💡Okay, so I have this problem: Given a triangle with all sides less than one, I need to prove that there exists an isosceles triangle containing it, where all sides are also less than one. Hmm, interesting. Let me try to break this down step by step.First, let me recall what an isosceles triangle is. It's a triangle with at least two sides equal in length. So, I need to find such a triangle that can contain the given triangle, and all its sides must still be less than one. The given triangle has all sides less than one, so it's a relatively small triangle.Maybe I can start by visualizing the problem. Imagine I have a triangle ABC with sides AB, BC, and CA, each less than one unit. I need to fit this triangle inside another triangle, say XYZ, which is isosceles, and all sides of XYZ are also less than one.I wonder if I can construct such an isosceles triangle by extending some sides or maybe by reflecting the given triangle. Reflection might help because reflecting a triangle over one of its sides can create symmetry, which is a characteristic of isosceles triangles.Let me think about reflection. If I reflect triangle ABC over one of its sides, say AB, I would get a new triangle ABD, where D is the reflection of C over AB. This new triangle ABD would be congruent to ABC, and together they form a kite shape. But is this kite shape an isosceles triangle? No, it's a quadrilateral. Hmm, maybe that's not the right approach.Alternatively, what if I construct an isosceles triangle by extending one of the sides of ABC? For example, if I extend side BC beyond point C, and then construct a point D such that BD equals BC, making triangle ABD isosceles with AB = AD. But I need to ensure that all sides of the new triangle are less than one. If AB is already less than one, then AD would also be less than one, but what about BD? If BC is less than one, and I extend it to make BD equal to BC, then BD would be less than one as well. Wait, no, because BD would actually be twice BC, right? If BC is less than one, then twice BC could be more than one. That might not work.Hmm, maybe extending isn't the way to go. Perhaps I need to scale the triangle somehow. If I can scale the given triangle up slightly to make it isosceles, but still keep all sides under one. But scaling up might make some sides exceed one, which isn't allowed.Wait, maybe I can use rotation. If I rotate the triangle around one of its vertices, I might be able to create an isosceles triangle. For example, if I rotate triangle ABC around point A by some angle, the new position of point C could form an isosceles triangle with AB and the rotated point. But I need to ensure that all sides remain less than one. I'm not sure if this guarantees that.Another idea: maybe the given triangle is already close to being isosceles. If two of its sides are almost equal, then perhaps I can adjust it slightly to make it isosceles without increasing any side beyond one. But this seems too vague.Let me think about the properties of triangles. In any triangle, the sum of any two sides must be greater than the third side. Since all sides are less than one, the triangle is quite small. Maybe I can use this property to construct the isosceles triangle.Suppose I take the longest side of the given triangle and make it the base of the isosceles triangle. Then, I can adjust the other two sides to be equal. But I need to ensure that the new sides are still less than one. If the original sides are less than one, maybe this is possible.Wait, let's consider specific lengths. Suppose the given triangle has sides of lengths a, b, and c, all less than one. Without loss of generality, assume a ≤ b ≤ c < 1. Then, to make an isosceles triangle containing this triangle, I can set two sides equal to c, and adjust the third side accordingly. But I need to make sure that the third side is also less than one.Alternatively, maybe I can construct an isosceles triangle by using the given triangle as a part of it. For example, if I take one side of the given triangle as the base of the isosceles triangle and then construct the other two sides to be equal. But I need to ensure that these new sides don't exceed one in length.I'm getting a bit stuck. Maybe I should look for a theorem or a lemma that relates to this problem. I recall that any triangle can be contained within an isosceles triangle, but I'm not sure about the specific conditions here.Wait, perhaps I can use the concept of affine transformations. If I can apply an affine transformation to the given triangle to make it