answer：Okay, so I have this problem where I need to find the remainder when ( x^3 + 7x^2 + 4 ) divides ( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 ). Hmm, polynomial division, right? I remember that when dividing polynomials, it's similar to long division with numbers, but with variables involved. Let me try to recall the steps.First, I need to set up the division. The divisor is ( x^3 + 7x^2 + 4 ), and the dividend is ( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 ). So, I guess I should write it like this:``` _______________________x^3 + 7x^2 + 4 | 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50```Alright, now, the first step is to divide the leading term of the dividend by the leading term of the divisor. The leading term of the dividend is ( 2x^5 ), and the leading term of the divisor is ( x^3 ). So, ( 2x^5 ) divided by ( x^3 ) is ( 2x^{5-3} = 2x^2 ). That's the first term of the quotient.Next, I need to multiply the entire divisor by this term ( 2x^2 ) and subtract the result from the dividend. Let's do that:Multiply ( 2x^2 ) by each term of the divisor:- ( 2x^2 times x^3 = 2x^5 )- ( 2x^2 times 7x^2 = 14x^4 )- ( 2x^2 times 4 = 8x^2 )So, the product is ( 2x^5 + 14x^4 + 8x^2 ).Now, subtract this from the original dividend:- ( (2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50) - (2x^5 + 14x^4 + 8x^2) )Let's perform the subtraction term by term:- ( 2x^5 - 2x^5 = 0 )- ( 11x^4 - 14x^4 = -3x^4 )- ( -48x^3 - 0 = -48x^3 )- ( -60x^2 - 8x^2 = -68x^2 )- ( 20x - 0 = 20x )- ( 50 - 0 = 50 )So, after the first subtraction, the new dividend is ( -3x^4 - 48x^3 - 68x^2 + 20x + 50 ).Now, I repeat the process. Divide the leading term of the new dividend by the leading term of the divisor. The leading term is ( -3x^4 ), and the divisor's leading term is still ( x^3 ). So, ( -3x^4 ) divided by ( x^3 ) is ( -3x^{4-3} = -3x ). That's the next term in the quotient.Multiply the entire divisor by ( -3x ):- ( -3x times x^3 = -3x^4 )- ( -3x times 7x^2 = -21x^3 )- ( -3x times 4 = -12x )So, the product is ( -3x^4 - 21x^3 - 12x ).Subtract this from the current dividend:- ( (-3x^4 - 48x^3 - 68x^2 + 20x + 50) - (-3x^4 - 21x^3 - 12x) )Subtracting term by term:- ( -3x^4 - (-3x^4) = 0 )- ( -48x^3 - (-21x^3) = -27x^3 )- ( -68x^2 - 0 = -68x^2 )- ( 20x - (-12x) = 32x )- ( 50 - 0 = 50 )So, now the new dividend is ( -27x^3 - 68x^2 + 32x + 50 ).I check the degree of this new dividend. It's 3, which is the same as the degree of the divisor. So, I can continue dividing.Divide the leading term ( -27x^3 ) by the leading term ( x^3 ) to get ( -27 ). That's the next term in the quotient.Multiply the entire divisor by ( -27 ):- ( -27 times x^3 = -27x^3 )- ( -27 times 7x^2 = -189x^2 )- ( -27 times 4 = -108 )So, the product is ( -27x^3 - 189x^2 - 108 ).Subtract this from the current dividend:- ( (-27x^3 - 68x^2 + 32x + 50) - (-27x^3 - 189x^2 - 108) )Subtracting term by term:- ( -27x^3 - (-27x^3) = 0 )- ( -68x^2 - (-189x^2) = 121x^2 )- ( 32x - 0 = 32x )- ( 50 - (-108) = 158 )So, the new dividend is ( 121x^2 + 32x + 158 ).Now, the degree of this dividend is 2, which is less than the degree of the divisor (which is 3). So, we can't continue dividing further. This means that ( 121x^2 + 32x + 158 ) is the remainder.Wait, but in the initial problem, the divisor is ( x^3 + 7x^2 + 4 ). Let me double-check my calculations to make sure I didn't make a mistake.In the first step, I had ( 2x^5 ) divided by ( x^3 ) giving ( 2x^2 ). Then, multiplying ( 2x^2 ) by the divisor gave ( 2x^5 + 14x^4 + 8x^2 ). Subtracting