💡Alright, so I've got this problem about two candles. They have the same thickness and material, which probably means they burn at the same rate. The lengths are in the ratio 21:16. They start burning at the same time, and after 18 minutes, the ratio of their lengths becomes 15:11. I need to find out how many more minutes the longer candle can burn.Okay, let's break this down. First, the initial lengths are in the ratio 21:16. Let me denote the lengths as 21x and 16x, where x is some common unit of length. So, the longer candle is 21x, and the shorter one is 16x.Now, both candles are burning for 18 minutes. Since they have the same thickness and material, they should burn at the same rate. Let's assume that the burning rate is r units per minute. So, after 18 minutes, the longer candle would have burned 18r units, and the shorter one would have burned 18r units as well.So, the remaining lengths after 18 minutes would be:- Longer candle: 21x - 18r- Shorter candle: 16x - 18rAccording to the problem, after 18 minutes, the ratio of their lengths is 15:11. So, we can set up the equation:(21x - 18r) / (16x - 18r) = 15 / 11Now, I need to solve this equation to find the relationship between x and r. Let's cross-multiply to eliminate the fractions:11(21x - 18r) = 15(16x - 18r)Let's expand both sides:11 * 21x - 11 * 18r = 15 * 16x - 15 * 18rCalculating each term:231x - 198r = 240x - 270rNow, let's bring like terms to one side:231x - 240x = -270r + 198rSimplifying:-9x = -72rDivide both sides by -9:x = 8rSo, x is 8 times the burning rate r.Now, I need to find out how much longer the longer candle can burn. The longer candle's remaining length after 18 minutes is 21x - 18r. Since x = 8r, let's substitute that in:21x - 18r = 21 * 8r - 18r = 168r - 18r = 150rSo, the remaining length is 150r units. Since the burning rate is r units per minute, the time it can burn is:150r / r = 150 minutesTherefore, the longer candle can burn for 150 more minutes.

💡First, identify the letters in "EQUALS": E, Q, U, A, L, S.The sequence must start with E, followed by Q, and end with S. This leaves the letters U, A, and L to be arranged in the middle two positions.For the third position, there are 3 choices: U, A, or L.After choosing the third letter, there are 2 remaining letters for the fourth position.Multiply the number of choices for each position: 3 × 2 = 6.Therefore, there are 6 distinct sequences that meet the given criteria.

💡Okay, so I have this hyperbola equation: (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where (a > 0) and (b > 0). I need to find the eccentricity of the hyperbola given a specific condition about the distance from a focus to an asymptote being one-fourth of the focal distance.First, let me recall some basic properties of hyperbolas. The standard form is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), which opens left and right. The foci of this hyperbola are located at ((pm c, 0)), where (c^2 = a^2 + b^2). The eccentricity (e) is defined as (e = frac{c}{a}), which is always greater than 1 for hyperbolas.The asymptotes of this hyperbola are the lines (y = pm frac{b}{a}x). These are the lines that the hyperbola approaches but never touches as (x) and (y) go to infinity.Now, the problem states that the distance from one of the foci to an asymptote is one-fourth of its focal distance. I need to parse this carefully.First, the "focal distance." I think this refers to the distance between the two foci. Since the foci are at ((pm c, 0)), the distance between them is (2c). So, the focal distance is (2c).Next, the distance from a focus to an asymptote. Let's pick one focus, say ((c, 0)), and one asymptote, say (y = frac{b}{a}x). I need to calculate the distance from the point ((c, 0)) to the line (y = frac{b}{a}x).The formula for the distance from a point ((x_0, y_0)) to the line (Ax + By + C = 0) is:[d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}]So, let's write the asymptote (y = frac{b}{a}x) in standard form. Subtracting (frac{b}{a}x) from both sides gives:[-frac{b}{a}x + y = 0]So, (A = -frac{b}{a}), (B = 1), and (C = 0).Now, plugging in