💡To find the sum of the numerical coefficients of all the terms in the expansion of ((2x - 3y)^{20}), we can use a simple technique. The sum of the coefficients in a polynomial is obtained by evaluating the polynomial at (x = 1) and (y = 1).So, substituting (x = 1) and (y = 1) into the expression, we get:[(2 cdot 1 - 3 cdot 1)^{20} = (-1)^{20}]Since ((-1)) raised to an even power is (1), we have:[(-1)^{20} = 1]Therefore, the sum of the numerical coefficients is (1).The final answer is (boxed{textbf{(A)} 1}).

💡Okay, so I have this problem about a right triangle where the ratio of the legs is 2:3. I need to find the ratio of the segments of the hypotenuse created by a perpendicular from the right angle. Hmm, let me try to visualize this.First, let me draw a right triangle. Let's say the right angle is at point B, and the two legs are AB and BC. The problem says the ratio of the legs is 2:3, so maybe AB is 2 units and BC is 3 units? Or maybe it's the other way around. Wait, actually, the ratio is 2:3, so I can represent the legs as 2k and 3k, where k is some positive real number. That way, the ratio remains 2:3 regardless of the actual lengths.Now, the hypotenuse would be AC. I can find the length of AC using the Pythagorean theorem. So, AC squared equals AB squared plus BC squared. Plugging in the values, AC squared equals (2k)^2 plus (3k)^2, which is 4k² + 9k², so that's 13k². Therefore, AC is the square root of 13k², which simplifies to k times the square root of 13. So, AC is k√13.Next, the problem mentions a perpendicular from the vertex of the right angle, which is point B, to the hypotenuse AC. Let's call the point where this perpendicular meets AC as point D. So, BD is perpendicular to AC, and it divides AC into two segments: AD and DC. I need to find the ratio of AD to DC.I remember that in a right triangle, when you draw an altitude from the right angle to the hypotenuse, it creates two smaller right triangles that are similar to each other and to the original triangle. So, triangles ABD, CBD, and ABC are all similar.Since these triangles are similar, the ratios of their corresponding sides should be equal. Let me think about how to apply this. The ratio of the legs in the original triangle is 2:3, so maybe that ratio is preserved in some way in the smaller triangles.Let me denote AD as x and DC as y. So, the entire hypotenuse AC is x + y, which we already found to be k√13. But I don't know if that helps directly.Wait, I remember there's a geometric mean theorem which states that in a right triangle, the length of the altitude drawn to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse. So, BD squared equals AD times DC. That is, BD² = AD * DC.But I don't know BD yet. Maybe I can find BD in terms of k. Alternatively, since the triangles are similar, I can set up proportions based on the sides.Looking at triangle ABD and triangle ABC, they are similar. So, the ratio of AB to AC should be equal to the ratio of AD to AB. So, AB/AC = AD/AB. Plugging in the values, (2k)/(k√13) = x/(2k). Simplifying, 2k divided by k√13 is 2/√13, and x/(2k) is x/(2k). So, 2/√13 = x/(2k). Solving for x, I get x = (2/√13) * 2k = 4k/√13.Similarly, looking at triangle CBD and triangle ABC, they are also similar. So, the ratio of BC to AC should be equal to the ratio of DC to BC. So, BC/AC = DC/BC. Plugging in the values, (3k)/(k√13) = y/(3k). Simplifying, 3k divided by k√13 is 3/√13, and y/(3k) is y/(3k). So, 3/√13 = y/(3k). Solving for y, I get y = (3/√13) * 3k = 9k/√13.Now, I have AD = 4k/√13 and DC = 9k/√13. So, the ratio of AD to DC is (4k/√13) : (9k/√13). The k and √13 cancel out, leaving 4:9.Wait, but the options given are 1:2, 2:3, 3:2, 9:4, and 4:9. So, 4:9 is one of the options, specifically option E. But I want to double-check my work because sometimes ratios can be tricky.Let me verify using another method. I remember that in similar triangles, the ratio of areas is the square of the ratio of corresponding sides. But