💡Okay, so I have this trapezium (ABCD) where (AD) is parallel to (BC). Points (K) and (L) are on sides (AB) and (CD) respectively, and it's given that (angle BAL = angle CDK). I need to prove that (angle BLA = angle CKD).Hmm, trapeziums can sometimes be tricky because of their parallel sides. Maybe I can use some properties of parallel lines or similar triangles here. Let me try to visualize the trapezium. Let me draw it mentally: (AD) is the top base, (BC) is the bottom base, and (AB) and (CD) are the non-parallel sides. Points (K) and (L) are somewhere on (AB) and (CD), respectively.Given that (angle BAL = angle CDK), maybe I can relate these angles somehow. Since (AD) is parallel to (BC), perhaps there are some corresponding angles or alternate interior angles that I can use.Let me label the angles. Let’s denote (angle BAL = alpha), so (angle CDK = alpha) as well. Now, I need to find a relationship between (angle BLA) and (angle CKD). Maybe if I can show that triangles (BLA) and (CKD) are similar or something like that.Wait, but triangles (BLA) and (CKD) might not necessarily be similar because I don't have enough information about their sides. Maybe I can look for cyclic quadrilaterals instead. If certain points lie on a circle, then opposite angles might be equal or supplementary.Let me think about quadrilateral (ABLK). If I can show that this quadrilateral is cyclic, then opposite angles would sum to 180 degrees. But how? I know that (angle BAL = angle CDK = alpha), but I'm not sure how that helps with (ABLK).Alternatively, maybe I can consider triangles (BAL) and (CDK). They both have an angle of (alpha), but without knowing more about their sides or other angles, it's hard to say.Wait, since (AD) is parallel to (BC), maybe the alternate interior angles can help. If I draw a line from (A) to (K), then (angle BAL) is equal to (angle AKL) because of the parallel lines. Hmm, not sure if that's correct.Let me try another approach. Maybe I can use the concept of similar triangles. If I can find two triangles that are similar, then their corresponding angles would be equal. Let's see, triangle (BAL) and triangle (CDK). They both have angle (alpha), but unless the sides are proportional, I can't say they are similar.Alternatively, maybe triangle (BLA) and triangle (CKD). If I can show that these triangles are similar, then their corresponding angles would be equal, which would give me (angle BLA = angle CKD). To show similarity, I need either AA, SAS, or SSS similarity.Let me check for AA similarity. If I can find two angles that are equal in both triangles, then they are similar. I know that (angle BAL = angle CDK = alpha). Maybe there's another pair of equal angles.Wait, since (AD) is parallel to (BC), the angles formed by a transversal might be equal. For example, (angle ABL) and (angle KCD) could be equal because they are alternate interior angles. Is that right?Let me think. If I consider line (AB) as a transversal cutting the parallel lines (AD) and (BC), then (angle ABL) and (angle KCD) are on opposite sides of the transversal and between the parallel lines. So yes, they should be equal.So, (angle ABL = angle KCD). That's another pair of equal angles. So in triangles (BLA) and (CKD), we have two angles equal: (angle BAL = angle CDK = alpha) and (angle ABL = angle KCD). Therefore, by AA