answer：First, I need to determine how many books to place on each shelf. Let’s denote the number of books on the second shelf as ( y ). According to the problem, the first shelf should have four more than twice the number of books on the second shelf, so the number of books on the first shelf is ( 2y + 4 ).Since the total number of books is 80, I can set up the equation:[(2y + 4) + y = 80]Combining like terms gives:[3y + 4 = 80]Subtracting 4 from both sides:[3y = 76]Dividing both sides by 3:[y = frac{76}{3} approx 25.overline{3}]Since the number of books must be a whole number, I'll round ( y ) to 25 books on the second shelf. Then, the first shelf will have:[2(25) + 4 = 54 text{ books}]This totals 79 books, which is one less than the total. To adjust, I'll add one more book to the first shelf, making it 55 books on the first shelf and 25 on the second shelf.Next, to find out how many pamphlets are needed for 10 attendees, I'll substitute ( x = 10 ) into the quadratic expression:[p(10) = 10^2 + 8(10) + 15 = 100 + 80 + 15 = 195]So, the librarian needs to prepare 195 pamphlets.

answer：Okay, so I have this geometry problem here about a square PQRS with side length 2. Points M and N are the midpoints of sides SR and RQ, respectively. I need to find the value of cos(angle MPN). The answer choices are given, so that might help if I get stuck.First, I should probably sketch the square to visualize the problem. Let me imagine square PQRS. Let's label the square so that P is at the top-left corner, Q is at the top-right, R is at the bottom-right, and S is at the bottom-left. So, PQ is the top side, QR is the right side, RS is the bottom side, and SP is the left side.Points M and N are midpoints. So, M is the midpoint of SR, which is the bottom side. Since the side length is 2, the midpoint M would be 1 unit from both S and R. Similarly, N is the midpoint of RQ, which is the right side. So, N is 1 unit from both R and Q.Now, I need to find cos(angle MPN). So, points M, P, and N form triangle MPN, and I need the cosine of the angle at point P.Hmm, maybe coordinate geometry would help here. Assigning coordinates to each point might make it easier to calculate the vectors or use the dot product formula to find the cosine of the angle.Let's set up a coordinate system. Let me place point P at (0, 2), Q at (2, 2), R at (2, 0), and S at (0, 0). That way, the square is in the first quadrant with side length 2.So, point M is the midpoint of SR. Since S is (0, 0) and R is (2, 0), the midpoint M would be at ((0+2)/2, (0+0)/2) = (1, 0).Similarly, point N is the midpoint of RQ. R is (2, 0) and Q is (2, 2), so midpoint N is at ((2+2)/2, (0+2)/2) = (2, 1).Now, point P is (0, 2). So, we have points M(1, 0), P(0, 2), and N(2, 1).I need to find angle MPN, which is the angle at point P between points M and N. So, vectors PM and PN are involved.Let me find the coordinates of vectors PM and PN.Vector PM goes from P(0, 2) to M(1, 0). So, the vector PM is (1 - 0, 0 - 2) = (1, -2).Vector PN goes from P(0, 2) to N(2, 1). So, the vector PN is (2 - 0, 1 - 2) = (2, -1).Now, to find the angle between vectors PM and PN, I can use the dot product formula:cos(theta) = (PM · PN) / (|PM| |PN|)First, compute the dot product PM · PN.PM · PN = (1)(2) + (-2)(-1) = 2 + 2 = 4.Next, compute the magnitudes of PM and PN.|PM| = sqrt(1^2 + (-2)^2) = sqrt(1 + 4) = sqrt(5).|PN| = sqrt(2^2 + (-1)^2) = sqrt(4 + 1) = sqrt(5).So, cos(theta) = 4 / (sqrt(5) * sqrt(5)) = 4 / 5.Wait, so cos(angle MPN) is 4/5. Looking at the answer choices, option A is 4/5. So, is that the answer?But wait, let me double-check my steps to make sure I didn't make a mistake.1. Assigning coordinates: P(0,2), Q(2,2), R(2,0), S(0,0). That seems correct.2. Midpoints: M is midpoint of SR, which is (1,0). N is midpoint of RQ, which is (2,1). That seems correct.3. Vectors PM and PN: PM is from P(0,2) to M(1,0): (1, -2). PN is from P(0,2) to N(2,1): (2, -1). Correct.4. Dot product: (1)(2) + (-2)(-1) = 2 + 2 = 4. Correct.5. Magnitudes: |PM| = sqrt(1 + 4) = sqrt(5). |PN| = sqrt(4 + 1) = sqrt(5). Correct.6. Cos(theta) = 4 / (sqrt(5)*sqrt(5)) = 4/5. Correct.Hmm, seems solid. So, the answer should be 4/5, which is option A.Wait, but let me think again. Sometimes, in geometry problems, especially with squares, there might be a different approach or a trick that could give a different answer. Maybe using triangle properties or something else.Alternatively, perhaps using coordinates is the straightforward way, and I think I did that correctly. Maybe I can try another method to verify.Another