💡Okay, so I have this quadratic function f(x) = (x + a)(bx + 2a), and I need to find its expression given that it's an even function and its range is (-∞, 4]. Hmm, let's break this down step by step.First, I remember that an even function satisfies the condition f(x) = f(-x) for all x. That means the function is symmetric about the y-axis. For a quadratic function, which is a parabola, being even implies that it doesn't have a linear term; otherwise, it would be shifted left or right, making it not symmetric about the y-axis.So, let me expand the given function to see what it looks like. Multiplying out (x + a)(bx + 2a), I get:f(x) = x*(bx) + x*(2a) + a*(bx) + a*(2a) = bx² + 2a x + ab x + 2a²Combining like terms, the linear terms are 2a x and ab x, so together they become (2a + ab)x. Therefore, the expanded form is:f(x) = bx² + (2a + ab)x + 2a²Since f(x) is even, the coefficient of the x term must be zero. That gives me the equation:2a + ab = 0I can factor out an 'a' from this equation:a(2 + b) = 0So, either a = 0 or 2 + b = 0. If a = 0, then the function simplifies to f(x) = bx² + 0x + 0, which is f(x) = bx². But then, the range of this function would depend on the value of b. If b is positive, the range would be [0, ∞), and if b is negative, it would be (-∞, 0]. However, the given range is (-∞, 4], which suggests that the maximum value of the function is 4. So, if a = 0, the function can't have a maximum at 4 because it would either go to infinity or negative infinity. Therefore, a = 0 is not a valid solution here.So, the other possibility is 2 + b = 0, which means b = -2. Now, substituting b = -2 back into the function:f(x) = (-2)x² + (2a + a*(-2))x + 2a²Simplify the linear term:2a + (-2a) = 0So, the linear term cancels out, which is consistent with the function being even. Now, the function becomes:f(x) = -2x² + 2a²Now, we know that the range of this quadratic function is (-∞, 4]. Since the coefficient of x² is -2, which is negative, the parabola opens downward, meaning it has a maximum point. The maximum value occurs at the vertex of the parabola.For a quadratic function in the form f(x) = ax² + c, the vertex is at (0, c). So, in this case, the maximum value is 2a², and according to the problem, this maximum value is 4. Therefore:2a² = 4Divide both sides by 2:a² = 2Taking the square root of both sides:a = √2 or a = -√2But since a is squared in the function, both values of a will give the same function. So, substituting back into f(x):f(x) = -2x² + 2*(√2)² = -2x² + 2*2 = -2x² + 4Alternatively, if a = -√2:f(x) = -2x² + 2*(-√2)² = -2x² + 2*2 = -2x² + 4Either way, the function is the same.So, putting it all together, the quadratic function is f(x) = -2x² + 4.**Final Answer**The expression for ( f(x) ) is boxed{-2x^2 + 4}.

💡First, I recognize that William shared 3 marbles with Theresa and was left with 7 marbles.To find out how many marbles William had initially, I need to add the number of marbles he shared with Theresa to the number of marbles he has left.So, I add 3 marbles (shared with Theresa) to 7 marbles (left with William), which equals 10 marbles.Therefore, William initially had 10 marbles.