similarity, triangles (BLA) and (CKD) are similar.If triangles (BLA) and (CKD) are similar, then their corresponding angles are equal. Therefore, (angle BLA = angle CKD), which is what we needed to prove.Wait, did I just do that correctly? Let me double-check. I used the fact that (AD parallel BC) to say that (angle ABL = angle KCD). Then, since (angle BAL = angle CDK), triangles (BLA) and (CKD) have two equal angles, so they are similar by AA. Therefore, their corresponding angles are equal, so (angle BLA = angle CKD). Yeah, that seems right.I think that makes sense. So, the key was to recognize that the parallel sides give us equal angles, and then using the given equal angles to establish similarity between the triangles. Once similarity is established, the corresponding angles are equal, which gives the desired result.**Final Answer**boxed{angle BLA = angle CKD}

💡Okay, so I have this problem here where I need to find the value of x, which is the ratio of two sums: the numerator is the sum of cosines from 1 degree up to 90 degrees, and the denominator is the sum of sines from 1 degree up to 90 degrees. Then, I have to find the greatest integer that doesn't exceed 100 times x. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me write down what x is:x = [Σ (cos n°) from n=1 to 90] / [Σ (sin n°) from n=1 to 90]So, I need to compute both the sum of cosines and the sum of sines from 1° to 90°, and then take their ratio.I remember that there are formulas for the sum of sines and cosines in an arithmetic sequence. Maybe I can use those here. Let me recall them.For the sum of sines, the formula is:Σ (sin a + (n-1)d) from n=1 to N = [sin(Nd/2) / sin(d/2)] * sin(a + (N-1)d/2)Similarly, for the sum of cosines, it's:Σ (cos a + (n-1)d) from n=1 to N = [sin(Nd/2) / sin(d/2)] * cos(a + (N-1)d/2)In this problem, the angle starts at 1° and goes up by 1° each time, so the common difference d is 1°, and the number of terms N is 90. The first term a is 1°, so let me plug these into the formulas.First, let's compute the sum of cosines:Σ cos n° from n=1 to 90 = [sin(90 * 1° / 2) / sin(1° / 2)] * cos(1° + (90 - 1) * 1° / 2)Simplify that:= [sin(45°) / sin(0.5°)] * cos(1° + 44.5°)= [sin(45°) / sin(0.5°)] * cos(45.5°)Similarly, the sum of sines:Σ sin n° from n=1 to 90 = [sin(90 * 1° / 2) / sin(1° / 2)] * sin(1° + (90 - 1) * 1° / 2)Simplify:= [sin(45°) / sin(0.5°)] * sin(45.5°)So, both sums have the same factor [sin(45°) / sin(0.5°)], and then multiplied by either cos(45.5°) or sin(45.5°). Therefore, when I take the ratio x, this common factor will cancel out.So, x = [cos(45.5°)] / [sin(45.5°)] = cot(45.5°)Hmm, cotangent of 45.5 degrees. I know that cotangent is the reciprocal of tangent, so x = 1 / tan(45.5°)Now, 45.5 degrees is halfway between 45° and 46°, so maybe I can approximate tan(45.5°). I know that tan(45°) is 1, and tan(46°) is approximately 1.0355. Since 45.5° is halfway, maybe tan(45.5°) is approximately 1.01775? Wait, that might not be accurate. Maybe I should use a better approximation.Alternatively, I can use the identity that tan(a + b) = (tan a + tan b) / (1 - tan a tan b). Let me consider 45.5° as 45° + 0.5°. So,tan(45° + 0.5°) = [tan 45° + tan 0.5°] / [1 - tan 45° tan 0.5°]Since tan 45° = 1, this becomes:= [1 + tan 0.5°] / [1 - tan 0.5°]I need to find tan 0.5°. I know that tan θ ≈ θ (in radians) for