💡Okay, so I have this problem about sets and I need to solve two parts. Let me start by understanding what each part is asking.First, the universal set U is all real numbers, which is good to know. Then, set A is defined as all x such that 1 is less than or equal to x and x is less than 5. So, A is the interval [1, 5). Set B is all x such that 2 is less than x and x is less than 8, so B is the interval (2, 8).Part (1) asks me to find A union B and the complement of A intersected with B. Let me tackle A union B first. The union of two sets includes all elements that are in either set. So, A is from 1 to 5, and B is from 2 to 8. If I combine these, the smallest number is 1, and the largest number is 8. But I need to be careful about the endpoints. Since A includes 1 and excludes 5, and B excludes 2 and includes up to 8, the union should start at 1 (included) and go up to 8 (excluded). So, A union B is [1, 8).Next, I need to find the complement of A intersected with B. The complement of A, denoted as ∁U A, would be all real numbers not in A. Since A is [1, 5), the complement is everything less than 1 and everything greater than or equal to 5. So, ∁U A is (-∞, 1) union [5, ∞). Now, intersecting this with B, which is (2, 8). So, I need to find where (-∞, 1) union [5, ∞) overlaps with (2, 8). The (-∞, 1) part doesn't overlap with (2, 8) because 1 is less than 2. The [5, ∞) part does overlap with (2, 8), specifically from 5 to 8. But since B is up to 8 (excluded), the intersection is [5, 8). So, (∁U A) ∩ B is [5, 8).Okay, that seems straightforward. Let me double-check. A union B: starts at 1 (included) and goes up to 8 (excluded). Complement of A is everything outside [1,5), so less than 1 or greater than or equal to 5. Intersecting that with B, which is (2,8), so the overlap is from 5 to 8, not including 8. Yep, that makes sense.Moving on to part (2). We have set C defined as {x | a < x ≤ a + 3}. So, C is an interval (a, a + 3]. The condition given is that C intersect A equals C. That means C is a subset of A because when you intersect C with A, you get C itself. So, C must be entirely contained within A.Given that A is [1, 5), C must lie entirely within [1, 5). So, the interval (a, a + 3] must be a subset of [1, 5). Let me write that down: (a, a + 3] ⊆ [1, 5). To find the range of a, I need to ensure that the entire interval C is within A. So, the lower bound of C, which is a, must be greater than or equal to 1. Because if a were less than 1, then part of C would be less than 1, which is outside of A. Similarly, the upper bound of C, which is a + 3, must be less than 5 because A ends at 5 (not including 5). So, a + 3 < 5.Let me write these inequalities:1. a ≥ 1 (since the lower bound of C must be at least 1)2. a + 3 < 5 (since the upper bound of C must be less than 5)Solving the second inequality: a + 3 < 5 ⇒ a < 2.So, combining both inequalities, we have a ≥ 1 and a < 2. Therefore, a is in the interval [1, 2).Wait, let me verify. If a is 1, then C is (1, 4], which is entirely within A [1,5). If a is 2, then C would be (2, 5], but since a must be less than 2, a approaches 2 from the left. So, the upper limit is just below 2, ensuring that a + 3 is just below 5. That seems correct.Is there any other condition I might have missed? Let me think. The interval C is (a, a + 3]. So, the lower bound is a, which needs to be ≥1, and the upper bound is a + 3, which needs to be ≤5. Wait, actually, the upper bound is a + 3, which must be less than 5 because A is up to 5, not including 5. So, a + 3 must be strictly less than 5, hence a < 2.But what about the lower bound? Since C is (a, a + 3], the lower bound is a, which is not included in C. But A includes 1. So, as long as a is ≥1, the lower bound of C is