💡Alright, so I've got this problem about a gambler who bought 15,000 worth of chips in denominations of 25, 75, and 250. The gambler lost some chips and then cashed in the remaining ones. The problem gives me some relationships between the number of chips lost in each denomination and tells me that the total number of lost chips is 40. Also, the gambler had at least one chip left of each denomination after the losses. I need to find the largest amount of money the gambler could have received back.Okay, let's break this down. First, I need to figure out how many chips of each denomination the gambler lost. The problem says that the number of 25 chips lost was twice the number of 75 chips lost, and the number of 75 chips lost was half the number of 250 chips lost. So, if I let the number of 75 chips lost be x, then the number of 25 chips lost would be 2x, and the number of 250 chips lost would be 2x as well because x is half of that.Wait, let me make sure I got that right. If the number of 75 chips lost is x, then the number of 25 chips lost is twice that, so 2x. And the number of 250 chips lost is twice the number of 75 chips lost, which would also be 2x. So, total lost chips would be 2x (for 25) + x (for 75) + 2x (for 250) = 5x. But the problem says the total number of lost chips is 40, so 5x = 40, which means x = 8.So, the gambler lost 8 75 chips, 16 25 chips, and 16 250 chips. Now, I need to calculate the total value lost. Let's do that:- 25 chips: 16 chips * 25 = 400- 75 chips: 8 chips * 75 = 600- 250 chips: 16 chips * 250 = 4,000Adding those up: 400 + 600 + 4,000 = 5,000So, the gambler lost 5,000 worth of chips. Since the gambler started with 15,000, the remaining amount would be 15,000 - 5,000 = 10,000.But wait, the problem also mentions that the gambler had at least one chip left of each denomination. I need to make sure that after losing those chips, there's at least one of each left. So, I need to check how many chips were originally bought and ensure that after losing 16 25, 8 75, and 16 250 chips, there's at least one of each left.Let me denote the original number of 25 chips as A, 75 as B, and 250 as C. So, the total value is:25A + 75B + 250C = 15,000We also know that after losing 16 25, 8 75, and 16 250 chips, the remaining chips are:A - 16 ≥ 1B - 8 ≥ 1C - 16 ≥ 1Which simplifies to:A ≥ 17B ≥ 9C ≥ 17Now, I need to find the values of A, B, and C that satisfy the total value equation and the inequalities above.Let me express the total value equation in terms of A, B, and C:25A + 75B + 250C = 15,000I can divide the entire equation by 25 to simplify:A + 3B + 10C = 600Now, I have:A + 3B + 10C = 600A ≥ 17B ≥ 9C ≥ 17I need to find integer values of A, B, and C that satisfy these conditions.Let me try to express A in terms of B and C:A = 600 - 3B - 10CSince A must be at least 17, we have:600 - 3B - 10C ≥ 17-3B - 10C ≥ -5833B + 10C ≤ 583Similarly, B must be at least 9, and C must be at least 17.Let me try to find values of C first, since it has the highest coefficient in the equation.Let's start with C = 17:A = 600 - 3B - 10*17 = 600 - 3B - 170 = 430 - 3BNow, B must be at least 9, so let's plug B = 9:A = 430 - 27 = 403So, A = 403, B = 9, C = 17Check if A ≥ 17: 403 ≥ 17, yes.Now, let's see if we can increase C to get a higher value of C, which would mean more 250 chips, potentially allowing for more flexibility.Let's try C = 18:A = 600 - 3B - 10*18 = 600 - 3B - 180 = 420 - 3BAgain, B must be at least 9:A = 420 - 27 = 393So, A = 393, B = 9, C = 18Still, A is well above 17.Let's try C = 19:A = 600 - 3B - 190 = 410 - 3BB = 9:A = 410 - 27 = 383Continuing this way, I can see that as C increases, A decreases by 10 for each increment in C, but B is kept at its minimum of 9.However, I need to ensure that A remains at least 17. Let's see how high C can go before A drops below 17.Set A = 17:17 = 600 - 3B - 10CBut B must be at least 9, so:17 = 600 - 3*9 - 10C17 = 600 - 27 - 10C17 = 573 - 10C10C = 573 - 1710C = 556C = 55.6Since C must be an integer, the maximum