💡Okay, so I have this problem where I need to find the maximum value of the function h(x) = f(x) + g(x), where f(x) is 1/cos(x) and g(x) is f(x) shifted to the right by π/3 units. The interval for x is [π/12, π/4]. The answer choices are given, so I need to figure out which one is correct.First, let me understand what shifting a function to the right by π/3 means. If I have a function f(x), shifting it to the right by π/3 would replace x with (x - π/3). So, g(x) should be 1/cos(x - π/3). That makes sense.So, h(x) = f(x) + g(x) = 1/cos(x) + 1/cos(x - π/3). Hmm, okay. So, I need to find the maximum of this sum over the interval [π/12, π/4].I remember that 1/cos(x) is the secant function, which has vertical asymptotes where cos(x) is zero. But in the interval [π/12, π/4], cos(x) is positive and decreasing, so 1/cos(x) is increasing. Similarly, for g(x) = 1/cos(x - π/3), let's see what the interval becomes when shifted.If x is in [π/12, π/4], then x - π/3 is in [π/12 - π/3, π/4 - π/3]. Let me compute that:π/12 - π/3 = π/12 - 4π/12 = -3π/12 = -π/4.π/4 - π/3 = 3π/12 - 4π/12 = -π/12.So, x - π/3 is in [-π/4, -π/12]. Cosine is positive in this interval because cosine is positive in the fourth quadrant. So, 1/cos(x - π/3) is also positive here.So, both f(x) and g(x) are positive in this interval. Now, I need to find the maximum of their sum.I wonder if I can combine these two terms somehow. Maybe using a trigonometric identity? Let me think.I recall that 1/cos(A) + 1/cos(B) can be written as [cos(B) + cos(A)] / [cos(A)cos(B)]. But I'm not sure if that helps directly. Alternatively, maybe I can express both terms in terms of a common angle.Wait, another approach: maybe express both terms using the same angle shifted by some amount. Let me see.Let me consider the angle x - π/6. Because π/3 is 2π/6, so shifting by π/3 is like shifting by 2π/6. Maybe I can write both terms in terms of x - π/6.Let me try that. Let me write x as (x - π/6) + π/6, and x - π/3 as (x - π/6) - π/6.So, f(x) = 1/cos(x) = 1/cos[(x - π/6) + π/6]Similarly, g(x) = 1/cos(x - π/3) = 1/cos[(x - π/6) - π/6]Now, I can use the cosine addition and subtraction formulas.Recall that cos(A + B) = cos A cos B - sin A sin Band cos(A - B) = cos A cos B + sin A sin B.So, let me compute cos[(x - π/6) + π/6] and cos[(x - π/6) - π/6].First, cos[(x - π/6) + π/6] = cos(x - π/6 + π/6) = cos(x). That's just f(x).Similarly, cos[(x - π/6) - π/6] = cos(x - π/6 - π/6) = cos(x - π/3). That's g(x).Wait, maybe that's not helpful. Let me think differently.Alternatively, let me denote θ = x - π/6. Then, x = θ + π/6, and x - π/3 = θ - π/6.So, f(x) = 1/cos(θ + π/6) and g(x) = 1/cos(θ - π/6).So, h(x) = 1/cos(θ + π/6) + 1/cos(θ - π/6).Now, let me compute these two terms.Using the cosine addition formula:cos(θ + π/6) = cos θ cos π/6 - sin θ sin π/6Similarly, cos(θ - π/6) = cos θ cos π/6 + sin θ sin π/6We know that cos π/6 = √3/2 and sin π/6 = 1/2.So, cos(θ + π/6) = (√3/2) cos θ - (1/2) sin θcos(θ - π/6) = (√3/2) cos θ + (1/2) sin θTherefore, f(x) = 1 / [ (√3/2) cos θ - (1/2) sin θ ]g(x) = 1 / [ (√3/2) cos θ + (1/2) sin θ ]So, h(x) = 1 / [ (√3/2) cos θ - (1/2) sin θ ] + 1 / [ (√3/2) cos θ + (1/2) sin θ ]Hmm, this seems a bit complicated, but maybe I can combine these two fractions.Let me denote A = (√3/2) cos θ and B = (1/2) sin θ.So, f(x) = 1/(A - B) and g(x) = 1/(A + B).Therefore, h(x) = 1/(A - B) + 1/(A + B) = [ (A + B) + (A - B) ] / [ (A - B)(A + B) ] = [2A] / [A² - B²]So, h(x) = 2A / (A² - B²)Now, let's compute A² and B².A = (√3/2) cos θ, so A² = (3/4) cos² θB = (1/2) sin θ, so B² = (1/4) sin² θTherefore, A² - B² = (3/4) cos² θ - (1/4) sin² θSo, h(x) = 2*(√3/2 cos θ) / [ (3/4 cos² θ - 1/4 sin² θ) ]Simplify numerator: 2*(√3/2 cos θ) = √3 cos θDenominator: (3/4 cos² θ - 1/4 sin² θ) = (3 cos² θ - sin² θ)/4So, h(x) = √3 cos θ / [ (3 cos² θ - sin² θ)/4 ] = 4√3 cos θ / (3 cos² θ - sin² θ)Hmm, okay. Let me see if I can simplify this expression further.I know that sin² θ = 1 - cos² θ, so let's substitute that into the denominator:3 cos² θ - sin² θ = 3 cos² θ - (1 - cos² θ) = 3 cos² θ - 1 + cos² θ = 4 cos² θ - 1So, denominator becomes 4 cos² θ - 1.Therefore, h(x) = 4√3 cos θ / (4 cos² θ - 1)Hmm, that's a bit simpler.So, h(x) = 4√3 cos θ / (4 cos² θ - 1)But θ is x - π/6, so θ is in what interval?Given that x ∈ [π/12, π/4], then θ = x - π/6 ∈ [π/12 - π/6, π/4 - π/6] = [-π/12, π/12]So, θ is in [-π/12, π/12], which is a small interval around 0.Therefore, cos θ is positive and decreasing as θ increases from -π/12 to π/12.So, cos θ ∈ [cos(π/12), cos(-π/12)] = [cos(π/12), cos(π/12)] because cosine is even.Wait, no. Wait, cos(-π/12) = cos(π/12). So, cos θ is in [cos(π/12), 1], since θ is between -π/12 and π/12, and cosine is maximum at 0, which is 1, and minimum at π/12 and -π/12, which is cos(π/12).So, cos θ ∈ [cos(π/12), 1]Similarly, sin θ is in [-sin(π/12), sin(π/12)]But in our expression for h(x), we have cos θ in the numerator and denominator.So, let me denote t = cos θ. Then, t ∈ [cos(π/12), 1]So, h(x) can be written as:h(t) = 4√3 t / (4 t² - 1)Now, we need to find the maximum of h(t) over t ∈ [cos(π/12), 1]So, let's analyze this function h(t).First, let's compute h(t):h(t) = 4√3 t / (4 t² - 1)We can write this as:h(t) = 4√3 t / (4 t² - 1)To find the maximum, we can take the derivative of h(t) with respect to t and set it equal to zero.So, let's compute h'(t):h'(t) = [4√3 (4 t² - 1) - 4√3 t (8 t)] / (4 t² - 1)^2Wait, let me compute it step by step.Let me denote numerator as N = 4√3 t and denominator as D = 4 t² - 1.Then, h(t) = N / DSo, h'(t) = (N' D - N D') / D²Compute N' = 4√3Compute D' = 8 tSo, h'(t) = [4√3 (4 t² - 1) - 4√3 t (8 t)] / (4 t² - 1)^2Simplify numerator:4√3 (4 t² - 1) - 32√3 t² = 16√3 t² - 4√3 - 32√3 t² = (-16√3 t² - 4√3)So, numerator = -16√3 t² - 4√3 = -4√3 (4 t² + 1)Denominator = (4 t² - 1)^2So, h'(t) = [ -4√3 (4 t² + 1) ] / (4 t² - 1)^2Now, since denominator is always positive (squared term), the sign of h'(t) is determined by the numerator.The numerator is -4√3 (4 t² + 1), which is always negative because 4 t² + 1 is positive and multiplied by -4√3.Therefore, h'(t) is negative for all t in [cos(π/12), 1], meaning h(t) is decreasing on this interval.Therefore, the maximum of h(t) occurs at the left endpoint, which is t = cos(π/12)So, the maximum value is h(cos(π/12)) = 4√3 cos(π/12) / (4 cos²(π/12) - 1)Now, let's compute