answer：Alright, so I have this problem here: Let ( p ) be a prime number. If there exists a positive integer ( n ) such that ( p ) divides ( n^{2} + 7n + 23 ), then the minimum value of ( p ) is ______. Hmm, okay, I need to find the smallest prime number ( p ) that divides ( n^2 + 7n + 23 ) for some positive integer ( n ).First, I should understand what the problem is asking. It wants the smallest prime ( p ) such that when you plug in some positive integer ( n ) into the quadratic expression ( n^2 + 7n + 23 ), the result is divisible by ( p ). So, essentially, I need to find the smallest prime ( p ) for which the equation ( n^2 + 7n + 23 equiv 0 mod p ) has a solution.I think the best way to approach this is to check small prime numbers one by one and see if there's an integer ( n ) that satisfies the equation modulo each prime. If I can find such an ( n ) for a prime ( p ), then ( p ) is a candidate, and I need to find the smallest one.Let me start with the smallest primes: 2, 3, 5, 7, 11, etc.Starting with ( p = 2 ):I need to check if there's an ( n ) such that ( n^2 + 7n + 23 equiv 0 mod 2 ). Let's compute this modulo 2.Simplify each term modulo 2:- ( n^2 mod 2 ): Since any square modulo 2 is either 0 or 1. If ( n ) is even, ( n^2 equiv 0 mod 2 ); if ( n ) is odd, ( n^2 equiv 1 mod 2 ).- ( 7n mod 2 ): 7 is odd, so ( 7n equiv n mod 2 ). So, if ( n ) is even, ( 7n equiv 0 mod 2 ); if ( n ) is odd, ( 7n equiv 1 mod 2 ).- ( 23 mod 2 ): 23 is odd, so ( 23 equiv 1 mod 2 ).Putting it all together:If ( n ) is even:( n^2 + 7n + 23 equiv 0 + 0 + 1 equiv 1 mod 2 ).If ( n ) is odd:( n^2 + 7n + 23 equiv 1 + 1 + 1 equiv 3 equiv 1 mod 2 ).So, regardless of whether ( n ) is even or odd, ( n^2 + 7n + 23 equiv 1 mod 2 ). Therefore, ( p = 2 ) cannot divide ( n^2 + 7n + 23 ) because the result is always 1 modulo 2, which is never 0. So, ( p ) cannot be 2.Next, ( p = 3 ):Again, check if there's an ( n ) such that ( n^2 + 7n + 23 equiv 0 mod 3 ). Let's compute each term modulo 3.Simplify each term modulo 3:- ( n^2 mod 3 ): Squares modulo 3 can be 0, 1, or 1. Specifically, ( 0^2 equiv 0 ), ( 1^2 equiv 1 ), ( 2^2 equiv 4 equiv 1 mod 3 ).- ( 7n mod 3 ): 7 divided by 3 is 2 with a remainder of 1, so ( 7 equiv 1 mod 3 ). Therefore, ( 7n equiv n mod 3 ).- ( 23 mod 3 ): 23 divided by 3 is 7 with a remainder of 2, so ( 23 equiv 2 mod 3 ).Putting it all together:( n^2 + 7n + 23 equiv n^2 + n + 2 mod 3 ).Now, let's check for ( n = 0, 1, 2 mod 3 ) since modulo 3 cycles every 3 numbers.- If ( n equiv 0 mod 3 ): ( 0^2 + 0 + 2 equiv 0 + 0 + 2 equiv 2 mod 3 ). - If ( n equiv 1 mod 3 ): ( 1^2 + 1 + 2 equiv 1 + 1 + 2 equiv 4 equiv 1 mod 3 ). - If ( n equiv 2 mod 3 ): ( 2^2 + 2 + 2 equiv 4 + 2 + 2 equiv 8 equiv 2 mod 3 ).So, in all cases, ( n^2 + n + 2 mod 3 ) is either 1 or 2, never 0. Therefore, ( p = 3 ) cannot divide ( n^2 + 7n + 23 ) for any ( n ). So, ( p ) cannot be 3.Moving on to ( p = 5 ):Check if there's an ( n ) such that ( n^2 + 7n + 23 equiv 0 mod 5 ).Simplify each term modulo 5:- ( n^2 mod 5 ): Squares modulo 5 can be 0, 1, 4, 4, 1 for ( n = 0, 1, 2, 3, 4 ) respectively.- ( 7n mod 5 ): 7 divided by 5 is 1 with a remainder of 2, so ( 7 equiv 2 mod 5 ). Therefore, ( 7n equiv 2n mod 5 ).