C can be is 55.But wait, if C is 55, then:A = 600 - 3B - 10*55 = 600 - 3B - 550 = 50 - 3BBut B must be at least 9:A = 50 - 27 = 23Which is still above 17.But let's check if C can be higher than 55:C = 56:A = 600 - 3B - 560 = 40 - 3BWith B = 9:A = 40 - 27 = 13Which is below 17, so C cannot be 56.Therefore, the maximum C can be is 55.But wait, earlier when I tried C = 17, I got A = 403, B = 9And when C = 55, A = 23, B = 9So, the number of 25 chips decreases as C increases.But I'm trying to maximize the amount received back, which is 10,000, as calculated earlier.Wait, but the amount received back is fixed at 10,000 because the gambler lost 5,000.But maybe I'm missing something. Perhaps the way the chips are distributed affects the total amount received back.Wait, no, because the total value is fixed at 15,000, and the gambler lost 5,000, so regardless of how the chips are distributed, the remaining value is 10,000.But the problem says "the largest amount of money that the gambler could have received back." So, maybe there's a way to have a higher amount by adjusting the number of chips lost?Wait, no, because the total value lost is fixed at 5,000, so the remaining is 10,000.But perhaps I'm misunderstanding. Maybe the gambler could have different combinations of chips that result in a higher cash-back amount due to the denominations.Wait, no, because the total value is fixed. The cash-back amount is the total value of the remaining chips, which is 15,000 - 5,000 = 10,000.But the problem specifies that the gambler had at least one chip left of each denomination. So, perhaps the way the chips are distributed affects the total cash-back.Wait, no, because the total value is fixed. The denominations don't affect the total cash-back amount, just the number of chips.So, I think the largest amount the gambler could have received back is 10,000.But let me double-check.Total value lost: 5,000Total value initially: 15,000Remaining value: 10,000Yes, that seems correct.But wait, maybe I need to consider that the gambler could have different numbers of chips lost, but the relationships between the numbers of chips lost are fixed.Wait, the relationships are fixed: number of 25 lost is twice the number of 75 lost, and number of 75 lost is half the number of 250 lost.So, the ratios are fixed, and the total number of lost chips is 40.So, the number of 25 lost is 16, 75 is 8, and 250 is 16, as I calculated earlier.So, the total value lost is fixed at 5,000, so the remaining is 10,000.Therefore, the largest amount the gambler could have received back is 10,000.I think that's the answer.

💡Alright, so I have this geometry problem here about triangle ABC where angle A is 60 degrees. The question is asking which additional condition still doesn't allow us to determine that triangle ABC is equilateral. The options are A: AB equals AC, B: angle A equals angle B, C: AD is perpendicular to BC, and D: angle B equals angle C.Okay, let's start by recalling what makes a triangle equilateral. An equilateral triangle has all sides equal and all angles equal to 60 degrees. So, if we can show that all sides are equal or all angles are equal, then the triangle is equilateral.Given that angle A is already 60 degrees, let's see what each condition adds.**Condition A: AB = AC**If AB equals AC, then triangle ABC is isosceles with AB and AC as the equal sides. In an isosceles triangle, the base angles are equal. Since angle A is 60 degrees, the other two angles (B and C) must be equal. Let's call them both x degrees.So, angle A + angle B + angle C = 180 degrees60 + x + x = 18060 + 2x = 1802x = 120x = 60So, both angles B and C are 60 degrees. That means all angles are 60 degrees, which makes triangle ABC equilateral. So, condition A is sufficient to determine it's equilateral.**Condition B: Angle A = Angle B**Given angle A is 60 degrees, if angle A equals angle B, then angle B is also 60 degrees. Now, let's find angle C.Angle A + angle B + angle C = 18060 + 60 + angle C = 180120 + angle C = 180angle C = 60So, all angles are 60 degrees, making triangle ABC equilateral. So, condition B also determines it's equilateral.