💡Okay, so I have this problem here about a hyperbola and a parabola, and I need to find the length of AB where A and B are points where the asymptotes of the hyperbola intersect the parabola. The hyperbola is given by the equation (x^2 - y^2 = 1), and the parabola is (y^2 = 4x). The origin, O, is also a point of intersection. The answer choices are 4, 6, 8, and 16. First, I need to recall what asymptotes are. For a hyperbola, asymptotes are lines that the hyperbola approaches but never touches. For the hyperbola (x^2 - y^2 = 1), I think the asymptotes are the lines (y = x) and (y = -x). Let me confirm that. The standard form of a hyperbola is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), and its asymptotes are (y = pm frac{b}{a}x). In this case, (a^2 = 1) and (b^2 = 1), so (a = 1) and (b = 1), which means the asymptotes are indeed (y = x) and (y = -x). Okay, that part checks out.Now, I need to find where these asymptotes intersect the parabola (y^2 = 4x). So, I have two lines: (y = x) and (y = -x), and I need to find their points of intersection with the parabola. Let's start with the first asymptote, (y = x).Substituting (y = x) into the parabola equation (y^2 = 4x), we get:[(x)^2 = 4x]Simplifying that:[x^2 = 4x]Subtracting (4x) from both sides:[x^2 - 4x = 0]Factoring out an x:[x(x - 4) = 0]So, the solutions are (x = 0) and (x = 4). When (x = 0), substituting back into (y = x), we get (y = 0). That's the origin, O. When (x = 4), substituting back, we get (y = 4). So, one point of intersection is (4, 4).Now, let's do the same for the other asymptote, (y = -x). Substituting (y = -x) into the parabola equation:[(-x)^2 = 4x]Simplifying:[x^2 = 4x]Which is the same equation as before:[x^2 - 4x = 0]Factoring:[x(x - 4) = 0]So, again, (x = 0) and (x = 4). When (x = 0), (y = 0), which is the origin, O. When (x = 4), substituting back into (y = -x), we get (y = -4). So, the other point of intersection is (4, -4).Therefore, the points A and B are (4, 4) and (4, -4). Now, I need to find the distance between these two points to get the length of AB.The distance formula between two points ((x_1, y_1)) and ((x_2, y_2)) is:[sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]Plugging in the coordinates of A and B:[sqrt{(4 - 4)^2 + (-4 - 4)^2} = sqrt{0 + (-8)^2} = sqrt{64} = 8]So, the length of AB is 8 units. Looking back at the answer choices, 8 is option C. Wait, just to make sure I didn't make any mistakes. Let me double-check my steps.1. Found the asymptotes correctly: (y = x) and (y = -x).2. Substituted each into the parabola equation: - For (y = x): Got x = 0 and x = 4, leading to points (0,0) and (4,4). - For (y = -x): Similarly, got x = 0 and x = 4, leading to points (0,0) and (4, -4).3. Calculated the distance between (4,4) and (4,-4): The x-coordinates are the same, so the distance is just the difference in y-coordinates, which is 8.Yes, that all seems correct. I think I'm confident that the answer is 8, which is option C.**Final Answer**boxed{C}

💡Alright, so I have this geometry problem to prove that a line lying in one of two perpendicular planes and perpendicular to the line of intersection of these planes is perpendicular to the second plane. Hmm, okay, let me try to visualize this.First, let me recall what perpendicular planes are. If two planes are perpendicular, it means that they intersect at a right angle. So, imagine two walls meeting at a corner; they form a right angle where they meet. The line where they intersect is called the line of intersection.Now, the problem says there's a line lying in one of these planes, and this line is perpendicular to the line of intersection. I need to show that this line is also perpendicular to the second plane.Let me try to draw a mental picture. Let's say we have two perpendicular planes, α and β, intersecting along line c. There's a line a in plane α that's perpendicular to line c. I need to prove that line a is also perpendicular to plane β.Okay, so line a is in plane α and perpendicular to c. Since c is the intersection of α and β, and α and β are perpendicular, maybe there's a relationship between line a and plane β because of this perpendicularity.I remember that if a line is perpendicular to a plane, it is perpendicular to every line in that plane. But here, line a is only known to be perpendicular to line c, which is in both planes. So, does that mean it's perpendicular to the entire plane β?Wait, no, because line a is in plane α, and plane α is perpendicular to plane β. So, maybe there's a way to use the properties of perpendicular planes to relate line a to plane β.Let me think about the dihedral angle between the two planes. Since α and β are perpendicular, the dihedral angle between them is 90 degrees. Line c is the intersection, so it's like the hinge around which the two planes open up at a right angle.Line a is in plane α and perpendicular to c. So, in plane α, line a is sticking out from line c at a right angle. Since plane α and β are perpendicular, maybe line a is also sticking out from plane β at a right angle.But how can I formalize this? Maybe I can use some vector geometry or coordinate systems