isosceles while keeping all sides under one, that might work. But affine transformations can distort lengths, so I'm not sure.Another approach: maybe I can use coordinates. Let me place the given triangle in a coordinate system and try to construct the isosceles triangle around it. Suppose I place vertex A at (0,0), vertex B at (x,0), and vertex C at (p,q). Then, I can try to find points D and E such that triangle ADE is isosceles and contains triangle ABC, with all sides less than one.But this seems complicated. Maybe there's a simpler way. Let me think about the maximum distance between any two points in the given triangle. Since all sides are less than one, the maximum distance is less than one. So, if I can construct an isosceles triangle where the base is this maximum distance, and the other two sides are equal and also less than one, that might work.Wait, but the base would be less than one, and the other two sides would need to be equal and also less than one. So, as long as the height of the isosceles triangle is chosen appropriately, this should be possible.Let me try to formalize this. Let the given triangle have vertices A, B, and C, with AB < 1, BC < 1, and CA < 1. Let’s denote the maximum side length as c, so c < 1. Without loss of generality, assume AB = c. Then, I can construct an isosceles triangle with base AB and equal sides AD and BD, where D is a point such that AD = BD. I need to choose D such that AD = BD < 1.But how do I ensure that the entire triangle ABC is contained within the isosceles triangle ABD? I need to position D such that point C lies inside triangle ABD. Maybe by choosing D appropriately above the base AB, I can encompass point C.To find such a D, I can consider the circumcircle of triangle ABC. The circumradius R of triangle ABC can be calculated using the formula R = (a*b*c)/(4*Δ), where Δ is the area of the triangle. Since all sides are less than one, R will be less than some value. But I'm not sure if this helps directly.Alternatively, perhaps I can use the concept of the smallest enclosing circle. The smallest circle that contains triangle ABC will have a diameter at most equal to the longest side, which is less than one. So, the radius is less than 0.5. Then, constructing an isosceles triangle around this circle might work.Wait, if the smallest enclosing circle has a radius less than 0.5, then the diameter is less than one. So, if I construct an isosceles triangle with a base equal to the diameter of this circle and the other two sides equal, then all sides would be less than one. But I need to ensure that the original triangle is inside this isosceles triangle.Hmm, maybe I can place the base of the isosceles triangle along the diameter of the smallest enclosing circle and then have the apex of the isosceles triangle above this diameter. Since the original triangle is inside the circle, it should also be inside the isosceles triangle.But I'm not entirely sure if this construction guarantees that all sides of the isosceles triangle are less than one. The base is less than one, and the other two sides would be the distance from the apex to the ends of the diameter. If the apex is chosen such that these distances are also less than one, then it works.Let me try to calculate this. Suppose the diameter of the smallest enclosing circle is d < 1. Let’s place the base of the isosceles triangle along this diameter, from point (-d/2, 0) to (d/2, 0). Now, the apex of the isosceles triangle will be at some point (0, h). The length of the equal sides will be sqrt((d/2)^2 + h^2). I need this to be less than one.So, sqrt((d/2)^2 + h^2) < 1. Since d < 1, (d/2)^2 < 1/4. Therefore, h^2 < 1 - (d/2)^2. Since d < 1, 1 - (d/2)^2 > 3/4. So, h < sqrt(3/4) = sqrt(3)/2 ≈ 0.866.But the height h also needs to be such that the original triangle is inside the isosceles triangle. The height of the original triangle can be calculated, and we need to ensure that h is greater than this height.Wait, maybe I'm overcomplicating this. Perhaps there's a simpler way to construct the isosceles triangle. Let me think about the original triangle and its properties.If I take the original triangle and construct an isosceles triangle by extending one of its sides and making the other two sides equal, I might be able to contain the original triangle within it. For example, if I extend side BC beyond C and construct a point D such that BD = BC, then triangle ABD would be isosceles with AB = AD. But I need to ensure that all sides are less than one.Wait, if BC is less than one, then