💡Okay, so I have this problem here: Given the complex number ( z = frac{5}{3 + 4i} ), I need to find the absolute value of its conjugate, which is ( |overline{z}| ). Hmm, complex numbers can be a bit tricky, but let me try to break this down step by step.First, I remember that the conjugate of a complex number ( a + bi ) is ( a - bi ). So, if I have ( z = frac{5}{3 + 4i} ), then the conjugate of ( z ), which is ( overline{z} ), should be ( frac{5}{3 - 4i} ). Wait, is that right? Let me think. Actually, no, that's not quite accurate. The conjugate of ( z ) isn't just changing the sign of the imaginary part in the denominator; I need to consider the entire expression.Maybe I should first simplify ( z ) before finding its conjugate. To simplify ( z = frac{5}{3 + 4i} ), I can multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of ( 3 + 4i ) is ( 3 - 4i ). So, multiplying numerator and denominator by ( 3 - 4i ):[z = frac{5}{3 + 4i} times frac{3 - 4i}{3 - 4i} = frac{5(3 - 4i)}{(3 + 4i)(3 - 4i)}]Now, let's compute the denominator first. Multiplying ( (3 + 4i)(3 - 4i) ) is a difference of squares, so it should be ( 3^2 - (4i)^2 ). Calculating that:[3^2 = 9][(4i)^2 = 16i^2 = 16(-1) = -16]So, ( 9 - (-16) = 9 + 16 = 25 ). Okay, the denominator simplifies to 25.Now, the numerator is ( 5(3 - 4i) ). Let's distribute the 5:[5 times 3 = 15][5 times (-4i) = -20i]So, the numerator becomes ( 15 - 20i ).Putting it all together, ( z = frac{15 - 20i}{25} ). I can separate this into real and imaginary parts:[z = frac{15}{25} - frac{20}{25}i = frac{3}{5} - frac{4}{5}i]Alright, so ( z ) simplifies to ( frac{3}{5} - frac{4}{5}i ). Now, to find the conjugate ( overline{z} ), I just change the sign of the imaginary part:[overline{z} = frac{3}{5} + frac{4}{5}i]Now, I need to find the absolute value of ( overline{z} ), which is ( |overline{z}| ). The absolute value of a complex number ( a + bi ) is given by ( sqrt{a^2 + b^2} ). Applying this to ( overline{z} ):[|overline{z}| = sqrt{left(frac{3}{5}right)^2 + left(frac{4}{5}right)^2}]Calculating each term:[left(frac{3}{5}right)^2 = frac{9}{25}][left(frac{4}{5}right)^2 = frac{16}{25}]Adding them together:[frac{9}{25} + frac{16}{25} = frac{25}{25} = 1]Taking the square root of 1:[sqrt{1} = 1]So, the absolute value of the conjugate of ( z ) is 1. Wait a second, that seems too straightforward. Let me double-check my steps to make sure I didn't make a mistake.Starting from ( z = frac{5}{3 + 4i} ), I multiplied numerator and denominator by ( 3 - 4i ) to get rid of the imaginary part in the denominator. That gave me ( z = frac{15 - 20i}{25} ), which simplifies to ( frac{3}{5} - frac{4}{5}i ). Taking the conjugate changes the imaginary part's sign, resulting in ( frac{3}{5} + frac{4}{5}i ). Then, calculating the absolute value using ( sqrt{a^2 + b^2} ) gave me 1.Hmm, another thought: I remember that the absolute value of a complex number and its conjugate are the same. So, ( |z| = |overline{z}| ). Maybe I could have calculated ( |z| ) directly instead of finding ( overline{z} ) first. Let me try that approach to confirm.Calculating ( |z| ) where ( z = frac{5}{3 + 4i} ). The absolute value of a quotient is the quotient of the absolute values, so:[|z| = frac{|5|}{|3 + 4i|}]We know that ( |5| = 5 ), and ( |3 + 4i| = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5 ). Therefore:[|z| = frac{5}{5} = 1]Since ( |z| = |overline{z}| ), this confirms that ( |overline{z}| = 1 ). Okay, so both methods give me the same result, which is reassuring.I think I'm confident now that the absolute value of the conjugate of ( z ) is indeed 1.