small angles, so 0.5° is π/360 radians, which is approximately 0.0087265 radians. So, tan 0.5° ≈ 0.0087265.Plugging that in:tan(45.5°) ≈ [1 + 0.0087265] / [1 - 0.0087265] ≈ 1.0087265 / 0.9912735 ≈ 1.0177So, tan(45.5°) ≈ 1.0177, which means x ≈ 1 / 1.0177 ≈ 0.9827Therefore, 100x ≈ 98.27, so the greatest integer that does not exceed 100x is 98.Wait, but earlier I thought the sum of cosines might cancel out, but that doesn't seem right. Let me double-check that.I used the sum formulas correctly, right? For the sum of cosines, it's [sin(Nd/2)/sin(d/2)] * cos(a + (N-1)d/2). Plugging in N=90, d=1°, a=1°, so yes, that gives [sin(45°)/sin(0.5°)] * cos(45.5°). Similarly for sines. So, the ratio is cot(45.5°), which is approximately 0.9827.But wait, another thought: when summing from 1° to 90°, the terms from 1° to 44° and 46° to 89° can be paired with their complementary angles. For example, cos(1°) = sin(89°), cos(2°) = sin(88°), and so on, up to cos(44°) = sin(46°). Then, cos(45°) is equal to sin(45°), and cos(90°) is 0, while sin(90°) is 1.So, let's see: the numerator is sum_{n=1}^{90} cos n°, which is [sum_{n=1}^{44} cos n° + cos 45° + cos 90°]. Similarly, the denominator is [sum_{n=1}^{44} sin n° + sin 45° + sin 90°].But since cos n° = sin(90° - n°), the sum_{n=1}^{44} cos n° = sum_{n=46}^{89} sin n°. Therefore, the numerator becomes [sum_{n=46}^{89} sin n° + sin 45° + 0], and the denominator is [sum_{n=1}^{44} sin n° + sin 45° + 1].So, numerator = sum_{n=46}^{89} sin n° + sin 45°Denominator = sum_{n=1}^{44} sin n° + sin 45° + 1But sum_{n=46}^{89} sin n° = sum_{n=1}^{44} sin(90° - n°) = sum_{n=1}^{44} cos n°, which is equal to sum_{n=1}^{44} sin(90° - n°). Wait, that's the same as sum_{n=46}^{89} sin n°, which is the same as sum_{n=1}^{44} cos n°.But in the numerator, it's sum_{n=46}^{89} sin n° + sin 45°, and in the denominator, it's sum_{n=1}^{44} sin n° + sin 45° + 1.So, if I denote S = sum_{n=1}^{44} sin n°, then sum_{n=46}^{89} sin n° = sum_{n=1}^{44} cos n°, which is another sum, let's say C.Therefore, numerator = C + sin 45°Denominator = S + sin 45° + 1But I don't know if C and S are related. Wait, actually, from the earlier approach, using the sum formulas, we found that the ratio is cot(45.5°), which is approximately 0.9827, so 100x ≈ 98.27, so the greatest integer not exceeding that is 98.But let me check if there's another way to see this. Maybe using complex exponentials or something else.Alternatively, I can consider that the sum of cosines from 1° to 90° can be related to the imaginary part of a complex exponential sum, and similarly for sines. But that might complicate things more.Alternatively, I can think about the fact that the sum of cosines from 1° to 90° is equal to the imaginary part of the sum of e^{i n°} from n=1 to 90, and similarly for sines, it's the real part. But that might not necessarily help me directly.Wait, but actually, the sum of e^{i n°} from n=1 to 90 is a geometric series with ratio e^{i 1°}. So, the sum is e^{i 1°} (1 - e^{i 90°}) / (1 - e^{i 1°}). Then, the real part is the sum of cosines, and the imaginary part is the sum of sines.But maybe I can compute this sum and then take the ratio of imaginary to real parts.Let me try that.Let S = sum_{n=1}^{90} e^{i n°} = e^{i 1°} (1 - e^{i 90°}) / (1 - e^{i 1°})Simplify numerator: 1 - e^{i 90°} = 1 - iDenominator: 