BD would be BC + CD. If I set CD = BC, then BD = 2*BC. But if BC is less than one, 2*BC could be greater than one, which violates the condition. So that won't work.Hmm, maybe instead of extending, I can reflect the triangle over one of its sides. If I reflect triangle ABC over side AB, I get a new triangle ABD, which is congruent to ABC. The union of ABC and ABD forms a kite shape, which is a quadrilateral. But I need a triangle, so maybe I can use part of this kite to form an isosceles triangle.Alternatively, perhaps I can construct an isosceles triangle by using the given triangle as a base. If I take the base of the given triangle and construct two equal sides above it, making sure those sides are less than one.Wait, but the given triangle might not be positioned in a way that allows this easily. Maybe I need to use a different approach.Let me think about the concept of enclosing triangles. There's a theorem that says any triangle can be enclosed in an isosceles triangle, but I don't remember the exact conditions. Maybe I can use that theorem here.Alternatively, perhaps I can use the fact that any triangle can be transformed into an isosceles triangle by moving one vertex along a certain path. But I need to ensure that all sides remain less than one.Wait, maybe I can use the concept of the medial triangle. The medial triangle is formed by connecting the midpoints of the sides of the original triangle. It's similar to the original triangle and scaled down by a factor of 1/2. But I'm not sure if this helps directly.Alternatively, perhaps I can use the concept of the circumcircle. The circumcircle of the given triangle has a radius R, and I can construct an isosceles triangle inscribed in a circle of radius slightly larger than R, ensuring all sides are less than one.But I'm not sure about the exact construction.Wait, maybe I can use the following approach: Given triangle ABC, construct an isosceles triangle ABD such that AB is the base, and AD = BD. Then, adjust the position of D such that triangle ABD contains triangle ABC and all sides are less than one.To do this, I need to find a point D such that AD = BD < 1 and point C lies inside triangle ABD.Since AB < 1, I can choose D such that AD = BD = AB. Wait, but then triangle ABD would be equilateral, which is a special case of isosceles. But if AB is less than one, then AD and BD would also be less than one, satisfying the condition.But does triangle ABD contain triangle ABC? Not necessarily, because point C might lie outside of ABD. So, I need to adjust D such that C is inside ABD.Alternatively, maybe I can construct D such that it's above triangle ABC, forming a larger isosceles triangle that contains ABC.Wait, perhaps I can use the concept of the Fermat-Toricelli point. This point minimizes the total distance to the vertices of the triangle, but I'm not sure if that helps here.Alternatively, maybe I can use the concept of the centroid. If I construct an isosceles triangle with the centroid as the apex, but I'm not sure.Wait, maybe I can use the following method: Given triangle ABC, construct an isosceles triangle by extending the altitude from the longest side. For example, if AB is the longest side, construct the altitude from C to AB, and then extend it to a point D such that CD = CE, where E is the foot of the altitude. Then, triangle ABD would be isosceles with AD = BD.But I need to ensure that all sides are less than one. If AB is less than one, and AD = BD, then AD and BD would be equal to the length from A to D and B to D, which might be more than one if D is too far.Hmm, this is tricky. Maybe I need to think differently.Let me consider the maximum distance between any two points in the given triangle. Since all sides are less than one, the maximum distance is less than one. So, if I construct an isosceles triangle with this maximum distance as the base, and the other two sides equal and less than one, that should work.But how do I ensure that the original triangle is inside this isosceles triangle?Wait, maybe I can use the concept of the smallest enclosing isosceles triangle. There might be a way to compute the smallest isosceles triangle that can contain a given triangle, and then show that its sides are all less than one.But I'm not sure about the exact method to compute this.Alternatively, maybe I can use the fact that any triangle can be enclosed in an isosceles triangle by rotating it appropriately. For example, if I rotate the given triangle around one of its vertices, I can create an isosceles triangle that