the point ((c, 0)):[d = frac{|left(-frac{b}{a}right)c + 1 cdot 0 + 0|}{sqrt{left(-frac{b}{a}right)^2 + 1^2}} = frac{left|frac{-bc}{a}right|}{sqrt{frac{b^2}{a^2} + 1}} = frac{frac{bc}{a}}{sqrt{frac{b^2 + a^2}{a^2}}} = frac{frac{bc}{a}}{frac{sqrt{a^2 + b^2}}{a}} = frac{bc}{sqrt{a^2 + b^2}}]But since (c^2 = a^2 + b^2), then (sqrt{a^2 + b^2} = c). So, substituting that in:[d = frac{bc}{c} = b]So, the distance from the focus ((c, 0)) to the asymptote (y = frac{b}{a}x) is (b).According to the problem, this distance (d = b) is one-fourth of the focal distance. The focal distance is (2c), so:[b = frac{1}{4} times 2c = frac{c}{2}]So, (b = frac{c}{2}), which implies (c = 2b).Now, since (c^2 = a^2 + b^2), substituting (c = 2b) gives:[(2b)^2 = a^2 + b^2 implies 4b^2 = a^2 + b^2 implies 3b^2 = a^2 implies a = bsqrt{3}]Now, the eccentricity (e) is given by (e = frac{c}{a}). We have (c = 2b) and (a = bsqrt{3}), so:[e = frac{2b}{bsqrt{3}} = frac{2}{sqrt{3}} = frac{2sqrt{3}}{3}]So, the eccentricity is (frac{2sqrt{3}}{3}).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Calculated the distance from focus to asymptote correctly? Yes, using the distance formula, ended up with (d = b).2. Focal distance is (2c), correct.3. Given (d = frac{1}{4} times 2c), so (b = frac{c}{2}), correct.4. Then (c = 2b), substituted into (c^2 = a^2 + b^2), got (4b^2 = a^2 + b^2), so (3b^2 = a^2), correct.5. Then (a = bsqrt{3}), correct.6. Eccentricity (e = c/a = 2b / (bsqrt{3}) = 2/sqrt{3} = 2sqrt{3}/3), correct.Everything seems to check out. So, I think the eccentricity is (frac{2sqrt{3}}{3}).**Final Answer**The eccentricity of the hyperbola is boxed{dfrac{2sqrt{3}}{3}}.

💡Okay, so I have this formula from the theory of concave mirrors: x = r frac{sin^2 frac{alpha}{2}}{cos x}And I need to show that for small angles α, this formula approximates to:x = frac{r alpha^2}{4}Alright, let's start by understanding what each term represents. I know that in optics, especially with concave mirrors, 'r' usually stands for the radius of curvature of the mirror. 'x' is likely the distance from the mirror to the focal point or some other significant point, and α is probably the angle of incidence or something similar.Now, the problem mentions that for small α, the formula simplifies to x = (r α²)/4. So, I need to see how the original formula can be approximated for small angles.First, I recall that for small angles, trigonometric functions can be approximated. Specifically, sin(θ) ≈ θ and cos(θ) ≈ 1 - θ²/2 when θ is small (in radians). These approximations come from the Taylor series expansions of sine and cosine around θ = 0.Given that α is small, let's apply these approximations to the original formula.Starting with the numerator: sin²(α/2). Using the approximation sin(θ) ≈ θ, we can say sin(α/2) ≈ α/2. Therefore, sin²(α/2) ≈ (α/2)² = α²/4.Now, the denominator is cos(x). If x is also small (which it probably is if α is small, since x depends on α), then we can approximate cos(x) ≈ 1 - x²/2. However, if x is very small, maybe we can approximate cos(x) ≈ 1. But I need to be careful here because x is on both sides of the equation.Wait, the original equation is x = r * [sin²(α/2)] / cos(x). So, x is expressed in terms of cos(x). That seems a bit tricky because x appears on both sides. Maybe for small angles, x is also small, so cos(x) ≈ 1. Let's test that assumption.If cos(x) ≈ 1, then the equation simplifies to x ≈ r * sin²(α/2). And since sin(α/2) ≈ α/2, sin²(α/2) ≈ α²/4. Therefore, x ≈ r * (α²/4) = (r α²)/4.That seems to match the approximation given in the problem. But I need to make sure that this approximation is valid. Specifically, I need to check if x is indeed small enough for cos(x) ≈ 1 to be a good approximation.If x is proportional to α², and α is small, then x would be even smaller, making cos(x) ≈ 1 a better approximation. So, it seems reasonable to use cos(x) ≈ 1 in this case.But just to be thorough, let's consider the next term in the Taylor series for cos(x). If cos(x) ≈ 1 - x²/2, then the original equation becomes:x = r * [sin²(α/2)] / (1 - x²/2)If x is small, then x²/2 is very small, and