answer：Okay, so I have this problem where I need to find the largest integer ( m ) such that ( Q ) is divisible by ( 5^m ). Here, ( Q ) is the product of the first 50 positive odd integers. Hmm, let me try to break this down step by step.First, let's understand what ( Q ) is. The first 50 positive odd integers are 1, 3, 5, ..., up to the 50th odd number. I know that the ( n )-th odd number can be expressed as ( 2n - 1 ), so the 50th odd number is ( 2 times 50 - 1 = 99 ). Therefore, ( Q = 1 times 3 times 5 times cdots times 99 ).Now, I need to find the highest power of 5 that divides this product ( Q ). In other words, I need to determine how many times 5 is a factor in the prime factorization of ( Q ). To do this, I remember that for any factorial ( n! ), the exponent of a prime ( p ) in its prime factorization can be found using Legendre's formula:[text{Exponent of } p text{ in } n! = leftlfloor frac{n}{p} rightrfloor + leftlfloor frac{n}{p^2} rightrfloor + leftlfloor frac{n}{p^3} rightrfloor + cdots]But wait, ( Q ) isn't a factorial; it's a product of odd numbers. How can I relate this to factorials? Let me think. I recall that the product of the first ( n ) odd integers can be expressed as ( frac{(2n)!}{2^n n!} ). Let me verify this.Consider the product of the first ( n ) odd integers:[1 times 3 times 5 times cdots times (2n - 1)]This can be written as:[frac{(2n)!}{2^n n!}]Because ( (2n)! = 1 times 2 times 3 times cdots times (2n) ), which includes all even numbers up to ( 2n ). If we factor out the 2s from each even number, we get ( 2^n times n! ) times the product of the odd numbers. Therefore, the product of the odd numbers is ( frac{(2n)!}{2^n n!} ).In this problem, ( n = 50 ), so:[Q = frac{100!}{2^{50} times 50!}]Okay, so now ( Q ) is expressed in terms of factorials. To find the exponent of 5 in ( Q ), I can find the exponent of 5 in ( 100! ) and subtract the exponents of 5 in ( 2^{50} ) and ( 50! ). However, since ( 2^{50} ) doesn't have any factors of 5, the exponent of 5 in ( Q ) is simply the exponent of 5 in ( 100! ) minus the exponent of 5 in ( 50! ).Let me compute the exponent of 5 in ( 100! ) first. Using Legendre's formula:[leftlfloor frac{100}{5} rightrfloor + leftlfloor frac{100}{25} rightrfloor + leftlfloor frac{100}{125} rightrfloor + cdots]Calculating each term:- ( leftlfloor frac{100}{5} rightrfloor = 20 )- ( leftlfloor frac{100}{25} rightrfloor = 4 )- ( leftlfloor frac{100}{125} rightrfloor = 0 ) (since 125 > 100)Adding these up: ( 20 + 4 = 24 ). So, the exponent of 5 in ( 100! ) is 24.Next, let's find the exponent of 5 in ( 50! ):[leftlfloor frac{50}{5} rightrfloor + leftlfloor frac{50}{25} rightrfloor + leftlfloor frac{50}{125} rightrfloor + cdots]Calculating each term:- ( leftlfloor frac{50}{5} rightrfloor = 10 )- ( leftlfloor frac{50}{25} rightrfloor = 2 )- ( leftlfloor frac{50}{125} rightrfloor = 0 )Adding these up: ( 10 + 2 = 12 ). So, the exponent of 5 in ( 50! ) is 12.Therefore, the exponent of 5 in ( Q ) is ( 24 - 12 = 12 ). So, the largest integer ( m ) such that ( Q ) is divisible by ( 5^m ) is 12.Wait, let me double-check my reasoning. I expressed ( Q ) as ( frac{100!}{2^{50} times 50!} ). Since ( 2^{50} ) doesn't contribute any 5s, subtracting the exponents from ( 50! ) from ( 100! ) gives the exponent in ( Q ). That makes sense.Is there another way to approach this? Maybe by directly counting the number of multiples of 5, 25, etc., in the product of odd numbers. Let's try that.The product ( Q ) is ( 1 times 3 times 5 times 7 times cdots times 99 ). So, the numbers are 1, 3, 5, 7, ..., 99. Let's list the multiples of 5 in this sequence.Multiples of 5 are 5, 15, 25, 35, 45, 55, 65, 75, 85, 95. That's 10 numbers. Each contributes at least one 5.Now, among these, the multiples of 25 (since 25 = 5^2) contribute an extra 5. Let's see which numbers in the list are multiples of 25:25, 75. So, two numbers. Each contributes an additional 5.Multiples of 125 (5^3) would contribute yet another 5, but 125 is greater than 99, so there are none in this product.So, total number of 5s is 10 (from multiples of 5) + 2 (from multiples of 25) = 12. That's the same result as before.Therefore, both methods give me 12, which reinforces that the answer is correct.I think I've covered all the necessary steps and double-checked my work. It seems solid.**Final Answer**The largest integer ( m ) such that ( Q ) is divisible by ( 5^m ) is boxed{12}.

