💡Okay, so I have this problem where I need to construct an isosceles triangle. The given information is a median to one of the legs and the angle between this median and the other leg. Hmm, let me try to visualize this.First, let me recall what an isosceles triangle is. It's a triangle with at least two sides equal in length. Usually, we consider the two equal sides as the legs, and the third side as the base. The median to a leg would be a line segment from the vertex opposite the leg to the midpoint of that leg.Wait, so if I have an isosceles triangle ABC, with AB and AC as the equal legs, and BC as the base. Then, the median from A to BC would be the line segment from A to the midpoint of BC. But the problem says the median to one of the legs. So, maybe it's the median from B to AC or from C to AB?Let me clarify. If the triangle is isosceles with AB = AC, then the median from B would go to the midpoint of AC, and the median from C would go to the midpoint of AB. So, the problem is giving me one of these medians and the angle between this median and the other leg.Let me denote the triangle as ABC, with AB = AC. Let's say the median given is from B to the midpoint of AC, which we'll call D. So, BD is the median. The angle given is between BD and the other leg, which would be AB or BC? Wait, BD is a median to the leg AC, so the other leg is AB. So, the angle between BD and AB is given.Wait, but BD is a median, so it connects B to D, the midpoint of AC. So, the angle between BD and AB is the angle at point B between AB and BD. That makes sense.So, to construct triangle ABC, given the length of median BD and the angle between BD and AB, which is angle ABD.Okay, so I need to construct triangle ABC with AB = AC, given BD (the median from B to AC) and angle ABD.Let me think about how to approach this. Maybe I can use coordinate geometry or some geometric constructions.Let me try coordinate geometry. Let me place point B at the origin (0,0). Let me denote point D as the midpoint of AC. Since AB = AC, point A must lie somewhere such that AC = AB.Wait, if AB = AC, then triangle ABC is isosceles with AB = AC, so point A is equidistant from B and C. Hmm, but if I place B at (0,0), then point C must be somewhere on the x-axis, say at (c,0). Then point A would be somewhere in the plane such that AB = AC.Wait, but if I place B at (0,0) and C at (c,0), then the midpoint D of AC would be at ((a + c)/2, b/2), assuming point A is at (a,b). Then, the median BD connects B(0,0) to D((a + c)/2, b/2). The length of BD is given, and the angle between BD and AB is given.Hmm, this might get complicated, but maybe manageable.Alternatively, maybe I can use geometric constructions. Let me try that.First, I can draw the median BD with the given length. Then, from point B, I can construct an angle equal to the given angle between BD and AB. The intersection of this angle with the plane will give me the direction of AB. Then, since AB = AC, I can find point A such that AC = AB.Wait, but I need to ensure that D is the midpoint of AC. So, once I have point A, I can find point C as the reflection of A over D.Let me outline the steps:1. Draw the median BD with the given length.2. At point B, construct an angle equal to the given angle between BD and AB. This will give me the direction of AB.3. Along this direction, mark a point A such that AB is equal to AC. Wait, but AC is not known yet. Hmm, maybe I need another approach.Alternatively, since D is the midpoint of AC, and BD is given, perhaps I can use triangle properties.In triangle ABD, we know BD, angle ABD, and AB = AC. Wait, but AB is not necessarily equal to BD.Wait, maybe I can use the Law of Sines or Cosines in triangle ABD.In triangle ABD, we have side BD, angle ABD, and side AB. If I can express AB in terms of BD and the given angle, maybe I can find AB.Let me denote:- Let BD = m (given).