💡Alright, so I've got this problem here about two sets, E and F, defined using cosine and sine functions with different arguments. I need to figure out the relationship between them. The options are whether one is a subset of the other, if they're equal, or if they don't intersect at all.First, let's write down the sets again to make sure I have them right.Set E is defined as:[ E = left{ x mid x = cos frac{n pi}{3}, n in mathbb{Z} right} ]Set F is defined as:[ F = left{ x mid x = sin frac{(2m - 3) pi}{6}, m in mathbb{Z} right} ]Okay, so E consists of all cosines of multiples of π/3, and F consists of all sines of (2m - 3)π/6. I need to find out what these sets actually contain.Starting with set E. The cosine function has a period of 2π, but since the argument is nπ/3, the period for this specific cosine function would be when nπ/3 increases by 2π. So, solving for n:nπ/3 + 2π = (n + k)π/3Wait, maybe that's complicating things. Instead, since the cosine function repeats every 2π, the values of cos(nπ/3) will start repeating after a certain number of n. Let's find how many distinct values there are.If I plug in n = 0, 1, 2, 3, 4, 5, 6, etc., what do I get?For n = 0:cos(0) = 1n = 1:cos(π/3) = 0.5n = 2:cos(2π/3) = -0.5n = 3:cos(π) = -1n = 4:cos(4π/3) = -0.5n = 5:cos(5π/3) = 0.5n = 6:cos(2π) = 1So, after n = 6, it starts repeating the same values: 1, 0.5, -0.5, -1, -0.5, 0.5, 1, etc. So, the distinct elements in E are {1, 0.5, -0.5, -1}.Alright, so E = {1, 0.5, -0.5, -1}.Now, moving on to set F. It's defined using sine with the argument (2m - 3)π/6.Let me simplify that argument first. Let's write it as:(2m - 3)π/6 = (2mπ/6) - (3π/6) = (mπ/3) - (π/2)So, the argument is mπ/3 - π/2.So, set F is:[ F = left{ x mid x = sin left( frac{m pi}{3} - frac{pi}{2} right), m in mathbb{Z} right} ]Hmm, maybe I can use a trigonometric identity to simplify this sine function. Remember that sin(A - B) = sinA cosB - cosA sinB.So, let's apply that identity:sin(mπ/3 - π/2) = sin(mπ/3)cos(π/2) - cos(mπ/3)sin(π/2)We know that cos(π/2) = 0 and sin(π/2) = 1, so this simplifies to:= sin(mπ/3)*0 - cos(mπ/3)*1= -cos(mπ/3)So, sin((2m - 3)π/6) = -cos(mπ/3)Therefore, set F can be rewritten as:[ F = left{ x mid x = -cos frac{m pi}{3}, m in mathbb{Z} right} ]But wait, since m is any integer, and cosine is an even function, cos(-mπ/3) = cos(mπ/3). So, the negative sign just flips the sign of the cosine values.So, F is the set of all negative cosines of mπ/3.But from set E, we already have cos(nπ/3) for all integers n, which gives us {1, 0.5, -0.5, -1}. So, if we take the negative of those, we get {-1, -0.5, 0.5, 1}.But wait, that's the same set as E, just written in a different order. So, F = {-1, -0.5, 0.5, 1} which is the same as E = {1, 0.5, -0.5, -1}.Therefore, E and F have exactly the same elements, just listed in a different order. So, E equals F.But let me double-check to make sure I didn't make a mistake.Starting with set E:- n = 0: cos(0) = 1- n = 1: cos(π/3) = 0.5- n = 2: cos(2π/3) = -0.5- n = 3: cos(π) = -1- n = 4: cos(4π/3) = -0.5- n = 5: cos(5π/3) = 0.5- n = 6: cos(2π) = 1So, E = {1, 0.5, -0.5, -1}For set F:- m = 0: sin((2*0 - 3)π/6) = sin(-π/2) = -1- m = 1: sin((2*1 - 3)π/6) = sin(-π/6) = -0.5- m = 2: sin((2*2 - 3)π/6) = sin(π/6) = 0.5- m = 3: sin((2*3 - 3)π/6) = sin(3π/6) = sin(π/2) = 1- m = 4: sin((2*4 - 3)π/6) = sin(5π/6) = 0.5- m = 5: sin((2*5 - 3)π/6) = sin(7π/6) = -0.5- m = 6: sin((2*6 - 3)π/6) = sin(9π/6) = sin(3π/2) = -1So, F = {-1, -0.5, 0.5, 1}Yes, that's the same as E. So, E and F are equal.I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be that E equals F.