**Condition C: AD is perpendicular to BC**Hmm, this one is a bit trickier. AD is perpendicular to BC, which means AD is the altitude from A to side BC. Let's visualize this. In triangle ABC, if AD is perpendicular to BC, then AD is the height of the triangle.But does this necessarily make the triangle equilateral? Let's think. If angle A is 60 degrees, and AD is the altitude, then AD divides BC into two segments. However, unless we know more about the sides or other angles, we can't conclude that all sides are equal.For example, suppose triangle ABC is not equilateral but has angle A = 60 degrees and AD perpendicular to BC. It could still be an isosceles triangle with AB = AC, but not necessarily all sides equal. Wait, but earlier we saw that if AB = AC and angle A = 60, then it's equilateral. So, maybe in this case, since AD is the altitude, if AB = AC, then it's equilateral. But the problem is, condition C doesn't specify AB = AC; it just says AD is perpendicular to BC.So, without knowing that AB = AC or that angles B and C are equal, just having an altitude doesn't necessarily make the triangle equilateral. It could be any triangle with angle A = 60 degrees and an altitude from A to BC, but sides AB and AC could be different lengths, making it not equilateral.Therefore, condition C doesn't necessarily determine the triangle to be equilateral.**Condition D: Angle B = Angle C**If angle B equals angle C, then triangle ABC is isosceles with AB = AC. Given that angle A is 60 degrees, and angles B and C are equal, let's find their measures.Angle A + angle B + angle C = 18060 + x + x = 18060 + 2x = 1802x = 120x = 60So, angles B and C are both 60 degrees, making all angles 60 degrees, which means the triangle is equilateral. So, condition D also determines it's equilateral.**Conclusion**Out of all the conditions, only condition C (AD perpendicular to BC) doesn't necessarily make triangle ABC equilateral. The other conditions either make the triangle equilateral by ensuring all angles are 60 degrees or all sides are equal. But condition C only adds an altitude without providing enough information about the sides or other angles to conclude the triangle is equilateral.So, the answer should be condition C.

💡Okay, so I have this problem where I'm given four positive numbers a, b, c, d such that a > b > c > d > 0. Also, it's given that a + d = b + c. I need to prove that the sum of the square roots of d and a is less than the sum of the square roots of b and c. That is, √d + √a < √b + √c.Hmm, let me start by writing down what I know:1. a > b > c > d > 02. a + d = b + cI need to show that √d + √a < √b + √c.Maybe I can manipulate the given equation a + d = b + c. Since a is the largest and d is the smallest, and b and c are in between, perhaps I can express d in terms of a, b, and c. Let's see:From a + d = b + c, we can solve for d:d = b + c - aSo, d is equal to b + c minus a. Since a is larger than b and c, this makes sense because d is smaller than c.Now, I need to compare √d + √a with √b + √c. Maybe I can square both sides to eliminate the square roots. Let's try that.Let me denote S = √d + √a and T = √b + √c. I need to show that S < T.If I square both sides, I get:S² = (√d + √a)² = d + a + 2√(da)T² = (√b + √c)² = b + c + 2√(bc)Since a + d = b + c, let's substitute that into S²:S² = (b + c) + 2√(da)T² = (b + c) + 2√(bc)So, S² = T² + 2√(da) - 2√(bc)Therefore, to show that S < T, it suffices to show that 2√(da) < 2√(bc), which simplifies to √(da) < √(bc). Squaring both sides, this is equivalent to da < bc.So, now I need to prove that da < bc.Given that d = b + c - a, let's substitute that into da:da = a(b + c - a) = ab + ac - a²So, da = ab + ac - a²Similarly, bc is just bc.So, I need to show that ab + ac - a² < bcLet's rearrange this inequality:ab + ac - a² - bc < 0Factor terms:ab - a² + ac - bc < 0Factor a from the first two terms and