we can approximate 1 / (1 - x²/2) ≈ 1 + x²/2 using the binomial approximation (1/(1 - ε) ≈ 1 + ε for small ε).So, substituting back, we get:x ≈ r * sin²(α/2) * (1 + x²/2)But since x is small, x² is negligible compared to 1, so this term can be ignored for the first approximation. Therefore, x ≈ r * sin²(α/2), which again leads us to x ≈ (r α²)/4.So, even when considering the next term in the approximation, the leading term is still (r α²)/4, and higher-order terms are negligible for small α.Therefore, the approximation x ≈ (r α²)/4 is justified for small angles α.As for the units of measurement for angle α, in physics and optics, angles are typically measured in radians when dealing with such trigonometric approximations. This is because the small-angle approximations (sin θ ≈ θ, cos θ ≈ 1 - θ²/2) are only valid when θ is in radians. If α were in degrees, these approximations wouldn't hold as neatly, and the formula wouldn't simplify as nicely.So, α is expressed in radians.**Final Answer**The angle α is expressed in radians, and the approximation is justified as shown. The final result is boxed{x = frac{r alpha^2}{4}}.

💡Okay, so I have this problem about vectors. It says that the magnitude of vector a is 1, and the magnitude of vector b is 2. The angle between them is 60 degrees. There are two parts: first, I need to find the dot product of a and b, and the magnitude of a + 3b. Second, I have to prove that (a - b) is perpendicular to a.Alright, let's start with part (1). The dot product of two vectors is given by the formula:a · b = |a| |b| cosθWhere |a| is the magnitude of vector a, |b| is the magnitude of vector b, and θ is the angle between them. So, plugging in the values we have:|a| = 1, |b| = 2, θ = 60 degrees.First, I need to calculate cos(60 degrees). I remember that cos(60°) is 0.5. So, substituting these values into the formula:a · b = 1 * 2 * 0.5Let me compute that: 1 times 2 is 2, and 2 times 0.5 is 1. So, the dot product a · b is 1.Alright, that seems straightforward. Now, the next part is finding the magnitude of a + 3b. To find the magnitude of a vector, I can use the formula:|a + 3b| = sqrt( (a + 3b) · (a + 3b) )This is because the magnitude squared is equal to the dot product of the vector with itself. So, expanding this dot product:(a + 3b) · (a + 3b) = a · a + 2 * a · 3b + (3b) · (3b)Wait, no, that's not quite right. Let me correct that. When expanding (a + 3b) · (a + 3b), it should be:a · a + a · 3b + 3b · a + 3b · 3bBut since the dot product is commutative, a · 3b is the same as 3b · a, so we can simplify this:= a · a + 2 * (a · 3b) + (3b) · (3b)Wait, no, that's not correct either. Let me think again. The expansion should be:(a + 3b) · (a + 3b) = a · a + a · 3b + 3b · a + 3b · 3bWhich simplifies to:= |a|² + 3(a · b) + 3(a · b) + 9|b|²Because a · a is |a| squared, and 3b · 3b is 9|b| squared. Also, a · 3b is 3(a · b), and similarly for 3b · a.So, combining like terms:= |a|² + 6(a · b) + 9|b|²Now, substituting the known values:|a|² = 1² = 1a · b = 1 (from earlier)|b|² = 2² = 4So, plugging these in:= 1 + 6*1 + 9*4Let me compute each term:6*1 = 69*4 = 36So, adding them up:1 + 6 + 36 = 43Therefore, |a + 3b| = sqrt(43)Okay, that seems right. So, part (1) is done: a · b is 1, and |a + 3b| is sqrt(43).Now, moving on to part (2): Prove that (a - b) is perpendicular to a.I remember that two vectors are perpendicular if their dot product is zero. So, to prove (a - b) is perpendicular to a, I need to show that (a - b) · a = 0.Let's compute this dot product:(a - b) · a = a · a - b · aWhich is the same as:= |a|² - (a · b)We already know |a|² is 1, and a · b is 1.So, substituting these values:= 1 - 1 = 0Since the dot product is zero, this means that (a - b) is indeed perpendicular to a.Wait, that was straightforward. Let me double-check to make sure I didn't make any mistakes.First, the dot product formula for (a - b) · a:= a · a - b · a= |a|² - (a · b)Yes, that's correct.We have |a|² = 1, a · b = 1, so 1 - 1 = 0.Therefore, the dot product is zero, which confirms that (a - b) is perpendicular to a.So, both parts are done. I think I've got it right.