- ( 23 mod 5 ): 23 divided by 5 is 4 with a remainder of 3, so ( 23 equiv 3 mod 5 ).Putting it all together:( n^2 + 7n + 23 equiv n^2 + 2n + 3 mod 5 ).Now, let's check for ( n = 0, 1, 2, 3, 4 mod 5 ):- ( n = 0 ): ( 0 + 0 + 3 equiv 3 mod 5 ). - ( n = 1 ): ( 1 + 2 + 3 equiv 6 equiv 1 mod 5 ). - ( n = 2 ): ( 4 + 4 + 3 equiv 11 equiv 1 mod 5 ). - ( n = 3 ): ( 9 + 6 + 3 equiv 18 equiv 3 mod 5 ). - ( n = 4 ): ( 16 + 8 + 3 equiv 27 equiv 2 mod 5 ).So, the results modulo 5 are 3, 1, 1, 3, 2. None of these are 0. Therefore, ( p = 5 ) cannot divide ( n^2 + 7n + 23 ) for any ( n ). So, ( p ) cannot be 5.Next, ( p = 7 ):Check if there's an ( n ) such that ( n^2 + 7n + 23 equiv 0 mod 7 ).Simplify each term modulo 7:- ( n^2 mod 7 ): Squares modulo 7 can be 0, 1, 2, 4, 2, 1, 0 for ( n = 0, 1, 2, 3, 4, 5, 6 ) respectively.- ( 7n mod 7 ): 7 is 0 modulo 7, so ( 7n equiv 0 mod 7 ).- ( 23 mod 7 ): 23 divided by 7 is 3 with a remainder of 2, so ( 23 equiv 2 mod 7 ).Putting it all together:( n^2 + 7n + 23 equiv n^2 + 0 + 2 equiv n^2 + 2 mod 7 ).Now, let's check for ( n = 0, 1, 2, 3, 4, 5, 6 mod 7 ):- ( n = 0 ): ( 0 + 2 equiv 2 mod 7 ). - ( n = 1 ): ( 1 + 2 equiv 3 mod 7 ). - ( n = 2 ): ( 4 + 2 equiv 6 mod 7 ). - ( n = 3 ): ( 9 + 2 equiv 11 equiv 4 mod 7 ). - ( n = 4 ): ( 16 + 2 equiv 18 equiv 4 mod 7 ). - ( n = 5 ): ( 25 + 2 equiv 27 equiv 6 mod 7 ). - ( n = 6 ): ( 36 + 2 equiv 38 equiv 3 mod 7 ).So, the results modulo 7 are 2, 3, 6, 4, 4, 6, 3. None of these are 0. Therefore, ( p = 7 ) cannot divide ( n^2 + 7n + 23 ) for any ( n ). So, ( p ) cannot be 7.Moving on to ( p = 11 ):Check if there's an ( n ) such that ( n^2 + 7n + 23 equiv 0 mod 11 ).Simplify each term modulo 11:- ( n^2 mod 11 ): Squares modulo 11 can be 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1 for ( n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ) respectively.- ( 7n mod 11 ): 7 is 7 modulo 11, so ( 7n mod 11 ) will cycle through multiples of 7.- ( 23 mod 11 ): 23 divided by 11 is 2 with a remainder of 1, so ( 23 equiv 1 mod 11 ).Putting it all together:( n^2 + 7n + 23 equiv n^2 + 7n + 1 mod 11 ).Now, let's check for ( n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 mod 11 ):- ( n = 0 ): ( 0 + 0 + 1 equiv 1 mod 11 ). - ( n = 1 ): ( 1 + 7 + 1 equiv 9 mod 11 ). - ( n = 2 ): ( 4 + 14 + 1 equiv 19 equiv 8 mod 11 ). - ( n = 3 ): ( 9 + 21 + 1 equiv 31 equiv 9 mod 11 ). - ( n = 4 ): ( 16 + 28 + 1 equiv 45 equiv 1 mod 11 ). - ( n = 5 ): ( 25 + 35 + 1 equiv 61 equiv 6 mod 11 ). - ( n = 6 ): ( 36 + 42 + 1 equiv 79 equiv 2 mod 11 ). - ( n = 7 ): ( 49 + 49 + 1 equiv 99 equiv 0 mod 11 ). Oh, wait! When ( n = 7 mod 11 ), the expression ( n^2 + 7n + 1 equiv 0 mod 11 ). That means ( p = 11 ) divides ( n^2 + 7n + 23 ) when ( n = 7 ). So, ( p = 11 ) is a candidate.But just to be thorough, let's check the rest of the ( n ) values to see if there's a smaller ( n ) that also works.- ( n = 8 ): ( 64 + 56 + 1 equiv 121 equiv 0 mod 11 ). Wait, ( n = 8 ) also gives 0 modulo 11. Interesting, so both ( n = 7 ) and ( n = 8 ) satisfy the equation modulo 11.But regardless, since we found that ( p = 11 ) works, and we've already checked all smaller primes (2, 3, 5, 7) and none of them work, ( p = 11 ) must be the smallest prime that satisfies the condition.Just to double-check, let's plug ( n = 7 ) into the original expression:( 7^2 + 7*7 + 23 = 49 + 49 + 23 = 121 ).121 divided by 11 is 11, so yes, 11 divides 121. Similarly, for ( n = 8 ):( 8^2 + 7*8 + 23 = 64 + 56 + 23 = 143 ).143 divided by 11 is 13, so 11 divides 143 as well. So, both ( n = 7 ) and ( n = 8 ) work, confirming that ( p = 11 ) is indeed a prime that divides the expression.Since we've checked all primes smaller than 11 and none of them work, the minimum value of ( p ) is 11.**Final Answer**The minimum value of ( p ) is boxed{11}.

