answer：Okay, so I have this problem where the sides of a triangle are 35, 65, and 85 units. I need to find the length of the larger segment cut off on the side of length 85 when an altitude is dropped onto it. The options are A) 58, B) 59, C) 60, D) 61, and E) 62.Hmm, let's start by visualizing the triangle. It has sides of 35, 65, and 85. So, the side of length 85 is the longest side, which probably means it's the base of the triangle. When we drop an altitude onto this side, it will split the side into two segments. The question is asking for the length of the larger segment.I remember that in a triangle, the altitude creates two right-angled triangles. Each of these right triangles shares the altitude as one of their sides. So, if I denote the altitude as 'h', then I can use the Pythagorean theorem on both of these right triangles.Let me denote the two segments of the side of length 85 as 'a' and 'b', where 'a' is the shorter segment and 'b' is the longer segment. Since the entire side is 85, I know that a + b = 85. So, b = 85 - a.Now, applying the Pythagorean theorem to both right triangles:For the first triangle with sides 35, h, and a:35² = a² + h² ...(1)For the second triangle with sides 65, h, and b:65² = b² + h² ...(2)Since both equations equal h², I can set them equal to each other to eliminate h²:35² = a² + h²65² = b² + h²Subtracting equation (1) from equation (2):65² - 35² = b² - a²Calculating the left side:65² is 4225, and 35² is 1225. So, 4225 - 1225 = 3000.So, 3000 = b² - a²But I also know that b = 85 - a, so I can substitute that into the equation:3000 = (85 - a)² - a²Expanding (85 - a)²:85² - 2*85*a + a² - a²Simplifying:7225 - 170a + a² - a² = 7225 - 170aSo, 3000 = 7225 - 170aNow, solving for 'a':170a = 7225 - 3000170a = 4225a = 4225 / 170Calculating that:4225 divided by 170. Let me do this step by step.170 times 24 is 4080. 4225 - 4080 is 145. So, 145 / 170 is 0.8529 approximately.So, a ≈ 24.8529.Since a is approximately 24.85, the larger segment b is 85 - a ≈ 85 - 24.85 ≈ 60.15.Looking at the options, 60 is one of them, so the larger segment is approximately 60 units.Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting from 3000 = (85 - a)² - a²Expanding (85 - a)²:85² - 2*85*a + a² = 7225 - 170a + a²Subtracting a²:7225 - 170a + a² - a² = 7225 - 170aSo, 3000 = 7225 - 170aThen, 170a = 7225 - 3000 = 4225So, a = 4225 / 170Calculating 4225 / 170:170 * 24 = 40804225 - 4080 = 145145 / 170 = 0.8529So, a ≈ 24.8529, which is approximately 24.85Therefore, b = 85 - 24.85 ≈ 60.15Since the options are whole numbers, 60 is the closest, so the answer should be C) 60.I think that's correct. Let me see if there's another way to approach this problem to confirm.Alternatively, I could use the formula for the length of the altitude in terms of the area.First, find the area of the triangle using Heron's formula.The semi-perimeter, s = (35 + 65 + 85)/2 = (185)/2 = 92.5Area = √[s(s - a)(s - b)(s - c)] = √[92.5*(92.5 - 35)*(92.5 - 65)*(92.5 - 85)]Calculating each term:92.5 - 35 = 57.592.5 - 65 = 27.592.5 - 85 = 7.5So, Area = √[92.5 * 57.5 * 27.5 * 7.5]This seems a bit complicated, but let's compute it step by step.First, multiply 92.5 and 57.5:92.5 * 57.5Let me compute 90*57.5 = 51752.5*57.5 = 143.75So, total is 5175 + 143.75 = 5318.75Now, multiply 27.5 and 7.5:27.5 * 7.5 = 206.25Now, multiply 5318.75 * 206.25This is a bit tedious, but let's approximate.5318.75 * 200 = 1,063,7505318.75 * 6.25 = 33,242.1875Adding them together: 1,063,750 + 33,242.1875 ≈ 1,096,992.1875So, the area squared is approximately 1,096,992.1875Therefore, the area is √1,096,992.1875 ≈ 