answer：Okay, so I've got this problem about a quadratic function, and I need to figure out the values of c and m, and then find the equation of the quadratic function itself. Let me see how I can approach this step by step.First, let's understand what a quadratic function is. It's a function of the form y = ax² + bx + c, where a, b, and c are constants, and a is not zero. The graph of a quadratic function is a parabola, which can open upwards or downwards depending on the value of a.Looking at the table provided, it shows some x-values and their corresponding y-values. The x-values given are -3, -2, -1, 0, and 1, with corresponding y-values of 5/2, 4, 9/2, 4, and m. So, I need to find c and m, and then the entire quadratic function.Starting with part (1): Directly write down the values of c and m.I remember that in a quadratic function y = ax² + bx + c, the constant term c is the y-intercept, which is the value of y when x = 0. Looking at the table, when x = 0, y = 4. So, that should mean c = 4. That seems straightforward.Now, for m, which is the y-value when x = 1. I need to figure out what m is. Since the table shows y-values for x = -3, -2, -1, 0, and 1, I can see if there's a pattern or symmetry in the y-values that might help me find m.Looking at the y-values: when x = -3, y = 5/2; x = -2, y = 4; x = -1, y = 9/2; x = 0, y = 4; and x = 1, y = m. Hmm, I notice that the y-values seem to be symmetric around x = -1. Let me check that.If I consider the points x = -3 and x = 1, they are both 2 units away from x = -1. Similarly, x = -2 and x = 0 are both 1 unit away from x = -1. So, if the function is symmetric around x = -1, then the y-values at x = -3 and x = 1 should be the same. Since y = 5/2 at x = -3, then y should also be 5/2 at x = 1. Therefore, m should be 5/2.Wait, let me make sure that makes sense. If the parabola is symmetric around x = -1, then for every point (x, y), there should be a corresponding point (2*(-1) - x, y). So, for x = -3, the corresponding point should be x = 2*(-1) - (-3) = -2 + 3 = 1. So, yes, the y-value at x = 1 should be the same as at x = -3, which is 5/2. Therefore, m = 5/2.Okay, so part (1) seems done: c = 4 and m = 5/2.Now, moving on to part (2): Find the analytical expression of this quadratic function.To find the quadratic function, I need to determine the coefficients a, b, and c. I already know that c = 4, so I just need to find a and b.I can use the other points from the table to set up equations and solve for a and b. Let's use the points (x, y) = (-3, 5/2), (-2, 4), (-1, 9/2), and (1, 5/2). Since I already know c = 4, I can plug these points into the equation y = ax² + bx + 4 and solve for a and b.Let's start with x = -3, y = 5/2:5/2 = a*(-3)² + b*(-3) + 45/2 = 9a - 3b + 4Let's subtract 4 from both sides:5/2 - 4 = 9a - 3b5/2 - 8/2 = 9a - 3b(-3/2) = 9a - 3bLet's simplify this equation by dividing both sides by 3:(-1/2) = 3a - bSo, equation (1): 3a - b = -1/2Now, let's use another point, say x = -2, y = 4:4 = a*(-2)² + b*(-2) + 44 = 4a - 2b + 4Subtract 4 from both sides:0 = 4a - 2bSimplify by dividing both sides by 2:0 = 2a - bSo, equation (2): 2a - b = 0Now, we have two equations:1) 3a - b = -1/22) 2a - b = 0Let's subtract equation (2) from equation (1):(3a - b) - (2a - b) = (-1/2) - 03a - b - 2a + b = -1/2a = -1/2So, a = -1/2Now, plug a = -1/2 into equation (2):2*(-1/2) - b = 0-1 - b = 0-b = 1b = -1So, a = -1/2, b = -1, and c = 4.Therefore, the quadratic function is:y = (-1/2)x² - x + 4Let me double-check this with another point to make sure. Let's use x = -1, y = 9/2:y = (-1/2)*(-1)² - (-1) + 4y = (-1/2)*(1) + 1 + 4y = -1/2 + 1 + 4y = (-1/2 + 1) + 4y = 1/2 + 4y = 9/2Yes, that matches the table. Let's check x = 1, y = 5/2:y = (-1/2)*(1)² - (1) + 4y = (-1/2) - 1 + 4y = (-3/2) + 4y = (-3/2 + 8/2)y = 5/2That also matches. Okay, so it seems correct.Alternatively, since I noticed the axis of symmetry is x = -1, I could have used vertex form to find the equation. The vertex form of a quadratic is y = a(x - h)² + k, where (h, k) is the vertex.From the table, the vertex seems to be at x = -1, y = 9/2. So, h = -1, k = 9/2.So, y = a(x + 1)² + 9/2Now, to find a, I can use another point, say x = 0, y = 4:4 = a(0 + 1)² + 9/24 = a + 9/2Subtract 9/2 from both sides:4 - 9/2 = aConvert 4 to 8/2:8/2 - 9/2 = a(-1/2) = aSo, a = -1/2Therefore, the equation is:y = (-1/2)(x + 1)² + 9/2Let me expand this to standard form to confirm:First, expand (x + 1)²:(x + 1)² = x² + 2x + 1Multiply by -1/2:(-1/2)x² - x - 1/2Add 9/2:(-1/2)x² - x - 1/2 + 9/2Combine constants:-1/2 + 9/2 = 8/2 = 4So, y = (-1/2)x² - x + 4Which matches what I found earlier. So, that's consistent.Therefore, the quadratic function is y = (-1/2)x² - x + 4.I think that's it. I've found c = 4, m = 5/2, and the quadratic function is y = (-1/2)x² - x + 4.

answer：First, I need to determine how many meters of cloth the loom weaves every second. I know that the loom weaves 26 meters of cloth in 203.125 seconds.To find the weaving rate per second, I should divide the total amount of cloth woven by the total time taken.So, I'll set up the division: 26 meters divided by 203.125 seconds.Performing the division will give me the number of meters woven per second.After calculating, I'll round the result to three decimal places for simplicity.

answer：Alright, so I've got this math puzzle here, and I need to figure out what digits each letter represents. The problem says:"Let ( AB ), multiplied by ( C ), equal ( DE ). If ( DE ) is subtracted from ( FG ), the result is ( HI ). Each letter represents a distinct digit (1, 2, 3, 4, 5, 6, 7, 8, or 9). The digit 0 does not appear in the problem."Okay, so first, I need to understand what's being asked. There are several two-digit numbers here: ( AB ), ( DE ), ( FG ), and ( HI ). Each letter represents a unique digit from 1 to 9, and no digit is repeated. Also, 0 isn't used at all, which simplifies things a bit.The problem gives me two equations:1. ( AB times C = DE )2. ( FG - DE = HI )My goal is to assign digits to each letter such that both equations are satisfied, and all digits are unique.Let me start by breaking down the first equation: ( AB times C = DE ). Here, ( AB ) is a two-digit number, ( C ) is a single digit, and their product ( DE ) is another two-digit number. So, ( AB ) multiplied by ( C ) must result in a two-digit number. That means ( AB ) can't be too large, because if ( AB ) is, say, 99 and ( C ) is 9, the product would be 891, which is a three-digit number. So, ( AB ) must be small enough that when multiplied by ( C ), it stays within two digits.Similarly, for the second equation: ( FG - DE = HI ). Here, ( FG ) is a two-digit number, ( DE ) is another two-digit number, and their difference ( HI ) is also a two-digit number. So, ( FG ) must be larger than ( DE ) to ensure that the result is positive and a two-digit number.Given that all digits are unique and from 1 to 9, I need to make sure that no digit repeats across all the letters ( A, B, C, D, E, F, G, H, I ).Let me list out the letters and their possible digits:- ( A ): 1-9- ( B ): 1-9, not equal to ( A )- ( C ): 1-9, not equal to ( A ) or ( B )- ( D ): 1-9, not equal to ( A, B, C )- ( E ): 1-9, not equal to ( A, B, C, D )- ( F ): 1-9, not equal to ( A, B, C, D, E )- ( G ): 1-9, not equal to ( A, B, C, D, E, F )- ( H ): 1-9, not equal to ( A, B, C, D, E, F, G )- ( I ): 1-9, not equal to ( A, B, C, D, E, F, G, H )That's a lot of constraints! Maybe I can find some patterns or narrow down the possibilities.Starting with the first equation: ( AB times C = DE ). Let's think about possible values for ( AB ) and ( C ) such that their product is a two-digit number ( DE ).Since ( AB ) is a two-digit number, it ranges from 10 to 99. But considering that ( AB times C ) must also be a two-digit number, ( AB ) can't be too large. Let's see:If ( AB = 10 ), then ( C ) can be up to 9, making ( DE = 90 ).If ( AB = 11 ), ( C ) can be up to 8, making ( DE = 88 )....If ( AB = 32 ), ( C ) can be up to 2, making ( DE = 64 ).If ( AB = 33 ), ( C ) can be up to 2, making ( DE = 66 ).And so on.But since all digits must be unique, we have to ensure that the digits in ( AB ), ( C ), and ( DE ) don't repeat.This seems a bit broad. Maybe I can look for possible ( AB ) and ( C ) combinations where the product ( DE ) uses unique digits not used in ( AB ) or ( C ).Alternatively, maybe I can look at the second equation first: ( FG - DE = HI ). Since ( FG ) and ( DE ) are two-digit numbers, and their difference is also a two-digit number, ( FG ) must be larger than ( DE ).Let me think about possible values for ( FG ) and ( DE ) such that their difference is another two-digit number ( HI ), with all digits unique.This also seems quite broad. Maybe I can consider that ( FG ) is just a bit larger than ( DE ), so that ( HI ) is a reasonable two-digit number.Alternatively, perhaps I can look for ( FG ) and ( DE ) such that their difference ( HI ) uses digits not used in ( FG ) or ( DE ).This is getting a bit too vague. Maybe I need to find some overlapping constraints.Since both equations involve ( DE ), perhaps I can find a value for ( DE ) that satisfies both equations.Let me list possible values for ( DE ) based on the first equation ( AB times C = DE ). Since ( DE ) is a two-digit number, and ( AB ) and ( C ) are also digits, let's see:Possible ( AB ) values could be from 12 to 32, as higher numbers multiplied by even small ( C ) values would exceed two digits.Wait, actually, ( AB ) can be up to 99, but ( C ) has to be such that ( AB times C ) is still two digits. So, ( AB times C leq 99 ).Given that, ( AB ) can be up to 99, but ( C ) would have to be 1 in that case, which is unlikely since ( C ) needs to be unique and ( AB ) would already use two digits.So, more realistically, ( AB ) is probably in the range of 12 to 32, and ( C ) is from 2 to 9.Let me try some examples.Suppose ( AB = 12 ). Then ( C ) can be from 2 to 8, since ( 12 times 9 = 108 ), which is three digits.So, ( DE ) would be ( 12 times C ).Let's try ( C = 3 ): ( 12 times 3 = 36 ). So, ( DE = 36 ). Now, check if all digits are unique: ( A=1 ), ( B=2 ), ( C=3 ), ( D=3 ), ( E=6 ). Wait, ( C=3 ) and ( D=3 ) would repeat the digit 3, which is not allowed. So, this doesn't work.Next, ( C=4 ): ( 12 times 4 = 48 ). So, ( DE=48 ). Check uniqueness: ( A=1 ), ( B=2 ), ( C=4 ), ( D=4 ), ( E=8 ). Again, ( C=4 ) and ( D=4 ) repeat. Not good.