answer：Okay, so I have this problem here asking which of the given equations is a linear equation in two variables. The options are A, B, C, and D. I need to figure out which one is linear. Hmm, let me recall what a linear equation in two variables looks like. I think it's something like ax + by = c, where a, b, and c are constants, and x and y are the variables. So, the equation should be of degree one, meaning no variables are squared or cubed, and no variables are multiplied together.Alright, let's go through each option one by one.Starting with Option A: (x/3) - (2/y) = x. Hmm, this looks a bit complicated. There's a fraction with x in the numerator and another fraction with 2 over y. I'm not sure if this is linear because of the y in the denominator. If y is in the denominator, that might make it a non-linear equation because it's like 1/y, which is not a linear term. Also, if I try to rearrange this equation, I might end up with terms that are not linear. Let me try to rearrange it.So, (x/3) - (2/y) = x. If I subtract x from both sides, I get (x/3) - x - (2/y) = 0. Simplifying (x/3 - x) gives me (-2x/3) - (2/y) = 0. That still has both x and y, but y is in the denominator. I don't think this is linear because of the 1/y term. So, I'll probably rule out Option A.Moving on to Option B: 3x = 2y. This seems simpler. If I rearrange it, I can write it as 3x - 2y = 0. That definitely looks like a linear equation because it's in the form ax + by = c, where a is 3, b is -2, and c is 0. There are no squared terms or products of variables here. So, Option B seems like a good candidate for a linear equation.Let's check Option C: x - y² = 0. This equation has a y squared term. Since y is squared, that makes it a quadratic equation, not linear. Linear equations can't have variables raised to a power higher than one. So, Option C is out.Finally, Option D: 2x - 3y = xy. This one has a product of x and y on the right side. That means it's a quadratic term because it's x multiplied by y. Linear equations can't have variables multiplied together like that. So, Option D is also not linear.After going through all the options, it seems like only Option B fits the criteria of a linear equation in two variables. The other options either have variables in the denominator, squared terms, or products of variables, which make them non-linear.I think I'm confident now that Option B is the correct answer.

answer：To convert the decimal number 96 to its binary representation, I will start by identifying the largest power of 2 that is less than or equal to 96. The powers of 2 are as follows:- (2^0 = 1)- (2^1 = 2)- (2^2 = 4)- (2^3 = 8)- (2^4 = 16)- (2^5 = 32)- (2^6 = 64)- (2^7 = 128)Since (2^6 = 64) is the largest power of 2 that is less than or equal to 96, I will begin with that.Next, I will subtract 64 from 96 to find the remaining value:[ 96 - 64 = 32 ]Now, I will determine the largest power of 2 that fits into the remaining value of 32. This is (2^5 = 32).Subtracting 32 from 32 gives:[ 32 - 32 = 0 ]Since the remaining value is now 0, I have accounted for all the bits in the binary representation. The binary representation is constructed by placing a 1 in the positions corresponding to the powers of 2 used and 0s in the remaining positions. Therefore, the binary representation of 96 is:[ 1100000_2 ]

answer：First, I need to determine how many pieces of clothing Luke put in the first load.Luke has a total of 47 pieces of clothing.He split the remaining pieces into 5 equal loads, with each small load containing 6 pieces of clothing.To find the total number of pieces in the 5 small loads, I multiply the number of loads by the number of pieces per load:5 loads × 6 pieces per load = 30 pieces.Now, I subtract the number of pieces in the small loads from the total number of pieces to find out how many pieces were in the first load:47 total pieces - 30 pieces in small loads = 17 pieces.Therefore, Luke put 17 pieces of clothing in the first load.

