💡Okay, so I have this problem here about an even function f(x) defined on all real numbers. It says that when x is in (-∞, 0), the derivative of f(x) is less than 2x. Then, I need to solve the inequality f(3x - 1) - f(2) > (3x - 3)(3x + 1). The options are given as A, B, C, D, with A being (-∞, -1/3) union (1, ∞), and so on.First, I remember that an even function satisfies f(-x) = f(x) for all x. So, f is symmetric about the y-axis. That might come in handy later.Next, the derivative condition: when x is negative, f'(x) < 2x. Since x is negative, 2x is also negative. So, the derivative is negative in that region. That tells me something about the behavior of f(x) when x is negative. If the derivative is negative, the function is decreasing there. But since it's an even function, what does that say about the behavior when x is positive?Well, for even functions, the derivative at x is the negative of the derivative at -x. So, f'(-x) = -f'(x). So, if f'(x) < 2x for x < 0, then for x > 0, f'(x) = -f'(-x). Let me think: if x > 0, then -x < 0, so f'(-x) < 2*(-x) = -2x. Therefore, f'(x) = -f'(-x) > 2x. So, for x > 0, the derivative is greater than 2x. That means f(x) is increasing when x is positive, and the rate of increase is more than 2x.So, putting it together: f(x) is decreasing for x < 0, with derivative less than 2x (which is negative), and increasing for x > 0, with derivative greater than 2x (which is positive). So, the function has a minimum at x=0.Now, the inequality to solve is f(3x - 1) - f(2) > (3x - 3)(3x + 1). Let me first simplify the right-hand side.(3x - 3)(3x + 1) can be expanded as 9x² + 3x - 9x - 3 = 9x² - 6x - 3. Alternatively, maybe I can factor it differently or recognize it as a difference of squares? Wait, 3x - 3 is 3(x - 1), and 3x + 1 is just 3x + 1. Not sure if that helps.Alternatively, maybe I can write (3x - 3)(3x + 1) as (3x - 1 - 2)(3x - 1 + 2). Let me check: (3x - 1 - 2)(3x - 1 + 2) = (3x - 3)(3x + 1). Yes, that's correct. So, that's a difference of squares: (3x - 1)^2 - (2)^2. So, (3x - 1)^2 - 4.So, the right-hand side is (3x - 1)^2 - 4.Therefore, the inequality becomes f(3x - 1) - f(2) > (3x - 1)^2 - 4.Let me rearrange this inequality:f(3x - 1) - (3x - 1)^2 > f(2) - 4.Hmm, interesting. Let me define a new function g(x) = f(x) - x². Then, the inequality becomes g(3x - 1) > g(2).So, if I can analyze the function g(x), maybe I can solve this inequality.Given that f(x) is even, let's see what g(x) is like. Since f(x) is even, f(-x) = f(x). So, g(-x) = f(-x) - (-x)^2 = f(x) - x² = g(x). So, g(x) is also even.Now, let's find the derivative of g(x). g'(x) = f'(x) - 2x.We know that for x < 0, f'(x) < 2x. So, g'(x) = f'(x) - 2x < 0 for x < 0. That means g(x) is decreasing on (-∞, 0).Since g(x) is even, it's symmetric about the y-axis. So, for x > 0, the behavior of g(x) should mirror that of x < 0. Since g(x) is decreasing on (-∞, 0), it must be increasing on (0, ∞). Because the derivative for x > 0 would be g'(x) = f'(x) - 2x. Earlier, we saw that for x > 0, f'(x) > 2x, so g'(x) = f'(x) - 2x > 0. So, g(x) is increasing on (0, ∞).Therefore, g(x) has a minimum at x = 0, is decreasing on (-∞, 0), and increasing on (0, ∞).So, now, the inequality is g(3x - 1) > g(2).Since g(x) is even, g(3x - 1) = g(|3x - 1|). So, the inequality becomes g(|3x - 1|) > g(2).But since g(x) is increasing on [0, ∞), because for x > 0, g is increasing, then if |3x - 1| > 2, then g(|3x - 1|) > g(2). Similarly, if |3x - 1| < 2, then g(|3x - 1|) < g(2). And if |3x - 1| = 2, then g(|3x - 1|) = g(2).Therefore, the inequality g(|3x - 1|) > g(2) is equivalent to |3x - 1| > 2.So, solving |3x - 1| > 2.This absolute value inequality splits into two cases:1. 3x - 1 > 22. 3x - 1 < -2Let's solve the first inequality:3x - 1 > 2Add 1 to both sides: 3x > 3Divide by 3: x > 1Second inequality:3x - 1 < -2Add 1 to both sides: 3x < -1Divide by 3: x < -1/3So, the solution set is x < -1/3 or x > 1.Therefore, the solution set is (-∞, -1/3) union (1, ∞), which is option A.Wait, let me double-check if I missed anything. So, we transformed the original inequality into g(|3x - 1|) > g(2), and since g is increasing for x > 0, this reduces to |3x - 1| > 2. Solving that gives x < -1/3 or x > 1. So, yes, that seems correct.I think that's the solution.

