💡Okay, so I have this problem: I need to find the value of x that gives the minimum value for the quadratic expression x² - 6x + 5. Hmm, quadratic expressions, right? They graph as parabolas. Since the coefficient of x² is positive (which is 1 in this case), the parabola opens upwards. That means it has a minimum point, which is the vertex of the parabola. So, I need to find the vertex of this quadratic function.I remember there are a couple of ways to find the vertex of a quadratic function. One way is by completing the square, and another is using the formula for the x-coordinate of the vertex, which is -b/(2a). Let me try both methods to make sure I get the same answer.First, let me try completing the square. The expression is x² - 6x + 5. To complete the square, I need to take the coefficient of x, which is -6, divide it by 2 to get -3, and then square it to get 9. So, I can rewrite the expression as:x² - 6x + 9 - 9 + 5This simplifies to:(x - 3)² - 4So, the expression is now in the form (x - h)² + k, where h is 3 and k is -4. The vertex form of a quadratic function is (x - h)² + k, and the vertex is at (h, k). Therefore, the vertex is at (3, -4). Since the parabola opens upwards, this is the minimum point. So, the value of x that gives the minimum value is 3.Now, let me try using the formula for the x-coordinate of the vertex, which is -b/(2a). In the quadratic expression ax² + bx + c, a is 1, b is -6, and c is 5. Plugging these into the formula:x = -(-6)/(2*1) = 6/2 = 3So, again, I get x = 3 as the value where the minimum occurs. To find the minimum value itself, I can plug x = 3 back into the original expression:(3)² - 6*(3) + 5 = 9 - 18 + 5 = -4So, the minimum value of the expression is -4, occurring at x = 3.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For completing the square, I took x² - 6x and added and subtracted 9 to complete the square, which gives (x - 3)² - 4. That seems correct. Using the vertex formula, I correctly identified a, b, and c, and calculated x = 3. Plugging back into the original equation, I correctly calculated 9 - 18 + 5 to get -4. So, both methods confirm that x = 3 is the point where the expression reaches its minimum value.I think I'm confident with this answer. It makes sense because the parabola opens upwards, so the vertex is indeed the minimum point. If I were to graph this, I could plot a few points around x = 3 to see how the function behaves. For example, at x = 2, the value is 4 - 12 + 5 = -3; at x = 4, it's 16 - 24 + 5 = -3 as well. So, the function decreases as it approaches x = 3 and then increases after that, which aligns with x = 3 being the minimum point.Another way to think about this is using calculus, although I'm not sure if that's necessary here. If I take the derivative of the function f(x) = x² - 6x + 5, I get f'(x) = 2x - 6. Setting the derivative equal to zero to find critical points: 2x - 6 = 0, so x = 3. Then, checking the second derivative, f''(x) = 2, which is positive, confirming that it's a minimum point. So, calculus also leads me to the same conclusion.In summary, whether I complete the square, use the vertex formula, or apply calculus, I consistently find that the value of x which gives the minimum value of the expression x² - 6x + 5 is 3. The minimum value itself is -4, but the question specifically asks for the value of x, so the answer is 3.