💡Okay, so I need to figure out the smallest number n such that the product of all integers from 1 to n, which is n!, is divisible by 1029. Hmm, let's start by understanding what 1029 is made up of. Maybe I should factorize it into its prime factors because that will help me see what n! needs to include.Alright, let's try dividing 1029 by some small primes. I know that 1029 is an odd number, so it's not divisible by 2. Let's try 3. Adding the digits: 1 + 0 + 2 + 9 = 12, which is divisible by 3, so 1029 ÷ 3 should give me an integer. Let's do that: 1029 ÷ 3 = 343. Okay, so 3 is a prime factor.Now, I need to factorize 343. Hmm, 343 seems familiar. I think it's a power of 7 because 7 times 7 is 49, and 49 times 7 is 343. Let me check: 7 × 7 = 49, and 49 × 7 = 343. Yes, that's right. So, 343 is 7³. Therefore, the prime factorization of 1029 is 3 × 7³.So, for n! to be divisible by 1029, it needs to have at least one 3 and three 7s in its prime factors. I know that n! includes all the numbers from 1 up to n, so I need to make sure that there are enough multiples of 3 and 7 in that range.First, let's think about the number of 7s. Since 7 is a prime number, every multiple of 7 contributes at least one 7 to the factorial. But since we need three 7s, we need to have at least three multiples of 7 in the numbers from 1 to n.Let's list the multiples of 7: 7, 14, 21, 28, 35, 42, etc. Each of these contributes one 7. However, some numbers contribute more than one 7. For example, 49 is 7², so it contributes two 7s. Similarly, 147 is 7³, but that's probably beyond our needed range.But since we need three 7s, let's see how many multiples of 7 we need. The first multiple is 7, which gives one 7. The second is 14, which is 7×2, giving another 7. The third is 21, which is 7×3, giving another 7. So, up to 21, we have three 7s. But wait, 21 is 7×3, which is still just one 7. So, actually, to get three 7s, we need three multiples of 7.But hold on, 49 is 7×7, so it contributes two 7s. So, if n is 49, we would have more than three 7s, but we need the smallest n. So, maybe n doesn't need to be as high as 49.Wait, let's think again. If n is 21, then the multiples of 7 are 7, 14, and 21. Each contributes one 7, so that's three 7s. But 21 is 7×3, so does that mean it contributes an extra 3? Yes, it does, but we only need one 3, so that's fine.But wait, 21 is 7×3, so it's contributing both a 7 and a 3. So, in n=21, we have the necessary three 7s and at least one 3. Therefore, n=21 should be sufficient.But let me double-check. If n=21, then n! includes 7, 14, and 21, each contributing a 7, so that's three 7s. And it includes 3, 6, 9, 12, 15, 18, 21, which are multiples of 3, so definitely more than one 3. Therefore, n=21 should work.But wait, is there a smaller n? Let's see. If n=14, then the multiples of 7 are 7 and 14, which is only two 7s. We need three 7s, so n=14 is not enough. Similarly, n=21 is the next multiple of 7, which gives us three 7s.Therefore, the smallest n is 21.

💡Okay, so I have this problem: I need to prove that when taking the square root of two rational positive numbers, which are not perfect squares, the resulting decimal sequences cannot be identical. Hmm, that sounds interesting. Let me try to break it down step by step.First, let me make sure I understand the problem correctly. We have two positive rational numbers, say p and q, neither of which is a perfect square. Their square roots, sqrt(p) and sqrt(q), are irrational numbers because if they were rational, p and q would be perfect squares. Now, the claim is that the decimal expansions of sqrt(p) and sqrt(q) cannot be identical. So, if I write out the decimal expansions of both, they must differ somewhere.Alright, so I need to show that if p and q are distinct positive rational numbers that are not perfect squares, then sqrt(p) and sqrt(q) have different decimal expansions. That means there must be at least one digit where their decimal expansions differ.Let me think about how to approach this. Maybe I can assume the opposite and reach a contradiction. So, suppose that sqrt(p) and sqrt(q) have the same decimal expansion. That would mean that sqrt(p) equals sqrt(q), right? Because if their decimal expansions are identical, then they are the same number. But wait, if sqrt(p) equals sqrt(q), then squaring both sides would give p equals q. But p and q are distinct numbers, so that leads to a contradiction. Therefore, our assumption must be wrong, meaning that sqrt(p) and sqrt(q) cannot have the same decimal expansion.Wait, is that it? That seems too straightforward. Let me check if I missed something. I assumed that if two numbers have the same decimal expansion, they are equal. That is true because the decimal expansion uniquely represents a real number. So, if sqrt(p) and sqrt(q) have the same decimal expansion, they must be equal, which would imply p equals q. But since p and q are distinct, this is impossible. Therefore, their decimal expansions must differ.Hmm, maybe I should consider the case where p and q are not necessarily distinct. Wait, no, the problem states two rational positive numbers, which are not perfect squares. It doesn't specify they are distinct, but if they are the same number, then their square roots would obviously be the same. So, perhaps the problem assumes p and q are distinct. Let me read the problem again."Prove that when taking the square root of two rational positive numbers, which are not perfect squares, the resulting decimal sequences cannot be identical."Okay, so it's about two