at least 1, which is fine because A starts at 1. However, if a is exactly 1, then C starts just above 1, which is still within A. So, a can be equal to 1 because even though 1 is included in A, C doesn't include 1, but that's okay because C is a subset of A as long as all elements of C are in A. Since A includes 1, but C doesn't have to include 1. So, a can be 1.Wait, hold on. If a is 1, then C is (1, 4]. All elements of C are in A because A is [1,5). So, even though 1 is in A, C doesn't include 1, but that's fine. So, a can be 1.Similarly, if a is just below 2, say 1.999, then C is (1.999, 4.999], which is still within A [1,5). So, a can approach 2 from below, but can't be equal to 2 because then a + 3 would be 5, which is not included in A. So, a must be less than 2.Therefore, the range of a is [1, 2). That seems correct.Let me think if there's another way to approach this. Maybe by considering the endpoints. For C to be a subset of A, the lower bound of C must be ≥ the lower bound of A, and the upper bound of C must be ≤ the upper bound of A. Since A is [1,5), the lower bound is 1 (inclusive), and the upper bound is 5 (exclusive). C is (a, a + 3], so its lower bound is a (exclusive) and upper bound is a + 3 (inclusive). To have (a, a + 3] ⊆ [1,5), we need:1. a ≥ 1 (so that the lower bound of C is not less than 1)2. a + 3 ≤ 5 (so that the upper bound of C is not greater than 5)But wait, the upper bound of C is a + 3, which is included in C. However, A does not include 5. So, actually, a + 3 must be less than 5, not less than or equal to. Because if a + 3 were equal to 5, then C would include 5, which is not in A. Therefore, a + 3 must be strictly less than 5, so a < 2.Therefore, the conditions are:1. a ≥ 12. a < 2So, combining these, a is in [1, 2). That seems consistent with my earlier conclusion.I think that's solid. I don't see any mistakes in the reasoning. So, for part (2), the range of a is [1, 2).**Final Answer**(1) (A cup B = boxed{[1, 8)}) and ((complement_{U}A) cap B = boxed{[5, 8)});(2) The range of values for (a) is (boxed{[1, 2)}).

💡First, I recognize that the sum (1 + 2 + 3 + ldots + 100) is given as (a).Next, I need to find the sum (100 + 101 + 102 + ldots + 199). I can rewrite each term in this sequence by adding 100 to the corresponding term in the sequence (1 + 2 + 3 + ldots + 99).So, the sum becomes:[100 + (100 + 1) + (100 + 2) + ldots + (100 + 99)]This can be separated into two parts:[100 times 100 + (1 + 2 + 3 + ldots + 99)]I notice that (1 + 2 + 3 + ldots + 99) is almost the same as (a), but it's missing the last term, which is 100. Therefore, I can express it as:[100 times 99 + a]Calculating (100 times 99) gives me 9900. So, the final expression is:[9900 + a]Thus, the correct answer is (9900 + a), which corresponds to option C.

💡Okay, so I have this expression to compute: (frac{1}{cos 70^circ} - frac{sqrt{3}}{sin 70^circ}). Hmm, that looks a bit complicated, but maybe I can simplify it using some trigonometric identities. Let me think about how to approach this.First, I notice that both terms have denominators involving cosine and sine of the same angle, 70 degrees. Maybe I can combine them into a single fraction. To do that, I need a common denominator. The denominators are (cos 70^circ) and (sin 70^circ), so the common denominator would be (cos 70^circ sin 70^circ). Let me try rewriting the expression:[frac{1}{cos 70^circ} - frac{sqrt{3}}{sin 70^circ} = frac{sin 70^circ - sqrt{3} cos 70^circ}{cos 70^circ sin 70^circ}]Okay, so now I have a single fraction. The numerator is (sin 70^circ - sqrt{3} cos 70^circ). I remember there's a formula for combining sine and cosine terms into a single sine or cosine function. It's something like (a sin theta + b cos theta = R