💡Okay, so I have this problem about a geometric sequence. Let me try to understand what it's asking. It says that the geometric sequence {a_n} is increasing, and S_n is the sum of its first n terms. We are given two pieces of information: a₁ + a₅ = 17/2 and a₂a₄ = 4. We need to find S₆, which is the sum of the first six terms. The options are A: 27/16, B: 27/8, C: 63/4, and D: 63/2.Alright, so let's start by recalling what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted by q. So, the nth term of a geometric sequence can be written as a_n = a₁ * q^(n-1).Given that the sequence is increasing, that tells me that the common ratio q must be greater than 1 because each term is larger than the previous one.Now, let's write down the given information using the formula for the nth term.First, a₁ + a₅ = 17/2. Since a₅ is the fifth term, that would be a₁ * q^(5-1) = a₁ * q⁴. So, we can write:a₁ + a₁ * q⁴ = 17/2Similarly, the second piece of information is a₂ * a₄ = 4. Let's express a₂ and a₄ in terms of a₁ and q.a₂ = a₁ * q^(2-1) = a₁ * qa₄ = a₁ * q^(4-1) = a₁ * q³So, multiplying these together:a₂ * a₄ = (a₁ * q) * (a₁ * q³) = a₁² * q⁴ = 4So now we have two equations:1. a₁ + a₁ * q⁴ = 17/22. a₁² * q⁴ = 4Let me write these again for clarity:1. a₁(1 + q⁴) = 17/22. (a₁)²(q⁴) = 4Hmm, so we have two equations with two unknowns: a₁ and q. Maybe we can solve for one variable in terms of the other and substitute.Looking at equation 2: (a₁)²(q⁴) = 4. Let's solve for a₁²:a₁² = 4 / q⁴So, a₁ = sqrt(4 / q⁴) = 2 / q²But since the sequence is increasing and q > 1, a₁ must be positive, so we don't need to worry about the negative root.So, a₁ = 2 / q²Now, let's substitute this into equation 1:a₁(1 + q⁴) = 17/2Substituting a₁:(2 / q²)(1 + q⁴) = 17/2Let me simplify this equation:Multiply both sides by q² to eliminate the denominator:2(1 + q⁴) = (17/2) q²Multiply both sides by 2 to eliminate the fraction:4(1 + q⁴) = 17 q²Expand the left side:4 + 4 q⁴ = 17 q²Bring all terms to one side:4 q⁴ - 17 q² + 4 = 0Hmm, this is a quartic equation, but it's quadratic in terms of q². Let me let x = q², so the equation becomes:4x² - 17x + 4 = 0Now, let's solve this quadratic equation for x.Using the quadratic formula:x = [17 ± sqrt(17² - 4*4*4)] / (2*4)x = [17 ± sqrt(289 - 64)] / 8x = [17 ± sqrt(225)] / 8x = [17 ± 15] / 8So, two solutions:x = (17 + 15)/8 = 32/8 = 4x = (17 - 15)/8 = 2/8 = 1/4So, x = 4 or x = 1/4But x = q², so q² = 4 or q² = 1/4Since q > 1 (because the sequence is increasing), q² must be greater than 1. So, q² = 4, which gives q = 2 (since q is positive).So, q = 2.Now, let's find a₁. Earlier, we had a₁ = 2 / q²So, a₁ = 2 / (2²) = 2 / 4 = 1/2So, a₁ = 1/2 and q = 2.Now, we need to find S₆, the sum of the first six terms.The formula for the sum of the first n terms of a geometric sequence is:S_n = a₁ (1 - qⁿ) / (1 - q) when q ≠ 1So, plugging in the values:S₆ = (1/2) (1 - 2⁶) / (1 - 2)First, compute 2⁶: 2^6 = 64So, 1 - 64 = -63Denominator: 1 - 2 = -1So, S₆ = (1/2) * (-63) / (-1)Simplify:The negatives cancel out: (1/2) * 63 / 1 = 63/2So, S₆ = 63/2Looking back at the options, that's option D.Wait, let me double-check my calculations to make sure I didn't make a mistake.We had a₁ = 1/2, q = 2.Compute S₆:S₆ = a₁ (q⁶ - 1)/(q - 1)Wait, I think I might have written the formula incorrectly earlier. Let me confirm.Yes, the formula is S_n = a₁ (1 - qⁿ)/(1 - q) when q ≠ 1.But sometimes it's written as S_n = a₁ (qⁿ - 1)/(q - 1). Both are equivalent because (1 - qⁿ)/(1 - q) = (qⁿ - 1)/(q - 1).So, using the second form:S₆ = (1/2)(2⁶ - 1)/(2 - 1) = (1/2)(64 - 1)/1 = (1/2)(63) = 63/2Yes, that's correct. So, 63/2 is the sum of the first six terms.Therefore, the correct answer is D: 63/2.**Final Answer**boxed{D}

💡First, I need to determine the radius of the well from the given diameter. Since the diameter is 2 meters, the radius is half of that, which is 1 meter.Next, I'll use the formula for the volume of a cylinder, which is ( V = pi r^2 h ), where ( r ) is the radius and ( h ) is the height (or depth in this case).Plugging in the known values:- Radius (( r )) = 1 meter- Depth (( h )) = 14 metersSo, the volume becomes:[V = pi (1)^2 (14) = 14pi text{ cubic meters}]To provide a numerical approximation, I'll use ( pi approx 3.14159 ):[V approx 3.14159 times 14 approx 43.9823 text{ cubic meters}]Rounding this to one decimal place gives:[V approx 44.0 text{ cubic meters}]