💡First, I need to find four consecutive even numbers that add up to 140. I'll start by letting the smallest number be ( n ).Since the numbers are consecutive even numbers, the next three numbers will be ( n + 2 ), ( n + 4 ), and ( n + 6 ).Adding these together gives the equation:[n + (n + 2) + (n + 4) + (n + 6) = 140]Combining like terms simplifies the equation to:[4n + 12 = 140]Subtracting 12 from both sides:[4n = 128]Dividing both sides by 4:[n = 32]Now that I have the smallest number, the largest number is:[32 + 6 = 38]

💡Okay, so I have this geometry problem here. It says that inside a unit circle, there's a triangle ABC with an incircle center K. The condition given is that the product of the distances from K to each vertex, KA times KB times KC, equals 1. I need to prove that under this condition, the triangle ABC must be equilateral.First, let me make sure I understand all the terms. A unit circle has a radius of 1. The triangle ABC is inscribed in this circle, meaning all its vertices lie on the circumference. The incircle of a triangle is the largest circle that fits inside the triangle, tangent to all three sides. The center of this incircle is called the incenter, which is point K here.So, K is the incenter of triangle ABC. The incenter is the intersection of the angle bisectors of the triangle. It's equidistant from all sides of the triangle, and this distance is called the inradius.Given that KA * KB * KC = 1, and the circumradius (radius of the circumscribed circle) is 1, I need to show that ABC is equilateral.Hmm, okay. Let me think about the properties of the incenter and the distances from the incenter to the vertices.I remember that in a triangle, the inradius can be expressed in terms of the area and the semiperimeter. The formula is r = A / s, where A is the area and s is the semiperimeter. But I'm not sure if that's directly helpful here since we're dealing with distances from the incenter to the vertices, not the inradius itself.Wait, maybe I can express KA, KB, and KC in terms of the triangle's angles or sides. Let me recall if there's a formula for the distance from the incenter to a vertex.Yes, I think there is a formula for the distance from the incenter to a vertex. It involves the angles of the triangle and the inradius. Let me try to recall it.I think the distance from the incenter to vertex A can be expressed as KA = r / sin(A/2), where r is the inradius and A is the angle at vertex A. Similarly, KB = r / sin(B/2) and KC = r / sin(C/2). Is that correct?Let me verify this. In triangle ABC, the inradius r is related to the distances from the incenter to the vertices. If I consider triangle AKB, for example, it's a triangle with sides KA, KB, and AB. The angle at K is 90 degrees plus half of angle C, right? Because the incenter makes angles of 90 degrees plus half the opposite angle with each vertex.Wait, maybe I should use the Law of Sines on triangle AKB. Let's see.In triangle AKB, the angles at A and B are half of angles A and B of triangle ABC, respectively. The angle at K is 180 degrees minus (A/2 + B/2), which simplifies to 90 degrees plus C/2.So, using the Law of Sines on triangle AKB:KA / sin(angle at B) = KB / sin(angle at A) = AB / sin(angle at K)So, KA / sin(B/2) = KB / sin(A/2) = AB / sin(90 + C/2)Since sin(90 + C/2) is cos(C/2), we can write:KA = (AB * sin(B/2)) / cos(C/2)Similarly, KB = (AB * sin(A/2)) / cos(C/2)But AB is a side of triangle ABC. Since ABC is inscribed in a unit circle, AB can be expressed using the Law of Sines as AB = 2 * R * sin(C), where R is the circumradius, which is 1 here. So AB = 2 sin(C).Substituting back into KA:KA = (2 sin(C) * sin(B/2)) / cos(C/2)Similarly, KB = (2 sin(C) * sin(A/2)) / cos(C/2)Hmm, this seems a bit complicated. Maybe there's a better way to express KA, KB, and KC.Wait, earlier I thought KA = r / sin(A/2). Let me see if that's consistent with what I just derived.If KA = r / sin(A/2), then