to make this more concrete.Let's assign a coordinate system where line c is along the z-axis. Then, plane α can be the x-z plane, and plane β can be the y-z plane. These two planes are perpendicular, intersecting along the z-axis (line c).Now, line a is in plane α (the x-z plane) and is perpendicular to line c (the z-axis). So, in the x-z plane, a line perpendicular to the z-axis would be along the x-axis. So, line a would be along the x-axis.Now, is this line a (along the x-axis) perpendicular to plane β (the y-z plane)? Well, to check if a line is perpendicular to a plane, the line must be perpendicular to every line in that plane. But in plane β, we have lines along the y-axis and the z-axis.Is the x-axis line a perpendicular to both the y-axis and the z-axis? Yes, because in a coordinate system, the x, y, and z axes are all mutually perpendicular. So, the x-axis is perpendicular to both the y-axis and the z-axis, which are in plane β.Therefore, line a is perpendicular to plane β.Wait, but I used a coordinate system to make this argument. Is there a way to do this without coordinates, using just geometric properties?Let me think. If two planes are perpendicular, then any line in one plane that is perpendicular to the line of intersection must be perpendicular to the other plane. Because the other plane contains lines that are perpendicular to the line of intersection, and since the two planes are perpendicular, those lines must also be perpendicular to the first plane.But I'm not sure if that's rigorous enough. Maybe I need to use the definition of perpendicular planes and lines.I recall that two planes are perpendicular if their normal vectors are perpendicular. So, if plane α has a normal vector n₁ and plane β has a normal vector n₂, then n₁ · n₂ = 0.Now, line a is in plane α and perpendicular to line c. Line c is the intersection of α and β, so it lies in both planes. If line a is perpendicular to line c, and line c is in both planes, then line a must be aligned with the normal vector of plane β.Wait, is that right? If line a is perpendicular to line c, and line c is in both planes, then line a is perpendicular to plane β because it aligns with the normal vector of β.But I need to make sure that line a is indeed aligned with the normal vector of β. Since plane α and β are perpendicular, their normal vectors are perpendicular. So, if line a is in plane α and perpendicular to line c, which is in both planes, then line a must be parallel to the normal vector of β.Therefore, line a is perpendicular to plane β.Hmm, this seems to make sense, but I'm not entirely sure if I've covered all the bases. Maybe I should think about it in terms of projections.If line a is in plane α and perpendicular to line c, then its projection onto plane β would be a line perpendicular to line c as well. But since plane α and β are perpendicular, the projection of line a onto β would actually be line a itself, but rotated 90 degrees.Wait, no, projections work differently. If line a is in plane α and perpendicular to line c, then its projection onto plane β would be a line that's also perpendicular to line c, but in plane β.But since line a is already perpendicular to line c, and plane β contains line c, then line a must be perpendicular to plane β.I think I'm going in circles here. Maybe I should recall the theorem that states if a line is perpendicular to two intersecting lines at a point, then it is perpendicular to the plane containing those two lines.In this case, line a is perpendicular to line c, which is in both planes. If I can find another line in plane β that is also perpendicular to line a, then I can apply the theorem.Wait, but line a is in plane α, and plane α is perpendicular to plane β. So, any line in plane α that is perpendicular to line c must also be perpendicular to plane β.I think I'm repeating myself. Maybe I need to use the fact that in plane β, there are lines perpendicular to line c, and since line a is perpendicular to line c, it must be aligned with those lines in plane β.But I'm not entirely confident. Maybe I should look for a more formal proof.Let me try to write it out step by step.1. Let α and β be two perpendicular planes intersecting along line c.2. Let a be a line in plane α such that a is perpendicular to line c.3. Since α and β are perpendicular, their normal vectors are perpendicular.4. Line a is in plane α and perpendicular to line c, which is the intersection of α and β.5. Therefore, line a must be parallel to the normal vector of plane β.6. Hence, line a is perpendicular to plane β.Wait, that seems too quick. Let me elaborate.Since α and β are perpendicular, their normal vectors n₁ and n₂ satisfy n₁ · n₂ = 0.Line a is in plane α, so it is perpendicular to n₁.Line a is also perpendicular to line c, which lies in both planes.Since line c is in both planes, it is perpendicular to both normal vectors n₁ and n₂.Wait, no, line c is the intersection of α and β, so it is not necessarily perpendicular to the normal vectors. Instead, line c is parallel to both planes.Actually, line c lies in both planes, so it is parallel to the line of intersection, which