💡Okay, so I have this equation here: [frac{x^2}{49} - frac{(y-3)^2}{25} = 0.] Hmm, I remember that this looks like a hyperbola equation, but it's set equal to zero instead of one. The user mentioned it's a degenerate hyperbola, so I guess that means it's not the usual hyperbola we see. I need to figure out what this equation represents and then find the largest possible y-coordinate on its graph.First, let me recall what a hyperbola equation looks like. Normally, it's something like [frac{(x-h)^2}{a^2} - frac{(y-k)^2}{b^2} = 1] for a hyperbola that opens left and right, or [frac{(y-k)^2}{b^2} - frac{(x-h)^2}{a^2} = 1] for one that opens up and down. But in this case, the equation is equal to zero instead of one. So, that must mean it's degenerate.I think a degenerate conic section happens when the plane cuts through the apex of the cone, resulting in something simpler, like a line or a pair of lines. So, maybe this equation represents two lines instead of a hyperbola. Let me try to factor it or rearrange it to see.Starting with the given equation:[frac{x^2}{49} - frac{(y-3)^2}{25} = 0.]I can rewrite this as:[frac{x^2}{49} = frac{(y-3)^2}{25}.]Taking square roots on both sides, I get:[frac{x}{7} = pm frac{y - 3}{5}.]So, that gives me two equations:1. [frac{x}{7} = frac{y - 3}{5}]2. [frac{x}{7} = -frac{y - 3}{5}]Let me simplify both of these.For the first equation:[frac{x}{7} = frac{y - 3}{5}]Multiply both sides by 35 to eliminate denominators:[5x = 7(y - 3)][5x = 7y - 21]Bring all terms to one side:[5x - 7y + 21 = 0]Or, solving for y:[7y = 5x + 21][y = frac{5}{7}x + 3]For the second equation:[frac{x}{7} = -frac{y - 3}{5}]Again, multiply both sides by 35:[5x = -7(y - 3)][5x = -7y + 21]Bring all terms to one side:[5x + 7y - 21 = 0]Or, solving for y:[7y = -5x + 21][y = -frac{5}{7}x + 3]So, the original equation represents two straight lines:1. (y = frac{5}{7}x + 3)2. (y = -frac{5}{7}x + 3)These are two lines intersecting at the point where both equations are satisfied. Let me find that point by setting them equal:[frac{5}{7}x + 3 = -frac{5}{7}x + 3]Adding (frac{5}{7}x) to both sides:[frac{10}{7}x + 3 = 3]Subtracting 3 from both sides:[frac{10}{7}x = 0]So, (x = 0). Plugging back into one of the equations:(y = frac{5}{7}(0) + 3 = 3). So, the two lines intersect at (0, 3).Now, the question is asking for the largest possible y-coordinate on the graph of this equation. Since the graph consists of two straight lines, I need to see if there's a maximum y-value. However, straight lines extend infinitely in both directions, so unless there's a restriction, the y-values can go to infinity. But wait, in this case, both lines have a slope, so as x increases or decreases without bound, y will also increase or decrease without bound.But hold on, maybe I'm missing something. The original equation was a degenerate hyperbola, which in this case is two intersecting lines. So, the graph is just these two lines. Since they extend infinitely, the y-values can be as large or as small as possible. But the question is about the largest possible y-coordinate. If there's no restriction on x, then y can be infinitely large. But that doesn't make sense because the problem is asking for a specific value.Wait, maybe I made a mistake in interpreting the equation. Let me go back to the original equation:[frac{x^2}{49} - frac{(y-3)^2}{25} = 0.]If I think of this as a hyperbola, normally it would require the equation to be equal to 1, but here it's equal to 0. So, it's the set of points where the difference of the squares equals zero, which means each term must individually be zero. So, (frac{x^2}{49} = 0) and (-frac{(y-3)^2}{25} = 0). Therefore, x must be 0 and y must be 3. So, the only point is (0, 3).Wait, that contradicts my earlier conclusion where I thought it was two lines. Which one is correct?Let me check. If I set the equation equal to zero, it's equivalent to (frac{x^2}{49} = frac{(y-3)^2}{25}). So, that can be rewritten as (left(frac{x}{7}right)^2 = left(frac{y-3}{5}right)^2), which