💡Alright, so I have this problem where I need to find the smallest possible munificence for a monic cubic polynomial of the form ( p(x) = x^3 + px + q ). Munificence is defined as the maximum absolute value of the polynomial on the interval ([-1, 1]). First, I need to understand what exactly munificence means. It's the maximum of ( |p(x)| ) for ( x ) between -1 and 1. So, I need to find the polynomial ( p(x) ) with the smallest such maximum value. Since it's a monic cubic, the leading coefficient is 1, so the polynomial looks like ( x^3 + px + q ).I think the strategy here is to adjust the coefficients ( p ) and ( q ) such that the polynomial doesn't get too large in absolute value on the interval ([-1, 1]). Maybe I can make the polynomial oscillate between positive and negative values in a way that keeps the maximum absolute value as small as possible.Let me start by considering the endpoints of the interval, ( x = -1 ) and ( x = 1 ). Plugging these into the polynomial:- At ( x = -1 ): ( p(-1) = (-1)^3 + p(-1) + q = -1 - p + q )- At ( x = 1 ): ( p(1) = (1)^3 + p(1) + q = 1 + p + q )Also, at ( x = 0 ): ( p(0) = 0 + 0 + q = q )So, the values at these points are ( -1 - p + q ), ( q ), and ( 1 + p + q ). To minimize the maximum absolute value, I want these values to be as small as possible in absolute terms.Maybe I can set these values to be equal in absolute value but opposite in sign. That way, the polynomial doesn't go too high or too low. Let me try to set ( p(-1) = -M ), ( p(0) = M ), and ( p(1) = -M ) for some ( M ). This would create a sort of balance.So, setting up the equations:1. ( -1 - p + q = -M )2. ( q = M )3. ( 1 + p + q = -M )From equation 2, ( q = M ). Plugging this into equations 1 and 3:1. ( -1 - p + M = -M ) => ( -1 - p = -2M ) => ( p = 2M - 1 )2. ( 1 + p + M = -M ) => ( 1 + p = -2M ) => ( p = -2M - 1 )Now, I have two expressions for ( p ):- ( p = 2M - 1 )- ( p = -2M - 1 )Setting them equal:( 2M - 1 = -2M - 1 )Solving for ( M ):( 2M - 1 = -2M - 1 )( 4M = 0 )( M = 0 )Hmm, that's not helpful because if ( M = 0 ), then ( q = 0 ) and ( p = -1 ). Let me check what the polynomial becomes: ( x^3 - x ). Let's compute its values at -1, 0, and 1:- ( p(-1) = (-1)^3 - (-1) = -1 + 1 = 0 )- ( p(0) = 0 )- ( p(1) = 1 - 1 = 0 )So, all these points are zero, but what about in between? The polynomial ( x^3 - x ) factors as ( x(x^2 - 1) ), which has roots at -1, 0, and 1. The critical points are where the derivative is zero. Let's compute the derivative:( p'(x) = 3x^2 - 1 )Setting this equal to zero:( 3x^2 - 1 = 0 )( x^2 = 1/3 )( x = pm sqrt{1/3} approx pm 0.577 )So, the critical points are at approximately 0.577 and -0.577. Let's compute the value of the polynomial at these points:At ( x = sqrt{1/3} ):( p(sqrt{1/3}) = (sqrt{1/3})^3 - sqrt{1/3} = (1/3)^{3/2} - (1/3)^{1/2} = (1/3sqrt{3}) - (1/sqrt{3}) = (1 - 3)/3sqrt{3} = (-2)/3sqrt{3} approx -0.385 )Similarly, at ( x = -sqrt{1/3} ):( p(-sqrt{1/3}) = (-sqrt{1/3})^3 - (-sqrt{1/3}) = - (1/3)^{3/2} + (1/sqrt{3}) = (-1/3sqrt{3}) + (1/sqrt{3}) = ( -1 + 3 ) / 3sqrt{3} = 2 / 3sqrt{3} approx 0.385 )So, the maximum absolute value is approximately 0.385. That's better than 1, which was the case when ( p = 0 ) and ( q = 0 ). So, maybe