contains it.But I need to ensure that all sides of the resulting isosceles triangle are less than one.Wait, maybe I can use the following approach: Given triangle ABC, construct an isosceles triangle by reflecting one of its vertices over the perpendicular bisector of the opposite side. This might create an isosceles triangle that contains the original triangle.Let me try to visualize this. Suppose I reflect point C over the perpendicular bisector of AB. The reflection point, say D, would form an isosceles triangle ABD with AB as the base and AD = BD. Since AB < 1, and the reflection doesn't change the distance from A and B, AD and BD would also be less than one. Now, does triangle ABC lie inside triangle ABD? Well, point C is reflected to D, so the original triangle ABC would be on one side of the perpendicular bisector, and triangle ABD would be on the other side. So, maybe ABC is not inside ABD.Hmm, that might not work. Maybe I need to reflect over a different line.Alternatively, perhaps I can construct an isosceles triangle by extending the median of the given triangle. For example, if I extend the median from C to the midpoint of AB, and then construct a point D such that CD = CE, where E is the midpoint. Then, triangle ABD would be isosceles with AD = BD.But again, I need to ensure that all sides are less than one and that the original triangle is inside.Wait, maybe I can use the concept of the circumradius. The circumradius R of the given triangle is given by R = (a*b*c)/(4*Δ), where Δ is the area. Since all sides are less than one, R will be less than some value. If I can construct an isosceles triangle with circumradius slightly larger than R, then all sides would be less than one.But I'm not sure about the exact relationship.Alternatively, maybe I can use the fact that the maximum distance between any two points in the given triangle is less than one, so constructing an isosceles triangle with sides equal to this maximum distance would work.Wait, but the sides of the isosceles triangle need to be less than one, and the base can be this maximum distance, which is less than one. Then, the other two sides would be equal and also less than one.But how do I ensure that the original triangle is inside this isosceles triangle?Maybe by positioning the apex of the isosceles triangle such that it encompasses the original triangle.Wait, perhaps I can use the following method: Given triangle ABC, construct an isosceles triangle ABD where AB is the base, and D is chosen such that AD = BD and point C lies inside triangle ABD. Since AB < 1, and AD = BD, I need to choose D such that AD and BD are less than one.To ensure that C lies inside ABD, D must be positioned such that the height of ABD is greater than the height of ABC from C to AB.Let me formalize this. Let’s denote the length of AB as c < 1. The height h_c from C to AB can be calculated as h_c = (2Δ)/c, where Δ is the area of triangle ABC. Since all sides are less than one, Δ is also bounded.Now, to construct triangle ABD, I need to choose D such that AD = BD and the height h_d from D to AB is greater than h_c. The length AD can be calculated using the Pythagorean theorem: AD = sqrt((c/2)^2 + h_d^2). Since AD < 1, we have sqrt((c/2)^2 + h_d^2) < 1. Therefore, h_d < sqrt(1 - (c/2)^2).But h_d needs to be greater than h_c. So, we need sqrt(1 - (c/2)^2) > h_c.Is this always possible? Let's see. Since c < 1, (c/2)^2 < 1/4, so 1 - (c/2)^2 > 3/4. Therefore, sqrt(1 - (c/2)^2) > sqrt(3)/2 ≈ 0.866.Now, h_c = (2Δ)/c. The area Δ can be expressed using Heron's formula: Δ = sqrt(s(s - a)(s - b)(s - c)), where s = (a + b + c)/2. Since a, b, c < 1, s < 3/2. Therefore, Δ < sqrt(3/2*(3/2 - 1)^3) = sqrt(3/2*(1/2)^3) = sqrt(3/2*1/8) = sqrt(3/16) = sqrt(3)/4 ≈ 0.433.Thus, h_c = (2Δ)/c < (2*(sqrt(3)/4))/1 = sqrt(3)/2 ≈ 0.866.Therefore, h_d needs to be greater than h_c, but h_d < sqrt(1 - (c/2)^2) > sqrt(3)/2 ≈ 0.866.Wait, but h_c < sqrt(3)/2 and h_d < sqrt(1 - (c/2)^2) > sqrt(3)/2. So, as long as h_d is chosen between h_c and sqrt(1 - (c/2)^2), it's possible.Therefore, such a point D exists, making triangle ABD isosceles with AD = BD < 1 and containing triangle ABC.Hence, we've constructed an isosceles triangle containing the given triangle with all sides less than one.I think this approach works. By constructing an isosceles triangle with the base as the longest side of the given triangle and adjusting the height appropriately, we can ensure that the original triangle is contained within it and all sides of the isosceles triangle are less than one.