implies that (frac{x}{7} = pm frac{y-3}{5}), leading to the two lines I found earlier.But if I consider the equation as (frac{x^2}{49} - frac{(y-3)^2}{25} = 0), it's also equivalent to (frac{x^2}{49} = frac{(y-3)^2}{25}), which can be rearranged to (frac{x^2}{49} + frac{(3 - y)^2}{25} = 0). Since both terms are squares, they must each be zero. Therefore, x must be 0 and y must be 3. So, the only solution is the point (0, 3).Hmm, so which is it? Is it two lines or just a single point?I think it depends on how you interpret the equation. If you treat it as a hyperbola set to zero, it's a degenerate case where the hyperbola collapses to its asymptotes, which are the two lines I found. However, if you consider the equation as a sum of squares equal to zero, then each square must individually be zero, leading to just the point (0, 3).But wait, in the case of a hyperbola, the standard form is (frac{(x-h)^2}{a^2} - frac{(y-k)^2}{b^2} = 1). When it's equal to zero, it's not a hyperbola anymore but the asymptotes. So, in this case, the equation represents the asymptotes of the hyperbola, which are the two lines.However, if I consider the equation as a quadratic equation, it's equivalent to (frac{x^2}{49} = frac{(y-3)^2}{25}), which can be factored into two linear equations, giving the two lines. So, the graph is indeed two lines intersecting at (0, 3).But then, if the graph is two lines, they extend to infinity, so y can be as large as possible. But the problem is asking for the largest possible y-coordinate. That seems contradictory because y can be infinitely large. However, maybe I'm misunderstanding the nature of the degenerate hyperbola.Wait, perhaps in the context of the problem, the degenerate hyperbola is considered as the two lines, but the question is about the largest y-coordinate on the graph. If the graph is two lines, then technically, y can be any real number, so there's no maximum. But that can't be right because the problem is asking for a specific value.Alternatively, maybe the equation is intended to represent just the point (0, 3), making the largest y-coordinate 3. But that seems inconsistent with the idea of a degenerate hyperbola, which usually refers to the asymptotes.I think I need to clarify this. Let me check the definition of a degenerate hyperbola. A degenerate hyperbola occurs when the plane intersects the cone at the apex, resulting in two intersecting lines. So, in this case, the equation represents two lines, and the graph is those two lines. Therefore, the y-coordinate can be any value along those lines, which extend to infinity. So, there is no largest possible y-coordinate because y can be infinitely large.But the problem is asking for the largest possible y-coordinate, implying that there is a maximum. That suggests that maybe the equation is intended to represent just the point (0, 3), making the largest y-coordinate 3. But that contradicts the idea of a degenerate hyperbola being two lines.Wait, perhaps I'm overcomplicating this. Let me consider the equation again:[frac{x^2}{49} - frac{(y-3)^2}{25} = 0.]If I rearrange it:[frac{x^2}{49} = frac{(y-3)^2}{25}.]This implies that either both sides are positive or both are negative. But since squares are always non-negative, the only way for the equation to hold is if both sides are zero. Therefore, x must be 0 and y must be 3. So, the only solution is the point (0, 3). Therefore, the largest possible y-coordinate is 3.But then, why is it called a degenerate hyperbola? I thought a degenerate hyperbola is two lines. Maybe in this case, it's a degenerate hyperbola that has collapsed to a single point. That would make sense because if the hyperbola's asymptotes intersect at a single point, then it's just that point.So, in this case, the equation represents only the point (0, 3), making the largest y-coordinate 3.I think that's the correct interpretation. The equation is a degenerate hyperbola that has collapsed to a single point, so the only y-coordinate is 3, which is also the largest.