this is a better candidate.But wait, I set ( p(-1) = -M ), ( p(0) = M ), and ( p(1) = -M ), but in reality, ( p(-1) = 0 ), ( p(0) = 0 ), and ( p(1) = 0 ). So, actually, the maximum absolute value occurs at the critical points, not at the endpoints or zero.So, perhaps the minimal munificence is approximately 0.385, but I need an exact value. Let's compute it exactly.We had:( p(sqrt{1/3}) = -2 / (3sqrt{3}) )So, the absolute value is ( 2 / (3sqrt{3}) ). Let's rationalize the denominator:( 2 / (3sqrt{3}) = (2sqrt{3}) / 9 approx 0.385 )So, the maximum absolute value is ( 2sqrt{3}/9 ). Is this the minimal munificence? Maybe, but let me check if there's a way to make it even smaller.Alternatively, perhaps I can adjust ( p ) and ( q ) such that the maximum absolute value is minimized. Maybe instead of setting the endpoints and zero to specific values, I can set the maximum at the critical points to be equal in absolute value but opposite in sign.Wait, in the case of ( p(x) = x^3 - x ), the maximum absolute value is ( 2sqrt{3}/9 ) at the critical points. Is this the minimal possible?I recall that Chebyshev polynomials are known for minimizing the maximum deviation from zero over the interval ([-1, 1]). The third Chebyshev polynomial is ( T_3(x) = 4x^3 - 3x ). Let me see:( T_3(x) = 4x^3 - 3x )This polynomial oscillates between -1 and 1 at four points in the interval ([-1, 1]). If I scale it appropriately, maybe I can get a monic polynomial.Let me divide ( T_3(x) ) by 4 to make it monic:( (1/4)T_3(x) = x^3 - (3/4)x )So, ( p(x) = x^3 - (3/4)x ). Let's compute its maximum absolute value on ([-1, 1]).First, compute the derivative:( p'(x) = 3x^2 - 3/4 )Setting this equal to zero:( 3x^2 - 3/4 = 0 )( x^2 = 1/4 )( x = pm 1/2 )So, the critical points are at ( x = 1/2 ) and ( x = -1/2 ).Compute ( p(1/2) ):( (1/2)^3 - (3/4)(1/2) = 1/8 - 3/8 = -2/8 = -1/4 )Compute ( p(-1/2) ):( (-1/2)^3 - (3/4)(-1/2) = -1/8 + 3/8 = 2/8 = 1/4 )So, the maximum absolute value is ( 1/4 ). That's better than the previous ( 2sqrt{3}/9 approx 0.385 ), since ( 1/4 = 0.25 ).Wait, is this correct? Let me double-check.( p(1/2) = (1/8) - (3/4)(1/2) = 1/8 - 3/8 = -2/8 = -1/4 )Yes, that's correct. Similarly, ( p(-1/2) = 1/4 ). So, the maximum absolute value is ( 1/4 ).But wait, what about the endpoints?Compute ( p(1) = 1 - (3/4)(1) = 1 - 3/4 = 1/4 )Compute ( p(-1) = (-1)^3 - (3/4)(-1) = -1 + 3/4 = -1/4 )So, at the endpoints, the polynomial is also ( pm 1/4 ). So, the maximum absolute value is indeed ( 1/4 ) at four points: ( x = -1, -1/2, 1/2, 1 ).Therefore, the polynomial ( p(x) = x^3 - (3/4)x ) has a munificence of ( 1/4 ).But wait, is this the minimal possible? Because Chebyshev polynomials are known to minimize the maximum deviation, so scaling ( T_3(x) ) to be monic should give the minimal maximum.Let me confirm. The Chebyshev polynomial ( T_3(x) ) has a leading coefficient of 4, so dividing by 4 gives a monic polynomial with leading coefficient 1. The maximum absolute value of ( T_3(x) ) is 1, so dividing by 4 scales it down to ( 1/4 ).Therefore, the minimal munificence is ( 1/4 ).But wait, in the earlier case with ( p(x) = x^3 - x ), the maximum was ( 2sqrt{3}/9 approx 0.385 ), which is larger than ( 1/4 = 0.25 ). So, indeed, scaling the Chebyshev polynomial gives a better result.Therefore, the minimal munificence is ( 1/4 ).But let me think again. The problem is to find the minimal munificence for a monic cubic polynomial of the form ( x^3 + px + q ). The Chebyshev polynomial approach gives ( p = -3/4 ) and ( q = 0 ). So, ( p(x) = x^3 - (3/4)x ).Is there any way to get a smaller maximum? I don't think so because Chebyshev polynomials are minimal in the maximum norm on the interval ([-1, 1]).Therefore, the smallest possible munificence is ( 1/4 ).Wait, but in the initial problem, the polynomial is ( x^3 + px + q ). In the Chebyshev case, ( q = 0 ). Is there a way to adjust ( q ) to make the maximum even smaller? Let me see.Suppose I set ( q ) to some value. If I shift the polynomial up or down, it might affect the maximum absolute value. For example, if I set ( q ) such that the maximum positive and negative values are balanced.But in the case of ( p(x) = x^3 - (3/4)x ), the maximum absolute value is already achieved at multiple points, both at the endpoints and the critical points. Shifting it up or down would likely increase the maximum absolute value somewhere else.For example, if I set ( q = c ), then ( p(x) = x^3 - (3/4)x + c ). The maximum absolute value would be the maximum of ( |x^3 - (3/4)x + c| ) over ([-1, 1]). To minimize this, I would need to choose ( c ) such that the maximum deviation is minimized. However, since the original polynomial already oscillates between ( -1/4 ) and ( 1/4 ), adding a constant ( c ) would shift all values by ( c ), potentially increasing the maximum absolute value.Therefore, the minimal maximum occurs when ( c = 0 ), giving the munificence of ( 1/4 ).Thus, the smallest possible munificence is ( 1/4 ).But wait, let me double-check with another approach. Suppose I use the equioscillation theorem, which states that the minimal maximum deviation is achieved when the error function equioscillates between its extreme values at least ( n+1 ) times, where ( n ) is the degree of the polynomial. For a cubic, that would be 4 points.In our case, the polynomial ( p(x) = x^3 - (3/4)x ) equioscillates between ( -1/4 ) and ( 1/4 ) at four points: ( x = -1, -1/2, 1/2, 1 ). Therefore, by the equioscillation theorem, this is indeed the minimal maximum.So, I'm confident that the minimal munificence is ( 1/4 ).**Final Answer**The smallest possible munificence is boxed{dfrac{1}{4}}.

💡Okay, so I have these two expressions to solve. Let me start with the first one. It looks a bit complicated with all these exponents and roots, but maybe I can break it down step by step.The first expression is:[frac{(a^{frac{2}{3}} cdot b^{-1})^{-frac{1}{2}} cdot a^{frac{1}{2}} cdot b^{frac{1}{3}}}{sqrt[6]{a cdot b^{5}}} quad (a > 0, b > 0)]Alright, so I need to simplify this. Let me remember the rules of exponents. When you have an exponent raised to another exponent, you multiply them. Also, when you multiply like bases, you add exponents, and when you divide, you subtract exponents.First, let's simplify the numerator:The numerator is ((a^{frac{2}{3}} cdot b^{-1})^{-frac{1}{2}} cdot a^{frac{1}{2}} cdot b^{frac{1}{3}}).Let me handle the first part: ((a^{frac{2}{3}} cdot