💡Alright, so I have this problem about a hyperbola. Let me try to figure it out step by step. First, the problem says that the hyperbola has its foci on the x-axis, which means it's a horizontal hyperbola. The distance between its two vertices is 2, and the distance from a focus to an asymptote is √2. I need to find the standard equation of the hyperbola and then some other properties like the lengths of the real and imaginary axes, the coordinates of the foci, the eccentricity, and the equations of the asymptotes.Okay, starting with part (1). For a hyperbola, the standard equation when it's centered at the origin and opens left and right (since the foci are on the x-axis) is:[frac{x^2}{a^2} - frac{y^2}{b^2} = 1]Here, (a) is the distance from the center to each vertex, and (b) is related to the conjugate axis. The distance between the two vertices is given as 2, so that means the distance from the center to each vertex is 1. Therefore, (a = 1).Now, the foci are located at ((pm c, 0)), where (c) is the distance from the center to each focus. For hyperbolas, the relationship between (a), (b), and (c) is:[c^2 = a^2 + b^2]So, if I can find (b), I can find (c), and that will help me write the standard equation.The problem also mentions that the distance from a focus to an asymptote is √2. Hmm, okay. The asymptotes of a hyperbola are the lines that the hyperbola approaches but never touches. For a horizontal hyperbola, the asymptotes are given by:[y = pm frac{b}{a}x]Since (a = 1), the asymptotes simplify to:[y = pm b x]Now, I need to find the distance from one of the foci to one of these asymptotes. Let's pick the focus at ((c, 0)) and the asymptote (y = b x). The distance from a point ((x_0, y_0)) to the line (Ax + By + C = 0) is given by:[text{Distance} = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}]So, let's rewrite the asymptote equation in standard form. The asymptote is (y = b x), which can be rewritten as:[b x - y = 0]So, (A = b), (B = -1), and (C = 0). The focus is at ((c, 0)), so plugging into the distance formula:[text{Distance} = frac{|b cdot c + (-1) cdot 0 + 0|}{sqrt{b^2 + (-1)^2}} = frac{|b c|}{sqrt{b^2 + 1}}]We are told this distance is √2. So,[frac{|b c|}{sqrt{b^2 + 1}} = sqrt{2}]Since (b) and (c) are positive (they are distances), we can drop the absolute value:[frac{b c}{sqrt{b^2 + 1}} = sqrt{2}]Now, let's square both sides to eliminate the square roots:[left(frac{b c}{sqrt{b^2 + 1}}right)^2 = (sqrt{2})^2][frac{b^2 c^2}{b^2 + 1} = 2]But we also know from the hyperbola relationship that (c^2 = a^2 + b^2). Since (a = 1), this becomes:[c^2 = 1 + b^2]So, (c^2 = 1 + b^2). Let me substitute (c^2) in the previous equation:[frac{b^2 (1 + b^2)}{b^2 + 1} = 2]Simplify the denominator:[frac{b^2 (1 + b^2)}{1 + b^2} = 2]The (1 + b^2) terms cancel out:[b^2 = 2]So, (b = sqrt{2}). Since (b) is positive, we take the positive square root.Now that we have (b = sqrt{2}), we can find (c):[c^2 = 1 + (sqrt{2})^2 = 1 + 2 = 3][c = sqrt{3}]So, now we have all the necessary values to write the standard equation of the hyperbola:[frac{x^2}{1^2} - frac{y^2}{(sqrt{2})^2} = 1][frac{x^2}{1} - frac{y^2}{2} = 1][x^2 - frac{y^2}{2} = 1]So, that's the standard equation for part (1).Moving on to part (2). I need to find several properties:1. Length of the real axis2. Length of the imaginary axis3. Coordinates of the foci4. Eccentricity5. Equations of the asymptotesStarting with the length of the real axis. The real axis is the transverse axis, which is the distance between the two vertices. Since the distance between the vertices is given as 2, that's the length of the real axis. So, the length is 2.Next, the length of the imaginary axis. The imaginary axis is the conjugate axis, which is the distance between