answer：First, I substitute ( x = 2 ) into the expression ( ax^3 + bx - 7 ) and set it equal to (-19):[a(2)^3 + b(2) - 7 = -19]Simplifying, I get:[8a + 2b - 7 = -19]Adding 7 to both sides:[8a + 2b = -12]Dividing the entire equation by 2:[4a + b = -6]Next, I substitute ( x = -2 ) into the original expression:[a(-2)^3 + b(-2) - 7]Simplifying, this becomes:[-8a - 2b - 7]I notice that (-8a - 2b) is the negative of (8a + 2b), which from the earlier equation equals (-12). Therefore:[-8a - 2b = 12]Substituting back, the expression when ( x = -2 ) is:[12 - 7 = 5]

answer：Okay, so I have this geometry problem here. Let me try to visualize it first. There's a triangle ABC, and on side AC, points E and K are taken such that E is between A and K. The ratio of AE:EK:KC is 3:5:4. So, if I imagine side AC divided into three parts, the first part from A to E is 3 units, then from E to K is 5 units, and from K to C is 4 units. That makes the total length of AC equal to 3 + 5 + 4, which is 12 units. So, AC is divided into segments of 3, 5, and 4.Now, the median AD intersects segments BE and BK at points L and M respectively. I need to find the ratio of the areas of triangles BLM and ABC, and the answer is supposed to be 1/5. Hmm, okay.First, let me recall that a median of a triangle divides it into two smaller triangles of equal area. So, since AD is the median, it splits ABC into two triangles, ABD and ADC, each with half the area of ABC.But the problem is about triangles BLM and ABC. So, I need to figure out where exactly L and M are located on BE and BK, respectively, and then find the area of triangle BLM.Maybe coordinate geometry can help here. Let me assign coordinates to the triangle ABC. Let me place point A at (0, 0), point C at (12, 0) since AC is 12 units long. Then, point B can be somewhere; since it's a triangle, I can assign coordinates to B as (x, y). But maybe it's easier to place B at (0, b) so that the triangle is not degenerate. Wait, but if I place A at (0,0) and C at (12,0), then point B can be at (p, q). Hmm, but maybe it's better to place B at (0, b) so that the triangle is right-angled? Or maybe not necessarily right-angled, but just somewhere above the x-axis.Wait, perhaps it's better to set coordinates such that calculations are easier. Let me think. Let me place point A at (0, 0), point C at (12, 0). Then, point B can be at (6, h), so that the median AD goes from A(0,0) to D, the midpoint of BC. If B is at (6, h), then midpoint D of BC would be at ((6 + 12)/2, (h + 0)/2) = (9, h/2). Wait, no, if C is at (12, 0), then midpoint D is at ((6 + 12)/2, (h + 0)/2) = (9, h/2). Hmm, okay.Wait, but if I place B at (6, h), then the triangle is symmetric with respect to the median AD. Hmm, maybe that's a good idea because it might simplify calculations.So, let me define the coordinates:- A: (0, 0)- C: (12, 0)- B: (6, h)- D: midpoint of BC, so coordinates ((6 + 12)/2, (h + 0)/2) = (9, h/2)Now, points E and K are on AC. Since AC is from (0,0) to (12,0), and the ratio AE:EK:KC is 3:5:4. So, starting from A, moving 3 units to E, then 5 units to K, then 4 units to C. So, the coordinates of E and K can be calculated.Since AC is along the x-axis from (0,0) to (12,0), the points E and K will have coordinates:- E is 3 units from A, so E is at (3, 0)- K is 3 + 5 = 8 units from A, so K is at (8, 0)So, E is at (3, 0) and K is at (8, 0).Now, we need to find the equations of lines BE and BK, and then find their intersection points L and M with the median AD.First, let's find the equation of median AD. Since A is at (0,0) and D is at (9, h/2), the parametric equations for AD can be written as:x = 9ty = (h/2)twhere