💡First, I need to determine for how many positive integers ( x ) the expression ( x^2 + 4x + 4 ) falls between 24 and 64.I start by simplifying the expression:[x^2 + 4x + 4 = (x + 2)^2]Next, I set up the inequality to reflect that ( (x + 2)^2 ) should be between 24 and 64:[24 < (x + 2)^2 < 64]To solve for ( x ), I take the square root of each part of the inequality:[sqrt{24} < x + 2 < sqrt{64}][sqrt{24} approx 4.9 quad text{and} quad sqrt{64} = 8]This gives me the range for ( x + 2 ):[4.9 < x + 2 < 8]Subtracting 2 from all parts of the inequality to solve for ( x ):[4.9 - 2 < x < 8 - 2][2.9 < x < 6]Since ( x ) must be a positive integer, the possible values of ( x ) are 3, 4, 5, and 6. I verify each value to ensure ( (x + 2)^2 ) falls within the range:- For ( x = 3 ): ( (3 + 2)^2 = 25 )- For ( x = 4 ): ( (4 + 2)^2 = 36 )- For ( x = 5 ): ( (5 + 2)^2 = 49 )- For ( x = 6 ): ( (6 + 2)^2 = 64 )All these values (25, 36, 49, 64) are within the range [24, 64].Therefore, there are 4 positive integers ( x ) that satisfy the condition.

💡First, I need to find the common difference ( d ) of the arithmetic sequence. I know that ( a_1 = frac{1}{3} ) and ( a_2 + a_5 = 4 ).Using the formula for the ( n )-th term of an arithmetic sequence, ( a_n = a_1 + (n-1)d ), I can express ( a_2 ) and ( a_5 ) in terms of ( d ):[a_2 = a_1 + d = frac{1}{3} + d][a_5 = a_1 + 4d = frac{1}{3} + 4d]Adding these two equations gives:[a_2 + a_5 = left( frac{1}{3} + d right) + left( frac{1}{3} + 4d right) = frac{2}{3} + 5d = 4]Solving for ( d ):[5d = 4 - frac{2}{3} = frac{10}{3}][d = frac{2}{3}]Now that I have the common difference, I can write the general formula for the ( n )-th term:[a_n = a_1 + (n-1)d = frac{1}{3} + (n-1) times frac{2}{3} = frac{2}{3}n - frac{1}{3}]Given that ( a_n = 33 ), I set up the equation:[frac{2}{3}n - frac{1}{3} = 33][frac{2}{3}n = 33 + frac{1}{3} = frac{100}{3}][n = frac{100}{3} times frac{3}{2} = 50]Therefore, the value of ( n ) is 50.

💡Okay, so I have this triangle problem where one angle is three times another, and the sides opposite these angles are 21 and 14. I need to find the length of the third side. Hmm, let me think about how to approach this.First, let me visualize the triangle. Let's call it triangle ABC. I'll label the sides opposite angles A, B, and C as a, b, and c respectively. But wait, the problem says one angle is thrice another, and the sides opposite are 21 and 14. So, let me assign these sides to specific angles to make it clearer.Let me assume that angle B is three times angle C. So, angle B = 3 * angle C. Then, the side opposite angle B is 21, and the side opposite angle C is 14. That makes sense because the larger side is opposite the larger angle. So, side AC is 21, and side AB is 14. Therefore, side BC, which is opposite angle A, is the one we need to find. Let's denote it as 'a'.Now, I remember that in such problems, the Law of Sines and the Law of Cosines are usually helpful. The Law of Sines relates the sides and angles of a triangle, stating that a/sin A = b/sin B = c/sin C. The Law of Cosines relates the sides of a triangle with one of its angles, stating that c² = a² + b² - 2ab cos C.Since we have two angles related (angle B and angle C) and their opposite sides, maybe I can use the Law of Sines first. Let me write that down:14 / sin C = 21 / sin BBut since angle B is three times angle C, I can write sin B as sin(3C). So, substituting that in:14 / sin C = 21 / sin(3C)I need to express sin(3C) in terms of sin C. I recall the triple angle formula for