💡To find the equations of the asymptotes of the hyperbola given by ( x^{2} - frac{y^{2}}{3} = 1 ), we start by identifying the values of ( a^2 ) and ( b^2 ) from the standard form of the hyperbola equation.Given the equation of the hyperbola is ( x^{2} - frac{y^{2}}{3} = 1 ), we can directly see that:- ( a^{2} = 1 )- ( b^{2} = 3 )The next step is to find the value of ( c ), which is the distance from the center to a focus of the hyperbola. This is found using the formula ( c = sqrt{a^{2} + b^{2}} ). Substituting the known values of ( a^2 ) and ( b^2 ):[ c = sqrt{a^{2} + b^{2}} = sqrt{1 + 3} = sqrt{4} = 2 ]The equations of the asymptotes for a hyperbola in the form ( x^{2} - frac{y^{2}}{b^2} = 1 ) are given by ( y = pm frac{b}{a}x ). Substituting the values of ( a ) and ( b ):[ y = pm frac{sqrt{3}}{1}x ]which simplifies to:[ y = pm sqrt{3}x ]Therefore, the correct equations of the asymptotes are:[ boxed{y = pm sqrt{3}x} ]

💡Alright, so I have this geometry problem here. Let me try to visualize it first. There's a triangle ABC where AB is longer than AC. AD is the angle bisector of angle BAC. There's a point E inside the triangle such that EC is perpendicular to AD, and ED is parallel to AC. I need to prove that the ray AE bisects side BC. Hmm, okay.First, let me draw triangle ABC. Since AB > AC, I'll make AB longer than AC. Point D is on BC because AD is the angle bisector. So, AD splits angle BAC into two equal angles. Now, point E is inside the triangle. EC is perpendicular to AD, so EC forms a right angle with AD. Also, ED is parallel to AC. That should give us some similar triangles maybe?Let me label the points. A is the top vertex, B is the left, and C is the right. AD is the angle bisector, so D is somewhere on BC. E is somewhere inside the triangle. EC is perpendicular to AD, so if I draw AD, then EC will meet AD at a right angle. Also, ED is parallel to AC, so ED should be a line segment from E to somewhere on BC, parallel to AC.I think I need to use some properties of angle bisectors and parallel lines. Maybe similar triangles? Let me see.Since ED is parallel to AC, triangle EDA should be similar to triangle ACA? Wait, that doesn't make sense. Maybe triangle EDA is similar to triangle AC something. Let me think.Alternatively, since ED is parallel to AC, the corresponding angles should be equal. So, angle AED should be equal to angle ACD. Hmm, not sure if that helps directly.EC is perpendicular to AD. So, triangle ECD is a right triangle. Maybe I can use some right triangle properties here.Wait, maybe coordinate geometry could help. Let me assign coordinates to the triangle. Let me place point A at (0, 0), point B at (c, 0), and point C at (b, 0). Wait, no, that would make AB and AC on the same line. Maybe better to place A at (0, 0), B at (0, b), and C at (c, 0). Hmm, not sure.Alternatively, let me place A at (0, 0), B at (1, 0), and C at (k, 0) where k > 1 since AB > AC. Wait, no, AB is the side opposite to angle C, so maybe I need a different coordinate system.Alternatively, let me place A at (0, 0), B at (c, 0), and C at (b, 0). Wait, no, that would make AB and AC both on the x-axis. Maybe better to have A at (0, 0), B at (0, b), and C at (c, 0). That way, AB is vertical, AC is horizontal, and BC is the hypotenuse.So, AD is the angle bisector from A to BC. Let me find the coordinates of D. Since AD is the angle bisector, by the angle bisector theorem, BD/DC = AB/AC. AB is