b^{-1})^{-frac{1}{2}}).Using the rule ((x cdot y)^n = x^n cdot y^n), I can distribute the exponent:[(a^{frac{2}{3}})^{-frac{1}{2}} cdot (b^{-1})^{-frac{1}{2}}]Now, applying the exponent multiplication rule:For (a^{frac{2}{3}}) raised to (-frac{1}{2}):[a^{frac{2}{3} cdot -frac{1}{2}} = a^{-frac{1}{3}}]For (b^{-1}) raised to (-frac{1}{2}):[b^{-1 cdot -frac{1}{2}} = b^{frac{1}{2}}]So, the first part simplifies to (a^{-frac{1}{3}} cdot b^{frac{1}{2}}).Now, the entire numerator is:[a^{-frac{1}{3}} cdot b^{frac{1}{2}} cdot a^{frac{1}{2}} cdot b^{frac{1}{3}}]Combine like terms. For the (a) terms:[a^{-frac{1}{3} + frac{1}{2}} = a^{frac{-2}{6} + frac{3}{6}} = a^{frac{1}{6}}]For the (b) terms:[b^{frac{1}{2} + frac{1}{3}} = b^{frac{3}{6} + frac{2}{6}} = b^{frac{5}{6}}]So, the numerator simplifies to (a^{frac{1}{6}} cdot b^{frac{5}{6}}).Now, the denominator is (sqrt[6]{a cdot b^{5}}). Let me rewrite that using exponents:[(a cdot b^{5})^{frac{1}{6}} = a^{frac{1}{6}} cdot b^{frac{5}{6}}]So, the entire expression becomes:[frac{a^{frac{1}{6}} cdot b^{frac{5}{6}}}{a^{frac{1}{6}} cdot b^{frac{5}{6}}} = 1]Wait, that's interesting. So, the entire expression simplifies to 1? Let me double-check my steps.Starting from the numerator:1. ((a^{frac{2}{3}} cdot b^{-1})^{-frac{1}{2}}) becomes (a^{-frac{1}{3}} cdot b^{frac{1}{2}}).2. Multiply by (a^{frac{1}{2}} cdot b^{frac{1}{3}}), so combining exponents: - For (a): (-frac{1}{3} + frac{1}{2} = frac{-2 + 3}{6} = frac{1}{6}) - For (b): (frac{1}{2} + frac{1}{3} = frac{3 + 2}{6} = frac{5}{6})3. Denominator: (sqrt[6]{a cdot b^5} = a^{frac{1}{6}} cdot b^{frac{5}{6}})4. So, numerator and denominator are the same, hence 1.Looks correct. So, the first expression simplifies to 1.Now, moving on to the second expression:[frac{1}{2} lg frac{32}{49} - frac{4}{3} lg sqrt{8} + lg sqrt{245}]Hmm, logarithms. I need to simplify this expression. Let me recall the logarithm properties:1. (lg frac{a}{b} = lg a - lg b)2. (lg a^n = n lg a)3. (lg a + lg b = lg (a cdot b))4. (lg a - lg b = lg frac{a}{b})Let me break down each term.First term: (frac{1}{2} lg frac{32}{49})Using property 1:[frac{1}{2} (lg 32 - lg 49)]I know that 32 is (2^5) and 49 is (7^2), so:[frac{1}{2} (lg 2^5 - lg 7^2) = frac{1}{2} (5 lg 2 - 2 lg 7)]Second term: (-frac{4}{3} lg sqrt{8})First, (sqrt{8} = 8^{frac{1}{2}} = (2^3)^{frac{1}{2}} = 2^{frac{3}{2}}). So:[-frac{4}{3} lg 2^{frac{3}{2}} = -frac{4}{3} cdot frac{3}{2} lg 2 = -2 lg 2]Wait, let me verify that step. The exponent is (frac{3}{2}), so multiplying by (-frac{4}{3}):[-frac{4}{3} cdot frac{3}{2} = -2]Yes, that's correct.Third term: (lg sqrt{245})(sqrt{245} = 245^{frac{1}{2}}). Let me factor 245. 245 divided by 5 is 49, so 245 = 5 * 49 = 5 * 7^2. So:[lg (5 cdot 7^2)^{frac{1}{2}} = frac{1}{2} lg (5 cdot 7^2) = frac{1}{2} (lg 5 + lg 7^2) = frac{1}{2} (lg 5 + 2 lg 7)]So, putting it all together:First term: (frac{5}{2} lg 2 - lg 7)Second term: (-2 lg 2)Third term: (frac{1}{2} lg 5 + lg 7)So, combining all terms:[left( frac{5}{2} lg 2 - lg 7 right) + (-2 lg 2) + left( frac{1}{2} lg 5 + lg 7 right)]Let me combine like terms:For (lg 2):[frac{5}{2} lg 2 - 2 lg 2 = frac{5}{2} lg 2 - frac{4}{2} lg 2 = frac{1}{2} lg 2]For (lg 7):[- lg 7 + lg 7 = 0]And the remaining