polynomial constant, but then all outputs would be ( M ), which is prime, but the problem allows for varying outputs.Wait, hold on. If the polynomial is constant, say ( P(x) = M ), then yes, all outputs are ( M ), which is prime. But the problem specifies a polynomial of the form ( a_0 x^n + ldots + a_n ), which includes non-constant polynomials. So, if the polynomial is non-constant, it can't take the same value infinitely often. Therefore, for a non-constant polynomial, ( P(N + kM) ) can't be equal to ( M ) for all ( k ), which leads to a contradiction because we just saw that ( P(N + kM) ) must be divisible by ( M ) and hence either ( M ) or a multiple of ( M ). But since it can't be ( M ) infinitely often, it must be a multiple of ( M ) for some ( k ), which would make ( P(N + kM) ) composite (since it's a multiple of ( M ) and greater than ( M )), contradicting the assumption that all outputs are prime.Therefore, such a polynomial cannot exist because it would lead to a contradiction where some outputs are composite numbers.Another angle to consider is the density of primes. As numbers get larger, primes become less frequent. A polynomial of degree ( n geq 1 ) will grow without bound as ( x ) increases, and the gaps between primes grow as numbers increase. So, it's unlikely that a polynomial could consistently hit primes for every integer input.Additionally, using the fact that polynomials are determined by their values at sufficiently many points, if a polynomial were to take prime values at enough points, it would have to be a constant polynomial, which isn't the case here since we're considering non-constant polynomials.So, putting it all together, the key idea is that for any non-constant polynomial, evaluating it at points spaced by a prime value will eventually yield composite numbers, which contradicts the requirement that all outputs are prime. Hence, such a polynomial cannot exist.**Final Answer**boxed{text{There does not exist a polynomial where all its values } P(0), P(1), P(2), ldots text{are prime numbers.}}

💡Okay, so I have this problem where a projectile is launched, and its height in meters is given by the equation y = -4.9t² + 23.8t. I need to find out at what time t the projectile first reaches 28 meters in height. Hmm, let me think about how to approach this.First, I know that the equation is a quadratic in terms of t. Quadratic equations usually have two solutions, which in this context would mean two times when the projectile is at 28 meters: once going up and once coming down. Since the question asks for the first time it reaches 28 meters, I'm looking for the smaller value of t.So, I should set y equal to 28 and solve for t. Let me write that down:28 = -4.9t² + 23.8tNow, to solve for t, I need to rearrange this equation into the standard quadratic form, which is ax² + bx + c = 0. So, I'll subtract 28 from both sides:-4.9t² + 23.8t - 28 = 0Hmm, dealing with decimals can be a bit messy, so maybe I can simplify this equation by eliminating the decimals. If I multiply every term by 10, that should help:10*(-4.9t²) + 10*(23.8t) - 10*28 = 0-49t² + 238t - 280 = 0That looks better. Now, I have a quadratic equation: -49t² + 238t - 280 = 0. I can also multiply the entire equation by -1 to make the coefficient of t² positive, which might make factoring easier:49t² - 238t + 280 = 0Okay, now I have 49t² - 238t + 280 = 0. Let's see if I can factor this. Factoring quadratics can sometimes be tricky, especially with larger coefficients. Maybe I can factor out a common factor first. Let's check if all the coefficients have a common divisor.49, 238, and 280. Let's see:- 49 is 7².- 238 divided by 7 is 34.- 280 divided by 7 is 40.So, yes, all terms are divisible by 7. Let's factor that out:7(7t² - 34t + 40) = 0Now, the equation simplifies to 7t² - 34t + 40 = 0. Let's try to factor this quadratic. I'm looking for two numbers that multiply to 7*40 = 280 and add up to -34. Hmm, let's think:What pairs of numbers multiply to 280? Let's list some:- 1 and 280- 2 and 140- 4 and 70- 5 and 56- 7 and 40- 8 and 35- 10 and 28- 14 and 20Now, I need two numbers that add up to -34. Let's see:- 14 and 20 add up to 34. If both are negative, they add up to -34 and multiply to 280. Perfect!So, I can rewrite the middle term using these numbers:7t² - 14t - 20t + 40 = 0Now, let's group the terms:(7t² - 14t) + (-20t + 40) = 0Factor out the greatest common factor from each group:7t(t - 2) -20(t - 2) = 0Now, I can factor out the common binomial factor (t - 2):(7t - 20)(t - 2) = 0So, the factored form of the quadratic is (7t - 20)(t - 2) = 0. Now, set each factor equal to zero and solve for t:7t - 20 = 07t = 20t = 20/7 ≈ 2.857 secondst - 2 = 0t = 2 secondsSo, the two times when the projectile is at 28 meters are approximately 2.857 seconds and 2 seconds. Since the question asks for the first time it reaches 28 meters, the smaller value of t is the answer, which is 2 seconds.Wait a minute, let me double-check my factoring to make sure I didn't make a mistake. I had 7t² - 34t + 40, and I factored it as (7t - 20)(t - 2). Let's multiply it back out to verify:(7t - 20)(t - 2) = 7t*t + 7t*(-2) -20*t -20*(-2)= 7t² -14t -20t +40= 7t² -34t +40Yes, that matches. So, my factoring is correct. Therefore, the solutions t = 20/7 and t = 2 are correct.Just to be thorough, let's plug t = 2 back into the original equation to ensure it gives y = 28:y = -4.9*(2)² + 23.8*(2)= -4.9*4 + 47.6= -19.6 + 47.6= 28Perfect, that checks out. Now, let's check t = 20/7 ≈ 2.857:y = -4.9*(20/7)² + 23.8*(20/7)First, calculate (20/7)² = 400/49 ≈ 8.163So, -4.9*8.163 ≈ -4.9*8.163 ≈ -40Then, 23.8*(20/7) ≈ 23.8*2.857 ≈ 68So, y ≈ -40 + 68 = 28That also checks out. So, both times are correct, and since we're looking for the first time, t = 2 seconds is the answer.I think I've covered all the steps and double-checked my work. It makes sense that the projectile reaches 28 meters at 2 seconds on the way up and again at approximately 2.857 seconds on the way down. Since the question asks for the first time, 2 seconds is the correct answer.