that from the dividend gave ( -3x^4 - 48x^3 - 68x^2 + 20x + 50 ). That seems correct.Next, dividing ( -3x^4 ) by ( x^3 ) gave ( -3x ). Multiplying ( -3x ) by the divisor gave ( -3x^4 - 21x^3 - 12x ). Subtracting that from the dividend gave ( -27x^3 - 68x^2 + 32x + 50 ). That also looks correct.Then, dividing ( -27x^3 ) by ( x^3 ) gave ( -27 ). Multiplying ( -27 ) by the divisor gave ( -27x^3 - 189x^2 - 108 ). Subtracting that from the dividend gave ( 121x^2 + 32x + 158 ). Hmm, that seems a bit high, but let me check the subtraction:- ( -27x^3 - (-27x^3) = 0 )- ( -68x^2 - (-189x^2) = 121x^2 )- ( 32x - 0 = 32x )- ( 50 - (-108) = 158 )Yes, that's correct. So, the remainder is indeed ( 121x^2 + 32x + 158 ).Wait a second, but in the initial problem, the divisor is ( x^3 + 7x^2 + 4 ), and the dividend is ( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 ). Let me verify if my calculations are correct.Alternatively, maybe I made a mistake in the multiplication or subtraction steps. Let me go through them again.First step: ( 2x^2 times (x^3 + 7x^2 + 4) = 2x^5 + 14x^4 + 8x^2 ). Subtracting this from the dividend:( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 )minus( 2x^5 + 14x^4 + 8x^2 )equals( (2x^5 - 2x^5) + (11x^4 - 14x^4) + (-48x^3) + (-60x^2 - 8x^2) + 20x + 50 )which simplifies to( 0 - 3x^4 - 48x^3 - 68x^2 + 20x + 50 ). That's correct.Second step: ( -3x times (x^3 + 7x^2 + 4) = -3x^4 - 21x^3 - 12x ). Subtracting this from the current dividend:( -3x^4 - 48x^3 - 68x^2 + 20x + 50 )minus( -3x^4 - 21x^3 - 12x )equals( (-3x^4 + 3x^4) + (-48x^3 + 21x^3) + (-68x^2) + (20x + 12x) + 50 )which simplifies to( 0 - 27x^3 - 68x^2 + 32x + 50 ). Correct.Third step: ( -27 times (x^3 + 7x^2 + 4) = -27x^3 - 189x^2 - 108 ). Subtracting this from the current dividend:( -27x^3 - 68x^2 + 32x + 50 )minus( -27x^3 - 189x^2 - 108 )equals( (-27x^3 + 27x^3) + (-68x^2 + 189x^2) + 32x + (50 + 108) )which simplifies to( 0 + 121x^2 + 32x + 158 ). Correct.So, the remainder is indeed ( 121x^2 + 32x + 158 ). Wait, but in the initial problem, the user mentioned the remainder as ( -27x^3 - 68x^2 + 32x + 50 ). That seems different from what I got. Did I make a mistake?Wait, no. Let me see. The user's initial problem was to find the remainder when ( x^3 + 7x^2 + 4 ) divides ( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 ). They started the division and got a remainder of ( -27x^3 - 68x^2 + 32x + 50 ), but then continued further and got ( 121x^2 + 32x + 158 ).Wait, so in the initial problem, the user stopped at ( -27x^3 - 68x^2 + 32x + 50 ) as the remainder, but actually, since the degree of that polynomial is equal to the degree of the divisor, they should continue dividing. So, my process shows that the correct remainder is ( 121x^2 + 32x + 158 ).But wait, let me check the degrees. The divisor is degree 3, and the remainder after the first two steps was degree 3, so we should continue dividing until the remainder's degree is less than 3. So, in my process, after the third step, the remainder is degree 2, which is less than 3, so that's the final remainder.Therefore, the correct remainder is ( 121x^2 + 32x + 158 ).Wait, but let me verify this by another method, maybe using the Remainder Theorem or synthetic division. Hmm, the Remainder Theorem states that the remainder of dividing a polynomial ( f(x) ) by ( (x - a) ) is ( f(a) ). But here, the divisor is a cubic polynomial, not linear, so the Remainder Theorem doesn't directly apply. However, I can use polynomial equivalence.Alternatively, I can express the