1047.37Now, the area can also be expressed as (base * height)/2Here, the base is 85, and the height is h.So, 1047.37 ≈ (85 * h)/2Solving for h:h ≈ (1047.37 * 2)/85 ≈ 2094.74 / 85 ≈ 24.64So, the altitude h is approximately 24.64Wait, but earlier I found that a ≈ 24.85, which is very close to h. That makes sense because in the right triangle with sides 35, h, and a, h is the altitude, and a is the base segment.So, if h ≈ 24.64, then a can be found using Pythagoras:35² = a² + h²So, a² = 35² - h² ≈ 1225 - (24.64)²Calculating (24.64)²:24² = 5760.64² = 0.40962*24*0.64 = 30.72So, (24.64)² ≈ 576 + 30.72 + 0.4096 ≈ 607.1296Thus, a² ≈ 1225 - 607.1296 ≈ 617.8704So, a ≈ √617.8704 ≈ 24.86Which is consistent with my earlier calculation of a ≈ 24.85Therefore, the larger segment b = 85 - a ≈ 85 - 24.86 ≈ 60.14Again, rounding to the nearest whole number, it's 60.So, both methods give me the same result, which is reassuring.Therefore, the larger segment cut off on the side of length 85 is 60 units, which corresponds to option C.

answer：To determine how many gumballs Alex should buy, I need to find the smallest number that is at least 50 and leaves a remainder of 5 when divided by 7.First, I'll calculate how many gumballs Alex would need if he had exactly 50. Dividing 50 by 7 gives a quotient of 7 and a remainder of 1. This means 50 gumballs leave 1 gumball remaining.Since Alex wants 5 gumballs remaining, I need to find how many more gumballs are needed to increase the remainder from 1 to 5. The difference between 5 and 1 is 4, so Alex should buy 4 more gumballs than 50.Adding 4 to 50 gives a total of 54 gumballs. To confirm, dividing 54 by 7 results in a quotient of 7 and a remainder of 5, which meets Alex's requirement.Therefore, Alex should buy 54 gumballs.

answer：Okay, so I have this problem with four propositions about trigonometric functions, and I need to figure out which ones are correct. Let me take them one by one.Starting with proposition ①: "If θ is an angle in the second quadrant, then sin θ · tan θ < 0." Hmm, okay. I remember that in the second quadrant, which is between 90 degrees and 180 degrees (or π/2 to π radians), the sine of an angle is positive because sine corresponds to the y-coordinate on the unit circle, and it's above the x-axis there. Tangent, on the other hand, is sine over cosine. In the second quadrant, cosine is negative because it's to the left of the y-axis. So tan θ would be positive over negative, which is negative. Therefore, sin θ is positive and tan θ is negative, so their product should be negative. That makes sense. So proposition ① is correct.Moving on to proposition ②: "If sin θ · tan θ < 0, then θ is an angle in the second quadrant." Hmm, this seems like the converse of proposition ①. Let me think. If sin θ · tan θ is negative, that means either sin θ is positive and tan θ is negative, or sin θ is negative and tan θ is positive. In the second quadrant, sin θ is positive and tan θ is negative, so that gives a negative product. But what about the third quadrant? In the third quadrant, both sine and cosine are negative, so tan θ (which is sin θ / cos θ) is positive. So sin θ is negative and tan θ is positive, so their product would be negative as well. So if sin θ · tan θ < 0, θ could be in the second or third quadrant. Therefore, proposition ② is not necessarily true because θ could also be in the third quadrant. So proposition ② is incorrect.Now, proposition ③: "sin 1 · cos 2 · tan 3 > 0." Hmm, okay. This one is a bit trickier because it's not about quadrants but specific angles. Let me figure out the signs of each function at these