- Let angle ABD = θ (given).Then, in triangle ABD, using the Law of Cosines:AD² = AB² + BD² - 2 * AB * BD * cos(θ)But AD is half of AC because D is the midpoint. Since AB = AC, AD = (AB)/2.Wait, no. If AB = AC, then AC = AB, so AD = AC / 2 = AB / 2.So, AD = AB / 2.Therefore, substituting into the equation:(AB / 2)² = AB² + m² - 2 * AB * m * cos(θ)Simplify:AB² / 4 = AB² + m² - 2 * AB * m * cos(θ)Multiply both sides by 4:AB² = 4 * AB² + 4 * m² - 8 * AB * m * cos(θ)Bring all terms to one side:0 = 3 * AB² + 4 * m² - 8 * AB * m * cos(θ)This is a quadratic equation in AB:3 * AB² - 8 * m * cos(θ) * AB + 4 * m² = 0Let me denote AB = x for simplicity:3x² - 8m cosθ x + 4m² = 0Solving for x:x = [8m cosθ ± sqrt(64m² cos²θ - 48m²)] / 6Simplify the discriminant:sqrt(64m² cos²θ - 48m²) = m sqrt(64 cos²θ - 48) = m sqrt(16(4 cos²θ - 3)) = 4m sqrt(4 cos²θ - 3)So,x = [8m cosθ ± 4m sqrt(4 cos²θ - 3)] / 6Factor out 4m:x = [4m (2 cosθ ± sqrt(4 cos²θ - 3))] / 6 = [2m (2 cosθ ± sqrt(4 cos²θ - 3))] / 3Therefore,AB = [2m (2 cosθ ± sqrt(4 cos²θ - 3))] / 3Hmm, this gives two possible solutions for AB, depending on the ± sign.But for real solutions, the discriminant must be non-negative:4 cos²θ - 3 ≥ 0 => cos²θ ≥ 3/4 => |cosθ| ≥ sqrt(3)/2 ≈ 0.866Which implies that θ ≤ 30° or θ ≥ 150°, but since θ is an angle between two lines, it's between 0° and 180°, so θ ≤ 30° or θ ≥ 150°. But in a triangle, angles are less than 180°, so θ can be up to 180°, but in our case, since it's the angle between BD and AB, which are both sides connected at B, θ must be less than 180°, but more specifically, in a triangle, it's less than 180°, but also, since BD is a median, it's less than 90°, maybe?Wait, not necessarily. Depending on the triangle, the median can create angles greater than 90°, but in an isosceles triangle, it might be constrained.But regardless, the discriminant condition tells us that for real solutions, θ must satisfy cosθ ≥ sqrt(3)/2 or cosθ ≤ -sqrt(3)/2. Since θ is between 0° and 180°, cosθ ≤ -sqrt(3)/2 would correspond to θ ≥ 150°, which is possible, but in an isosceles triangle, maybe not.But let's proceed.So, AB can be expressed in terms of m and θ as above.Once we have AB, we can find AC since AB = AC.Then, knowing AC, we can find point C as the reflection of A over D.Wait, but how do we construct this geometrically?Alternatively, maybe I can use vector geometry or coordinate geometry.Let me try coordinate geometry again.Let me place point B at (0,0).Let me denote the median BD as a vector. Let me assume BD has length m, and angle θ between BD and AB.Let me set up coordinates such that point B is at (0,0), and the median BD is along the x-axis for simplicity. Wait, but the angle between BD and AB is θ, so if BD is along the x-axis, then AB makes an angle θ with the x-axis.Wait, that might complicate things because then AB is not necessarily along any axis.Alternatively, let me place point B at (0,0), and let me have BD along the positive x-axis, so point D is at (m,0). Then, the angle between BD (along x-axis) and AB is θ. So, point A lies somewhere in the plane such that the line BA makes an angle θ with BD (the x-axis).So, point A can be represented in polar coordinates from B as (r, θ), where r is the length AB.But since D is the midpoint of AC, we can express point C in terms of A and D.If D is the midpoint of AC, then coordinates of D are ((Ax + Cx)/2, (Ay + Cy)/2). Since D is at (m,0), we have:(Ax + Cx)/2 = m => Cx = 2m - Ax(Ay + Cy)/2 = 0 => Cy = -AySo, point C is (2m - Ax, -Ay)Now, since AB = AC, the distance from A to B equals the distance from A to C.Distance AB: sqrt((Ax - 0)^2 + (Ay - 0)^2) = sqrt(Ax² + Ay²)Distance AC: sqrt((Cx - Ax)^2 + (Cy - Ay)^2) = sqrt((2m - 2Ax)^2 + (-2Ay)^2) = sqrt(4(m - Ax)^2 + 4Ay²) = 2 sqrt((m - Ax)^2 + Ay²)Since AB = AC,sqrt(Ax² + Ay²) = 2 sqrt((m - Ax)^2 + Ay²)Square both sides:Ax² + Ay² = 4[(m - Ax)^2 + Ay²]Expand the right side:Ax² + Ay² = 4(m² - 2mAx + Ax² + Ay²)Bring all terms to left:Ax² + Ay² - 4m² + 8mAx - 4Ax² - 4Ay² = 0Combine like terms:-3Ax² + 8mAx - 4Ay² - 4m² = 0Multiply both sides by -1:3Ax² - 8mAx + 4Ay² + 4m² = 0Hmm, this is a