💡Alright, so I've got this problem here about a function and a triangle. Let me try to figure it out step by step. First, the function is given as f(x) = 2√3 sin x cos x - 2 cos²x. I need to find its range. Hmm, okay, trigonometric function. Maybe I can simplify it using some identities.I remember that sin(2x) = 2 sin x cos x, so maybe I can rewrite the first term. Let's see, 2√3 sin x cos x is the same as √3 sin(2x). That seems right. Now, the second term is -2 cos²x. I also recall that cos²x can be written as (1 + cos(2x))/2. So substituting that in, -2 cos²x becomes -2*(1 + cos(2x))/2, which simplifies to -1 - cos(2x). Okay, so putting it all together, f(x) = √3 sin(2x) - 1 - cos(2x).Hmm, that looks a bit messy, but maybe I can combine the sine and cosine terms into a single sine function. I remember something about expressing a sin x + b cos x as R sin(x + φ), where R is the amplitude and φ is the phase shift. Let me try that.So, for the terms √3 sin(2x) - cos(2x), I can think of this as a sin(2x) + b cos(2x), where a = √3 and b = -1. The amplitude R would be sqrt(a² + b²) = sqrt((√3)² + (-1)²) = sqrt(3 + 1) = sqrt(4) = 2. Okay, so R is 2.Now, to find the phase shift φ, I can use tan φ = b/a. Wait, actually, I think it's tan φ = b/a, but I might have to adjust the signs. Since a is positive and b is negative, φ should be in the fourth quadrant. Let's calculate φ: tan φ = (-1)/√3, which is -1/√3. So φ is -30 degrees or -π/6 radians. So, putting it all together, √3 sin(2x) - cos(2x) can be written as 2 sin(2x - π/6).Therefore, f(x) = 2 sin(2x - π/6) - 1. Now, since the sine function ranges between -1 and 1, multiplying by 2 gives us a range of -2 to 2, and then subtracting 1 shifts it down, so the range becomes from -3 to 1. So, the range of f(x) is [-3, 1]. That seems straightforward.Now, moving on to part (2). We have a triangle ABC with sides a, b, c opposite angles A, B, C respectively. We're told that f(A) = -2 and a = √3. We need to find the maximum area of triangle ABC.First, let's figure out what angle A is. Since f(A) = -2, and we have f(x) = 2 sin(2x - π/6) - 1, setting this equal to -2 gives us:2 sin(2A - π/6) - 1 = -2Adding 1 to both sides:2 sin(2A - π/6) = -1Divide both sides by 2:sin(2A - π/6) = -1/2Now, sin θ = -1/2 occurs at θ = 7π/6 and 11π/6 in the interval [0, 2π). But since A is an angle in a triangle, it must be between 0 and π. So, let's solve for 2A - π/6:2A - π/6 = 7π/6 + 2πk or 11π/6 + 2πk, where k is an integer.But since A is between 0 and π, 2A is between 0 and 2π, so 2A - π/6 is between -π/6 and 11π/6. So, the solutions within this interval are 7π/6 and 11π/6.Let's solve for A:Case 1: 2A - π/6 = 7π/62A = 7π/6 + π/6 = 8π/6 = 4π/3A = 2π/3Case 2: 2A - π/6 = 11π/62A = 11π/6 + π/6 = 12π/6 = 2πA = πBut A = π would mean the triangle is degenerate, so we discard that. Therefore, A = 2π/3.Okay, so angle A is 120 degrees. Now, we have side a = √3 opposite angle A. We need to find the maximum area of triangle ABC.The area of a triangle can be given by (1/2)ab sin C, but in this