I'm not sure if that applies here directly.Alternatively, I can use the fact that the segments of the hypotenuse are proportional to the squares of the adjacent sides. So, AD/DC = (AB)²/(BC)². Plugging in the values, (2k)²/(3k)² = 4k²/9k² = 4/9. So, AD:DC = 4:9.Hmm, that confirms my earlier result. So, the ratio is 4:9, which is option E.Wait a second, but I recall sometimes the ratio is taken as DC:AD instead of AD:DC. Let me check the problem statement again. It says the ratio of the corresponding segments of the hypotenuse created by the perpendicular from the vertex of the right angle. So, it depends on the order. If AD is closer to the shorter leg, then DC is closer to the longer leg. Since AB is 2k and BC is 3k, AD is closer to AB and DC is closer to BC. So, the ratio is AD:DC = 4:9.But let me think again. If the legs are in the ratio 2:3, then the segments of the hypotenuse should be in the ratio of the squares of the legs. So, (2)^2 : (3)^2 = 4:9. So, that's consistent.Alternatively, if I use coordinate geometry, placing point B at the origin, point A at (0, 2k), and point C at (3k, 0). Then, the hypotenuse AC goes from (0, 2k) to (3k, 0). The equation of AC can be found, and then the foot of the perpendicular from B(0,0) to AC can be calculated.Let me try that approach. The coordinates of A are (0, 2k) and C are (3k, 0). The slope of AC is (0 - 2k)/(3k - 0) = -2/3. Therefore, the equation of AC is y = (-2/3)x + 2k.The slope of the perpendicular from B to AC is the negative reciprocal of -2/3, which is 3/2. So, the equation of BD is y = (3/2)x.To find the intersection point D, solve the system:y = (-2/3)x + 2ky = (3/2)xSet them equal:(3/2)x = (-2/3)x + 2kMultiply both sides by 6 to eliminate denominators:9x = -4x + 12k9x + 4x = 12k13x = 12kx = (12k)/13Then, y = (3/2)x = (3/2)*(12k/13) = (36k)/26 = (18k)/13So, point D is at ((12k)/13, (18k)/13)Now, let's find the lengths of AD and DC.First, coordinates of A: (0, 2k), D: (12k/13, 18k/13)Distance AD:sqrt[(12k/13 - 0)^2 + (18k/13 - 2k)^2]= sqrt[(144k²/169) + (18k/13 - 26k/13)^2]= sqrt[(144k²/169) + (-8k/13)^2]= sqrt[(144k²/169) + (64k²/169)]= sqrt[(208k²)/169]= sqrt(208k²)/13= (4√13 k)/13Similarly, coordinates of C: (3k, 0), D: (12k/13, 18k/13)Distance DC:sqrt[(3k - 12k/13)^2 + (0 - 18k/13)^2]= sqrt[(39k/13 - 12k/13)^2 + (-18k/13)^2]= sqrt[(27k/13)^2 + (18k/13)^2]= sqrt[(729k²/169) + (324k²/169)]= sqrt[(1053k²)/169]= sqrt(1053k²)/13= (9√13 k)/13So, AD = (4√13 k)/13 and DC = (9√13 k)/13. Therefore, the ratio AD:DC is 4:9.This matches my earlier result. So, the ratio is 4:9, which is option E.Wait, but looking back at the problem, the options are:(A) 1:2(B) 2:3(C) 3:2(D) 9:4(E) 4:9So, 4:9 is option E. Therefore, the answer should be E.But just to make sure, let me think if there's another way to approach this. Maybe using trigonometry?In the original triangle, the legs are 2k and 3k, so the angles can be found using trigonometric ratios. Let's say angle at A is θ, then tanθ = opposite/adjacent = BC/AB = 3k/2k = 3/2. So, θ = arctan(3/2).When we draw the altitude BD, it creates two smaller right triangles, ABD and CBD. In triangle ABD, angle at D is the same as angle θ, so tanθ = BD/AD. Similarly, in triangle CBD, tan(angle at D) = BD/DC.But since tanθ = 3/2, we have BD/AD = 3/2 and BD/DC = 2/3. Wait, that seems contradictory. Let me clarify.Actually, in triangle ABD, angle at A is θ, so tanθ = BD/AD. In triangle CBD, angle at C is φ, which is complementary to θ, so tanφ = BD/DC.Since θ + φ = 90°, tanφ = cotθ = 1/tanθ = 2/3.So, tanθ = 3/2 = BD/AD => BD = (3/2)ADAnd tanφ = 2/3 = BD/DC => BD = (2/3)DCTherefore, (3/2)AD = (2/3)DCCross-multiplying, 9AD = 4DC => AD/DC = 4/9So, AD:DC = 4:9, which again confirms the ratio is 4:9.Alright, I think I've approached this problem in multiple ways: using similar triangles, coordinate geometry, and trigonometry, and all methods lead me to the same conclusion that the ratio is 4:9.Therefore, the correct answer is E) 4:9.