division as:( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 = (x^3 + 7x^2 + 4) times Q(x) + R(x) ),where ( Q(x) ) is the quotient and ( R(x) ) is the remainder with degree less than 3.From my division, I have:Quotient: ( 2x^2 - 3x - 27 )Remainder: ( 121x^2 + 32x + 158 )So, let's multiply ( (x^3 + 7x^2 + 4) ) by ( 2x^2 - 3x - 27 ) and add the remainder to see if we get back the original dividend.First, multiply ( 2x^2 times (x^3 + 7x^2 + 4) = 2x^5 + 14x^4 + 8x^2 )Then, multiply ( -3x times (x^3 + 7x^2 + 4) = -3x^4 - 21x^3 - 12x )Then, multiply ( -27 times (x^3 + 7x^2 + 4) = -27x^3 - 189x^2 - 108 )Now, add all these together:( 2x^5 + 14x^4 + 8x^2 )( -3x^4 - 21x^3 - 12x )( -27x^3 - 189x^2 - 108 )Plus the remainder ( 121x^2 + 32x + 158 )Adding term by term:- ( 2x^5 )- ( 14x^4 - 3x^4 = 11x^4 )- ( 0x^3 - 21x^3 - 27x^3 = -48x^3 )- ( 8x^2 - 189x^2 + 121x^2 = (8 - 189 + 121)x^2 = (-181 + 121)x^2 = -60x^2 )- ( 0x - 12x + 32x = 20x )- ( 0 - 108 + 158 = 50 )So, combining all these, we get ( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 ), which matches the original dividend. Therefore, my calculations are correct, and the remainder is indeed ( 121x^2 + 32x + 158 ).Wait, but in the initial problem, the user had a remainder of ( -27x^3 - 68x^2 + 32x + 50 ). That must be because they stopped too early, not continuing the division until the remainder's degree was less than the divisor's degree. So, the correct remainder is ( 121x^2 + 32x + 158 ).Alternatively, maybe I made a mistake in the multiplication or subtraction steps. Let me double-check the third step where I multiplied ( -27 ) by the divisor.( -27 times x^3 = -27x^3 )( -27 times 7x^2 = -189x^2 )( -27 times 4 = -108 )Yes, that's correct.Subtracting ( -27x^3 - 189x^2 - 108 ) from ( -27x^3 - 68x^2 + 32x + 50 ):( -27x^3 - (-27x^3) = 0 )( -68x^2 - (-189x^2) = 121x^2 )( 32x - 0 = 32x )( 50 - (-108) = 158 )Yes, that's correct.So, the remainder is ( 121x^2 + 32x + 158 ).Wait, but let me check if the user's initial problem had a different remainder. They mentioned:"Conclusion: The resultant remainder is (boxed{-27x^3 - 68x^2 + 32x + 50})."But according to my calculations, that's not the final remainder because the degree is still 3, equal to the divisor. So, they should have continued dividing. Therefore, the correct remainder is ( 121x^2 + 32x + 158 ).Alternatively, maybe the user made a mistake in their initial problem, or perhaps I misread it. Let me check the original problem again.The original problem was: "What is the remainder when ( x^3 + 7x^2 + 4 ) divides ( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 )?"So, the user's initial answer was ( -27x^3 - 68x^2 + 32x + 50 ), but that's not the final remainder because its degree is equal to the divisor's degree. Therefore, the correct remainder is ( 121x^2 + 32x + 158 ).Wait, but let me see if there's another way to approach this problem, maybe using modular arithmetic or something else. Alternatively, perhaps I can use the fact that the remainder must be of degree less than 3, so it should be a quadratic or lower. Therefore, my result makes sense.Alternatively, maybe I can use polynomial equivalence to find the remainder. Let me set up the equation:( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 = (x^3 + 7x^2 + 4) times Q(x) + R(x) )where ( R(x) = ax^2 + bx + c ).Then, I can express ( Q(x) ) as ( 2x^2 + dx + e ), since the leading term of the dividend is ( 2x^5 ) and the divisor is ( x^3 ), so the quotient starts with ( 2x^2 ).Wait, but in my previous division, the