r = KA * sin(A/2). Similarly, r = KB * sin(B/2) and r = KC * sin(C/2). So, KA * sin(A/2) = KB * sin(B/2) = KC * sin(C/2) = r.But in my previous step, I have KA = (2 sin(C) * sin(B/2)) / cos(C/2). If I set this equal to r / sin(A/2), then:(2 sin(C) * sin(B/2)) / cos(C/2) = r / sin(A/2)So, r = (2 sin(C) * sin(B/2) * sin(A/2)) / cos(C/2)Hmm, not sure if this helps directly. Maybe I should consider that in a triangle, the sum of angles is 180 degrees, so A + B + C = 180. Therefore, A/2 + B/2 + C/2 = 90 degrees.Also, since the triangle is inscribed in a unit circle, the circumradius R is 1. The formula for the inradius r is r = (a + b - c)/2 * tan(C/2), but I might be mixing things up.Wait, actually, the formula for the inradius is r = (Area) / s, where s is the semiperimeter. The area can also be expressed as (a b c) / (4 R), where R is the circumradius. Since R = 1, the area is (a b c)/4.So, r = (a b c) / (4 s), where s = (a + b + c)/2.But I'm not sure if this is helpful yet.Let me go back to the original condition: KA * KB * KC = 1.If I can express KA, KB, KC in terms of the angles or sides, maybe I can find a relationship that forces the triangle to be equilateral.Earlier, I thought KA = r / sin(A/2). If that's the case, then:KA * KB * KC = (r / sin(A/2)) * (r / sin(B/2)) * (r / sin(C/2)) = r^3 / (sin(A/2) sin(B/2) sin(C/2)) = 1So, r^3 / (sin(A/2) sin(B/2) sin(C/2)) = 1Therefore, r^3 = sin(A/2) sin(B/2) sin(C/2)Hmm, interesting. So, the inradius cubed equals the product of the sines of half-angles.I wonder if there's a known formula for sin(A/2) sin(B/2) sin(C/2). Let me recall.Yes, in triangle geometry, there's a formula that relates sin(A/2) sin(B/2) sin(C/2) to the inradius and the circumradius.Specifically, sin(A/2) sin(B/2) sin(C/2) = r / (4 R)Since R = 1 here, this simplifies to sin(A/2) sin(B/2) sin(C/2) = r / 4So, substituting back into our earlier equation:r^3 = r / 4Multiply both sides by 4:4 r^3 = rSubtract r from both sides:4 r^3 - r = 0Factor out r:r (4 r^2 - 1) = 0So, either r = 0 or 4 r^2 - 1 = 0r = 0 is not possible since the inradius can't be zero in a valid triangle.So, 4 r^2 - 1 = 0 => r^2 = 1/4 => r = 1/2So, the inradius r is 1/2.Therefore, from sin(A/2) sin(B/2) sin(C/2) = r / 4 = (1/2) / 4 = 1/8So, sin(A/2) sin(B/2) sin(C/2) = 1/8Now, I need to find the angles A, B, C such that their halves' sines multiply to 1/8.Also, since A + B + C = π, we have A/2 + B/2 + C/2 = π/2.Let me denote x = A/2, y = B/2, z = C/2. Then x + y + z = π/2, and sin x sin y sin z = 1/8.I need to find x, y, z such that their sum is π/2 and their sines multiply to 1/8.I recall that for positive numbers x, y, z with x + y + z = constant, the product sin x sin y sin z is maximized when x = y = z.Is that true? Let me think.Yes, by the AM-GM inequality, the product is maximized when all variables are equal, given a fixed sum.So, if x = y = z, then x = y = z = π/6.So, sin(π/6) = 1/2, so sin x sin y sin z = (1/2)^3 = 1/8.Which is exactly the value we have.Therefore, the maximum of sin x sin y sin z is 1/8, achieved when x = y = z = π/6.Hence, the only solution is x = y = z = π/6, which implies A = B = C = π/3.Therefore, triangle ABC is equilateral.Wait, let me make sure I didn't skip any steps.We started with KA * KB * KC = 1.Expressed KA, KB, KC in terms of r and the half-angles.Used the formula sin(A/2) sin(B/2) sin(C/2) = r / (4 R) = r / 4.Substituted into the equation to get r^3 = r / 4, leading to r = 1/2.Then, using the identity sin(A/2) sin(B/2) sin(C/2) = 1/8, and knowing that the maximum of this product is 1/8 when all angles are equal, concluded that A = B = C = 60 degrees.Therefore, triangle ABC is equilateral.Yes, that seems solid.I think I covered all the necessary steps and didn't make any logical jumps that are unclear. I used known formulas and properties of triangle centers and trigonometric identities to arrive at the conclusion.So, I'm confident that the proof is correct.