quadratic equation in Ax and Ay.But we also know that point A lies at a distance AB from B, which is r, and the angle between BA and BD is θ.Since BD is along the x-axis from B(0,0) to D(m,0), and BA makes an angle θ with BD, point A can be represented as (r cosθ, r sinθ), where r = AB.So, Ax = r cosθ, Ay = r sinθ.Substitute into the equation:3(r cosθ)^2 - 8m(r cosθ) + 4(r sinθ)^2 + 4m² = 0Simplify:3r² cos²θ - 8mr cosθ + 4r² sin²θ + 4m² = 0Factor r²:r²(3 cos²θ + 4 sin²θ) - 8mr cosθ + 4m² = 0Let me compute 3 cos²θ + 4 sin²θ:= 3 cos²θ + 4 sin²θ = 3 cos²θ + 4(1 - cos²θ) = 3 cos²θ + 4 - 4 cos²θ = -cos²θ + 4So, the equation becomes:r²(-cos²θ + 4) - 8mr cosθ + 4m² = 0Let me write it as:(4 - cos²θ) r² - 8m cosθ r + 4m² = 0This is a quadratic in r:(4 - cos²θ) r² - 8m cosθ r + 4m² = 0Let me solve for r:r = [8m cosθ ± sqrt(64m² cos²θ - 16m²(4 - cos²θ))]/[2(4 - cos²θ)]Simplify the discriminant:sqrt(64m² cos²θ - 16m²(4 - cos²θ)) = sqrt(64m² cos²θ - 64m² + 16m² cos²θ) = sqrt(80m² cos²θ - 64m²) = m sqrt(80 cos²θ - 64)Factor out 16:= m sqrt(16(5 cos²θ - 4)) = 4m sqrt(5 cos²θ - 4)So,r = [8m cosθ ± 4m sqrt(5 cos²θ - 4)]/[2(4 - cos²θ)] = [4m cosθ ± 2m sqrt(5 cos²θ - 4)]/(4 - cos²θ)Factor out 2m:= 2m [2 cosθ ± sqrt(5 cos²θ - 4)]/(4 - cos²θ)Hmm, this seems similar to what I got earlier but with different coefficients.Wait, earlier I had AB = [2m (2 cosθ ± sqrt(4 cos²θ - 3))]/3, but now it's different. Maybe I made a mistake in the coordinate setup.Wait, in the coordinate setup, I assumed BD is along the x-axis from B(0,0) to D(m,0). Then, point A is at (r cosθ, r sinθ). Then, point C is at (2m - r cosθ, -r sinθ). Then, AB = AC leads to the equation above.But in the earlier approach, I considered triangle ABD and used the Law of Cosines, leading to a different expression. Maybe I need to reconcile these.Wait, perhaps I made a mistake in the coordinate approach. Let me double-check.In the coordinate approach, I set BD along the x-axis, so D is at (m,0). Point A is at (r cosθ, r sinθ). Then, point C is at (2m - r cosθ, -r sinθ). Then, AB = AC.Calculating AB: sqrt(r² cos²θ + r² sin²θ) = r.Calculating AC: sqrt[(2m - 2r cosθ)^2 + (-2r sinθ)^2] = sqrt[4(m - r cosθ)^2 + 4r² sin²θ] = 2 sqrt[(m - r cosθ)^2 + r² sin²θ]Setting AB = AC:r = 2 sqrt[(m - r cosθ)^2 + r² sin²θ]Square both sides:r² = 4[(m - r cosθ)^2 + r² sin²θ]Expand:r² = 4[m² - 2m r cosθ + r² cos²θ + r² sin²θ]Simplify inside the brackets:= 4[m² - 2m r cosθ + r² (cos²θ + sin²θ)] = 4[m² - 2m r cosθ + r²]So,r² = 4m² - 8m r cosθ + 4r²Bring all terms to left:r² - 4m² + 8m r cosθ - 4r² = 0 => -3r² + 8m r cosθ - 4m² = 0Multiply by -1:3r² - 8m r cosθ + 4m² = 0Ah, this is the same quadratic as before! So, my earlier step was correct, but I must have miscalculated when I tried to solve it.So, quadratic equation:3r² - 8m r cosθ + 4m² = 0Solutions:r = [8m cosθ ± sqrt(64m² cos²θ - 48m²)] / 6= [8m cosθ ± 4m sqrt(4 cos²θ - 3)] / 6= [4m cosθ ± 2m sqrt(4 cos²θ - 3)] / 3= (2m/3)[2 cosθ ± sqrt(4 cos²θ - 3)]So, AB = r = (2m/3)[2 cosθ ± sqrt(4 cos²θ - 3)]This matches my earlier result.So, for real solutions, the discriminant must be non-negative:4 cos²θ - 3 ≥ 0 => cos²θ ≥ 3/4 => |cosθ| ≥ sqrt(3)/2 ≈ 0.866So, θ ≤ 30° or θ ≥ 150°, but in a triangle, θ is between 0° and 180°, so θ ≤ 30° or θ ≥ 150°.But in an isosceles triangle, the angle between the median and the leg can't be too large, so θ ≤ 30° is the feasible solution.So, if θ > 30°, there is no solution.If θ = 30°, then sqrt(4 cos²θ - 3) = sqrt(4*(sqrt(3)/2)^2 - 3) = sqrt(4*(3/4) - 3) = sqrt(3 - 3) = 0, so r = (2m/3)(2*(sqrt(3)/2)) = (2m/3)(sqrt(3)) = (2m sqrt(3))/3.If θ < 30°, then we have two solutions for r, corresponding to the ± in the expression.So, depending on θ, we can have two possible triangles, one triangle, or no triangle.Now, how do we construct this?Given m (length of BD) and θ (angle between BD and AB), we can compute AB using the above formula, then construct the triangle.But since this is a construction problem, perhaps using compass and straightedge, let me