the length from A(0,0) to B(0,b), which is b. AC is from A(0,0) to C(c,0), which is c. So, BD/DC = b/c.Since BC is from (0,b) to (c,0), the coordinates of D can be found using the ratio BD/DC = b/c. So, using the section formula, D divides BC in the ratio b:c. So, coordinates of D are ((c*b)/(b + c), (0*c)/(b + c)) = (cb/(b + c), 0). Wait, that can't be right because D is on BC, which goes from (0,b) to (c,0). So, the x-coordinate should be (c*b)/(b + c) and y-coordinate should be (0*b + b*c)/(b + c)? Wait, no, the section formula is ( (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n) ). So, since BD/DC = b/c, m = b, n = c. So, D is ( (b*c + c*0)/(b + c), (b*0 + c*b)/(b + c) ) = (bc/(b + c), cb/(b + c)). Wait, that would be (bc/(b + c), cb/(b + c)) which simplifies to (bc/(b + c), bc/(b + c)). So, D is at (bc/(b + c), bc/(b + c)). Interesting, so D is equidistant along both axes.Now, point E is inside the triangle such that EC is perpendicular to AD and ED is parallel to AC. Let me find the coordinates of E.First, let me find the equation of AD. A is (0,0), D is (bc/(b + c), bc/(b + c)). So, the slope of AD is (bc/(b + c) - 0)/(bc/(b + c) - 0) = 1. So, AD has a slope of 1, meaning it's the line y = x.Since EC is perpendicular to AD, which has a slope of 1, the slope of EC must be -1. So, EC has a slope of -1.Also, ED is parallel to AC. AC is from A(0,0) to C(c,0), so AC is horizontal. Therefore, ED must also be horizontal, meaning it has a slope of 0.So, ED is horizontal, so the y-coordinate of E is equal to the y-coordinate of D. Since D is at (bc/(b + c), bc/(b + c)), the y-coordinate of E is bc/(b + c). Therefore, E is at (x, bc/(b + c)) for some x.Also, EC has a slope of -1. Let me write the equation of EC. Point C is at (c,0), and E is at (x, bc/(b + c)). The slope between E and C is (0 - bc/(b + c))/(c - x) = (-bc/(b + c))/(c - x) = -1.So, (-bc/(b + c))/(c - x) = -1. Let's solve for x:(-bc/(b + c))/(c - x) = -1Multiply both sides by (c - x):-bc/(b + c) = -1*(c - x)Simplify:-bc/(b + c) = -c + xMultiply both sides by -1:bc/(b + c) = c - xThen, x = c - bc/(b + c) = (c(b + c) - bc)/(b + c) = (cb + c^2 - bc)/(b + c) = c^2/(b + c)So, x = c^2/(b + c). Therefore, E is at (c^2/(b + c), bc/(b + c)).Now, I need to find the equation of AE and see where it intersects BC. If it bisects BC, then the intersection point should be the midpoint of BC.Midpoint of BC: B is at (0,b), C is at (c,0). Midpoint M is at ((0 + c)/2, (b + 0)/2) = (c/2, b/2).So, if AE passes through M, then AE bisects BC. Let's find the equation of AE.Point A is (0,0), and E is (c^2/(b + c), bc/(b + c)). So, the slope of AE is (bc/(b + c) - 0)/(c^2/(b + c) - 0) = (bc/(b + c))/(c^2/(b + c)) = bc/c^2 = b/c.So, the equation of AE is y = (b/c)x.Now, let's see if point M(c/2, b/2) lies on AE. Plugging x = c/2 into AE's equation: y = (b/c)*(c/2) = b/2. So, yes, M lies on AE.Therefore, AE passes through the midpoint of BC, so AE bisects BC. Hence, proved.Wait, that seems straightforward. Did I miss something? Let me double-check.I used coordinate geometry, assigned coordinates to the triangle, found D using the angle bisector theorem, determined E's coordinates based on the given conditions, found the equation of AE, and checked if it passes through the midpoint of BC. It does, so AE bisects BC. Yeah, that seems correct.I think this approach works. Maybe there's a more geometric proof without coordinates, but this seems solid.