this value.First, compute cos(π/12). I know that π/12 is 15 degrees, so cos(15°) = (√6 + √2)/4 ≈ 0.9659Similarly, cos²(π/12) = [(√6 + √2)/4]^2 = (6 + 2 + 2√12)/16 = (8 + 4√3)/16 = (2 + √3)/4So, 4 cos²(π/12) = 4*(2 + √3)/4 = 2 + √3Therefore, denominator = 4 cos²(π/12) - 1 = (2 + √3) - 1 = 1 + √3So, h(cos(π/12)) = 4√3 * cos(π/12) / (1 + √3)Now, let's compute 4√3 * cos(π/12):cos(π/12) = (√6 + √2)/4So, 4√3 * cos(π/12) = 4√3 * (√6 + √2)/4 = √3 (√6 + √2) = √18 + √6 = 3√2 + √6Wait, let me compute that again:√3 * √6 = √(3*6) = √18 = 3√2√3 * √2 = √6So, yes, √3 (√6 + √2) = 3√2 + √6So, numerator is 3√2 + √6Denominator is 1 + √3So, h(cos(π/12)) = (3√2 + √6) / (1 + √3)Hmm, can we simplify this expression?Let me factor numerator:3√2 + √6 = √6 (3√2/√6 + 1) = √6 (3/√3 + 1) = √6 (√3 + 1)Wait, let me see:Wait, 3√2 + √6 = √6 * (3√2 / √6 + 1) = √6 (3 / √3 + 1) = √6 (√3 + 1)Because √2 / √6 = 1/√3So, yes, 3√2 + √6 = √6 (√3 + 1)Therefore, h(cos(π/12)) = √6 (√3 + 1) / (1 + √3) = √6Because (√3 + 1) cancels out in numerator and denominator.So, h(cos(π/12)) = √6Therefore, the maximum value of h(x) is √6.Let me double-check my steps to make sure I didn't make any mistakes.1. Expressed g(x) as 1/cos(x - π/3).2. Expressed h(x) as 1/cos(x) + 1/cos(x - π/3).3. Used substitution θ = x - π/6 to rewrite h(x) in terms of θ.4. Applied cosine addition and subtraction formulas to express f(x) and g(x) in terms of θ.5. Combined the two terms into a single fraction, resulting in h(x) = 4√3 cos θ / (4 cos² θ - 1).6. Recognized that θ ∈ [-π/12, π/12], so cos θ ∈ [cos(π/12), 1].7. Let t = cos θ, transformed h(x) into h(t) = 4√3 t / (4 t² - 1).8. Took derivative h'(t), found it was negative, meaning h(t) is decreasing in t.9. Therefore, maximum occurs at t = cos(π/12).10. Calculated h(cos(π/12)) step by step, simplifying to √6.Everything seems to check out. So, the maximum value is √6, which is option B.**Final Answer**boxed{B}

💡First, I need to determine the area of both the small and large pizzas. The area of a square is calculated using the formula ( A = s^2 ), where ( s ) is the length of a side.For the small pizza with a side length of 12 inches:[A_{text{small}} = 12^2 = 144 text{ square inches}]For the large pizza with a side length of 18 inches:[A_{text{large}} = 18^2 = 324 text{ square inches}]Next, I'll calculate how much pizza each friend can buy individually with 30. Since each small pizza costs 10, a friend can buy:[frac{30}{10} = 3 text{ small pizzas}]The total area of pizza each friend can buy is:[3 times 144 = 432 text{ square inches}]Together, without pooling their money, they can buy:[2 times 432 = 864 text{ square inches}]Now, if they pool their money, they have a total of 60. Let ( x ) be the cost of one large pizza. With 60, they can buy:[frac{60}{x} text{ large pizzas}]The total area of pizza they can buy by pooling their money is:[frac{60}{x} times 324 text{ square inches}]According to the problem, pooling their money allows them to buy 36 more square inches of pizza:[frac{60}{x} times 324 = 864 + 36][frac{60}{x} times 324 = 900]Solving for ( x ):[60 times 324 = 900x][19440 = 900x][x = frac{19440}{900} = 21.6]Therefore, the large pizza costs 21.60.