( C=5 ): ( 12 times 5 = 60 ). But 0 is not allowed, so discard.( C=6 ): ( 12 times 6 = 72 ). ( DE=72 ). Check uniqueness: ( A=1 ), ( B=2 ), ( C=6 ), ( D=7 ), ( E=2 ). Wait, ( B=2 ) and ( E=2 ) repeat. Not allowed.( C=7 ): ( 12 times 7 = 84 ). ( DE=84 ). Check uniqueness: ( A=1 ), ( B=2 ), ( C=7 ), ( D=8 ), ( E=4 ). All unique so far. Okay, this is a possibility.So, ( AB=12 ), ( C=7 ), ( DE=84 ).Now, moving to the second equation: ( FG - DE = HI ). So, ( FG - 84 = HI ).We need ( FG ) and ( HI ) to be two-digit numbers, with all digits unique and not repeating any digits from ( AB ), ( C ), or ( DE ).Digits used so far: 1, 2, 7, 8, 4.Available digits: 3, 5, 6, 9.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 84 ) because ( HI ) is positive.Possible ( FG ) values could be 93, 95, 96, etc., but using only the available digits.Let's see:If ( FG=93 ), then ( HI = 93 - 84 = 09 ). But 0 is not allowed, so discard.If ( FG=95 ), ( HI=95-84=11 ). But 1 is already used, and digits must be unique. Also, 11 repeats digits, which is not allowed.If ( FG=96 ), ( HI=96-84=12 ). But 1 and 2 are already used.If ( FG=53 ), ( HI=53-84=-31 ). Negative, not allowed.If ( FG=59 ), ( HI=59-84=-25 ). Negative.If ( FG=63 ), ( HI=63-84=-21 ). Negative.If ( FG=65 ), ( HI=65-84=-19 ). Negative.If ( FG=69 ), ( HI=69-84=-15 ). Negative.If ( FG=35 ), ( HI=35-84=-49 ). Negative.If ( FG=36 ), ( HI=36-84=-48 ). Negative.If ( FG=39 ), ( HI=39-84=-45 ). Negative.If ( FG=56 ), ( HI=56-84=-28 ). Negative.If ( FG=59 ), already checked.If ( FG=65 ), already checked.If ( FG=95 ), already checked.Hmm, none of these seem to work. Maybe ( FG ) is a larger number, but we only have digits 3,5,6,9 left.Wait, maybe I made a mistake. Let's check ( FG=93 ) again. ( HI=93-84=09 ). But 0 is not allowed, so that's out.What about ( FG=53 )? ( HI=53-84=-31 ). Negative, so no.Is there any other combination? Maybe ( FG=63 ), but that gives a negative ( HI ).Alternatively, maybe ( FG=83 ), but 8 is already used in ( DE=84 ).Wait, maybe I need to try a different ( AB ) and ( C ) combination.Let's go back to the first equation.Next, ( AB=13 ).( C=2 ): ( 13 times 2 = 26 ). ( DE=26 ). Check uniqueness: ( A=1 ), ( B=3 ), ( C=2 ), ( D=2 ), ( E=6 ). ( C=2 ) and ( D=2 ) repeat. Not good.( C=3 ): ( 13 times 3 = 39 ). ( DE=39 ). Check uniqueness: ( A=1 ), ( B=3 ), ( C=3 ), ( D=3 ), ( E=9 ). ( B=3 ), ( C=3 ), ( D=3 ) all repeat. Not good.( C=4 ): ( 13 times 4 = 52 ). ( DE=52 ). Check uniqueness: ( A=1 ), ( B=3 ), ( C=4 ), ( D=5 ), ( E=2 ). All unique so far.So, ( AB=13 ), ( C=4 ), ( DE=52 ).Now, moving to the second equation: ( FG - 52 = HI ).Digits used so far: 1,3,4,5,2.Available digits: 6,7,8,9.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 52 ).Possible ( FG ) values: 67, 68, 69, 76, 78, 79, 86, 87, 89, 96, 97, 98.Let's try ( FG=67 ): ( HI=67-52=15 ). But 1 and 5 are already used.( FG=68 ): ( HI=68-52=16 ). 1 is used.( FG=69 ): ( HI=69-52=17 ). 1 is used.( FG=76 ): ( HI=76-52=24 ). 2 and 4 are used.( FG=78 ): ( HI=78-52=26 ). 2 is used.( FG=79 ): ( HI=79-52=27 ). 2 is used.( FG=86 ): ( HI=86-52=34 ). 3 and 4 are used.( FG=87 ): ( HI=87-52=35 ). 3 and 5 are used.( FG=89 ): ( HI=89-52=37 ). 3 is used.( FG=96 ): ( HI=96-52=44 ). Repeats 4, and 4 is used.( FG=97 ): ( HI=97-52=45 ). 4 and 5 are used.( FG=98 ): ( HI=98-52=46 ). 4 and 6 are used.Hmm, none of these seem to work either. Maybe I need to try a different ( AB ) and ( C ) combination.Let's try ( AB=14 ).( C=2 ): ( 14 times 2 = 28 ). ( DE=28 ). Check uniqueness: ( A=1 ), ( B=4 ), ( C=2 ), ( D=2 ), ( E=8 ). ( C=2 ) and ( D=2 ) repeat. Not good.( C=3 ): ( 14 times 3 = 42 ). ( DE=42 ). Check uniqueness: ( A=1 ), ( B=4 ), ( C=3 ), ( D=4 ), ( E=2 ). ( B=4 ) and ( D=4 ) repeat. Not good.( C=4 ): ( 14 times 4 = 56 ). ( DE=56 ). Check uniqueness: ( A=1 ), ( B=4 ), ( C=4 ), ( D=5 ), ( E=6 ). ( B=4 ) and ( C=4 ) repeat. Not good.