quotient was ( 2x^2 - 3x - 27 ). So, let me assume ( Q(x) = 2x^2 + dx + e ).Then, expanding ( (x^3 + 7x^2 + 4)(2x^2 + dx + e) ):First, multiply ( x^3 times 2x^2 = 2x^5 )( x^3 times dx = dx^4 )( x^3 times e = ex^3 )Then, ( 7x^2 times 2x^2 = 14x^4 )( 7x^2 times dx = 7dx^3 )( 7x^2 times e = 7ex^2 )Then, ( 4 times 2x^2 = 8x^2 )( 4 times dx = 4dx )( 4 times e = 4e )So, combining all terms:( 2x^5 + (d + 14)x^4 + (e + 7d)x^3 + (7e + 8)x^2 + 4dx + 4e )Now, adding the remainder ( R(x) = ax^2 + bx + c ), the entire expression becomes:( 2x^5 + (d + 14)x^4 + (e + 7d)x^3 + (7e + 8 + a)x^2 + (4d + b)x + (4e + c) )This must equal the original dividend ( 2x^5 + 11x^4 - 48x^3 - 60x^2 + 20x + 50 ).Therefore, we can set up equations by equating coefficients:1. ( 2x^5 ): Coefficient is 2 on both sides, so that's fine.2. ( x^4 ): ( d + 14 = 11 ) => ( d = 11 - 14 = -3 )3. ( x^3 ): ( e + 7d = -48 ). Since ( d = -3 ), ( e + 7(-3) = -48 ) => ( e - 21 = -48 ) => ( e = -48 + 21 = -27 )4. ( x^2 ): ( 7e + 8 + a = -60 ). Plugging ( e = -27 ): ( 7(-27) + 8 + a = -60 ) => ( -189 + 8 + a = -60 ) => ( -181 + a = -60 ) => ( a = -60 + 181 = 121 )5. ( x ): ( 4d + b = 20 ). Plugging ( d = -3 ): ( 4(-3) + b = 20 ) => ( -12 + b = 20 ) => ( b = 32 )6. Constant term: ( 4e + c = 50 ). Plugging ( e = -27 ): ( 4(-27) + c = 50 ) => ( -108 + c = 50 ) => ( c = 158 )So, the remainder ( R(x) = ax^2 + bx + c = 121x^2 + 32x + 158 ), which matches my earlier result.Therefore, the correct remainder is ( 121x^2 + 32x + 158 ).Wait, but in the initial problem, the user had a remainder of ( -27x^3 - 68x^2 + 32x + 50 ). That must be because they stopped the division process too early, not continuing until the remainder's degree was less than the divisor's degree. So, the correct remainder is indeed ( 121x^2 + 32x + 158 ).I think that's thorough enough. I've double-checked my work using two methods: polynomial long division and setting up equations based on polynomial equivalence. Both methods lead to the same remainder, so I'm confident that's the correct answer.

answer：First, recognize that a power function has the form ( y = f(x) = x^a ).Given that the function passes through the point ( left( frac{1}{2}, 8 right) ), substitute ( x = frac{1}{2} ) and ( y = 8 ) into the equation:[left( frac{1}{2} right)^a = 8]To solve for ( a ), rewrite 8 as a power of 2:[left( frac{1}{2} right)^a = 2^3]Since ( frac{1}{2} = 2^{-1} ), the equation becomes:[(2^{-1})^a = 2^3][2^{-a} = 2^3]Equating the exponents:[-a = 3 quad Rightarrow quad a = -3]Thus, the function is ( f(x) = x^{-3} ).Finally, calculate ( f(2) ):[f(2) = 2^{-3} = frac{1}{8}]

answer：Okay, so I have this sequence problem here. The sequence is defined by ( a_1 = a_2 = 1 ) and ( a_3 = 2 ). There's also a condition that for any natural number ( n ), the product ( a_n a_{n+1} a_{n+2} neq 1 ). Additionally, there's this equation that relates four consecutive terms: ( a_n a_{n+1} a_{n+2} a_{n+3} = a_n + a_{n+1} + a_{n+2} + a_{n+3} ). I need to find the sum of the first 100 terms of this sequence.Alright, let's start by writing down what we know. The first three terms are given: ( a_1 = 1 ), ( a_2 = 1 ), and ( a_3 = 2 ). I need to find ( a_4 ) using the given equation. Let me plug in ( n = 1 ) into the equation:( a_1 a_2 a_3 a_4 = a_1 + a_2 + a_3 + a_4 )Substituting the known values:( 1 times 1 times 2 times a_4 = 1 + 1 + 2 + a_4 )Simplifying the left side: ( 2a_4 )Simplifying the right side: ( 4 + a_4 )So, ( 2a_4 = 4 + a_4 )Subtracting ( a_4 ) from both sides: ( a_4 = 4 )Okay, so ( a_4 = 4 ). Now, let's