💡Okay, so I have this problem where I need to select a real number x from the interval [0, 2π], and I need to find the probability that cos x is greater than or equal to √2/2. The options are A: 1/3, B: 1/4, C: 1/6, D: 1/8. Hmm, let me think about how to approach this.First, I remember that the cosine function oscillates between -1 and 1. So, cos x = √2/2 is a specific value where the cosine function is positive. I think √2/2 is approximately 0.707, which is a common value in trigonometry, especially related to 45-degree angles or π/4 radians.So, I need to find all the x in [0, 2π] where cos x is at least √2/2. That means I need to find the intervals where cos x is greater than or equal to √2/2.I recall that cos x is positive in the first and fourth quadrants, which correspond to angles between 0 to π/2 and 3π/2 to 2π. But since we're dealing with cos x ≥ √2/2, it's not just any positive value, but specifically above √2/2.Let me visualize the cosine graph. It starts at 1 when x=0, goes down to -1 at π, and back to 1 at 2π. So, the points where cos x = √2/2 are the points where the graph intersects the line y=√2/2.I think these points occur at x = π/4 and x = 7π/4. Let me verify that. Cos(π/4) is indeed √2/2, and cos(7π/4) is also √2/2 because cosine is positive in the fourth quadrant.So, the intervals where cos x is greater than or equal to √2/2 are from 0 to π/4 and from 7π/4 to 2π. That makes sense because cosine is decreasing from 0 to π, so it starts at 1 and goes down to -1, crossing √2/2 at π/4. Similarly, on the way back up from π to 2π, it crosses √2/2 again at 7π/4.Now, to find the probability, I need to calculate the length of these intervals where cos x ≥ √2/2 and then divide by the total length of the interval [0, 2π], which is 2π.So, the length of the interval from 0 to π/4 is π/4, and the length from 7π/4 to 2π is also π/4. Adding these together gives π/4 + π/4 = π/2.Therefore, the probability is (π/2) divided by (2π). Let me compute that: (π/2) / (2π) = (π/2) * (1/(2π)) = 1/4.Wait, that seems straightforward. So, the probability is 1/4, which is option B.But just to make sure I didn't make a mistake, let me double-check. The intervals where cos x is ≥ √2/2 are indeed [0, π/4] and [7π/4, 2π]. Each of these intervals has a length of π/4, so together they sum up to π/2. Dividing by the total interval length of 2π gives π/2 divided by 2π, which simplifies to 1/4. Yep, that seems correct.I was a bit confused at first about whether I was considering the right quadrants, but since cosine is positive in the first and fourth quadrants, and we're looking for where it's above √2/2, those intervals make sense. I also thought about whether there are other points where cos x equals √2/2, but no, in the interval [0, 2π], it only occurs at π/4 and 7π/4.Another way to think about it is using the unit circle. At π/4, the angle is 45 degrees, and the cosine value is √2/2. As we move away from 0 towards π/2, the cosine decreases, so it's only above √2/2 from 0 to π/4. Similarly, as we approach 2π from 7π/4, the cosine increases back to 1, so it's above √2/2 from 7π/4 to 2π.So, combining these two intervals, the total favorable length is π/2, and the probability is 1/4. Therefore, the correct answer is B: 1/4.**Final Answer**boxed{B}