💡Alright, so I have this sequence problem here, and I need to figure out two parts. Let me try to understand what's going on step by step.First, the sequence is defined as {a_n} with a_1 = 1. Then, for each subsequent term, a_{n+1} is defined differently depending on whether n is odd or even. If n is odd, then a_{n+1} = a_n + 1, and if n is even, then a_{n+1} = a_n + 2. Okay, so the rule alternates between adding 1 and adding 2 based on the position of n.Part (1) asks me to define b_n = a_{2n}, which means b_n is the sequence of even-indexed terms of a_n. I need to find b_1 and b_2, and then come up with a general formula for b_n.Let me start by computing the first few terms of a_n to get a sense of the pattern.Given a_1 = 1.Now, to find a_2: since n=1 is odd, a_2 = a_1 + 1 = 1 + 1 = 2.Next, a_3: n=2 is even, so a_3 = a_2 + 2 = 2 + 2 = 4.Then, a_4: n=3 is odd, so a_4 = a_3 + 1 = 4 + 1 = 5.a_5: n=4 is even, so a_5 = a_4 + 2 = 5 + 2 = 7.a_6: n=5 is odd, so a_6 = a_5 + 1 = 7 + 1 = 8.a_7: n=6 is even, so a_7 = a_6 + 2 = 8 + 2 = 10.a_8: n=7 is odd, so a_8 = a_7 + 1 = 10 + 1 = 11.a_9: n=8 is even, so a_9 = a_8 + 2 = 11 + 2 = 13.a_{10}: n=9 is odd, so a_{10} = a_9 + 1 = 13 + 1 = 14.Okay, so writing these out:n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10a_n: 1, 2, 4, 5, 7, 8, 10, 11, 13, 14Now, b_n = a_{2n}, so let's see what that is.b_1 = a_{2*1} = a_2 = 2b_2 = a_{2*2} = a_4 = 5b_3 = a_{2*3} = a_6 = 8b_4 = a_{2*4} = a_8 = 11b_5 = a_{2*5} = a_{10} = 14Looking at these b_n terms: 2, 5, 8, 11, 14,...Hmm, this seems like an arithmetic sequence where each term increases by 3. Let me check:5 - 2 = 38 - 5 = 311 - 8 = 314 - 11 = 3Yes, so the common difference is 3. Therefore, the general formula for b_n should be linear, something like b_n = 3n + c, where c is a constant.Let's test this with n=1: b_1 = 2 = 3*1 + c => c = -1So, the formula would be b_n = 3n - 1.Let me verify this with the known terms:For n=1: 3*1 -1 = 2 ✔️n=2: 3*2 -1 =5 ✔️n=3: 3*3 -1=8 ✔️n=4: 3*4 -1=11 ✔️n=5: 3*5 -1=14 ✔️Looks good. So, the general formula for b_n is 3n -1.Alright, that was part (1). Now, part (2) asks for the sum of the first 20 terms of the sequence {a_n}.So, I need to compute S = a_1 + a_2 + a_3 + ... + a_{20}.Given that I have the first 10 terms already:a_1=1, a_2=2, a_3=4, a_4=5, a_5=7, a_6=8, a_7=10, a_8=11, a_9=13, a_{10}=14But I need up to a_{20}. Maybe I can find a pattern or a formula for a_n so I don't have to compute each term individually.Looking back at the sequence:a_1=1a_2=2a_3=4a_4=5a_5=7a_6=8a_7=10a_8=11a_9=13a_{10}=14a_{11}=16a_{12}=17a_{13}=19a_{14}=20a_{15}=22a_{16}=23a_{17}=25a_{18}=26a_{19}=28a_{20}=29Wait, let me compute up to a_{20} step by step:From a_{10}=14,a_{11}: n=10 is even, so a_{11}=a_{10}+2=14+2=16a_{12}: n=11 is odd, so a_{12}=a_{11}+1=16+1=17a_{13}: n=12 is even, so a_{13}=a_{12}+2=17+2=19a_{14}: n=13 is odd, so a_{14}=a_{13}+1=19+1=20a_{15}: n=14 is even, so a_{15}=a_{14}+2=20+2=22a_{16}: n=15 is odd, so a_{16}=a_{15}+1=22+1=23a_{17}: n=16 is even, so a_{17}=a_{16}+2=23+2=25a_{18}: n=17 is odd, so a_{18}=a_{17}+1=25+1=26a_{19}: n=18 is even, so a_{19}=a_{18}+2=26+2=28a_{20}: n=19 is odd, so a_{20}=a_{19}+1=28+1=29So, the first 20 terms are:1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29Now, to find the sum S = sum_{n=1}^{20} a_n.One approach is to add them all up directly, but that might be time-consuming. Alternatively, maybe I can find a pattern or use the formula for b_n to help.Looking at the sequence, it alternates between adding 1 and adding 2. So, for odd n, a_{n+1} = a_n +1, and for even n, a_{n+1}=a_n +2.This alternation suggests that the sequence increases by 1 and 2 alternately.Looking at the even and odd terms separately:Odd-indexed terms: a_1, a_3, a_5, ..., a_{19}Even-indexed terms: a_2, a_4, a_6, ..., a_{20}From part (1), we have b_n = a_{2n} = 3n -1. So, the even terms follow the formula 3n -1.Similarly, maybe the odd terms also follow a linear pattern.Looking at the odd terms:a_1=1a_3=4a_5=7a_7=10a_9=13a_{11}=16a_{13}=19a_{15}=22a_{17}=25a_{19}=28So, let's see the pattern:1, 4, 7, 10, 13, 16, 19, 22, 25, 28This is also an arithmetic sequence with a common difference of 3.Let me check:4 -1=37 -4=310 -7=3Yes, so the odd terms also form an arithmetic sequence with first term 1 and common difference 3.So, the general formula for a_{2n-1} would be a_{2n-1} = 1 + (n-1)*3 = 3n -2.Let me verify:For n=1: 3*1 -2=1 ✔️n=2: 3*2 -2=4 ✔️n=3: 3*3 -2=7 ✔️n=4: 3*4 -2=10 ✔️n=5: 3*5 -2=13 ✔️And so on. Perfect.So, now, to compute the sum of the first 