💡First, I start with 1 gallon of paint.On the first day, half of the paint is used:[frac{1}{2} times 1 = frac{1}{2} text{ gallon used}]So, the remaining paint after the first day is:[1 - frac{1}{2} = frac{1}{2} text{ gallon}]On the second day, one fourth of the remaining paint is used:[frac{1}{4} times frac{1}{2} = frac{1}{8} text{ gallon used}]So, the remaining paint after the second day is:[frac{1}{2} - frac{1}{8} = frac{4}{8} - frac{1}{8} = frac{3}{8} text{ gallon}]On the third day, one third of the remaining paint is used:[frac{1}{3} times frac{3}{8} = frac{3}{24} = frac{1}{8} text{ gallon used}]So, the remaining paint after the third day is:[frac{3}{8} - frac{1}{8} = frac{2}{8} = frac{1}{4} text{ gallon}]Therefore, the fraction of the original amount of paint available to use on the fourth day is (frac{1}{4}).

💡Okay, so I have this algebra problem here. It says that a + b + c equals 3, and then there's another equation: (a - 1)^3 + (b - 1)^3 + (c - 1)^3 equals 0. Also, they tell me that a is 2. I need to find the value of a^2 + b^2 + c^2. Hmm, let me see how to approach this.First, since a is given as 2, I can substitute that into the first equation. So, a + b + c = 3 becomes 2 + b + c = 3. If I subtract 2 from both sides, that gives me b + c = 1. Okay, so the sum of b and c is 1. That's one equation down.Now, looking at the second equation: (a - 1)^3 + (b - 1)^3 + (c - 1)^3 = 0. Let me substitute a = 2 here as well. So, (2 - 1)^3 + (b - 1)^3 + (c - 1)^3 = 0. Simplifying that, (1)^3 + (b - 1)^3 + (c - 1)^3 = 0, which is 1 + (b - 1)^3 + (c - 1)^3 = 0. If I subtract 1 from both sides, I get (b - 1)^3 + (c - 1)^3 = -1.Alright, so now I have two equations:1. b + c = 12. (b - 1)^3 + (c - 1)^3 = -1I need to find a^2 + b^2 + c^2, which is 2^2 + b^2 + c^2 = 4 + b^2 + c^2. So, I just need to find b^2 + c^2.I remember that (b + c)^2 = b^2 + 2bc + c^2. Since I know b + c is 1, then (1)^2 = b^2 + 2bc + c^2, which is 1 = b^2 + 2bc + c^2. Therefore, b^2 + c^2 = 1 - 2bc. So, if I can find the value of bc, I can find b^2 + c^2.Looking back at the second equation: (b - 1)^3 + (c - 1)^3 = -1. I think there's a formula for the sum of cubes: x^3 + y^3 = (x + y)(x^2 - xy + y^2). Maybe I can apply that here.Let me set x = (b - 1) and y = (c - 1). Then, x + y = (b - 1) + (c - 1) = (b + c) - 2. Since b + c is 1, this becomes 1 - 2 = -1. So, x + y = -1.Now, applying the sum of cubes formula: x^3 + y^3 = (x + y)(x^2 - xy + y^2). We know that x^3 + y^3 is -1, and x + y is -1. So, substituting in, we have -1 = (-1)(x^2 - xy + y^2). Dividing both sides by -1, we get 1 = x^2 - xy + y^2.But x^2 - xy + y^2 can also be expressed in terms of (x + y)^2 and xy. Let me recall that (x + y)^2 = x^2 + 2xy + y^2. So, x^2 + y^2 = (x + y)^2 - 2xy. Therefore, x^2 - xy + y^2 = (x^2 + y^2) - xy = [(x + y)^2 - 2xy] - xy = (x + y)^2 - 3xy.Substituting back, we have 1 = (x + y)^2 - 3xy. We know that x + y is -1, so (x + y)^2 is (-1)^2 = 1. Therefore, 1 = 1 - 3xy. Subtracting 1 from both sides, we get 0 = -3xy. Dividing both sides by -3, we find that xy = 0.But x = (b - 1) and y = (c - 1), so xy = (b - 1)(c - 1) = 0. Therefore, either (b - 1) = 0 or (c - 1) = 0. That means either b = 1 or c = 1.So, let's consider two cases:Case 1: b = 1. Then, since b + c = 1, c = 0.Case 2: c = 1. Then, since b + c = 1, b = 0.So, either (b, c) = (1, 0) or (0, 1). Let's