is itself.Hmm, maybe I need to think differently.Let me consider vectors. Let’s assign coordinates such that line c is along the z-axis, plane α is the x-z plane, and plane β is the y-z plane.Then, line a is in the x-z plane and perpendicular to the z-axis, so it must be along the x-axis.Now, plane β is the y-z plane, and the x-axis is perpendicular to the y-z plane because the x-axis is orthogonal to both the y and z axes.Therefore, line a, being along the x-axis, is perpendicular to plane β.This coordinate-based approach seems to confirm the result.But I need to make sure this isn't just a special case. What if the planes are not aligned with the coordinate axes?Well, in general, if two planes are perpendicular, we can always choose a coordinate system where one plane is the x-y plane and the other is the x-z plane, with their intersection along the x-axis.Then, a line in the x-y plane perpendicular to the x-axis would be along the y-axis, which is perpendicular to the x-z plane.Similarly, a line in the x-z plane perpendicular to the x-axis would be along the z-axis, which is perpendicular to the x-y plane.So, regardless of the orientation, the property holds.Therefore, I think the proof is solid. A line in one of two perpendicular planes, perpendicular to their line of intersection, must be perpendicular to the other plane.

💡Okay, so I need to figure out the starting number such that between this number and 30, there's exactly one number that has two of its digits as 3. Hmm, let's break this down step by step.First, I should understand what it means for a number to have two digits as 3. That would mean the number has a 3 in both the tens place and the ones place. For example, 33 has two 3s. But wait, the problem says "between a certain number and 30," so I'm assuming we're dealing with numbers less than 30. But 33 is greater than 30, so maybe I'm misunderstanding something.Wait, perhaps the starting number could be greater than 30, and we're looking at numbers between that starting number and 30. But that doesn't make much sense because 30 is the upper limit. Maybe the starting number is less than 30, and we're looking at numbers between the starting number and 30, inclusive or exclusive? The problem isn't entirely clear on that.Let me consider both possibilities. If the starting number is less than 30, and we're looking at numbers between that starting number and 30, then the numbers we're considering are from the starting number up to 30. If the starting number is greater than 30, then we're looking at numbers from the starting number down to 30. But since 33 is greater than 30 and has two 3s, maybe the starting number is 33, and we're looking at numbers between 33 and 30. But that seems a bit odd because 30 is less than 33.Alternatively, maybe the starting number is less than 30, and we're looking for a number between the starting number and 30 that has two 3s. But since 33 is greater than 30, it wouldn't be in that range. So perhaps there's no number between the starting number and 30 that has two 3s, which contradicts the problem statement that says there is exactly one such number.Wait, maybe I'm misinterpreting the problem. Perhaps the starting number is part of a range where numbers are considered, and within that range, exactly one number has two 3s. If the starting number is 32, then the numbers between 32 and 30 would be 31 and 32, but neither of those have two 3s. That doesn't seem right.Hold on, maybe the starting number is 33. If we consider numbers between 33 and 30, that would include 33, 32, 31, and 30. Among these, only 33 has two 3s. So that fits the condition of exactly one number having two 3s. Therefore, the starting number would be 33.But wait, the problem says "between a certain number and 30," which might imply that 30 is not included. If we exclude 30, then the numbers between 33 and 30 would be 32 and 31, which don't have two 3s. That doesn't work. So maybe including 30 is necessary.Alternatively, if the starting number is 31, then the numbers between 31 and 30 would be just 31, which doesn't have two 3s. That doesn't fit. If the starting number is 30, then the numbers between 30 and 30 would be none, which also doesn't fit.I'm getting a bit confused here. Let me try a different approach. Let's list out numbers around 30 and see which ones have two 3s. The number 33 has two 3s, but it's greater than 30. The number 30 itself has a 3 in the tens place and a 0 in the ones place, so it doesn't have two 3s. The number 23 has a 2 and a 3, so only one 3. Similarly, 13, 43, etc., all have only one 3.Wait, maybe the starting number is 33, and we're considering numbers from 33 down to 30. In that case, 33 is included and has two 3s, and the other numbers (32, 31, 30) don't have two 3s. So that would satisfy the condition of exactly one number having two 3s between the starting number and 30.But I'm not sure if the problem expects the starting number to be less than 30. If so, then there might be no solution because numbers less than 30 don't have two 3s. Unless we consider numbers like 33, but that's greater than 30.Maybe the problem is phrased