the two endpoints of the conjugate axis. The length is (2b). Since (b = sqrt{2}), the length is:[2 times sqrt{2} = 2sqrt{2}]So, the length of the imaginary axis is (2sqrt{2}).Coordinates of the foci. The foci are located at ((pm c, 0)). We found (c = sqrt{3}), so the coordinates are:[(pm sqrt{3}, 0)]Eccentricity. The eccentricity (e) of a hyperbola is given by:[e = frac{c}{a}]We have (c = sqrt{3}) and (a = 1), so:[e = frac{sqrt{3}}{1} = sqrt{3}]Equations of the asymptotes. Earlier, we had the asymptotes as (y = pm b x). Since (b = sqrt{2}), the equations are:[y = pm sqrt{2} x]Let me just double-check all these results to make sure I didn't make any mistakes.1. Standard equation: (x^2 - frac{y^2}{2} = 1). That seems correct because (a = 1) and (b = sqrt{2}).2. Length of the real axis: 2. That's correct because the distance between the vertices is 2.3. Length of the imaginary axis: (2sqrt{2}). Correct, since (2b = 2sqrt{2}).4. Coordinates of the foci: ((pm sqrt{3}, 0)). Correct, since (c = sqrt{3}).5. Eccentricity: (sqrt{3}). Correct, since (e = c/a = sqrt{3}/1 = sqrt{3}).6. Asymptotes: (y = pm sqrt{2}x). Correct, since (b/a = sqrt{2}/1 = sqrt{2}).Everything seems to check out. I think I've got it right.**Final Answer**(1) The standard equation of the hyperbola is boxed{x^2 - frac{y^2}{2} = 1}.(2) The length of the real axis is boxed{2}, the length of the imaginary axis is boxed{2sqrt{2}}, the coordinates of the foci are boxed{(pm sqrt{3}, 0)}, the eccentricity is boxed{sqrt{3}}, and the equations of the asymptotes are boxed{y = pm sqrt{2}x}.

💡Alright, so I've got this math problem here with four options, labeled A through D. Each option presents a different algebraic calculation, and I need to figure out which one is correct. Let me go through each option one by one and see if I can spot any mistakes or confirm if they're right.Starting with **Option A**: ( a^{3} + a^{2} = 2a^{5} ). Hmm, okay. So, this is saying that if I add ( a^{3} ) and ( a^{2} ), I get ( 2a^{5} ). That doesn't seem right to me. When you add terms with exponents, you can only combine like terms, meaning the exponents have to be the same. Here, ( a^{3} ) and ( a^{2} ) have different exponents, so they can't be combined into a single term. Also, adding them wouldn't result in a term with a higher exponent like ( a^{5} ). So, I think Option A is incorrect.Moving on to **Option B**: ( (-ab^{3})^{2} = a^{3}b^{6} ). Let's break this down. When you square a product, you square each factor individually. So, ( (-ab^{3})^{2} ) should be ( (-a)^{2} times (b^{3})^{2} ). Calculating each part, ( (-a)^{2} ) is ( a^{2} ) because the negative sign gets squared away, and ( (b^{3})^{2} ) is ( b^{6} ). So, multiplying those together, we get ( a^{2}b^{6} ). But the option says ( a^{3}b^{6} ), which means there's an extra ( a ) in the exponent. That doesn't seem right. So, Option B is also incorrect.Next is **Option C**: ( 2a(1 - a) = 2a - 2a^{2} ). Alright, let's distribute the ( 2a ) across the parentheses. Multiplying ( 2a ) by 1 gives ( 2a ), and multiplying ( 2a ) by ( -a ) gives ( -2a^{2} ). So, putting it together, we have ( 2a - 2a^{2} ), which matches exactly what's given in the option. That seems correct to me. So, Option C looks like the right one.But just to be thorough, let's check **Option D**: ( (a + b)^{2} = a^{2} + b^{2} ). I remember that squaring a binomial involves using the formula ( (a + b)^{2} = a^{2} + 2ab + b^{2} ). So, the given option is missing the middle term ( 2ab ). That means it's not fully expanded and is therefore incorrect.So, after going through all four options, it seems like only **Option C** is correct. The others either incorrectly combine like terms, mishandle exponents, or miss terms in their expansions.

