answer：Alright, so I have this problem here about a number line with points P and V at 3 and 33 respectively. The segment between them is divided into six equal parts by points Q, R, S, T, U. I need to find the sum of the lengths of PS and TV. Hmm, okay, let me try to visualize this.First, let me sketch a number line in my mind. Point P is at 3, and point V is at 33. So the distance between P and V is 33 - 3 = 30 units. That makes sense. Now, this segment PV is divided into six equal parts. So each part should be 30 divided by 6, which is 5 units long. So each segment PQ, QR, RS, ST, TU, and UV is 5 units.Wait, hold on, the points dividing PV are Q, R, S, T, U. So starting from P, the first division is Q, then R, then S, then T, then U, and finally V. So each of these points is spaced 5 units apart. Let me write down the positions of each point.Starting at P, which is 3. Then Q is 3 + 5 = 8. Then R is 8 + 5 = 13. Then S is 13 + 5 = 18. Then T is 18 + 5 = 23. Then U is 23 + 5 = 28. And finally, V is 28 + 5 = 33. Okay, so the positions are:P: 3Q: 8R: 13S: 18T: 23U: 28V: 33Alright, so now I need to find the lengths of PS and TV.First, PS. P is at 3, and S is at 18. So the length of PS is 18 - 3 = 15 units.Next, TV. T is at 23, and V is at 33. So the length of TV is 33 - 23 = 10 units.Now, the problem asks for the sum of these two lengths. So 15 + 10 = 25 units. Hmm, 25 is one of the answer choices, option A.Wait, but let me double-check my calculations to make sure I didn't make a mistake.Starting with the total distance between P and V: 33 - 3 = 30. Divided into six equal parts, each part is 5. So each segment is 5 units. So the positions are correct: 3, 8, 13, 18, 23, 28, 33.Calculating PS: from 3 to 18. 18 - 3 = 15. That seems right.Calculating TV: from 23 to 33. 33 - 23 = 10. That also seems correct.Adding them together: 15 + 10 = 25. So the answer should be 25, which is option A.But wait, hold on, let me think again. Is there another way to interpret the problem? Maybe the points are not equally spaced in terms of length, but in terms of intervals? No, the problem says the number line between 3 and 33 is divided into six equal parts, so each part is 5 units. So my initial interpretation is correct.Alternatively, maybe I misread the points. It says points Q, R, S, T, U divide the segment into six equal parts. So that means there are five points dividing the segment into six equal segments. So that would mean each segment is 5 units, as I calculated.So PS is from P to S, which skips over Q and R, so that's two segments, which is 10 units? Wait, hold on, no. Wait, if each segment is 5 units, then from P to Q is 5, Q to R is another 5, R to S is another 5. So from P to S is three segments, which is 15 units. That's correct.Similarly, TV is from T to V. T is at 23, V is at 33. So that's one segment of 10 units? Wait, no, from T to U is 5, and U to V is another 5. So from T to V is two segments, which is 10 units. So that's correct.So 15 + 10 is 25. So yeah, 25 is the answer, which is option A.But wait, let me think again. Maybe the question is asking for the sum of the lengths of PS and TV, but PS is from P to S, which is 15, and TV is from T to V, which is 10. So 15 + 10 is 25. So yeah, that's correct.Alternatively, maybe the problem is considering the distance from P to S as the number of intervals, but no, the question says the sum of the lengths, so it's the actual distance on the number line.So I think my answer is correct, 25, which is option A.Wait, but hold on, looking back at the problem statement, it says "the number line between 3 and 33 is divided into six equal parts by the points Q, R, S, T, U." So that means the total number of intervals is six, each of length 5. So each point is 5 units apart.So, P is at 3, Q at 8, R at 13, S at 18, T at 23, U at 28, and V at 33.So PS is from 3 to 18, which is 15 units, and TV is from 23 to 33, which is 10 units. So 15 + 10 is 25.Therefore, the answer is 25, which is option A.Wait, but the options are (A) 25, (B) 23, (C) 24, (D) 21, (E) 27. So 25 is an option, so that's correct.I think I've double-checked enough. I don't see any mistakes in my reasoning.**Final Answer**boxed{A}

answer：Okay, so I have this problem about carbon emissions modeled by a polynomial function. Let me try to figure out how to approach both parts.Starting with part 1: I need to calculate the difference in emissions between the original model E(x) and the reduced model R(x) after 5 years. Hmm, so that means I have to find E(5) and R(5) and then subtract them. Let me write down the functions again to make sure I have them right.E(x) is given by 2x³ - 5x² + 3x + 100. And R(x) is a quadratic function: -x² + 2x + 30. So, to find the difference, I need to compute E(5) - R(5).Let me calculate E(5) first. Plugging x = 5 into E(x):E(5) = 2*(5)^3 - 5*(5)^2 + 3*(5) + 100.Breaking it down step by step:5³ is 125, so 2*125 is 250.5² is 25, so 5*25 is 125. But it's subtracted, so that's -125.3*5 is 15.And then there's the constant term, 100.So adding all those together: 250 - 125 + 15 + 100.250 - 125 is 125. 125 + 15 is 140. 140 + 100 is 240. So E(5) is 240 metric tons.Now, R(5): plugging x = 5 into R(x):R(5) = -(5)^2 + 2*(5) + 30.Calculating each term:5² is 25, so with the negative sign, that's -25.2*5 is 10.And the constant term is 30.Adding them up: -25 + 10 + 30.-25 + 10 is -15. -15 + 30 is 15. So R(5) is 15 metric tons.Now, the difference is E(5) - R(5) = 240 - 15 = 225. So after 5 years, the difference in emissions is 225 metric tons.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For E(5):2*(125) = 250-5*(25) = -1253*5 = 15+100250 - 125 = 125; 125 + 15 = 140; 140 + 100 = 240. That seems right.For R(5):-25 + 10 + 30 = 15. That also looks correct.So the difference is indeed 225 metric tons. Okay, part 1 seems done.Moving on to part 2: The scientist wants to show that without the reduction strategy, emissions will exceed 500 metric tons per year. So I need to solve the inequality E(x) > 500.Given E(x) = 2x³ - 5x² + 3x + 100. So the inequality is:2x³ - 5x² + 3x + 100 > 500.Let me subtract 500 from both sides to set it to zero:2x³ - 5x² + 3x + 100 - 500 > 0Simplify that:2x³ - 5x² + 3x - 400 > 0.So I need to solve 2x³ - 5x² + 3x - 400 > 0.Hmm, solving a cubic inequality. I remember that for polynomials, the sign can change at the roots, so I need to find the real roots of the equation 2x³ - 5x² + 3x - 400 = 0, and then determine the intervals where the polynomial is positive.But solving a cubic equation can be tricky. Maybe I can try to find rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term divided by factors of the leading coefficient.The constant term is -400, and the leading coefficient is 2. So possible rational roots are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±25, ±40, ±50, ±80, ±100, ±200, ±400, and each divided by 1 or 2. So possible roots include ±1, ±1/2, ±2, etc.Let me test x = 5 first because in part 1, x=5 gave E(x)=240, which is less than 500, so maybe the root is beyond that.Wait, actually, let me compute E(5) again: 240, which is less than 500. So when does E(x) exceed 500? Probably at some x greater than 5.Let me try x=10:E(10) = 2*(1000) -5*(100) +3*(10) +100 = 2000 -500 +30 +100 = 2000 -500 is 1500, 1500 +30 is 1530, 1530 +100 is 1630. So E(10) is 1630, which is way above 500.So between x=5 and x=10, E(x) crosses 500. So maybe the root is somewhere in between.Let me try x=6:E(6) = 2*(216) -5*(36) +3*(6) +100.2*216 is 432.-5*36 is -180.3*6 is 18.+100.So 432 -180 = 252; 252 +18 = 270; 270 +100 = 370. So E(6)=370 <500.x=7:2*(343) -5*(49) +3*7 +100.2*343=686-5*49=-2453*7=21+100686 -245=441; 441 +21=462; 462 +100=562. So E(7)=562>500.So between x=6 and x=7, E(x) crosses 500.So the real root is between 6 and7.But since we need to solve the inequality, we can say that for x > root, the polynomial is positive because the leading coefficient is positive (2), so as x approaches infinity, E(x) tends to infinity.Therefore, the inequality E(x) >500 holds for x > root.But since we need to find the range of x where E(x) >500, we can say x > approximately 6.something.But the question is to solve for x in the inequality E(x) >500. So we need to find the exact value or approximate.Alternatively, maybe we can factor the cubic equation.Let me try to see if x=5 is a root:2*(125) -5*(25) +3*5 -400 = 250 -125 +15 -400 = (250 -125)=125; 125 +15=140; 140 -400= -260 ≠0.x=6: 2*216 -5*36 +3*6 -400=432 -180 +18 -400= (432-180)=252; 252+18=270; 270-400=-130≠0.x=7: 2*343 -5*49 +3*7 -400=686 -245 +21 -400= (686-245)=441; 441+21=462; 462-400=62≠0.x=8: 2*512 -5*64 +3*8 -400=1024 -320 +24 -400= (1024-320)=704; 704+24=728; 728-400=328≠0.Hmm, not a root. Maybe x=4:2*64 -5*16 +3*4 -400=128 -80 +12 -400= (128-80)=48; 48+12=60; 60-400=-340≠0.x=3: 2*27 -5*9 +3*3 -400=54 -45 +9 -400= (54-45)=9; 9+9=18; 18-400=-382≠0.x=2: 2*8 -5*4 +3*2 -400=16 -20 +6 -400= (16-20)=-4; -4+6=2; 2-400=-398≠0.x=1: 2*1 -5*1 +3*1 -400=2 -5 +3 -400= (2-5)=-3; -3+3=0; 0-400=-400≠0.x=0: 0 -0 +0 -400=-400≠0.x=-1: 2*(-1)^3 -5*(-1)^2 +3*(-1) -400= -2 -5 -3 -400=-410≠0.So none of these are roots. Maybe I need to use the method of trial and error or use the cubic formula, which is complicated.Alternatively, since we know that between x=6 and x=7, the function crosses 500, we can approximate the root.Let me use the Intermediate Value Theorem. At x=6, E(x)=370; at x=7, E(x)=562.We can approximate the root using linear approximation.The difference between E(7) and E(6) is 562 - 370 = 192 over an interval of 1 year.We need to find x where E(x)=500. The difference from E(6)=370 to 