💡Okay, so I have this problem here about a rational function, and I need to figure out the minimum degree that the denominator polynomial p(x) must have for the function to have a horizontal asymptote. The function is given as (-3x^6 + 5x^4 - 4x^3 + 2)/p(x). Hmm, let me think about this step by step.First, I remember that a rational function is just a ratio of two polynomials. In this case, the numerator is a polynomial of degree 6 because the highest power of x is x^6. The denominator is p(x), which is another polynomial, but I don't know its degree yet. That's what I need to find.Now, what's a horizontal asymptote? If I recall correctly, a horizontal asymptote is a horizontal line that the graph of the function approaches as x goes to positive or negative infinity. The rules for horizontal asymptotes depend on the degrees of the numerator and the denominator.Let me try to remember the rules. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, the horizontal asymptote is y = a/b, where a and b are the leading coefficients of the numerator and denominator, respectively. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote; instead, there might be an oblique or slant asymptote.So, applying this to my problem, the numerator has a degree of 6. I need the function to have a horizontal asymptote, so I need to make sure that either the degree of the denominator is equal to or greater than the degree of the numerator.Wait, if the degree of the denominator is equal to the numerator, we get a horizontal asymptote at y = a/b. If the denominator's degree is higher, the horizontal asymptote is y = 0. But in both cases, there is a horizontal asymptote. So, the minimum degree that p(x) must have is 6 because if p(x) is degree 6, then the degrees are equal, and we have a horizontal asymptote. If p(x) is of a higher degree, say 7 or more, we still have a horizontal asymptote, but the question asks for the minimum degree, so 6 is the answer.Let me double-check. If p(x) is degree 5, then the numerator is degree 6, which is higher. That would mean there's no horizontal asymptote, right? Instead, there would be an oblique asymptote. So, p(x) needs to be at least degree 6 to ensure that the function has a horizontal asymptote.Another way to think about it is by looking at the behavior of the function as x approaches infinity. For very large x, the highest degree terms dominate. So, the function behaves like (-3x^6)/p(x). If p(x) is degree 6, then it behaves like (-3x^6)/(ax^6) = -3/a, which is a constant, hence a horizontal asymptote. If p(x) is degree 7, it behaves like (-3x^6)/(ax^7) = -3/(a x), which tends to 0 as x approaches infinity, so y=0 is the horizontal asymptote. If p(x) is degree 5, it behaves like (-3x^6)/(ax^5) = -3x/a, which goes to infinity, so no horizontal asymptote.Therefore, the minimum degree p(x) must have is 6. I think that makes sense. So, p(x) needs to be at least a 6th-degree polynomial to ensure the function has a horizontal asymptote.Just to be thorough, let me consider specific examples. If p(x) is x^6, then the function simplifies to (-3x^6 + ...)/x^6, which as x approaches infinity, the lower degree terms become negligible, and it behaves like -3x^6/x^6 = -3, so y = -3 is the horizontal asymptote. If p(x) is x^7, then it behaves like (-3x^6)/x^7 = -3/x, which approaches 0, so y=0 is the horizontal asymptote. If p(x) is x^5, then it behaves like (-3x^6)/x^5 = -3x, which goes to infinity, so no horizontal asymptote.So, yeah, p(x) needs to be at least degree 6. That seems to cover all the cases. I don't think I missed anything. The key is that the degree of the denominator must be greater than or equal to the degree of the numerator for a horizontal asymptote to exist. Since the numerator is degree 6, p(x) must be degree 6 or higher. The minimum is 6.I wonder if there's a way to generalize this. Suppose the numerator is degree n, then the denominator needs to be degree m where m ≥ n for a horizontal asymptote. If m < n, there's no horizontal asymptote. If m = n, the horizontal asymptote is the ratio of leading coefficients. If m > n, the horizontal asymptote is y=0. So, in this problem, n=6, so m must be at least 6.Another thought: sometimes people confuse horizontal asymptotes with vertical asymptotes. Vertical asymptotes occur where the denominator is zero and the numerator isn't, but that's a different topic. Here, we're specifically talking about horizontal asymptotes, which depend on the degrees of the polynomials.Also, it's interesting to note that even if p(x) has a higher degree, say 7, 8, etc., the function will still have a horizontal asymptote, but it will just be y=0 instead of some non-zero constant. So, the minimum degree is 6, but higher degrees are also acceptable if you want a different horizontal asymptote.Wait, could there be