💡Alright, so I have this problem where I need to construct a triangle given the midline ( AA_1 ) and the two angles that this midline makes with the sides ( AB ) and ( AC ). Hmm, okay, let me try to visualize this.First, let me recall what a midline is. In a triangle, a midline is a segment that connects a vertex to the midpoint of the opposite side. So, in this case, ( A_1 ) is the midpoint of side ( BC ). That means ( BA_1 = A_1C ). Got it.Now, the problem gives me the midline ( AA_1 ) and the two angles that ( AA_1 ) makes with sides ( AB ) and ( AC ). Let's denote these angles as ( angle BAA_1 = alpha ) and ( angle CA A_1 = beta ). So, I have the length of ( AA_1 ), and I know the angles ( alpha ) and ( beta ).My goal is to construct triangle ( ABC ) using this information. Okay, let's think about how to approach this.First, I can start by drawing the midline ( AA_1 ). Let's assume I have a straightedge and a compass for this construction. I'll draw segment ( AA_1 ) with the given length.Next, at point ( A ), I need to construct the angles ( alpha ) and ( beta ) with respect to ( AA_1 ). So, I'll use my protractor to draw these angles. One angle ( alpha ) will be on one side of ( AA_1 ), and the other angle ( beta ) will be on the opposite side.Wait, actually, since ( AA_1 ) is a midline, it's inside the triangle, so both angles ( alpha ) and ( beta ) should be on the same side of ( AA_1 ). Hmm, maybe I need to clarify that.Let me think again. If ( AA_1 ) is the midline, then it connects vertex ( A ) to the midpoint ( A_1 ) of side ( BC ). So, in triangle ( ABC ), ( AA_1 ) is a median. The angles ( alpha ) and ( beta ) are the angles between the median ( AA_1 ) and the sides ( AB ) and ( AC ), respectively.So, at point ( A ), the median ( AA_1 ) splits the angle at ( A ) into two angles: ( alpha ) between ( AB ) and ( AA_1 ), and ( beta ) between ( AA_1 ) and ( AC ). Therefore, the total angle at vertex ( A ) is ( alpha + beta ).Okay, so I can construct point ( A ), draw segment ( AA_1 ), and then construct angles ( alpha ) and ( beta ) at ( A ) to get the directions of sides ( AB ) and ( AC ).But how do I determine the lengths of ( AB ) and ( AC )? I only know the length of ( AA_1 ) and the angles ( alpha ) and ( beta ). Hmm, maybe I can use some triangle properties or trigonometry here.Let me recall that in a triangle, the length of a median can be related to the lengths of the sides using the formula:[AA_1 = frac{1}{2} sqrt{2AB^2 + 2AC^2 - BC^2}]But wait, I don't know ( BC ) either. Hmm, this might not be directly helpful.Alternatively, maybe I can use the Law of Sines or Cosines in triangle ( ABA_1 ) and triangle ( ACA_1 ).In triangle ( ABA_1 ), I know side ( AA_1 ), angle ( alpha ) at ( A ), and angle at ( A_1 ). Similarly, in triangle ( ACA_1 ), I know side ( AA_1 ), angle ( beta ) at ( A ), and angle at ( A_1 ).Wait, but I don't know the angles at ( A_1 ) in these smaller triangles. Hmm, maybe I need another approach.Let me think about the coordinates. Maybe assigning coordinates to points ( A ), ( A_1 ), ( B ), and ( C ) can help.Let's place point ( A ) at the origin ((0, 0)). Let me denote the coordinates of ( A_1 ) as ((x, y)). Since ( AA_1 ) is a median, ( A_1 ) is the midpoint of ( BC ).Now, I need to express the angles ( alpha ) and ( beta ) in terms of coordinates. The angle ( alpha ) is between ( AB ) and ( AA_1 ), and ( beta ) is between ( AC ) and ( AA_1 ).Using vectors, the angle between two vectors can be found using the dot product formula:[cos theta = frac{vec{u} cdot vec{v}}{|vec{u}| |vec{v}|}]So, if I can express vectors ( AB ) and ( AA_1 ), and vectors ( AC ) and ( AA_1 ), I can set up