500 is 130.So the fraction is 130/192 ≈0.677.So the root is approximately at x=6 +0.677≈6.677.So approximately 6.68 years.Therefore, the inequality E(x) >500 holds for x > approximately 6.68 years.But since x represents years since the start of operations, and we can't have a fraction of a year in this context, we might say that emissions exceed 500 metric tons starting in the 7th year.But the question says "solve for x in the inequality E(x) >500". So we need to express the solution in terms of x.Since it's a cubic, and we found that the real root is approximately 6.68, the solution is x >6.68.But to write it more precisely, maybe we can express it as x > (the real root). But since it's a cubic, it's not factorable easily, so we can write the solution as x > approximately 6.68 years.Alternatively, if we can find an exact expression, but that might be complicated.Alternatively, maybe we can factor by grouping or use synthetic division, but since none of the rational roots worked, it's probably an irrational root.So, in conclusion, the solution to E(x) >500 is x > approximately 6.68 years.But let me check if I can write it more accurately.Alternatively, maybe I can use the Newton-Raphson method to approximate the root.Let me try that.Let f(x) =2x³ -5x² +3x -400.We know f(6)= -130, f(7)=62.Let me take x₀=6.5.f(6.5)=2*(6.5)^3 -5*(6.5)^2 +3*(6.5) -400.Calculate 6.5³: 6.5*6.5=42.25; 42.25*6.5=274.625.So 2*274.625=549.25.6.5²=42.25; 5*42.25=211.25.3*6.5=19.5.So f(6.5)=549.25 -211.25 +19.5 -400.549.25 -211.25=338; 338 +19.5=357.5; 357.5 -400= -42.5.So f(6.5)= -42.5.Now, f(6.5)= -42.5, f(7)=62.Let me compute f(6.75):6.75³: 6.75*6.75=45.5625; 45.5625*6.75≈308.59375.2*308.59375≈617.1875.6.75²=45.5625; 5*45.5625=227.8125.3*6.75=20.25.So f(6.75)=617.1875 -227.8125 +20.25 -400.617.1875 -227.8125=389.375; 389.375 +20.25=409.625; 409.625 -400=9.625.So f(6.75)=9.625.So between x=6.5 and x=6.75, f(x) crosses zero.At x=6.5, f=-42.5; at x=6.75, f=9.625.Let me use linear approximation between these two points.The change in x is 0.25, and the change in f is 9.625 - (-42.5)=52.125.We need to find x where f(x)=0.Starting at x=6.5, f=-42.5.The fraction needed is 42.5/52.125≈0.815.So the root is approximately x=6.5 +0.815*0.25≈6.5 +0.20375≈6.70375.So approximately 6.704.Let me check f(6.7):6.7³=6.7*6.7=44.89; 44.89*6.7≈301.973.2*301.973≈603.946.6.7²=44.89; 5*44.89≈224.45.3*6.7=20.1.So f(6.7)=603.946 -224.45 +20.1 -400.603.946 -224.45≈379.496; 379.496 +20.1≈399.596; 399.596 -400≈-0.404.So f(6.7)≈-0.404.Close to zero.Now, f(6.7)= -0.404, f(6.75)=9.625.Let me use linear approximation between x=6.7 and x=6.75.Change in x=0.05, change in f=9.625 - (-0.404)=10.029.We need to find x where f(x)=0, starting from x=6.7, f=-0.404.The fraction needed is 0.404/10.029≈0.0403.So the root is approximately x=6.7 +0.0403*0.05≈6.7 +0.002≈6.702.Wait, that seems contradictory. Wait, no, the change in x is 0.05 (from 6.7 to 6.75), and the change in f is 10.029.We need to cover 0.404 to reach zero from f=-0.404.So the fraction is 0.404/10.029≈0.0403.So the required x is 6.7 +0.0403*(6.75 -6.7)=6.7 +0.0403*0.05≈6.7 +0.002≈6.702.Wait, that seems too small. Alternatively, perhaps I should set up the linear approximation formula.Let me denote x₁=6.7, f(x₁)=-0.404; x₂=6.75, f(x₂)=9.625.We can write the linear approximation as:f(x) ≈ f(x₁) + (f(x₂) - f(x₁))/(x₂ -x₁)*(x -x₁).We set f(x)=0:0 ≈ -0.404 + (9.625 - (-0.404))/(6.75 -6.7)*(x -6.7).Calculate the slope:(9.625 +0.404)/0.05=10.029/0.05=200.58.So:0 ≈ -0.404 +200.58*(x -6.7).Adding 0.404 to both sides:0.404 ≈200.58*(x -6.7).Divide both sides by 200.58:x -6.7 ≈0.404/200.58≈0.002013.So x≈6.7 +0.002013≈6.702013.So approximately 6.702 years.So the root is approximately 6.702.Therefore, the solution to E(x) >500 is x >6.702.Since x represents years, and we can't have a fraction of a year in this context, the emissions will exceed 500 metric tons starting in the 7th year.But the question says "solve for x in the inequality E(x) >500", so we need to express it as x > approximately 6.702.Alternatively, if we can write it as x > (the real root), but since it's a cubic, it's not easily expressible without approximation.So, to summarize, the difference after 5 years is 225 metric tons, and the emissions exceed 500 metric tons starting around 6.7 years, so x >6.702.But let me check if I did the calculations correctly.For part 2, E(x)=2x³ -5x² +3x +100 >500.So 2x³ -5x² +3x -400 >0.We found that the root is approximately 6.702, so x >6.702.Yes, that seems correct.I think that's it. So the answers are:1. 225 metric tons.2. x > approximately 6.702 years.But since the problem might expect an exact form, but since it's a cubic, it's not easily factorable, so the approximate value is acceptable.Alternatively, maybe I can write it in terms of the cubic root, but that would be complicated.So, I think the answers are:1. 225 metric tons.2. x > approximately 6.7 years.But to be precise, maybe I can write it as x > (the real root of 2x³ -5x² +3x -400=0), but since it's not a nice number, the approximate decimal is better.So, final answers:1. The difference is 225 metric tons.2. The emissions exceed 500 metric tons when x > approximately 6.7 years.I think that's it.

answer：Okay, so I'm trying to solve this problem about the productivity of a study group. The function given is a polynomial: P(x) = -2x³ + 12x² - 20x + 5. I need to find the number of hours, x, that maximizes the productivity, and then solve the inequality P(x) > 10 to find the range of hours where productivity exceeds 10 units.Starting with part 1: finding the maximum productivity. Since this is a polynomial function, specifically a cubic, I know that its graph will have a certain shape. The leading coefficient is -2, which is negative, so as x approaches infinity, P(x) will go to negative infinity, and as x approaches negative infinity, it will go to positive infinity. But since x represents hours studied, it can't be negative, so we're only concerned with x ≥ 0.To find the maximum productivity, I need to find the critical points of the function. Critical points occur where the derivative is zero or undefined. Since this is a polynomial, the derivative will exist everywhere, so I just need to find where the derivative equals zero.Let me compute the derivative of P(x). The derivative of -2x³ is -6x², the derivative of 12x² is 24x, the derivative of -20x is -20, and the derivative of the constant 5 is 0. So, P'(x) = -6x² + 24x - 20.Now, I need to solve P'(x) = 0:-6x² + 24x - 20 = 0.This is a quadratic equation. I can try to factor it, but the coefficients are a bit messy. Maybe I should use the quadratic formula. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). Here, a = -6, b = 24, c = -20.Plugging in the values:x = [-24 ± sqrt(24² - 4*(-6)*(-20))] / (2*(-6)).Calculating the discriminant first: 24² = 576, and 4*6*20 = 480. So, discriminant is 576 - 480 = 96.So, sqrt(96) is sqrt(16*6) = 4*sqrt(6). So, sqrt(96) ≈ 4*2.449 ≈ 9.796.Therefore, x = [-24 ± 9.796] / (-12).Let me compute both roots:First root: (-24 + 9.796)/(-12) = (-14.204)/(-12) ≈ 1.1837.Second root: (-24 - 9.796)/(-12) = (-33.796)/(-12) ≈ 2.8163.So, the critical points are approximately x ≈ 1.1837 and x ≈ 2.8163.Now, since we're dealing with a cubic function, and the leading coefficient is negative, the function will have a local maximum and a local minimum. The critical point with the smaller x-value is likely the local maximum, and the larger one is the local minimum. Let me confirm this by using the second derivative test.Compute the second derivative P''(x). The first derivative was P'(x) = -6x² + 24x - 20, so the second derivative is P''(x) = -12x + 24.Now, evaluate P''(x) at each critical point.First, at x ≈ 1.1837:P''(1.1837) = -12*(1.1837) + 24 ≈ -14.204 + 24 ≈ 9.796. Since this is positive, the function is concave up at this point, meaning it's a local minimum. Wait, that contradicts my initial thought. Hmm, maybe I got it backwards.Wait, no, actually, the second derivative test says that if P''(x) > 0, the function is concave up, so it's a local minimum. If P''(x) < 0, it's concave down, so it's a local maximum.So, at x ≈ 1.1837, P''(x) ≈ 9.796 > 0, so it's a local minimum.At x ≈ 2.8163:P''(2.8163) = -12*(2.8163) + 24 ≈ -33.796 + 24 ≈ -9.796. Since this is negative, the function is concave down, so it's a local maximum.Therefore, the productivity is maximized at x ≈ 2.8163 hours.But since the problem says to assume the productivity peak occurs at a real number of hours, and since we have a local maximum at x ≈ 2.8163, that's our answer for part 1.But maybe I should express it more precisely. Let me compute the exact value instead of the approximate decimal.Going back to the quadratic equation:-6x² + 24x - 20 = 0.Let me write it as 6x² - 24x + 20 = 0 (multiplying both sides by -1 to make it positive).Then, divide all terms by 2: 3x² - 12x + 10 = 0.Now, applying the quadratic formula:x = [12 ± sqrt(144 - 120)] / 6.Because a = 3, b = -12, c = 10.Wait, hold on, if I have 3x² - 12x + 10 = 0, then a = 3, b = -12, c = 10.So discriminant is b² - 4ac = (-12)² - 4*3*10 = 144 - 120 = 24.So sqrt(24) = 2*sqrt(6).Thus, x = [12 ± 2sqrt(6)] / 6 = [12/6] ± [2sqrt(6)/6] = 2 ± (sqrt(6)/3).So, the critical points are x = 2 + sqrt(6)/3 and x = 2 - sqrt(6)/3.Compute sqrt(6): approximately 2.449.So, sqrt(6)/3 ≈ 0.816.Therefore, x ≈ 2 + 0.816 ≈ 2.816 and x ≈ 2 - 0.816 ≈ 1.184.So, the exact critical points are x = 2 ± sqrt(6)/3.Since we found that x = 2 + sqrt(6)/3 is a local maximum, that's our point of maximum productivity.So, the number of hours that maximizes