angles.First, sin 1: 1 radian is approximately 57 degrees, which is in the first quadrant. In the first quadrant, all trigonometric functions are positive, so sin 1 is positive.Next, cos 2: 2 radians is approximately 114 degrees, which is in the second quadrant. In the second quadrant, cosine is negative because it's to the left of the y-axis. So cos 2 is negative.Lastly, tan 3: 3 radians is approximately 171 degrees, which is still in the second quadrant. In the second quadrant, tangent is negative because sine is positive and cosine is negative, so tan θ = sin θ / cos θ is negative. So tan 3 is negative.Now, multiplying them together: positive (sin 1) times negative (cos 2) times negative (tan 3). Multiplying two negatives gives a positive, so positive times positive is positive. So the product sin 1 · cos 2 · tan 3 is positive. Therefore, proposition ③ is correct.Finally, proposition ④: "If θ ∈ (3π/2, 2π), then sin(π + θ) > 0." Let me parse this. θ is between 3π/2 and 2π, which is the fourth quadrant. So θ is in the fourth quadrant.Now, sin(π + θ). Let me recall the sine addition formula: sin(π + θ) = sin π cos θ + cos π sin θ. But I also remember that sin(π + θ) is equal to -sin θ because it's the sine of an angle shifted by π, which flips the sign. So sin(π + θ) = -sin θ.Given that θ is in the fourth quadrant, sin θ is negative because sine corresponds to the y-coordinate, which is below the x-axis there. So -sin θ would be positive because sin θ is negative. Therefore, sin(π + θ) is positive. So proposition ④ is correct.Wait, let me double-check that. If θ is in the fourth quadrant, then π + θ would be in the fifth quadrant? Wait, no, angles are typically considered modulo 2π, so π + θ where θ is between 3π/2 and 2π would be between 5π/2 and 3π, which is the same as between π/2 and π when subtracting 2π. So π + θ is in the second quadrant. In the second quadrant, sine is positive, so sin(π + θ) is positive. Alternatively, using the identity sin(π + θ) = -sin θ, and since θ is in the fourth quadrant, sin θ is negative, so -sin θ is positive. Either way, it's positive. So proposition ④ is correct.Wait, hold on, if θ is in the fourth quadrant, then π + θ is in the fifth quadrant, but since we usually consider angles between 0 and 2π, π + θ would be equivalent to π + θ - 2π = θ - π, which would be in the fourth quadrant minus π, which is in the third quadrant? Hmm, maybe I confused myself.Let me think differently. Let's take θ = 7π/4, which is in the fourth quadrant. Then π + θ = π + 7π/4 = 11π/4, which is equivalent to 11π/4 - 2π = 3π/4, which is in the second quadrant. In the second quadrant, sine is positive. So sin(11π/4) = sin(3π/4) = √2/2, which is positive. So yes, sin(π + θ) is positive when θ is in the fourth quadrant. So proposition ④ is correct.So to recap:①: Correct.②: Incorrect, because θ could be in the third quadrant as well.③: Correct.④: Correct.Therefore, the correct propositions are ①, ③, and ④.**Final Answer**The correct propositions are boxed{①③④}.

answer：First, I need to convert the binary number (111011010010_2) to its decimal equivalent. I'll do this by expanding each bit multiplied by its corresponding power of 2 and then summing the results.Next, I will divide the decimal value by 4. Dividing by 4 in binary is equivalent to shifting the number to the right by two positions, which effectively removes the last two bits. This operation will give me a new decimal value.Finally, I will double the result obtained from the division step. Doubling means multiplying the number by 2, which will give me the final result in decimal form.