approach could be to calculate the lengths of PM, PN, and MN, then use the Law of Cosines to find the angle at P.Let's try that.First, compute the lengths:PM: distance from P(0,2) to M(1,0).Distance formula: sqrt[(1-0)^2 + (0-2)^2] = sqrt[1 + 4] = sqrt(5).Similarly, PN: distance from P(0,2) to N(2,1).Distance formula: sqrt[(2-0)^2 + (1-2)^2] = sqrt[4 + 1] = sqrt(5).Now, distance MN: from M(1,0) to N(2,1).Distance formula: sqrt[(2-1)^2 + (1-0)^2] = sqrt[1 + 1] = sqrt(2).So, triangle MPN has sides PM = sqrt(5), PN = sqrt(5), and MN = sqrt(2).Now, using the Law of Cosines on triangle MPN:MN^2 = PM^2 + PN^2 - 2*PM*PN*cos(theta)Where theta is angle MPN.Plugging in the values:(sqrt(2))^2 = (sqrt(5))^2 + (sqrt(5))^2 - 2*(sqrt(5))*(sqrt(5))*cos(theta)Simplify:2 = 5 + 5 - 2*5*cos(theta)2 = 10 - 10*cos(theta)Subtract 10 from both sides:2 - 10 = -10*cos(theta)-8 = -10*cos(theta)Divide both sides by -10:(-8)/(-10) = cos(theta)8/10 = cos(theta)Simplify:4/5 = cos(theta)Same result as before. So, that confirms it.Therefore, the value of cos(angle MPN) is 4/5, which is option A.I think that's solid. I used two different methods: coordinate vectors with dot product and Law of Cosines. Both gave me the same answer, so I feel confident.**Final Answer**boxed{A}

answer：Okay, so I have this problem about an arithmetic sequence. The sum of the first n terms is given as S_n. They told me that S_5 is 35 and S_10 is 120, and I need to find S_20. Hmm, arithmetic sequences... I remember they have a common difference, right? So each term increases by the same amount.First, let me recall the formula for the sum of the first n terms of an arithmetic sequence. I think it's S_n = n/2 times (2a_1 + (n-1)d), where a_1 is the first term and d is the common difference. Alternatively, it can also be written as S_n = n times (a_1 + a_n)/2, since the average of the first and nth term multiplied by the number of terms gives the sum. Maybe both formulas can be useful here.Given that, I have two equations:1. S_5 = 352. S_10 = 120Let me write these out using the sum formula.For S_5:35 = 5/2 [2a_1 + (5 - 1)d]Simplify that:35 = (5/2)(2a_1 + 4d)Multiply both sides by 2 to eliminate the denominator:70 = 5(2a_1 + 4d)Divide both sides by 5:14 = 2a_1 + 4dI can simplify this equation by dividing both sides by 2:7 = a_1 + 2dSo that's equation (1): a_1 + 2d = 7.Now for S_10:120 = 10/2 [2a_1 + (10 - 1)d]Simplify:120 = 5(2a_1 + 9d)Divide both sides by 5:24 = 2a_1 + 9dSo that's equation (2): 2a_1 + 9d = 24.Now I have a system of two equations:1. a_1 + 2d = 72. 2a_1 + 9d = 24I need to solve for a_1 and d. Let me use substitution or elimination. Maybe elimination is easier here.If I multiply equation (1) by 2, I get:2a_1 + 4d = 14Now subtract this from equation (2):(2a_1 + 9d) - (2a_1 + 4d) = 24 - 14Simplify:5d = 10So d = 2.Now plug d = 2 back into equation (1):a_1 + 2*2 = 7a_1 + 4 = 7a_1 = 3.Okay, so the first term is 3 and the common difference is 2. Let me verify these values with the original sums.For S_5:Using the formula S_5 = 5/2 [2*3 + (5-1)*2] = 5/2 [6 + 8] = 5/2 *14 = 5*7 = 35. That's correct.For S_10:S_10 = 10/2 [2*3 + (10-1)*2] = 5 [6 + 18] = 5*24 = 120. That's also correct.Great, so a_1 = 3 and d = 2. Now, I need to find S_20.Using the sum formula again:S_20 = 20/2 [2*3 + (20 - 1)*2]Simplify:S_20 = 10 [6 + 38]Wait, 20 - 1 is 19, so 19*2 is 38. Then 6 + 38 is 44.So S_20 = 10 * 44 = 440.Let me double-check my calculations. 20/2 is 10. 2*3 is 6, 19*2 is 38. 6 + 38 is 44. 10*44 is indeed 440. Hmm, that seems straightforward.Alternatively, I can think about the sequence itself. The first term is 3, common difference is 2, so the terms go 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, and so on. The 20th term would be a_1 + 19d = 3 + 38 = 41. Then, the sum S_20 is 20*(first term + last term)/2 = 20*(3 + 41)/2 = 20*44/2 = 20*22 = 440. Yep, same result.Wait, just to make sure, maybe I can compute S_20 another way. Since I know S_10 is 120, maybe I can find the sum from 11th to 20th term and add it to S_10.The 11th term is a_11 = a_1 + 10d = 3 + 20 = 23. The 20th term is 41 as before. So the sum from 11th to 20th term is 10 terms starting at 23 and ending at 41. The average of 23 and 41 is (23 + 41)/2 = 64/2 = 32. So the sum is 10*32 = 320. Then S_20 = S_10 + 320 = 120 + 320 = 440. Same answer again. So I feel confident that 440 is correct.I think that's solid. I used two different methods to compute S_20 and both gave me the same result. So I don't think I made a mistake here.**Final Answer**The sum of the first 20 terms is boxed{440}.