see if we can find a pattern or a recurrence relation. The given equation involves four consecutive terms, so maybe we can find a relationship between terms that are four apart.Let me consider the equation for ( n ) and ( n+1 ):For ( n ):( a_n a_{n+1} a_{n+2} a_{n+3} = a_n + a_{n+1} + a_{n+2} + a_{n+3} )For ( n+1 ):( a_{n+1} a_{n+2} a_{n+3} a_{n+4} = a_{n+1} + a_{n+2} + a_{n+3} + a_{n+4} )If I divide these two equations, I might get a useful relationship. Let's try that:( frac{a_{n+1} a_{n+2} a_{n+3} a_{n+4}}{a_n a_{n+1} a_{n+2} a_{n+3}}} = frac{a_{n+1} + a_{n+2} + a_{n+3} + a_{n+4}}{a_n + a_{n+1} + a_{n+2} + a_{n+3}} )Simplifying the left side: ( frac{a_{n+4}}{a_n} )So, we have:( frac{a_{n+4}}{a_n} = frac{a_{n+1} + a_{n+2} + a_{n+3} + a_{n+4}}{a_n + a_{n+1} + a_{n+2} + a_{n+3}} )Hmm, this seems a bit complicated. Maybe I can rearrange the original equation differently. Let's look back at the original equation:( a_n a_{n+1} a_{n+2} a_{n+3} = a_n + a_{n+1} + a_{n+2} + a_{n+3} )I can factor out ( a_n ) on the right side:( a_n (a_{n+1} a_{n+2} a_{n+3} - 1) = a_{n+1} + a_{n+2} + a_{n+3} )But I'm not sure if that helps. Maybe instead, I can express ( a_{n+3} ) in terms of the previous terms. Let's try solving for ( a_{n+3} ):( a_{n+3} = frac{a_n + a_{n+1} + a_{n+2}}{a_n a_{n+1} a_{n+2} - 1} )Wait, that might be useful. Let me write that down:( a_{n+3} = frac{a_n + a_{n+1} + a_{n+2}}{a_n a_{n+1} a_{n+2} - 1} )Given that ( a_n a_{n+1} a_{n+2} neq 1 ), this division is valid.Okay, so with this formula, I can compute ( a_{n+3} ) if I know ( a_n ), ( a_{n+1} ), and ( a_{n+2} ). Let's try computing the next few terms to see if a pattern emerges.We already have ( a_1 = 1 ), ( a_2 = 1 ), ( a_3 = 2 ), ( a_4 = 4 ).Let's compute ( a_5 ):Using ( n = 2 ):( a_5 = frac{a_2 + a_3 + a_4}{a_2 a_3 a_4 - 1} = frac{1 + 2 + 4}{1 times 2 times 4 - 1} = frac{7}{8 - 1} = frac{7}{7} = 1 )So, ( a_5 = 1 ).Now, ( a_6 ):Using ( n = 3 ):( a_6 = frac{a_3 + a_4 + a_5}{a_3 a_4 a_5 - 1} = frac{2 + 4 + 1}{2 times 4 times 1 - 1} = frac{7}{8 - 1} = frac{7}{7} = 1 )So, ( a_6 = 1 ).Next, ( a_7 ):Using ( n = 4 ):( a_7 = frac{a_4 + a_5 + a_6}{a_4 a_5 a_6 - 1} = frac{4 + 1 + 1}{4 times 1 times 1 - 1} = frac{6}{4 - 1} = frac{6}{3} = 2 )So, ( a_7 = 2 ).Continuing, ( a_8 ):Using ( n = 5 ):( a_8 = frac{a_5 + a_6 + a_7}{a_5 a_6 a_7 - 1} = frac{1 + 1 + 2}{1 times 1 times 2 - 1} = frac{4}{2 - 1} = frac{4}{1} = 4 )So, ( a_8 = 4 ).Wait a minute, this looks familiar. Let me list out the terms I have so far:( a_1 = 1 )( a_2 = 1 )( a_3 = 2 )( a_4 = 4 )( a_5 = 1 )( a_6 = 1 )( a_7 = 2 )( a_8 = 4 )It seems like the sequence is repeating every four terms: 1, 1, 2, 4, 1, 1, 2, 4, ...To confirm, let's compute ( a_9 ):Using ( n = 6 ):( a_9 = frac{a_6 + a_7 + a_8}{a_6 a_7 a_8 - 1} = frac{1 + 2 + 4}{1 times 2 times 4 - 1} = frac{7}{8 - 1} = frac{7}{7} = 1 )Yes, ( a_9 = 1 ), which continues the pattern.Therefore, the sequence is periodic with period 4: 1, 1, 2, 4, 1, 1, 2, 4, ...So, every four terms, the sequence repeats. To find the sum of the first 100 terms, I can calculate how many complete periods of four terms fit into 100 terms and then multiply by the sum of one period.First, let's find the sum of one period:( a_1 + a_2 + a_3 + a_4 = 1 + 1 + 2 + 4 = 8 )Now, how many complete periods are there in 100 terms?Since each period is 4 terms, the number of periods is ( frac{100}{4} = 25 ).Therefore, the total sum is ( 25 times 8 = 200 ).So, the sum of the first 100 terms is 200.**Final Answer**The value of ( a_{1} + a_{2} + cdots + a_{100} ) is boxed{200}.