💡Alright, so I've got this question about oblique projection method and its effect on different shapes. Let me try to understand what's being asked here. The question lists four statements about how certain shapes appear when using oblique projection, and I need to determine which of these statements are correct.First, let me recall what oblique projection is. From what I remember, oblique projection is a method used in technical drawing and engineering to represent 3D objects in 2D. It's a type of parallel projection where the projectors are parallel but not perpendicular to the projection plane. This gives a somewhat realistic view of the object, especially useful for showing both front and side views in a single drawing.Now, the question mentions four specific shapes: triangle, parallelogram, square, and rhombus. It asks whether their intuitive diagrams (which I assume are their projections) under oblique projection remain the same shape. The options are:① The intuitive diagram of a triangle is a triangle;② The intuitive diagram of a parallelogram is a parallelogram;③ The intuitive diagram of a square is a square;④ The intuitive diagram of a rhombus is a rhombus.And the multiple-choice answers are:A: ①② B: ① C: ③④ D: ①②③④So, I need to figure out which of these four statements are correct when applying oblique projection.Let me start by thinking about each shape one by one.First, a triangle. In oblique projection, since it's a parallel projection, the shape should retain its general form, right? So, a triangle should still look like a triangle, just perhaps skewed or foreshortened depending on the angle of projection. So, statement ① seems correct.Next, a parallelogram. Parallelograms have opposite sides that are equal and parallel. In oblique projection, since parallel lines remain parallel (as it's a parallel projection), the projected image should still have opposite sides equal and parallel. Therefore, the intuitive diagram of a parallelogram should still be a parallelogram. So, statement ② also seems correct.Now, a square. Squares have all sides equal and all angles right angles. In oblique projection, while parallel lines remain parallel, the angles can change. Specifically, the projection might cause the right angles to become acute or obtuse, depending on the projection angle. Moreover, the lengths might not remain equal if the projection foreshortens one axis more than the other. Therefore, the intuitive diagram of a square might not remain a square; it could become a parallelogram or another shape. So, statement ③ might not be correct.Lastly, a rhombus. Rhombuses have all sides equal but angles that are not necessarily right angles. In oblique projection, since the sides are all equal and parallel, the projection should maintain the equal lengths and parallelism. However, the angles might change, but since a rhombus is defined by equal sides and opposite angles, as long as the sides remain equal and parallel, the shape should still be a rhombus. Wait, but if the angles change, does it still qualify as a rhombus? Hmm, I'm a bit confused here. Let me think again. A rhombus is a quadrilateral with all sides equal and opposite angles equal. If the projection changes the angles, but keeps the sides equal and parallel, then it should still be a rhombus. So, maybe statement ④ is correct.But wait, earlier I thought that a square might not remain a square because the angles could change. But a rhombus is more general; it doesn't require right angles. So, if the projection changes the angles but keeps the sides equal and parallel, then the rhombus remains a rhombus. So, statement ④ might be correct.But now I'm a bit torn because I initially thought that statement ③ was incorrect, but statement ④ might still be correct. Let me double-check.In oblique projection, the key properties are that parallel lines remain parallel, and the scale along the direction of projection is usually reduced (foreshortened). So, for a square, which has equal sides and right angles, the projection might make the sides unequal or the angles non-right, thus turning