productivity is x = 2 + sqrt(6)/3. To write it neatly, that's 2 + (√6)/3.Alternatively, we can write it as (6 + √6)/3, but 2 + (√6)/3 is also fine.So, that's part 1 done.Moving on to part 2: solving the inequality P(x) > 10.So, we have P(x) = -2x³ + 12x² - 20x + 5 > 10.Subtract 10 from both sides: -2x³ + 12x² - 20x + 5 - 10 > 0 => -2x³ + 12x² - 20x - 5 > 0.Let me write that as: -2x³ + 12x² - 20x - 5 > 0.It's often easier to deal with polynomials when the leading coefficient is positive, so let me multiply both sides by -1. But remember, multiplying an inequality by a negative number reverses the inequality sign.So, multiplying both sides by -1:2x³ - 12x² + 20x + 5 < 0.So, now the inequality is 2x³ - 12x² + 20x + 5 < 0.We need to find the values of x where this cubic is less than zero.To solve this, I need to find the roots of the cubic equation 2x³ - 12x² + 20x + 5 = 0, and then determine the intervals where the cubic is negative.Finding roots of a cubic can be tricky. Maybe I can try rational root theorem to see if there are any rational roots.The possible rational roots are factors of the constant term (5) over factors of the leading coefficient (2). So possible roots are ±1, ±5, ±1/2, ±5/2.Let me test x = 1:2(1)^3 - 12(1)^2 + 20(1) + 5 = 2 - 12 + 20 + 5 = 15 ≠ 0.x = -1:2(-1)^3 - 12(-1)^2 + 20(-1) + 5 = -2 -12 -20 +5 = -29 ≠ 0.x = 5:2(125) - 12(25) + 20(5) +5 = 250 - 300 + 100 +5 = 55 ≠ 0.x = -5:2(-125) - 12(25) + 20(-5) +5 = -250 - 300 -100 +5 = -645 ≠ 0.x = 1/2:2*(1/8) - 12*(1/4) + 20*(1/2) +5 = 0.25 - 3 + 10 +5 = 12.25 ≠ 0.x = -1/2:2*(-1/8) - 12*(1/4) + 20*(-1/2) +5 = -0.25 - 3 -10 +5 = -8.25 ≠ 0.x = 5/2:2*(125/8) - 12*(25/4) + 20*(5/2) +5.Calculate each term:2*(125/8) = 250/8 = 31.25-12*(25/4) = -300/4 = -7520*(5/2) = 50+5So total: 31.25 -75 +50 +5 = (31.25 +50 +5) -75 = 86.25 -75 = 11.25 ≠ 0.x = -5/2:2*(-125/8) -12*(25/4) +20*(-5/2) +5.Calculate each term:2*(-125/8) = -250/8 = -31.25-12*(25/4) = -300/4 = -7520*(-5/2) = -50+5Total: -31.25 -75 -50 +5 = (-31.25 -75 -50) +5 = -156.25 +5 = -151.25 ≠ 0.So none of the rational roots work. That means the cubic doesn't factor nicely, so I might need to use the rational root theorem didn't help, so perhaps I should use the method of depressed cubic or try to find approximate roots.Alternatively, maybe I can graph the function or use calculus to find where it crosses zero.Alternatively, since it's a cubic, it will have at least one real root. Let me try to estimate it.Let me evaluate the cubic at some points.Let me define Q(x) = 2x³ - 12x² + 20x + 5.Compute Q(0): 0 -0 +0 +5 =5.Q(1): 2 -12 +20 +5=15.Q(2): 16 - 48 +40 +5=13.Q(3): 54 - 108 +60 +5=11.Q(4): 128 - 192 +80 +5=21.Wait, so at x=0, Q=5; x=1, Q=15; x=2, Q=13; x=3, Q=11; x=4, Q=21.Wait, so it's positive at all these points. Hmm, but since it's a cubic with positive leading coefficient, as x approaches infinity, Q(x) approaches infinity, and as x approaches negative infinity, Q(x) approaches negative infinity. So, it must cross the x-axis somewhere for x < 0.But since x represents hours, we're only interested in x ≥0. So, if Q(x) is positive at x=0, and remains positive for all x ≥0, then the inequality Q(x) < 0 would have no solution in x ≥0.But wait, that can't be, because the original function P(x) = -2x³ +12x² -20x +5. At x=0, P(0)=5. As x increases, the function goes up to a maximum and then decreases.But when we set P(x) >10, we subtract 10, so P(x) -10 >0, which is equivalent to -2x³ +12x² -20x -5 >0, which we multiplied by -1 to get 2x³ -12x² +20x +5 <0.But if Q(x) is always positive for x ≥0, then the inequality Q(x) <0 has no solution, meaning P(x) >10 never occurs. But that seems unlikely because P(x) is a cubic with negative leading coefficient, so it goes to negative infinity as x increases, so it must cross y=10 somewhere.Wait, perhaps I made a mistake in the algebra.Let me double-check:Original inequality: P(x) >10So, -2x³ +12x² -20x +5 >10Subtract 10: -2x³ +12x² -20x -5 >0Multiply by -1: 2x³ -12x² +20x +5 <0Yes, that seems correct.But when I plug in x=0, Q(0)=5>0; x=1, Q(1)=15>0; x=2, Q(2)=13>0; x=3, Q(3)=11>0; x=4, Q(4)=21>0.So, Q(x) is positive at x=0,1,2,3,4. Let me check x=5:Q(5)=2*125 -12*25 +20*5 +5=250 -300 +100 +5=55>0.x=6: 2*216 -12*36 +20*6 +5=432 -432 +120 +5=125>0.Hmm, so it's positive at x=0,1,2,3,4,5,6.Wait, but as x approaches infinity, Q(x) approaches positive infinity because the leading term is 2x³. So, it's positive at x=0 and goes to positive infinity as x increases. But since it's a cubic, it must have a local maximum and minimum somewhere.Wait, let's compute Q(x) at some negative x, even though x can't be negative, just to see.x=-1: Q(-1)=2*(-1)^3 -12*(-1)^2 +20*(-1) +5= -2 -12 -20 +5=-29<0.So, Q(x) is negative at x=-1, positive at x=0, so it crosses the x-axis somewhere between x=-1 and x=0.But since x can't be negative, in the domain x ≥0, Q(x) is always positive. Therefore, the inequality Q(x) <0 has no solution for x ≥0.Wait, that can't be right because P(x) is a cubic with negative leading coefficient, so it must go to negative infinity as x increases. Therefore, P(x) must cross y=10 somewhere.Wait, let me check P(x) at x=0: P(0)=5.At x=1: P(1)=-2+12-20+5=-5.Wait, that's interesting. So, P(1)=-5, which is less than 10. Wait, but earlier I thought P(x) was positive at x=1, but no, P(1)= -2 +12 -20 +5= (-2-20) + (12+5)= -22 +17=-5.Wait, so P(1)=-5, which is less than 10. So, P(x) starts at 5 when x=0, goes down to -5 at x=1, then?Wait, but earlier, when I computed Q(x)=2x³ -12x² +20x +5, which is related to P(x)-10, but maybe I made a mistake in the transformation.Wait, let me re-examine:Original inequality: P(x) >10.P(x) = -2x³ +12x² -20x +5.So, P(x) >10 => -2x³ +12x² -20x +5 >10.Subtract 10: -2x³ +12x² -20x -5 >0.Multiply both sides by -1 (inequality flips): 2x³ -12x² +20x +5 <0.So, that's correct.But when I plug in x=0, Q(0)=5>0.x=1: Q(1)=2 -12 +20 +5=15>0.x=2: 16 -48 +40 +5=13>0.x=3:54 -108 +60 +5=11>0.x=4:128 - 192 +80 +5=21>0.x=5:250 - 300 +100 +5=55>0.So, Q(x) is always positive for x ≥0, meaning 2x³ -12x² +20x +5 <0 has no solution for x ≥0.Therefore, the inequality P(x) >10 has no solution.But wait, that contradicts the fact that P(x) is a cubic with negative leading coefficient, so it must go to negative infinity as x increases, meaning it must cross y=10 somewhere.Wait, but P(x) at x=0 is 5, which is less than 10. At x=1, it's -5, which is even less. Then, as x increases beyond that, it goes to negative infinity. So, it never exceeds 10? That can't be right because the maximum productivity was at x≈2.816, which we found earlier.Wait, let me compute P(x) at x=2.816.Compute P(2.816):First, x≈2.816.Compute each term:-2x³: -2*(2.816)^3.2.816^3: 2.816*2.816=7.929; 7.929*2.816≈22.33.So, -2*22.33≈-44.66.12x²: 12*(2.816)^2≈12*7.929≈95.15.-20x: -20*2.816≈-56.32.+5.So, total P(x)≈-44.66 +95.15 -56.32 +5≈(-44.66 -56.32) + (95.15 +5)= (-100.98) +100.15≈-0.83.Wait, that's strange. So, at the local maximum, P(x) is approximately -0.83, which is less than 10. So, the maximum productivity is negative? That can't be right because P(0)=5, which is positive.Wait, maybe I made a mistake in calculating P(2.816). Let me compute it more accurately.Compute x=2 + sqrt(6)/3. Let's compute sqrt(6)≈2.449, so sqrt(6)/3≈0.816. So, x≈2.816.Compute P(x):P(x) = -2x³ +12x² -20x +5.Let me compute each term step by step.First, x=2.816.Compute x³: 2.816^3.2.816*2.816: Let's compute 2.8*2.8=7.84, 2.8*0.016=0.0448, 0.016*2.8=0.0448, 0.016*0.016=0.000256.So, (2.8 +0.016)^2=2.8² + 2*2.8*0.016 +0.016²=7.84 +0.0896 +0.000256≈7.929856.Then, x³= x² *x≈7.929856 *2.816.Compute 7.929856 *2=15.859712.7.929856 *0.8=6.3438848.7.929856 *0.016≈0.1268777.Add them up:15.859712 +6.3438848≈22.2035968 +0.1268777≈22.3304745.So, x³≈22.3305.Then, -2x³≈-44.661.12x²: 12*7.929856≈95.158272.-20x: -20*2.816≈-56.32.+5.So, P(x)= -44.661 +95.158272 -56.32 +5.Compute step by step:-44.661 +95.158272≈50.497272.50.497272 -56.32≈-5.822728.-5.822728 +5≈-0.822728.So, P(x)≈-0.82 at x≈2.816.Wait, so the maximum productivity is negative? That seems odd because at x=0, P(x)=5, which is positive. So, the function goes from 5 at x=0, decreases to -5 at x=1, then increases to a local maximum of approximately -0.82 at x≈2.816, and then decreases again to negative infinity.So, the function never exceeds 5, which is its value at x=0. Therefore, P(x) >10 is never true because the maximum P(x) is about -0.82, which is still less than 10.Wait, that can't be right because P(x) at x=0 is 5, which is less than 10, but the function might have a peak above 10 somewhere else.Wait, but we found that the local maximum is at x≈2.816, and P(x) there is≈-0.82, which is still below 10. So, the function never reaches 10.Therefore, the inequality P(x) >10 has no solution.But that seems counterintuitive because the function is a cubic, so it should cross y=10 somewhere. Wait, but since the leading coefficient is negative, it goes to negative infinity as x increases, but it starts at P(0)=5, which is less than 10. So, maybe it never reaches 10.Wait, let me check P(x) at x=2:P(2)= -2*(8) +12*(4) -20*(2) +5= -16 +48 -40 +5= (-16-40)+(48+5)= (-56)+53=-3.At x=3:P(3)= -2*27 +12*9 -20*3 +5= -54 +108 -60 +5= (-54-60)+(108+5)= (-114)+113=-1.At x=4:P(4)= -2*64 +12*16 -20*4 +5= -128 +192 -80 +5= (-128-80)+(192+5)= (-208)+197=-11.So, P(x) is decreasing after x≈2.816, but it's still negative.Wait, but what about for x <0? Well, x can't be negative, so we don't consider that.Therefore, P(x) starts at 5 when x=0, goes down to -5 at x=1, then up to a local maximum of≈-0.82 at x≈2.816, then decreases again to negative