answer：Okay, so I have this problem where seven students are standing in a row for a graduation photo. There are some specific conditions: student A must be in the very center, and students B and C must stand together. I need to figure out how many different arrangements there are. The options are A. 240, B. 192, C. 96, D. 48.Alright, let's break this down step by step. First, there are seven students in total, and they need to stand in a row. So, normally, without any restrictions, the number of arrangements would be 7 factorial, which is 7! = 5040. But here, we have some constraints, so the number will be much less.The first condition is that student A must stand in the very center. Since there are seven positions in a row, the center position is the fourth one. So, position number 4 is fixed for student A. That means we don't have to worry about where A is; he's already placed. So, now we have six remaining positions to arrange the other six students.But wait, actually, no. Wait, there are seven students, and one is already placed in the center, so there are six remaining students to arrange in the remaining six positions. But hold on, no, actually, the problem says seven students, so including A, so when A is fixed in the center, we have six other students to arrange in the remaining six spots.But there's another condition: students B and C must stand together. So, B and C have to be next to each other somewhere in the row. So, how do we handle that?I remember that when two people must stand together, we can treat them as a single unit or "block." So, instead of thinking of B and C as two separate individuals, we can think of them as one "BC" block. This reduces the number of units we need to arrange.So, if we treat B and C as a single block, then along with the other four students (since A is already placed), we have five units to arrange: the "BC" block and the four other students. So, the number of ways to arrange these five units is 5 factorial, which is 5! = 120.But wait, within the "BC" block, B and C can switch places, right? So, for each of these arrangements, there are two possibilities: B is first and C is second, or C is first and B is second. So, we need to multiply by 2 to account for these internal arrangements. So, that gives us 5! * 2 = 120 * 2 = 240.But hold on, the options don't have 240 as the correct answer. Wait, no, option A is 240. Hmm. But I think I might have made a mistake here because the initial assumption was that the "BC" block can be placed anywhere in the remaining six positions, but actually, since A is fixed in the center, the positions to the left and right of A are separate.Let me think again. So, we have seven positions, with A fixed in position 4. The remaining six positions are positions 1, 2, 3 on the left and positions 5, 6, 7 on the right.Now, the "BC" block needs to be placed somewhere in these six positions. But since B and C must stand together, they can occupy either positions 1 and 2, 2 and 3, 5 and 6, or 6 and 7. So, there are four possible places for the "BC" block.Wait, is that correct? Let's see:Positions 1 and 2: that's one possibility.Positions 2 and 3: another.Positions 5 and 6: another.Positions 6 and 7: another.So, four possible positions for the "BC" block.But actually, if we consider the "BC" block as a single unit, we can place it in the remaining six positions, but since it takes up two adjacent positions, the number of possible positions is actually five: positions 1-2, 2-3, 3-4, 5-6, 6-7. But wait, position 3-4 would include position 4, which is already occupied by A. So, we can't place the "BC" block across position 4. So, actually, the "BC" block can only be placed in positions 1-2, 2-3, 5-6, or 6-7. So, four possible positions.So, for each of these four positions, we can have the "BC" block, and then arrange the remaining four students in the remaining four positions.But wait, no. Because the "BC" block takes up two positions, so if we fix the "BC" block in one of these four positions, then we have four remaining positions to arrange the other four students.But actually, the "BC" block is treated as a single unit, so when we place the "BC" block in one of these four positions, we effectively have five units to arrange: the "BC" block and the