answer：First, I'll define your favorite number as ( x ).According to the problem, Alexander's favorite number is 5 more than three times your favorite number, so Alexander's number is ( 3x + 5 ).The product of these two numbers is given as 60. Therefore, I can set up the equation:[x(3x + 5) = 60]Expanding this equation, I get:[3x^2 + 5x - 60 = 0]Next, I'll solve this quadratic equation using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]where ( a = 3 ), ( b = 5 ), and ( c = -60 ).Calculating the discriminant:[b^2 - 4ac = 5^2 - 4(3)(-60) = 25 + 720 = 745]Now, applying the quadratic formula:[x = frac{-5 pm sqrt{745}}{6}]This gives two potential solutions:[x = frac{-5 + sqrt{745}}{6} quad text{and} quad x = frac{-5 - sqrt{745}}{6}]Since a favorite number should be positive, I'll discard the negative solution:[x = frac{-5 + sqrt{745}}{6}]Finally, I'll verify the additional clue provided by Alexander. The difference between Alexander's favorite number and your favorite number is:[(3x + 5) - x = 2x + 5]Doubling this difference:[2(2x + 5) = 4x + 10]According to the clue, this equals 20:[4x + 10 = 20]Solving for ( x ):[4x = 10 quad Rightarrow quad x = frac{10}{4} = 2.5]However, this value of ( x ) does not match the solution found earlier, indicating that the additional clue is inconsistent with the initial problem.

answer：To solve the quadratic equation (2x^2 - 3x - 5 = 0), I'll use the quadratic formula. First, I'll identify the coefficients: (a = 2), (b = -3), and (c = -5). Plugging these into the formula, I get:[x = frac{-(-3) pm sqrt{(-3)^2 - 4 cdot 2 cdot (-5)}}{2 cdot 2}]Simplifying the discriminant:[sqrt{9 + 40} = sqrt{49} = 7]So the roots are:[x = frac{3 + 7}{4} = frac{10}{4} = frac{5}{2}][x = frac{3 - 7}{4} = frac{-4}{4} = -1]Next, to solve the inequality (2x^2 - 3x - 5 < 0), I'll analyze the quadratic function. Since the coefficient of (x^2) is positive, the parabola opens upwards. The inequality (2x^2 - 3x - 5 < 0) holds between the roots of the equation. Therefore, the solution is all real numbers between (-1) and (frac{5}{2}).