sine, which is sin(3θ) = 3 sin θ - 4 sin³ θ. So, applying that here:sin(3C) = 3 sin C - 4 sin³ CSubstituting back into the equation:14 / sin C = 21 / (3 sin C - 4 sin³ C)Let me denote sin C as x to simplify the equation:14 / x = 21 / (3x - 4x³)Cross-multiplying:14 * (3x - 4x³) = 21 * xExpanding the left side:42x - 56x³ = 21xSubtracting 21x from both sides:21x - 56x³ = 0Factor out 7x:7x(3 - 8x²) = 0So, either 7x = 0 or 3 - 8x² = 0Since x = sin C and angles in a triangle can't be zero, we discard x = 0. Therefore:3 - 8x² = 0Solving for x²:8x² = 3x² = 3/8x = sqrt(3/8) = (√6)/4 ≈ 0.612So, sin C = √6 / 4Now, let's find cos C using the Pythagorean identity:cos² C = 1 - sin² C = 1 - (6/16) = 10/16 = 5/8Therefore, cos C = sqrt(5/8) = (√10)/4 ≈ 0.7906Now, I can use the Law of Cosines to find side a. Let me recall that the Law of Cosines states:a² = b² + c² - 2bc cos ABut in this case, I need to relate the sides and angles I have. Wait, angle A is the remaining angle in the triangle. Since the sum of angles in a triangle is 180 degrees, angle A = 180 - angle B - angle C = 180 - 3C - C = 180 - 4C.Hmm, that might complicate things. Alternatively, maybe I can use the Law of Cosines on angle C or angle B.Wait, let me think. I know sides AB = 14, AC = 21, and BC = a. I also know angle B = 3C and angle C.Alternatively, maybe using the Law of Cosines on angle C. Let's see:In triangle ABC, angle C is opposite side AB = 14, angle B is opposite side AC = 21, and angle A is opposite side BC = a.So, using the Law of Cosines on angle C:AB² = AC² + BC² - 2 * AC * BC * cos CPlugging in the known values:14² = 21² + a² - 2 * 21 * a * cos CCalculating the squares:196 = 441 + a² - 42a * (√10 / 4)Simplify the equation:196 = 441 + a² - (42a * √10)/4Simplify (42/4) to (21/2):196 = 441 + a² - (21a√10)/2Bring all terms to one side:a² - (21a√10)/2 + 441 - 196 = 0Simplify 441 - 196:441 - 196 = 245So, the equation becomes:a² - (21a√10)/2 + 245 = 0This is a quadratic equation in terms of a. Let me write it as:a² - (21√10 / 2) a + 245 = 0To solve for a, I can use the quadratic formula:a = [ (21√10 / 2) ± sqrt( (21√10 / 2)^2 - 4 * 1 * 245 ) ] / (2 * 1)First, calculate the discriminant:D = (21√10 / 2)^2 - 4 * 1 * 245Calculate (21√10 / 2)^2:(21)^2 = 441(√10)^2 = 10So, (21√10 / 2)^2 = (441 * 10) / 4 = 4410 / 4 = 1102.5Now, calculate 4 * 1 * 245 = 980So, D = 1102.5 - 980 = 122.5Therefore, sqrt(D) = sqrt(122.5) ≈ 11.07But let me express 122.5 as a fraction to keep it exact. 122.5 = 245/2So, sqrt(245/2) = (√245)/√2 = (7√5)/√2 = (7√10)/2Wait, let me check that:245 = 49 * 5, so sqrt(245) = 7√5Therefore, sqrt(245/2) = (7√5)/√2 = (7√10)/2 after rationalizing the denominator.So, sqrt(D) = (7√10)/2Now, plug back into the quadratic formula:a = [ (21√10 / 2) ± (7√10)/2 ] / 2Factor out √10 / 2:a = [ √10 / 2 (21 ± 7) ] / 2Calculate 21 ± 7:21 + 7 = 2821 - 7 = 14So, two possible solutions:a = [ √10 / 2 * 28 ] / 2 = (28√10 / 2) / 2 = (14√10) / 2 = 7√10ora = [ √10 / 2 * 14 ] / 2 = (14√10 / 2) / 2 = (7√10) / 2Now, we need to determine which solution is valid. Since side lengths must be positive, both are positive, but we need to check if they satisfy the triangle inequality.Given sides 14, 21, and a.If a = 7√10 ≈ 7 * 3.16 ≈ 22.12Check triangle inequalities:14 + 21 > 22.12: 35 > 22.12, which is true.14 + 22.12 > 21: 36.12 > 21, true.21 + 22.12 > 14: 43.12 > 14, true.So, a = 7√10 is valid.If a = 7√10 / 2 ≈ 7 * 3.16 / 2 ≈ 11.06Check triangle inequalities:14 + 21 > 11.06: 35 > 11.06, true.14 + 11.06 > 21: 25.06 > 21, true.21 + 11.06 > 