answer：Okay, so I have this problem about a perpendicular equilibrium plane, which I think is like a rectangular box or something similar. The problem says that the diagonal of this plane forms angles α₁, α₂, α₃ with the side-diagonals from the common vertex. Then, there are two equations given:sin²α₁ + sin²α₂ + sin²α₃ = 1andcos²α₁ + cos²α₂ + cos²α₃ = 2I need to understand why these equations hold true. Let me try to visualize this. Imagine a rectangular box with edges of lengths a, b, and c meeting at a common vertex. The space diagonal of this box would form angles with each of the edges. These angles are α₁, α₂, and α₃.First, I recall that in a rectangular box, the space diagonal d can be found using the Pythagorean theorem in three dimensions: d = √(a² + b² + c²). This diagonal makes angles with each of the edges a, b, and c. The sine of these angles would be the ratio of the opposite side to the hypotenuse, which in this case is the space diagonal d.So, sinα₁ = opposite/hypotenuse = a/d, similarly sinα₂ = b/d, and sinα₃ = c/d. Therefore, sin²α₁ = a²/d², sin²α₂ = b²/d², and sin²α₃ = c²/d². If I add these up:sin²α₁ + sin²α₂ + sin²α₃ = (a² + b² + c²)/d²But since d² = a² + b² + c², this simplifies to:(a² + b² + c²)/(a² + b² + c²) = 1So that's why the first equation holds: sin²α₁ + sin²α₂ + sin²α₃ = 1.Now, for the cosine part. I know that cos²θ + sin²θ = 1 for any angle θ. So, cos²α₁ = 1 - sin²α₁, cos²α₂ = 1 - sin²α₂, and cos²α₃ = 1 - sin²α₃. If I add these up:cos²α₁ + cos²α₂ + cos²α₃ = (1 - sin²α₁) + (1 - sin²α₂) + (1 - sin²α₃)This simplifies to:3 - (sin²α₁ + sin²α₂ + sin²α₃)From the first equation, we know that sin²α₁ + sin²α₂ + sin²α₃ = 1, so substituting that in:3 - 1 = 2Therefore, cos²α₁ + cos²α₂ + cos²α₃ = 2.Wait, but does this make sense? Cosine squared of angles adding up to 2? Let me think. Each cosine squared term is less than or equal to 1, so three of them adding up to 2 means that on average, each is about 2/3, which seems reasonable because the angles are between 0 and 90 degrees in a rectangular box.Let me check with a cube, where a = b = c. In a cube, the space diagonal makes equal angles with each edge. So, sinα = a/d, and since a = b = c, d = a√3. So sinα = a/(a√3) = 1/√3, so sin²α = 1/3. Therefore, sin²α₁ + sin²α₂ + sin²α₃ = 3*(1/3) = 1, which matches the first equation.For cosine, cosα = √(1 - sin²α) = √(1 - 1/3) = √(2/3). So cos²α = 2/3. Therefore, cos²α₁ + cos²α₂ + cos²α₃ = 3*(2/3) = 2, which matches the second equation.Okay, that makes sense. So in the case of a cube, the equations hold true. What about a different rectangular box? Let's say a = 3, b = 4, c = 12. Then d = √(9 + 16 + 144) = √169 = 13.So sinα₁ = 3/13, sin²α₁ = 9/169sinα₂ = 4/13, sin²α₂ = 16/169sinα₃ = 12/13, sin²α₃ = 144/169Adding them up: 9 + 16 + 144 = 169, so 169/169 = 1. That works.For cosine, cos²α₁ = 1 - 9/169 = 160/169cos²α₂ = 1 - 16/169 = 153/169cos²α₃ = 1 - 144/169 = 25/169Adding them up: 160 + 153 + 25 = 338, which is 2*169. So 338/169 = 2. That also works.Alright, so