equations based on the given angles ( alpha ) and ( beta ).But this might get complicated. Maybe there's a simpler geometric construction.Let me try to reconstruct the triangle step by step.1. Draw segment ( AA_1 ) with the given length.2. At point ( A ), construct angle ( alpha ) on one side of ( AA_1 ) and angle ( beta ) on the other side.3. The sides ( AB ) and ( AC ) will lie along these constructed angles.4. Now, I need to find points ( B ) and ( C ) such that ( A_1 ) is the midpoint of ( BC ).Hmm, so if I can find points ( B ) and ( C ) such that ( A_1 ) is their midpoint, and lines ( AB ) and ( AC ) make angles ( alpha ) and ( beta ) with ( AA_1 ), then I can complete the triangle.Maybe I can use the concept of similar triangles or parallelograms here.Wait, if ( A_1 ) is the midpoint of ( BC ), then ( BA_1 = A_1C ). So, if I can construct points ( B ) and ( C ) such that they are symmetric with respect to ( A_1 ), that might work.Let me try this approach:1. Draw segment ( AA_1 ).2. At point ( A ), construct angles ( alpha ) and ( beta ) with ( AA_1 ).3. Along the angle ( alpha ), draw a line from ( A ) to some point ( B ).4. Along the angle ( beta ), draw a line from ( A ) to some point ( C ).5. Now, ( A_1 ) is supposed to be the midpoint of ( BC ). So, if I can adjust the lengths of ( AB ) and ( AC ) such that ( A_1 ) is exactly halfway between ( B ) and ( C ), then I have my triangle.But how do I ensure that ( A_1 ) is the midpoint? Maybe I can use the concept of midpoints in coordinate geometry.Let me assign coordinates again for clarity.Let’s place point ( A ) at ((0, 0)) and point ( A_1 ) at ((m, n)). Since ( A_1 ) is the midpoint of ( BC ), if I can find coordinates for ( B ) and ( C ) such that:[A_1 = left( frac{B_x + C_x}{2}, frac{B_y + C_y}{2} right)]So, if I can express ( B ) and ( C ) in terms of ( A_1 ), that might help.Also, the angles ( alpha ) and ( beta ) can be expressed in terms of slopes or vectors.Let me denote vector ( AB ) as ((b_x, b_y)) and vector ( AC ) as ((c_x, c_y)). Then, the angle between ( AB ) and ( AA_1 ) is ( alpha ), and the angle between ( AC ) and ( AA_1 ) is ( beta ).Using the dot product formula:[cos alpha = frac{(b_x)(m) + (b_y)(n)}{sqrt{b_x^2 + b_y^2} cdot sqrt{m^2 + n^2}}]Similarly,[cos beta = frac{(c_x)(m) + (c_y)(n)}{sqrt{c_x^2 + c_y^2} cdot sqrt{m^2 + n^2}}]But this seems quite involved. Maybe there's a simpler geometric construction without coordinates.Let me think about using triangle similarity or some proportionality.Since ( A_1 ) is the midpoint, perhaps I can construct a parallelogram where ( AA_1 ) is a diagonal, and then use the angles to find the other vertices.Wait, if I construct a parallelogram with ( AA_1 ) as one diagonal, then the other diagonal would be ( BC ). But I'm not sure if that helps directly.Alternatively, maybe I can use the fact that the median divides the triangle into two smaller triangles of equal area.But I'm not sure how that helps in the construction.Wait, perhaps I can use the concept of the median and the given angles to determine the direction of sides ( AB ) and ( AC ), and then find points ( B ) and ( C ) such that ( A_1 ) is their midpoint.Let me try to outline the steps:1. Draw segment ( AA_1 ) with the given length.2. At point ( A ), construct angle ( alpha ) on one side of ( AA_1 ) and angle ( beta ) on the other side.3. From point ( A ), draw two lines making angles ( alpha ) and ( beta ) with ( AA_1 ).4. These lines will intersect some points ( B ) and ( C ), but I need to ensure that ( A_1 ) is the midpoint of ( BC ).Hmm, so maybe I can use the concept of similar triangles or some proportionality to find the correct