numbers, not necessarily distinct. But if they are the same number, then their square roots are the same, so their decimal expansions are identical. But the problem says "two rational positive numbers," so maybe they are considering distinct numbers. Or perhaps the problem is just stating that for any two such numbers, their square roots cannot have identical decimal expansions unless they are the same number.Wait, maybe I need to clarify that. Let me think again.Suppose p and q are two distinct positive rational numbers, neither of which is a perfect square. Then sqrt(p) and sqrt(q) are irrational. I need to show that their decimal expansions cannot be identical.So, if I assume that sqrt(p) and sqrt(q) have the same decimal expansion, then sqrt(p) = sqrt(q), which implies p = q. But p and q are distinct, so this is a contradiction. Therefore, their decimal expansions must differ.Is that the proof? It seems too simple, but maybe it's correct. Let me see if there's another way to think about it.Alternatively, maybe I can think about the decimal expansions themselves. If two numbers have the same decimal expansion, they are equal. Since p and q are distinct, their square roots must be distinct, so their decimal expansions must differ.Wait, but maybe the problem is more about the decimal expansions being non-repeating and non-terminating, so they can't be identical in some other way. But no, if two numbers have the same decimal expansion, they are equal, regardless of whether it's repeating or terminating.So, perhaps the key point is that if two numbers have the same decimal expansion, they must be equal, and since p and q are distinct, their square roots must be distinct, hence their decimal expansions must differ.But let me think about whether there could be two different irrational numbers with the same decimal expansion. I don't think so, because the decimal expansion uniquely defines a real number. So, if two numbers have the same decimal expansion, they are the same number.Therefore, if sqrt(p) and sqrt(q) have the same decimal expansion, they must be equal, which would imply p = q, contradicting the assumption that p and q are distinct.Wait, but the problem didn't specify that p and q are distinct. It just says two rational positive numbers, not perfect squares. So, if p and q are the same, then their square roots are the same, so their decimal expansions are identical. But if p and q are different, then their square roots must be different, hence their decimal expansions must differ.So, maybe the problem is implicitly assuming that p and q are distinct, or perhaps it's considering that even if they are the same, the decimal expansions are identical, but the main point is that for different p and q, the decimal expansions of their square roots must differ.But let me think again. Suppose p and q are different, but their square roots have the same decimal expansion. Then, as I said, sqrt(p) = sqrt(q), so p = q, which contradicts the assumption that p and q are different. Therefore, their decimal expansions must differ.So, maybe that's the proof. It seems straightforward, but perhaps I'm missing something. Let me try to formalize it.Assume, for contradiction, that there exist two distinct positive rational numbers p and q, neither of which is a perfect square, such that sqrt(p) and sqrt(q) have the same decimal expansion. Then, sqrt(p) = sqrt(q), which implies p = q. But this contradicts the assumption that p and q are distinct. Therefore, no such p and q exist, meaning that the decimal expansions of sqrt(p) and sqrt(q) must differ.Yes, that seems correct. So, the key idea is that if two numbers have the same decimal expansion, they must be equal, and since p and q are distinct, their square roots must be distinct, hence their decimal expansions must differ.Alternatively, if p and q are the same, then their square roots are the same, so their decimal expansions are identical. But the problem is about two numbers, not necessarily distinct, so perhaps the conclusion is that if p ≠ q, then sqrt(p) and sqrt(q) have different decimal expansions.Wait, but the problem statement says "two rational positive numbers, which are not perfect squares," and asks to prove that their decimal sequences cannot be identical. So, it's about two numbers, not necessarily distinct, but if they are distinct, their square roots have different decimal expansions.But if p = q, then their square roots are the same, so their decimal expansions are identical. So, the statement is that for two distinct positive rational numbers that are not perfect squares, their square roots have different decimal expansions.Therefore, the proof is as follows: Assume p and q are distinct positive rational numbers, neither of which is a perfect square. Suppose, for contradiction, that sqrt(p) and sqrt(q) have the same decimal expansion. Then, sqrt(p) = sqrt(q), which implies p = q, contradicting the assumption that p and q are distinct. Therefore, sqrt(p) and sqrt(q) must have different decimal expansions.Yes, that seems to cover it. So, the main point is that the decimal expansion uniquely identifies a real number, so if two numbers have the same decimal expansion, they must be equal. Since p and q are distinct, their square roots must be distinct, hence their decimal expansions must differ.I think that's the essence of the proof. It's a bit of a straightforward argument, but it relies on the uniqueness of decimal expansions for real numbers. So, unless I'm missing some nuance, that should be the proof.