answer：Okay, so I have this problem about a rectangle with length ( a ) and width ( b ). The area is given as 6, and the perimeter is 12. I need to find the value of ( a^2 + b^2 ). Hmm, let me think about how to approach this.First, I remember that for a rectangle, the area is calculated by multiplying the length and the width. So, that gives me the equation:[ a times b = 6 ]And the perimeter of a rectangle is calculated by adding up all the sides, which is twice the length plus twice the width. So, the perimeter equation is:[ 2a + 2b = 12 ]I can simplify that perimeter equation by dividing both sides by 2, which gives me:[ a + b = 6 ]Alright, so now I have two equations:1. ( ab = 6 )2. ( a + b = 6 )I need to find ( a^2 + b^2 ). Hmm, I recall that there's a formula that relates ( a^2 + b^2 ) to ( (a + b)^2 ) and ( ab ). Let me try to remember it. I think it goes like this:[ (a + b)^2 = a^2 + 2ab + b^2 ]Yes, that's right. So, if I can find ( (a + b)^2 ) and subtract ( 2ab ), I should get ( a^2 + b^2 ). Let me write that down:[ a^2 + b^2 = (a + b)^2 - 2ab ]Great, so I already know ( a + b ) is 6, and ( ab ) is 6. Let me plug those values into the equation.First, calculate ( (a + b)^2 ):[ (6)^2 = 36 ]Then, calculate ( 2ab ):[ 2 times 6 = 12 ]Now, subtract the second result from the first:[ 36 - 12 = 24 ]So, ( a^2 + b^2 = 24 ). Let me check the options given:A: 30B: 24C: 25D: 13So, 24 is option B. Hmm, that seems straightforward, but let me just verify my steps to make sure I didn't make a mistake.Starting with the area and perimeter equations:1. ( ab = 6 )2. ( a + b = 6 )Using the identity ( a^2 + b^2 = (a + b)^2 - 2ab ), plugging in the known values:[ (6)^2 - 2(6) = 36 - 12 = 24 ]Yep, that looks correct. So, I think the answer is B: 24.But wait, just to be thorough, let me solve for ( a ) and ( b ) individually to see if they make sense. Maybe that will help solidify my answer.From the equation ( a + b = 6 ), I can express ( b ) in terms of ( a ):[ b = 6 - a ]Now, substitute this into the area equation ( ab = 6 ):[ a(6 - a) = 6 ]Expanding this:[ 6a - a^2 = 6 ]Rearranging terms:[ -a^2 + 6a - 6 = 0 ]Multiply both sides by -1 to make it a standard quadratic equation:[ a^2 - 6a + 6 = 0 ]Now, I can use the quadratic formula to solve for ( a ). The quadratic formula is:[ a = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]Wait, hold on, that might be confusing because the variable here is also ( a ). Let me rewrite the quadratic equation with different coefficients to avoid confusion.Let me denote the quadratic equation as:[ x^2 - 6x + 6 = 0 ]So, here, ( x ) is the variable, and the coefficients are:- ( A = 1 )- ( B = -6 )- ( C = 6 )Applying the quadratic formula:[ x = frac{-B pm sqrt{B^2 - 4AC}}{2A} ]Plugging in the values:[ x = frac{-(-6) pm sqrt{(-6)^2 - 4 times 1 times 6}}{2 times 1} ]Simplify step by step:First, calculate the numerator:- ( -(-6) = 6 )- ( (-6)^2 = 36 )- ( 4 times 1 times 6 = 24 )- So, the discriminant ( sqrt{36 - 24} = sqrt{12} )Simplify ( sqrt{12} ):[ sqrt{12} = sqrt{4 times 3} = 2sqrt{3} ]So, putting it back into the formula:[ x = frac{6 pm 2sqrt{3}}{2} ]Divide numerator and denominator by 2:[ x = 3 pm sqrt{3} ]So, the solutions are:[ x = 3 + sqrt{3} quad text{and} quad x = 3 - sqrt{3} ]Therefore, the length and width are ( 3 + sqrt{3} ) and ( 3 - sqrt{3} ). Let me verify if these satisfy the original equations.First, check the perimeter:[ a + b = (3 + sqrt{3}) + (3 - sqrt{3}) = 6 ]Yes, that's correct.Now, check the area:[ a times b = (3 + sqrt{3})(3 - sqrt{3}) ]Multiply using the difference of squares formula:[ (3)^2 - (sqrt{3})^2 = 9 - 3 = 6 ]Perfect, that matches the area given.Now, let's compute ( a^2 + b^2 ) using these values.First, compute ( a^2 ):[ (3 + sqrt{3})^2 = 3^2 + 2 times 3 times sqrt{3} + (sqrt{3})^2 = 9 + 6sqrt{3} + 3 = 12 + 6sqrt{3} ]Then, compute ( b^2 ):[ (3 - sqrt{3})^2 = 3^2 - 2 times 3 times sqrt{3} + (sqrt{3})^2 = 9 - 6sqrt{3} + 3 = 12 - 6sqrt{3} ]Now, add ( a^2 + b^2 ):[ (12 + 6sqrt{3}) + (12 - 6sqrt{3}) = 24 ]Yes, the ( 6sqrt{3} ) terms cancel out, leaving 24. So, that confirms it.Alternatively, I could have just used the identity ( a^2 + b^2 = (a + b)^2 - 2ab ) without solving for ( a ) and ( b ) individually, which is a quicker method. But going through the process of solving for ( a ) and ( b ) helped me verify that my initial answer was correct.So, after double-checking, I'm confident that the value of ( a^2 + b^2 ) is 24, which is option B.**Final Answer**boxed{B}