infinity.So, P(x) never exceeds 5, which is its value at x=0. Therefore, P(x) >10 is never true.But that contradicts the initial thought that as x increases, the function goes to negative infinity, but it's possible that it never crosses y=10 because it's already below 10 at x=0 and goes further down.Wait, let me check P(x) at x=0.5:P(0.5)= -2*(0.125) +12*(0.25) -20*(0.5) +5= -0.25 +3 -10 +5= (-0.25 -10) + (3 +5)= (-10.25)+8=-2.25.So, P(0.5)=-2.25.At x=0.25:P(0.25)= -2*(0.015625) +12*(0.0625) -20*(0.25) +5≈-0.03125 +0.75 -5 +5≈(-0.03125 -5) + (0.75 +5)= (-5.03125)+5.75≈0.71875.So, P(0.25)≈0.71875.Wait, so P(x) at x=0.25 is≈0.72, which is less than 5.Wait, so P(x) starts at 5 when x=0, goes down to≈0.72 at x=0.25, then further down to -2.25 at x=0.5, then to -5 at x=1, then up to≈-0.82 at x≈2.816, then down again.So, the maximum value of P(x) is≈-0.82, which is still less than 10.Therefore, P(x) never exceeds 10, so the inequality P(x) >10 has no solution.But that seems strange because the function is a cubic, so it should go to positive infinity as x approaches negative infinity, but since x can't be negative, we're only looking at x ≥0.Therefore, the answer to part 2 is that there is no solution; the productivity never exceeds 10 units.But wait, let me double-check my calculations because I might have made a mistake.Wait, when I computed P(2.816), I got≈-0.82, but maybe I should compute it more accurately.Let me use exact values.We have x=2 + sqrt(6)/3.Compute P(x)= -2x³ +12x² -20x +5.Let me compute x=2 + sqrt(6)/3.Let me denote sqrt(6)=a, so x=2 + a/3.Compute x³:x³=(2 + a/3)^3=8 + 3*(2)^2*(a/3) + 3*(2)*(a/3)^2 + (a/3)^3.Compute each term:=8 + 3*4*(a/3) + 3*2*(a²/9) + (a³/27)=8 + (12a)/3 + (6a²)/9 + (a³)/27=8 +4a + (2a²)/3 + (a³)/27.Similarly, x²=(2 + a/3)^2=4 + (4a)/3 + (a²)/9.Now, compute each term of P(x):-2x³= -2*(8 +4a + (2a²)/3 + (a³)/27)= -16 -8a - (4a²)/3 - (2a³)/27.12x²=12*(4 + (4a)/3 + (a²)/9)=48 +16a + (12a²)/9=48 +16a + (4a²)/3.-20x= -20*(2 + a/3)= -40 - (20a)/3.+5.Now, sum all terms:-16 -8a - (4a²)/3 - (2a³)/27 +48 +16a + (4a²)/3 -40 - (20a)/3 +5.Combine like terms:Constants: -16 +48 -40 +5= (-16-40)+(48+5)= (-56)+53=-3.a terms: -8a +16a - (20a)/3= (8a) - (20a)/3= (24a/3 -20a/3)=4a/3.a² terms: - (4a²)/3 + (4a²)/3=0.a³ terms: - (2a³)/27.So, P(x)= -3 + (4a)/3 - (2a³)/27.Now, substitute a=sqrt(6):P(x)= -3 + (4sqrt(6))/3 - (2*(sqrt(6))³)/27.Compute (sqrt(6))³=6*sqrt(6).So, (2*(sqrt(6))³)/27= (2*6*sqrt(6))/27= (12sqrt(6))/27= (4sqrt(6))/9.Therefore, P(x)= -3 + (4sqrt(6))/3 - (4sqrt(6))/9.Combine the terms with sqrt(6):(4sqrt(6))/3 - (4sqrt(6))/9= (12sqrt(6) -4sqrt(6))/9= (8sqrt(6))/9.So, P(x)= -3 + (8sqrt(6))/9.Compute numerically:sqrt(6)≈2.449.So, 8*2.449≈19.592.Divide by 9:≈2.177.So, P(x)= -3 +2.177≈-0.823.So, exact value is P(x)= -3 + (8sqrt(6))/9≈-0.823.So, indeed, the maximum productivity is≈-0.823, which is less than 10.Therefore, the inequality P(x) >10 has no solution.But wait, that seems odd because the function is a cubic, so it should go to positive infinity as x approaches negative infinity, but since x can't be negative, we're only considering x ≥0. So, in the domain x ≥0, P(x) never exceeds 10.Therefore, the answer to part 2 is that there are no real numbers x ≥0 for which P(x) >10.But let me check P(x) at x=0. Let me see if P(x) ever exceeds 10.Wait, P(0)=5, which is less than 10.As x increases from 0, P(x) decreases to P(1)=-5, then increases to a local maximum of≈-0.82 at x≈2.816, then decreases again.So, P(x) never exceeds 5, which is its value at x=0. Therefore, P(x) >10 is never true.Therefore, the solution to the inequality is the empty set.But the problem says "the productivity peak occurs at a real number of hours," which we found to be≈2.816, but the maximum productivity is≈-0.82, which is still less than 10.Therefore, the answer to part 2 is that there are no hours x for which P(x) >10.But let me double-check my calculations because it's unusual for a cubic to have a maximum below 10 when it starts at 5.Wait, perhaps I made a mistake in the derivative.Wait, P(x)= -2x³ +12x² -20x +5.P'(x)= -6x² +24x -20.Set to zero: -6x² +24x -20=0.Multiply by -1:6x² -24x +20=0.Divide by 2:3x² -12x +10=0.Solutions: x=(12±sqrt(144-120))/6=(12±sqrt(24))/6=(12±2sqrt(6))/6=2±sqrt(6)/3.Yes, that's correct.Then, P''(x)= -12x +24.At x=2 + sqrt(6)/3, P''(x)= -12*(2 + sqrt(6)/3) +24= -24 -4sqrt(6) +24= -4sqrt(6)<0, so it's a local maximum.So, the calculations are correct.Therefore, the conclusion is that the maximum productivity is≈-0.82, which is less than 10, so P(x) >10 has no solution.Therefore, the answer to part 2 is that there are no real numbers x for which P(x) >10.But the problem says "the productivity peak occurs at a real number of hours," which we found, but the peak is below 10.Therefore, the range of hours where productivity exceeds 10 units is empty.So, summarizing:1. The number of hours that maximizes productivity is x=2 + sqrt(6)/3.2. There are no hours x for which P(x) >10.But let me write the exact form for part 1.x=2 + sqrt(6)/3.Alternatively, x=(6 + sqrt(6))/3.But 2 + sqrt(6)/3 is also acceptable.So, final answers:1. x=2 + (√6)/3 hours.2. No solution; the productivity never exceeds 10 units.

answer：Alright, so I've got these two math problems to solve, and I need to figure them out step by step. Let's start with the first one.**Problem 1:** Solve for ( x ) when the record makes exactly 500 revolutions. The polynomial expression given is ( r(x) = 33.33x + 2x^2 - 5 ). So, I need to set this equal to 500 and solve for ( x ).Okay, let's write that equation out:( 33.33x + 2x^2 - 5 = 500 )Hmm, that looks like a quadratic equation. I remember quadratic equations are in the form ( ax^2 + bx + c = 0 ), so I need to rearrange this equation to match that form.First, subtract 500 from both sides to get everything on one side:( 33.33x + 2x^2 - 5 - 500 = 0 )Simplify the constants:( 33.33x + 2x^2 - 505 = 0 )Let me write it in standard form, which is usually ( ax^2 + bx + c = 0 ). So, I'll rearrange the terms:( 2x^2 + 33.33x - 505 = 0 )Alright, now I have a quadratic equation. To solve for ( x ), I can use the quadratic formula:( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Where ( a = 2 ), ( b = 33.33 ), and ( c = -505 ).Let me plug these values into the formula.First, calculate the discriminant ( D = b^2 - 4ac ):( D = (33.33)^2 - 4 * 2 * (-505) )Calculating ( (33.33)^2 ):33.33 squared is approximately 1110.8889.Then, ( 4 * 2 * (-505) = 8 * (-505) = -4040 ). But since it's subtracting this, it becomes:( D = 1110.8889 - (-4040) = 1110.8889 + 4040 = 5150.8889 )So, the discriminant is approximately 5150.8889. That's a positive number, so we have two real solutions.Now, compute the square root of the discriminant:( sqrt{5150.8889} approx 71.77 )Wait, let me check that. 71.77 squared is approximately 5150.8889? Let me compute 71.77 * 71.77:70 * 70 = 490070 * 1.77 = 123.91.77 * 70 = 123.91.77 * 1.77 ≈ 3.13So, adding up: 4900 + 123.9 + 123.9 + 3.13 ≈ 4900 + 247.8 + 3.13 ≈ 5150.93. That's very close to 5150.8889, so yes, the square root is approximately 71.77.Now, plug back into the quadratic formula:( x = frac{-33.33 pm 71.77}{2 * 2} )Compute both solutions:First, the positive case:( x = frac{-33.33 + 71.77}{4} = frac{38.44}{4} = 9.61 )Second, the negative case:( x = frac{-33.33 - 71.77}{4} = frac{-105.1}{4} = -26.275 )Since time cannot be negative, we discard the negative solution. So, ( x approx 9.61 ) minutes.Wait, let me double-check my calculations because 33.33 is a repeating decimal, which is actually 100/3. Maybe using fractions would be more precise.Let me try that. So, 33.33 is approximately 100/3, so let's write the equation as:( frac{100}{3}x + 2x^2 - 5 = 500 )Subtract 500:( 2x^2 + frac{100}{3}x - 505 = 0 )Multiply all terms by 3 to eliminate the fraction:( 6x^2 + 100x - 1515 = 0 )Now, use the quadratic formula with ( a = 6 ), ( b = 100 ), ( c = -1515 ):Discriminant ( D = 100^2 - 4*6*(-1515) = 10000 + 4*6*1515 )Compute 4*6 = 24, then 24*1515:1515 * 24: Let's compute 1500*24 = 36,000 and 15*24=360, so total is 36,000 + 360 = 36,360.So, discriminant D = 10,000 + 36,360 = 46,360.Square root of 46,360. Let's see:215^2 = 46,225, which is close. 215^2 = 46,225, so 46,360 - 46,225 = 135. So, sqrt(46,360) ≈ 215 + 135/(2*215) ≈ 215 + 0.312 ≈ 215.312.So, sqrt(46,360) ≈ 215.312.Now, compute x:( x = frac{-100 pm 215.312}{12} )First solution:( x = frac{-100 + 215.312}{12} = frac{115.312}{12} ≈ 9.6093 )Second solution:( x = frac{-100 - 215.312}{12} = frac{-315.312}{12} ≈ -26.276 )Again, negative time doesn't make sense, so x ≈ 9.6093 minutes, which is approximately 9.61 minutes.So, both methods give the same result, which is reassuring. Therefore, the solution is approximately 9.61 minutes.But let me check if 9.61 minutes actually gives 500 revolutions.Compute ( r(9.61) = 33.33*9.61 + 2*(9.61)^2 - 5 )First, 33.33 * 9.61:33.33 * 9 = 299.9733.33 * 0.61 ≈ 20.3313Total ≈ 299.97 + 20.3313 ≈ 320.3013Next, 2*(9.61)^2:9.61^2 ≈ 92.35212*92.3521 ≈ 184.7042Now, add them together and subtract 5:320.3013 + 184.7042 ≈ 505.0055505.0055 - 5 ≈ 500.0055Wow, that's very close to 500. So, x ≈ 9.61 is correct.Therefore, the answer is approximately 9.61 minutes.**Problem 2:** Solve the inequality ( 400 leq 33.33x + 2x^2 - 5 leq 600 ) to find the range of time ( x ) that Ms. Carter can enjoy her music while grading.So, this is a compound inequality. I need to solve both parts:1. ( 33.33x + 2x^2 - 5 geq 400 )2. ( 33.33x + 2x^2 - 5 leq 600 )Let me handle each inequality separately.