term is (frac{1}{2} lg 5).So, altogether:[frac{1}{2} lg 2 + frac{1}{2} lg 5]Factor out (frac{1}{2}):[frac{1}{2} (lg 2 + lg 5) = frac{1}{2} lg (2 cdot 5) = frac{1}{2} lg 10]Since (lg 10 = 1), this simplifies to (frac{1}{2}).Wait, but in the original problem, it's written as (frac{1}{2} lg 10). Depending on the context, sometimes (lg) is base 10, so (lg 10 = 1), hence (frac{1}{2}). But maybe I should leave it as (frac{1}{2} lg 10) unless specified otherwise.Let me double-check my steps:1. First term: (frac{1}{2} lg frac{32}{49}) becomes (frac{5}{2} lg 2 - lg 7)2. Second term: (-frac{4}{3} lg sqrt{8}) becomes (-2 lg 2)3. Third term: (lg sqrt{245}) becomes (frac{1}{2} lg 5 + lg 7)4. Combining: (frac{5}{2} lg 2 - lg 7 - 2 lg 2 + frac{1}{2} lg 5 + lg 7)5. Simplify: (frac{1}{2} lg 2 + frac{1}{2} lg 5 = frac{1}{2} lg 10)Yes, that seems correct. So, the second expression simplifies to (frac{1}{2} lg 10).But wait, if (lg) is base 10, then (lg 10 = 1), so it's just (frac{1}{2}). Maybe the answer is (frac{1}{2}), but the problem didn't specify the base, so perhaps it's better to leave it as (frac{1}{2} lg 10).Alternatively, if (lg) is natural logarithm, but usually (lg) is base 10 and (ln) is natural. So, I think (lg 10 = 1), so the answer is (frac{1}{2}).But let me check the original problem. It says:[frac{1}{2} lg frac{32}{49} - frac{4}{3} lg sqrt{8} + lg sqrt{245}]Yes, so (lg) is likely base 10. So, (lg 10 = 1), so the expression simplifies to (frac{1}{2}).Wait, but in my earlier step, I had (frac{1}{2} lg 10), which is (frac{1}{2}) if (lg 10 = 1). So, the final answer is (frac{1}{2}).But let me make sure I didn't make any mistakes in the coefficients.First term: (frac{1}{2} lg frac{32}{49} = frac{1}{2} (5 lg 2 - 2 lg 7) = frac{5}{2} lg 2 - lg 7)Second term: (-frac{4}{3} lg sqrt{8} = -frac{4}{3} cdot frac{1}{2} lg 8 = -frac{2}{3} lg 8). Wait, hold on, I think I made a mistake here earlier.Wait, (sqrt{8} = 8^{1/2}), so (lg sqrt{8} = frac{1}{2} lg 8). So, (-frac{4}{3} cdot frac{1}{2} lg 8 = -frac{2}{3} lg 8).But 8 is (2^3), so (lg 8 = 3 lg 2). Therefore:(-frac{2}{3} cdot 3 lg 2 = -2 lg 2). So, that part was correct.Third term: (lg sqrt{245} = frac{1}{2} lg 245). Since 245 = 5 * 49 = 5 * 7^2, so:(frac{1}{2} (lg 5 + 2 lg 7) = frac{1}{2} lg 5 + lg 7). That was correct.So, combining all terms:(frac{5}{2} lg 2 - lg 7 - 2 lg 2 + frac{1}{2} lg 5 + lg 7)Simplify:(frac{5}{2} lg 2 - 2 lg 2 = frac{5}{2} lg 2 - frac{4}{2} lg 2 = frac{1}{2} lg 2)(- lg 7 + lg 7 = 0)And (frac{1}{2} lg 5)So, total: (frac{1}{2} lg 2 + frac{1}{2} lg 5 = frac{1}{2} (lg 2 + lg 5) = frac{1}{2} lg (2*5) = frac{1}{2} lg 10)Since (lg 10 = 1), this is (frac{1}{2}).Wait, but in the initial problem, the expression is:[frac{1}{2} lg frac{32}{49} - frac{4}{3} lg sqrt{8} + lg sqrt{245}]So, all steps are correct, and the final result is (frac{1}{2}).But in the first problem, I got 1, and in the second, (frac{1}{2}). Let me just make sure I didn't make any calculation errors.For the first problem:Numerator after simplifying: (a^{frac{1}{6}} b^{frac{5}{6}})Denominator: (a^{frac{1}{6}} b^{frac{5}{6}})So, ratio is 1. Correct.Second problem:All steps led to (frac{1}{2} lg 10 = frac{1}{2}). Correct.So, I think both answers are correct.