a case where p(x) is degree 6 but the horizontal asymptote isn't just a simple ratio? For example, if p(x) has multiple terms, does that affect the horizontal asymptote? Let me think. No, because as x approaches infinity, the highest degree term dominates, so regardless of the other terms in p(x), the leading term will determine the behavior. So, even if p(x) is something like x^6 + x^5 + x^4 + ... + 1, as x becomes very large, the x^6 term will dominate, and the function will behave like (-3x^6)/x^6 = -3, so y = -3 is still the horizontal asymptote.Similarly, if p(x) is 2x^6 + 3x^5 + 4x^4 + ..., the leading term is 2x^6, so the function behaves like (-3x^6)/(2x^6) = -3/2, so y = -3/2 is the horizontal asymptote. So, the specific coefficients don't affect whether a horizontal asymptote exists, just its value.Therefore, as long as p(x) is degree 6 or higher, the function will have a horizontal asymptote. The exact value of the asymptote depends on the leading coefficients, but the question only asks about the existence of a horizontal asymptote, not its specific value.So, to sum up, the minimum degree that p(x) must have is 6. If p(x) is degree 6, the function will have a horizontal asymptote at y = (leading coefficient of numerator)/(leading coefficient of denominator). If p(x) is higher than 6, the horizontal asymptote will be y=0. Either way, there is a horizontal asymptote, so the minimum degree is 6.I think I've covered all the bases here. I've considered different scenarios, checked the behavior at infinity, thought about specific examples, and even considered how the coefficients affect the asymptote. All of this leads me to conclude that the minimum degree for p(x) is indeed 6.Just to recap:- Numerator degree: 6- Denominator degree: m- For horizontal asymptote: m ≥ 6- Minimum m: 6Yep, that's solid. I don't think I made any mistakes in my reasoning. It all lines up with what I know about rational functions and their asymptotes.One last thing: sometimes people might confuse horizontal asymptotes with other types of asymptotes, like slant or curved asymptotes. But in this case, since we're dealing with polynomials, slant asymptotes only occur when the degree of the numerator is exactly one more than the denominator. So, if p(x) were degree 5, the function would have a slant asymptote, not a horizontal one. That reinforces the idea that p(x) needs to be at least degree 6 for a horizontal asymptote.Alright, I think I'm confident now. The minimum degree of p(x) is 6.**Final Answer**The minimum degree that ( p(x) ) must have is boxed{6}.

💡Let's solve the problem step by step.1. **Understanding the Problem:** - There are three candidates: Devin, Elena, and Fiona. - Fiona received (frac{4}{15}) of the total votes. - Elena received 15 votes more than Fiona. - Devin received twice as many votes as Elena. - We need to find the total number of votes cast.2. **Setting Up Variables:** - Let the total number of votes be ( V ). - Votes received by Fiona: ( frac{4}{15}V ). - Votes received by Elena: ( frac{4}{15}V + 15 ). - Votes received by Devin: ( 2 times left( frac{4}{15}V + 15 right) ).3. **Summing Up the Votes:** - Total votes ( V ) is the sum of votes received by all three candidates: [ V = frac{4}{15}V + left( frac{4}{15}V + 15 right) + 2 times left( frac{4}{15}V + 15 right) ]4. **Simplifying the Equation:** - Combine like terms: [ V = frac{4}{15}V + frac{4}{15}V + 15 + frac{8}{15}V + 30 ] [ V = left( frac{4}{15} + frac{4}{15} + frac{8}{15} right)V + 45 ] [ V = frac{16}{15}V + 45 ]5. **Solving for ( V ):** - Subtract (frac{16}{15}V) from both sides: [ V - frac{16}{15}V = 45 ] [ frac{15}{15}V - frac{16}{15}V = 45 ] [ -frac{1}{15}V = 45 ] - Multiply both sides by (-15): [ V = 45 times (-15) = -675 ] - Since the number of votes can't be negative, we take the absolute value: [ V = 675 ]6. **Conclusion:** - The total number of votes cast in the election is ( 675 ).**Final Answer:**(boxed{textbf{(C)} 675})

💡To determine how many whole numbers are between (sqrt{50}) and (sqrt{200}), we start by estimating the values of these square roots.1. **Estimate (sqrt{50}):** - We know that (sqrt{49} = 7) and (sqrt{64} = 8). - Therefore, (sqrt{50}) is slightly more than 7.2. **Estimate (sqrt{200}):** - We know that (sqrt{196} = 14) and (sqrt{225} = 15). - Therefore, (sqrt{200}) is slightly more than 14 but less than 15.3. **Identify the range of whole numbers:** - The smallest whole number greater than (sqrt{50}) is 8. - The largest whole number less than (sqrt{200}) is 14.4. **Count the whole numbers from 8 to 14:** - The whole numbers in this range are 8, 9, 10, 11, 12, 13, and 14. - To find the number of these integers, we subtract the smallest from the largest and add one: [ 14 - 8 + 1 = 7 ]Thus, there are 7 whole numbers between (sqrt{50}) and (sqrt{200}).(boxed{text{C}})