💡Okay, so I need to figure out what tan 45 degrees is. Hmm, I remember that tangent has something to do with sine and cosine. Let me think... Oh yeah, tan θ is equal to sin θ divided by cos θ. So, tan 45° should be sin 45° divided by cos 45°. Wait, what are the values of sin 45° and cos 45°? I think they're both the same because in a 45-45-90 triangle, the legs are equal. So, if the legs are both 1, then the hypotenuse would be sqrt(2). That means sin 45° is 1 over sqrt(2), and cos 45° is also 1 over sqrt(2). So, if I plug those into the tangent formula: tan 45° = (1/sqrt(2)) divided by (1/sqrt(2)). Dividing these two should give me 1 because the same numbers cancel out. Let me double-check. Maybe I can think about the unit circle. At 45 degrees, both the x and y coordinates are equal, which makes sense why sin and cos are the same. So, their ratio should be 1. Yeah, that makes sense. I guess another way to look at it is that the tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. In a 45-45-90 triangle, both sides are equal, so their ratio is 1. Wait, is there any chance I'm mixing up sine and cosine? No, because for 45 degrees, both are equal, so their ratio is still 1. I think I'm confident that tan 45° is 1. Just to make sure, maybe I can recall some special angles. I know that tan 30° is 1/sqrt(3) and tan 60° is sqrt(3). So, tan 45° being exactly 1 fits right in between them, which seems logical because 45 degrees is halfway between 30 and 60. Yeah, I think I've got it. Tan 45° is definitely 1.

💡Okay, so I have this problem about four pairwise intersecting lines, and I need to figure out if it's possible to choose one point on each line such that these points form a trapezoid and then a parallelogram. Hmm, let's start by understanding the problem better.First, what does it mean for four lines to be pairwise intersecting? I think it means that every pair of lines intersects at some point. So, if I have four lines, each one intersects with the other three. That should mean that all four lines lie in the same plane, right? Because if they are all intersecting each other, they can't be skew lines in three-dimensional space. So, I can assume all four lines are in a single plane.Now, part a) asks if it's possible to choose points on each line to form a trapezoid. A trapezoid is a quadrilateral with at least one pair of parallel sides. So, I need to see if I can pick points on these four lines such that two of the sides are parallel.Since all four lines are in the same plane, I can use properties of plane geometry here. Let me think about how lines can intersect and form parallel sides. If I have four lines, each intersecting the others, maybe I can find two lines that are parallel? Wait, but if all four lines are pairwise intersecting, they can't be parallel because parallel lines don't intersect. So, none of the four lines can be parallel to each other. That complicates things because if none of the original lines are parallel, how can I get a trapezoid?Maybe I don't need the original lines to be parallel, but just the sides of the trapezoid. So, even though the original lines aren't parallel, by choosing specific points on them, I can create sides that are parallel. That makes sense. So, perhaps I can choose points such that two sides of the trapezoid are parallel even if the original lines aren't.Let me visualize this. Imagine four lines intersecting each other, forming a sort of star shape. If I pick points on each line, I can connect them in a way that two sides are parallel. Maybe by choosing points that are proportionally spaced along the lines, I can create parallel sides.I remember something about similar triangles and intercept theorems. Maybe I can use that here. If I can find two pairs of points on two different lines such that the segments between them are parallel, that could work. Let me try to formalize this.Suppose I have lines L1, L2, L3, and L4, all intersecting each other. Let's pick a point A on L1 and a point B on L2. Then, I need to find points C on L3 and D on L4 such that AB is parallel to CD. Since all lines are in the same plane, I can use the concept of slopes. If I assign coordinates to the points, I can calculate the slopes