💡First, I identify the initial amount of sugar Jack has, which is 65 pounds.Next, I let ( x ) represent the amount of sugar Jack will use tomorrow.After using ( x ) pounds, Jack will have ( 65 - x ) pounds of sugar left.The following day, Jack buys an additional 50 pounds of sugar, bringing his total to ( (65 - x) + 50 ) pounds.According to the problem, the final amount of sugar Jack has is 97 pounds. This allows me to set up the equation:[(65 - x) + 50 = 97]Simplifying the equation:[115 - x = 97]Solving for ( x ):[-x = 97 - 115][-x = -18][x = 18]Therefore, Jack will use 18 pounds of sugar tomorrow.

💡First, I notice that both terms in the expression ( x^3 - 16x ) have a common factor of ( x ). I'll factor this out to simplify the expression.After factoring out ( x ), I'm left with ( x^2 - 16 ). This is a difference of squares, which I know can be factored using the formula ( a^2 - b^2 = (a + b)(a - b) ).Applying this formula to ( x^2 - 16 ), I identify ( a = x ) and ( b = 4 ). This allows me to factor ( x^2 - 16 ) into ( (x + 4)(x - 4) ).Finally, I combine the factored terms to get the complete factorization of the original expression.

💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation: (cos A - 2 cos C) / cos B = (2c - a) / b. I need to find the value of sin C / sin A for part (I), and then in part (II), given that cos B = 1/4 and the area of the triangle is sqrt(15)/4, I need to find the perimeter.Alright, starting with part (I). Let me write down the given equation:(cos A - 2 cos C) / cos B = (2c - a) / b.Hmm, this looks a bit complicated, but maybe I can use the Law of Sines or the Law of Cosines to relate these terms. Let me recall that the Law of Sines says that a / sin A = b / sin B = c / sin C. Maybe I can express everything in terms of sine functions.First, let me rewrite the equation:(cos A - 2 cos C) / cos B = (2c - a) / b.I can write 2c - a as 2c - a, and b is just b. Maybe I can express c and a in terms of sine functions. Since a = 2R sin A and c = 2R sin C, where R is the circumradius. So, 2c - a = 2*(2R sin C) - 2R sin A = 4R sin C - 2R sin A. Similarly, b = 2R sin B.So, substituting back into the equation:(cos A - 2 cos C) / cos B = (4R sin C - 2R sin A) / (2R sin B).Simplify the right-hand side: (4R sin C - 2R sin A) / (2R sin B) = (2 sin C - sin A) / sin B.So, now the equation becomes:(cos A - 2 cos C) / cos B = (2 sin C - sin A) / sin B.Hmm, that's interesting. Let me cross-multiply to eliminate the denominators:(cos A - 2 cos C) * sin B = (2 sin C - sin A) * cos B.Expanding both sides:cos A sin B - 2 cos C sin B = 2 sin C cos B - sin A cos B.Let me rearrange terms to group similar ones:cos A sin B + sin A cos B = 2 sin C cos B + 2 cos C sin B.Wait, on the left side, cos A sin B + sin A cos B is equal to sin(A + B) because sin(A + B) = sin A cos B + cos A sin B.Similarly, on the right side, 2 sin C cos B + 2 cos C sin B can be factored as 2(sin C cos B + cos C sin B) = 2 sin(B + C).So, now the equation becomes:sin(A + B) = 2 sin(B + C).But in a triangle, A + B + C = π, so A + B = π - C and B + C = π - A.Therefore, sin(A + B) = sin(π - C) = sin C, and sin(B + C) = sin(π - A) = sin A.Substituting back:sin C = 2 sin A.So, sin C / sin A = 2.Alright, that seems straightforward. So, part (I) is solved, and the value is 2.Moving on to part (II). Given that cos B = 1/4 and the area of the triangle is sqrt(15)/4, find the perimeter.First, from part (I), we know that sin C = 2 sin A. So, from the Law of Sines, c = 2a because a / sin A = c / sin C, so c = a * (sin C / sin A) = 2a.So, c = 2a.Given that cos B = 1/4, we can find sin B since sin^2 B + cos^2 B = 1. So, sin B = sqrt(1 - (1/4)^2) = sqrt(1 - 1/16) = sqrt(15/16) = sqrt(15)/4.Now, the area of the triangle is given by (1/2)ab sin C, but wait, actually, the area can be expressed in multiple ways. Let me think.Alternatively, using sides a, b, c and angles, the area can be expressed as (1/2)ab sin C, but since we have sides a, b, c, and angles, maybe it's better to use another formula.Wait, actually, the area can also be expressed as (1/2)ac sin B because sides a and c with included angle B. Let me confirm:Yes, the area is (1/2)*a*c*sin B.Given that the area is sqrt(15)/4, and sin B is sqrt(15)/4, so:(1/2)*a*c*(sqrt(15)/4) = sqrt(15)/4.Simplify:(1/2)*a*c*(sqrt(15)/4) = sqrt(15)/4.Multiply both sides by 4:(1/2)*a*c*sqrt(15) = sqrt(15).Divide both sides by sqrt(15):(1/2)*a*c = 1.So, a*c = 2.But from earlier, we have c = 2a, so substituting:a*(2a) = 2 => 2a^2 = 2 => a^2 = 1 => a = 1.So, a = 1, then c = 2a = 2.Now, we need to find b. Let's use the Law of Cosines on angle B.Law of Cosines: b^2 = a^2 + c^2 - 2ac cos B.Substituting the known values:b^2 = 1^2 + 2^2 - 2*1*2*(1/4).Calculate:b^2 = 1 + 4 - 4*(1/4) = 5 - 1 = 4.So, b = 2.Therefore, the sides are a = 1, b = 2, c = 2.So, the perimeter is a + b + c = 1 + 2 + 2 = 5.Wait, that seems straightforward. Let me double-check.Given a = 1, c = 2, b = 2.Check the area: (1/2)*a*c*sin B = (1/2)*1*2*(sqrt(15)/4) = (1)*sqrt(15)/4 = sqrt(15)/4, which matches the given area.Also, check the given equation in part (I):(cos A - 2 cos C)/cos B = (2c - a)/b.We have a = 1, c = 2, b = 2, cos B = 1/4.Compute the right-hand side: (2*2 - 1)/2 = (4 - 1)/2 = 3/2.Now, compute the left-hand side: (cos A - 2 cos C)/cos B.First, find angles A and C.Using the Law of Sines: a / sin A = c / sin C = 1 / sin A = 2 / sin C.From part (I), sin C = 2 sin A, so 1 / sin A = 2 / (2 sin A) = 1 / sin A, which is consistent.So, let's compute angles A and C.We know that A + B + C = π.We have cos B = 1/4, so angle B is arccos(1/4). Let me compute sin A and sin C.From sin C = 2 sin A.Also, using the Law of Sines: a / sin A = b / sin B.So, 1 / sin A = 2 / (sqrt(15)/4).Thus, 1 / sin A = 8 / sqrt(15).So, sin A = sqrt(15)/8.Therefore, sin C = 2 sin A = 2*sqrt(15)/8 = sqrt(15)/4.Now, compute cos A and cos C.cos A = sqrt(1 - sin^2 A) = sqrt(1 - (15/64)) = sqrt(49/64) = 7/8.Similarly, cos C = sqrt(1 - sin^2 C) = sqrt(1 - (15/16)) = sqrt(1/16) = 1/4.So, cos A = 7/8, cos C = 1/4.Therefore, (cos A - 2 cos C)/cos B = (7/8 - 2*(1/4)) / (1/4) = (7/8 - 1/2) / (1/4) = (7/8 - 4/8) / (1/4) = (3/8) / (1/4) = (3/8)*(4/1) = 3/2.Which matches the right-hand side, so the equation holds.Therefore, all the computations are consistent.So, the perimeter is 5.**Final Answer**Ⅰ. boxed{2}Ⅱ. boxed{5}

💡Okay, so I have this problem about two ellipses, and I need to figure out what they have in common. The equations are given as:First ellipse: (x²/a²) + (y²/b²) = 1Second ellipse: (x²/a²) + (y²/b²) = k, where k > 0And the options are:A: The same eccentricityB: The same fociC: The same verticesD: The same major and minor axesAlright, let's start by recalling what an ellipse is. An ellipse is a curve on a plane surrounding two focal points such that the sum of the distances to the two focal points is constant for every point on the curve. The standard equation of an ellipse centered at the origin is (x²/a²) + (y²/b²) = 1, where 'a' is the semi-major axis and 'b' is the semi-minor axis.Now, the first ellipse is already in the standard form, so I can directly identify its major and minor axes, foci, and eccentricity.For the second ellipse, the equation is (x²/a²) + (y²/b²) = k. This isn't in the standard form because the right-hand side isn't 1. To convert it to standard