think about how to do it geometrically.1. Draw the median BD with length m.2. At point B, construct an angle θ with BD. This will give two possible directions for AB, one on each side of BD.3. From point B, draw two lines making angle θ with BD, one on each side.4. On each of these lines, mark points A1 and A2 such that BA1 = BA2 = r, where r is computed as above.5. For each Ai, construct point Ci as the reflection of Ai over D (since D is the midpoint of AC).6. Connect points Ai and Ci to form triangles ABC1 and ABC2.But wait, in the coordinate approach, once we have point A, point C is determined as (2m - Ax, -Ay). So, if we have point A, we can find point C.But in construction, how do we ensure that AC = AB?Alternatively, since D is the midpoint of AC, and BD is given, perhaps we can use the property that in triangle ABC, the centroid divides the median in the ratio 2:1.Wait, but in this case, BD is a median, and the centroid would be located at 2/3 of BD from B.But I'm not sure if that helps directly.Alternatively, maybe I can use the fact that in triangle ABD, we have sides AB, BD, and AD, with AD = AB/2.Wait, AD is half of AC, and since AB = AC, AD = AB/2.So, in triangle ABD, we have sides AB, BD, and AD = AB/2, with angle ABD = θ.So, using the Law of Cosines in triangle ABD:AD² = AB² + BD² - 2 * AB * BD * cosθBut AD = AB/2, so:(AB/2)² = AB² + BD² - 2 * AB * BD * cosθWhich simplifies to:AB²/4 = AB² + BD² - 2 AB BD cosθMultiply both sides by 4:AB² = 4 AB² + 4 BD² - 8 AB BD cosθBring all terms to left:0 = 3 AB² + 4 BD² - 8 AB BD cosθWhich is the same quadratic as before.So, solving for AB gives us the two possible lengths.Therefore, the construction steps would involve:1. Draw the median BD with length m.2. At point B, construct an angle θ with BD.3. Along this angle, mark two points A1 and A2 such that BA1 and BA2 are the two possible lengths of AB computed from the quadratic equation.4. For each Ai, construct point Ci as the reflection over D, ensuring AC = AB.5. Connect the points to form the triangles.But how do we construct points A1 and A2 without knowing AB in advance?Perhaps we can use the quadratic solution geometrically.Alternatively, maybe we can use the intersecting circles method.Let me think.Given BD of length m, and angle θ at B, we can construct two lines from B making angle θ with BD. Then, the intersection of these lines with a circle centered at D with radius AD = AB/2 will give us point A.Wait, but AD = AB/2, so if we can express AD in terms of AB, which is variable, it's tricky.Alternatively, since AD = AB/2, and BD is given, perhaps we can use similar triangles or some proportion.Wait, in triangle ABD, AD = AB/2, BD = m, angle ABD = θ.So, triangle ABD has sides AB, BD = m, AD = AB/2, and angle θ between AB and BD.This is similar to having a triangle with sides x, m, x/2, and angle θ between x and m.So, using the Law of Cosines:(x/2)² = x² + m² - 2 x m cosθWhich simplifies to:x²/4 = x² + m² - 2 x m cosθMultiply by 4:x² = 4x² + 4m² - 8 x m cosθBring terms to left:0 = 3x² + 4m² - 8 x m cosθWhich is the same quadratic.So, perhaps we can construct this geometrically by creating a triangle where one side is m, another side is x, and the included angle is θ, with the third side being x/2.But how?Alternatively, maybe we can use the concept of similar triangles or some proportion.Wait, if AD = AB/2, then triangle ABD has sides AB, BD, AD with AD = AB/2.So, if I can construct a triangle where one side is m, another side is x, and the third side is x/2, with the angle between x and m being θ.This seems like a specific case, but perhaps we can use a geometric construction.Let me try to outline the steps:1. Draw segment BD with length m.2. At point B, construct an angle θ with BD. Let this be the direction of BA.3. Along this angle, mark a point A such that AD = AB/2.But how do we ensure AD = AB/2?Wait, since D is the midpoint of AC, and AB = AC, then AD = AB/2.So, if we can construct point A such that AD = AB/2, then we can