20 terms, I can split the sum into the sum of the first 10 odd-indexed terms and the sum of the first 10 even-indexed terms.Sum = (Sum of a_1, a_3, ..., a_{19}) + (Sum of a_2, a_4, ..., a_{20})We have formulas for both:For the odd terms: a_{2n-1} = 3n -2For the even terms: a_{2n} = 3n -1So, let's compute the sum of the odd terms first.Sum of odd terms: sum_{n=1}^{10} (3n -2)Similarly, sum of even terms: sum_{n=1}^{10} (3n -1)Let me compute these separately.First, sum of odd terms:sum_{n=1}^{10} (3n -2) = 3*sum_{n=1}^{10} n - 2*10We know that sum_{n=1}^{10} n = (10)(10+1)/2 = 55So, 3*55 = 1652*10=20Therefore, sum of odd terms = 165 -20 = 145Now, sum of even terms:sum_{n=1}^{10} (3n -1) = 3*sum_{n=1}^{10} n - 1*10Again, sum_{n=1}^{10} n =55So, 3*55=1651*10=10Therefore, sum of even terms =165 -10=155Now, total sum S = sum of odd terms + sum of even terms =145 +155=300So, the sum of the first 20 terms is 300.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Sum of odd terms:sum_{n=1}^{10} (3n -2) =3*sum(n) -2*10=3*55 -20=165-20=145 ✔️Sum of even terms:sum_{n=1}^{10} (3n -1)=3*sum(n) -1*10=3*55 -10=165-10=155 ✔️Total sum=145+155=300 ✔️Yes, that seems correct.Alternatively, I could have added all the terms one by one, but this method using the arithmetic sequences is much more efficient.So, to recap:- Identified that the even terms form an arithmetic sequence with a_n=3n -1- Identified that the odd terms form another arithmetic sequence with a_n=3n -2- Calculated the sum of each sequence separately and added them together to get the total sum.Everything checks out. I think I'm confident that the sum is 300.

💡Alright, so I've got these two equations to solve. Let's take them one at a time. The first one is a quadratic equation: 5x² + 2x - 1 = 0. Quadratic equations usually have two solutions, and I remember there's a formula for that. It's called the quadratic formula, right? It goes like x equals negative b plus or minus the square root of b squared minus four a c, all over 2a. So, in this equation, a is 5, b is 2, and c is -1. Let me plug those values into the formula.First, I need to calculate the discriminant, which is b squared minus four a c. So, that would be 2 squared minus four times 5 times (-1). 2 squared is 4, and four times five is twenty, and twenty times negative one is negative twenty. So, 4 minus negative twenty is 4 plus twenty, which is twenty-four. Okay, so the discriminant is twenty-four. Since it's positive, I know there are two real solutions.Now, applying the quadratic formula: x equals negative 2 plus or minus the square root of twenty-four over 2 times 5. Let's simplify that. The square root of twenty-four can be simplified because twenty-four is four times six, and the square root of four is two. So, square root of twenty-four is two times the square root of six. So, x equals negative 2 plus or minus two times the square root of six over ten. I can factor out a two from the numerator: two times (negative one plus or minus the square root of six) over ten. Simplifying that, I divide both numerator and denominator by two, which gives me negative one plus or minus the square root of six over five.So, the two solutions are x equals (-1 + sqrt(6))/5 and x equals (-1 - sqrt(6))/5. That seems right. I think I did that correctly.Now, moving on to the second equation: x(x - 3) - 4(3 - x) = 0. Hmm, let's see. First, I can try to simplify this equation. Let's distribute the terms. So, x times (x - 3) is x² - 3x, and then -4 times (3 - x) is -12 + 4x. So, putting it all together, the equation becomes x² - 3x - 12 + 4x = 0.Combining like terms: -3x plus 4x is x, so the equation simplifies to x² + x - 12 = 0. Okay, that's a simpler quadratic equation. Now, I can try to factor this. I need two numbers that multiply to -12 and add up to 1. Let's think: 4 and -3. 4 times -3 is -12, and 4 plus (-3) is 1. Perfect! So, the equation factors to (x + 4)(x - 3) = 0.Setting each factor equal to zero gives the solutions: x + 4 = 0 or x - 3 = 0. So, x equals -4 or x equals 3. That seems straightforward.Wait a minute, let me double-check my factoring. If I expand (x + 4)(x - 3), I get x² - 3x + 4x - 12, which is x² + x - 12. Yes, that matches the simplified equation. So, my solutions are correct: x = -4 and x = 3.Okay, so to summarize, for the first equation, I used the quadratic formula and found two solutions involving square roots, and for the second equation, I simplified and factored it to find two integer solutions. I think I've got both equations solved correctly.