four other students. But since the "BC" block is fixed in one of these four positions, the number of ways to arrange the remaining four students is 4!.But wait, no. Because the "BC" block is already placed in a specific position, so the remaining four students can be arranged in the remaining four positions, which is 4!.But also, within the "BC" block, B and C can switch places, so we need to multiply by 2.So, the total number of arrangements would be:Number of positions for "BC" block * arrangements of the remaining students * arrangements within "BC" block.That is:4 (positions) * 4! (arrangements of the other students) * 2 (arrangements within "BC" block).Calculating that:4 * 24 * 2 = 4 * 48 = 192.So, that gives us 192, which is option B.Wait, but earlier I thought it was 240, but that was without considering the fixed position of A and the restriction on where the "BC" block can be placed.So, I think the correct answer is 192, which is option B.But let me double-check to make sure I didn't make a mistake.So, A is fixed in position 4.We have six remaining positions: 1, 2, 3, 5, 6, 7.We need to place the "BC" block in two adjacent positions among these six, but not crossing position 4.So, possible "BC" block positions:- Positions 1 and 2- Positions 2 and 3- Positions 5 and 6- Positions 6 and 7So, four possible positions for the "BC" block.For each of these four positions, we have the "BC" block, and then the remaining four students can be arranged in the remaining four positions.Since the "BC" block can be in two different orders (B first or C first), we multiply by 2.So, total arrangements:4 (positions) * 4! (arrangements of the other students) * 2 (arrangements within "BC" block) = 4 * 24 * 2 = 192.Yes, that seems correct.Alternatively, another way to think about it is:Treat "BC" as a single unit. So, along with the other four students, we have five units to arrange.But since A is fixed in the center, we have to consider the positions relative to A.The "BC" block can be placed either entirely to the left of A or entirely to the right of A.If we consider the left side, there are three positions: 1, 2, 3. The "BC" block can be placed in positions 1-2 or 2-3, so two possibilities.Similarly, on the right side, positions 5, 6, 7, the "BC" block can be placed in positions 5-6 or 6-7, so two possibilities.So, total of four positions for the "BC" block.Then, for each of these four positions, we have five units: "BC" block and the four other students.Wait, no. If we fix the "BC" block in one of these four positions, then we have four remaining positions to arrange the other four students.So, the number of ways is:Number of positions for "BC" block * arrangements of the remaining students * arrangements within "BC" block.Which is 4 * 4! * 2 = 4 * 24 * 2 = 192.Yes, that's consistent.So, I think 192 is the correct answer, which is option B.

answer：Okay, so I need to find the equation of the slant asymptote for the rational function (frac{3x^2 - 2x - 4}{2x - 5}). Hmm, I remember that slant asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator. In this case, the numerator is a quadratic (degree 2) and the denominator is linear (degree 1), so yes, there should be a slant asymptote.I think the way to find it is by performing polynomial long division. I've heard about this method before, where you divide the numerator by the denominator to see how it behaves as (x) gets really large. The quotient from this division should give me the equation of the slant asymptote, right?Alright, let me set up the division. The numerator is (3x^2 - 2x - 4) and the denominator is (2x - 5). I need to divide (3x^2 - 2x - 4) by (2x - 5).First, I'll divide the leading term of the numerator, which is (3x^2), by the leading term of the denominator, (2x). So, (3x^2 ÷ 2x) is (frac{3}{2}x). That's the first term of my quotient.Now, I need to multiply the entire denominator (2x - 5) by this term (frac{3}{2}x). Let's do that:[frac{3}{2}x times (2x - 5) = 3x^2 - frac{15}{2}x]Next, I subtract this result from the original numerator:[(3x^2 - 2x - 4) - (3x^2 - frac{15}{2}x) = 0x^2 + left(-2x + frac{15}{2}xright) - 4]Simplifying the (x) terms:[-2x + frac{15}{2}x = left(-frac{4}{2} + frac{15}{2}right)x = frac{11}{2}x]So now, the new polynomial after