case, since we know one side and its opposite angle, maybe we can use the formula involving the sine of angle A.Alternatively, we can use the formula:Area = (1/2) * b * c * sin ABut we don't know sides b and c. However, we can relate them using the Law of Cosines. The Law of Cosines states that:a² = b² + c² - 2bc cos AWe know a = √3 and A = 2π/3, so cos A = cos(2π/3) = -1/2.Plugging in:(√3)² = b² + c² - 2bc*(-1/2)3 = b² + c² + bcSo, we have the equation b² + c² + bc = 3.We need to maximize the area, which is (1/2) * b * c * sin(2π/3). Sin(2π/3) is √3/2, so the area becomes:Area = (1/2) * b * c * (√3/2) = (√3/4) * b * cSo, to maximize the area, we need to maximize the product b * c, given that b² + c² + bc = 3.Hmm, how can we maximize b * c under this constraint? Maybe using some inequality or optimization technique.Let me consider that b and c are positive real numbers. Let's denote x = b and y = c. Then, we have:x² + y² + xy = 3We need to maximize xy.This seems like a constrained optimization problem. Maybe I can use Lagrange multipliers, but that might be overkill. Alternatively, I can express one variable in terms of the other and then find the maximum.Let me try expressing y in terms of x. From the equation:x² + y² + xy = 3This is a quadratic in y:y² + x y + (x² - 3) = 0Using the quadratic formula:y = [-x ± sqrt(x² - 4*(x² - 3))]/2= [-x ± sqrt(x² - 4x² + 12)]/2= [-x ± sqrt(-3x² + 12)]/2Since y must be real, the discriminant must be non-negative:-3x² + 12 ≥ 03x² ≤ 12x² ≤ 4|x| ≤ 2Since x is a length, x > 0, so 0 < x ≤ 2.Now, to maximize xy, let's denote P = xy. We can express P in terms of x.From the equation x² + y² + xy = 3, we can write y² + xy = 3 - x².But y² + xy = y(y + x) = (3 - x²)Hmm, not sure if that helps. Alternatively, since we have y expressed in terms of x, we can write P = x * y, where y is given by the quadratic formula above.But that might get complicated. Maybe another approach.Let me consider that x and y are positive real numbers satisfying x² + y² + xy = 3. I want to maximize P = xy.I recall that for expressions involving x² + y² + xy, sometimes substituting variables or using inequalities like AM-GM can help.Let me try to use the AM-GM inequality. But the expression x² + y² + xy is a bit tricky.Alternatively, I can consider that x² + y² + xy = 3 can be rewritten as x² + y² = 3 - xy.We know that x² + y² ≥ 2xy by AM-GM, so 3 - xy ≥ 2xyWhich implies 3 ≥ 3xySo, xy ≤ 1Therefore, the maximum value of xy is 1.So, the maximum area is (√3/4)*1 = √3/4.Wait, but does equality hold? When does x² + y² = 2xy? That happens when x = y.So, if x = y, then from x² + x² + x*x = 3 => 3x² = 3 => x² = 1 => x = 1So, x = y = 1, which satisfies the equation. Therefore, the maximum product xy is indeed 1, achieved when b = c = 1.Therefore, the maximum area is √3/4.So, putting it all together, the range of f(x) is [-3, 1], and the maximum area of triangle ABC is √3/4.**Final Answer**(1) The range of the function is boxed{[-3, 1]}.(2) The maximum area of triangle (ABC) is boxed{dfrac{sqrt{3}}{4}}.