positions of ( B ) and ( C ).Alternatively, I can use the method of intersection. If I can find points ( B ) and ( C ) such that lines ( AB ) and ( AC ) make angles ( alpha ) and ( beta ) with ( AA_1 ), and ( A_1 ) is the midpoint of ( BC ), then the triangle is constructed.Let me try to formalize this.Suppose I have segment ( AA_1 ). I construct angles ( alpha ) and ( beta ) at ( A ). Then, I draw lines ( l ) and ( m ) from ( A ) making angles ( alpha ) and ( beta ) with ( AA_1 ).Now, I need to find points ( B ) on line ( l ) and ( C ) on line ( m ) such that ( A_1 ) is the midpoint of ( BC ).This seems like a system of equations. If I can express the coordinates of ( B ) and ( C ) in terms of parameters and set ( A_1 ) as their midpoint, I can solve for those parameters.But maybe there's a more straightforward geometric approach.Wait, perhaps I can use the concept of reflection. If ( A_1 ) is the midpoint of ( BC ), then reflecting ( B ) over ( A_1 ) gives ( C ), and vice versa.So, if I can construct point ( B ) such that when I reflect it over ( A_1 ), I get point ( C ), and both ( B ) and ( C ) lie on lines making angles ( alpha ) and ( beta ) with ( AA_1 ), then I can complete the triangle.Let me try to outline this approach:1. Draw segment ( AA_1 ).2. At point ( A ), construct angles ( alpha ) and ( beta ) with ( AA_1 ).3. From point ( A ), draw lines ( l ) and ( m ) along these angles.4. Choose a point ( B ) on line ( l ).5. Reflect ( B ) over ( A_1 ) to get point ( C ).6. Check if ( C ) lies on line ( m ).7. If not, adjust the position of ( B ) until ( C ) lies on line ( m ).This seems like a viable method, but it might require some trial and error.Alternatively, I can use the concept of similar triangles to find the correct position of ( B ) and ( C ).Let me think about triangle ( ABA_1 ) and triangle ( ACA_1 ). Both share the side ( AA_1 ), and I know the angles at ( A ). Maybe I can use the Law of Sines in these triangles to find the ratios of the sides.In triangle ( ABA_1 ):[frac{AB}{sin angle AA_1 B} = frac{AA_1}{sin alpha}]Similarly, in triangle ( ACA_1 ):[frac{AC}{sin angle AA_1 C} = frac{AA_1}{sin beta}]But I don't know the angles at ( A_1 ) in these triangles. Hmm, maybe I can relate them somehow.Wait, since ( A_1 ) is the midpoint of ( BC ), triangles ( ABA_1 ) and ( ACA_1 ) have a relationship in terms of their areas and side lengths.But I'm not sure how to proceed from here.Maybe I can consider the coordinates again. Let's assign coordinates to make it more concrete.Let’s place point ( A ) at ((0, 0)) and point ( A_1 ) at ((m, 0)) for simplicity, since the actual position can be adjusted later.Now, the angles ( alpha ) and ( beta ) are with respect to ( AA_1 ), which is along the x-axis in this coordinate system.So, the line ( AB ) makes an angle ( alpha ) above the x-axis, and the line ( AC ) makes an angle ( beta ) below the x-axis.Wait, but in reality, both angles ( alpha ) and ( beta ) are on the same side of ( AA_1 ), since ( AA_1 ) is inside the triangle. Hmm, maybe I need to adjust this.Actually, in triangle ( ABC ), the median ( AA_1 ) is inside the triangle, so both angles ( alpha ) and ( beta ) are on the same side of ( AA_1 ). Therefore, in my coordinate system, both lines ( AB ) and ( AC ) would be on the same side of ( AA_1 ).Wait, that doesn't make sense because ( AB ) and ( AC ) are two different sides of the triangle, so they should be on opposite sides of ( AA_1 ).Hmm, maybe I need to reconsider the coordinate system.Let me place point ( A ) at ((0, 0)) and point ( A_1 ) at ((m, 0)). Then, line ( AA_1 ) is along the x-axis.Now, the angle ( alpha ) is between ( AB ) and ( AA_1 ), so