**First inequality:** ( 2x^2 + 33.33x - 5 geq 400 )Subtract 400:( 2x^2 + 33.33x - 405 geq 0 )Again, this is a quadratic inequality. Let's find the roots of the equation ( 2x^2 + 33.33x - 405 = 0 ).Using the quadratic formula:( x = frac{-33.33 pm sqrt{(33.33)^2 - 4*2*(-405)}}{2*2} )Compute discriminant:( D = (33.33)^2 - 4*2*(-405) )33.33 squared is approximately 1110.8889.4*2*405 = 8*405 = 3240.So, D = 1110.8889 + 3240 = 4350.8889Square root of 4350.8889:Let me approximate. 66^2 = 4356, which is very close. So sqrt(4350.8889) ≈ 65.96.So, sqrt(D) ≈ 65.96.Now, compute x:( x = frac{-33.33 pm 65.96}{4} )First solution:( x = frac{-33.33 + 65.96}{4} = frac{32.63}{4} ≈ 8.1575 )Second solution:( x = frac{-33.33 - 65.96}{4} = frac{-99.29}{4} ≈ -24.8225 )Again, negative time is irrelevant, so the critical point is at approximately 8.1575 minutes.Since the quadratic opens upwards (coefficient of x^2 is positive), the inequality ( 2x^2 + 33.33x - 405 geq 0 ) is satisfied when ( x leq -24.8225 ) or ( x geq 8.1575 ). Since time can't be negative, the solution is ( x geq 8.1575 ) minutes.**Second inequality:** ( 2x^2 + 33.33x - 5 leq 600 )Subtract 600:( 2x^2 + 33.33x - 605 leq 0 )Again, solve the quadratic equation ( 2x^2 + 33.33x - 605 = 0 ).Quadratic formula:( x = frac{-33.33 pm sqrt{(33.33)^2 - 4*2*(-605)}}{4} )Compute discriminant:( D = 1110.8889 + 4*2*605 = 1110.8889 + 4840 = 5950.8889 )Square root of 5950.8889:Approximate. 77^2 = 5929, 78^2=6084. So sqrt(5950.8889) is between 77 and 78.Compute 77.1^2 = 5945.4177.2^2 = 5959.84So, 5950.8889 is between 77.1 and 77.2.Compute 77.1^2 = 5945.41Difference: 5950.8889 - 5945.41 = 5.4789Each 0.1 increase in x increases x^2 by approximately 2*77.1*0.1 + 0.1^2 ≈ 15.42 + 0.01 ≈ 15.43 per 0.1.Wait, actually, the derivative of x^2 is 2x, so at x=77.1, the approximate increase per 0.1 is 2*77.1*0.1 = 15.42.So, to get 5.4789, how much delta x?delta x ≈ 5.4789 / 15.42 ≈ 0.355So, sqrt(5950.8889) ≈ 77.1 + 0.355 ≈ 77.455Let me check 77.455^2:77^2 = 59290.455^2 ≈ 0.207Cross term: 2*77*0.455 ≈ 2*77*0.455 ≈ 154*0.455 ≈ 70.17So total ≈ 5929 + 70.17 + 0.207 ≈ 5999.377, which is a bit low. Wait, that can't be right.Wait, no, 77.455 is the sqrt of 5950.8889, not 5999.377. Maybe my approximation was off.Alternatively, perhaps I should use a calculator-like approach.But for the sake of time, let's say sqrt(5950.8889) ≈ 77.16.Wait, let me compute 77.16^2:77^2 = 59290.16^2 = 0.0256Cross term: 2*77*0.16 = 24.64So, total is 5929 + 24.64 + 0.0256 ≈ 5953.6656But we need 5950.8889, which is less than that. So, 77.16^2 ≈ 5953.6656, which is higher than 5950.8889.So, let's try 77.1:77.1^2 = 5945.41Difference: 5950.8889 - 5945.41 = 5.4789So, how much more than 77.1 do we need?Each 0.01 increase in x increases x^2 by approximately 2*77.1*0.01 + (0.01)^2 ≈ 1.542 + 0.0001 ≈ 1.5421.So, to get 5.4789, we need approximately 5.4789 / 1.5421 ≈ 3.55 units of 0.01, which is 0.0355.So, sqrt(5950.8889) ≈ 77.1 + 0.0355 ≈ 77.1355.Let me check 77.1355^2:77^2 = 59290.1355^2 ≈ 0.01836Cross term: 2*77*0.1355 ≈ 154*0.1355 ≈ 20.867Total ≈ 5929 + 20.867 + 0.01836 ≈ 5949.885, which is still a bit low.Wait, 5949.885 vs 5950.8889. The difference is 1.0039.So, we need a bit more. Each additional 0.01 adds approximately 1.5421 to the square.So, 1.0039 / 1.5421 ≈ 0.651, so 0.00651.So, total sqrt ≈ 77.1355 + 0.00651 ≈ 77.142.Check 77.142^2:77^2 = 59290.142^2 ≈ 0.020164Cross term: 2*77*0.142 ≈ 154*0.142 ≈ 21.868Total ≈ 5929 + 21.868 + 0.020164 ≈ 5950.888, which is very close to 5950.8889.So, sqrt(5950.8889) ≈ 77.142.Therefore, back to the quadratic formula:( x = frac{-33.33 pm 77.142}{4} )First solution:( x = frac{-33.33 + 77.142}{4} = frac{43.812}{4} ≈ 10.953 )Second solution:( x = frac{-33.33 - 77.142}{4} = frac{-110.472}{4} ≈ -27.618 )Again, negative time is irrelevant, so the critical point is at approximately 10.953 minutes.Since the quadratic opens upwards, the inequality ( 2x^2 + 33.33x - 605 leq 0 ) is satisfied between the roots, i.e., ( -27.618 leq x leq 10.953 ). Since time can't be negative, the solution is ( 0 leq x leq 10.953 ) minutes.**Combining both inequalities:**From the first inequality, ( x geq 8.1575 ).From the second inequality, ( x leq 10.953 ).Therefore, the range of ( x ) is ( 8.1575 leq x leq 10.953 ) minutes.But let me check if these endpoints are correct.At x ≈ 8.1575, r(x) should be 400.Compute ( r(8.1575) = 33.33*8.1575 + 2*(8.1575)^2 -5 )33.33 * 8.1575 ≈ 33.33*8 + 33.33*0.1575 ≈ 266.64 + 5.25 ≈ 271.892*(8.1575)^2 ≈ 2*(66.54) ≈ 133.08Total ≈ 271.89 + 133.08 -5 ≈ 400. So, correct.Similarly, at x ≈ 10.953, r(x) should be 600.Compute ( r(10.953) = 33.33*10.953 + 2*(10.953)^2 -5 )33.33*10.953 ≈ 33.33*10 + 33.33*0.953 ≈ 333.3 + 31.76 ≈ 365.062*(10.953)^2 ≈ 2*(119.96) ≈ 239.92Total ≈ 365.06 + 239.92 -5 ≈ 600. So, correct.Therefore, the range is approximately 8.16 minutes to 10.95 minutes.But let me express these numbers more precisely.From the first inequality, x ≈ 8.1575, which is approximately 8.16 minutes.From the second inequality, x ≈ 10.953, which is approximately 10.95 minutes.So, the range is approximately 8.16 ≤ x ≤ 10.95 minutes.But to be more precise, let's use the exact fractions.Earlier, when I converted 33.33 to 100/3, perhaps I can solve the inequalities using fractions for more accuracy.Let's try that.**First inequality:** ( 2x^2 + (100/3)x - 5 geq 400 )Subtract 400:( 2x^2 + (100/3)x - 405 geq 0 )Multiply all terms by 3 to eliminate fractions:( 6x^2 + 100x - 1215 geq 0 )Quadratic equation: ( 6x^2 + 100x - 1215 = 0 )Discriminant D = 100^2 - 4*6*(-1215) = 10,000 + 4*6*1215Compute 4*6=24, 24*1215:1215*24: 1200*24=28,800; 15*24=360; total=29,160.So, D=10,000 +29,160=39,160.sqrt(39,160). Let's compute:197^2=38,809198^2=39,204So, sqrt(39,160) is between 197 and 198.Compute 197.8^2:197^2=38,8090.8^2=0.64Cross term: 2*197*0.8=315.2Total: 38,809 + 315.2 + 0.64=39,124.84Still less than 39,160.197.9^2:197^2=38,8090.9^2=0.81Cross term: 2*197*0.9=354.6Total: 38,809 + 354.6 + 0.81=39,164.41Which is more than 39,160.So, sqrt(39,160) is between 197.8 and 197.9.Compute 197.8^2=39,124.84Difference: 39,160 -39,124.84=35.16Each 0.1 increase in x adds approximately 2*197.8*0.1 +0.1^2=39.56 +0.01=39.57 per 0.1.So, 35.16 /39.57≈0.888, so 0.0888.So, sqrt(39,160)≈197.8 +0.0888≈197.8888.Therefore, sqrt(D)=197.8888.Now, compute x:( x = frac{-100 pm 197.8888}{12} )First solution:( x = frac{-100 + 197.8888}{12} = frac{97.8888}{12} ≈8.1574 )Second solution:( x = frac{-100 -197.8888}{12}= frac{-297.8888}{12}≈-24.824 )So, same as before, x≈8.1574 minutes.**Second inequality:** ( 2x^2 + (100/3)x -5 leq 600 )Subtract 600:( 2x^2 + (100/3)x -605 leq 0 )Multiply by 3:( 6x^2 + 100x -1815 leq 0 )Quadratic equation: ( 6x^2 + 100x -1815 =0 )Discriminant D=100^2 -4*6*(-1815)=10,000 +4*6*18154*6=24, 24*1815:1815*24: 1800*24=43,200; 15*24=360; total=43,560.So, D=10,000 +43,560=53,560.sqrt(53,560). Let's compute:231^2=53,361232^2=53,824So, sqrt(53,560) is between 231 and 232.Compute 231.5^2=231^2 +2*231*0.5 +0.5^2=53,361 +231 +0.25=53,592.25Which is higher than 53,560.Compute 231.4^2:231^2=53,3610.4^2=0.16Cross term:2*231*0.4=184.8Total:53,361 +184.8 +0.16=53,545.96Still less than 53,560.Difference:53,560 -53,545.96=14.04Each 0.1 increase in x adds approximately 2*231.4*0.1 +0.1^2=46.28 +0.01=46.29 per 0.1.So, 14.04 /46.29≈0.303, so 0.0303.Thus, sqrt(53,560)≈231.4 +0.0303≈231.4303.Therefore, sqrt(D)=231.4303.Now, compute x:( x = frac{-100 pm231.4303}{12} )First solution:( x = frac{-100 +231.4303}{12}= frac{131.4303}{12}≈10.9525 )Second solution:( x = frac{-100 -231.4303}{12}= frac{-331.4303}{12}≈-27.619 )So, same as before, x≈10.9525 minutes.Therefore, combining both, the range is approximately 8.1574 ≤x ≤10.9525 minutes.Rounding to two decimal places, it's approximately 8.16 ≤x ≤10.95 minutes.But let me check if the endpoints are inclusive.Since the inequalities are "at least 400" and "not more than 600", the endpoints where r(x)=400 and r(x)=600 are included. So, the range is inclusive.Therefore, the range of time Ms. Carter can enjoy her music while grading is from approximately 8.16 minutes to 10.95 minutes.But to express these as exact values, perhaps we can write them as fractions or decimals.Alternatively, since the quadratic solutions were approximately 8.16 and 10.95, we can write them as such.Alternatively, if we want to express them more precisely, we can write them as:x ≈8.16 minutes and x≈10.95 minutes.But since the problem didn't specify the form, decimal is fine.So, summarizing:1. When the record makes exactly 500 revolutions, x≈9.61 minutes.2. The range of time for revolutions between 400 and 600 is approximately 8.16 ≤x ≤10.95 minutes.I think that's it. I've double-checked the calculations, and they seem consistent.**Final Answer**1. boxed{9.61}2. boxed{[8.16, 10.95]}

answer：To determine the maximum number of hours the child actor can spend on set while meeting their schoolwork and free time requirements, I'll start by defining the total hours in a week and subtracting the time allocated to schoolwork and free time.Total weekly hours: 168 hoursSchoolwork requirement: At least 20 hoursFree time requirement: At least 10 hoursThis leaves the remaining hours for acting on set.So, the inequality becomes:x ≤ 168 - 20 - 10x ≤ 138Therefore, the maximum number of hours the child actor can spend on set is 138 hours per week.For the second part, to find the possible values of x where the time required for the special project is zero, I'll set the quadratic equation equal to zero and solve for x.x² - 5x + 6 = 0Factoring the quadratic equation:(x - 2)(x - 3) = 0So, the solutions are x = 2 and x = 3.These are the values of x for which the time required for the project is zero.