and ensure they are equal.Wait, assigning coordinates might be a good approach. Let me set up a coordinate system where one of the lines is the x-axis. Let's say L1 is the x-axis. Then, I can assign coordinates to the intersection points of the other lines with L1. Let's say L2 intersects L1 at (a, 0), L3 at (b, 0), and L4 at (c, 0). Now, I can express the equations of L2, L3, and L4 in terms of their slopes.Let me denote the slopes of L2, L3, and L4 as m2, m3, and m4 respectively. So, the equations of the lines would be:- L2: y = m2(x - a)- L3: y = m3(x - b)- L4: y = m4(x - c)Now, I need to choose points A, B, C, D on L1, L2, L3, L4 respectively such that AB is parallel to CD. Let's pick a point A on L1, say at (x1, 0). Then, point B on L2 would be at (x2, m2(x2 - a)). Similarly, point C on L3 would be at (x3, m3(x3 - b)), and point D on L4 would be at (x4, m4(x4 - c)).For AB to be parallel to CD, the slope of AB should equal the slope of CD. The slope of AB is (m2(x2 - a) - 0)/(x2 - x1) = m2(x2 - a)/(x2 - x1). The slope of CD is (m4(x4 - c) - m3(x3 - b))/(x4 - x3). So, we need:m2(x2 - a)/(x2 - x1) = (m4(x4 - c) - m3(x3 - b))/(x4 - x3)This seems complicated, but maybe I can choose points in such a way that this equation holds. Perhaps by choosing x1, x2, x3, x4 appropriately.Alternatively, maybe there's a simpler way. Since all lines intersect, maybe I can use the concept of projective geometry or something like that. But I'm not too familiar with that.Wait, maybe I can use the fact that in a plane, given four lines, I can always find a trapezoid by selecting appropriate points. Since none of the lines are parallel, but by choosing points, I can create parallel sides.Let me think about it differently. If I fix two lines, say L1 and L2, and choose points A on L1 and B on L2, then I can find points C and D on L3 and L4 such that AB is parallel to CD. Is that always possible?I think it is, because for any two lines, you can find a pair of points that create a segment parallel to another pair. But I'm not entirely sure. Maybe I need to construct it.Suppose I have lines L1 and L2 intersecting at point O. Then, L3 and L4 also intersect at O. Wait, no, pairwise intersecting doesn't mean all intersect at the same point. They could form a complete quadrilateral, where each pair intersects at a different point.So, actually, four pairwise intersecting lines in a plane form a complete quadrilateral, which has six intersection points. So, in that case, maybe I can use those intersection points to form a trapezoid.But I need to choose one point on each line, not necessarily the intersection points. Hmm.Wait, maybe I can use the concept of similar triangles. If I can find two triangles that are similar, their corresponding sides would be parallel.Alternatively, maybe I can use affine transformations. Since affine transformations preserve parallelism, I can transform the figure into a simpler one where it's easier to see.But I'm not sure if that's necessary. Maybe I can just use coordinate geometry.Let me try assigning coordinates again. Let me set L1 as the x-axis, L2 as some line through the origin with slope m, L3 as another line through the origin with slope n, and L4 as a line through the origin with slope p. Wait, but if all four lines pass through the origin, then they all intersect at the origin, but pairwise intersecting doesn't require them to intersect at the same point.Wait, no, pairwise intersecting just means each pair intersects somewhere, not necessarily all at the same point. So, maybe I should assign them to intersect at different points.This is getting complicated. Maybe I should look for a theorem or something that states whether this is possible.I recall that in projective geometry, given four lines, you can always find a trapezoid, but I'm not sure. Maybe I should try to construct it.Suppose I have four lines: L1, L2, L3, L4. Let me pick a point A on L1. Then, I need to pick a point B on L2 such that the line AB can be made parallel to some line CD, where C is on L3 and D is on L4.Alternatively, maybe I can fix two points and then find the other two. Let me pick A on L1 and C on L3. Then, I need to find B on L2 and D on L4 such that AB is parallel to CD.To ensure AB is parallel to CD, the vectors AB and CD must be scalar multiples of