it seems consistent. The key idea is that in a rectangular box, the space diagonal forms angles with each edge, and the sum of the squares of the sines of these angles equals 1, while the sum of the squares of the cosines equals 2. This is because the space diagonal relates to the edges through the Pythagorean theorem, and the trigonometric identities tie the sine and cosine parts together.I think I've got it. The equations hold because of the properties of the space diagonal in a rectangular box and the fundamental trigonometric identity.

answer：First, I need to determine the number of new workers hired, which is 25% of the original 852 employees.Calculating 25% of 852:25% of 852 = 0.25 × 852 = 213Next, I add the number of new hires to the original number of employees to find the total number of employees now:852 + 213 = 1065Therefore, the factory now has 1065 employees.

answer：First, I need to understand the equation (Diamond4_7 = Diamond1_8). The subscript numbers indicate the base of the numeral system being used. This means that (Diamond4_7) is a number in base 7, and (Diamond1_8) is a number in base 8.To solve for (Diamond), I'll convert both sides of the equation to base 10. For the base 7 number (Diamond4_7), the conversion to base 10 is calculated as:[Diamond times 7^1 + 4 times 7^0 = 7Diamond + 4]Similarly, for the base 8 number (Diamond1_8), the conversion to base 10 is:[Diamond times 8^1 + 1 times 8^0 = 8Diamond + 1]Now, I can set up the equation in base 10:[7Diamond + 4 = 8Diamond + 1]To isolate (Diamond), I'll subtract (7Diamond) from both sides:[4 = Diamond + 1]Finally, subtracting 1 from both sides gives:[Diamond = 3]Therefore, the value of (Diamond) that satisfies the equation is 3.

answer：Alright, so I've got this problem here about inserting n real numbers between 1 and 2 to form an increasing geometric sequence. Then, I need to find the product of these n+2 numbers, which is called A_n, and then use that to find a_n, which is the logarithm base 2 of A_n. Part (1) asks for the sum of the first n terms of the sequence {A_n}, denoted as S_n. Hmm, okay. Let me try to break this down step by step.First, I know that in a geometric sequence, each term is multiplied by a common ratio r. So, if I have n+2 terms starting from 1 and ending at 2, the sequence would look like 1, r, r^2, ..., r^{n+1} = 2. So, the last term is 2, which is equal to r^{n+1}. Therefore, r = 2^{1/(n+1)}. That makes sense because if you raise r to the power of n+1, you get 2.Now, the product A_n is the product of all these n+2 terms. So, A_n = 1 * r * r^2 * ... * r^{n+1}. That's a geometric series product. I remember that the product of a geometric sequence can be expressed as (r^{(n+2)(n+1)/2}) because each term is r^k where k goes from 0 to n+1. So, multiplying them all together gives r^{0+1+2+...+(n+1)} which is r^{(n+1)(n+2)/2}.Since r = 2^{1/(n+1)}, substituting that in, A_n becomes [2^{1/(n+1)}]^{(n+1)(n+2)/2} = 2^{(n+2)/2}. So, A_n = 2^{(n+2)/2}. Wait, that seems a bit too straightforward. Let me verify. If I have n+2 terms, the exponents of r go from 0 to n+1, so the sum of exponents is (n+1)(n+2)/2. Then, since r = 2^{1/(n+1)}, raising it to that exponent gives 2^{(n+2)/2}. Yeah, that seems correct.Now, a_n is log base 2 of A_n, so a_n = log_2(2^{(n+2)/2}) = (n+2)/2. That simplifies things