14: 32.06 > 14, true.So, both solutions satisfy the triangle inequality. Hmm, that's interesting. So, does that mean there are two possible triangles? Or is one of them not valid due to the angle constraints?Wait, let's think about the angles. We assumed angle B = 3 angle C. If a is shorter, does that affect the angles?Alternatively, maybe both solutions are valid, resulting in two different triangles satisfying the given conditions. But the problem doesn't specify whether the triangle is acute or obtuse, so perhaps both are possible.But let me check the Law of Sines again. If a = 7√10, then angle A would be opposite this side, which is the largest side, so angle A would be the largest angle. Since angle B is 3 angle C, and angles sum to 180, angle A = 180 - 4 angle C.If a is larger, angle A is larger, which makes sense. If a is smaller, angle A is smaller, but still, angle B is still 3 angle C.Wait, but let's calculate angle C in both cases.From earlier, sin C = √6 / 4 ≈ 0.612, so angle C ≈ arcsin(0.612) ≈ 37.7 degrees.Therefore, angle B = 3 * 37.7 ≈ 113.1 degrees.Then, angle A = 180 - 37.7 - 113.1 ≈ 29.2 degrees.So, if angle A is 29.2 degrees, the side opposite, which is a, should be the smallest side. But in our solutions, a is either approximately 22.12 or 11.06. Wait, 11.06 is smaller than 14 and 21, so that would make sense for angle A being the smallest angle.But 22.12 is larger than 14 and 21? Wait, 22.12 is larger than 21, which is side AC. So, if a is 22.12, then side BC is the largest side, meaning angle A is the largest angle. But angle A was calculated as 29.2 degrees, which is not the largest angle. That seems contradictory.Wait, that suggests that a cannot be 22.12 because angle A would have to be larger than angle B, which is 113.1 degrees, but 29.2 is less than 113.1. Therefore, a cannot be 22.12 because that would imply angle A is larger, which contradicts the angle measures.Therefore, the correct solution is a = 7√10 / 2 ≈ 11.06, which is smaller than 14 and 21, making angle A the smallest angle, consistent with our earlier calculation.Wait, but let me double-check this reasoning. If a = 7√10 / 2 ≈ 11.06, then it's the smallest side, opposite angle A, which is 29.2 degrees, the smallest angle. That makes sense.If a were 7√10 ≈ 22.12, it would be the largest side, opposite angle A, which should be the largest angle. But angle A is only 29.2 degrees, which is less than angle B (113.1 degrees). That's a contradiction because the largest side should be opposite the largest angle.Therefore, a = 7√10 / 2 is the valid solution.Wait, but let me verify this with the Law of Sines.If a = 7√10 / 2, then:a / sin A = 14 / sin CSo, (7√10 / 2) / sin(29.2°) ≈ (7 * 3.16 / 2) / 0.487 ≈ (22.12 / 2) / 0.487 ≈ 11.06 / 0.487 ≈ 22.7And 14 / sin C ≈ 14 / 0.612 ≈ 22.87These are approximately equal, considering rounding errors, so that checks out.If a were 7√10 ≈ 22.12, then:22.12 / sin(29.2°) ≈ 22.12 / 0.487 ≈ 45.4But 14 / sin C ≈ 22.87, which is not equal to 45.4, so that doesn't satisfy the Law of Sines. Therefore, a = 7√10 is invalid.So, the correct length of the third side is 7√10 / 2.Wait, but let me express this in a simplified radical form. 7√10 / 2 is already simplified, so that's the answer.Alternatively, if I rationalize or present it differently, but I think 7√10 / 2 is the simplest form.So, after going through the Law of Sines, triple angle formula, solving the quadratic, and verifying the solutions, I conclude that the length of the third side is 7√10 divided by 2.