💡Alright, so I've got this vector problem here. Let me try to figure it out step by step. First, the problem says that vector OA is (1, -3). So, if I imagine this on a coordinate plane, point A is at (1, -3). Then, it mentions that the magnitude of OA is equal to the magnitude of OB. So, OB must be another vector starting from the origin, just like OA, but with the same length. Let me calculate the magnitude of OA first because that will help me understand what OB's magnitude should be. The magnitude of a vector (a, b) is given by the square root of (a² + b²). So, for OA, that's sqrt(1² + (-3)²) = sqrt(1 + 9) = sqrt(10). So, |OA| is sqrt(10), which means |OB| is also sqrt(10). Now, the problem also says that the magnitude of vector AB is 2*sqrt(5). Vector AB is the vector from point A to point B, right? So, if OA is (1, -3) and OB is some vector (x, y), then vector AB would be OB - OA, which is (x - 1, y - (-3)) = (x - 1, y + 3). The magnitude of AB is given as 2*sqrt(5), so let's write that equation out. The magnitude of AB is sqrt[(x - 1)² + (y + 3)²] = 2*sqrt(5). If I square both sides to eliminate the square root, I get (x - 1)² + (y + 3)² = (2*sqrt(5))² = 4*5 = 20. So, now I have two equations:1. x² + y² = 10 (since |OB| = sqrt(10))2. (x - 1)² + (y + 3)² = 20I need to solve these two equations to find the values of x and y, which will give me vector OB. Once I have OB, I can compute the dot product of OA and OB, which is what the problem is asking for.Let me expand the second equation to make it easier to work with. Expanding (x - 1)² gives x² - 2x + 1, and expanding (y + 3)² gives y² + 6y + 9. So, putting it all together, the second equation becomes:x² - 2x + 1 + y² + 6y + 9 = 20Simplify that:x² + y² - 2x + 6y + 10 = 20But from the first equation, I know that x² + y² = 10. So, I can substitute that into the equation above:10 - 2x + 6y + 10 = 20Combine like terms:10 + 10 - 2x + 6y = 2020 - 2x + 6y = 20Subtract 20 from both sides:-2x + 6y = 0I can simplify this equation by dividing both sides by -2:x - 3y = 0So, x = 3yNow, I can substitute x = 3y back into the first equation, which is x² + y² = 10.Substituting x:(3y)² + y² = 109y² + y² = 1010y² = 10Divide both sides by 10:y² = 1So, y = sqrt(1) or y = -sqrt(1), which means y = 1 or y = -1.Now, let's find the corresponding x values using x = 3y.If y = 1, then x = 3*1 = 3. So, one possible vector OB is (3, 1).If y = -1, then x = 3*(-1) = -3. So, the other possible vector OB is (-3, -1).So, there are two possible vectors for OB: (3, 1) and (-3, -1). Now, the problem asks for the dot product of OA and OB. Let's compute that for both cases.First, OA is (1, -3). Let's compute the dot product with OB = (3, 1):OA · OB = (1)(3) + (-3)(1) = 3 - 3 = 0Now, let's compute the dot product with OB = (-3, -1):OA · OB = (1)(-3) + (-3)(-1) = -3 + 3 = 0So, in both cases, the dot product is 0.Therefore, regardless of which vector OB we choose, the dot product OA · OB is 0.**Final Answer**The dot product of vectors OA and OB is boxed{0}.