c from the last two terms:a(b - a) + c(a - b) < 0Notice that (b - a) is negative because a > b, and (a - b) is positive. So, let's factor out (a - b):= (a - b)(c - a) < 0Because a > b, (a - b) is positive. Also, since a > c, (c - a) is negative. Therefore, the product of a positive and a negative number is negative, which is less than zero.Hence, (a - b)(c - a) < 0, which implies that ab + ac - a² - bc < 0, which in turn implies that da < bc.Since da < bc, it follows that √(da) < √(bc), and thus 2√(da) < 2√(bc). Therefore, S² < T², which implies that S < T because square roots are positive.So, √d + √a < √b + √c.I think that makes sense. Let me double-check my steps to make sure I didn't make any errors.1. Started with the given a + d = b + c and expressed d in terms of a, b, c.2. Expressed S and T as the sums of square roots and squared both sides.3. Substituted a + d with b + c in S².4. Compared S² and T², leading to the need to show that da < bc.5. Substituted d into da and simplified the expression.6. Factored the resulting expression and showed that it's negative because of the ordering of a, b, c.7. Concluded that da < bc, hence √(da) < √(bc), leading to S < T.Everything seems to check out. I don't see any mistakes in the reasoning. The key step was recognizing that after squaring both sides, the inequality reduces to showing that da < bc, which follows from the given ordering and the relationship a + d = b + c.Another way to think about it might be to consider the function f(x) = √x, which is concave. Maybe using Jensen's inequality or something related? But since the variables are ordered, perhaps that's overcomplicating it.Alternatively, I could consider specific numbers to test the inequality. For example, let me choose a = 4, b = 3, c = 2, d = 1. Then a + d = 5 and b + c = 5, so the condition holds.Compute √d + √a = √1 + √4 = 1 + 2 = 3Compute √b + √c = √3 + √2 ≈ 1.732 + 1.414 ≈ 3.146Indeed, 3 < 3.146, so the inequality holds in this case.Let me try another set: a = 5, b = 4, c = 3, d = 2. Then a + d = 7 and b + c = 7.Compute √d + √a = √2 + √5 ≈ 1.414 + 2.236 ≈ 3.650Compute √b + √c = √4 + √3 = 2 + 1.732 ≈ 3.732Again, 3.650 < 3.732, so the inequality holds.One more example: a = 6, b = 5, c = 4, d = 3. a + d = 9, b + c = 9.√d + √a = √3 + √6 ≈ 1.732 + 2.449 ≈ 4.181√b + √c = √5 + √4 ≈ 2.236 + 2 ≈ 4.236Again, 4.181 < 4.236.These examples seem to support the inequality. It gives me more confidence that the proof is correct.I think the key takeaway is that because a is the largest and d is the smallest, their product da is less than bc, which are in the middle. This is because when you have two numbers with a fixed sum, their product is maximized when they are equal. Here, since a and d are further apart than b and c, their product is smaller.Wait, actually, in the case of a fixed sum, the product is maximized when the numbers are equal. So, if a + d = b + c, then the product ad will be less than bc if b and c are closer to each other than a and d are. Since a > b > c > d, a and d are further apart, so their product is smaller.This is another way to see why ad < bc. Because b and c are closer together, their product is larger than the product of a and d, which are further apart, given the same sum.So, this provides another perspective on why the inequality holds. It's related to the arithmetic mean-geometric mean inequality, where for positive numbers with a fixed sum, the product is maximized when the numbers are equal.In this case, since a and d are not equal and are further apart, their product is less than that of b and c, which are closer together.Therefore, combining these insights, the proof holds.**Final Answer**boxed{sqrt{d} + sqrt{a} < sqrt{b} + sqrt{c}}