💡Okay, so Peter went to the market with 500, and he bought some stuff. I need to figure out how much money he has left after his shopping. Let's see, he bought potatoes, tomatoes, cucumbers, and bananas. Each of these has a different price per kilo, and he bought different amounts of each. First, I think I need to calculate how much he spent on each item individually. That means multiplying the number of kilos he bought by the price per kilo for each item. Then, I can add up all those amounts to get the total he spent. Finally, I'll subtract that total from the 500 he started with to find out how much he has left.Alright, let's start with the potatoes. He bought 6 kilos at 2 per kilo. So, 6 times 2 is... let me do that math. 6 times 2 is 12. So, he spent 12 on potatoes.Next, the tomatoes. He bought 9 kilos at 3 per kilo. So, 9 times 3 is... 27. Okay, so he spent 27 on tomatoes.Then, the cucumbers. He bought 5 kilos at 4 per kilo. 5 times 4 is 20. So, 20 on cucumbers.Lastly, the bananas. He bought 3 kilos at 5 per kilo. 3 times 5 is 15. So, 15 on bananas.Now, I need to add up all these amounts to find the total he spent. Let's see: 12 on potatoes, 27 on tomatoes, 20 on cucumbers, and 15 on bananas. Adding those together: 12 plus 27 is 39, plus 20 is 59, plus 15 is 74. So, the total amount he spent is 74.He started with 500, so to find out how much he has left, I subtract the total spent from the initial amount. That would be 500 minus 74. Let me do that subtraction: 500 minus 74 is... 426. So, Peter has 426 left after his purchases.Wait, let me double-check my calculations to make sure I didn't make a mistake. For the potatoes: 6 times 2 is definitely 12. Tomatoes: 9 times 3 is 27. Cucumbers: 5 times 4 is 20. Bananas: 3 times 5 is 15. Adding those up: 12 plus 27 is 39, plus 20 is 59, plus 15 is 74. Yes, that seems right. And 500 minus 74 is indeed 426. Okay, I think that's correct.

💡First, I need to determine the work rates of one woman and one child per day.Given that 10 women can complete the work in 5 days, the work rate of one woman per day is ( frac{1}{10 times 5} = frac{1}{50} ) of the total work.Similarly, since 10 children can complete the work in 10 days, the work rate of one child per day is ( frac{1}{10 times 10} = frac{1}{100} ) of the total work.Next, I calculate the combined work rate of 5 women and 10 children.The work done by 5 women in one day is ( 5 times frac{1}{50} = frac{1}{10} ) of the total work.The work done by 10 children in one day is ( 10 times frac{1}{100} = frac{1}{10} ) of the total work.Adding these together, the combined work rate is ( frac{1}{10} + frac{1}{10} = frac{2}{10} = frac{1}{5} ) of the total work per day.Finally, to find out how many days they need to complete the entire work, I take the reciprocal of the combined work rate: ( frac{1}{frac{1}{5}} = 5 ) days.