t ranges from 0 to 1.Similarly, let's find the equation of BE. Point B is at (6, h) and point E is at (3, 0). So, the parametric equations for BE can be written as:x = 6 - 3sy = h - hswhere s ranges from 0 to 1.Similarly, the equation of BK. Point B is at (6, h) and point K is at (8, 0). So, the parametric equations for BK can be written as:x = 6 + 2ty = h - htwhere t ranges from 0 to 1.Wait, actually, to avoid confusion with parameter t used in AD, maybe I should use different parameters for BE and BK. Let me adjust that.For BE:x = 6 - 3sy = h - hswhere s ∈ [0,1].For BK:x = 6 + 2uy = h - huwhere u ∈ [0,1].Now, to find point L, which is the intersection of AD and BE.So, we have parametric equations for AD:x = 9ty = (h/2)tand for BE:x = 6 - 3sy = h - hsWe need to find t and s such that:9t = 6 - 3s ...(1)(h/2)t = h - hs ...(2)From equation (2):(h/2)t = h(1 - s)Divide both sides by h (assuming h ≠ 0):(1/2)t = 1 - sSo, s = 1 - (t/2) ...(3)From equation (1):9t = 6 - 3sSubstitute s from equation (3):9t = 6 - 3(1 - t/2) = 6 - 3 + (3t)/2 = 3 + (3t)/2So, 9t = 3 + (3t)/2Multiply both sides by 2 to eliminate the fraction:18t = 6 + 3tSubtract 3t:15t = 6So, t = 6/15 = 2/5Then, from equation (3):s = 1 - (2/5)/2 = 1 - (1/5) = 4/5So, point L is on AD at t = 2/5, so its coordinates are:x = 9*(2/5) = 18/5 = 3.6y = (h/2)*(2/5) = h/5So, L is at (18/5, h/5)Similarly, let's find point M, which is the intersection of AD and BK.Parametric equations for AD:x = 9ty = (h/2)tParametric equations for BK:x = 6 + 2uy = h - huSet them equal:9t = 6 + 2u ...(4)(h/2)t = h - hu ...(5)From equation (5):(h/2)t = h(1 - u)Divide both sides by h:(1/2)t = 1 - uSo, u = 1 - (t/2) ...(6)From equation (4):9t = 6 + 2uSubstitute u from equation (6):9t = 6 + 2(1 - t/2) = 6 + 2 - t = 8 - tSo, 9t + t = 810t = 8t = 8/10 = 4/5Then, from equation (6):u = 1 - (4/5)/2 = 1 - (2/5) = 3/5So, point M is on AD at t = 4/5, so its coordinates are:x = 9*(4/5) = 36/5 = 7.2y = (h/2)*(4/5) = (2h)/5So, M is at (36/5, 2h/5)Now, we have points B, L, and M. Let's write their coordinates:- B: (6, h)- L: (18/5, h/5)- M: (36/5, 2h/5)We need to find the area of triangle BLM.To find the area of triangle BLM, we can use the shoelace formula.First, let's list the coordinates:B: (6, h)L: (18/5, h/5)M: (36/5, 2h/5)Let me write them in order:Point 1: B (6, h)Point 2: L (18/5, h/5)Point 3: M (36/5, 2h/5)Shoelace formula:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Plugging in the values:x1 = 6, y1 = hx2 = 18/5, y2 = h/5x3 = 36/5, y3 = 2h/5Compute each term:First term: x1(y2 - y3) = 6*(h/5 - 2h/5) = 6*(-h/5) = -6h/5Second term: x2(y3 - y1) = (18/5)*(2h/5 - h) = (18/5)*(-3h/5) = -54h/25Third term: x3(y1 - y2) = (36/5)*(h - h/5) = (36/5)*(4h/5) = 144h/25Now, sum these terms:-6h/5 -54h/25 + 144h/25Convert -6h/5 to -30h/25 to have a common denominator:-30h/25 -54h/25 + 144h/25 = (-30 -54 + 144)h/25 = (60)h/25 = 12h/5Take the absolute value and divide by 2:Area = |12h/5| / 2 = (12h/5)/2 = 6h/5So, the area of triangle BLM is 6h/5.Now, let's find the area of triangle ABC.Since ABC is a triangle with base AC = 12 units and height h (since B is at (6, h)), the area is (base * height)/2 = (12 * h)/2 = 6h.So, the area of ABC is 6h.Therefore, the ratio of the areas of triangles BLM to ABC is (6h/5) / (6h) = (6h/5) * (1/(6h)) = 1/5.So, the ratio is 1/5, which matches the given answer.Wait, let me just double-check my calculations to make sure I didn't make any mistakes.First, coordinates:- A(0,0), C(12,0), B(6,h), D(9, h/2). That seems correct.Points E and K:- E at 3 units from A: (3,0)- K at 8 units from A: (8,0). Correct.Equations for AD, BE, BK:- AD: parametric from (0,0) to (9, h/2): x=9t, y=(h/2)t. Correct.