💡Alright, so I have this inequality: (t)/(t² + 9) ≤ a ≤ (t + 2)/t², and it's supposed to hold for all t in the interval (0, 2]. I need to find the range of a. Hmm, okay. Let me try to break this down step by step.First, I think I need to analyze both sides of the inequality separately. That is, I need to find the maximum value of the left side, (t)/(t² + 9), and the minimum value of the right side, (t + 2)/t², over the interval (0, 2]. Because if a has to be greater than or equal to the left side for all t, then a must be at least as big as the maximum of the left side. Similarly, a has to be less than or equal to the right side for all t, so a must be at most the minimum of the right side.So, let me start with the left side: f(t) = t/(t² + 9). I need to find its maximum on (0, 2]. Maybe I can take the derivative to find critical points.Let's compute f'(t). Using the quotient rule: if f(t) = numerator/denominator, then f'(t) = (num’ * den - num * den’)/den².So, numerator is t, so num’ is 1. Denominator is t² + 9, so den’ is 2t.Therefore, f'(t) = [1*(t² + 9) - t*(2t)] / (t² + 9)².Simplify numerator: (t² + 9 - 2t²) = (-t² + 9).So, f'(t) = (-t² + 9)/(t² + 9)².Now, to find critical points, set f'(t) = 0: (-t² + 9) = 0 => t² = 9 => t = ±3. But since t is in (0, 2], the only critical point near this interval is t = 3, which is outside of (0, 2]. So, within (0, 2], f'(t) is always positive or negative?Let me test a value in (0, 2], say t = 1. Plug into f'(1): (-1 + 9)/(1 + 9)² = 8/100 = 0.08 > 0. So, f'(t) is positive on (0, 2]. That means f(t) is increasing on (0, 2]. Therefore, its maximum occurs at t = 2.So, f(t) maximum is f(2) = 2/(4 + 9) = 2/13.Alright, so a has to be at least 2/13.Now, moving on to the right side: g(t) = (t + 2)/t². I need to find its minimum on (0, 2]. Again, maybe take the derivative.Compute g'(t). Let's rewrite g(t) as (t + 2)/t² = t/t² + 2/t² = 1/t + 2/t². So, g(t) = t^{-1} + 2t^{-2}.Taking derivative: g'(t) = -1*t^{-2} + (-4)*t^{-3} = -1/t² - 4/t³.Alternatively, using the original form: g(t) = (t + 2)/t². So, using the quotient rule: numerator = t + 2, num’ = 1; denominator = t², den’ = 2t.Thus, g'(t) = [1*(t²) - (t + 2)*(2t)] / t^4.Simplify numerator: t² - 2t(t + 2) = t² - 2t² - 4t = -t² - 4t.So, g'(t) = (-t² - 4t)/t^4 = (-t(t + 4))/t^4 = -(t + 4)/t³.Since t is in (0, 2], t³ is positive, and t + 4 is always positive, so g'(t) is negative on (0, 2]. That means g(t) is decreasing on (0, 2]. Therefore, its minimum occurs at t = 2.So, g(t) minimum is g(2) = (2 + 2)/4 = 4/4 = 1.Therefore, a has to be at most 1.Putting it all together, a must satisfy 2/13 ≤ a ≤ 1. So, looking at the options, that's option B.Wait, just to make sure I didn't make any mistakes. Let me double-check the derivatives.For f(t) = t/(t² + 9):f'(t) = [1*(t² + 9) - t*(2t)] / (t² + 9)^2 = (t² + 9 - 2t²)/denominator = (9 - t²)/denominator.So, critical points at t = ±3, which is outside our interval. Since f'(t) is positive in (0, 2], f(t) is increasing, so maximum at t=2: 2/(4 + 9) = 2/13. That seems correct.For g(t) = (t + 2)/t²:g'(t) = [1*t² - (t + 2)*2t]/t^4 = (t² - 2t² - 4t)/t^4 = (-t² - 4t)/t^4 = -t(t + 4)/t^4 = -(t + 4)/t³.Which is negative for t > 0, so g(t) is decreasing on (0, 2], so minimum at t=2: (4)/4 = 1. That seems correct too.So, yeah, the range of a is [2/13, 1], which is option B.**Final Answer**boxed{B}

💡First, I recognize that Kevin Kangaroo hops (frac{1}{4}) of the remaining distance each time. This means that after each hop, the remaining distance is reduced by a factor of (frac{3}{4}).The total distance Kevin hops after six hops can be calculated using the sum of the first six terms of a geometric series. The first term (a) of the series is (frac{1}{4}), and the common ratio (r) is (frac{3}{4}).The formula for the sum of the first (n) terms of a geometric series is:[S_n = a frac{1 - r^n}{1 - r}]Substituting the known values:[S_6 = frac{1}{4} frac{1 - left(frac{3}{4}right)^6}{1 - frac{3}{4}}]Next, I calculate (left(frac{3}{4}right)^6):[left(frac{3}{4}right)^6 = frac{729}{4096}]Substituting this back into the sum formula:[S_6 = frac{1}{4} frac{1 - frac{729}{4096}}{frac{1}{4}} = 1 - frac{729}{4096} = frac{4096 - 729}{4096} = frac{3367}{4096}]Therefore, Kevin has hopped a total distance of (frac{3367}{4096}) after six hops.