each other. So, if I can express the coordinates of B and D in terms of A and C, I can set up equations to satisfy this condition.This seems doable, but I need to make sure that such points B and D exist on L2 and L4 respectively.Alternatively, maybe I can use the concept of harmonic division or something like that. But I'm not too familiar with that.Wait, maybe I can use the fact that in a plane, given two lines, you can always find a transversal that intersects them at points such that the segments are parallel. So, if I have L1 and L2, I can find a transversal that intersects them at points A and B such that AB is parallel to some direction. Then, I can do the same for L3 and L4.But I need to ensure that the direction is the same for both pairs. So, maybe I can choose a direction, and then find points on L1 and L2 such that AB is in that direction, and similarly points on L3 and L4 such that CD is in the same direction.But how do I choose the direction? Maybe I can use the slopes of the lines to determine a suitable direction.Alternatively, maybe I can use the concept of midpoints. If I can find midpoints such that the midline is parallel, but I'm not sure.Wait, maybe I can use the concept of translation. If I can translate one line to coincide with another, but I'm not sure if that helps.This is getting a bit tangled. Maybe I should look for a simpler approach. Let's think about the dual problem. Instead of choosing points on lines, maybe think about choosing lines through points, but I'm not sure.Wait, another idea: since all four lines are in a plane, and they are pairwise intersecting, they form a complete quadrilateral. A complete quadrilateral has three diagonals, and the midpoints of the diagonals are collinear (Newton-Gauss line). Maybe this property can help in constructing a trapezoid.But I'm not sure how to connect that to forming a trapezoid. Maybe if I can find midpoints that are aligned in a certain way, I can create parallel sides.Alternatively, maybe I can use the concept of similar triangles formed by the intersections. If I can find two triangles that are similar, their corresponding sides would be parallel.But I'm not sure. Maybe I need to take a step back.Let me consider specific cases. Suppose I have four lines in a plane, no two parallel, no three concurrent. So, they form a complete quadrilateral with six intersection points. Now, can I choose one point on each line such that they form a trapezoid?I think yes. Because I can choose points such that two sides are parallel. For example, pick two lines, and on each, pick a point such that the segment between them is parallel to a segment on the other two lines.But how to formalize this?Maybe I can use the concept of projective correspondence. If I can establish a correspondence between points on two lines such that the connecting lines are parallel, then I can extend this to four lines.Alternatively, maybe I can use the fact that in a plane, given two lines, there exists a unique line parallel to a given direction through a point. So, if I fix a direction, I can draw lines parallel to that direction through each of the four lines, and their intersection points would form a trapezoid.Wait, that might work. Let me try to think about it.Suppose I choose a direction, say horizontal. Then, for each line, I can draw a horizontal line through it, and take the intersection point as the vertex. Since the original lines are not parallel, the horizontal lines will intersect them at different points, and the quadrilateral formed by these four points will have two sides parallel (the horizontal ones). So, that would form a trapezoid.But wait, the original lines are not necessarily aligned in any particular direction, so choosing a horizontal direction might not work. But since I can choose any direction, I can pick a direction that is not parallel to any of the original lines.So, in general, for any four lines in a plane, I can choose a direction, draw lines parallel to that direction through each of the four lines, and the intersection points will form a trapezoid.Therefore, the answer to part a) is yes, it's possible to choose points on each line to form a trapezoid.Now, part b) asks if it's possible to form a parallelogram. A parallelogram requires both pairs of opposite sides to be