answer：Okay, so I have this problem here about special fractions. Let me try to understand it step by step. The problem says:"A fraction a/b, not necessarily in the simplest form, is special if a and b are positive integers whose sum is 15. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?"And the answer choices are from A to E, with E being 13. Hmm, okay.First, let me parse this. A special fraction is a/b where a and b are positive integers, and a + b = 15. So, for example, if a is 1, then b is 14; if a is 2, b is 13, and so on, up to a = 14, b = 1. So, all these fractions have denominators and numerators that add up to 15.Now, the question is asking how many distinct integers can be written as the sum of two such special fractions. The fractions don't have to be different, so we can use the same fraction twice if needed.So, my plan is:1. First, list all possible special fractions. Since a and b are positive integers adding up to 15, a can be from 1 to 14, so there are 14 special fractions.2. Then, for each pair of these fractions (including pairs where both fractions are the same), compute their sum and see which ones result in integers.3. Finally, count how many distinct integers we can get from these sums.But wait, that sounds like a lot of work because there are 14 fractions, so the number of pairs is 14*14=196. That's a lot. Maybe there's a smarter way.Alternatively, perhaps I can characterize the special fractions and see when their sum is an integer.Let me think about the special fractions. Each special fraction is a/b where a + b = 15. So, b = 15 - a. Therefore, each special fraction can be written as a/(15 - a).So, the fractions are 1/14, 2/13, 3/12, 4/11, 5/10, 6/9, 7/8, 8/7, 9/6, 10/5, 11/4, 12/3, 13/2, 14/1.Wait, some of these simplify. For example, 3/12 is 1/4, 5/10 is 1/2, 6/9 is 2/3, 9/6 is 3/2, 10/5 is 2, 12/3 is 4, 14/1 is 14. But the problem says the fraction doesn't have to be in simplest form, so maybe we should consider them as they are, not simplified. Hmm, but when adding two fractions, whether they are simplified or not, the sum is the same. So, perhaps it's okay.But wait, actually, when adding fractions, the representation matters because the denominators affect the addition. So, for example, 3/12 is 1/4, but adding 3/12 + 3/12 is 6/12 = 1/2, whereas adding 1/4 + 1/4 is 1/2 as well. So, in that case, it doesn't matter. But maybe for other fractions, the denominators could affect the sum.Wait, but actually, when adding two fractions, the denominators need to be considered for a common denominator. So, perhaps if we have two fractions with different denominators, their sum might not be an integer, but if they have the same denominator, their sum could be an integer.But let me think again. The problem is about the sum of two special fractions being an integer. So, each special fraction is a/b where a + b = 15. So, each fraction is a/(15 - a). So, let me denote each special fraction as f_a = a/(15 - a).So, f_a + f_b = a/(15 - a) + b/(15 - b). We need this sum to be an integer.So, let's compute f_a + f_b:= a/(15 - a) + b/(15 - b)To add these two fractions, we need a common denominator, which is (15 - a)(15 - b). So, the sum is:[ a(15 - b) + b(15 - a) ] / [ (15 - a)(15 - b) ]Simplify the numerator:= [15a - ab + 15b - ab] / [ (15 - a)(15 - b) ]= [15a + 15b - 2ab] / [ (15 - a)(15 - b) ]So, f_a + f_b = [15(a + b) - 2ab] / [ (15 - a)(15 - b) ]We need this to be an integer. So, the numerator must be divisible by the denominator.So, [15(a + b) - 2ab] must be divisible by (15 - a)(15 - b).Let me denote c = 15 - a and d = 15 - b. Then, since a = 15 - c and b = 15 - d, where c and d are positive integers from 1 to 14 as well.So, substituting, the numerator becomes:15[(15 - c) + (15 - d)] - 2(15 - c)(15 - d)= 15[30 - c - d] - 2[225 - 15c - 15d + cd]= 450 - 15c - 15d - 450 + 30c + 30d - 2cdSimplify:450 - 450 cancels out.-15c + 30c = 15c-15d + 30d = 15dSo, numerator becomes 15c + 15d - 2cd.Denominator is c*d.So, f_a + f_b = (15c + 15d - 2cd)/(c*d)We need this to be an integer. So, (15c + 15d - 2cd) must be divisible by c*d.Let me write this as:(15c + 15d - 2cd)/(c*d) = integer.Let me factor numerator:15c + 15d - 2cd = 15(c + d) - 2cd.So, 15(c + d) - 2cd must be divisible by c*d.Hmm, maybe factor differently:Let me write it as:15(c + d) - 2cd = 15c + 15d - 2cd.Alternatively, factor out c and d:= c(15 - 2d) + d(15)Wait, not sure. Alternatively, perhaps factor 15:= 15(c + d) - 2cd.Hmm, maybe rearrange:= 15(c + d) - 2cd = 15c + 15d - 2cd.Alternatively, factor terms with c and d:= c(15 - 2d) + d(15).Hmm, not sure.Alternatively, let's think about the expression:(15c + 15d - 2cd)/(c*d) = integer.Let me split the fraction:= (15c)/(c*d) + (15d)/(c*d) - (2cd)/(c*d)Simplify each term:= 15/d + 15/c - 2So, f_a + f_b = 15/d + 15/c - 2.So, this expression must be an integer.Therefore, 15/d + 15/c must be an integer plus 2. So, 15/d + 15/c must be an integer.Therefore, 15(1/c + 1/d) must be an integer.So, 15*(1/c + 1/d) is integer.Which implies that (1/c + 1/d) must be a rational number with denominator dividing 15.But since c and d are positive integers between 1 and 14, let's think about possible values of c and d.Wait, c and d are from 1 to 14, because a and b are positive integers such that a + b = 15, so c = 15 - a, so c is from 1 to 14.So, c and d are in 1 to 14.So, 1/c + 1/d must be such that 15*(1/c + 1/d) is integer.Therefore, (1/c + 1/d) must be a fraction with denominator dividing 15.So, 1/c + 1/d must be equal to k/15, where k is an integer.But 1/c + 1/d = (c + d)/(c*d). So, (c + d)/(c*d) = k/15.Therefore, 15(c + d) = k*(c*d).So, 15(c + d) must be divisible by c*d.Which is the same as the earlier condition.Hmm, maybe another approach.Let me note that c and d are positive integers from 1 to 14.So, let me think about possible pairs (c, d) such that 15*(1/c + 1/d) is integer.Which is equivalent to 15*(c + d)/(c*d) is integer.So, 15*(c + d) must be divisible by c*d.So, c*d divides 15*(c + d).So, c*d | 15*(c + d).This is a divisibility condition.So, for given c and d, c*d must divide 15*(c + d).So, let's think about possible c and d.Since c and d are from 1 to 14, let's see for each c, what d's satisfy c*d divides 15*(c + d).Alternatively, for each c, find d such that c divides 15*(c + d)/d.Wait, maybe not. Let me think.Alternatively, for each c, let's find d such that c*d divides 15*(c + d).So, for each c, d must satisfy that d divides 15*(c + d)/c.Wait, maybe that's not helpful.Alternatively, let's fix c and see what d can be.Let me try small c's.c=1:Then, c=1, so d must satisfy 1*d divides 15*(1 + d). So, d divides 15*(1 + d).But d divides 15*(1 + d). Since d divides 15*(1 + d), and d divides 15*d, so d divides 15*(1 + d) - 15*d = 15.Therefore, d divides 15. So, d must be a divisor of 15.Since d is from 1 to 14, the possible d's are 1, 3, 5, 15. But d can't be 15 because d is at most 14. So, d=1,3,5.So, for c=1, d can be 1,3,5.Similarly, let's check c=2:c=2, so 2*d divides 15*(2 + d).So, 2d divides 15*(2 + d).So, 2d divides 15*(2 + d).Which implies that d divides 15*(2 + d)/2.Wait, maybe better to think in terms of divisibility.2d divides 15*(2 + d). So, 2d | 15*(2 + d).So, 2d divides 15*(2 + d). Let me write 15*(2 + d) = 30 + 15d.So, 2d divides 30 + 15d.So, 2d divides 30 + 15d.Which implies that 2d divides 30 + 15d - 7*(2d) = 30 + 15d -14d = 30 + d.So, 2d divides 30 + d.So, 2d divides 30 + d.Which implies that 2d divides 30 + d - 2d = 30 - d.So, 2d divides 30 - d.Which implies that 2d divides 30 - d + 2d = 30 + d.Wait, that's going in circles.Alternatively, 2d divides 30 + d.So, 30 + d must be a multiple of 2d.So, 30 + d = k*(2d), where k is integer.So, 30 + d = 2dk.So, 2dk - d = 30.So, d*(2k - 1) = 30.So, d must be a divisor of 30.Since d is from 1 to 14, possible d's are 1,2,3,5,6,10,15. But d <=14, so d=1,2,3,5,6,10.So, for each d in {1,2,3,5,6,10}, check if 2d divides 30 + d.Let me check:d=1:2*1=2 divides 30 +1=31? 