💡First, I need to determine how many nieces Uncle Dave has based on the total number of ice cream sandwiches and the number each niece receives.Uncle Dave has a total of 143 ice cream sandwiches.Each niece receives 13 ice cream sandwiches.To find the number of nieces, I will divide the total number of ice cream sandwiches by the number each niece gets.So, I divide 143 by 13.143 ÷ 13 equals 11.Therefore, Uncle Dave has 11 nieces.

💡First, I recognize that 15 degrees can be expressed as the difference between 45 degrees and 30 degrees. This allows me to use the sine difference identity: sin(A - B) = sin A cos B - cos A sin B.Next, I substitute A = 45 degrees and B = 30 degrees into the identity:sin(15°) = sin(45°)cos(30°) - cos(45°)sin(30°).I recall the exact values of the sine and cosine functions for these angles:sin(45°) = √2/2, cos(30°) = √3/2, cos(45°) = √2/2, and sin(30°) = 1/2.Substituting these values into the equation:sin(15°) = (√2/2)(√3/2) - (√2/2)(1/2).I then perform the multiplication:sin(15°) = (√6)/4 - (√2)/4.Finally, I combine the terms to obtain the simplified expression:sin(15°) = (√6 - √2)/4.

💡Okay, so I have this problem about planting trees along five different streets in a housing development. Each tree takes up one square foot of sidewalk space, and each street has a specific length and spacing between the trees. I need to figure out how many trees can be planted in total on all five streets.First, I need to understand the problem properly. Each street has a certain length, and between each tree, there's a specific spacing. So, for example, Street A is 151 feet long, and the spacing between each tree is 14 feet. That means between the first tree and the second tree, there's 14 feet of sidewalk, and so on.I think the key here is to figure out how many intervals of spacing fit into the length of each street. Since each interval corresponds to the space between two trees, the number of intervals will be one less than the number of trees. So, if I can find the number of intervals, I can add one to get the total number of trees.Let me take Street A as an example. It's 151 feet long with a spacing of 14 feet between each tree. So, if I divide the length of the street by the spacing, I should get the number of intervals. That would be 151 divided by 14. Let me do that calculation: 151 ÷ 14 equals approximately 10.79. Hmm, that's not a whole number. Since I can't have a fraction of an interval, I need to round down to the nearest whole number, which is 10.But wait, if there are 10 intervals, that means there are 11 trees because the number of trees is always one more than the number of intervals. So, for Street A, I can plant 11 trees.Let me apply the same logic to Street B. It's 210 feet long with a spacing of 18 feet. Dividing 210 by 18 gives me 11.67. Again, rounding down gives me 11 intervals, which means 12 trees.Moving on to Street C, which is 275 feet long with a spacing of 12 feet. Dividing 275 by 12 equals approximately 22.92. Rounding down gives me 22 intervals, so 23 trees.For Street D, it's 345 feet long with a spacing of 20 feet. Dividing 345 by 20 gives me 17.25. Rounding down, I get 17 intervals, which means 18 trees.Finally, Street E is 475 feet long with a spacing of 22 feet. Dividing 475 by 22 equals approximately 21.59. Rounding down gives me 21 intervals, so 22 trees.Now, to find the total number of trees, I need to add up the number of trees from each street. That would be 11 (Street A) + 12 (Street B) + 23 (Street C) + 18 (Street D) + 22 (Street E). Let me add these up step by step:11 + 12 = 2323 + 23 = 4646 + 18 = 6464 + 22 = 86So, the total number of trees that can be planted on all five streets is 86.Wait a minute, let me double-check my calculations to make sure I didn't make any mistakes. Starting with Street A: 151 ÷ 14 ≈ 10.79, which rounds down to 10 intervals, so 11 trees. That seems correct.Street B: 210 ÷ 18 ≈ 11.67, rounds down to 11 intervals, so 12 trees. Correct.Street C: 275 ÷ 12 ≈ 22.92, rounds down to 22 intervals, so 23 trees. Correct.Street D: 345 ÷ 20 = 17.25, rounds down to 17 intervals, so 18 trees. Correct.Street E: 475 ÷ 22 ≈ 21.59, rounds down to 21 intervals, so 22 trees. Correct.Adding them up: 11 + 12 + 23 + 18 + 22 = 86. Yep, that checks out.I think I've got it right. The total number of trees that can be planted on these five streets is 86.