💡First, determine the regular ticket price. Since the senior ticket costs 7.00 after a 30% discount, it represents 70% of the regular price. Let ( x ) be the regular ticket price:[0.7x = 7 implies x = frac{7}{0.7} = 10]The regular ticket price is 10.00. Next, calculate the cost of a child's ticket with a 40% discount, which means paying 60% of the regular price:[text{Child's ticket} = 0.6 times 10 = 6]Now, calculate the total cost for all tickets:[2 times 7 , (text{senior tickets}) + 2 times 10 , (text{regular tickets}) + 2 times 6 , (text{child tickets}) = 14 + 20 + 12 = 46]Therefore, the total amount Grandfather Huang must pay is 46.00.

💡Okay, so I have this problem here where I'm given a function f(x) = 2^x + log₂x, and I know that 'a' is a root of this function. That means when I plug in 'a' into the function, it equals zero: f(a) = 0. The question is asking me about the value of f(x) when x is between 0 and a. Specifically, it wants to know if f(x) is greater than zero, equal to zero, less than zero, or if the sign is uncertain.Alright, let's break this down. First, I need to understand the behavior of the function f(x) = 2^x + log₂x. Since both 2^x and log₂x are functions I've dealt with before, maybe I can analyze their properties separately and then see how they interact.Starting with 2^x: I know that 2^x is an exponential function. It's always positive for any real x, and it increases as x increases. When x is 0, 2^0 is 1. As x becomes more positive, 2^x grows rapidly, and as x becomes negative, 2^x approaches zero but never actually reaches zero. So, 2^x is a strictly increasing function that's always positive.Next, log₂x: This is the logarithm function with base 2. Logarithms are only defined for positive real numbers, so x has to be greater than 0 here. The logarithm function increases as x increases, but it does so more slowly than linear functions. At x = 1, log₂1 is 0. For x between 0 and 1, log₂x is negative, and as x approaches 0 from the right, log₂x approaches negative infinity. For x greater than 1, log₂x is positive and increases as x increases.So, putting these together, f(x) = 2^x + log₂x is the sum of an always-positive, increasing exponential function and a logarithm function that can be negative or positive depending on whether x is less than or greater than 1.Now, since 'a' is a root of f(x), f(a) = 0. That means 2^a + log₂a = 0. So, 2^a = -log₂a. Since 2^a is always positive, that implies that log₂a must be negative. Therefore, a must be less than 1 because log₂a is negative only when a is between 0 and 1.So, we know that a is somewhere between 0 and 1. Now, the question is about the value of f(x) when x is between 0 and a. So, x is less than a, which is already less than 1.Let me think about how f(x) behaves as x approaches 0 from the right. As x approaches 0+, 2^x approaches 1, and log₂x approaches negative infinity. So, f(x) approaches 1 + (-∞), which is negative infinity. That means near x = 0, f(x) is very negative.On the other hand, at x = a, f(a) = 0. So, somewhere between 0 and a, f(x) goes from negative infinity up to 0. Since f(x) is continuous (both 2^x and log₂x are continuous on their domains), by the Intermediate Value Theorem, there must be some point where f(x) crosses zero, which is our root 'a'.But the question is about the value of f(x) when x is less than a. So, if I pick an x that's between 0 and a, what can I say about f(x)?Since f(x) is increasing or decreasing? Let me check the derivative to see if the function is increasing or decreasing.The derivative of f(x) is f'(x) = d/dx [2^x] + d/dx [log₂x]. The derivative of 2^x is ln(2) * 2^x, which is always positive because ln(2) is