1 - e^{i 1°} = 1 - cos 1° - i sin 1°So, S = e^{i 1°} (1 - i) / (1 - cos 1° - i sin 1°)Multiply numerator and denominator by the conjugate of the denominator:= e^{i 1°} (1 - i) (1 - cos 1° + i sin 1°) / [(1 - cos 1°)^2 + (sin 1°)^2]Simplify denominator:(1 - cos 1°)^2 + (sin 1°)^2 = 1 - 2 cos 1° + cos² 1° + sin² 1° = 2(1 - cos 1°)Because cos² + sin² = 1, so 1 - 2 cos 1° + 1 = 2(1 - cos 1°)So, denominator is 2(1 - cos 1°)Now, numerator:e^{i 1°} (1 - i)(1 - cos 1° + i sin 1°)Let me compute (1 - i)(1 - cos 1° + i sin 1°):= (1)(1 - cos 1°) + (1)(i sin 1°) - i(1 - cos 1°) - i(i sin 1°)= (1 - cos 1°) + i sin 1° - i + i cos 1° + sin 1°Because i(i sin 1°) = - sin 1°Wait, let me compute term by term:First term: 1*(1 - cos 1°) = 1 - cos 1°Second term: 1*(i sin 1°) = i sin 1°Third term: -i*(1 - cos 1°) = -i + i cos 1°Fourth term: -i*(i sin 1°) = -i² sin 1° = sin 1°, since i² = -1So, combining all terms:Real parts: (1 - cos 1°) + sin 1°Imaginary parts: sin 1° - 1 + cos 1°So, numerator becomes:e^{i 1°} [ (1 - cos 1° + sin 1°) + i (sin 1° - 1 + cos 1°) ]Now, e^{i 1°} = cos 1° + i sin 1°, so multiplying this with the above:= [cos 1° + i sin 1°] [ (1 - cos 1° + sin 1°) + i (sin 1° - 1 + cos 1°) ]This will get complicated, but let's try to compute it.Let me denote A = 1 - cos 1° + sin 1°, and B = sin 1° - 1 + cos 1°.So, the expression becomes:(cos 1° + i sin 1°)(A + i B) = cos 1° A - sin 1° B + i (sin 1° A + cos 1° B)So, the real part is cos 1° A - sin 1° B, and the imaginary part is sin 1° A + cos 1° B.Therefore, the sum S = [Real part] + i [Imaginary part]But S is also equal to sum_{n=1}^{90} e^{i n°}, whose real part is the sum of cosines, and imaginary part is the sum of sines.Therefore, the sum of cosines is Real part = cos 1° A - sin 1° BAnd the sum of sines is Imaginary part = sin 1° A + cos 1° BSo, x = (Real part) / (Imaginary part) = [cos 1° A - sin 1° B] / [sin 1° A + cos 1° B]Now, let's compute A and B:A = 1 - cos 1° + sin 1°B = sin 1° - 1 + cos 1° = (sin 1° + cos 1°) - 1So, let's compute A and B numerically.First, compute cos 1° and sin 1°:cos 1° ≈ 0.9998477sin 1° ≈ 0.0174524So,A = 1 - 0.9998477 + 0.0174524 ≈ 1 - 0.9998477 = 0.0001523 + 0.0174524 ≈ 0.0176047B = 0.0174524 + 0.9998477 - 1 ≈ (0.0174524 + 0.9998477) = 1.0173001 - 1 ≈ 0.0173001So, A ≈ 0.0176047, B ≈ 0.0173001Now, compute Real part:cos 1° * A - sin 1° * B ≈ 0.9998477 * 0.0176047 - 0.0174524 * 0.0173001Compute each term:0.9998477 * 0.0176047 ≈ 0.0175980.0174524 * 0.0173001 ≈ 0.000302So, Real part ≈ 0.017598 - 0.000302 ≈ 0.017296Imaginary part:sin 1° * A + cos 1° * B ≈ 0.0174524 * 0.0176047 + 0.9998477 * 0.0173001Compute each term:0.0174524 * 0.0176047 ≈ 0.0003070.9998477 * 0.0173001 ≈ 0.017294So, Imaginary part ≈ 0.000307 + 0.017294 ≈ 0.017601Therefore, x ≈ Real part / Imaginary part ≈ 0.017296 / 0.017601 ≈ 0.9827Which matches our earlier approximation. So, x ≈ 0.9827, so 100x ≈ 98.27, so the greatest integer not exceeding 100x is 98.Wait, but earlier I thought the sum of cosines might cancel out, but that seems incorrect. Let me check that again.I thought that cos(45° + k) = cos(45° - k), but actually, cos(45° + k) = cos(45° - k) because cosine is even around 45°, but when k approaches 45°, cos(90° - k) = sin k, which is positive, but cos(90° + k) would be negative, but in our case, we're only summing up to 90°, so cos(90°) is 0.Wait, actually, when