( C=5 ): ( 14 times 5 = 70 ). 0 is not allowed.( C=6 ): ( 14 times 6 = 84 ). ( DE=84 ). Check uniqueness: ( A=1 ), ( B=4 ), ( C=6 ), ( D=8 ), ( E=4 ). ( B=4 ) and ( E=4 ) repeat. Not good.( C=7 ): ( 14 times 7 = 98 ). ( DE=98 ). Check uniqueness: ( A=1 ), ( B=4 ), ( C=7 ), ( D=9 ), ( E=8 ). All unique so far.So, ( AB=14 ), ( C=7 ), ( DE=98 ).Now, moving to the second equation: ( FG - 98 = HI ).Digits used so far: 1,4,7,9,8.Available digits: 2,3,5,6.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 98 ), but the available digits are only 2,3,5,6. The largest possible ( FG ) is 65, which is less than 98. So, this doesn't work. Therefore, ( AB=14 ), ( C=7 ) is invalid.Let's try ( AB=15 ).( C=2 ): ( 15 times 2 = 30 ). 0 is not allowed.( C=3 ): ( 15 times 3 = 45 ). ( DE=45 ). Check uniqueness: ( A=1 ), ( B=5 ), ( C=3 ), ( D=4 ), ( E=5 ). ( B=5 ) and ( E=5 ) repeat. Not good.( C=4 ): ( 15 times 4 = 60 ). 0 is not allowed.( C=5 ): ( 15 times 5 = 75 ). ( DE=75 ). Check uniqueness: ( A=1 ), ( B=5 ), ( C=5 ), ( D=7 ), ( E=5 ). Multiple repeats. Not good.( C=6 ): ( 15 times 6 = 90 ). 0 is not allowed.( C=7 ): ( 15 times 7 = 105 ). Three digits, not allowed.So, ( AB=15 ) doesn't seem to work.Let's try ( AB=16 ).( C=2 ): ( 16 times 2 = 32 ). ( DE=32 ). Check uniqueness: ( A=1 ), ( B=6 ), ( C=2 ), ( D=3 ), ( E=2 ). ( C=2 ) and ( E=2 ) repeat. Not good.( C=3 ): ( 16 times 3 = 48 ). ( DE=48 ). Check uniqueness: ( A=1 ), ( B=6 ), ( C=3 ), ( D=4 ), ( E=8 ). All unique so far.So, ( AB=16 ), ( C=3 ), ( DE=48 ).Now, moving to the second equation: ( FG - 48 = HI ).Digits used so far: 1,6,3,4,8.Available digits: 2,5,7,9.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 48 ).Possible ( FG ) values: 25, 27, 29, 52, 57, 59, 72, 75, 79, 92, 95, 97.Let's try ( FG=52 ): ( HI=52-48=04 ). 0 is not allowed.( FG=57 ): ( HI=57-48=09 ). 0 is not allowed.( FG=59 ): ( HI=59-48=11 ). Repeats 1, which is used.( FG=72 ): ( HI=72-48=24 ). 2 and 4 are used.( FG=75 ): ( HI=75-48=27 ). 2 and 7 are used.( FG=79 ): ( HI=79-48=31 ). 3 is used.( FG=92 ): ( HI=92-48=44 ). Repeats 4, which is used.( FG=95 ): ( HI=95-48=47 ). 4 is used.( FG=97 ): ( HI=97-48=49 ). 4 and 9 are used.Hmm, none of these seem to work. Maybe I need to try a different ( AB ) and ( C ) combination.Let's try ( AB=17 ).( C=2 ): ( 17 times 2 = 34 ). ( DE=34 ). Check uniqueness: ( A=1 ), ( B=7 ), ( C=2 ), ( D=3 ), ( E=4 ). All unique so far.So, ( AB=17 ), ( C=2 ), ( DE=34 ).Now, moving to the second equation: ( FG - 34 = HI ).Digits used so far: 1,7,2,3,4.Available digits: 5,6,8,9.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 34 ).Possible ( FG ) values: 56, 58, 59, 65, 68, 69, 85, 86, 89, 95, 96, 98.Let's try ( FG=56 ): ( HI=56-34=22 ). Repeats 2, which is used.( FG=58 ): ( HI=58-34=24 ). 2 and 4 are used.( FG=59 ): ( HI=59-34=25 ). 2 is used.( FG=65 ): ( HI=65-34=31 ). 3 is used.( FG=68 ): ( HI=68-34=34 ). 3 and 4 are used.( FG=69 ): ( HI=69-34=35 ). 3 is used.( FG=85 ): ( HI=85-34=51 ). 5 and 1 are used.( FG=86 ): ( HI=86-34=52 ). 5 and 2 are used.( FG=89 ): ( HI=89-34=55 ). Repeats 5.( FG=95 ): ( HI=95-34=61 ). 6 and 1 are used.( FG=96 ): ( HI=96-34=62 ). 6 and 2 are used.( FG=98 ): ( HI=98-34=64 ). 6 and 4 are used.Hmm, none of these seem to work either. Maybe I need to try a different ( AB ) and ( C ) combination.Let's try ( AB=18 ).( C=2 ): ( 18 times 2 = 36 ). ( DE=36 ). Check uniqueness: ( A=1 ), ( B=8 ), ( C=2 ), ( D=3 ), ( E=6 ). All unique so far.So, ( AB=18 ), ( C=2 ), ( DE=36 ).Now, moving to the second equation: ( FG - 36 = HI ).Digits used so far: 1,8,2,3,6.Available digits: 4,5,7,9.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 36 ).Possible ( FG ) values: 45, 47, 49, 54, 57, 59, 74, 75, 79, 94, 95, 97.Let's try ( FG=45 ): ( HI=45-36=09 ). 0 is not allowed.( FG=47 ): ( HI=47-36=11 ). Repeats 1, which is used.( FG=49 ): ( HI=49-36=13 ). 1 is used.( FG=54 ): ( HI=54-36=18 ). 1 and 8 are used.( FG=57 ): ( HI=57-36=21 ). 2 and 1 are used.( FG=59 ): ( HI=59-36=23 ). 2 and 3 are used.( FG=74 ): ( HI=74-36=38 ). 3 and 8 are used.( FG=75 ): ( HI=75-36=39 ). 3 is used.( FG=79 ): ( HI=79-36=43 ). 4 and 3 are used.( FG=94 ): ( HI=94-36=58 ). 5 and 8 are used.( FG=95 ): ( HI=95-36=59 ). 5 and 9 are used.( FG=97 ): ( HI=97-36=61 ). 6 and 1 are used.Hmm, none of these seem to work. Maybe I need to try a different ( AB ) and ( C ) combination.Let's try ( AB=19 ).