💡First, I need to understand the given ratio of boys to girls, which is 1:2. This means for every 1 boy, there are 2 girls.Next, I'll define variables to represent the number of boys and girls. Let B be the number of boys and G be the number of girls.According to the ratio, B:G = 1:2, which implies that B = G/2.The total number of students in the school is given as 90. Therefore, B + G = 90.Substituting B with G/2 in the total equation, I get:G/2 + G = 90Combining like terms:(1/2)G + G = 90To eliminate the fraction, I'll multiply every term by 2:G + 2G = 180Combining like terms again:3G = 180Finally, I'll solve for G by dividing both sides by 3:G = 60So, there are 60 girls in the school.

💡Okay, so I need to find the equation of the tangent line to the curve y = 2x tan(x) at the point where x = π/4. Hmm, I remember that to find the equation of a tangent line, I need two things: the slope of the tangent at that point and a point through which the tangent passes. First, let me recall the general formula for the tangent line. If I have a function y = f(x), the equation of the tangent line at a point x = a is given by:y = f(a) + f’(a)(x - a)So, in this case, a is π/4. That means I need to compute f(π/4) and f’(π/4).Alright, let's start by finding f(π/4). The function is f(x) = 2x tan(x). So, plugging in x = π/4:f(π/4) = 2*(π/4)*tan(π/4)I know that tan(π/4) is 1 because tan(45 degrees) is 1. So, simplifying:f(π/4) = 2*(π/4)*1 = (2π)/4 = π/2Okay, so the point on the curve is (π/4, π/2). Got that.Now, I need to find the derivative f’(x) to get the slope of the tangent at x = π/4. The function is f(x) = 2x tan(x). This is a product of two functions: 2x and tan(x). So, I should use the product rule for differentiation. The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is u’(x)v(x) + u(x)v’(x).Let me set u(x) = 2x and v(x) = tan(x). Then, u’(x) is the derivative of 2x, which is 2. And v’(x) is the derivative of tan(x), which is sec²(x). So, applying the product rule:f’(x) = u’(x)v(x) + u(x)v’(x) = 2*tan(x) + 2x*sec²(x)Alright, so f’(x) = 2 tan(x) + 2x sec²(x). Now, I need to evaluate this at x = π/4.Let me compute each term separately. First, tan(π/4) is 1, as I already know. So, the first term is 2*1 = 2.Next, the second term is 2x sec²(x). Let's compute sec(π/4). Since sec(x) is 1/cos(x), and cos(π/4) is √2/2, so sec(π/4) is 2/√2, which simplifies to √2. Therefore, sec²(π/4) is (√2)² = 2.So, the second term becomes 2*(π/4)*2. Let's compute that:2*(π/4) = π/2, and then π/2 * 2 = π.Wait, hold on. Let me double-check that. 2*(π/4) is indeed π/2, and then multiplying by 2 gives π. So, the second term is π.Therefore, f’(π/4) = 2 + π.So, the slope of the tangent line at x = π/4 is 2 + π.Now, with the point (π/4, π/2) and the slope 2 + π, I can write the equation of the tangent line using the point-slope form:y - y1 = m(x - x1)Plugging in the values:y - (π/2) = (2 + π)(x - π/4)Now, let's simplify this equation to get it into a more standard form.First, distribute the slope on the right-hand side:y - π/2 = (2 + π)x - (2 + π)*(π/4)Let me compute (2 + π)*(π/4):(2 + π)*(π/4) = (2π)/4 + (π²)/4 = π/2 + π²/4So, the equation becomes:y - π/2 = (2 + π)x - π/2 - π²/4Now, add π/2 to both sides to isolate y:y = (2 + π)x - π²/4So, the equation of the tangent line is:y = (2 + π)x - π²/4Alternatively, if I want to write it in standard form (Ax + By + C = 0), I can rearrange it:(2 + π)x - y - π²/4 = 0Let me just double-check my calculations to make sure I didn't make any mistakes.Starting from f’(x) = 2 tan(x) + 2x sec²(x). At x = π/4, tan(π/4) = 1, so first term is 2*1 = 2. Then, sec(π/4) = √2, so sec²(π/4) = 2. Then, 2x sec²(x) is 2*(π/4)*2 = (π/2)*2 = π. So, f’(π/4) = 2 + π. That seems correct.Then, f(π/4) = 2*(π/4)*1 = π/2. Correct.So, point is (π/4, π/2), slope is 2 + π.Equation: y - π/2 = (2 + π)(x - π/4). Expanding gives y = (2 + π)x - (2 + π)(π/4) + π/2.Compute (2 + π)(π/4) = 2π/4 + π²/4 = π/2 + π²/4. So, subtracting that and adding π/2 cancels out the π/2 terms, leaving y = (2 + π)x - π²/4. That seems right.Alternatively, in standard form: (2 + π)x - y - π²/4 = 0.I think that's correct. Let me see if there's another way to write it or if I can simplify it further, but I don't think so. The coefficients are already simplified.So, the equation of the tangent line is y = (2 + π)x - π²/4, or in standard form, (2 + π)x - y - π²/4 = 0.I think that's the final answer.