answer：Okay, so I have this math problem here: "The opposite of 2023 is ( )", and there are four options: A is 1/2023, B is -1/2023, C is 2023, and D is -2023. I need to figure out which one is the correct answer.First, I should understand what the question is asking. It's asking for the "opposite" of 2023. Now, the term "opposite" in math can sometimes be a bit ambiguous because it can refer to different things depending on the context. But in this case, since it's a number, I think it's referring to the additive inverse, which is the number that, when added to the original number, gives zero. So, the additive inverse of 2023 would be the number that, when you add it to 2023, you get zero.Let me write that down to visualize it better. If I have 2023, and I want to find its opposite, which is x, then:2023 + x = 0To solve for x, I can subtract 2023 from both sides:x = -2023Okay, so that makes sense. The opposite of 2023 is -2023. That would be option D.But wait, let me double-check to make sure I'm not missing anything. Sometimes, "opposite" can also refer to the multiplicative inverse, which is the reciprocal of the number. The reciprocal of 2023 would be 1/2023, which is option A. So, I need to be clear about which kind of opposite they're asking for.Looking back at the question, it just says "the opposite of 2023." Without any additional context, in most cases, especially in basic math problems, "opposite" refers to the additive inverse rather than the multiplicative inverse. The additive inverse is more commonly referred to as the opposite because it's the number that cancels out the original number in addition.To confirm, let's think about what each option represents:- Option A: 1/2023 is the reciprocal or multiplicative inverse of 2023. If you multiply 2023 by 1/2023, you get 1.- Option B: -1/2023 is the negative reciprocal. It's not commonly referred to as the opposite.- Option C: 2023 is the number itself, so that's not the opposite.- Option D: -2023 is the additive inverse, which is the number that, when added to 2023, gives zero.Given that, it seems clear that the intended answer is D: -2023.But just to be thorough, let's consider if there's any other interpretation of "opposite" that might lead to a different answer. For example, in some contexts, especially in more advanced math, "opposite" could refer to the negative of a number, which is essentially the same as the additive inverse. So, in that sense, -2023 is still the opposite.Another thought: sometimes, in geometry, opposite can refer to something like the opposite side of a polygon or the opposite angle, but that doesn't apply here because we're dealing with a number, not a geometric figure.Also, in terms of vectors, the opposite vector would point in the exact opposite direction, which again would be the negative of the original vector. So, even in that context, the opposite would be -2023.Therefore, after considering different interpretations and confirming through various mathematical contexts, I'm confident that the opposite of 2023 is -2023, which is option D.

answer：Okay, so I have this problem where two parabolas intersect, and I need to find the difference between their x-coordinates of intersection, specifically r - p where r is greater than or equal to p. The equations given are y = 3x² - 6x + 3 and y = -x² - 3x + 3. Hmm, let me think about how to approach this.First, I remember that to find the points where two graphs intersect, I need to set their equations equal to each other because at the points of intersection, both y-values are the same for the same x-value. So, I can set 3x² - 6x + 3 equal to -x² - 3x + 3. That should give me an equation that I can solve for x.Let me write that down:3x² - 6x + 3 = -x² - 3x + 3Now, I need to solve for x. To do that, I should bring all the terms to one side so that I can have a quadratic equation equal to zero. Let me add x² + 3x - 3 to both sides to move those terms over. Wait, actually, maybe it's easier if I subtract the right side from both sides. Let me try that:3x² - 6x + 3 - (-x² - 3x + 3) = 0Simplifying that, I distribute the negative sign:3x² - 6x + 3 + x² + 3x - 3 = 0Now, combine like terms. The x² terms: 3x² + x² is 4x². The x terms: -6x + 3x is -3x. The constants: 3 - 3 is 0. So, putting it all together:4x² - 3x = 0Hmm, that's a quadratic equation. I can factor this to find the solutions for x. Let me factor out