💡First, I need to evaluate the expression (3 cdot 8 cdot 9 + 18 div 3 - 2^3) by following the order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).1. **Exponents**: Calculate (2^3). [ 2^3 = 8 ]2. **Multiplication**: Compute (3 cdot 8 cdot 9). [ 3 cdot 8 = 24 ] [ 24 cdot 9 = 216 ]3. **Division**: Solve (18 div 3). [ 18 div 3 = 6 ]4. **Addition and Subtraction**: Combine the results from the previous steps. [ 216 + 6 - 8 = 222 - 8 = 214 ]By carefully following each step in the order of operations, I arrive at the final answer.

💡Alright, so I have this function here: f(x) = sqrt(x² - 2x - 3). And I need to figure out where this function is monotonically increasing. The options are A: (-∞, 1], B: [3, +∞), C: (-∞, -1], and D: [1, +∞). Hmm, okay, let me break this down step by step.First, I remember that for a function to be increasing, its derivative should be positive in that interval. So, maybe I should start by finding the derivative of f(x). But before that, I should also consider the domain of the function because the square root requires the expression inside to be non-negative. So, let's solve x² - 2x - 3 ≥ 0.To solve x² - 2x - 3 ≥ 0, I can factor the quadratic. Let me see, what two numbers multiply to -3 and add to -2? Hmm, that would be -3 and +1. So, the quadratic factors as (x - 3)(x + 1) ≥ 0. Now, to find where this product is non-negative, I can use a sign chart.The critical points are x = 3 and x = -1. So, the intervals to test are (-∞, -1), (-1, 3), and (3, +∞). Let's pick test points in each interval:1. For x < -1, say x = -2: (-2 - 3)(-2 + 1) = (-5)(-1) = 5, which is positive.2. For -1 < x < 3, say x = 0: (0 - 3)(0 + 1) = (-3)(1) = -3, which is negative.3. For x > 3, say x = 4: (4 - 3)(4 + 1) = (1)(5) = 5, which is positive.So, the expression x² - 2x - 3 is non-negative in (-∞, -1] and [3, +∞). Therefore, the domain of f(x) is (-∞, -1] ∪ [3, +∞).Now, to find where f(x) is increasing, I need to find where its derivative is positive. Let's compute f'(x).f(x) = sqrt(x² - 2x - 3) can be written as (x² - 2x - 3)^(1/2). Using the chain rule, the derivative f'(x) is (1/2)(x² - 2x - 3)^(-1/2) * (2x - 2). Simplifying, f'(x) = (2x - 2)/(2 sqrt(x² - 2x - 3)) = (x - 1)/sqrt(x² - 2x - 3).So, f'(x) = (x - 1)/sqrt(x² - 2x - 3). Now, for f'(x) to be positive, the numerator and denominator must have the same sign. Since the denominator sqrt(x² - 2x - 3) is always non-negative (because square roots are non-negative), the sign of f'(x) depends on the numerator (x - 1).So, f'(x) > 0 when x - 1 > 0, which is when x > 1. But wait, we have to consider the domain of f(x), which is (-∞, -1] ∪ [3, +∞). So, x > 1 intersects with the domain at [3, +∞). Therefore, f'(x) is positive in [3, +∞), meaning f(x) is increasing there.But hold on, what about the interval (-∞, -1]? In that interval, x < -1, so x - 1 < -2, which is negative. So, f'(x) would be negative there, meaning f(x) is decreasing in (-∞, -1].So, putting it all together, f(x) is increasing on [3, +∞) and decreasing on (-∞, -1]. Therefore, the interval of monotonic increase is [3, +∞), which is option B.Wait, let me double-check. If I plug in x = 4, which is in [3, +∞), into f'(x): (4 - 1)/sqrt(16 - 8 - 3) = 3/sqrt(5), which is positive. And if I plug in x = -2, which is in (-∞, -1], into f'(x): (-2 - 1)/sqrt(4 + 4 - 3) = (-3)/sqrt(5), which is negative. So, yes, that confirms it.I think that's solid. So, the function increases from 3 onwards and decreases before -1. Therefore, the correct answer should be B: [3, +∞).**Final Answer**boxed{B}