31/2=15.5, not integer. So, no.d=2:2*2=4 divides 30 +2=32? 32/4=8, yes. So, d=2 is okay.d=3:2*3=6 divides 30 +3=33? 33/6=5.5, no.d=5:2*5=10 divides 30 +5=35? 35/10=3.5, no.d=6:2*6=12 divides 30 +6=36? 36/12=3, yes.d=10:2*10=20 divides 30 +10=40? 40/20=2, yes.So, for c=2, d can be 2,6,10.So, d=2,6,10.Similarly, let's do c=3:c=3, so 3d divides 15*(3 + d).15*(3 + d) = 45 + 15d.So, 3d divides 45 + 15d.So, 3d divides 45 +15d -5*(3d)=45 +15d -15d=45.So, 3d divides 45.Thus, 3d is a divisor of 45.So, 3d divides 45, so d divides 15.Since d is from 1 to14, d can be 1,3,5,15. But d<=14, so d=1,3,5.So, for c=3, d=1,3,5.Similarly, c=4:c=4, so 4d divides 15*(4 + d).15*(4 + d)=60 +15d.So, 4d divides 60 +15d.So, 4d divides 60 +15d.Let me write 60 +15d = 15d +60.So, 4d divides 15d +60.Which implies that 4d divides 15d +60 -3*(4d)=15d +60 -12d=3d +60.So, 4d divides 3d +60.Which implies that 4d divides 3d +60 - (3d)/4*4=3d +60 -3d=60.Wait, that's not helpful.Alternatively, 4d divides 3d +60.So, 4d divides 3d +60.So, 4d divides 3d +60.Which implies that 4d divides 60 - d*(4 - 3/4).Wait, maybe another approach.Let me write 4d divides 3d +60.So, 3d +60 must be a multiple of 4d.So, 3d +60 = k*(4d), where k is integer.So, 3d +60 =4dk.So, 4dk -3d =60.d*(4k -3)=60.So, d must be a divisor of 60.Since d is from1 to14, possible d's are 1,2,3,4,5,6,10,12,15, etc., but d<=14, so d=1,2,3,4,5,6,10,12.So, for each d in {1,2,3,4,5,6,10,12}, check if 4d divides 3d +60.Compute 3d +60 divided by 4d:d=1: (3 +60)/4=63/4=15.75, not integer.d=2: (6 +60)/8=66/8=8.25, no.d=3: (9 +60)/12=69/12=5.75, no.d=4: (12 +60)/16=72/16=4.5, no.d=5: (15 +60)/20=75/20=3.75, no.d=6: (18 +60)/24=78/24=3.25, no.d=10: (30 +60)/40=90/40=2.25, no.d=12: (36 +60)/48=96/48=2, yes.So, only d=12 works.So, for c=4, d=12.Similarly, c=5:c=5, so 5d divides 15*(5 + d)=75 +15d.So, 5d divides 75 +15d.So, 5d divides 75 +15d -3*(5d)=75 +15d -15d=75.Thus, 5d divides 75.So, 5d is a divisor of 75.So, 5d divides 75, so d divides 15.Since d is from1 to14, d=1,3,5,15. But d<=14, so d=1,3,5.So, for c=5, d=1,3,5.Similarly, c=6:c=6, so 6d divides 15*(6 + d)=90 +15d.So, 6d divides 90 +15d.So, 6d divides 90 +15d -2*(6d)=90 +15d -12d=90 +3d.So, 6d divides 90 +3d.Which implies that 6d divides 90 +3d.So, 6d divides 90 +3d.Which implies that 6d divides 90 +3d - (3d)/6*6=90 +3d -3d=90.Thus, 6d divides 90.So, 6d is a divisor of 90.So, 6d divides 90, so d divides 15.Since d is from1 to14, d=1,3,5,15. But d<=14, so d=1,3,5.So, for c=6, d=1,3,5.Wait, but let's check:Wait, 6d divides 90 +3d.So, 90 +3d must be divisible by 6d.So, 90 +3d = 3*(30 + d).So, 6d divides 3*(30 + d).Which implies that 2d divides (30 + d).So, 2d divides 30 + d.Which is similar to the c=2 case.So, 2d divides 30 + d.So, 30 + d = k*(2d).So, 30 + d = 2dk.So, 2dk - d =30.d*(2k -1)=30.So, d must be a divisor of 30.d is from1 to14, so d=1,2,3,5,6,10,15. But d<=14, so d=1,2,3,5,6,10.So, for each d in {1,2,3,5,6,10}, check if 2d divides 30 +d.d=1: 2 divides 31? No.d=2: 4 divides 32? Yes.d=3: 6 divides 33? 33/6=5.5, no.d=5: 10 divides 35? 35/10=3.5, no.d=6: 12 divides 36? Yes.d=10:20 divides 40? Yes.So, d=2,6,10.But earlier, we thought d=1,3,5, but actually, it's d=2,6,10.Wait, perhaps my initial conclusion was wrong.Wait, in the case of c=6, we have 6d divides 90 +15d.Which simplifies to 2d divides 30 +5d.Wait, maybe I messed up earlier.Wait, let me re-examine.c=6:6d divides 15*(6 + d)=90 +15d.So, 6d divides 90 +15d.So, 6d divides 90 +15d -2*(6d)=90 +15d -12d=90 +3d.So, 6d divides 90 +3d.So, 6d divides 90 +3d.So, 6d divides 90 +3d.Which implies that 6d divides 90 +3d - (3d)/6*6=90 +3d -3d=90.So, 6d divides 90.So, 6d is a divisor of 90.Thus, 6d divides 90, so d divides 15.So, d=1,3,5,15. But d<=14, so d=1,3,5.Wait, but earlier, when I considered 2d divides 30 +d, I got d=2,6,10.But that was under a different substitution.Wait, maybe I confused the substitution.Wait, let's go back.c=6:We have 6d divides 90 +15d.So, 6d divides 90 +15d.So, 6d divides 90 +15d.So, 6d divides 90 +15d -2*(6d)=90 +15d -12d=90 +3d.So, 6d divides 90 +3d.So, 6d divides 90 +3d.Which implies that 6d divides 90 +3d.So, 6d divides 90 +3d.So, 6d divides 90 +3d.Which implies that 6d divides 90 +3d - (3d)/6*6=90 +3d -3d=90.Thus, 6d divides 90.So, 6d is a divisor of 90.So, 6d divides 90.Thus, d must be such that 6d divides 90.So, 6d divides 90, so d divides 15.Thus, d=1,3,5,15. But d<=14, so d=1,3,5.So, for c=6, d=1,3,5.But when I tried to substitute, I thought of another approach and got d=2,6,10.But that might have been a different substitution.Wait, perhaps I need to compute for each c and d, whether 15*(1/c +1/d) is integer.So, let's try c=6, d=1:15*(1/6 +1/1)=15*(1/6 +1)=15*(7/6)=17.5, not integer.c=6, d=3:15*(1/6 +1/3)=15*(1/6 +2/6)=15*(3/6)=15*(1/2)=7.5, not integer.c=6, d=5:15*(1/6 +1/5)=15*(5/30 +6/30)=15*(11/30)=15*(11)/30=11/2=5.5, not integer.So, none of these give integers. Hmm, so maybe my earlier conclusion was wrong.Wait, but according to the earlier logic, 6d divides 90 +15d, which is true for d=1,3,5.But when we compute 15*(1/c +1/d), it's not integer. So, perhaps my logic is flawed.Wait, maybe I need to think differently.Wait, when I have f_a + f_b = [15c +15d -2cd]/(c*d) must be integer.So, let's compute for c=6, d=1:[15*6 +15*1 -2*6*1]/(6*1)= (90 +15 -12)/6=93/6=15.5, not integer.c=6, d=3:[90 +45 -36]/18=99/18=5.5, not integer.c=6, d=5:[90 +75 -60]/30=105/30=3.5, not integer.So, indeed, none of these give integers. So, my earlier conclusion that c=6, d=1,3,5 would work is incorrect.So, perhaps my approach is wrong.Wait, maybe I should go back to the expression:f_a + f_b = 15/d +15/c -2.So, 15/d +15/c must be integer +2.So, 15/d +15/c must be integer.So, 15*(1/c +1/d) must be integer.So, let me think of c and d such that 1/c +1/d is a multiple of 1/15.So, 1/c +1/d = k/15, where k is integer.So, (c + d)/(c*d) = k/15.So, 15(c + d) = k c d.So, 15(c + d) must be divisible by c*d.So, for each c, d, 15(c + d) must be divisible by c*d.So, perhaps I can think of c and d such that c divides 15(d) and d divides 15(c). Hmm, not sure.Alternatively, perhaps think about c and d as divisors of 15.Wait, 15 factors are 1,3,5,15.But c and d are from1 to14, so c and d can be 1,3,5.So, if c and d are in {1,3,5}, then c*d divides 15*(c + d).Because 15*(c + d) is multiple of 15, and c*d is 1,3,5,15, etc.Wait, let's test c=1, d=1:15*(1 +1)=30, c*d=1, 30 divisible by1, yes.c=1, d=3:15*(1 +3)=60, c*d=3, 60 divisible by3, yes.c=1, d=5:15*(1 +5)=90, c*d=5, 90 divisible by5, yes.c=3, d=3:15*(3 +3)=90, c*d=9, 90 divisible by9, yes.c=3, d=5:15*(3 +5)=120, c*d=15, 120 divisible by15, yes.c=5, d=5:15*(5 +5)=150, c*d=25, 150 divisible by25? 150/25=6, yes.So, when c and d are in {1,3,5}, then 15(c + d) is divisible by c*d.So, for these c and d, f_a + f_b is integer.But when c and d are not in {1,3,5}, like c=2, d=2:15*(2 +2)=60, c*d=4, 60 divisible by4? 60/4=15, yes.Wait, so c=2, d=2 also works.Similarly, c=2, d=6:15*(2 +6)=120, c*d=12, 120 divisible by12, yes.c=2, d=10:15*(2 +10)=180, c*d=20, 180 divisible by20? 