💡Okay, so I have this matrix B, which is a 2x2 matrix:[mathbf{B} = begin{pmatrix} 2 & 4 3 & 2 end{pmatrix}]And I need to compute the determinant of (mathbf{B}^2 - 3mathbf{B}). Hmm, determinants can be tricky, but let's break it down step by step.First, I remember that for any square matrix, the determinant of a product is the product of the determinants. So, if I can express (mathbf{B}^2 - 3mathbf{B}) as a product of two matrices, maybe I can use that property. Let me see:[mathbf{B}^2 - 3mathbf{B} = mathbf{B}(mathbf{B} - 3mathbf{I})]Where (mathbf{I}) is the identity matrix. That seems right because when you distribute (mathbf{B}), you get (mathbf{B}^2 - 3mathbf{B}). So, now I can write:[det(mathbf{B}^2 - 3mathbf{B}) = det(mathbf{B}(mathbf{B} - 3mathbf{I})) = det(mathbf{B}) cdot det(mathbf{B} - 3mathbf{I})]Alright, so now I just need to find the determinants of (mathbf{B}) and (mathbf{B} - 3mathbf{I}) separately and then multiply them together.Let's start with (det(mathbf{B})). For a 2x2 matrix (begin{pmatrix} a & b c & d end{pmatrix}), the determinant is (ad - bc). Applying that to matrix B:[det(mathbf{B}) = (2)(2) - (4)(3) = 4 - 12 = -8]Okay, so (det(mathbf{B}) = -8). Got that.Next, I need to find (det(mathbf{B} - 3mathbf{I})). Let's compute (mathbf{B} - 3mathbf{I}) first. The identity matrix (mathbf{I}) is:[mathbf{I} = begin{pmatrix} 1 & 0 0 & 1 end{pmatrix}]So, multiplying by 3:[3mathbf{I} = begin{pmatrix} 3 & 0 0 & 3 end{pmatrix}]Subtracting this from (mathbf{B}):[mathbf{B} - 3mathbf{I} = begin{pmatrix} 2 - 3 & 4 - 0 3 - 0 & 2 - 3 end{pmatrix} = begin{pmatrix} -1 & 4 3 & -1 end{pmatrix}]Now, let's find the determinant of this resulting matrix:[det(mathbf{B} - 3mathbf{I}) = (-1)(-1) - (4)(3) = 1 - 12 = -11]So, (det(mathbf{B} - 3mathbf{I}) = -11).Now, going back to the original expression:[det(mathbf{B}^2 - 3mathbf{B}) = det(mathbf{B}) cdot det(mathbf{B} - 3mathbf{I}) = (-8) times (-11) = 88]Wait, is that it? It seems straightforward, but let me double-check my calculations to make sure I didn't make any mistakes.First, (det(mathbf{B})):- (2 times 2 = 4)- (4 times 3 = 12)- (4 - 12 = -8)That's correct.Next, (mathbf{B} - 3mathbf{I}):- (2 - 3 = -1)- (4 - 0 = 4)- (3 - 0 = 3)- (2 - 3 = -1)So, the matrix is correct:[begin{pmatrix} -1 & 4 3 & -1 end{pmatrix}]Then, determinant:- ((-1) times (-1) = 1)- (4 times 3 = 12)- (1 - 12 = -11)That's correct too.Multiplying the two determinants:- (-8 times -11 = 88)Yes, that seems right.Alternatively, I could have computed (mathbf{B}^2 - 3mathbf{B}) directly and then found its determinant. Maybe I should try that method to confirm.First, compute (mathbf{B}^2):[mathbf{B}^2 = mathbf{B} times mathbf{B} = begin{pmatrix} 2 & 4 3 & 2 end{pmatrix} times begin{pmatrix} 2 & 4 3 & 2 end{pmatrix}]Calculating each element:- First row, first column: (2 times 2 + 4 times 3 = 4 + 12 = 16)- First row, second column: (2 times 4 + 4 times 2 = 8 + 8 = 16)- Second row, first column: (3 times 2 + 2 times 3 = 6 + 6 = 12)- Second row, second column: (3 times 4 + 2 times 2 = 12 + 4 = 16)So,[mathbf{B}^2 = begin{pmatrix} 16 & 16 12 & 16 end{pmatrix}]Now, compute (3mathbf{B}):[3mathbf{B} = 3 times begin{pmatrix} 2 & 4 3 & 2 end{pmatrix} = begin{pmatrix} 6 & 12 9 & 6 end{pmatrix}]Subtracting (3mathbf{B}) from (mathbf{B}^2):[mathbf{B}^2 - 3mathbf{B} = begin{pmatrix} 16 - 6 & 16 - 12 12 - 9 & 16 - 6 end{pmatrix} = begin{pmatrix} 10 & 4 3 & 10 end{pmatrix}]Now, find the determinant of this matrix:[detleft(begin{pmatrix} 10 & 4 3 & 10 end{pmatrix}right) = (10)(10) - (4)(3) = 100 - 12 = 88]So, that also gives me 88. Great, both methods lead to the same result, which gives me more confidence that 88 is the correct answer.I think I've covered all the steps and double-checked my work. It seems solid. I don't see any mistakes in my calculations, so I'm pretty sure the determinant of (mathbf{B}^2 - 3mathbf{B}) is 88.