line ( AB ) makes an angle ( alpha ) above the x-axis.Similarly, the angle ( beta ) is between ( AC ) and ( AA_1 ), so line ( AC ) makes an angle ( beta ) below the x-axis.Wait, but in reality, both ( AB ) and ( AC ) are on the same side of ( AA_1 ) because ( AA_1 ) is a median inside the triangle. So, maybe both angles ( alpha ) and ( beta ) are on the same side.Hmm, I'm getting confused. Let me clarify.In triangle ( ABC ), the median ( AA_1 ) connects vertex ( A ) to the midpoint ( A_1 ) of side ( BC ). Therefore, ( AA_1 ) is inside the triangle, and both sides ( AB ) and ( AC ) are on the same side of ( AA_1 ).Wait, no. Actually, ( AB ) and ( AC ) are on opposite sides of ( AA_1 ) because ( AA_1 ) is a median, splitting the triangle into two smaller triangles.So, in my coordinate system, if ( AA_1 ) is along the x-axis from ((0, 0)) to ((m, 0)), then ( AB ) would be on the upper half-plane and ( AC ) on the lower half-plane.Therefore, angle ( alpha ) is the angle between ( AB ) and ( AA_1 ) above the x-axis, and angle ( beta ) is the angle between ( AC ) and ( AA_1 ) below the x-axis.Okay, that makes sense.So, let's define:- Line ( AB ) makes an angle ( alpha ) above the x-axis.- Line ( AC ) makes an angle ( beta ) below the x-axis.Now, I need to find points ( B ) and ( C ) such that ( A_1 ) is the midpoint of ( BC ).Let me denote the coordinates:- Point ( A ): ((0, 0))- Point ( A_1 ): ((m, 0))- Point ( B ): ((x_b, y_b))- Point ( C ): ((x_c, y_c))Since ( A_1 ) is the midpoint of ( BC ):[m = frac{x_b + x_c}{2}, quad 0 = frac{y_b + y_c}{2}]So,[x_c = 2m - x_b, quad y_c = -y_b]Now, line ( AB ) has a slope of ( tan alpha ), so:[frac{y_b - 0}{x_b - 0} = tan alpha implies y_b = x_b tan alpha]Similarly, line ( AC ) has a slope of ( -tan beta ) (since it's below the x-axis), so:[frac{y_c - 0}{x_c - 0} = -tan beta implies y_c = -x_c tan beta]But from earlier, ( y_c = -y_b ), so:[-y_b = -x_c tan beta implies y_b = x_c tan beta]But ( x_c = 2m - x_b ), so:[y_b = (2m - x_b) tan beta]But we also have ( y_b = x_b tan alpha ), so:[x_b tan alpha = (2m - x_b) tan beta]Let me solve for ( x_b ):[x_b tan alpha + x_b tan beta = 2m tan beta][x_b (tan alpha + tan beta) = 2m tan beta][x_b = frac{2m tan beta}{tan alpha + tan beta}]Okay, so now I can find ( x_b ) in terms of ( m ), ( alpha ), and ( beta ). Then, ( y_b = x_b tan alpha ), and ( x_c = 2m - x_b ), ( y_c = -y_b ).Once I have ( x_b ) and ( y_b ), I can plot points ( B ) and ( C ), and thus complete the triangle ( ABC ).Let me summarize the steps:1. Draw segment ( AA_1 ) with length ( m ).2. At point ( A ), construct angles ( alpha ) above ( AA_1 ) and ( beta ) below ( AA_1 ).3. Along the angle ( alpha ), draw line ( AB ).4. Along the angle ( beta ), draw line ( AC ).5. Using the relationship ( x_b = frac{2m tan beta}{tan alpha + tan beta} ), find the x-coordinate of point ( B ).6. Calculate ( y_b = x_b tan alpha ).7. Determine ( x_c = 2m - x_b ) and ( y_c = -y_b ).8. Plot points ( B ) and ( C ) based on these coordinates.9. Connect points ( B ) and ( C ) to complete triangle ( ABC ).This seems like a solid approach. Let me verify it with an example.Suppose ( AA_1 = 2 ) units, ( alpha = 30^circ ), and ( beta = 45^circ ).Then,[x_b = frac{2 times 2 times tan 45^circ}{tan 30^circ + tan 45^circ} = frac{4 times 1}{frac{sqrt{3}}{3} + 1} = frac{4}{frac{sqrt{3} + 3}{3}} = frac{12}{sqrt{3} + 3}]Rationalizing the denominator:[x_b = frac{12 (sqrt{3} - 3)}{(sqrt{3} + 3)(sqrt{3} - 3)} = frac{12 (sqrt{3} - 3)}{3 - 9} = frac{12 (sqrt{3} - 3)}{-6} = -2 (sqrt{3} - 3) = 6 - 2sqrt{3}]So, ( x_b = 6 - 2sqrt{3} ).Then,[y_b = x_b tan 30^circ = (6 - 2sqrt{3}) times frac{sqrt{3}}{3} = 2sqrt{3} - 2]And,[x_c = 2 times 2 - (6 - 2sqrt{3}) = 4 - 6 + 2sqrt{3} = -2 + 2sqrt{3}][y_c = -y_b = -2sqrt{3} + 2]So, points ( B ) and ( C ) are:- ( B ): ((6 - 2sqrt{3}, 2sqrt{3} - 2))- ( C ): ((-2 + 2sqrt{3}, -2sqrt{3} + 2))Let me check if ( A_1 ) is indeed the midpoint of ( BC ):Midpoint ( M ):[M_x = frac{(6 - 2sqrt{3}) + (-2 + 2sqrt{3})}{2} = frac{4}{2} = 2][M_y = frac{(2sqrt{3} - 2) + (-2sqrt{3} + 2)}{2} = frac{0}{2} = 0]So, midpoint ( M ) is ((2, 0)), which is indeed ( A_1 ). Great, it checks out.Therefore, this method works. By using the given midline ( AA_1 ) and the angles ( alpha ) and ( beta ), I can construct triangle ( ABC ) by determining the coordinates of points ( B ) and ( C ) such that ( A_1 ) is their midpoint.In summary, the steps are:1. Draw the given midline ( AA_1 ).2. Construct angles ( alpha ) and ( beta ) at point ( A ) with respect to ( AA_1 ).3. Use the relationship derived from the midpoint condition to find the coordinates of points ( B ) and ( C ).4. Plot points ( B ) and ( C ) and connect them to complete the triangle.This method ensures that the constructed triangle satisfies the given conditions of having the specified midline and angles.

💡Okay, so I have this problem where I'm given two functions: f(x) = x² - 14x + 52 and g(x) = ax + b. I need to find the value of 'a' given that f(g(-5)) = 3 and f(g(0)) = 103. Both 'a' and 'b' are positive. Hmm, let's break this down step by step.First, I should probably figure out what g(-5) and g(0) are. Since g(x) is a linear function, g(-5) would be a*(-5) + b, which simplifies to -5a + b. Similarly, g(0) is just b because when x is 0, the term with 'a' disappears.So, f(g(-5)) means I plug g(-5) into f(x). That would be f(-5a + b). Let me write that out:f(-5a + b) = (-5a + b)² - 14*(-5a + b) + 52.And this equals 3, according to the problem. So, I can set up the equation:(-5a + b)² - 14*(-5a + b) + 52 = 3.Similarly, f(g(0)) is f(b), so:f(b) = b² - 14b + 52 = 103.Alright, so I have two equations here. Let me write them down clearly:1. (-5a + b)² - 14*(-5a + b) + 52 = 32. b² - 14b + 52 = 103Maybe I can solve the second equation first because it only involves 'b' and might be simpler. Let's do that.Starting with equation 2:b² - 14b + 52 = 103Subtract 103 from both sides:b² - 14b + 52 - 103 = 0Simplify:b² - 14b - 51 = 0Okay, so I have a quadratic equation in terms of 'b'. Let me try to factor this or use the quadratic formula. Let's see if it factors. I'm looking for two numbers that multiply to -51 and add up to -14. Hmm, -17 and 3? Because -17 * 3 = -51 and -17 + 3 = -14. Perfect!So, the equation factors as:(b - 17)(b + 3) = 0Setting each factor equal to zero:b - 17 = 0 => b = 17b + 3 = 0 => b = -3But since 'b' is positive, we discard b = -3. So, b = 17.Great, now that I have b, I can plug this back into equation 1 to find 'a'. Let's write equation 1 again:(-5a + b)² - 14*(-5a + b) + 52 = 3Substituting b = 17:(-5a + 17)² - 14*(-5a + 17) + 52 = 3Let me expand this step by step.First, expand (-5a + 17)²:(-5a + 17)² = (-5a)² + 2*(-5a)*(17) + 17² = 25a² - 170a + 289Next, expand -14*(-5a + 17):-14*(-5a) + (-14)*17 = 70a - 238Now, putting it all back into the equation:25a² - 170a + 289 + 70a - 238 + 52 = 3Combine like terms:25a² + (-170a + 70a) + (289 - 238 + 52) = 3Simplify each part:-170a + 70a = -100a289 - 238 = 51; 51 + 52 = 103So, the equation becomes:25a² - 100a + 103 = 3Subtract 3 from both sides:25a² - 100a + 100 = 0Hmm, this is another quadratic equation. Let me see if I can factor this or use the quadratic formula. Let's try factoring first.25a² - 100a + 100 = 0I notice that all coefficients are divisible by 25, so let's factor that out:25(a² - 4a + 4) = 0So, a² - 