( C=2 ): ( 19 times 2 = 38 ). ( DE=38 ). Check uniqueness: ( A=1 ), ( B=9 ), ( C=2 ), ( D=3 ), ( E=8 ). All unique so far.So, ( AB=19 ), ( C=2 ), ( DE=38 ).Now, moving to the second equation: ( FG - 38 = HI ).Digits used so far: 1,9,2,3,8.Available digits: 4,5,6,7.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 38 ).Possible ( FG ) values: 45, 46, 47, 54, 56, 57, 64, 65, 67, 74, 75, 76.Let's try ( FG=45 ): ( HI=45-38=07 ). 0 is not allowed.( FG=46 ): ( HI=46-38=08 ). 0 is not allowed.( FG=47 ): ( HI=47-38=09 ). 0 is not allowed.( FG=54 ): ( HI=54-38=16 ). 1 is used.( FG=56 ): ( HI=56-38=18 ). 1 and 8 are used.( FG=57 ): ( HI=57-38=19 ). 1 and 9 are used.( FG=64 ): ( HI=64-38=26 ). 2 is used.( FG=65 ): ( HI=65-38=27 ). 2 is used.( FG=67 ): ( HI=67-38=29 ). 2 is used.( FG=74 ): ( HI=74-38=36 ). 3 is used.( FG=75 ): ( HI=75-38=37 ). 3 is used.( FG=76 ): ( HI=76-38=38 ). 3 and 8 are used.Hmm, none of these seem to work either. Maybe I need to try a different ( AB ) and ( C ) combination.Let's try ( AB=21 ).( C=2 ): ( 21 times 2 = 42 ). ( DE=42 ). Check uniqueness: ( A=2 ), ( B=1 ), ( C=2 ), ( D=4 ), ( E=2 ). Repeats 2. Not good.( C=3 ): ( 21 times 3 = 63 ). ( DE=63 ). Check uniqueness: ( A=2 ), ( B=1 ), ( C=3 ), ( D=6 ), ( E=3 ). ( C=3 ) and ( E=3 ) repeat. Not good.( C=4 ): ( 21 times 4 = 84 ). ( DE=84 ). Check uniqueness: ( A=2 ), ( B=1 ), ( C=4 ), ( D=8 ), ( E=4 ). ( C=4 ) and ( E=4 ) repeat. Not good.( C=5 ): ( 21 times 5 = 105 ). Three digits, not allowed.So, ( AB=21 ) doesn't seem to work.Let's try ( AB=23 ).( C=2 ): ( 23 times 2 = 46 ). ( DE=46 ). Check uniqueness: ( A=2 ), ( B=3 ), ( C=2 ), ( D=4 ), ( E=6 ). ( A=2 ) and ( C=2 ) repeat. Not good.( C=3 ): ( 23 times 3 = 69 ). ( DE=69 ). Check uniqueness: ( A=2 ), ( B=3 ), ( C=3 ), ( D=6 ), ( E=9 ). ( B=3 ) and ( C=3 ) repeat. Not good.( C=4 ): ( 23 times 4 = 92 ). ( DE=92 ). Check uniqueness: ( A=2 ), ( B=3 ), ( C=4 ), ( D=9 ), ( E=2 ). ( A=2 ) and ( E=2 ) repeat. Not good.( C=5 ): ( 23 times 5 = 115 ). Three digits, not allowed.So, ( AB=23 ) doesn't seem to work.Let's try ( AB=24 ).( C=2 ): ( 24 times 2 = 48 ). ( DE=48 ). Check uniqueness: ( A=2 ), ( B=4 ), ( C=2 ), ( D=4 ), ( E=8 ). ( A=2 ) and ( C=2 ) repeat; ( B=4 ) and ( D=4 ) repeat. Not good.( C=3 ): ( 24 times 3 = 72 ). ( DE=72 ). Check uniqueness: ( A=2 ), ( B=4 ), ( C=3 ), ( D=7 ), ( E=2 ). ( A=2 ) and ( E=2 ) repeat. Not good.( C=4 ): ( 24 times 4 = 96 ). ( DE=96 ). Check uniqueness: ( A=2 ), ( B=4 ), ( C=4 ), ( D=9 ), ( E=6 ). ( B=4 ) and ( C=4 ) repeat. Not good.( C=5 ): ( 24 times 5 = 120 ). Three digits, not allowed.So, ( AB=24 ) doesn't seem to work.Let's try ( AB=25 ).( C=2 ): ( 25 times 2 = 50 ). 0 is not allowed.( C=3 ): ( 25 times 3 = 75 ). ( DE=75 ). Check uniqueness: ( A=2 ), ( B=5 ), ( C=3 ), ( D=7 ), ( E=5 ). ( B=5 ) and ( E=5 ) repeat. Not good.( C=4 ): ( 25 times 4 = 100 ). Three digits, not allowed.So, ( AB=25 ) doesn't seem to work.Let's try ( AB=26 ).( C=2 ): ( 26 times 2 = 52 ). ( DE=52 ). Check uniqueness: ( A=2 ), ( B=6 ), ( C=2 ), ( D=5 ), ( E=2 ). ( A=2 ), ( C=2 ), ( E=2 ) repeat. Not good.( C=3 ): ( 26 times 3 = 78 ). ( DE=78 ). Check uniqueness: ( A=2 ), ( B=6 ), ( C=3 ), ( D=7 ), ( E=8 ). All unique so far.So, ( AB=26 ), ( C=3 ), ( DE=78 ).Now, moving to the second equation: ( FG - 78 = HI ).Digits used so far: 2,6,3,7,8.Available digits: 1,4,5,9.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 78 ).Possible ( FG ) values: 91, 94, 95.Let's try ( FG=91 ): ( HI=91-78=13 ). 1 and 3 are used.( FG=94 ): ( HI=94-78=16 ). 1 and 6 are used.( FG=95 ): ( HI=95-78=17 ). 1 and 7 are used.Hmm, none of these seem to work. Maybe I need to try a different ( AB ) and ( C ) combination.Let's try ( AB=27 ).( C=2 ): ( 27 times 2 = 54 ). ( DE=54 ). Check uniqueness: ( A=2 ), ( B=7 ), ( C=2 ), ( D=5 ), ( E=4 ). ( A=2 ) and ( C=2 ) repeat. Not good.( C=3 ): ( 27 times 3 = 81 ). ( DE=81 ). Check uniqueness: ( A=2 ), ( B=7 ), ( C=3 ), ( D=8 ), ( E=1 ). All unique so far.So, ( AB=27 ), ( C=3 ), ( DE=81 ).Now, moving to the second equation: ( FG - 81 = HI ).Digits used so far: 2,7,3,8,1.Available digits: 4,5,6,9.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 81 ).Possible ( FG ) values: 94, 95, 96.Let's try ( FG=94 ): ( HI=94-81=13 ). 1 is used.( FG=95 ): ( HI=95-81=14 ). 1 is used.( FG=96 ): ( HI=96-81=15 ). 1 is used.Hmm, none of these seem to work. Maybe I need to try a different ( AB ) and ( C ) combination.Let's try ( AB=28 ).( C=2 ): ( 28 times 2 = 56 ). ( DE=56 ). Check uniqueness: ( A=2 ), ( B=8 ), ( C=2 ), ( D=5 ), ( E=6 ). ( A=2 ) and ( C=2 ) repeat. Not good.