subtraction is (frac{11}{2}x - 4).Now, I repeat the process. I divide the leading term of this new polynomial, which is (frac{11}{2}x), by the leading term of the denominator, (2x):[frac{11}{2}x ÷ 2x = frac{11}{4}]This is the next term in my quotient.Multiply the denominator by this term:[frac{11}{4} times (2x - 5) = frac{11}{2}x - frac{55}{4}]Subtract this from the current polynomial:[left(frac{11}{2}x - 4right) - left(frac{11}{2}x - frac{55}{4}right) = 0x + left(-4 + frac{55}{4}right)]Simplify the constant terms:[-4 + frac{55}{4} = -frac{16}{4} + frac{55}{4} = frac{39}{4}]So, the remainder is (frac{39}{4}).Putting it all together, the division gives me:[frac{3x^2 - 2x - 4}{2x - 5} = frac{3}{2}x + frac{11}{4} + frac{39}{4(2x - 5)}]As (x) approaches infinity, the term (frac{39}{4(2x - 5)}) becomes negligible because the denominator grows without bound, making the whole fraction approach zero. Therefore, the slant asymptote is given by the quotient part of the division, which is:[y = frac{3}{2}x + frac{11}{4}]Wait, let me double-check my calculations to make sure I didn't make any mistakes. I divided (3x^2) by (2x) to get (frac{3}{2}x), then multiplied back and subtracted. Then I took the new term (frac{11}{2}x) divided by (2x) to get (frac{11}{4}), multiplied back, and subtracted. The remainder was (frac{39}{4}). That seems correct.So, the equation of the slant asymptote should be (y = frac{3}{2}x + frac{11}{4}). To write it as a decimal, (frac{3}{2}) is 1.5 and (frac{11}{4}) is 2.75, so (y = 1.5x + 2.75). But I think it's better to leave it as fractions for exactness.Wait, hold on, in the initial problem statement, the assistant had a different answer: (y = 1.5x + 4.25). Did I do something wrong? Let me check my steps again.Starting over, dividing (3x^2 - 2x - 4) by (2x - 5).First term: (3x^2 ÷ 2x = frac{3}{2}x).Multiply (2x - 5) by (frac{3}{2}x): (3x^2 - frac{15}{2}x).Subtract: ((3x^2 - 2x - 4) - (3x^2 - frac{15}{2}x) = 0x^2 + (-2x + frac{15}{2}x) - 4 = frac{11}{2}x - 4).Next term: (frac{11}{2}x ÷ 2x = frac{11}{4}).Multiply (2x - 5) by (frac{11}{4}): (frac{11}{2}x - frac{55}{4}).Subtract: ((frac{11}{2}x - 4) - (frac{11}{2}x - frac{55}{4}) = 0x + (-4 + frac{55}{4}) = frac{39}{4}).So, the division is correct. Therefore, the slant asymptote is (y = frac{3}{2}x + frac{11}{4}), which is (y = 1.5x + 2.75). But the assistant had (y = 1.5x + 4.25). Hmm, that's different.Wait, maybe I made a mistake in the subtraction step. Let me check that again.After the first subtraction, I had (frac{11}{2}x - 4). Then I divided (frac{11}{2}x) by (2x) to get (frac{11}{4}). Then I multiplied (2x - 5) by (frac{11}{4}) to get (frac{11}{2}x - frac{55}{4}). Subtracting that from (frac{11}{2}x - 4):[frac{11}{2}x - 4 - frac{11}{2}x + frac{55}{4} = (-4 + frac{55}{4}) = (-frac{16}{4} + frac{55}{4}) = frac{39}{4}]Yes, that's correct. So the remainder is (frac{39}{4}), and the quotient is (frac{3}{2}x + frac{11}{4}). Therefore, the slant asymptote is (y = frac{3}{2}x + frac{11}{4}), which is (1.5x + 2.75).But the assistant's answer was (1.5x + 4.25). That must be a mistake. Let me see if I can figure out where they went wrong.Looking at the assistant's work, they had:1.5x + 17/4 + 39/4 over (2x -5). So their quotient was 1.5x + 17/4, which is 1.5x + 4.25. But according to my calculation, the quotient should be 1.5x + 11/4, which is 1.5x + 2.75.Wait, maybe they made a mistake in their division. Let me check their steps.They wrote:3x^2 -2x -4 divided by 2x -5.First term: 3x^2 ÷ 2x = 1.5x.Multiply 2x -5 by 1.5x: 3x^2 -7.5x.Subtract: (3x^2 -2x -4) - (3x^2 -7.5x) = 0 +5.5x -4.Then, 5.5x ÷ 2x = 2.75, which is 11/4.Multiply 2x -5 by 2.75: 5.5x -13.75.Subtract: (5.5x -4) - (5.5x -13.75) = 0 +9.75.Wait, 9.75 is 39/4, which matches my calculation. So their quotient is 1.5x + 2.75, but they wrote 1.5x +17/4, which is 4.25. That's incorrect.So, the assistant made a mistake in their quotient. They added 1.5x and 17/4, but according to the division, it should be 1.5x + 11/4.Therefore, the correct slant asymptote is (y = frac{3}{2}x + frac{11}{4}), or (y = 1.5x + 2.75).I think that's the correct answer. Maybe the assistant confused the remainder with the quotient term. Anyway, I'm confident now that the slant asymptote is (y = frac{3}{2}x + frac{11}{4}).