form, I can divide both sides by k:(x²/(a²k)) + (y²/(b²k)) = 1So, now it's in the standard form, where the semi-major axis becomes sqrt(a²k) and the semi-minor axis becomes sqrt(b²k). Wait, is that right? Let me think.Actually, if I have (x²/A²) + (y²/B²) = 1, then A is the semi-major axis and B is the semi-minor axis. So, in this case, A² = a²k and B² = b²k. Therefore, A = a*sqrt(k) and B = b*sqrt(k).So, the second ellipse is just a scaled version of the first ellipse by a factor of sqrt(k). That means all the lengths related to the ellipse—like the major axis, minor axis, vertices, and foci—are scaled by sqrt(k).Now, let's go through each option one by one.Option A: The same eccentricity.Eccentricity (e) of an ellipse is given by e = sqrt(1 - (b²/a²)) for the first ellipse. For the second ellipse, since it's scaled by sqrt(k), the new semi-major axis is a*sqrt(k) and the new semi-minor axis is b*sqrt(k). So, the eccentricity would be sqrt(1 - ( (b²k)/(a²k) )) = sqrt(1 - (b²/a²)), which is the same as the first ellipse. So, they have the same eccentricity. That seems correct.Option B: The same foci.The foci of an ellipse are located at a distance of c from the center, where c = sqrt(a² - b²). For the first ellipse, c = sqrt(a² - b²). For the second ellipse, the semi-major axis is a*sqrt(k), so c' = sqrt( (a²k) - (b²k) ) = sqrt(k(a² - b²)) = sqrt(k)*sqrt(a² - b²). So, the distance from the center to the foci is scaled by sqrt(k). Therefore, the foci are not the same; they are further apart if k > 1 or closer if k < 1. So, they don't have the same foci.Option C: The same vertices.Vertices are the endpoints of the major axis. For the first ellipse, the vertices are at (±a, 0). For the second ellipse, the vertices are at (±a*sqrt(k), 0). So, unless k = 1, the vertices are different. Since k is just a positive number, not necessarily 1, the vertices are different. So, they don't have the same vertices.Option D: The same major and minor axes.The major axis length is 2a for the first ellipse and 2a*sqrt(k) for the second. Similarly, the minor axis is 2b for the first and 2b*sqrt(k) for the second. So, unless k = 1, the major and minor axes are different. Therefore, they don't have the same major and minor axes.So, summarizing:- A: Same eccentricity - Yes- B: Same foci - No- C: Same vertices - No- D: Same major and minor axes - NoTherefore, the correct answer should be A: The same eccentricity.But wait, let me double-check about the foci. If both ellipses are scaled by sqrt(k), then their foci are scaled by sqrt(k) as well. So, if the first ellipse has foci at (±c, 0), the second has them at (±c*sqrt(k), 0). So, unless k = 1, they are different. So, yes, they don't have the same foci.Similarly, vertices are scaled, so they are different. Major and minor axes are scaled, so they are different.Eccentricity, however, is a ratio that doesn't depend on the scaling factor. Since both ellipses have the same ratio of b/a, their eccentricities are the same. So, A is correct.I think that's solid. So, the answer is A.

💡First, I need to simplify the rotation of 630 degrees. Since a full rotation is 360 degrees, I'll subtract 360 from 630 to find the remaining degrees after one full rotation.630 - 360 = 270 degreesNext, I'll consider the initial angle of 60 degrees. Rotating an additional 270 degrees counterclockwise from this position will give me the new position of ray CA.60 + 270 = 330 degreesHowever, 330 degrees is a reflex angle, and we're interested in the acute angle. To find the acute angle, I'll subtract 330 degrees from 360 degrees.360 - 330 = 30 degreesTherefore, the new acute angle ACB measures 30 degrees.