( C=3 ): ( 28 times 3 = 84 ). ( DE=84 ). Check uniqueness: ( A=2 ), ( B=8 ), ( C=3 ), ( D=8 ), ( E=4 ). ( B=8 ) and ( D=8 ) repeat. Not good.( C=4 ): ( 28 times 4 = 112 ). Three digits, not allowed.So, ( AB=28 ) doesn't seem to work.Let's try ( AB=29 ).( C=2 ): ( 29 times 2 = 58 ). ( DE=58 ). Check uniqueness: ( A=2 ), ( B=9 ), ( C=2 ), ( D=5 ), ( E=8 ). ( A=2 ) and ( C=2 ) repeat. Not good.( C=3 ): ( 29 times 3 = 87 ). ( DE=87 ). Check uniqueness: ( A=2 ), ( B=9 ), ( C=3 ), ( D=8 ), ( E=7 ). All unique so far.So, ( AB=29 ), ( C=3 ), ( DE=87 ).Now, moving to the second equation: ( FG - 87 = HI ).Digits used so far: 2,9,3,8,7.Available digits: 1,4,5,6.So, ( FG ) and ( HI ) must be formed from these digits.Also, ( FG > 87 ).Possible ( FG ) values: 91, 94, 95, 96.Let's try ( FG=91 ): ( HI=91-87=04 ). 0 is not allowed.( FG=94 ): ( HI=94-87=07 ). 0 is not allowed.( FG=95 ): ( HI=95-87=08 ). 0 is not allowed.( FG=96 ): ( HI=96-87=09 ). 0 is not allowed.Hmm, none of these seem to work. Maybe I need to try a different ( AB ) and ( C ) combination.Let's try ( AB=31 ).( C=2 ): ( 31 times 2 = 62 ). ( DE=62 ). Check uniqueness: ( A=3 ), ( B=1 ), ( C=2 ), ( D=6 ), ( E=2 ). ( C=2 ) and ( E=2 ) repeat. Not good.( C=3 ): ( 31 times 3 = 93 ). ( DE=93 ). Check uniqueness: ( A=3 ), ( B=1 ), ( C=3 ), ( D=9 ), ( E=3 ). Multiple repeats. Not good.( C=4 ): ( 31 times 4 = 124 ). Three digits, not allowed.So, ( AB=31 ) doesn't seem to work.Let's try ( AB=32 ).( C=2 ): ( 32 times 2 = 64 ). ( DE=64 ). Check uniqueness: ( A=3 ), ( B=2 ), ( C=2 ), ( D=6 ), ( E=4 ). ( B=2 ) and ( C=2 ) repeat. Not good.( C=3 ): ( 32 times 3 = 96 ). ( DE=96 ). Check uniqueness: ( A=3 ), ( B=2 ), ( C=3 ), ( D=9 ), ( E=6 ). ( A=3 ) and ( C=3 ) repeat. Not good.( C=4 ): ( 32 times 4 = 128 ). Three digits, not allowed.So, ( AB=32 ) doesn't seem to work.Hmm, this is getting frustrating. Maybe I need to try a different approach. Let me think about the second equation again: ( FG - DE = HI ). Since ( FG ) and ( DE ) are two-digit numbers, and their difference is also a two-digit number, the difference must be at least 10. So, ( FG - DE geq 10 ).Also, since all digits are unique, ( HI ) must be a two-digit number with digits not used in ( FG ) or ( DE ).Wait, maybe I can look for ( FG ) and ( DE ) such that their difference ( HI ) uses digits not used in ( FG ) or ( DE ).Let me list possible ( FG ) and ( DE ) pairs where ( FG - DE = HI ), and all digits are unique.For example:- ( FG=93 ), ( DE=68 ), ( HI=25 ). Check uniqueness: 9,3,6,8,2,5. All unique.This seems promising. Let's see if this fits with the first equation.So, ( DE=68 ). Then, from the first equation: ( AB times C = 68 ).So, we need ( AB times C = 68 ).Let's find possible ( AB ) and ( C ) such that their product is 68.Possible factor pairs of 68:- 1 × 68 (but 68 is two digits, so ( AB=68 ), ( C=1 ))- 2 × 34 ( ( AB=34 ), ( C=2 ))- 4 × 17 ( ( AB=17 ), ( C=4 ))- 17 × 4 (same as above)- 34 × 2 (same as above)- 68 × 1 (same as above)But ( AB ) must be a two-digit number, so possible ( AB ) values are 17, 34, 68.Let's check each:1. ( AB=68 ), ( C=1 ). But ( DE=68 ), so ( D=6 ), ( E=8 ). Check uniqueness: ( A=6 ), ( B=8 ), ( C=1 ), ( D=6 ), ( E=8 ). ( A=6 ) and ( D=6 ) repeat; ( B=8 ) and ( E=8 ) repeat. Not good.2. ( AB=34 ), ( C=2 ). So, ( A=3 ), ( B=4 ), ( C=2 ), ( D=6 ), ( E=8 ). Check uniqueness: 3,4,2,6,8. All unique so far.Now, moving to the second equation: ( FG=93 ), ( DE=68 ), ( HI=25 ).Check uniqueness: ( F=9 ), ( G=3 ), ( H=2 ), ( I=5 ). All unique so far.Now, let's check all digits used:From ( AB=34 ): 3,4From ( C=2 ): 2From ( DE=68 ): 6,8From ( FG=93 ): 9,3From ( HI=25 ): 2,5Wait, ( C=2 ) and ( H=2 ) both use 2. That's a repeat. Not allowed.So, this combination doesn't work.3. ( AB=17 ), ( C=4 ). So, ( A=1 ), ( B=7 ), ( C=4 ), ( D=6 ), ( E=8 ). Check uniqueness: 1,7,4,6,8. All unique so far.Now, moving to the second equation: ( FG=93 ), ( DE=68 ), ( HI=25 ).Check uniqueness: ( F=9 ), ( G=3 ), ( H=2 ), ( I=5 ). All unique so far.Now, let's check all digits used:From ( AB=17 ): 1,7From ( C=4 ): 4From ( DE=68 ): 6,8From ( FG=93 ): 9,3From ( HI=25 ): 2,5All digits are unique: 1,7,4,6,8,9,3,2,5. Perfect!So, the assignments are:- ( A=1 )- ( B=7 )- ( C=4 )- ( D=6 )- ( E=8 )- ( F=9 )- ( G=3 )- ( H=2 )- ( I=5 )Let me double-check the equations:1. ( AB times C = 17 times 4 = 68 = DE ). Correct.2. ( FG - DE = 93 - 68 = 25 = HI ). Correct.All digits are unique and within 1-9. Great!So, the solution is:( AB=17 ), ( C=4 ), ( DE=68 ), ( FG=93 ), ( HI=25 ).