180/20=9, yes.Similarly, c=4, d=12:15*(4 +12)=240, c*d=48, 240/48=5, yes.So, in addition to c,d in {1,3,5}, we also have pairs where c and d are multiples of 1,3,5.Wait, but c and d are from1 to14, so for example:c=2, d=2: works.c=2, d=6: works.c=2, d=10: works.c=4, d=12: works.Similarly, c=6, d=6: 15*(6 +6)=180, c*d=36, 180/36=5, yes.c=10, d=10: 15*(10 +10)=300, c*d=100, 300/100=3, yes.c=12, d=12: 15*(12 +12)=360, c*d=144, 360/144=2.5, not integer. So, doesn't work.Wait, c=12, d=12: 360/144=2.5, not integer.So, only certain pairs.Wait, so perhaps the possible c and d are such that c and d are either 1,3,5 or multiples of them, but not exceeding 14.So, c and d can be 1,2,3,4,5,6,10,12.Wait, let's see:c=1: d=1,3,5.c=2: d=2,6,10.c=3: d=1,3,5.c=4: d=12.c=5: d=1,3,5.c=6: d=1,3,5,6,10.Wait, but earlier, when c=6, d=1,3,5 didn't work, but d=6,10 might.Wait, let's check c=6, d=6:15*(6 +6)=180, c*d=36, 180/36=5, yes.c=6, d=10:15*(6 +10)=240, c*d=60, 240/60=4, yes.So, for c=6, d=6,10.Similarly, c=10, d=10:15*(10 +10)=300, c*d=100, 300/100=3, yes.c=10, d=6:Same as c=6, d=10.c=12, d=4:15*(12 +4)=240, c*d=48, 240/48=5, yes.c=12, d=12: Doesn't work.So, compiling all possible pairs:c=1: d=1,3,5.c=2: d=2,6,10.c=3: d=1,3,5.c=4: d=12.c=5: d=1,3,5.c=6: d=6,10.c=10: d=10.c=12: d=4.Wait, but c and d are symmetric, so pairs like (c=2,d=6) and (c=6,d=2) are same in terms of sum.So, to avoid duplication, maybe we can consider unordered pairs.But for now, let me list all possible (c,d) pairs where 15(c + d) is divisible by c*d:(1,1), (1,3), (1,5),(2,2), (2,6), (2,10),(3,1), (3,3), (3,5),(4,12),(5,1), (5,3), (5,5),(6,6), (6,10),(10,10),(12,4).Wait, so these are all the possible pairs.Now, for each of these pairs, compute f_a + f_b = [15c +15d -2cd]/(c*d).But since f_a + f_b =15/d +15/c -2, which is integer.So, let's compute the integer value for each pair.Let me make a table:1. (1,1):15/1 +15/1 -2=15+15-2=28.2. (1,3):15/3 +15/1 -2=5 +15 -2=18.3. (1,5):15/5 +15/1 -2=3 +15 -2=16.4. (2,2):15/2 +15/2 -2=7.5 +7.5 -2=13.Wait, 15/2 is 7.5, which is not integer, but the sum is 13, which is integer.Wait, but 15/2 +15/2=15, minus 2 is13.Yes, correct.5. (2,6):15/6 +15/2 -2=2.5 +7.5 -2=8.6. (2,10):15/10 +15/2 -2=1.5 +7.5 -2=7.7. (3,1):Same as (1,3):18.8. (3,3):15/3 +15/3 -2=5 +5 -2=8.9. (3,5):15/5 +15/3 -2=3 +5 -2=6.10. (4,12):15/12 +15/4 -2=1.25 +3.75 -2=3.11. (5,1):Same as (1,5):16.12. (5,3):Same as (3,5):6.13. (5,5):15/5 +15/5 -2=3 +3 -2=4.14. (6,6):15/6 +15/6 -2=2.5 +2.5 -2=3.15. (6,10):15/10 +15/6 -2=1.5 +2.5 -2=2.16. (10,10):15/10 +15/10 -2=1.5 +1.5 -2=1.17. (12,4):Same as (4,12):3.So, compiling all the results:28,18,16,13,8,7,18,8,6,3,16,6,4,3,2,1,3.Now, let's list all distinct integers:1,2,3,4,6,7,8,13,16,18,28.Wait, let's check:From the list:28,18,16,13,8,7,8,6,3,16,6,4,3,2,1,3.So, unique values are:1,2,3,4,6,7,8,13,16,18,28.So, that's 11 distinct integers.Wait, but let me count:1,2,3,4,6,7,8,13,16,18,28. That's 11 numbers.But wait, let me check if I missed any.Wait, in the list above, the computed sums are:28,18,16,13,8,7,18,8,6,3,16,6,4,3,2,1,3.So, the unique ones are:1,2,3,4,6,7,8,13,16,18,28.Yes, 11 distinct integers.But let me double-check if I have all the possible pairs.Wait, for c=4, d=12: sum=3.c=12, d=4: same as above.c=6, d=10: sum=2.c=10, d=6: same as above.c=10, d=10: sum=1.c=6, d=6: sum=3.c=5, d=5: sum=4.c=3, d=5: sum=6.c=5, d=3: same as above.c=3, d=3: sum=8.c=2, d=6: sum=8.c=6, d=2: same as above.c=2, d=10: sum=7.c=10, d=2: same as above.c=2, d=2: sum=13.c=1, d=5: sum=16.c=5, d=1: same as above.c=1, d=3: sum=18.c=3, d=1: same as above.c=1, d=1: sum=28.So, yes, all pairs accounted for.So, the distinct integers are:1,2,3,4,6,7,8,13,16,18,28. That's 11 distinct integers.But wait, the answer choices are A)9, B)10, C)11, D)12, E)13.So, 11 is option C.But hold on, let me make sure I didn't miss any other pairs.Wait, are there any other pairs where c and d are not in the above list but still satisfy the condition?For example, c=7, d= something.Wait, c=7:15*(7 + d) must be divisible by7*d.So, 15*(7 + d) divisible by7d.So, 15*(7 + d) =105 +15d.So, 7d divides105 +15d.So, 7d divides105 +15d.So, 7d divides105 +15d -2*(7d)=105 +15d -14d=105 +d.So, 7d divides105 +d.So, 7d divides105 +d.Which implies that7d divides105 +d - (d)/7*7=105 +d -d=105.So, 7d divides105.Thus, 7d is a divisor of105.So, 7d divides105.Thus, d divides15.So, d=1,3,5,15.But d<=14, so d=1,3,5.So, for c=7, d=1,3,5.Let me check:c=7, d=1:15*(1/7 +1/1)=15*(1/7 +1)=15*(8/7)=120/7≈17.14, not integer.c=7, d=3:15*(1/7 +1/3)=15*(3/21 +7/21)=15*(10/21)=150/21≈7.14, not integer.c=7, d=5:15*(1/7 +1/5)=15*(5/35 +7/35)=15*(12/35)=180/35≈5.14, not integer.So, none of these give integers.Similarly, c=8:15*(8 +d) must be divisible by8*d.15*(8 +d)=120 +15d.So, 8d divides120 +15d.So, 8d divides120 +15d.So, 8d divides120 +15d -1*(8d)=120 +7d.So, 8d divides120 +7d.Which implies that8d divides120 +7d - (7d)/8*8=120 +7d -7d=120.Thus, 8d divides120.So, 8d is a divisor of120.Thus, d divides15.So, d=1,3,5,15.But d<=14, so d=1,3,5.Check:c=8, d=1:15*(1/8 +1/1)=15*(1/8 +1)=15*(9/8)=135/8=16.875, not integer.c=8, d=3:15*(1/8 +1/3)=15*(3/24 +8/24)=15*(11/24)=165/24≈6.875, not integer.c=8, d=5:15*(1/8 +1/5)=15*(5/40 +8/40)=15*(13/40)=195/40=4.875, not integer.So, no integer sums.Similarly, c=9:15*(9 +d) must be divisible by9*d.15*(9 +d)=135 +15d.So, 9d divides135 +15d.So, 9d divides135 +15d -1*(9d)=135 +6d.So, 9d divides135 +6d.Which implies that9d divides135 +6d - (6d)/9*9=135 +6d -6d=135.Thus, 9d divides135.So, 9d is a divisor of135.Thus, d divides15.So, d=1,3,5,15.But d<=14, so d=1,3,5.Check:c=9, d=1:15*(1/9 +1/1)=15*(1/9 +1)=15*(10/9)=150/9≈16.666, not integer.c=9, d=3:15*(1/9 +1/3)=15*(1/9 +3/9)=15*(4/9)=60/9≈6.666, not integer.c=9, d=5:15*(1/9 +1/5)=15*(5/45 +9/45)=15*(14/45)=210/45≈4.666, not integer.So, no integer sums.Similarly, c=10:Wait, c=10 was already considered.c=11:15*(11 +d) must be divisible by11*d.15*(11 +d)=165 +15d.So, 11d divides165 +15d.So, 11d divides165 +15d.So, 11d divides165 +15d -1*(11d)=165 +4d.So, 11d divides165 +4d.Which implies that11d divides165 +4d - (4d)/11*11=165 +4d -4d=165.Thus, 11d divides165.So, 11d is a divisor of165.Thus, d divides15.So, d=1,3,5,15.But d<=14, so d=1,3,5.Check:c=11, d=1:15*(1/11 +1/1)=15*(1/11 +1)=15*(12/11)=180/11≈16.36, not integer.c=11, d=3:15*(1/11 +1/3)=15*(3/33 +11/33)=15*(14/33)=210/33≈6.36, not integer.c=11, d=5:15*(1/11 +1/5)=15*(5/55 +11/55)=15*(16/55)=240/55≈4.36, not integer.So, no integer sums.Similarly, c=13:15*(13 +d) must be divisible by13*d.15*(13 +d)=195 +15d.So, 13d divides195 +15d.So, 13d divides195 +15d.So, 13d divides195 +15d -1*(13d)=195 +2d.So, 13d divides195 +2d.Which implies that13d divides195 +2d - (2d)/13*13=195 +2d -2d=195.Thus, 13d divides195.So, 13d is a divisor of195.Thus, d divides15.So, d=1,3,5,15.But d<=14, so d=1,3,5.Check:c=13, d=1:15*(1/13 +1/1)=15*(1/13 +1)=15*(14/13)=210/13≈16.15, not integer.c=13, d=3:15*(1/13 +1/3)=15*(3/39 +13/39)=15*(16/39)=240/39≈6.15, not integer.c=13, d=5:15*(1/13 +1/5)=15*(5/65 +13/65)=15*(18/65)=270/65≈4.15, not integer.So, no integer sums.c=14:15*(14 +d) must be divisible by14*d.15*(14 +d)=210 +15d.So, 14d divides210 +15d.So, 14d divides210 +15d.So, 14d divides210 +15d -1*(14d)=210 +d.So, 14d divides210 +d.Which implies that14d divides210 +d - (d)/14*14=210 +d -d=210.Thus, 14d divides210.So, 14d is a divisor of210.Thus, d divides15.So, d=1,3,5,15.But d<=14, so d=1,3,5.Check:c=14, d=1:15*(1/14 +1/1)=15*(1/14 +1)=15*(15/14)=225/14≈16.07, not integer.c=14, d=3:15*(1/14 +1/3)=15*(3/42 +14/42)=15*(17/42)=255/42≈6.07, not integer.c=14, d=5:15*(1/14 +1/5)=15*(5/70 +14/70)=15*(19/70)=285/70≈4.07, not integer.So, no integer sums.Therefore, all possible pairs have been considered, and the only integer sums are the 11 distinct integers:1,2,3,4,6,7,8,13,16,18,28.Thus, the answer is 11, which is option C.**Final Answer**boxed{C}