sin(theta + phi)) or (R cos(theta + phi)), where (R = sqrt{a^2 + b^2}) and (phi) is some angle. Let me try that.In this case, my numerator is (sin 70^circ - sqrt{3} cos 70^circ). So, comparing to (a sin theta + b cos theta), I have (a = 1) and (b = -sqrt{3}). Let me compute (R):[R = sqrt{a^2 + b^2} = sqrt{1^2 + (-sqrt{3})^2} = sqrt{1 + 3} = sqrt{4} = 2]Okay, so (R = 2). Now, I need to find the angle (phi) such that:[sin phi = frac{b}{R} = frac{-sqrt{3}}{2} quad text{and} quad cos phi = frac{a}{R} = frac{1}{2}]Wait, (sin phi = -sqrt{3}/2) and (cos phi = 1/2). Hmm, what angle has cosine 1/2 and sine -√3/2? That would be 60 degrees below the x-axis, so 360 - 60 = 300 degrees, but since sine is negative and cosine is positive, it's in the fourth quadrant. Alternatively, in terms of reference angles, it's 60 degrees. But maybe I can express it as a negative angle: -60 degrees. Let me check:[sin(-60^circ) = -sin 60^circ = -sqrt{3}/2 quad text{and} quad cos(-60^circ) = cos 60^circ = 1/2]Yes, that works. So, (phi = -60^circ). Therefore, the numerator can be written as:[sin 70^circ - sqrt{3} cos 70^circ = 2 sin(70^circ - 60^circ) = 2 sin 10^circ]Wait, let me make sure. The formula is (a sin theta + b cos theta = R sin(theta + phi)), but in this case, since (b) is negative, it's actually (R sin(theta - |phi|)). So, yes, it becomes (2 sin(70^circ - 60^circ) = 2 sin 10^circ). That seems correct.So, the numerator simplifies to (2 sin 10^circ). Now, let's look at the denominator: (cos 70^circ sin 70^circ). I remember there's a double-angle identity for sine: (sin 2theta = 2 sin theta cos theta). Maybe I can use that here.If I let (theta = 70^circ), then:[sin 140^circ = 2 sin 70^circ cos 70^circ]So, (sin 70^circ cos 70^circ = frac{1}{2} sin 140^circ). Therefore, the denominator becomes:[cos 70^circ sin 70^circ = frac{1}{2} sin 140^circ]But (sin 140^circ) is the same as (sin(180^circ - 40^circ) = sin 40^circ), because sine is positive in the second quadrant and (sin(180^circ - theta) = sin theta). So, (sin 140^circ = sin 40^circ). Therefore, the denominator simplifies to:[frac{1}{2} sin 40^circ]Putting it all together, the original expression becomes:[frac{2 sin 10^circ}{frac{1}{2} sin 40^circ} = frac{4 sin 10^circ}{sin 40^circ}]Hmm, now I have (frac{4 sin 10^circ}{sin 40^circ}). I wonder if I can simplify this further. Maybe using more trigonometric identities. Let me think about the relationship between (sin 40^circ) and (sin 10^circ).I recall that (sin 40^circ) can be expressed using double-angle identities. Specifically, (sin 40^circ = 2 sin 20^circ cos 20^circ). Let me verify that:[sin 40^circ = 2 sin 20^circ cos 20^circ]Yes, that's correct because (sin 2theta = 2 sin theta cos theta). So, substituting this into the denominator:[frac{4 sin 10^circ}{2 sin 20^circ cos 20^circ} = frac{2 sin 10^circ}{sin 20^circ cos 20^circ}]Now, I have (frac{2 sin 10^circ}{sin 20^circ cos 20^circ}). I notice that (sin 20^circ) can also be expressed using a double-angle identity. Specifically, (sin 20^circ = 2 sin 10^circ cos 10^circ). Let me check:[sin 20^circ = 2 sin 10^circ cos 10^circ]Yes, that's correct. So, substituting this into the denominator:[frac{2 sin 10^circ}{2 sin 10^circ cos 10^circ cos 20^circ} = frac{1}{cos 10^circ cos 20^circ}]Wait, let me make sure I did that correctly. The numerator is (2 sin 10^circ), and the denominator is (2 sin 10^circ cos 10^circ cos 20^circ). So, the (2 sin 10^circ) cancels out, leaving:[frac{1}{cos 10^circ cos 20^circ}]That seems right. So, the entire expression simplifies down to (frac{1}{cos 10^circ cos 20^circ}). Hmm, is there a way to simplify this even further? Maybe using product-to-sum identities or something else?I know that (cos A cos B = frac{1}{2} [cos(A + B) + cos(A - B)]). Let me apply that here:[cos 10^circ cos 20^circ = frac{1}{2} [cos(30^circ) + cos(-10^circ)] = frac{1}{2} [cos 30^circ + cos 10^circ]]Since cosine is even, (cos(-10^circ) = cos 10^circ). So, substituting back:[frac{1}{cos 10^circ cos 20^circ} = frac{2}{cos 30^circ + cos 10^circ}]But I'm not sure if this is any simpler. Maybe it's better to leave it as (frac{1}{cos 10^circ cos 20^circ}). Alternatively, I could compute the numerical value to check if it's a known value or if it simplifies to something else.Let me compute (cos 10^circ) and (cos 20^circ) approximately:- (cos 10^circ approx 0.9848)- (cos 20^circ approx 0.9397)Multiplying them together:[0.9848 times 0.9397 approx 0.925]So, (frac{1}{0.925} approx 1.081). Let me check what the original expression evaluates to numerically to see if it matches.Compute each term:- (frac{1}{cos 70^circ} approx frac{1}{0.3420} approx 2.9238)- (frac{sqrt{3}}{sin 70^circ} approx frac{1.732}{0.9397} approx 1.847)Subtracting them:[2.9238 - 1.847 approx 1.0768]Hmm, that's approximately 1.0768, which is close to 1.081, but not exact. Maybe my approximations are a bit off, or perhaps the exact value is a known constant. Let me think.Wait, I remember that (cos 10^circ cos 20^circ) can be related to other angles. Alternatively, maybe I can express it in terms of sine or cosine of other angles. Let me try another approach.I recall that (cos 10^circ cos 20^circ) can be expressed using product-to-sum identities as I did before, but perhaps another identity can help. Alternatively, maybe using the fact that (10^circ), (20^circ), and (30^circ) are related.Wait, another idea: perhaps using the identity for (sin 3theta). I know that (sin 3theta = 3 sin theta - 4 sin^3 theta). But I'm not sure if that directly helps here.Alternatively, maybe using the identity for (cos 3theta). Let me recall:[cos 3theta = 4 cos^3 theta - 3 cos theta]If I set (theta = 10^circ), then:[cos 30^circ = 4 cos^3 10^circ - 3 cos 10^circ]But I'm not sure if that helps me here. Maybe another approach.Wait, going back to the expression (frac{1}{cos 10^circ cos 20^circ}), perhaps I can write it as (sec 10^circ sec 20^circ). But I don't know if that's helpful.Alternatively, maybe I can express it in terms of tangent. Let me think:I know that (tan theta = frac{sin theta}{cos theta}), but I'm not sure how that would help here.Wait, another idea: perhaps using the identity for (sin(A - B)). Earlier, I used that to combine the numerator, but maybe there's another way.Alternatively, maybe I can relate this to the original expression. Let me see:We had:[frac{1}{cos 70^circ} - frac{sqrt{3}}{sin 70^circ} = frac{1}{cos 10^circ cos 20^circ}]Is there a way to express (frac{1}{cos 10^circ cos 20^circ}) in terms of tangent or something else? Let me try.I know that (tan theta = frac{sin theta}{cos theta}), so maybe:[frac{1}{cos 10^circ cos 20^circ} = frac{sin 10^circ}{sin 10^circ cos 10^circ cos 20^circ}]But that seems more complicated. Alternatively, maybe using the identity for (sin(A + B)) or (sin(A - B)).Wait, another approach: perhaps using the identity for (sin 30^circ), since 30 degrees is a common angle. Let me recall that (sin 30^circ = frac{1}{2}).But I'm not sure how that connects. Alternatively, maybe using multiple-angle identities.Wait, let me think differently. Since I have (frac{1}{cos 10^circ cos 20^circ}), maybe I can write it as (sec 10^circ sec 20^circ). But I don't know if that's helpful.Alternatively, perhaps using the identity for (cos(A - B)) or (cos(A + B)). Let me try:[cos(A - B) = cos A cos B + sin A sin B][cos(A + B) = cos A cos B - sin A sin B]If I add these two equations:[cos(A - B) + cos(A + B) = 2 cos A cos B]So,[cos A cos B = frac{1}{2} [cos(A - B) + cos(A + B)]]In our case, (A = 10^circ) and (B = 20^circ), so:[cos 10^circ cos 20^circ = frac{1}{2} [cos(-10^circ) + cos 30^circ] = frac{1}{2} [cos 10^circ + cos 30^circ]]Since (cos(-10^circ) = cos 10^circ). Therefore,[frac{1}{cos 10^circ cos 20^circ} = frac{2}{cos 10^circ + cos 30^circ}]But I'm not sure if this helps. Let me compute (cos 10^circ + cos 30^circ):- (cos 10^circ approx 0.9848)- (cos 30^circ approx 0.8660)Adding them together:[0.9848 + 0.8660 approx 1.8508]So,[frac{2}{1.8508} approx 1.081]Which matches the earlier approximation. So, it seems that (frac{1}{cos 10^circ cos 20^circ} approx 1.081), which is close to the numerical value of the original expression, which was approximately 1.0768. The slight difference is likely due to rounding errors in the approximations.Therefore, it seems that the expression simplifies to (frac{1}{cos 10^circ cos 20^circ}). I don't think it simplifies further using standard trigonometric identities, so this might be the simplest form.Alternatively, I could express this in terms of tangent functions or other identities, but I don't see an immediate way to make it simpler. Therefore, I think the final simplified form is (frac{1}{cos 10^circ cos 20^circ}).**Final Answer**boxed{frac{1}{cos 10^circ cos 20^circ}}

💡First, I need to recall the formula for calculating the length of an arc in a sector. The formula is ( l = alpha r ), where ( l ) is the arc length, ( alpha ) is the central angle in radians, and ( r ) is the radius of the sector.Next, I will identify the given values from the problem. The radius ( r ) is 2 units, and the central angle ( alpha ) is ( frac{pi}{3} ) radians.Now, I will substitute these values into the arc length formula:[l = alpha r = frac{pi}{3} times 2 = frac{2pi}{3}]Therefore, the length of the arc corresponding to the given central angle is ( frac{2pi}{3} ) units.

💡First, I need to find the greatest common divisor (GCD) of 1729 and 867. The most efficient method for this is the Euclidean algorithm.I'll start by dividing the larger number, 1729, by the smaller number, 867. The remainder from this division will be used in the next step.Next, I'll take the previous divisor, 867, and divide it by the remainder obtained from the first division. I'll continue this process, replacing the larger number with the smaller number and the smaller number with the remainder from each step.I'll repeat these divisions until the remainder is zero. The last non-zero remainder will be the GCD of the two numbers.By following this systematic approach, I'll be able to determine the greatest common divisor of 1729 and 867.

💡Let's solve the problem step by step.**Problem Statement:**A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is (frac{1}{3}). When 4 black cards are added to the deck, the probability of choosing red becomes (frac{1}{4}). How many cards were in the deck originally?**Solution:**1. **Define Variables:** - Let ( r ) be the number of red cards. - Let ( b ) be the number of black cards.2. **Initial Probability:** - The probability of choosing a red card is (frac{1}{3}). - This gives the equation: [ frac{r}{r + b} = frac{1}{3} ] - Cross-multiplying: [ 3r = r + b implies 2r = b ] - So, ( b = 2r ).3. **After Adding 4 Black Cards:** - The number of black cards becomes ( b + 4 ). - The new probability of choosing a red card is (frac{1}{4}). - This gives the equation: [ frac{r}{r + (b + 4)} = frac{1}{4} ] - Substitute ( b = 2r ): [ frac{r}{r + 2r + 4} = frac{1}{4} ] [ frac{r}{3r + 4} = frac{1}{4} ] - Cross-multiplying: [ 4r = 3r + 4 implies r = 4 ]4. **Find the Number of Black Cards:** - Using ( b = 2r ): [ b = 2 times 4 = 8 ]5. **Total Number of Cards Originally:** - Total cards = ( r + b = 4 + 8 = 12 )**Final Answer:**The original deck had (boxed{12}) cards.