💡Alright, so I have this problem here: the function f(x) is given as sin x plus cos x, and it's derivative satisfies f'(x) equals 3 times f(x). I need to find the value of tan 2x. The options are A) -4/3, B) 4/3, C) -3/4, and D) 3/4.Okay, let's start by understanding what's given. The function f(x) is sin x plus cos x. That seems straightforward. Now, the derivative of f(x) is given as f'(x) = 3f(x). So, f'(x) equals 3 times sin x plus 3 times cos x.First, I should find the derivative of f(x). The derivative of sin x is cos x, and the derivative of cos x is -sin x. So, f'(x) should be cos x minus sin x. Wait, but the problem says f'(x) equals 3f(x), which is 3 times (sin x plus cos x). So, setting these equal: cos x minus sin x equals 3 sin x plus 3 cos x.Hmm, okay, let's write that out:cos x - sin x = 3 sin x + 3 cos x.Now, I need to solve for x or maybe find a relationship between sin x and cos x. Let's bring all terms to one side:cos x - sin x - 3 sin x - 3 cos x = 0.Combining like terms:(cos x - 3 cos x) + (-sin x - 3 sin x) = 0Which simplifies to:-2 cos x - 4 sin x = 0.Wait, that doesn't seem right. Let me check my steps again. Starting from f'(x) = cos x - sin x, and f'(x) = 3f(x) = 3 sin x + 3 cos x. So:cos x - sin x = 3 sin x + 3 cos x.Subtracting 3 sin x and 3 cos x from both sides:cos x - sin x - 3 sin x - 3 cos x = 0.So, cos x - 3 cos x is -2 cos x, and -sin x - 3 sin x is -4 sin x. So, -2 cos x - 4 sin x = 0.Yes, that's correct. So, -2 cos x - 4 sin x = 0. Let's factor out a common factor, which is -2:-2 (cos x + 2 sin x) = 0.Dividing both sides by -2:cos x + 2 sin x = 0.So, cos x = -2 sin x.Now, I can divide both sides by cos x to get tan x:tan x = sin x / cos x = sin x / (-2 sin x) = -1/2.Wait, that gives tan x = -1/2. Hmm, but the question asks for tan 2x. Okay, so I need to find tan 2x given that tan x is -1/2.I remember there's a double angle formula for tangent:tan 2x = 2 tan x / (1 - tan² x).So, plugging in tan x = -1/2:tan 2x = 2*(-1/2) / (1 - (-1/2)²) = (-1) / (1 - 1/4) = (-1) / (3/4) = -4/3.So, tan 2x is -4/3, which corresponds to option A.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from f'(x) = cos x - sin x, set equal to 3f(x) = 3 sin x + 3 cos x. Then, moving all terms to one side: cos x - sin x - 3 sin x - 3 cos x = 0, which simplifies to -2 cos x - 4 sin x = 0. Factoring out -2 gives cos x + 2 sin x = 0, so cos x = -2 sin x. Dividing both sides by cos x gives tan x = -1/2. Then, using the double angle formula for tangent: tan 2x = 2*(-1/2)/(1 - (1/4)) = (-1)/(3/4) = -4/3. Yes, that seems correct.I think I did everything right. So, the answer should be A) -4/3.