summing cos n° from 1° to 90°, the terms from 1° to 44° and 46° to 89° can be paired as cos k° and cos(90° - k°) = sin k°, but that doesn't necessarily cancel out. Instead, each pair sums to cos k° + sin k°, which isn't zero.Similarly, the sum of sines from 1° to 90° can be paired as sin k° and sin(90° - k°) = cos k°, so each pair sums to sin k° + cos k°, which is the same as the cosine pair.But since both numerator and denominator have the same number of such pairs, their ratio would be 1, but that's not the case because the middle term in the numerator is cos 45°, and in the denominator, it's sin 45°, which are equal, but the last term in the numerator is cos 90° = 0, while in the denominator, it's sin 90° = 1.So, actually, the sums are:Numerator: sum_{k=1}^{44} (cos k° + cos(90° - k°)) + cos 45° + cos 90°= sum_{k=1}^{44} (cos k° + sin k°) + cos 45° + 0Denominator: sum_{k=1}^{44} (sin k° + sin(90° - k°)) + sin 45° + sin 90°= sum_{k=1}^{44} (sin k° + cos k°) + sin 45° + 1So, both sums have the same sum_{k=1}^{44} (sin k° + cos k°), but the numerator has an extra cos 45°, and the denominator has an extra 1.Therefore, the ratio x is:[sum_{k=1}^{44} (sin k° + cos k°) + cos 45°] / [sum_{k=1}^{44} (sin k° + cos k°) + sin 45° + 1]Let me denote T = sum_{k=1}^{44} (sin k° + cos k°). Then,x = (T + cos 45°) / (T + sin 45° + 1)Since cos 45° = sin 45° = √2/2 ≈ 0.7071, so:x = (T + 0.7071) / (T + 0.7071 + 1) = (T + 0.7071) / (T + 1.7071)Now, if T is a large number, then x would approach (T / T) = 1, but since T is the sum of sin and cos from 1° to 44°, which are all positive and less than 1, T is actually a moderate number, not extremely large.But from our earlier calculation using the sum formulas, we found that x ≈ 0.9827, which is close to 1 but slightly less.Alternatively, perhaps I can compute T numerically.Compute T = sum_{k=1}^{44} (sin k° + cos k°)This would require summing each term, which is tedious, but maybe I can approximate it.Alternatively, I can use the sum formulas again for T.Sum_{k=1}^{44} sin k° = [sin(44*1°/2) / sin(1°/2)] * sin(1° + (44-1)*1°/2)= [sin(22°) / sin(0.5°)] * sin(22.5°)Similarly, sum_{k=1}^{44} cos k° = [sin(22°) / sin(0.5°)] * cos(22.5°)Therefore, T = [sin(22°) / sin(0.5°)] * [sin(22.5°) + cos(22.5°)]Compute sin(22°) ≈ 0.3746, sin(0.5°) ≈ 0.0087265, sin(22.5°) ≈ 0.3827, cos(22.5°) ≈ 0.9239So,T ≈ [0.3746 / 0.0087265] * (0.3827 + 0.9239)≈ [42.92] * 1.3066≈ 42.92 * 1.3066 ≈ 56.0So, T ≈ 56.0Therefore, x ≈ (56.0 + 0.7071) / (56.0 + 0.7071 + 1) ≈ (56.7071) / (57.7071) ≈ 0.9827Which again gives us x ≈ 0.9827, so 100x ≈ 98.27, so the greatest integer not exceeding 100x is 98.Therefore, the answer is 98.

💡First, I need to calculate the price of the gadget after applying the first discount of 10%. The original price is 350.00, so the price after the first discount is 350.00 multiplied by 0.90, which equals 315.00.Next, I'll apply the second discount of 12% to the already discounted price of 315.00. This means multiplying 315.00 by 0.88, resulting in a final sale price of 277.20.To find out what percentage of the original price this final sale price represents, I'll divide 277.20 by the original price of 350.00 and then multiply by 100 to convert it to a percentage. This calculation gives approximately 79.2%.Therefore, the final sale price is 79.2% of the original 350.00.