differently. Perhaps it's asking for the starting number such that in the range from that number up to 30, there's exactly one number with two 3s. If that's the case, then the starting number would be 33 because 33 is the only number in that range with two 3s.But I'm still a bit uncertain. Let me think again. If the starting number is 33, and we're looking at numbers between 33 and 30, which would be 33, 32, 31, 30, then only 33 has two 3s. So that works. If the starting number were lower, like 32, then between 32 and 30, we have 32, 31, 30, none of which have two 3s, which doesn't satisfy the condition.Alternatively, if the starting number is 34, then between 34 and 30, we have 34, 33, 32, 31, 30. Here, 33 has two 3s, and 34 has one 3. So that's still only one number with two 3s. So starting at 34 would also satisfy the condition.Wait, but the problem says "exactly one number has two of the digits as 3." So if the starting number is 34, then 34 has one 3, and 33 has two 3s. So that's still only one number with two 3s. So starting at 34 would also work.But the problem asks for the starting number, implying a specific answer. Maybe the smallest starting number that satisfies the condition is 33 because starting at 33, the only number with two 3s is 33 itself. Starting at 34, you still have 33 as the only number with two 3s. So perhaps the starting number is 33.But I'm not entirely confident. Maybe I should look for numbers less than 30 that have two 3s, but I don't think any two-digit numbers less than 30 have two 3s because the tens digit would have to be 3, making the number 30 or above. So numbers like 33, 34, etc., are the ones with two 3s.Therefore, the starting number must be 33 because it's the smallest number with two 3s, and between 33 and 30, only 33 has two 3s. So the starting number is 33.

💡Okay, so I need to solve this inequality: (2x - 1)/(3x + 1) > 1. Hmm, I remember that when dealing with inequalities involving fractions, I have to be careful about the denominator because it can't be zero, and also, if I multiply both sides by the denominator, I have to consider whether it's positive or negative because that affects the inequality sign.First, let me write down the inequality again to make sure I have it right: (2x - 1)/(3x + 1) > 1. Alright, so I need to find all x values where this fraction is greater than 1.Maybe I should start by subtracting 1 from both sides to get everything on one side. That would give me (2x - 1)/(3x + 1) - 1 > 0. Now, I need to combine these terms into a single fraction. To do that, I'll express 1 as (3x + 1)/(3x + 1) so that both terms have the same denominator.So, subtracting these fractions: (2x - 1 - (3x + 1))/(3x + 1) > 0. Let me simplify the numerator: 2x - 1 - 3x - 1. Combining like terms, that's (2x - 3x) + (-1 - 1) = (-x - 2). So now the inequality is (-x - 2)/(3x + 1) > 0.Hmm, I have a negative coefficient for x in the numerator. Maybe I can factor out a negative sign to make it easier. If I factor out -1 from the numerator, it becomes -1*(x + 2)/(3x + 1) > 0. Now, if I multiply both sides by -1, I have to remember to flip the inequality sign. So, (x + 2)/(3x + 1) < 0.Alright, now I have (x + 2)/(3x + 1) < 0. To solve this inequality, I need to find where the expression is negative. That happens when the numerator and denominator have opposite signs.First, let's find the critical points where the expression equals zero or is undefined. The numerator is zero when x + 2 = 0, so x = -2. The denominator is zero when 3x + 1 = 0, so x = -1/3. These points divide the real number line into intervals that I can test.So, the critical points are x = -2 and x = -1/3. Let me plot these on a number line:<---(-2)----(-1/3)--->Now, I'll test each interval:1. When x < -2: Let's pick x = -3. Plugging into (x + 2)/(3x + 1): (-3 + 2)/(3*(-3) + 1) = (-1)/(-9 + 1) = (-1)/(-8) = 1/8, which is positive. So, this interval doesn't satisfy the inequality.2. When -2 < x < -1/3: Let's pick x = -1. Plugging in: (-1 + 2)/(3*(-1) + 1) = (1)/(-3 + 1) = 1/(-2) = -1/2, which is negative. So, this interval satisfies the inequality.3. When x > -1/3: Let's pick x = 0. Plugging in: (0 + 2)/(3*0 + 1) = 2/1 = 2, which is positive. So, this interval doesn't satisfy the inequality.Therefore, the solution is the interval where the expression is negative, which is -2 < x < -1/3.But wait, I should also consider the points where the expression is zero or undefined. At x = -2, the expression is zero, but the inequality is strict (less than zero), so x = -2 is not included. At x = -1/3, the expression is undefined, so x = -1/3 is also not included.So, putting it all together, the solution set is all real numbers between -2 and -1/3, not including the endpoints.I think that's it. Let me just double-check my steps to make sure I didn't make any mistakes.1. Started with (2x - 1)/(3x + 1) > 1.2. Subtracted 1 to get (-x - 2)/(3x + 1) > 0.3. Factored out -1 to get (x + 2)/(3x + 1) < 0.4. Found critical points at x = -2 and x = -1/3.5. Tested intervals and found that only -2 < x < -1/3 satisfies the inequality.Everything seems to check out. I don't think I missed anything.