answer：Alright, so I have this problem to tackle, and it's about quadratic reciprocity and the Legendre symbol. Let me try to understand what's being asked here.Part (a) says: "Prove the quadratic reciprocity law using problem 31.39." Hmm, I don't have problem 31.39 in front of me, but I know quadratic reciprocity is a fundamental theorem in number theory that relates the solvability of quadratic equations modulo different primes. The standard statement is that for distinct odd primes ( p ) and ( q ), the Legendre symbols satisfy:[left( frac{p}{q} right) left( frac{q}{p} right) = (-1)^{frac{(p-1)(q-1)}{4}}]So, I need to prove this using problem 31.39. Since I don't have the exact content of problem 31.39, I'll assume it might involve some counting or combinatorial argument, maybe using permutations or something related to the Chinese Remainder Theorem. Quadratic reciprocity can be proven using various methods, including Gauss's original proof involving counting points on a circle, or using properties of Gauss sums. Maybe problem 31.39 is about permutations and their signs?Moving on to part (b): "Let ( m ) be an odd prime number. Prove that[left( frac{2}{m} right) = (-1)^{frac{m^2 - 1}{8}}]Okay, so this is about the Legendre symbol ( left( frac{2}{m} right) ). I remember that there's a specific formula for the Legendre symbol ( left( frac{2}{p} right) ) where ( p ) is an odd prime. It depends on ( p ) modulo 8. Specifically, ( left( frac{2}{p} right) ) is 1 if ( p equiv 1 ) or ( 7 mod 8 ), and -1 if ( p equiv 3 ) or ( 5 mod 8 ). So, the exponent ( frac{m^2 - 1}{8} ) must capture this behavior.Let me think about ( m^2 - 1 ). That factors as ( (m - 1)(m + 1) ). Since ( m ) is an odd prime, both ( m - 1 ) and ( m + 1 ) are even, so their product is divisible by 4. Therefore, ( frac{m^2 - 1}{8} ) is an integer because ( m ) is odd, so ( m equiv 1 mod 2 ), making ( m^2 equiv 1 mod 8 ), hence ( m^2 - 1 equiv 0 mod 8 ). So, the exponent is indeed an integer.Now, if I can show that ( (-1)^{frac{m^2 - 1}{8}} ) equals ( left( frac{2}{m} right) ), that would solve part (b). Let me test this with some small primes.Take ( m = 3 ). Then, ( frac{3^2 - 1}{8} = frac{9 - 1}{8} = 1 ), so ( (-1)^1 = -1 ). The Legendre symbol ( left( frac{2}{3} right) ) is indeed -1 because 2 is not a quadratic residue modulo 3.Next, ( m = 5 ). ( frac{25 - 1}{8} = 3 ), so ( (-1)^3 = -1 ). The Legendre symbol ( left( frac{2}{5} right) ) is also -1 since 2 is not a quadratic residue modulo 5.How about ( m = 7 )? ( frac{49 - 1}{8} = 6 ), so ( (-1)^6 = 1 ). The Legendre symbol ( left( frac{2}{7} right) ) is 1 because 2 is a quadratic residue modulo 7 (since ( 3^2 = 9 equiv 2 mod 7 )).Another one: ( m = 11 ). ( frac{121 - 1}{8} = 15 ), so ( (-1)^{15} = -1 ). The Legendre symbol ( left( frac{2}{11} right) ) is -1 because 2 is not a quadratic residue modulo 11.Okay, so this seems to hold for these examples. To generalize, I need to show that ( (-1)^{frac{m^2 - 1}{8}} ) gives the correct Legendre symbol. Since ( m ) is an odd prime, ( m ) is congruent to 1, 3, 5, or 7 modulo 8. Let's compute ( frac{m^2 - 1}{8} ) for each case:1. If ( m equiv 1 mod 8 ), then ( m = 8k + 1 ). So, ( m^2 = (8k + 1)^2 = 64k^2 + 16k + 1 ). Thus, ( m^2 - 1 = 64k^2 + 16k ), and ( frac{m^2 - 1}{8} = 8k^2 + 2k ), which is even. Hence, ( (-1)^{text{even}} = 1 ).2. If ( m equiv 3 mod 8 ), then ( m = 8k + 3 ). ( m^2 = (8k + 3)^2 = 64k^2 + 48k + 9 ). So, ( m^2 - 1 = 64k^2 + 48k + 8 ), and ( frac{m^2 - 1}{8} = 8k^2 + 6k + 1 ), which is odd. Thus, ( (-1)^{text{odd}} = -1 ).3. If ( m equiv 5 mod 8 ), then ( m = 8k + 5 ). ( m^2 = (8k + 5)^2 = 64k^2 + 80k + 25 ). So, ( m^2 - 1 = 64k^2 + 80k + 24 ), and ( frac{m^2 - 1}{8} = 8k^2 + 10k + 3 ), which is odd. Thus, ( (-1)^{text{odd}} = -1 ).4. If ( m equiv 7 mod 8 ), then ( m = 8k + 7 ). ( m^2 = (8k + 7)^2 = 64k^2 + 112k + 49 ). So, ( m^2 - 1 = 64k^2 + 112k + 48 ), and ( frac{m^2 - 1}{8} = 8k^2 + 14k + 6 ), which is even. Thus, ( (-1)^{text{even}} = 1 ).So, summarizing:- If ( m equiv 1 ) or ( 7 mod 8 ), ( (-1)^{frac{m^2 - 1}{8}} = 1 ).