💡Okay, so I have this problem here about the harmonic series. Let me try to understand it step by step. The function f(n) is defined as the sum of reciprocals from 1 to n, right? So, f(n) = 1 + 1/2 + 1/3 + ... + 1/n. They've given me some specific values and inequalities:- f(2) = 3/2- f(4) > 2- f(8) > 5/2- f(16) > 3- f(32) > 7/2And I need to figure out which of the options A, B, C, or D is correct based on these. The options are:A: f(2n) > (2n + 1)/2B: f(n²) ≥ (n + 2)/2C: f(2ⁿ) ≥ (n + 2)/2D: None of the aboveHmm, okay. Let me start by looking at the given information and see if I can spot a pattern or a relationship that might help me guess the correct option.First, let's list out what we know:1. f(2) = 3/22. f(4) > 23. f(8) > 5/24. f(16) > 35. f(32) > 7/2Looking at these, I notice that the arguments of f are powers of 2: 2, 4, 8, 16, 32, which are 2¹, 2², 2³, 2⁴, 2⁵. So, the input to f is 2ⁿ where n is 1, 2, 3, 4, 5 respectively.Now, let's look at the corresponding inequalities:- For n=1: f(2¹) = 3/2- For n=2: f(2²) > 2- For n=3: f(2³) > 5/2- For n=4: f(2⁴) > 3- For n=5: f(2⁵) > 7/2Wait a second, let me write these in terms of n:- When the input is 2¹, the output is 3/2, which is (1 + 2)/2 = 3/2- When the input is 2², the output is greater than 2, which is (2 + 2)/2 = 2- When the input is 2³, the output is greater than 5/2, which is (3 + 2)/2 = 5/2- When the input is 2⁴, the output is greater than 3, which is (4 + 2)/2 = 3- When the input is 2⁵, the output is greater than 7/2, which is (5 + 2)/2 = 7/2So, it seems like for each n, f(2ⁿ) is greater than or equal to (n + 2)/2. That is, f(2ⁿ) ≥ (n + 2)/2.Looking at the options, that's exactly option C. So, based on the given information, it looks like option C is correct.But wait, let me make sure I'm not jumping to conclusions. Maybe I should check each option one by one to see if they fit.Option A: f(2n) > (2n + 1)/2Let's test this with the given values. For n=1, f(2*1)=f(2)=3/2. (2*1 +1)/2=3/2. So, f(2)=3/2 which is equal to (2n +1)/2, but the option says f(2n) > (2n +1)/2. So, for n=1, it's not strictly greater, it's equal. That might be a problem.Let's check n=2: f(4)=f(2²)=f(4) > 2. (2*2 +1)/2=5/2=2.5. But f(4) is only greater than 2, which is less than 2.5. So, f(4) > 2 does not imply f(4) > 2.5. Therefore, option A is not necessarily true.Option B: f(n²) ≥ (n + 2)/2Hmm, let's see. For n=1, f(1²)=f(1)=1. (1 + 2)/2=1.5. So, 1 ≥ 1.5? That's not true. Therefore, option B is incorrect.Option C: f(2ⁿ) ≥ (n + 2)/2From the given data, we saw that for n=1, f(2)=3/2=(1 + 2)/2=1.5, which is equal. For n=2, f(4)=2, which is equal to (2 + 2)/2=2. For n=3, f(8)=5/2=2.5, which is equal to (3 + 2)/2=2.5. Wait, but in the given information, f(8) > 5/2. So, actually, f(8) is greater than 5/2, which is (3 + 2)/2=2.5. So, in that case, it's greater, not just equal. Similarly, for n=4, f(16) > 3, which is equal to (4 + 2)/2=3. So, f(16) is greater than 3. For n=5, f(32) > 7/2=3.5, which is equal to (5 + 2)/2=3.5. So, f(32) is greater than 3.5.Wait, so for n=1, f(2)=3/2 which is equal to (1 + 2)/2. For n=2, f(4)=2 which is equal to (2 + 2)/2. But for n=3, f(8) > 5/2, which is greater than (3 + 2)/2=2.5. Similarly, for n=4, f(16) > 3, which is greater than (4 + 2)/2=3. For n=5, f(32) > 7/2, which is greater than (5 + 2)/2=3.5.So, actually, for n ≥ 3, f(2ⁿ) is strictly greater than (n + 2)/2, but for n=1 and n=2, it's equal. So, the inequality f(2ⁿ) ≥ (n + 2)/2 holds for all n, because for n=1 and n=2, it's equal, and for n ≥3, it's greater. So, option C is correct.Option D: None of the aboveBut since option C is correct, D is incorrect.Wait, but let me think again about option A. For n=1, f(2)=3/2, which is equal to (2*1 +1)/2=3/2. So, f(2n)=f(2)=3/2 which is equal to (2n +1)/2. But the option says f(2n) > (2n +1)/2. So, it's not strictly greater, it's equal for n=1. Therefore, option A is not correct.Similarly, for n=2, f(4)=2, which is less than (2*2 +1)/2=5/2=2.5. So, f(4)=2 < 2.5, which contradicts option A. Therefore, option A is incorrect.Option B: f(n²) ≥ (n + 2)/2As I checked earlier, for n=1, f(1)=1 < 1.5=(1 + 2)/2. So, it's not true. Therefore, option B is incorrect.Therefore, the correct answer is option C.But wait, let me think about whether the inequality in option C is always true for all n. The given data shows that for n=1,2,3,4,5, f(2ⁿ) is equal or greater than (n + 2)/2. But does this hold for all n?I know that the harmonic series grows logarithmically, specifically, f(n) ≈ ln(n) + γ, where γ is the Euler-Mascheroni constant (~0.5772). So, f(2ⁿ) ≈ ln(2ⁿ) + γ = n ln(2) + γ.Now, let's compare this to (n + 2)/2.So, we need to check if n ln(2) + γ ≥ (n + 2)/2.Let's rearrange this inequality:n ln(2) + γ ≥ (n + 2)/2Multiply both sides by 2:2n ln(2) + 2γ ≥ n + 2Bring all terms to one side:2n ln(2) - n + 2γ - 2 ≥ 0Factor out n:n(2 ln(2) - 1) + (2γ - 2) ≥ 0Compute the coefficients:2 ln(2) ≈ 2 * 0.6931 ≈ 1.3862So, 2 ln(2) - 1 ≈ 0.38622γ - 2 ≈ 2*0.5772 - 2 ≈ 1.1544 - 2 ≈ -0.8456So, the inequality becomes:n(0.3862) - 0.8456 ≥ 0Solve for n:0.3862 n ≥ 0.8456n ≥ 0.8456 / 0.3862 ≈ 2.188So, for n ≥ 3, the inequality holds. For n=1 and n=2, we saw that f(2ⁿ) equals (n + 2)/2. So, for n=1: f(2)=3/2=1.5, which is equal to (1 + 2)/2=1.5. For n=2: f(4)=2, which is equal to (2 + 2)/2=2. For n=3: f(8)=5/2=2.5, which is equal to (3 + 2)/2=2.5, but actually f(8) > 5/2, so it's greater. Similarly, for n=4 and above, f(2ⁿ) is greater than (n + 2)/2.Therefore, the inequality f(2ⁿ) ≥ (n + 2)/2 holds for all n ≥1, with equality only for n=1 and n=2, and strict inequality for n ≥3.So, option C is correct.Just to be thorough, let me check for n=6:f(64) should be greater than (6 + 2)/2=4.I know that f(64) is approximately ln(64) + γ ≈ 4.1589 + 0.5772 ≈ 4.7361, which is indeed greater than 4. So, it holds.Similarly, for n=0, but n is in N*, so n starts at 1.Therefore, I'm confident that option C is correct.

💡Okay, so I need to find the point where the given line intersects the plane. Hmm, let me see. The line is given in a symmetrical form, which I think is like the ratio of the differences in coordinates over the direction ratios. The plane is given by an equation, so I need to figure out where they meet.First, the line is written as (x - 3)/1 = (y + 2)/(-1) = (z - 8)/0. Wait, the z-component has a denominator of 0. That seems a bit odd. Maybe that means z is constant? Because if the denominator is zero, it can't change, right? So z must be equal to 8 all the time. That simplifies things a bit.So, if I let the parameter be t, then I can write the parametric equations for the line. For x, it's (x - 3)/1 = t, so x = 3 + t. For y, it's (y + 2)/(-1) = t, so y = -2 - t. And for z, since the denominator is 0, z must be 8. So, the parametric equations are x = 3 + t, y = -2 - t, z = 8.Now, the plane equation is 5x + 9y + 4z - 25 = 0. I need to substitute the parametric equations into this plane equation to find the value of t where they intersect.Let me plug in x, y, and z into the plane equation:5*(3 + t) + 9*(-2 - t) + 4*8 - 25 = 0.Let me compute each term step by step:5*(3 + t) = 15 + 5t,9*(-2 - t) = -18 - 9t,4*8 = 32.So, putting it all together:15 + 5t - 18 - 9t + 32 - 25 = 0.Now, let's combine like terms:First, the constants: 15 - 18 + 32 - 25.15 - 18 is -3,-3 + 32 is 29,29 - 25 is 4.Then, the t terms: 5t - 9t = -4t.So, the equation simplifies to:4 - 4t = 0.Solving for t:-4t = -4,t = 1.Okay, so t is 1. Now, plug t = 1 back into the parametric equations to find the point of intersection.x = 3 + 1 = 4,y = -2 - 1 = -3,z = 8.So, the point where the line intersects the plane is (4, -3, 8).Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting with the parametric equations:x = 3 + t,y = -2 - t,z = 8.Substituting into the plane equation:5x + 9y + 4z - 25 = 0,5*(3 + t) + 9*(-2 - t) + 4*8 - 25 = 0,15 + 5t - 18 - 9t + 32 - 25 = 0,Combine constants: 15 - 18 = -3,-3 + 32 = 29,29 - 25 = 4,Combine t terms: 5t - 9t = -4t,So, 4 - 4t = 0,-4t = -4,t = 1.Yes, that seems correct. Plugging t = 1 back in gives x = 4, y = -3, z = 8.I think that's the correct point of intersection.