it into a parallelogram or another shape, not necessarily a square. So, statement ③ is likely incorrect.For a rhombus, which has equal sides but angles that are not right angles, the projection would still maintain equal sides and parallelism, so it should remain a rhombus. Therefore, statement ④ is correct.Wait, but earlier I thought that statement ④ might be correct, but now I'm more confident that it is correct. So, statements ①, ②, and ④ are correct, while statement ③ is incorrect.But looking back at the options, option A is ①②, option B is ①, option C is ③④, and option D is all four. So, if statements ①, ②, and ④ are correct, but ③ is incorrect, then none of the options match exactly. Hmm, that's confusing.Wait, maybe I made a mistake in assessing statement ④. Let me think again. A rhombus has all sides equal and opposite angles equal. In oblique projection, the sides remain equal and parallel, but the angles might change. However, as long as the sides are equal and the opposite angles are equal, it's still a rhombus. So, yes, statement ④ should be correct.But then, according to my analysis, statements ①, ②, and ④ are correct, but ③ is incorrect. However, none of the options include ①②④. The closest is option A: ①②, but that excludes ④.Wait, maybe I'm overcomplicating this. Let me try to recall the specific rules of oblique projection. In oblique projection, typically, one axis is foreshortened, usually the depth axis, while the other two axes are kept at full scale. This means that shapes perpendicular to the projection direction might be foreshortened, but shapes parallel to the projection plane remain unchanged.So, for a triangle, if it's parallel to the projection plane, it remains a triangle. If it's not, it might be foreshortened, but still, a triangle remains a triangle because the projection preserves the number of sides and the parallelism of lines.Similarly, a parallelogram, being a four-sided figure with two pairs of parallel sides, would still have those properties preserved under oblique projection, so it remains a parallelogram.A square, which is a special case of a parallelogram with equal sides and right angles, might lose its right angles under projection, turning it into a general parallelogram, not necessarily a square.A rhombus, which is a parallelogram with equal sides, would still have equal sides and parallel sides under projection, so it remains a rhombus.Therefore, statements ①, ②, and ④ are correct, while statement ③ is incorrect.But again, looking at the options, there's no option that includes ①, ②, and ④. The closest is option A: ①②, which excludes ④.Wait, maybe I'm misunderstanding the question. It says "the intuitive diagram of a square is a square." If the square is parallel to the projection plane, then its projection would still be a square. But if it's not, then it would be a parallelogram. So, the statement is not universally true; it depends on the orientation of the square relative to the projection direction.Similarly, for a rhombus, if it's parallel to the projection plane, it remains a rhombus. If not, it might become a different shape.But the question doesn't specify the orientation of the shapes. It just says "the intuitive diagram of a square is a square," implying generally. So, unless specified otherwise, we can't assume the orientation.Therefore, statements ① and ② are always correct because triangles and parallelograms, regardless of their orientation, will project to triangles and parallelograms, respectively, under oblique projection.Statements ③ and ④ are not necessarily correct because their specific properties (equal sides and right angles for squares, equal sides and angles for rhombuses) might not be preserved unless the shape is parallel to the projection plane.Therefore, the correct statements are ① and ②, which corresponds to option A.I think I confused myself earlier by considering the possibility of rhombus remaining a rhombus, but without knowing the orientation, we can't be certain. So, it's safer to stick with statements that are always true under oblique projection, which are ① and ②.