parallel. So, not only do I need one pair of sides to be parallel, but the other pair as well.Given that the original lines are not parallel, and we're in a plane, can we choose points such that both pairs of opposite sides are parallel?This seems trickier. For a parallelogram, not only do we need two sides to be parallel, but the other two sides must also be parallel, and the opposite sides must be equal in length.Given that the original lines are not parallel, it's not immediately clear if we can always find such points.Let me think about it. If I can choose points such that both pairs of opposite sides are parallel, then it's a parallelogram. But how?Maybe I can use the same idea as before, but now choosing two directions. First, choose a direction for one pair of sides, and then another direction for the other pair.But I need to ensure that both pairs are parallel. This might not always be possible because the original lines might not allow for such a configuration.Alternatively, maybe I can use the concept of midpoints. If I can find midpoints such that the midline is parallel, but I'm not sure.Wait, another idea: if I can find two pairs of points such that the vectors between them are equal, then it's a parallelogram. So, if I can find points A, B, C, D such that vector AB equals vector DC and vector AD equals vector BC, then it's a parallelogram.But how to ensure that such points exist on the four lines.This seems complicated. Maybe it's not always possible. In fact, I think it's not always possible to form a parallelogram from four arbitrary pairwise intersecting lines.Wait, but in the plane, given four lines, can we always find a parallelogram? I'm not sure. Maybe not.Let me think of a specific example. Suppose I have four lines: two pairs of perpendicular lines, like the x and y axes and two other lines at 45 degrees. Can I form a parallelogram?Yes, I think so. For example, pick points on the x-axis, y-axis, and the two 45-degree lines such that the sides are parallel.But what if the four lines are arranged differently? Suppose three lines are concurrent, and the fourth line intersects them at different points. Can I still form a parallelogram?Hmm, maybe not. If three lines meet at a point, and the fourth line intersects them elsewhere, it might be difficult to find points that form a parallelogram.Wait, but in the problem statement, it's four pairwise intersecting lines, not necessarily all intersecting at the same point. So, they form a complete quadrilateral with six intersection points.In that case, maybe it's possible to form a parallelogram by choosing appropriate points.Wait, I recall that in a complete quadrilateral, the midpoints of the three diagonals are collinear (Newton-Gauss line). Maybe this can help in forming a parallelogram.If I can find midpoints that are aligned, maybe I can use them to form a parallelogram. But I'm not sure.Alternatively, maybe I can use the concept of harmonic division or something like that to find points that form a parallelogram.But I'm not sure. Maybe it's not always possible. In fact, I think it's not always possible to form a parallelogram from four arbitrary pairwise intersecting lines.Wait, but in the plane, given four lines, you can always find a parallelogram by choosing appropriate points. Is that true? I'm not sure.Let me think of a specific case. Suppose I have four lines: L1, L2, L3, L4, all intersecting each other, but not all concurrent, and no two parallel.Can I always find points A, B, C, D on L1, L2, L3, L4 respectively such that ABCD is a parallelogram?I think yes, but I'm not entirely sure. Maybe I can use the concept of translation. If I can translate one line to coincide with another, but I'm not sure.Wait, another idea: if I can find two pairs of points such that the midpoints of the diagonals coincide, then it's a parallelogram. So, if I can choose points such that the midpoint of AC is the same as the midpoint of BD, then ABCD is a parallelogram.So, maybe I can set up equations to find such points.Let me assign coordinates again. Let me set L1 as the x-axis, L2 as some line through (a, 0) with slope m, L3 as some line through (b, 0) with slope n, and L4 as some line through (c, 0) with slope p.Let me pick points A(x1, 0) on L1, B(x2, m(x2 - a)) on L2, C(x3, n(x3 - b)) on