find point C as the reflection of A over D.But how?Alternatively, since AD = AB/2, we can use a circle centered at D with radius AB/2. But AB is variable, so we need another approach.Wait, perhaps we can use the fact that in triangle ABD, AD = AB/2, so triangle ABD is such that AD = (1/2) AB.This is similar to constructing a triangle with a given side, angle, and a ratio of sides.Maybe we can use the concept of Apollonius circles.Wait, Apollonius circle is the locus of points such that the ratio of distances to two fixed points is constant.In this case, we have point A such that AD = (1/2) AB.So, the locus of A is the Apollonius circle with ratio 1/2 with respect to points B and D.So, to construct this circle:1. Draw segment BD with length m.2. Find the Apollonius circle where the ratio of distances from A to B and A to D is 1/2.The Apollonius circle for ratio k = 1/2 can be constructed as follows:- The center lies on the line BD.- Let me denote the center as O.- The ratio of distances is AB/AD = 2.So, the Apollonius circle for ratio 2 has center O on BD such that BO/OD = 2.So, BO = 2 OD.Since BD = m, BO + OD = m, and BO = 2 OD, so 2 OD + OD = 3 OD = m => OD = m/3, BO = 2m/3.So, the center O is located at 2m/3 from B towards D.The radius R of the Apollonius circle is given by:R = (k / (1 + k)) * BD, where k = 2.Wait, no, the formula is R = (k / (1 + k)) * distance between B and D.Wait, actually, the radius can be calculated as follows:In Apollonius circle, the radius R is given by:R = (AB * AD) / (AB + AD)But since AB = 2 AD, let AD = x, then AB = 2x.So, R = (2x * x) / (2x + x) = (2x²)/(3x) = (2x)/3.But x is AD, which is variable.Alternatively, since we know the center O is at 2m/3 from B, and the radius can be found using the distance from O to A.Wait, since A lies on the Apollonius circle, OA = R.But OA can also be expressed in terms of AB and AD.Wait, maybe it's better to compute the radius using coordinates.Let me place B at (0,0), D at (m,0). Then, center O is at (2m/3, 0).The Apollonius circle has center O(2m/3, 0) and radius R.Any point A on this circle satisfies AB/AD = 2.So, for point A(x,y), distance AB = sqrt(x² + y²), distance AD = sqrt((x - m)^2 + y²).So, AB/AD = 2 => sqrt(x² + y²) / sqrt((x - m)^2 + y²) = 2Square both sides:(x² + y²) / ((x - m)^2 + y²) = 4Multiply both sides by denominator:x² + y² = 4[(x - m)^2 + y²]Expand:x² + y² = 4(x² - 2m x + m² + y²)Bring all terms to left:x² + y² - 4x² + 8m x - 4m² - 4y² = 0Combine like terms:-3x² + 8m x - 4m² - 3y² = 0Multiply by -1:3x² - 8m x + 4m² + 3y² = 0Divide by 3:x² - (8m/3) x + (4m²)/3 + y² = 0Complete the square for x:x² - (8m/3)x = (x - 4m/3)^2 - (16m²/9)So,(x - 4m/3)^2 - 16m²/9 + 4m²/3 + y² = 0Simplify constants:-16m²/9 + 12m²/9 = (-16 + 12)m²/9 = (-4m²)/9So,(x - 4m/3)^2 + y² = 4m²/9Thus, the Apollonius circle has center at (4m/3, 0) and radius 2m/3.Wait, but earlier I thought the center was at 2m/3 from B, but according to this, it's at 4m/3 from B, which is beyond point D.Wait, that makes sense because the ratio is AB/AD = 2, so the center is beyond D.So, the Apollonius circle is centered at (4m/3, 0) with radius 2m/3.So, to construct this circle:1. Draw BD with length m.2. Extend BD beyond D to a point O such that BO = 4m/3.3. Draw a circle centered at O with radius 2m/3.Then, the intersection points of this circle with the angle θ from B will give us points A1 and A2.So, the construction steps are:1. Draw segment BD with length m.2. Extend BD beyond D to a point O such that BO = 4m/3.3. Draw a circle centered at O with radius 2m/3.4. At point B, construct an angle θ with BD, creating two rays.5. The intersections of these rays with the Apollonius circle are points A1 and A2.6. For each Ai, construct point Ci as the reflection over D, i.e., Ci = 2D - Ai.7. Connect points Ai and Ci to form triangles ABC1 and ABC2.This should give the required isosceles triangles.But let me verify this.Given that A lies on the Apollonius circle, AB/AD = 2, which implies AD = AB/2.Since D