4a + 4 = 0This is a perfect square trinomial:(a - 2)² = 0Therefore, a = 2Since 'a' is positive, this is our solution.Let me double-check my steps to make sure I didn't make any mistakes.1. Found g(-5) = -5a + b and g(0) = b.2. Plugged g(-5) into f(x) and set it equal to 3.3. Plugged g(0) into f(x) and set it equal to 103, which gave me a quadratic equation in 'b'.4. Solved the quadratic equation for 'b' and found b = 17.5. Substituted b = 17 back into the first equation and solved for 'a', resulting in a = 2.Everything seems to check out. I don't see any calculation errors, so I'm confident that a = 2 is the correct answer.**Final Answer**boxed{2}

💡First, I need to determine the daily swimming distances for each person based on the information provided.Julien swims 50 meters per day.Sarah swims twice the distance Julien swims, so Sarah swims 2 * 50 = 100 meters per day.Jamir swims 20 meters more per day than Sarah, so Jamir swims 100 + 20 = 120 meters per day.Next, I'll calculate the total distance each person swims in a week by multiplying their daily distance by 7.Julien's weekly distance: 50 * 7 = 350 metersSarah's weekly distance: 100 * 7 = 700 metersJamir's weekly distance: 120 * 7 = 840 metersFinally, I'll add up the weekly distances of all three to find the combined total.350 + 700 + 840 = 1890 meters

💡Okay, so I'm trying to figure out the three basic structures of an algorithm. I remember from my computer science class that algorithms have different ways of organizing steps, and these structures are fundamental. Let me think about what I learned.First, there's the sequential structure. That's pretty straightforward—it means that the algorithm does one thing after another in a specific order. Like, if I have steps 1, 2, and 3, it does step 1 first, then step 2, and finally step 3. No branches or loops here; it's just a straight line of execution.Next, I think about the conditional structure. This is where the algorithm makes decisions based on certain conditions. For example, if a condition is true, it does one set of steps, and if it's false, it does another set or skips some steps altogether. This introduces branching into the algorithm, allowing it to handle different scenarios.Then there's the loop structure. This is where the algorithm repeats a set of steps multiple times until a certain condition is met. Like, if I want to sum numbers from 1 to 10, I can use a loop to add each number in sequence until I reach 10. Loops are essential for handling repetitive tasks efficiently.Now, looking at the options provided:A: Sequential structure, Module structure, Conditional structure B: Sequential structure, Loop structure, Module structure C: Sequential structure, Conditional structure, Loop structure D: Module structure, Conditional structure, Loop structure Option A includes Module structure, which I'm not sure is one of the basic structures. I think Module structure might refer to breaking down the algorithm into functions or procedures, which is more about organization rather than a basic structure.Option B also includes Module structure, so it's similar to A but swaps Conditional for Module. Again, I'm not sure Module is a basic structure.Option C lists Sequential, Conditional, and Loop structures. These align with what I remember from my studies—sequential for order, conditional for decisions, and loop for repetition.Option D starts with Module structure, which again I don't think is a basic structure, and then includes Conditional and Loop. So it's missing Sequential, which is fundamental.Therefore, the correct answer should be Option C: Sequential structure, Conditional structure, Loop structure.