💡Alright, so I have this problem about a sine function, f(x) = sin(2x + φ), and I need to find φ given that the graph has an axis of symmetry at x = π/8. Then, I need to figure out where the function is increasing, and finally, prove that a certain line can't be a tangent to the graph. Hmm, okay, let's take it step by step.Starting with part (I): Find φ. The function is f(x) = sin(2x + φ), and the axis of symmetry is x = π/8. I remember that for a sine function, the axis of symmetry would correspond to a peak or a trough of the sine wave. So, at x = π/8, the function should reach either its maximum or minimum value. That means f(π/8) should be equal to 1 or -1 because those are the maximum and minimum values of the sine function.Let me write that down: sin(2*(π/8) + φ) = ±1. Simplifying inside the sine function, 2*(π/8) is π/4. So, sin(π/4 + φ) = ±1. Now, when does sin(θ) equal to ±1? That's when θ is an odd multiple of π/2. So, π/4 + φ = π/2 + kπ, where k is any integer. Solving for φ, we get φ = π/2 - π/4 + kπ = π/4 + kπ.But wait, the problem states that φ is between -π and 0. So, let's find the value of k that makes φ fall into that interval. If k = 0, φ = π/4, which is positive, so that's not in our interval. If k = -1, φ = π/4 - π = -3π/4. That's within -π and 0. If k = -2, φ would be π/4 - 2π = -7π/4, which is less than -π, so that's out of our interval. So, the only valid solution is φ = -3π/4.Okay, that seems solid. So, part (I) is done, φ is -3π/4.Moving on to part (II): Determine the intervals where y = f(x) is monotonically increasing. Since we found φ, the function is now f(x) = sin(2x - 3π/4). To find where it's increasing, I need to look at its derivative.The derivative of sin(u) is cos(u)*u', so f'(x) = 2cos(2x - 3π/4). The function is increasing when the derivative is positive, so when cos(2x - 3π/4) > 0.Let me solve the inequality cos(2x - 3π/4) > 0. The cosine function is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for any integer k. So, setting up the inequality:-π/2 + 2πk < 2x - 3π/4 < π/2 + 2πk.Let me solve for x. First, add 3π/4 to all parts:-π/2 + 3π/4 + 2πk < 2x < π/2 + 3π/4 + 2πk.Simplify the left side: -π/2 is -2π/4, so -2π/4 + 3π/4 = π/4. Similarly, the right side: π/2 is 2π/4, so 2π/4 + 3π/4 = 5π/4.So, π/4 + 2πk < 2x < 5π/4 + 2πk.Divide all parts by 2:π/8 + πk < x < 5π/8 + πk.Therefore, the function is increasing on the intervals (π/8 + πk, 5π/8 + πk) for any integer k. Since the question asks for the intervals, I can write them as [π/8 + πk, 5π/8 + πk] because the function is increasing at the endpoints as well (since the derivative is zero at those points, but the function is still increasing in the neighborhood around them).So, the intervals where f(x) is increasing are [π/8 + πk, 5π/8 + πk] for all integers k.Alright, that seems to make sense. The sine function has a period of π because of the 2x inside, so the increasing intervals repeat every π units.Now, part (III): Prove that the line 5x - 2y + c = 0 does not form a tangent to the graph of y = f(x). Hmm, okay. So, first, let's write the line in slope-intercept form to find its slope.Starting with 5x - 2y + c = 0, let's solve for y:-2y = -5x - cDivide both sides by -2:y = (5/2)x + c/2.So, the slope of the line is 5/2. Now, to check if this line can be a tangent to the graph of f(x), we need to see if there exists some x where the derivative of f(x) equals 5/2.Earlier, we found that f'(x) = 2cos(2x - 3π/4). The maximum value of cos is 1, so the maximum slope of the tangent line is 2*1 = 2, and the minimum slope is -2. Therefore, the slope of any tangent line to f(x) must be between -2 and 2.But the slope of our given line is 5/2, which is 2.5, greater than 2. Therefore, there is no x where the derivative equals 5/2, so the line cannot be tangent to the graph of f(x).Wait, let me make sure. Is there any other way the line could be tangent? Maybe if we consider different c values? But the slope is fixed at 5/2, which is outside the range of possible slopes for the tangent lines. So, regardless of c, the slope is too steep to be a tangent. Therefore, the line cannot be tangent to the graph.Yeah, that seems correct. So, part (III) is proven.Overall, I think I've got all parts figured out. φ is -3π/4, the function increases on intervals [π/8 + πk, 5π/8 + πk], and the line can't be tangent because its slope is too steep.**Final Answer**(I) boxed{-dfrac{3pi}{4}}(II) The function is monotonically increasing on the intervals boxed{left[ dfrac{pi}{8} + kpi, dfrac{5pi}{8} + kpi right]} for all integers ( k ).(III) The line ( 5x - 2y + c = 0 ) does not form a tangent to the graph of ( y = f(x) ) as shown in the reasoning above.