is the midpoint of AC, AC = 2 AD = AB, so AB = AC, making triangle ABC isosceles.Thus, this construction should work.So, summarizing the construction steps:1. Draw the given median BD with length m.2. Extend BD beyond D to a point O such that BO = (4/3)m.3. Draw a circle centered at O with radius (2/3)m.4. At point B, construct an angle θ with BD, creating two rays.5. The intersections of these rays with the circle are points A1 and A2.6. For each Ai, construct point Ci as the reflection of Ai over D.7. Connect Ai to Ci to form the isosceles triangles ABC1 and ABC2.Depending on the value of θ, there may be two solutions, one solution, or no solution.If θ > 30°, no solution exists.If θ = 30°, one solution exists.If θ < 30°, two solutions exist.Thus, the construction is possible under these conditions.

💡First, I need to determine the total number of workers in the workshop. The problem states that the total number is 21.11111111111111, which is not a whole number. Since the number of workers must be a whole number, I will round it to 21.Next, I identify the number of technicians and the rest of the workers. There are 10 technicians, so the remaining workers are 21 minus 10, which equals 11.I calculate the total salary for the technicians by multiplying the number of technicians by their average salary: 10 multiplied by 1200 equals 12,000.Then, I calculate the total salary for the rest of the workers by multiplying the number of these workers by their average salary: 11 multiplied by 820 equals 9,020.I add the total salaries of the technicians and the rest of the workers to find the overall total salary: 12,000 plus 9,020 equals 21,020.Finally, I determine the average salary per head of all the workers by dividing the total salary by the total number of workers: 21,020 divided by 21 equals approximately 1001.9047619047619. Rounding this to the nearest whole number gives an average salary of 1002.

💡First, I need to determine how much Yulia earned from babysitting. I know her total net profit for the week is 44, and she spent 34 operating her lemonade stand. To find her total gross revenue from both the lemonade stand and babysitting, I add the net profit to the operating costs:44 + 34 = 78Next, I subtract the gross revenue from the lemonade stand (47) from the total gross revenue to find out how much she earned from babysitting:78 - 47 = 31Therefore, Yulia made 31 from babysitting.

💡Okay, so I need to find the domain of this function: (x⁴ - 4x³ + 7x² - 4x + 2)/(x³ - 3x² + 2x). Hmm, the domain of a function refers to all the real numbers x for which the function is defined. Since this is a rational function (a polynomial divided by another polynomial), the main thing I need to worry about is division by zero. So, the function will be undefined wherever the denominator is zero. First, let me write down the denominator separately: x³ - 3x² + 2x. I need to find the values of x that make this equal to zero. To do that, I can factor the denominator. Let me try factoring out the greatest common factor first. I see that each term has an x, so I can factor out an x:x(x² - 3x + 2) = 0.Now, I have a quadratic inside the parentheses: x² - 3x + 2. I can try to factor this quadratic. I'm looking for two numbers that multiply to 2 and add up to -3. Let's see, -1 and -2 multiply to 2 and add up to -3. So, the quadratic factors as:x(x - 1)(x - 2) = 0.So, the denominator is zero when x = 0, x = 1, or x = 2. That means the function is undefined at these points. Therefore, the domain of the function will be all real numbers except x = 0, x = 1, and x = 2.To express this in interval notation, I need to break the real number line at these points. The critical points are 0, 1, and 2. So, the intervals will be:(-∞, 0), (0, 1), (1, 2), and (2, ∞).Since the function is undefined at 0, 1, and 2, these points are excluded from the domain. Therefore, the domain is the union of these open intervals.Wait, just to make sure I didn't make any mistakes, let me double-check