💡Alright, so I have this problem here: Set A is {-1, 1}, set B is defined as {x | mx = 1}, and it's given that the union of A and B is equal to A. I need to find the possible values of m. Hmm, okay, let's break this down step by step.First off, I know that the union of two sets A and B, denoted as A ∪ B, includes all elements that are in A, in B, or in both. So if A ∪ B = A, that means every element in B must already be in A. In other words, B is a subset of A. That makes sense because if you take everything in A and add something from B, and the result is still A, then B can't have any elements that aren't already in A.So, B is a subset of A. Now, what is set B? It's defined as {x | mx = 1}. That means B contains all real numbers x such that when you multiply x by m, you get 1. So, solving for x, we get x = 1/m, provided that m is not zero. If m is zero, then the equation mx = 1 becomes 0 * x = 1, which simplifies to 0 = 1, which is impossible. So, if m is zero, B would be an empty set because there's no solution.Okay, so let's consider two cases: when m is zero and when m is not zero.**Case 1: m = 0**If m is zero, as I just thought, B is the empty set. Now, the empty set is a subset of every set, including A. So, in this case, A ∪ B would just be A, which satisfies the condition A ∪ B = A. Therefore, m = 0 is a valid solution.**Case 2: m ≠ 0**If m is not zero, then B contains exactly one element, which is x = 1/m. For B to be a subset of A, this element x must be in A. Since A is {-1, 1}, 1/m must be either -1 or 1.So, let's set up the equations:1. If 1/m = 1, then m = 1.2. If 1/m = -1, then m = -1.These are the only possibilities because A only has two elements, and B can only contain one element (since m ≠ 0). Therefore, m can be either 1 or -1.**Putting it all together**From Case 1, m can be 0. From Case 2, m can be 1 or -1. Therefore, the possible values of m are 0, 1, and -1.Wait, let me double-check to make sure I haven't missed anything.- If m = 0: B is empty, so A ∪ B = A. That works.- If m = 1: B = {1}, which is a subset of A. So, A ∪ B = A. That works.- If m = -1: B = {-1}, which is also a subset of A. So, A ∪ B = A. That works too.Is there any other value of m that could satisfy the condition? Let's think about it.Suppose m is some other number, say m = 2. Then B would be {1/2}. But 1/2 is not in A, so A ∪ B would be {-1, 1, 1/2}, which is not equal to A. Similarly, if m = -2, B would be {-1/2}, which is also not in A. So, A ∪ B would again not equal A.What if m is a fraction, like m = 1/2? Then B would be {2}, which is not in A. So, A ∪ B would be {-1, 1, 2}, which is not A. Same issue.What about m being a decimal, like m = 0.5? Then B would be {2}, same as above.So, it seems that only m = 0, 1, and -1 satisfy the condition that B is a subset of A, ensuring that A ∪ B = A.Just to be thorough, let's consider m being a very large number or a very small number. For example, m approaching infinity or zero. But in reality, m is just a real number, and unless it's exactly 0, 1, or -1, B will contain an element not in A, which violates the condition.Therefore, I'm confident that the possible values of m are 0, 1, and -1.