because a_n is just a linear function of n.Moving on to part (1), I need to find the sum S_n of the first n terms of {A_n}. So, S_n = A_1 + A_2 + ... + A_n. Since each A_k = 2^{(k+2)/2}, this is a geometric series with the first term A_1 = 2^{(1+2)/2} = 2^{3/2} = 2*sqrt(2). The common ratio is the ratio of A_{k+1}/A_k = [2^{(k+3)/2}]/[2^{(k+2)/2}] = 2^{1/2} = sqrt(2). So, S_n is the sum of a geometric series with first term 2*sqrt(2) and common ratio sqrt(2) for n terms. The formula for the sum of a geometric series is S_n = a1*(1 - r^n)/(1 - r). Plugging in the values, S_n = 2*sqrt(2)*(1 - (sqrt(2))^n)/(1 - sqrt(2)). But 1 - sqrt(2) is negative, so to make it look nicer, I can factor out a negative sign from numerator and denominator: S_n = 2*sqrt(2)*[(sqrt(2))^n - 1]/(sqrt(2) - 1). Then, rationalizing the denominator, multiply numerator and denominator by (sqrt(2) + 1): Denominator becomes (sqrt(2) - 1)(sqrt(2) + 1) = 2 - 1 = 1.Numerator becomes 2*sqrt(2)*[(sqrt(2))^n - 1]*(sqrt(2) + 1). Wait, hold on. Let me recast that. If I have S_n = 2*sqrt(2)*(1 - (sqrt(2))^n)/(1 - sqrt(2)) = 2*sqrt(2)*[(sqrt(2))^n - 1]/(sqrt(2) - 1). Multiplying numerator and denominator by (sqrt(2) + 1):Numerator: 2*sqrt(2)*[(sqrt(2))^n - 1]*(sqrt(2) + 1)Denominator: (sqrt(2) - 1)(sqrt(2) + 1) = 1So, S_n = 2*sqrt(2)*(sqrt(2) + 1)*[(sqrt(2))^n - 1]Calculating 2*sqrt(2)*(sqrt(2) + 1):First, sqrt(2)*sqrt(2) = 2, so 2*sqrt(2)*sqrt(2) = 2*2 = 4Then, 2*sqrt(2)*1 = 2*sqrt(2)So, altogether, 4 + 2*sqrt(2). Therefore, S_n = (4 + 2*sqrt(2))*[(sqrt(2))^n - 1]So, that's the sum for part (1). Moving on to part (2), I need to find T_n = tan(a_2)*tan(a_4) + tan(a_4)*tan(a_6) + ... + tan(a_{2n})*tan(a_{2n+2})From part (1), we have a_n = (n + 2)/2. So, a_{2k} = (2k + 2)/2 = k + 1. Therefore, tan(a_{2k}) = tan(k + 1). So, the terms in T_n are tan(2)*tan(3) + tan(3)*tan(4) + ... + tan(n+1)*tan(n+2)Hmm, so T_n is the sum from k=1 to k=n of tan(k+1)*tan(k+2). Let me denote m = k+1, so T_n = sum_{m=2}^{n+1} tan(m)*tan(m+1)I remember that there's a trigonometric identity involving tan(A)tan(B). Let me recall: tan(A - B) = (tan A - tan B)/(1 + tan A tan B). If I rearrange this, tan A tan B = (tan A - tan B)/tan(A - B) - 1. In our case, if I set A = m+1 and B = m, then A - B = 1. So, tan(m+1 - m) = tan(1) = (tan(m+1) - tan(m))/(1 + tan(m+1)tan(m)). Therefore, tan(m+1)tan(m) = [tan(m+1) - tan(m)] / tan(1) - 1.So, each term tan(m)tan(m+1) can be written as [tan(m+1) - tan(m)] / tan(1) - 1.Therefore, T_n = sum_{m=2}^{n+1} [ (tan(m+1) - tan(m)) / tan(1) - 1 ]This can be split into two sums: sum_{m=2}^{n+1} [ (tan(m+1) - tan(m)) / tan(1) ] - sum_{m=2}^{n+1} 1The first sum is a telescoping series. Let's write out the terms:For m=2: (tan(3) - tan(2))/tan(1)For m=3: (tan(4) - tan(3))/tan(1)...For m=n+1: (tan(n+2) - tan(n+1))/tan(1)Adding these up, all the intermediate terms cancel out, leaving [tan(n+2) - tan(2)] / tan(1)The second sum is just the number of terms, which is from m=2 to m=n+1, so that's n terms. Therefore, the second sum is n*1 = n.Putting it all together, T_n = [tan(n+2) - tan(2)] / tan(1) - nSo, that's the expression for T_n.Let me recap to make sure I didn't make any mistakes. Starting from the product A_n, I found it's 2^{(n+2)/2}, then a_n is (n+2)/2. For S_n, recognizing it's a geometric series with ratio sqrt(2), and summing it up gives (4 + 2sqrt(2))[(sqrt(2))^n - 1]. For T_n, using the identity for tan(A)tan(B), I expressed each term as a difference involving tan(m+1) - tan(m), which telescopes, and then subtracted n from the result. So, T_n ends up being [tan(n+2) - tan(2)] / tan(1) - n.I think that's solid. I don't see any errors in the steps. The telescoping worked out nicely, and the trigonometric identity was applied correctly. So, I feel confident about these answers.