💡Okay, so I need to prove that the square of the bisector of an angle of a triangle is equal to the product of the adjacent sides minus the product of the segments into which the bisector divides the opposite side. Hmm, that sounds a bit complicated, but let me break it down step by step.First, let me visualize the problem. Imagine a triangle ABC, where AD is the angle bisector of angle A. So, point D is on side BC, and AD divides BC into two segments, BD and DC. Let me denote the lengths of AB as c, AC as b, BD as m, and DC as n. So, the angle bisector AD is splitting BC into segments of length m and n.The goal is to show that the square of the length of AD (let's call it ℓ) is equal to the product of the adjacent sides (which are AB and AC, so that's c and b) minus the product of the segments BD and DC (which are m and n). In mathematical terms, I need to prove that:ℓ² = b * c - m * nAlright, so how can I approach this? I remember there's a theorem called the Angle Bisector Theorem which might be useful here. The Angle Bisector Theorem states that the angle bisector of a triangle divides the opposite side into segments proportional to the adjacent sides. So, in this case, BD/DC = AB/AC, which translates to m/n = c/b. That's a useful relationship.But how does that help me find ℓ²? Maybe I can use the Stewart's Theorem. Stewart's Theorem relates the lengths of the sides of a triangle to the length of a cevian (which is a line from a vertex to the opposite side). The formula is:b² * m + c² * n = a * (d² + m * n)Where a is the length of the side opposite the vertex, d is the length of the cevian, and m and n are the lengths into which the cevian divides the opposite side. In our case, the cevian is AD, so d = ℓ, and the side opposite is BC, which has length a = m + n.Wait, let me make sure I have the variables right. In Stewart's Theorem, it's usually written as:man + dad = bmb + cncWhere a, b, c are the lengths of the sides of the triangle, m and n are the segments of the side divided by the cevian, and d is the length of the cevian. So, in our case, the side being divided is BC, which is opposite angle A, so its length is a. The other sides are AB = c and AC = b. The cevian is AD, so d = ℓ, and the segments are BD = m and DC = n.So, plugging into Stewart's Theorem:a * ℓ² + m * n * a = b² * m + c² * nWait, that doesn't seem right. Let me double-check the formula. Stewart's Theorem is:b² * m + c² * n = a * (ℓ² + m * n)Yes, that's correct. So, rearranged, it gives:ℓ² = (b² * m + c² * n) / a - m * nBut I need to express ℓ² in terms of b, c, m, and n without a. Since a = m + n, maybe I can substitute that in.So, ℓ² = (b² * m + c² * n) / (m + n) - m * nHmm, that seems a bit messy. Is there a way to simplify this expression? Maybe I can factor it differently or find a common denominator.Alternatively, maybe there's another approach without using Stewart's Theorem. Let me think about using the Angle Bisector Theorem more directly. Since AD is the angle bisector, we know that m/n = c/b. So, m = (c/b) * n. And since a = m + n, we can write a = (c/b) * n + n = n(c/b + 1) = n(c + b)/b. Therefore, n = (a * b)/(b + c). Similarly, m = (a * c)/(b + c).So, now I can express m and n in terms of a, b, and c. Maybe substituting these back into the expression for ℓ² from Stewart's Theorem will help.From Stewart's Theorem, we had:ℓ² = (b² * m + c² * n) / (m + n) - m * nBut m + n = a, so:ℓ² = (b² * m + c² * n) / a - m * nNow, substituting m = (a * c)/(b + c) and n = (a * b)/(b + c):ℓ² = [b² * (a * c)/(b + c) + c² * (a * b)/(b + c)] / a - [(a * c)/(b + c)] * [(a * b)/(b + c)]Simplify the numerator of the first term:= [ (a * b² * c + a * b * c²) / (b + c) ] / a - [ (a² * b * c) / (b + c)² ]Factor out a * b * c from the numerator:= [ a * b * c (b + c) / (b + c) ] / a - [ (a² * b * c) / (b + c)² ]Simplify:= [ a * b * c ] / a - [ (a² * b * c) / (b + c)² ]= b * c - [ (a² * b * c) / (b + c)² ]Hmm, so ℓ² = b * c - [ (a² * b * c) / (b + c)² ]But I need to show that ℓ² = b * c - m * n. So, is [ (a² * b * c) / (b + c)² ] equal to m * n?Wait, let's compute m * n:m = (a * c)/(b + c)n = (a * b)/(b + c)So, m * n = (a² * b * c)/(b + c)²Which is exactly the term we have. Therefore,ℓ² = b * c - m * nWhich is what we wanted to prove!Wait, let me recap to make sure I didn't make any mistakes. I used Stewart's Theorem, expressed m and n in terms of a, b, c using the Angle Bisector Theorem, substituted them back into the expression for ℓ², and simplified to find that ℓ² equals b * c minus m * n. That seems correct.Alternatively, is there another way to approach this without Stewart's Theorem? Maybe using the formula for the length of an angle bisector. I recall there's a direct formula for the length of an angle bisector in a triangle.The formula is:ℓ = (2 * b * c * cos(A/2)) / (b + c)But that involves trigonometric functions, and I'm not sure if that would help me directly to get to ℓ² = b * c - m * n. Maybe I can square both sides:ℓ² = (4 * b² * c² * cos²(A/2)) / (b + c)²But I don't see an immediate way to relate this to m * n. Maybe it's better to stick with the Stewart's Theorem approach since it directly led me to the desired result.Another thought: perhaps using the Law of Cosines on some triangles. Let me consider triangles ABD and ADC. Since AD is the angle bisector, angles BAD and CAD are equal. Maybe I can apply the Law of Cosines on these two triangles and then combine the equations.In triangle ABD:BD² = AB² + AD² - 2 * AB * AD * cos(BAD)Similarly, in triangle ADC:DC² = AC² + AD² - 2 * AC * AD * cos(CAD)But since BAD = CAD, let's denote this angle as θ. So,m² = c² + ℓ² - 2 * c * ℓ * cosθn² = b² + ℓ² - 2 * b * ℓ * cosθNow, subtract the two equations:m² - n² = c² - b² - 2 * ℓ * cosθ (c - b)Factor the left side:(m - n)(m + n) = (c - b)(c + b) - 2 * ℓ * cosθ (c - b)Factor out (c - b) on the right side:(m - n)(m + n) = (c - b)(c + b - 2 * ℓ * cosθ)But from the Angle Bisector Theorem, we know that m/n = c/b, so m = (c/b) * n. Let me denote n = k, so m = (c/b) * k. Then, m + n = (c/b + 1) * k = (c + b)/b * k. Also, m - n = (c/b - 1) * k = (c - b)/b * k.Substituting back into the equation:[(c - b)/b * k] * [(c + b)/b * k] = (c - b)(c + b - 2 * ℓ * cosθ)Simplify the left side:[(c - b)(c + b)/b²] * k² = (c - b)(c + b - 2 * ℓ * cosθ)We can cancel out (c - b) from both sides (assuming c ≠ b, which is generally true unless the triangle is isoceles, but the formula should still hold):[(c + b)/b²] * k² = c + b - 2 * ℓ * cosθBut k = n = (a * b)/(b + c), so k² = (a² * b²)/(b + c)²Substitute back:[(c + b)/b²] * (a² * b²)/(b + c)² = c + b - 2 * ℓ * cosθSimplify:(a² * (c + b)) / (b + c)² = c + b - 2 * ℓ * cosθWhich simplifies to:a² / (b + c) = c + b - 2 * ℓ * cosθBut from earlier, we have ℓ² = b * c - m * n, and m * n = (a² * b * c)/(b + c)². So, perhaps this approach is getting too convoluted. Maybe sticking with Stewart's Theorem was the better path.Alternatively, let's consider using the formula for the length of the angle bisector. The formula is:ℓ = (2 * b * c * cos(A/2)) / (b + c)But I also know that in triangle ABC, by the Law of Cosines:cos A = (b² + c² - a²)/(2 * b * c)So, cos(A/2) can be expressed using the half-angle formula:cos(A/2) = sqrt[(1 + cos A)/2] = sqrt[(1 + (b² + c² - a²)/(2 * b * c))/2] = sqrt[(2 * b * c + b² + c² - a²)/(4 * b * c)] = sqrt[(b + c)² - a²)/(4 * b * c)]Therefore, cos(A/2) = sqrt[(b + c + a)(b + c - a)]/(2 * sqrt(b * c))Wait, that seems a bit messy, but let's proceed.So, plugging back into the formula for ℓ:ℓ = (2 * b * c * sqrt[(b + c + a)(b + c - a)]/(2 * sqrt(b * c))) / (b + c)Simplify:ℓ = (b * c * sqrt[(b + c + a)(b + c - a)] / sqrt(b * c)) / (b + c)= (sqrt(b * c) * sqrt[(b + c + a)(b + c - a)]) / (b + c)= sqrt[ b * c * (b + c + a)(b + c - a) ] / (b + c)Now, let's square both sides to get ℓ²:ℓ² = [ b * c * (b + c + a)(b + c - a) ] / (b + c)²Simplify the numerator:= b * c * [ (b + c)² - a² ] / (b + c)²= b * c * [ (b + c)² - a² ] / (b + c)²= b * c * [1 - a² / (b + c)² ]But from earlier, we have ℓ² = b * c - m * n, and m * n = (a² * b * c)/(b + c)². So,ℓ² = b * c - (a² * b * c)/(b + c)²Which matches the expression we derived earlier. Therefore, this confirms that ℓ² = b * c - m * n.So, whether I use Stewart's Theorem or the angle bisector length formula, I arrive at the same conclusion. Therefore, the square of the angle bisector is indeed equal to the product of the adjacent sides minus the product of the segments into which the bisector divides the opposite side.I think I've covered multiple approaches and verified the result through different methods, so I'm confident that the proof is correct.

💡First, I need to determine the initial range of the number of shares Mr. Sami owns. The range is the difference between the largest and smallest numbers of shares.The initial shares are:- Stock v: 68- Stock w: 112- Stock x: 56- Stock y: 94- Stock z: 45The largest number of shares is 112 (Stock w), and the smallest is 45 (Stock z). Therefore, the initial range is:112 - 45 = 67Next, Mr. Sami sells some shares of Stock x and buys 23 shares of Stock y. After this transaction:- Stock y increases to 94 + 23 = 117 shares.The new range increases by 14, so the new range is:67 + 14 = 81With the new range, the smallest number of shares can be calculated as:117 (new largest) - 81 (new range) = 36This means that after selling shares of Stock x, the number of shares of Stock x becomes 36. To find out how many shares were sold:56 (initial shares of Stock x) - 36 (new shares of Stock x) = 20Therefore, Mr. Sami sold 20 shares of Stock x.

💡First, I need to understand the problem. We have a group of 20 people, and we're told that the index for females exceeds the index for males by 0.3. The index is calculated as (n - k)/n, where n is the total number of people and k is the number of individuals of a specific sex.Let's denote the number of females as F and the number of males as M. Since the total number of people is 20, we have:F + M = 20The index for females is:Index_females = (20 - F) / 20Similarly, the index for males is:Index_males = (20 - M) / 20We are given that:Index_females - Index_males = 0.3Substituting the expressions for the indices:((20 - F) / 20) - ((20 - M) / 20) = 0.3Since M = 20 - F, we can substitute M in the equation:((20 - F) / 20) - (F / 20) = 0.3Combining the terms:(20 - 2F) / 20 = 0.3Multiplying both sides by 20:20 - 2F = 6Solving for F:-2F = 6 - 20-2F = -14F = 7Therefore, there are 7 females in the group.

💡First, I need to determine the typing rate of one typist. If some typists can type 20 letters in 20 minutes, I'll let ( T ) represent the number of typists in the initial group. This means each typist can type ( frac{20}{T} ) letters in 20 minutes.Next, I'll calculate how many letters one typist can type in 60 minutes (1 hour). Since 60 minutes is three times 20 minutes, each typist can type ( 3 times frac{20}{T} = frac{60}{T} ) letters in 60 minutes.Now, with 40 typists working together, the total number of letters they can type in 60 minutes is ( 40 times frac{60}{T} = frac{2400}{T} ) letters.According to the problem, 40 typists can type 240 letters in 1 hour. Therefore, I can set up the equation:[frac{2400}{T} = 240]Solving for ( T ):[T = frac{2400}{240} = 10]So, there were 10 typists in the initial group.