💡Okay, so I'm trying to figure out this problem where a word is the antonym of 26, and it means "infrequently" and "not often." Hmm, that's a bit confusing because 26 is a number, and numbers don't usually have antonyms like words do. Antonyms are words that have opposite meanings, like "hot" and "cold." So, how can a number have an antonym?Maybe the question is playing on something else. Perhaps it's not about the number 26 itself but about something related to it. Let me think about the number 26. In the alphabet, 26 corresponds to the letter "Z." So, maybe the antonym is related to the letter "Z." But "Z" doesn't really have an antonym either.Wait, the question says the word means "infrequently" and "not often." So, maybe it's looking for a word that means the opposite of something that happens often. If 26 is representing something that happens often, then the antonym would be something that happens rarely.But how does 26 relate to something happening often? Maybe it's a code or a cipher where 26 stands for a word. For example, in the alphabet, each number corresponds to a letter. So, 26 would be "Z." But "Z" doesn't mean "often" or "infrequently."Alternatively, maybe it's about the word "twenty-six." Does "twenty-six" have an antonym? I don't think so. It's just a number.Perhaps the question is using "26" as a stand-in for something else. Maybe it's a riddle where 26 represents something, and the antonym is another word. For example, in some contexts, 26 could represent "the end" because it's the last letter of the alphabet. So, the antonym might be "the beginning," which is "A."But "the beginning" doesn't mean "infrequently" or "not often." That doesn't seem to fit.Wait, maybe it's about the frequency of letters. The letter "E" is the most frequent in the English language, while "Z" is one of the least frequent. So, if 26 represents "Z," which is infrequent, then the antonym would be "E," which is frequent.But the question is asking for a word that means "infrequently" and "not often," so maybe the word is "rarely." That makes sense because "rarely" means something happens infrequently.So, putting it all together, even though 26 is a number and doesn't have a direct antonym, if we interpret it as representing "Z," which is infrequent, then the antonym would be "rarely," meaning something happens not often.I think that's the best approach. It's a bit of a stretch, but considering the context and the meaning provided, "rarely" seems to fit.

💡Okay, so I need to expand the expression 16 times (2x plus 5). Hmm, expanding an expression usually means I have to distribute the multiplication over the addition inside the parentheses. I remember something about the distributive property from math class. It's like, if I have a number multiplied by a sum, I can multiply that number by each term inside the sum separately and then add the results together.So, in this case, the number outside the parentheses is 16, and the sum inside is 2x plus 5. That means I need to multiply 16 by 2x and then 16 by 5, and then add those two products together. Let me write that down to visualize it better.First, I'll multiply 16 by 2x. Okay, 16 times 2 is 32, and then I still have the x, so that gives me 32x. Got that part. Now, the second part is multiplying 16 by 5. Well, 16 times 5 is... let me think, 10 times 5 is 50, and 6 times 5 is 30, so 50 plus 30 is 80. So, 16 times 5 is 80.Now, I need to add those two results together. So, 32x plus 80. That should be the expanded form of the original expression. Wait, is there any like terms I can combine here? I don't think so because 32x is a term with a variable, and 80 is a constant. They can't be combined further.Let me double-check my calculations to make sure I didn't make any mistakes. 16 times 2x is indeed 32x, and 16 times 5 is 80. Adding them together gives 32x plus 80. Yeah, that seems right. I don't think I missed anything here. It looks like I've successfully expanded the expression.I wonder if there's another way to approach this problem. Maybe by factoring or something? But no, since the question specifically asks to expand, distributing the multiplication is the correct method here. Factoring would be the opposite process, where you take an expanded expression and write it as a product of factors, which isn't what's needed here.Also, just to make sure, I can plug in a value for x and see if both the original expression and the expanded form give the same result. Let's say x is 1. Then the original expression would be 16 times (2*1 plus 5), which is 16 times (2 plus 5) = 16 times 7 = 112. The expanded form would be 32*1 plus 80 = 32 plus 80 = 112. Okay, that matches. Let's try another value, like x = 0. Original expression: 16 times (0 plus 5) = 16 times 5 = 80. Expanded form: 32*0 plus 80 = 0 plus 80 = 80. That also matches. Good, so my expansion seems correct.I think I've got it. Expanding 16(2x + 5) gives me 32x plus 80. It all checks out with the calculations and the test values I used. I feel confident about this solution now.