💡Okay, so I have this geometry problem here about a right-angled triangle, triangle ABC, where point B is the right angle. They tell me that the tangent of angle A is 4/3, and the length of side AB is 3. I need to find the length of side BC. Hmm, let me see how to approach this.First, I remember that in a right-angled triangle, the tangent of an angle is equal to the length of the opposite side divided by the adjacent side. So, for angle A, the opposite side would be BC, and the adjacent side would be AB. They've given me that tan A is 4/3, and AB is 3. So, if I set up the equation tan A = opposite/adjacent, that should help me find BC.Let me write that down: tan A = BC / AB. Plugging in the values I have, that would be 4/3 = BC / 3. To solve for BC, I can multiply both sides of the equation by 3. So, BC = (4/3) * 3. The 3s cancel out, so BC = 4. That seems straightforward.Wait, but just to make sure I'm not making a mistake, let me visualize the triangle. Since B is the right angle, sides AB and BC are the legs, and AC is the hypotenuse. So, angle A is at vertex A, between sides AB and AC. The side opposite angle A is BC, and the side adjacent is AB. Yes, that makes sense.So, using tan A = opposite/adjacent = BC/AB, and plugging in the known values, I correctly found BC to be 4. But maybe I should double-check using the Pythagorean theorem to find the hypotenuse and see if everything adds up.If AB is 3 and BC is 4, then the hypotenuse AC should be sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5. That checks out because 3-4-5 is a classic Pythagorean triple. So, the sides are 3, 4, and 5, which makes sense.I think I've got it right. But just to go through the steps again: in triangle ABC, right-angled at B, tan A is 4/3. AB is 3, which is adjacent to angle A. So, opposite over adjacent is BC/AB = 4/3, so BC is 4. Then, using Pythagoras, the hypotenuse is 5. Everything seems consistent.I don't think I made any mistakes here. The key was recognizing which sides correspond to the tangent ratio for angle A. Sometimes, it's easy to mix up opposite and adjacent, but since angle A is at vertex A, the sides opposite and adjacent are relative to that angle. So, opposite is BC, and adjacent is AB. That was the crucial step.Another way to think about it is labeling the triangle. Let me imagine point A, point B, and point C. Since B is the right angle, AB and BC are the legs. Angle A is at point A, so the sides relative to angle A are: opposite side BC, adjacent side AB, and hypotenuse AC. So, tan A is BC/AB, which is 4/3, leading to BC = 4.I guess another thing to consider is whether the triangle is labeled correctly. Sometimes, the labeling can be confusing, but in this case, since B is the right angle, it's clear that AB and BC are the legs, and AC is the hypotenuse.So, just to recap: tan A = opposite/adjacent = BC/AB = 4/3. AB is 3, so BC = 4. Using Pythagoras, AC is 5. All sides satisfy the Pythagorean theorem, so the calculations are correct.I think I'm confident with this answer. It makes sense, and the steps are logical. I don't see any errors in my reasoning. So, the length of side BC is 4.**Final Answer**The length of side BC is boxed{4}.