💡To determine the total cost of carpeting the room, follow these steps:1. **Calculate the area of the room:** - Length of the room = 15 meters - Breadth of the room = 6 meters - Area = Length × Breadth = 15 m × 6 m = 90 square meters2. **Determine the width of the carpet in meters:** - Carpet width = 75 cm = 0.75 meters3. **Calculate the length of carpet required:** - Length of carpet = Area / Width of carpet = 90 m² / 0.75 m = 120 meters4. **Compute the total cost:** - Cost per meter = 30 paise = ₹0.30 - Total cost = Length of carpet × Cost per meter = 120 m × ₹0.30/m = ₹36Therefore, the total cost of carpeting the room is ₹36.

💡Okay, so I have these five indefinite integrals to solve. Let me take them one by one and try to figure them out. I remember that integration can sometimes be tricky, but with the right techniques like substitution or integration by parts, I can handle them.Starting with the first one: ∫ sin x cos³ x dx. Hmm, I think substitution might work here. If I let u be cos x, then du would be -sin x dx. That seems promising because I have a sin x term in the integrand. So, substituting u = cos x, du = -sin x dx. Then, the integral becomes -∫ u³ du. That's straightforward. Integrating u³ gives (u⁴)/4, so the integral should be -(cos⁴ x)/4 + C. I think that's correct.Moving on to the second integral: ∫ dx / [(1 + √x)√x]. This looks a bit more complicated. Maybe substitution again? Let me set u = 1 + √x. Then, du/dx = 1/(2√x), which means du = dx/(2√x). So, rearranging, dx/√x = 2 du. That transforms the integral into 2 ∫ du / u, which is 2 ln |u| + C. Substituting back, it becomes 2 ln(1 + √x) + C. That seems right.Third integral: ∫ x² √(x³ + 1) dx. This one might also use substitution. Let me let u = x³ + 1. Then, du = 3x² dx, which means (1/3) du = x² dx. So, the integral becomes (1/3) ∫ √u du. The integral of √u is (2/3)u^(3/2), so multiplying by 1/3 gives (2/9)u^(3/2) + C. Substituting back, it's (2/9)(x³ + 1)^(3/2) + C. That looks good.Fourth integral: ∫ (e²ˣ - 3eˣ)/eˣ dx. Let me simplify the integrand first. Dividing both terms by eˣ gives eˣ - 3. So, the integral becomes ∫ (eˣ - 3) dx. Integrating term by term, the integral of eˣ is eˣ, and the integral of 3 is 3x. So, putting it together, it's eˣ - 3x + C. That seems straightforward.Fifth integral: ∫ (1 - x²)eˣ dx. This one looks like it might require integration by parts. Let me recall the formula: ∫ u dv = uv - ∫ v du. Let me set u = 1 - x², so du = -2x dx. Then, dv = eˣ dx, so v = eˣ. Applying the formula, we get (1 - x²)eˣ - ∫ eˣ(-2x) dx. Simplifying, that's (1 - x²)eˣ + 2 ∫ x eˣ dx. Now, I need to compute ∫ x eˣ dx, which again uses integration by parts. Let me set u = x, so du = dx, and dv = eˣ dx, so v = eˣ. Then, ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C. Putting it all back together, the original integral becomes (1 - x²)eˣ + 2(x eˣ - eˣ) + C. Simplifying, that's (1 - x²)eˣ + 2x eˣ - 2eˣ + C. Combining like terms, it's eˣ(1 - x² + 2x - 2) + C, which simplifies to eˣ(-x² + 2x -1) + C. Factoring out the negative, it's -eˣ(x² - 2x +1) + C, and since x² - 2x +1 is (x -1)², the final answer is -eˣ(x -1)² + C.Let me double-check each result to make sure I didn't make any mistakes.1. For the first integral, substitution worked well, and the result seems correct.2. The second integral also used substitution, and the steps look good. The natural logarithm makes sense here.3. The third integral was another substitution, and the algebra checks out.4. The fourth integral was simplified before integrating, and the result is straightforward.5. The fifth integral required integration by parts twice, and the algebra steps seem correct. The final expression simplifies nicely.I think all the solutions are correct. Let me write them down clearly.