an x:x(4x - 3) = 0So, according to the zero product property, either x = 0 or 4x - 3 = 0. Solving 4x - 3 = 0 gives x = 3/4.Therefore, the x-coordinates where the parabolas intersect are x = 0 and x = 3/4. Since r is greater than or equal to p, I can say that p = 0 and r = 3/4.Now, the question asks for r - p, which is 3/4 - 0, so that's just 3/4.Wait, let me double-check my steps to make sure I didn't make a mistake. I set the equations equal, moved all terms to one side, combined like terms, factored, and solved for x. It seems correct. The quadratic simplified nicely, and the solutions are straightforward. So, I think I did it right.Just to be thorough, maybe I should plug these x-values back into one of the original equations to find the corresponding y-values and ensure that they are indeed points of intersection.Let's take x = 0. Plugging into the first equation: y = 3(0)² - 6(0) + 3 = 0 - 0 + 3 = 3. So, the point is (0, 3).Now, plugging x = 0 into the second equation: y = -(0)² - 3(0) + 3 = 0 - 0 + 3 = 3. Yep, same y-value, so (0, 3) is correct.Now, x = 3/4. Plugging into the first equation: y = 3*(3/4)² - 6*(3/4) + 3. Let's calculate that step by step.First, (3/4)² is 9/16. Multiply by 3: 27/16.Next, 6*(3/4) is 18/4, which simplifies to 9/2 or 4.5.So, putting it all together: 27/16 - 9/2 + 3.Convert everything to sixteenths to add them up:27/16 - 72/16 + 48/16 = (27 - 72 + 48)/16 = (3)/16.So, y = 3/16 when x = 3/4.Now, let's check the second equation with x = 3/4: y = -(3/4)² - 3*(3/4) + 3.Calculating each term:(3/4)² is 9/16, so negative of that is -9/16.3*(3/4) is 9/4.So, y = -9/16 - 9/4 + 3.Convert to sixteenths:-9/16 - 36/16 + 48/16 = (-9 - 36 + 48)/16 = (3)/16.Same result, so y = 3/16. That confirms that (3/4, 3/16) is indeed another point of intersection.Therefore, my solutions for x are correct, and the difference r - p is 3/4 - 0 = 3/4.I think I've covered all the steps and double-checked my work. I don't see any mistakes, so I feel confident that the answer is 3/4.**Final Answer**The value of ( r - p ) is boxed{dfrac{3}{4}}.

answer：Alright, so I have this problem here: Expand (2 - x)^5 and express it as a polynomial a0 + a1x + a2x² + a3x³ + a4x⁴ + a5x⁵. Then, I need to find the sum of the coefficients a1 through a5. Hmm, okay, let me think about how to approach this.First off, I remember that when you have a polynomial expressed in terms of x, the coefficients can be found using the binomial theorem. The binomial theorem allows us to expand expressions like (a + b)^n into a sum of terms involving binomial coefficients. In this case, our expression is (2 - x)^5, so a is 2 and b is -x, and n is 5.But wait, do I really need to expand the entire polynomial to find the coefficients? That might take some time, especially since it's to the fifth power. Maybe there's a shortcut. I recall that if you plug in x = 1 into a polynomial, you get the sum of all its coefficients. Let me verify that.If I set x = 1 in (2 - x)^5, I get (2 - 1)^5 = 1^5 = 1. On the other side, plugging x = 1 into the polynomial gives a0 + a1*1 + a2*1² + a3*1³ + a4*1⁴ + a5*1⁵, which simplifies to a0 + a1 + a2 + a3 + a4 + a5. So, that means a0 + a1 + a2 + a3 + a4 + a5 = 1.But the problem is asking for a1 + a2 + a3 + a4 + a5, which is just the sum of all coefficients except a0. So, if I can find a0 separately, I can subtract it from the total sum to get the desired result.How do I find a0? Well, a0 is the constant term of the polynomial, which is the term when x = 0. So, if I plug x = 0 into (2 - x)^5, I get (2 - 0)^5 = 2^5 = 32. Therefore, a0 = 32.Now, going back to the earlier result where a0 + a1 + a2 + a3 + a4 + a5 = 1, I can substitute a0 with 32. That gives me 32 + a1 + a2 + a3 + a4 + a5 = 1. To find a1 + a2 + a3 + a4 + a5, I just subtract 32 from both sides:a1 + a2 + a3 + a4 + a5 = 1 - 32 = -31.So, the sum of the coefficients from a1 to a5 is -31. That seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.1. Plugging x = 1 into (2 - x)^5 gives the sum of all coefficients, which is 1.2. Plugging x = 0 into (2 - x)^5 gives the constant term a0, which is 32.3. Subtracting a0 from the total sum gives the sum of the remaining coefficients: 1 - 32 = -31.Everything seems to check out. I don't think I made any calculation errors. Just to be thorough, maybe I can expand (2 - x)^5 using the binomial theorem and verify the coefficients individually.The binomial