**Final Answer**(1) The sum of the first ( n ) terms of the sequence ( {A_n} ) is ( boxed{(4 + 2sqrt{2})[(sqrt{2})^n - 1]} ).(2) The value of ( T_n ) is ( boxed{frac{tan(n+2) - tan 2}{tan 1} - n} ).

answer：Okay, so I'm trying to figure out the variable cost per book for this publishing company. Let's see, they have a fixed cost of 56,430 for editing and printing. Then there's this variable cost per book, which I guess is like the cost to produce each additional book after the initial setup. They sell each book to the store for 21.75, and they need to print and sell 4,180 books to break even, meaning the production costs equal the money they make from sales.Alright, so first, I need to understand what "break even" means here. It means that the total revenue from selling the books equals the total production costs, which include both the fixed costs and the variable costs. So, total revenue equals fixed costs plus variable costs.Let me write that down:Total Revenue = Fixed Costs + Variable CostsI know the fixed costs are 56,430. The selling price per book is 21.75, and they need to sell 4,180 books to break even. So, I can calculate the total revenue by multiplying the selling price by the number of books sold.Total Revenue = Selling Price per Book × Number of BooksTotal Revenue = 21.75 × 4,180Let me do that multiplication. Hmm, 21.75 times 4,180. Let me break it down:First, 20 times 4,180 is 83,600.Then, 1.75 times 4,180. Let's see, 1 times 4,180 is 4,180, and 0.75 times 4,180 is 3,135.So, 4,180 + 3,135 = 7,315.Adding that to 83,600 gives me 90,915.So, Total Revenue is 90,915.Now, according to the break-even point, Total Revenue equals Fixed Costs plus Variable Costs. So, I can write:90,915 = 56,430 + Variable CostsTo find Variable Costs, I need to subtract Fixed Costs from Total Revenue.Variable Costs = Total Revenue - Fixed CostsVariable Costs = 90,915 - 56,430Let me do that subtraction. 90,915 minus 56,430. Let's see, 90,915 minus 50,000 is 40,915, and then minus 6,430 more would be 34,485.So, Variable Costs = 34,485.Now, this is the total variable cost for all 4,180 books. To find the variable cost per book, I need to divide the total variable cost by the number of books.Variable Cost per Book = Total Variable Costs / Number of BooksVariable Cost per Book = 34,485 / 4,180Let me calculate that. 34,485 divided by 4,180. Hmm, let's see. 4,180 times 8 is 33,440. Subtracting that from 34,485 gives me 1,045. Then, 4,180 times 0.25 is 1,045. So, 8 plus 0.25 is 8.25.Therefore, the variable cost per book is 8.25.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting with the total revenue: 21.75 times 4,180 equals 90,915. That seems right. Then, subtracting the fixed costs: 90,915 minus 56,430 equals 34,485. That also looks correct. Finally, dividing 34,485 by 4,180 gives me 8.25. Yeah, that all adds up.So, the variable cost per book is 8.25.