💡Okay, so I have this problem here about averages of two samples and then the average of the combined sample. Let me try to understand what it's asking.First, there's a sample of numbers: x₁, x₂, ..., xₙ, and their average is x̄. Then, there's another sample: y₁, y₂, ..., yₘ, and their average is ȳ. It's given that x̄ is not equal to ȳ, which is important because if they were equal, the combined average would just be the same as both, and the problem would be trivial.Now, when we combine these two samples, the new average is given as z = a x̄ + (1 - a) ȳ, where a is between 0 and 1/2. So, a is a weight that's less than 1/2, meaning that the weight on x̄ is less than the weight on ȳ. That probably tells us something about the sizes of the two samples.I need to figure out the relationship between n and m, the sizes of the two samples. The options are:A: n = m B: n ≥ m C: n < m D: n > mSo, I need to figure out whether n is equal to, greater than, or less than m, based on the given information.Let me recall how the average of combined samples works. The combined average z should be the total sum of all the x's and y's divided by the total number of elements, which is n + m.So, the total sum of the x's is n x̄, and the total sum of the y's is m ȳ. Therefore, the combined average z is:z = (n x̄ + m ȳ) / (n + m)But it's given that z is also equal to a x̄ + (1 - a) ȳ. So, I can set these two expressions equal to each other:(n x̄ + m ȳ) / (n + m) = a x̄ + (1 - a) ȳHmm, okay. Let me write this out:(n x̄ + m ȳ) / (n + m) = a x̄ + (1 - a) ȳI can rearrange this equation to solve for a or to find a relationship between n and m.Let me multiply both sides by (n + m) to eliminate the denominator:n x̄ + m ȳ = (a x̄ + (1 - a) ȳ) * (n + m)Expanding the right side:n x̄ + m ȳ = a x̄ (n + m) + (1 - a) ȳ (n + m)Let me distribute the terms:n x̄ + m ȳ = a n x̄ + a m x̄ + (1 - a) n ȳ + (1 - a) m ȳNow, let's collect like terms. Let's bring all the x̄ terms to one side and ȳ terms to the other side.n x̄ - a n x̄ - a m x̄ = (1 - a) n ȳ + (1 - a) m ȳ - m ȳFactor out x̄ on the left and ȳ on the right:x̄ (n - a n - a m) = ȳ [(1 - a) n + (1 - a) m - m]Simplify the expressions inside the parentheses.On the left side:n - a n - a m = n(1 - a) - a mOn the right side:(1 - a) n + (1 - a) m - m = (1 - a) n + (1 - a) m - mFactor m from the last two terms:= (1 - a) n + m [(1 - a) - 1] = (1 - a) n + m (-a) = (1 - a) n - a mSo, now we have:x̄ [n(1 - a) - a m] = ȳ [(1 - a) n - a m]Since x̄ ≠ ȳ, the coefficients of x̄ and ȳ must be equal for the equation to hold. Therefore:n(1 - a) - a m = (1 - a) n - a mWait, that seems redundant. Let me double-check my steps.Wait, actually, on the right side, it's [(1 - a) n - a m]. So, both sides have the same expression: n(1 - a) - a m.So, x̄ times that expression equals ȳ times the same expression.But since x̄ ≠ ȳ, the only way this can be true is if the coefficient is zero.So, n(1 - a) - a m = 0So, n(1 - a) = a mLet me write that:n(1 - a) = a mSo, solving for n/m:n/m = a / (1 - a)So, n/m = a / (1 - a)Given that 0 < a < 1/2, so a is less than 1/2, which means 1 - a is greater than 1/2.Therefore, a / (1 - a) is less than (1/2) / (1 - 1/2) = (1/2) / (1/2) = 1.So, a / (1 - a) < 1, which implies that n/m < 1, so n < m.Wait, that's interesting. So, n/m is less than 