positive and 2^x is always positive. The derivative of log₂x is 1/(x ln 2), which is also positive because x is positive (since we're considering x > 0) and ln 2 is positive. Therefore, f'(x) = ln(2) * 2^x + 1/(x ln 2) is always positive for x > 0. This means that f(x) is strictly increasing on the interval (0, ∞).So, f(x) is increasing. That means as x increases, f(x) increases. Since f(a) = 0, and f(x) is increasing, for any x less than a, f(x) must be less than f(a), which is 0. Therefore, f(x) < 0 when x < a.Wait, but let me make sure I'm not making a mistake here. Since f(x) is increasing, moving from left to right on the x-axis, the function's value increases. So, if I have a point x1 < x2, then f(x1) < f(x2). In our case, since a is a root, and we're looking at x < a, then f(x) < f(a) = 0. So, yes, f(x) is negative in that interval.Is there any chance that f(x) could be positive or zero in that interval? Well, since f(x) is strictly increasing, once it crosses zero at x = a, it can't come back down. So, before x = a, it's negative, and after x = a, it's positive. Therefore, for x < a, f(x) < 0.Let me also consider specific values to test this. Let's say a is 0.5, just as an example. Then, at x = 0.25, which is less than 0.5, what is f(0.25)?f(0.25) = 2^0.25 + log₂0.25. 2^0.25 is the fourth root of 2, which is approximately 1.1892. log₂0.25 is log₂(1/4) which is -2. So, f(0.25) ≈ 1.1892 - 2 ≈ -0.8108, which is negative.At x = a = 0.5, f(0.5) = 2^0.5 + log₂0.5 ≈ 1.4142 - 1 ≈ 0.4142, which is positive. Wait, that contradicts my earlier conclusion. Hmm, maybe my assumption about a being 0.5 is incorrect.Wait, no, because if a is a root, then f(a) = 0. So, if I take a = 0.5, f(a) ≈ 1.4142 - 1 ≈ 0.4142, which is not zero. So, a is not 0.5. Let me find a better approximation for a.Let me try to solve f(x) = 0 numerically. So, 2^x + log₂x = 0.Let me pick x = 0.5: f(0.5) ≈ 1.4142 - 1 ≈ 0.4142 > 0.x = 0.25: f(0.25) ≈ 1.1892 - 2 ≈ -0.8108 < 0.So, between 0.25 and 0.5, f(x) crosses zero. Let's try x = 0.3.f(0.3) = 2^0.3 + log₂0.3 ≈ 1.2311 - 1.73696 ≈ -0.5058 < 0.x = 0.4:2^0.4 ≈ 1.3195, log₂0.4 ≈ -1.3219.So, f(0.4) ≈ 1.3195 - 1.3219 ≈ -0.0024 < 0.Almost zero. x = 0.41:2^0.41 ≈ e^{0.41 ln2} ≈ e^{0.41*0.6931} ≈ e^{0.2842} ≈ 1.329.log₂0.41 ≈ ln0.41 / ln2 ≈ (-0.8918)/0.6931 ≈ -1.286.So, f(0.41) ≈ 1.329 - 1.286 ≈ 0.043 > 0.So, between 0.4 and 0.41, f(x) crosses zero. Therefore, a is approximately 0.405.So, if a is around 0.405, then for x < a, say x = 0.4, f(x) ≈ -0.0024 < 0, and for x = 0.41, f(x) ≈ 0.043 > 0.So, indeed, for x < a, f(x) is negative, and for x > a, f(x) is positive.Therefore, the correct answer is C: f(x) < 0.But wait, let me think again. The function is increasing, so it goes from negative infinity at x approaches 0, increases, and crosses zero at x = a, then continues increasing to positive infinity as x increases. So, yes, for x < a, f(x) is negative.Is there any possibility that f(x) could be positive or zero for some x < a? Since the function is strictly increasing, once it crosses zero at x = a, it can't go back. So, for all x < a, f(x) must be less than zero.Therefore, the answer should be C: f(x) < 0.

💡First, we note that the first initial must be 'A'. This leaves us with 25 remaining letters of the alphabet (B through Z) to choose from for the middle and last initials.Since the monogram must be in alphabetical order and no letters can be repeated, we need to choose 2 distinct letters from the remaining 25 letters. The number of ways to choose 2 letters from 25 is given by the combination formula:[binom{25}{2} = frac{25 times 24}{2 times 1} = 300]Thus, there are 300 possible monograms that meet the given criteria.