my factoring of the denominator. I had x³ - 3x² + 2x, factored out an x to get x(x² - 3x + 2). Then factored the quadratic into (x - 1)(x - 2). Multiplying it back out: x*(x - 1)*(x - 2) = x*(x² - 3x + 2) = x³ - 3x² + 2x. Yep, that matches the original denominator. So, my factoring is correct.Also, I should consider if there are any common factors between the numerator and the denominator that could simplify the function, potentially removing some points of discontinuity. Let me look at the numerator: x⁴ - 4x³ + 7x² - 4x + 2. Maybe I can factor this as well or see if it shares any common factors with the denominator.Trying to factor the numerator might be a bit more complicated since it's a quartic polynomial. Let me see if it can be factored by grouping or if it has any rational roots. Using the Rational Root Theorem, possible rational roots are ±1, ±2. Let me test x=1:1⁴ - 4*1³ + 7*1² - 4*1 + 2 = 1 - 4 + 7 - 4 + 2 = 2. Not zero.Testing x=2:2⁴ - 4*2³ + 7*2² - 4*2 + 2 = 16 - 32 + 28 - 8 + 2 = 6. Not zero.Testing x= -1:(-1)⁴ - 4*(-1)³ + 7*(-1)² - 4*(-1) + 2 = 1 + 4 + 7 + 4 + 2 = 18. Not zero.Testing x= -2:(-2)⁴ - 4*(-2)³ + 7*(-2)² - 4*(-2) + 2 = 16 + 32 + 28 + 8 + 2 = 86. Not zero.So, it seems like the numerator doesn't have any rational roots, at least. Maybe it's irreducible over the rationals. If that's the case, then there are no common factors between the numerator and the denominator, so the function doesn't simplify further, and the domain remains all real numbers except x=0, x=1, and x=2.Wait, just to be thorough, maybe I can try polynomial long division or synthetic division to see if the numerator can be expressed in terms of the denominator. Let me try dividing the numerator by the denominator.So, dividing x⁴ - 4x³ + 7x² - 4x + 2 by x³ - 3x² + 2x.First term: x⁴ divided by x³ is x. Multiply the denominator by x: x*(x³ - 3x² + 2x) = x⁴ - 3x³ + 2x².Subtract this from the numerator:(x⁴ - 4x³ + 7x² - 4x + 2) - (x⁴ - 3x³ + 2x²) = (-4x³ + 3x³) + (7x² - 2x²) + (-4x) + 2 = (-x³) + 5x² - 4x + 2.Now, the new dividend is -x³ + 5x² - 4x + 2. The degree is still 3, same as the denominator. So, divide -x³ by x³, which is -1. Multiply the denominator by -1: -1*(x³ - 3x² + 2x) = -x³ + 3x² - 2x.Subtract this from the current dividend:(-x³ + 5x² - 4x + 2) - (-x³ + 3x² - 2x) = (0x³) + (5x² - 3x²) + (-4x + 2x) + 2 = 2x² - 2x + 2.So, the remainder is 2x² - 2x + 2. Since the remainder is of lower degree than the denominator, we stop here. So, the numerator can be expressed as:(x - 1)*(denominator) + (2x² - 2x + 2).Therefore, the function can be written as:(x - 1) + (2x² - 2x + 2)/(x³ - 3x² + 2x).But this doesn't seem to help in simplifying the function further because the remainder doesn't share any common factors with the denominator. So, my initial conclusion stands: the function is undefined at x=0, x=1, and x=2, and defined everywhere else.Just to recap, the steps I took were:1. Identify that the function is rational, so the domain excludes values where the denominator is zero.2. Factored the denominator to find its roots: x=0, x=1, x=2.3. Checked if the numerator shares any common factors with the denominator by attempting to factor it and performing polynomial division, which indicated no common factors.4. Concluded that the domain is all real numbers except x=0, x=1, and x=2.I think that's thorough enough. I don't see any mistakes in my reasoning, so I feel confident about the domain.

💡First, I need to determine the lengths of the sides of triangle ( DEF ) based on the given ratio ( 3:4:5 ). Since the sides are in this ratio and the longest side ( DF ) is ( 12 ) inches, I can set up the proportion to find the scaling factor.Next, I'll apply the Angle Bisector Theorem, which states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. This will help me find the lengths of ( DG ) and ( GF ).Finally, I'll calculate the length of the longer subsegment ( GF ) by solving the proportion and substituting the scaling factor back into the expression.