L3, and D(x4, p(x4 - c)) on L4.For ABCD to be a parallelogram, the midpoint of AC must equal the midpoint of BD. So,Midpoint of AC: ((x1 + x3)/2, (0 + n(x3 - b))/2)Midpoint of BD: ((x2 + x4)/2, (m(x2 - a) + p(x4 - c))/2)Setting these equal:(x1 + x3)/2 = (x2 + x4)/2and(n(x3 - b))/2 = (m(x2 - a) + p(x4 - c))/2So, simplifying:x1 + x3 = x2 + x4andn(x3 - b) = m(x2 - a) + p(x4 - c)Now, I have two equations with four variables: x1, x2, x3, x4. So, in principle, I can choose two variables freely and solve for the other two.For example, I can choose x1 and x2, then solve for x3 and x4.From the first equation:x3 = x2 + x4 - x1But that still leaves me with one equation. Wait, no, the first equation is x1 + x3 = x2 + x4, so x3 = x2 + x4 - x1.Then, substitute into the second equation:n((x2 + x4 - x1) - b) = m(x2 - a) + p(x4 - c)Simplify:n(x2 + x4 - x1 - b) = m(x2 - a) + p(x4 - c)This is a linear equation in x2 and x4. So, I can choose x2 and x4 such that this equation holds, given x1.But I need to ensure that x3 and x4 are such that points C and D lie on L3 and L4 respectively.Wait, but x3 is determined by x2 and x4, and x4 is a variable I can choose. So, as long as I can find x2 and x4 such that the above equation holds, and x3 is such that C lies on L3, then it's possible.But I'm not sure if this is always possible. It depends on the slopes and the positions of the lines.Alternatively, maybe I can choose x1 and x3, then solve for x2 and x4.From the first equation:x2 + x4 = x1 + x3From the second equation:n(x3 - b) = m(x2 - a) + p(x4 - c)Substitute x4 = x1 + x3 - x2 into the second equation:n(x3 - b) = m(x2 - a) + p(x1 + x3 - x2 - c)Simplify:n(x3 - b) = m(x2 - a) + p(x1 + x3 - x2 - c)This is a linear equation in x2. So, I can solve for x2:n(x3 - b) - p(x1 + x3 - c) = m(x2 - a) - p(x2)Bring all terms to one side:n(x3 - b) - p(x1 + x3 - c) - m(x2 - a) + p(x2) = 0Factor x2:(n - p)x3 - n b - p x1 + p c - m x2 + m a + p x2 = 0Combine like terms:(n - p)x3 - n b - p x1 + p c + ( - m + p )x2 + m a = 0This is a linear equation in x2 and x3. So, for given x1, I can choose x3 and solve for x2, or vice versa.Therefore, as long as the coefficients are not all zero, which they aren't because the lines are not parallel, I can find solutions for x2 and x3 given x1.Thus, it seems that it's possible to find such points, meaning that a parallelogram can be formed.Wait, but earlier I thought it might not be possible. Maybe I was wrong.Alternatively, maybe it's always possible because we have enough degrees of freedom in choosing the points.So, in conclusion, for part a), it's possible to form a trapezoid, and for part b), it's also possible to form a parallelogram.But wait, I'm not entirely sure. Maybe there are cases where it's not possible. For example, if the four lines are arranged in such a way that the equations don't have a solution.But given that the lines are in general position (no two parallel, no three concurrent), and we have four variables (x1, x2, x3, x4) with two equations, it seems that there are infinitely many solutions, so it should be possible.Therefore, my final answer is:a) Yes, it's possible to form a trapezoid.b) Yes, it's possible to form a parallelogram.But wait, earlier I thought it might not be possible for the parallelogram, but after setting up the equations, it seems possible. Maybe I was overcomplicating it.So, I think both a) and b) are possible.

💡First, I need to solve the congruence (50n equiv 47 pmod{53}). To simplify this, I notice that (50) is congruent to (-3) modulo (53) because (50 + 3 = 53), which is a multiple of (53). This allows me to rewrite the original congruence as (-3n equiv 47 pmod{53}).Next, I want to isolate (n), so I multiply both sides of the congruence by the multiplicative inverse of (-3) modulo (53). The multiplicative inverse of (-3) is a number (x) such that (-3x equiv 1 pmod{53}). Through trial or using the Extended Euclidean Algorithm, I find that (x = 18) because (-3 times 18 = -54 equiv 1 pmod{53}).Multiplying both sides of (-3n equiv 47 pmod{53}) by (18) gives:[n equiv 47 times 18 pmod{53}]Calculating (47 times 18) results in (846). To find (846 mod 53), I divide (846) by (53) and find the remainder, which is (2). Therefore, the smallest non-negative integer (n) that satisfies the original congruence is (2).