💡Alright, so I've got this problem here about Xiaoming playing a chess game on a Cartesian coordinate system. The piece starts at the origin, and there are specific movement rules for each step. I need to figure out where the piece will be after the 100th step.First, let me try to understand the movement rules. It says that on the 1st step, move 1 unit to the right. On the 2nd step, move 2 units to the right. On the 3rd step, move 1 unit up. Then on the 4th step, it goes back to moving 1 unit to the right, and so on. So, it seems like there's a repeating pattern every three steps.Let me break it down:- When the step number (n) is divisible by 3, move 1 unit up.- When n divided by 3 leaves a remainder of 1, move 1 unit to the right.- When n divided by 3 leaves a remainder of 2, move 2 units to the right.So, for example:- Step 1: 1 mod 3 = 1 → move 1 right.- Step 2: 2 mod 3 = 2 → move 2 right.- Step 3: 3 mod 3 = 0 → move 1 up.- Step 4: 4 mod 3 = 1 → move 1 right.- Step 5: 5 mod 3 = 2 → move 2 right.- Step 6: 6 mod 3 = 0 → move 1 up.And this pattern continues. So every three steps, the piece moves 1 right, 2 right, and 1 up. That means in each cycle of three steps, the piece moves a total of 3 units to the right and 1 unit up.Okay, so if I can figure out how many complete cycles there are in 100 steps, and then account for any remaining steps, I can calculate the final coordinates.Let me calculate how many complete cycles are in 100 steps. Since each cycle is 3 steps, I'll divide 100 by 3.100 ÷ 3 = 33 with a remainder of 1.So, there are 33 complete cycles and 1 additional step.Now, let's calculate the movement for the complete cycles.In each cycle:- Right movement: 1 + 2 = 3 units- Up movement: 1 unitSo, over 33 cycles:- Total right movement: 33 cycles × 3 units = 99 units- Total up movement: 33 cycles × 1 unit = 33 unitsNow, we have 1 additional step after the 33 cycles. Since 100 steps = 33 cycles × 3 steps + 1 step.What does the 100th step entail? Let's see:Step 100: 100 mod 3 = 1 (since 100 = 3×33 + 1). So, according to the rules, when the remainder is 1, move 1 unit to the right.Therefore, the 100th step adds 1 more unit to the right.So, total right movement becomes 99 + 1 = 100 units.Total up movement remains 33 units.Therefore, the final coordinates after the 100th step are (100, 33).Wait, let me double-check my calculations to make sure I didn't make a mistake.Total cycles: 33, each contributing 3 right and 1 up. So, 33×3=99 right, 33×1=33 up. Then, step 100 is 1 right, so 99+1=100 right. That seems correct.Is there any chance I miscounted the cycles or the remainder? Let me verify:100 divided by 3 is 33.333..., so 33 full cycles, and 1 extra step. Yes, that's correct.And since the extra step is step 100, which is 100 mod 3=1, so it's a remainder of 1, meaning move 1 right. Correct.So, yeah, the final coordinates should be (100, 33).I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**The coordinates of the piece after the 100th step are boxed{(100, 33)}.