answer：Okay, so I have this problem where I need to find the minimum value of the expression:[frac{x^2 + y^2}{x + y} + frac{x^2 + z^2}{x + z} + frac{y^2 + z^2}{y + z}]given that (x), (y), and (z) are positive integers and their sum is 9. Hmm, let me try to figure this out step by step.First, I recall that for any two positive numbers, there's an inequality called the Quadratic Mean - Arithmetic Mean (QM-AM) inequality. It states that:[sqrt{frac{a^2 + b^2}{2}} geq frac{a + b}{2}]If I square both sides to get rid of the square root, it becomes:[frac{a^2 + b^2}{2} geq left(frac{a + b}{2}right)^2]Multiplying both sides by 2 gives:[frac{a^2 + b^2}{a + b} geq frac{a + b}{2}]So, applying this to each term in the given expression, I can say:[frac{x^2 + y^2}{x + y} geq frac{x + y}{2}][frac{x^2 + z^2}{x + z} geq frac{x + z}{2}][frac{y^2 + z^2}{y + z} geq frac{y + z}{2}]If I add all these inequalities together, I get:[frac{x^2 + y^2}{x + y} + frac{x^2 + z^2}{x + z} + frac{y^2 + z^2}{y + z} geq frac{x + y}{2} + frac{x + z}{2} + frac{y + z}{2}]Simplifying the right side, let's see:[frac{x + y}{2} + frac{x + z}{2} + frac{y + z}{2} = frac{2x + 2y + 2z}{2} = x + y + z]But we know that (x + y + z = 9), so the right side simplifies to 9. Therefore, the expression on the left is greater than or equal to 9. So, the minimum value should be 9.Wait, but is this achievable? Because sometimes inequalities give a lower bound, but it might not actually be attainable. Let me check if equality holds.In the QM-AM inequality, equality holds when all the numbers are equal. So, for each pair (x = y), (x = z), and (y = z). That would mean (x = y = z). But since (x + y + z = 9), each would have to be 3. So, if (x = y = z = 3), let's plug that into the original expression:[frac{3^2 + 3^2}{3 + 3} + frac{3^2 + 3^2}{3 + 3} + frac{3^2 + 3^2}{3 + 3} = 3 times frac{9 + 9}{6} = 3 times frac{18}{6} = 3 times 3 = 9]So, yes, when (x = y = z = 3), the expression equals 9. Therefore, the minimum value is indeed 9.But wait, the problem specifies that (x), (y), and (z) are positive integers. So, 3 is a positive integer, and their sum is 9, so that's valid. Therefore, the minimum value is 9.Just to make sure, let me try another set of numbers. Suppose (x = 1), (y = 4), (z = 4). Then:First term: (frac{1 + 16}{1 + 4} = frac{17}{5} = 3.4)Second term: (frac{1 + 16}{1 + 4} = 3.4)Third term: (frac{16 + 16}{4 + 4} = frac{32}{8} = 4)Total: (3.4 + 3.4 + 4 = 10.8), which is more than 9.Another example: (x = 2), (y = 3), (z = 4).First term: (frac{4 + 9}{5} = frac{13}{5} = 2.6)Second term: (frac{4 + 16}{6} = frac{20}{6} approx 3.333)Third term: (frac{9 + 16}{7} = frac{25}{7} approx 3.571)Total: (2.6 + 3.333 + 3.571 approx 9.504), which is still more than 9.Wait, so maybe 9 is indeed the minimum. Let me try another one: (x = 3), (y = 3), (z = 3). As before, each term is 3, so total is 9.What if I try (x = 4), (y = 4), (z = 1). That's similar to the first example, and the total was 10.8.Alternatively, (x = 5), (y = 2), (z = 2):First term: (frac{25 + 4}{7} = frac{29}{7} approx 4.142)Second term: (frac{25 + 4}{7} approx 4.142)Third term: (frac{4 + 4}{4} = 2)Total: (4.142 + 4.142 + 2 approx 10.284), still higher.Another test case: (x = 6), (y = 1), (z = 2):First term: (frac{36 + 1}{7} = frac{37}{7} approx 5.285)Second term: (frac{36 + 4}{8} = frac{40}{8} = 5)Third term: (frac{1 + 4}{3} = frac{5}{3} approx 1.666)Total: (5.285 + 5 + 1.666 approx 11.951), which is way higher.So, from these examples, it seems that when the numbers are equal, the expression reaches its minimum value of 9. When the numbers are unequal, the value increases.Therefore, I can confidently say that the minimum value is 9, achieved when (x = y = z = 3).**Final Answer**The minimum value is boxed{9}.