💡Okay, so I've got this problem here: "How many apples did two men and three women buy at a certain store if the two men each bought 30 apples, 20 less than the number of apples bought by each woman?" Alright, let's try to figure this out step by step.First, I need to understand what the problem is asking. It's about finding the total number of apples bought by two men and three women. The information given is that each man bought 30 apples, and this number is 20 less than what each woman bought. So, I need to find out how many apples each woman bought and then calculate the total.Let me break it down. Each man bought 30 apples. There are two men, so together, they bought 2 times 30 apples. That should be straightforward. 2 times 30 is 60. So, the two men bought a total of 60 apples.Now, the tricky part is figuring out how many apples each woman bought. The problem says that the number of apples each man bought is 20 less than the number each woman bought. So, if each man bought 30 apples, and that's 20 less than what each woman bought, then each woman must have bought 30 plus 20 apples. That would be 50 apples per woman.Wait, let me make sure I'm interpreting this correctly. It says "20 less than the number of apples bought by each woman." So, if the man bought 30, and that's 20 less than the woman's amount, then the woman bought 30 plus 20, which is indeed 50. Okay, that makes sense.Now, there are three women, and each bought 50 apples. So, the total number of apples bought by the women is 3 times 50. Let's calculate that. 3 times 50 is 150. So, the three women bought 150 apples in total.Now, to find the total number of apples bought by both the men and the women, I need to add the apples bought by the men and the apples bought by the women together. That would be 60 apples from the men plus 150 apples from the women. Adding those together gives me 210 apples in total.Let me double-check my calculations to make sure I didn't make any mistakes. Two men buying 30 apples each: 2 times 30 is 60. Each woman bought 20 more apples than each man, so 30 plus 20 is 50. Three women buying 50 apples each: 3 times 50 is 150. Adding 60 and 150 gives 210. Yep, that seems correct.I think I've got it. The total number of apples bought by the two men and three women is 210.

💡First, I need to determine the total points Terry would have earned if he got all 25 problems correct. Since each correct answer is worth 4 points, the total points for all correct answers would be 25 multiplied by 4, which equals 100 points.Terry got 3 problems wrong, so he got 22 problems correct. The points earned from the correct answers are 22 multiplied by 4, resulting in 88 points.Terry's actual score is 85 points. To find out how many points were subtracted due to the wrong answers, I subtract the actual score from the points earned from correct answers: 88 points minus 85 points equals 3 points subtracted.Since Terry got 3 problems wrong, I divide the total subtracted points by the number of wrong answers to find out how many points were subtracted per wrong answer: 3 points divided by 3 wrong answers equals 1 point subtracted per wrong answer.

💡To determine the probability of rolling a divisor of 12 on a fair 12-sided die, I first identify all the divisors of 12. These are 1, 2, 3, 4, 6, and 12, making a total of 6 divisors.Next, I consider the total number of possible outcomes when rolling a 12-sided die, which is 12.Finally, I calculate the probability by dividing the number of favorable outcomes (divisors of 12) by the total number of possible outcomes. This gives me 6 divided by 12, which simplifies to 1/2.