💡First, I need to determine the percentage of students who were female in Dana's senior high school class. Let's denote this percentage as ( F ).Since there are 200 students in total, the number of female students can be calculated as:[text{Number of female students} = frac{F}{100} times 200]Next, according to the problem, 50% of the female students were brunettes. Therefore, the number of female brunettes is:[text{Number of female brunettes} = 0.5 times left( frac{F}{100} times 200 right)]Additionally, 50% of these female brunettes were under 5 feet tall. Thus, the number of female brunettes under 5 feet tall is:[text{Number of female brunettes under 5 feet} = 0.5 times left( 0.5 times frac{F}{100} times 200 right)]We are given that there were 30 female brunettes under 5 feet tall. Setting up the equation:[0.5 times 0.5 times frac{F}{100} times 200 = 30]Simplifying the equation:[0.25 times frac{F}{100} times 200 = 30][0.25 times F times 2 = 30][0.5 times F = 30][F = frac{30}{0.5} = 60]Therefore, 60% of the students were female.

💡Alright, so I have this problem here: A car travels at a certain speed and covers 275 miles in 5 hours. I need to find out what the speed of the car is in miles per hour. Hmm, okay, let's break this down step by step.First off, I remember that speed is generally calculated by dividing the distance traveled by the time it took to travel that distance. So, the basic formula for speed is:[ text{Speed} = frac{text{Distance}}{text{Time}} ]In this problem, the distance is 275 miles, and the time is 5 hours. So, plugging those numbers into the formula, it would look like:[ text{Speed} = frac{275 text{ miles}}{5 text{ hours}} ]Now, doing the division, 275 divided by 5. Let me see, 5 times 50 is 250, and 275 minus 250 is 25. So, 5 goes into 25 five times. That means 50 plus 5 is 55. So, the speed is 55 miles per hour.Wait a second, let me double-check that to make sure I didn't make a mistake. If the car is going 55 miles per hour, then in one hour, it would cover 55 miles. Over 5 hours, that would be 55 times 5, which is 275 miles. Yeah, that matches up with the distance given in the problem. So, that seems correct.I guess another way to think about it is to see how many miles the car travels each hour. If it goes 275 miles in 5 hours, then each hour it's covering a portion of that distance. Dividing 275 by 5 gives me that hourly rate, which is 55 miles per hour.Is there another method to verify this? Maybe using unit analysis. The distance is in miles, and the time is in hours, so miles per hour is the desired unit for speed. When I divide miles by hours, I get miles per hour, which is exactly what I need.Let me also consider if there's any possibility of the car changing speed during the trip. The problem says it travels at a certain speed, implying that the speed is constant. So, I don't need to worry about varying speeds or anything like that. It's a straightforward calculation.Just to be thorough, I'll write out the calculation again:[ frac{275 text{ miles}}{5 text{ hours}} = 55 text{ miles per hour} ]Yep, that checks out. I think I've got it right. The car's speed is 55 miles per hour.