check both cases.Case 1: b = 1, c = 0.Compute (b - 1)^3 + (c - 1)^3 = (1 - 1)^3 + (0 - 1)^3 = 0 + (-1)^3 = -1. That works.Case 2: b = 0, c = 1.Compute (b - 1)^3 + (c - 1)^3 = (0 - 1)^3 + (1 - 1)^3 = (-1)^3 + 0 = -1. That also works.So, both cases satisfy the second equation. Therefore, the possible pairs for (b, c) are (1, 0) and (0, 1).Now, let's compute b^2 + c^2 for both cases.Case 1: b = 1, c = 0.b^2 + c^2 = 1^2 + 0^2 = 1 + 0 = 1.Case 2: b = 0, c = 1.b^2 + c^2 = 0^2 + 1^2 = 0 + 1 = 1.So, in both cases, b^2 + c^2 = 1.Therefore, a^2 + b^2 + c^2 = 2^2 + 1 = 4 + 1 = 5.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Substituted a = 2 into the first equation: 2 + b + c = 3 => b + c = 1. That seems correct.2. Substituted a = 2 into the second equation: (2 - 1)^3 + (b - 1)^3 + (c - 1)^3 = 0 => 1 + (b - 1)^3 + (c - 1)^3 = 0 => (b - 1)^3 + (c - 1)^3 = -1. That also looks correct.3. Applied the sum of cubes formula: x^3 + y^3 = (x + y)(x^2 - xy + y^2). Then, x + y = (b - 1) + (c - 1) = (b + c) - 2 = 1 - 2 = -1. So, x + y = -1. Then, substituted into the formula: -1 = (-1)(x^2 - xy + y^2) => 1 = x^2 - xy + y^2. Then, expressed x^2 - xy + y^2 in terms of (x + y)^2 and xy: (x + y)^2 - 3xy. Substituted (x + y)^2 = 1, so 1 = 1 - 3xy => 0 = -3xy => xy = 0. So, (b - 1)(c - 1) = 0 => b = 1 or c = 1. That seems solid.4. Then, considered both cases, found b and c, and computed b^2 + c^2 in each case, both giving 1. Then, added a^2 = 4, so total is 5.I think all the steps are correct. So, the value of a^2 + b^2 + c^2 is 5.**Final Answer**The value of the algebraic expression (a^2 + b^2 + c^2) is boxed{5}.

💡First, I need to understand the relationships between the volumes of the cups and the barrels.The small cup to large cup volume ratio is 2:3. This means that if the small cup has a volume of 2 units, the large cup has a volume of 3 units.Next, the volume ratio of juice in barrel A to barrel B is 4:5. This indicates that for every 4 units of juice in barrel A, there are 5 units in barrel B.Given that 120 small cups can be filled with the juice from barrel A, I can calculate the total volume of juice in barrel A by multiplying the number of cups by the volume of each small cup.Using the ratio of the volumes of barrels A and B, I can determine the total volume of juice in barrel B.Finally, to find out how many large cups can be filled with the juice from barrel B, I divide the total volume of juice in barrel B by the volume of one large cup.

💡First, I recognize that in an isosceles triangle, two sides are equal in length. Given the side lengths of 2 and 4, I need to determine which sides are the equal ones.I consider two possible configurations:1. The two equal sides are both 4 units long, with the base being 2 units.2. The two equal sides are both 2 units long, with the base being 4 units.For the first configuration, I check the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the remaining side. In this case, 4 + 4 > 2 and 4 + 2 > 4, so this configuration is valid.For the second configuration, I again apply the triangle inequality theorem. Here, 2 + 2 is not greater than 4, which means this configuration does not form a valid triangle.Therefore, the only valid configuration is the first one, where the two equal sides are 4 units each, and the base is 2 units. Adding these together gives the perimeter of the triangle.