answer：To solve the problem of expressing (0.overline{2} + 0.overline{02}) as a fraction in its lowest terms, I will follow a systematic approach.First, I will convert each repeating decimal into a fraction individually.Starting with (0.overline{2}):1. Let (x = 0.overline{2}).2. Multiplying both sides by 10 to shift the decimal point: (10x = 2.overline{2}).3. Subtracting the original equation from this new equation to eliminate the repeating part: (10x - x = 2.overline{2} - 0.overline{2}), which simplifies to (9x = 2).4. Solving for (x): (x = frac{2}{9}).Next, converting (0.overline{02}) into a fraction:1. Let (y = 0.overline{02}).2. Multiplying both sides by 100 to shift the decimal point two places: (100y = 2.overline{02}).3. Subtracting the original equation from this new equation: (100y - y = 2.overline{02} - 0.overline{02}), which simplifies to (99y = 2).4. Solving for (y): (y = frac{2}{99}).Now, I will add the two fractions:1. The fractions to add are (frac{2}{9}) and (frac{2}{99}).2. To add these, I need a common denominator. The least common multiple of 9 and 99 is 99.3. Converting (frac{2}{9}) to have a denominator of 99: (frac{2 times 11}{9 times 11} = frac{22}{99}).4. Adding the two fractions: (frac{22}{99} + frac{2}{99} = frac{24}{99}).5. Simplifying (frac{24}{99}) by dividing both numerator and denominator by their greatest common divisor, which is 3: (frac{24 div 3}{99 div 3} = frac{8}{33}).Therefore, the sum of (0.overline{2} + 0.overline{02}) expressed as a fraction in its lowest terms is (frac{8}{33}).

answer：Alright, so I have this multiple-choice question with four options: A, B, C, and D. I need to figure out which one is correct. Let me go through each option one by one and analyze them carefully.Starting with **Option A**: It says, "In triangle ABC, if sin A < 1/2, then A < π/6." Hmm, okay. I remember that in a triangle, the sine of an angle relates to the length of the opposite side. But wait, sine is also periodic and has a maximum value of 1. So, sin A < 1/2 means that angle A could be in two different ranges: either between 0 and π/6 or between 5π/6 and π. Because sine is positive in both the first and second quadrants. So, if sin A is less than 1/2, angle A could be less than π/6 or greater than 5π/6. But in a triangle, the sum of angles is π, so having an angle greater than 5π/6 would mean the other two angles would have to be less than π/6 each, which might not always be the case. So, the contrapositive of this statement would be: "If A ≥ π/6, then sin A ≥ 1/2." But wait, if A is between π/6 and 5π/6, sin A is actually greater than or equal to 1/2. So, the contrapositive seems to hold true because if A is not less than π/6, then sin A is not less than 1/2. So, maybe A is correct? I'm a bit confused here. Let me think again. The original statement says that if sin A is less than 1/2, then A is less than π/6. But as I thought earlier, A could also be greater than 5π/6, which is more than π/6. So, the original statement is not entirely true because there's another possibility where A is greater than 5π/6, and sin A is still less than 1/2. Therefore, the contrapositive would also not be entirely true because it's based on a flawed original statement. So, I think A is incorrect.Moving on to **Option B**: It says, "Given non-coincident lines a, b and plane α, where a is parallel to α, and b is parallel to α, if q: a is not parallel to b, then p is a sufficient condition for q." Wait, this is a bit confusing. Let me parse it again. We have two lines, a and b, that are both parallel to the same plane α. The statement is saying that if a is not parallel to b (q), then p is a sufficient condition for q. But what is p? It's not defined here. Maybe p is some condition related to a and b? Or is p another statement? The wording is unclear. Maybe it's saying that if a and b are both parallel to α, then a not being parallel to b is a sufficient condition for something else. I'm not sure. Maybe I need to think about the relationship between lines parallel to the same plane. If two lines are both parallel to the same plane, they can either be parallel to each other or skew. So, if a and b are both parallel to α, they might not necessarily be parallel to each other. So, if q is a not parallel to b, then is p a sufficient condition for q? Without knowing what p is, it's hard to evaluate. Maybe p is that the lines are not parallel? But then p would be the same as q. I'm not sure. This option is unclear, so I'll tentatively say B is incorrect, but I'm not entirely certain.Next is **Option C**: It states, "If the dot product of vectors a and b is greater than 0, then the angle between them is acute; if the dot product is less than 0, then the angle is obtuse." Okay, I remember that the dot product formula is a · b = |a||b|cosθ, where θ is the angle between them. So, if a · b > 0, then cosθ > 0, which means θ is between 0 and π/2, which is acute. If a · b < 0, then cosθ < 0, which means θ is between π/2 and π, which is obtuse. So, this seems correct. But wait, what if one of the vectors is the zero vector? Then the dot product is zero, but the angle isn't defined. But the statement says "if a · b > 0" and "if a · b < 0," so it doesn't cover the case where the dot product is zero. So, as long as neither vector is the zero vector, the statement holds. So, I think C is correct.Lastly, **Option D**: It says, "Given the proposition p: There exists an x in real numbers such that 2^x < x^2 holds. Then, the negation of p: For all x in real numbers, 2^x ≥ x^2 is true." Hmm, okay. So, p is saying that there is at least one real number x where 2^x is less than x squared. The negation of p would be that for every real number x, 2^x is greater than or equal to x squared. Now, is this negation true? Let's think about the functions 2^x and x^2. For x = 2, 2^2 = 4 and 2^2 = 4, so they're equal. For x = 3, 2^3 = 8 and 3^2 = 9, so 8 < 9, meaning 2^x < x^2. So, there exists an x (like x=3) where 2^x < x^2. Therefore, p is true. But the negation of p is claiming that for all x, 2^x ≥ x^2. But since we found an x where 2^x < x^2, the negation is false. So, D is incorrect.Wait, but the question is asking which statement is correct. So, A is incorrect, B is unclear but likely incorrect, C is correct, and D is incorrect. Therefore, the correct answer should be C.But hold on, earlier I thought A might be correct because the contrapositive seemed true, but then realized that the original statement wasn't entirely true because angle A could be greater than 5π/6. So, the contrapositive would also not hold because it's based on a false premise. So, A is incorrect.As for B, since it's unclear what p is, I think it's safer to say it's incorrect. So, yeah, C is the correct answer.