💡Okay, so I have this geometry problem here about dihedral angles and lines in different planes. Let me try to understand what it's asking.First, the problem says that we have a dihedral angle α-l-β. I remember that a dihedral angle is the angle between two planes, right? So, α and β are two planes intersecting along the line l. Got that.Now, there are two lines, a and b. Line a is in plane α, and line b is in plane β. Both of these lines are not perpendicular to l. The question is asking about the possible relationships between lines a and b. Specifically, can they be perpendicular, parallel, or neither?The options are:A: a and b may be perpendicular, but cannot be parallelB: a and b may be perpendicular or parallelC: a and b cannot be perpendicular, but may be parallelD: a and b cannot be perpendicular, nor can they be parallelAlright, so I need to figure out if a and b can be perpendicular, parallel, or neither, given that they're not perpendicular to l.Let me start by visualizing the dihedral angle. Imagine two planes intersecting along a line l. Plane α is one side, and plane β is the other. Line a is somewhere in α, and line b is somewhere in β. Neither a nor b is perpendicular to l.First, can a and b be parallel? For two lines to be parallel, they must lie in the same plane and never intersect. But here, a is in α and b is in β, which are two different planes intersecting along l. So, unless a and b are both parallel to l, they might not be parallel.Wait, if a and b are both parallel to l, then they would be parallel to each other because they're both parallel to the same line. But the problem says that a and b are not perpendicular to l. It doesn't say anything about them not being parallel to l. So, maybe a and b can be parallel if they're both parallel to l.Let me think. If a is parallel to l, then in plane α, a doesn't intersect l. Similarly, if b is parallel to l in plane β, then b doesn't intersect l either. Since both a and b are parallel to l, they should be parallel to each other. So, yes, a and b can be parallel.Now, can a and b be perpendicular? For two lines to be perpendicular, they need to form a 90-degree angle. Since a is in α and b is in β, and the dihedral angle between α and β is some angle, maybe they can form a right angle.But wait, the dihedral angle itself might affect whether a and b can be perpendicular. If the dihedral angle is 90 degrees, then the planes α and β are perpendicular. In that case, it's easier for lines in each plane to be perpendicular. But the problem doesn't specify the dihedral angle, so it could be any angle.However, the problem states that neither a nor b is perpendicular to l. So, even if the dihedral angle is 90 degrees, if a and b are not perpendicular to l, can they still be perpendicular to each other?Let me consider the case where the dihedral angle is 90 degrees. If a is in α and not perpendicular to l, and b is in β and not perpendicular to l, can a and b still be perpendicular?Imagine plane α is the floor and plane β is a wall, intersecting along line l which is a corner. If I draw a line a on the floor not perpendicular to the corner, and a line b on the wall not perpendicular to the corner, can these two lines be perpendicular?Yes, I think they can. For example, if line a is at some angle to l on the floor, and line b is at some angle to l on the wall, their projections might still form a right angle. So, even if neither is perpendicular to l, they can still be perpendicular to each other.But wait, is that always possible? Or does it depend on the specific angles?I think it's possible in some cases, but not always. The dihedral angle might influence whether they can be perpendicular. If the dihedral angle is not 90 degrees, maybe it's harder for a and b to be perpendicular.But the problem doesn't specify the dihedral angle, so we have to consider the general case. So, in some dihedral angles, a and b can be perpendicular, and in others, maybe not. But since the problem is asking if they "may" be perpendicular, meaning it's possible in some cases, then yes, they may be perpendicular.So, putting it together, a and b can be parallel if they're both parallel to l, and they can be perpendicular in some cases. Therefore, the answer should be B: a and b may be perpendicular or parallel.Wait, but the initial thought was that if the dihedral angle is 90 degrees, then a and b can be perpendicular. But if the dihedral angle is not 90 degrees, can a and b still be perpendicular?Let me think again. Suppose the dihedral angle is θ, which is not 90 degrees. If I have line a in α and line b in β, both not perpendicular to l, can they still form a 90-degree angle?I think yes, because the angle between a and b depends on both their angles with l and the dihedral angle θ. So, even if θ is not 90 degrees, by choosing appropriate angles for a and b with respect to l, they can still be perpendicular.For example, if θ is acute, say 60 degrees, then if a and b are each at 30 degrees to l, their projections might still form a right angle. I'm not entirely sure, but I think it's possible.Therefore, regardless of the dihedral angle, as long as a and b are not perpendicular to l, they can still be arranged to be perpendicular to each other. So, yes, they may be perpendicular.And as for being parallel, as I thought earlier, if both are parallel to l, they are parallel to each other. So, that's possible too.Hence, the correct answer is B: a and b may be perpendicular or parallel.