💡Okay, so I need to find the equations of the asymptotes for the hyperbola given by the equation (frac{y^2}{3} - x^2 = 1). Hmm, I remember that asymptotes are lines that the hyperbola approaches but never touches. They help in sketching the hyperbola and understanding its behavior as (x) and (y) get very large.First, let me recall the standard form of a hyperbola. There are two standard forms: one that opens up and down, and another that opens left and right. The given equation is (frac{y^2}{3} - x^2 = 1). Comparing this to the standard form, which is (frac{(y - k)^2}{a^2} - frac{(x - h)^2}{b^2} = 1) for a hyperbola centered at ((h, k)) opening upwards and downwards, I can see that this hyperbola is centered at the origin since there are no shifts in the equation.In this case, (a^2 = 3) and (b^2 = 1). So, (a = sqrt{3}) and (b = 1). I remember that for hyperbolas, the asymptotes are lines that pass through the center and have slopes related to (a) and (b). Specifically, for a hyperbola that opens up and down, the equations of the asymptotes are (y = pm frac{a}{b}x).Wait, let me make sure. If the hyperbola is of the form (frac{y^2}{a^2} - frac{x^2}{b^2} = 1), then the asymptotes are indeed (y = pm frac{a}{b}x). So, plugging in the values I have, (a = sqrt{3}) and (b = 1), the slopes should be (pm frac{sqrt{3}}{1}), which simplifies to (pm sqrt{3}).Therefore, the equations of the asymptotes should be (y = pm sqrt{3}x). Looking at the options given:A: (y = pm sqrt{3}x)B: (y = pm frac{sqrt{3}}{3}x)C: (y = pm 2x)D: (y = pm frac{2sqrt{3}}{3}x)So, option A matches my calculation. But wait, let me double-check my steps to make sure I didn't make a mistake.Starting again, the standard form is (frac{y^2}{a^2} - frac{x^2}{b^2} = 1). The asymptotes are found by setting the right-hand side to zero: (frac{y^2}{a^2} - frac{x^2}{b^2} = 0). This can be factored as (left(frac{y}{a} - frac{x}{b}right)left(frac{y}{a} + frac{x}{b}right) = 0), which gives the two equations (y = pm frac{a}{b}x).Plugging in (a^2 = 3) and (b^2 = 1), so (a = sqrt{3}) and (b = 1), so the slopes are (pm sqrt{3}). Therefore, the asymptotes are (y = pm sqrt{3}x), which is option A.Wait, but let me think again. Sometimes, I get confused between the slopes for horizontal and vertical hyperbolas. For a horizontal hyperbola, which is of the form (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), the asymptotes are (y = pm frac{b}{a}x). But in this case, since the hyperbola is vertical (because the positive term is with (y^2)), the asymptotes are (y = pm frac{a}{b}x). So, yes, that would be (sqrt{3}/1 = sqrt{3}).Alternatively, another way to find the asymptotes is to solve for (y) when the right-hand side is zero. So, starting from (frac{y^2}{3} - x^2 = 0), we can rearrange this as (frac{y^2}{3} = x^2), which leads to (y^2 = 3x^2). Taking square roots on both sides, we get (y = pm sqrt{3}x). So, that confirms it again.I think I'm confident now that option A is correct. But just to be thorough, let me consider if any of the other options could make sense.Option B is (y = pm frac{sqrt{3}}{3}x). That would be the case if the hyperbola was horizontal, but since it's vertical, the slope should be larger, not smaller. So, option B is probably for a different hyperbola.Option C is (y = pm 2x). That would be if (a/b = 2). But since (a = sqrt{3}) and (b = 1), (sqrt{3}) is approximately 1.732, which is less than 2, so option C is steeper than our asymptotes.Option D is (y = pm frac{2sqrt{3}}{3}x). Let me calculate that: (frac{2sqrt{3}}{3}) is approximately (1.1547), which is still less than (sqrt{3}) (approximately 1.732). So, that's not the slope we're looking for either.Therefore, after checking all options, option A is the correct one. I think I've covered all possible angles here, so I can be confident in my answer.**Final Answer**boxed{A}