💡Okay, so I have this problem here: I need to find the sum of (-1)^1 + (-1)^2 + ... + (-1)^{2007}. Hmm, that looks like a sequence where each term alternates between -1 and 1. Let me think about how to approach this.First, I remember that when you have exponents with -1, the result alternates depending on whether the exponent is odd or even. Specifically, (-1) raised to an odd power is -1, and (-1) raised to an even power is 1. So, in this sequence, the terms will alternate between -1 and 1.Let me write out the first few terms to see the pattern:- (-1)^1 = -1- (-1)^2 = 1- (-1)^3 = -1- (-1)^4 = 1- And so on...So, the sequence goes -1, 1, -1, 1, ..., up to the 2007th term. Since 2007 is an odd number, the last term will be -1.Now, I need to find the sum of all these terms. One way to approach this is to pair the terms. Let's see:If I pair the first term with the second term: (-1) + 1 = 0Then the third term with the fourth term: (-1) + 1 = 0And so on...Each pair of terms adds up to zero. So, if I can figure out how many such pairs there are, I can determine the total sum.But wait, the total number of terms is 2007, which is odd. That means there will be an odd number of terms, so one term will be left unpaired. Specifically, since we're pairing two terms at a time, the number of pairs will be 2007 divided by 2, but since 2007 is odd, it won't divide evenly.Let me calculate the number of pairs:2007 divided by 2 is 1003.5. But since we can't have half a pair, we'll have 1003 full pairs and one leftover term.Each pair sums to zero, so the total sum contributed by the pairs is 1003 * 0 = 0.Now, the leftover term is the last term, which is (-1)^{2007}. Since 2007 is odd, this term is -1.Therefore, the total sum is 0 (from the pairs) plus (-1) from the leftover term, which equals -1.Wait, let me double-check that. If I have 2007 terms, and I pair them as (-1)^1 + (-1)^2, (-1)^3 + (-1)^4, ..., up to (-1)^{2005} + (-1)^{2006}, that's 1003 pairs, each summing to zero. Then the last term, (-1)^{2007}, is indeed -1. So, adding all those zeros and the final -1 gives me -1.Is there another way to think about this? Maybe using the formula for the sum of a geometric series. The sequence is a geometric series with first term a = (-1)^1 = -1 and common ratio r = (-1).The formula for the sum of the first n terms of a geometric series is S_n = a * (1 - r^n) / (1 - r), provided that r ≠ 1.Let's apply that here:a = -1r = -1n = 2007So, S_{2007} = (-1) * (1 - (-1)^{2007}) / (1 - (-1))First, calculate (-1)^{2007}. Since 2007 is odd, this is -1.So, S_{2007} = (-1) * (1 - (-1)) / (1 + 1)= (-1) * (1 + 1) / 2= (-1) * 2 / 2= (-1) * 1= -1Okay, that confirms the earlier result. The sum is indeed -1.Just to make sure, let me consider a smaller example. Suppose I have the sum from k=1 to k=3: (-1)^1 + (-1)^2 + (-1)^3.Calculating each term:- (-1)^1 = -1- (-1)^2 = 1- (-1)^3 = -1Adding them up: -1 + 1 + (-1) = -1.Using the pairing method: Pair (-1)^1 + (-1)^2 = 0, then add (-1)^3 = -1. Total sum: -1.Using the geometric series formula:a = -1, r = -1, n = 3S_3 = (-1)*(1 - (-1)^3)/(1 - (-1)) = (-1)*(1 - (-1))/2 = (-1)*(2)/2 = -1.Consistent again.Another example: sum from k=1 to k=4: (-1)^1 + (-1)^2 + (-1)^3 + (-1)^4.Calculating each term:- (-1)^1 = -1- (-1)^2 = 1- (-1)^3 = -1- (-1)^4 = 1Adding them up: -1 + 1 + (-1) + 1 = 0.Using the pairing method: Pair (-1)^1 + (-1)^2 = 0, and (-1)^3 + (-1)^4 = 0. Total sum: 0.Using the geometric series formula:a = -1, r = -1, n = 4S_4 = (-1)*(1 - (-1)^4)/(1 - (-1)) = (-1)*(1 - 1)/2 = (-1)*(0)/2 = 0.Consistent as well.So, in the case of an even number of terms, the sum is 0, and in the case of an odd number of terms, the sum is -1. Since 2007 is odd, the sum should be -1.I think I've thoroughly checked this from multiple angles, so I'm confident that the sum is -1.