- If ( m equiv 3 ) or ( 5 mod 8 ), ( (-1)^{frac{m^2 - 1}{8}} = -1 ).This matches exactly with the known values of ( left( frac{2}{m} right) ). Therefore, the equality holds.Going back to part (a), since I don't have the exact content of problem 31.39, I'll assume it's about using permutations or some counting argument to establish the quadratic reciprocity law. Quadratic reciprocity can be proven using Gauss's lemma, which involves counting the number of inversions in a certain permutation. Alternatively, it can be proven using properties of Gauss sums or by considering the multiplicative properties of the Legendre symbol.Given that part (b) relies on properties of the Legendre symbol and quadratic residues, and part (a) is about proving quadratic reciprocity, it's likely that problem 31.39 involves a counting argument or permutation-based proof of quadratic reciprocity.In summary, for part (b), by evaluating ( (-1)^{frac{m^2 - 1}{8}} ) for different residues of ( m ) modulo 8, we see that it matches the Legendre symbol ( left( frac{2}{m} right) ), thus proving the required identity.For part (a), without the exact problem statement, I can outline that quadratic reciprocity can be proven using various methods, including counting inversions in permutations, leveraging properties of Gauss sums, or using the Chinese Remainder Theorem. Each method involves showing that the product of the Legendre symbols ( left( frac{p}{q} right) ) and ( left( frac{q}{p} right) ) equals ( (-1)^{frac{(p-1)(q-1)}{4}} ), which is the essence of the quadratic reciprocity law.Overall, both parts involve deep properties of quadratic residues and the Legendre symbol, with part (a) being a foundational theorem in number theory and part (b) providing a specific evaluation of the Legendre symbol for the prime 2.

answer：Okay, so I have this problem here about the positional relationship between two lines: y = -x and y = -x + 6. Hmm, I remember from my math class that when dealing with lines, their slopes and y-intercepts can tell us a lot about how they relate to each other. Let me try to recall.First, I think the general form of a line is y = mx + b, where m is the slope and b is the y-intercept. So, for the first line, y = -x, I can rewrite that as y = -1x + 0. That means the slope (m) is -1 and the y-intercept (b) is 0. Now, the second line is y = -x + 6. Let me write that in the same form: y = -1x + 6. So, the slope here is also -1, and the y-intercept is 6. Wait, so both lines have the same slope of -1. I remember that if two lines have the same slope, they are parallel. But I also recall that if two lines have the same slope and the same y-intercept, they are actually the same line, meaning they coincide. But in this case, the y-intercepts are different: one is 0 and the other is 6. So, since they have the same slope but different y-intercepts, they must be parallel lines. That makes sense because they never intersect; they just run alongside each other. Let me visualize it: one line crosses the y-axis at (0,0) and the other at (0,6), both going downwards at the same steepness. Yeah, they don't meet anywhere, so they're definitely parallel.I think that's it. They can't be intersecting because their slopes are the same, and they aren't the same line because their y-intercepts differ. So, the positional relationship between y = -x and y = -x + 6 is that they are parallel.