- BE: from (6,h) to (3,0): x=6-3s, y=h - hs. Correct.- BK: from (6,h) to (8,0): x=6+2u, y=h - hu. Correct.Intersection L:Solved equations for AD and BE:From y: (h/2)t = h - hs => s = 1 - t/2.From x: 9t = 6 - 3s => 9t = 6 - 3(1 - t/2) = 3 + 1.5t => 7.5t = 3 => t = 3 / 7.5 = 0.4 = 2/5. Correct.Then, s = 1 - (2/5)/2 = 1 - 1/5 = 4/5. Correct.Coordinates of L: (18/5, h/5). Correct.Intersection M:Solved equations for AD and BK:From y: (h/2)t = h - hu => u = 1 - t/2.From x: 9t = 6 + 2u => 9t = 6 + 2(1 - t/2) = 8 - t => 10t = 8 => t = 4/5. Correct.Then, u = 1 - (4/5)/2 = 1 - 2/5 = 3/5. Correct.Coordinates of M: (36/5, 2h/5). Correct.Shoelace formula:Points B(6,h), L(18/5, h/5), M(36/5, 2h/5).Compute:First term: 6*(h/5 - 2h/5) = 6*(-h/5) = -6h/5Second term: (18/5)*(2h/5 - h) = (18/5)*(-3h/5) = -54h/25Third term: (36/5)*(h - h/5) = (36/5)*(4h/5) = 144h/25Sum: -6h/5 -54h/25 + 144h/25Convert -6h/5 to -30h/25: -30h/25 -54h/25 + 144h/25 = (60h)/25 = 12h/5Area: |12h/5| /2 = 6h/5. Correct.Area of ABC: 6h. Correct.Ratio: (6h/5)/6h = 1/5. Correct.So, all steps seem correct. I think the answer is indeed 1/5.**Final Answer**The ratio of the areas of triangles (BLM) and (ABC) is boxed{dfrac{1}{5}}.

answer：Okay, so I have this problem where I need to find ( f_{2010}(17) ). Let me try to understand what this means. The function ( f(n) ) is defined as the sum of the digits of ( n^2 + 1 ). Then, ( f_1(n) = f(n) ), ( f_2(n) = f(f_1(n)) ), and so on, up to ( f_{2010}(n) ). So, essentially, I need to apply the function ( f ) repeatedly 2010 times starting from 17.First, let me make sure I understand how ( f(n) ) works. For example, they gave ( f(14) = 17 ) because ( 14^2 + 1 = 197 ) and the sum of the digits ( 1 + 9 + 7 = 17 ). So, for any natural number ( n ), I square it, add 1, and then sum the digits of the result.Alright, so starting with ( n = 17 ), I need to compute ( f(17) ), then ( f(f(17)) ), and keep doing this until I've applied ( f ) 2010 times. That seems like a lot, but maybe there's a pattern or cycle that emerges, which would make it easier.Let me start by computing the first few iterations manually to see if I can spot a pattern.First, compute ( f_1(17) ):- ( 17^2 + 1 = 289 + 1 = 290 )- Sum of digits: ( 2 + 9 + 0 = 11 )So, ( f_1(17) = 11 )Next, compute ( f_2(17) = f(11) ):- ( 11^2 + 1 = 121 + 1 = 122 )- Sum of digits: ( 1 + 2 + 2 = 5 )So, ( f_2(17) = 5 )Now, compute ( f_3(17) = f(5) ):- ( 5^2 + 1 = 25 + 1 = 26 )- Sum of digits: ( 2 + 6 = 8 )So, ( f_3(17) = 8 )Next, ( f_4(17) = f(8) ):- ( 8^2 + 1 = 64 + 1 = 65 )- Sum of digits: ( 6 + 5 = 11 )So, ( f_4(17) = 11 )Wait a minute, that's the same as ( f_1(17) ). So, ( f_4(17) = 11 ), which is equal to ( f_1(17) ). That suggests that starting from 11, the sequence cycles every 3 steps: 11 → 5 → 8 → 11 → 5 → 8 → ...Let me confirm that:Compute ( f_5(17) = f(11) = 5 ) as before.Compute ( f_6(17) = f(5) = 8 ).Compute ( f_7(17) = f(8) = 11 ).Yep, so it cycles every 3 steps: 11, 5, 8, 11, 5, 8, etc.So, starting from ( f_1(17) = 11 ), the sequence is 11, 5, 8, 11, 5, 8, and so on. So, the cycle length is 3.Therefore, the sequence of ( f_k(17) ) for ( k geq 1 ) is periodic with period 3: 11, 5, 8, 11, 5, 8, ...So, to find ( f_{2010}(17) ), I need to determine where 2010 falls in this cycle.Since the cycle starts at ( k = 1 ) with 11, the positions in the cycle correspond to ( k ) modulo 3.Let me see:- ( k = 1 ): 11 (position 1)- ( k = 2 ): 5 (position 2)- ( k = 3 ): 8 (position 0, since 3 mod 3 is 0)- ( k = 4 ): 11 (position 1)- ( k = 5 ): 5 (position 2)- ( k = 6 ): 8 (position 0)- And so on.Wait, so for ( k ) modulo 