expansion of (a + b)^n is the sum from k=0 to n of (n choose k) * a^(n - k) * b^k. In this case, a = 2, b = -x, and n = 5.So, expanding (2 - x)^5:= (5 choose 0)*2^5*(-x)^0 + (5 choose 1)*2^4*(-x)^1 + (5 choose 2)*2^3*(-x)^2 + (5 choose 3)*2^2*(-x)^3 + (5 choose 4)*2^1*(-x)^4 + (5 choose 5)*2^0*(-x)^5Calculating each term:1. (5 choose 0) = 1, 2^5 = 32, (-x)^0 = 1. So, first term is 32.2. (5 choose 1) = 5, 2^4 = 16, (-x)^1 = -x. So, second term is 5*16*(-x) = -80x.3. (5 choose 2) = 10, 2^3 = 8, (-x)^2 = x². So, third term is 10*8*x² = 80x².4. (5 choose 3) = 10, 2^2 = 4, (-x)^3 = -x³. So, fourth term is 10*4*(-x³) = -40x³.5. (5 choose 4) = 5, 2^1 = 2, (-x)^4 = x⁴. So, fifth term is 5*2*x⁴ = 10x⁴.6. (5 choose 5) = 1, 2^0 = 1, (-x)^5 = -x⁵. So, sixth term is 1*1*(-x⁵) = -x⁵.Putting it all together:(2 - x)^5 = 32 - 80x + 80x² - 40x³ + 10x⁴ - x⁵.So, the coefficients are:- a0 = 32- a1 = -80- a2 = 80- a3 = -40- a4 = 10- a5 = -1Adding a1 through a5: -80 + 80 - 40 + 10 - 1.Let me compute that step by step:- Start with -80 + 80 = 0- 0 - 40 = -40- -40 + 10 = -30- -30 - 1 = -31Yep, that's the same result as before. So, both methods confirm that a1 + a2 + a3 + a4 + a5 = -31. I feel confident that this is the correct answer.I guess the key takeaway here is that instead of expanding the entire polynomial, which can be time-consuming, especially for higher powers, you can use substitution to find the sum of coefficients. Plugging in x = 1 gives the total sum, and plugging in x = 0 gives the constant term. Subtracting the constant term from the total sum gives the sum of the remaining coefficients. That's a useful shortcut to remember for similar problems.Another thing to note is that the binomial coefficients alternate in sign here because of the negative x term. So, every other term in the expansion is negative, which affects the coefficients. But since we're only summing the coefficients, regardless of their sign, the negative signs just factor into the total sum.It's also interesting to see how the coefficients relate to each other. For example, a1 is -80, which is quite a large negative number, but a2 is +80, which cancels it out when you add them together. Then, a3 is -40, which brings the sum down, and a4 is +10, which slightly increases it, and finally, a5 is -1, bringing the total to -31. It's like a balancing act between the positive and negative coefficients.I wonder if there's another way to approach this problem without expanding or substituting. Maybe using calculus? For example, taking derivatives to find coefficients, but that seems more complicated for such a straightforward problem. I think the substitution method is the most efficient here.Also, thinking about the graph of the polynomial, (2 - x)^5 is a fifth-degree polynomial, which means it will have a certain shape with up to five real roots. But since all the roots are at x = 2, it's actually a single root with multiplicity five. So, the graph will touch the x-axis at x = 2 and turn around, but since it's an odd degree, it will go to negative infinity as x approaches positive infinity and positive infinity as x approaches negative infinity.But maybe that's getting a bit off track. The main point is that by using substitution, I can efficiently find the sum of the coefficients without having to compute each one individually. It's a neat trick that saves time, especially in exams or when dealing with higher powers.In summary, the steps I took were:1. Recognize that substituting x = 1 gives the sum of all coefficients.2. Recognize that substituting x = 0 gives the constant term.3. Subtract the constant term from the total sum to get the desired sum of coefficients.4. Verified the result by expanding the polynomial using the binomial theorem.This problem reinforces the importance of understanding polynomial properties and the binomial theorem. It also highlights the value of looking for shortcuts and efficient methods to solve problems, rather than always resorting to brute-force calculations.I think I'm satisfied with my understanding of this problem. It makes sense, and I can see how these concepts apply in different contexts. Maybe next time, I can explore similar problems where substitution isn't as straightforward or where more complex manipulations are required. But for now, I feel confident in my solution.**Final Answer**The value of (a_{1} + a_{2} + a_{3} + a_{4} + a_{5}) is boxed{-31}.