1, meaning n is less than m.But let me think again.Wait, n(1 - a) = a mSo, n = [a / (1 - a)] mSince a < 1/2, then a / (1 - a) < (1/2) / (1 - 1/2) = 1So, n = [something less than 1] * mTherefore, n < mSo, that would suggest that n is less than m, which is option C.But wait, let me check my reasoning again because I might have messed up.Wait, in the equation:n(1 - a) = a mSo, n = [a / (1 - a)] mGiven that a is between 0 and 1/2, so a is less than 1 - a because 1 - a is greater than 1/2.Therefore, a / (1 - a) is less than 1, so n is less than m.Therefore, n < m, which is option C.But wait, in the initial problem, the combined average is z = a x̄ + (1 - a) ȳ, with 0 < a < 1/2.So, since a is less than 1/2, the weight on x̄ is less than the weight on ȳ.Which would mean that the sample with the smaller weight has more influence? Wait, no, actually, the weight corresponds to the proportion of the sample.Wait, maybe I got that backwards.Wait, let's think about it differently.If the combined average is closer to ȳ than to x̄, because a is less than 1/2, so (1 - a) is greater than a.Therefore, the combined average is closer to ȳ, which would mean that the sample with average ȳ has more elements, right?Because if you have more elements from ȳ, the combined average would be closer to ȳ.So, if the combined average is closer to ȳ, then m must be greater than n.Which again suggests that n < m.So, that's consistent with what I found earlier.Therefore, the correct answer is C: n < m.Wait, but let me check with an example to make sure.Suppose n = 1 and m = 2.So, x̄ = let's say 10, ȳ = 20.Then, combined average z = (1*10 + 2*20)/(1 + 2) = (10 + 40)/3 = 50/3 ≈ 16.666...Expressed as a x̄ + (1 - a) ȳ, so:16.666... = a*10 + (1 - a)*20Let's solve for a:16.666... = 10a + 20 - 20a 16.666... = 20 - 10a 10a = 20 - 16.666... 10a = 3.333... a = 0.333...Which is 1/3, which is less than 1/2. So, in this case, n = 1, m = 2, and a = 1/3 < 1/2, which is consistent with n < m.Another example: n = 2, m = 3.x̄ = 10, ȳ = 20.Combined average z = (2*10 + 3*20)/(2 + 3) = (20 + 60)/5 = 80/5 = 16.Expressed as a x̄ + (1 - a) ȳ:16 = 10a + 20(1 - a) 16 = 10a + 20 - 20a 16 = 20 - 10a 10a = 4 a = 0.4Which is still less than 1/2, and n = 2 < m = 3.So, this seems to hold.Conversely, if n > m, let's see.Let n = 3, m = 2.x̄ = 10, ȳ = 20.Combined average z = (3*10 + 2*20)/(3 + 2) = (30 + 40)/5 = 70/5 = 14.Expressed as a x̄ + (1 - a) ȳ:14 = 10a + 20(1 - a) 14 = 10a + 20 - 20a 14 = 20 - 10a 10a = 6 a = 0.6Which is greater than 1/2, which contradicts the given condition that a < 1/2.Therefore, if n > m, a would be greater than 1/2, which is not allowed in this problem.Therefore, in order for a to be less than 1/2, n must be less than m.So, the correct answer is C: n < m.**Final Answer**boxed{C}

💡First, calculate the total number of tickets Amanda needs to sell, which is 80.On the first day, Amanda sells 5 friends 4 tickets each. So, the number of tickets sold on the first day is 5 multiplied by 4, which equals 20.On the second day, she sells 32 tickets.Adding the tickets sold on the first and second days gives a total of 20 + 32 = 52 tickets sold in the first two days.To meet her goal of 80 tickets, Amanda needs to sell 80 - 52 = 28 tickets on the third day.Therefore, Amanda has 3 days to sell all the tickets.