3:- If ( k equiv 1 mod 3 ), then ( f_k(17) = 11 )- If ( k equiv 2 mod 3 ), then ( f_k(17) = 5 )- If ( k equiv 0 mod 3 ), then ( f_k(17) = 8 )So, to find ( f_{2010}(17) ), I need to compute ( 2010 mod 3 ).Let me compute that:2010 divided by 3 is 670 with no remainder, because 3 × 670 = 2010. So, 2010 mod 3 is 0.Therefore, ( f_{2010}(17) = 8 ).Wait, but let me double-check my reasoning.Starting from ( f_1(17) = 11 ), which is position 1.So, for ( k = 1 ): 11( k = 2 ): 5( k = 3 ): 8( k = 4 ): 11( k = 5 ): 5( k = 6 ): 8So, yes, the cycle is 11,5,8 repeating every 3 steps.So, if I consider ( k = 1 ) as the first step, then:- ( k = 1 ): 11 (1 mod 3)- ( k = 2 ): 5 (2 mod 3)- ( k = 3 ): 8 (0 mod 3)- ( k = 4 ): 11 (1 mod 3)- etc.So, for ( k ) values, their mod 3 gives:1: 1 mod 3 → 112: 2 mod 3 → 53: 0 mod 3 → 84: 1 mod 3 → 115: 2 mod 3 → 56: 0 mod 3 → 8So, yes, 2010 mod 3 is 0, so it's the third element in the cycle, which is 8.Therefore, ( f_{2010}(17) = 8 ).But just to make sure, let me compute a few more terms to confirm the cycle.Compute ( f_7(17) = f(11) = 5 )( f_8(17) = f(5) = 8 )( f_9(17) = f(8) = 11 )( f_{10}(17) = f(11) = 5 )Yes, the cycle continues as 11,5,8,11,5,8,...So, every time ( k ) is a multiple of 3, like 3,6,9,..., ( f_k(17) = 8 ).Since 2010 is a multiple of 3, it falls into this category, so the result is 8.Therefore, I can confidently say that ( f_{2010}(17) = 8 ).**Final Answer**boxed{8}

answer：First, I recognize that the perimeter of a triangle is the sum of its three sides. Given the sides are 7, (x + 4), and (2x + 1), and the perimeter is 36, I can set up the equation:7 + (x + 4) + (2x + 1) = 36Next, I'll simplify and solve for (x):7 + x + 4 + 2x + 1 = 36 (7 + 4 + 1) + (x + 2x) = 36 12 + 3x = 36 3x = 24 x = 8Now that I have the value of (x), I can find the lengths of all sides:First side: 7 Second side: (x + 4 = 8 + 4 = 12) Third side: (2x + 1 = 2(8) + 1 = 17)Finally, I'll identify the longest side, which is 17.

answer：First, I'll tackle the first expression: √18 - √32 + √2.I know that √18 can be simplified by breaking it down into √(9×2), which is √9 × √2, resulting in 3√2.Similarly, √32 can be simplified as √(16×2), which is √16 × √2, giving 4√2.Now, substituting these back into the original expression, it becomes 3√2 - 4√2 + √2.Combining like terms, 3√2 - 4√2 is -√2, and adding √2 gives 0.Next, I'll solve the second expression: √6 ÷ √18 × √27.I'll start by simplifying each square root. √18 can be written as √(9×2), which is 3√2. Similarly, √27 is √(9×3), which is 3√3.Substituting these into the expression, it becomes √6 ÷ (3√2) × (3√3).The 3 in the denominator and the 3 in the numerator cancel each other out, leaving √6 ÷ √2 × √3.Simplifying √6 ÷ √2 gives √(6/2) which is √3. Multiplying √3 by √3 results in √9, which is 3.So, the final answers are 0 for the first expression and 3 for the second.

answer：First, I need to determine the total number of boxes distributed. There are 11 families receiving a box each, plus one additional box given to the community center, making a total of 12 boxes.Next, I'll calculate the number of each type of decoration in one box:- Tinsel: 4 pieces- Christmas trees: 1 tree- Snow globes: 5 globesTo find the total number of each decoration type distributed, I'll multiply the number of decorations in one box by the total number of boxes:- Total tinsel: 4 pieces/box * 12 boxes = 48 pieces- Total Christmas trees: 1 tree/box * 12 boxes = 12 trees- Total snow globes: 5 globes/box * 12 boxes = 60 globesFinally, I'll add up all the decorations to get the total number handed out:48 pieces of tinsel + 12 Christmas trees + 60 snow globes = 120 decorations