💡Alright, so I have this problem where there's a drawer with 4 shirts, 5 pairs of shorts, and 6 pairs of socks. I need to find the probability that if I randomly take out three articles of clothing, I'll end up with one shirt, one pair of shorts, and one pair of socks. Okay, let's break this down step by step.First, I need to understand the total number of articles of clothing in the drawer. There are 4 shirts, 5 pairs of shorts, and 6 pairs of socks. So, adding those up: 4 + 5 + 6 equals 15 articles in total. That means there are 15 different items I could potentially pull out.Now, I need to figure out how many ways I can choose 3 articles out of these 15. This sounds like a combination problem because the order in which I pick the clothes doesn't matter. Whether I pick a shirt first or last, it's still just one shirt. So, the total number of possible ways to choose 3 articles out of 15 is given by the combination formula, which is:[text{Total combinations} = binom{15}{3} = frac{15!}{3!(15-3)!}]Calculating that, I get:[binom{15}{3} = frac{15 times 14 times 13}{3 times 2 times 1} = 455]So, there are 455 different ways to pull out any 3 articles of clothing.Next, I need to find out how many of these combinations will give me exactly one shirt, one pair of shorts, and one pair of socks. To do this, I'll calculate the number of ways to choose one shirt from the 4 available, one pair of shorts from the 5, and one pair of socks from the 6.The number of ways to choose one shirt is simply 4, since there are 4 shirts. Similarly, the number of ways to choose one pair of shorts is 5, and the number of ways to choose one pair of socks is 6. To find the total number of favorable combinations, I multiply these together:[text{Favorable combinations} = 4 times 5 times 6 = 120]So, there are 120 ways to choose one shirt, one pair of shorts, and one pair of socks.Now, to find the probability, I'll divide the number of favorable combinations by the total number of possible combinations:[text{Probability} = frac{120}{455}]Simplifying this fraction by dividing both numerator and denominator by 5, I get:[frac{120 div 5}{455 div 5} = frac{24}{91}]So, the probability of pulling out one shirt, one pair of shorts, and one pair of socks is (frac{24}{91}).Wait a second, let me double-check my calculations to make sure I didn't make a mistake. The total number of articles is definitely 15, and the combination formula for choosing 3 out of 15 gives 455. For the favorable outcomes, multiplying 4 shirts by 5 shorts by 6 socks gives 120, which seems right. Dividing 120 by 455 simplifies to 24/91. Yeah, that seems correct.I think I've got it. The probability is (frac{24}{91}).

💡Alright, so I have this polynomial equation: (x^3 + ax^2 + bx + c = 0), where (a), (b), and (c) are rational numbers. I know that (3 - sqrt{5}) is one of the roots, and there are two other roots, one of which is an integer. I need to find that integer root.First, I remember that for polynomials with rational coefficients, if there's an irrational root like (3 - sqrt{5}), its conjugate (3 + sqrt{5}) must also be a root. That makes sense because the irrational parts have to cancel out when you perform operations with rational coefficients.So, now I know two of the roots: (3 - sqrt{5}) and (3 + sqrt{5}). Let's call the third root (r), which is an integer. Using Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots, I can write down some equations. For a cubic polynomial (x^3 + ax^2 + bx + c = 0), the sum of the roots is (-a), the sum of the products of the roots taken two at a time is (b), and the product of the roots is (-c).Let me write down the sum of the roots first:[(3 - sqrt{5}) + (3 + sqrt{5}) + r = -a]Simplifying the left side:[3 - sqrt{5} + 3 + sqrt{5} + r = 6 + r]So, (6 + r = -a). Since (a) is a rational number, (r) must also be rational. But we already know (r) is an integer, so that's consistent.Now, I don't know the value of (a), but maybe I can find another equation involving (r). Let's look at the product of the roots:[(3 - sqrt{5})(3 + sqrt{5})r = -c]First, calculate ((3 - sqrt{5})(3 + sqrt{5})). That's a difference of squares:[(3)^2 - (sqrt{5})^2 = 9 - 5 = 4]So, the product becomes:[4r = -c]Since (c) is rational, (4r) must be rational, which it is because (r) is an integer.Hmm, but this doesn't directly help me find (r). Maybe I should consider the sum of the products of the roots taken two at a time, which is equal to (b). Let's compute that:[(3 - sqrt{5})(3 + sqrt{5}) + (3 - sqrt{5})r + (3 + sqrt{5})r = b]We already know that ((3 - sqrt{5})(3 + sqrt{5}) = 4), so:[4 + (3 - sqrt{5})r + (3 + sqrt{5})r = b]Combine the terms with (r):[4 + [ (3 - sqrt{5}) + (3 + sqrt{5}) ]r = b]Simplify inside the brackets:[(3 - sqrt{5} + 3 + sqrt{5}) = 6]So, the equation becomes:[4 + 6r = b]Again, since (b) is rational, this is consistent because both 4 and (6r) are rational.But I still don't have enough information to find (r). Maybe I need to think differently. Since (a), (b), and (c) are rational, and we have expressions involving (r), perhaps I can find (r) by considering that the polynomial must factor nicely with integer coefficients.Let me try to write the polynomial as:[(x - (3 - sqrt{5}))(x - (3 + sqrt{5}))(x - r) = 0]First, multiply the two conjugate factors:[(x - (3 - sqrt{5}))(x - (3 + sqrt{5})) = [x - 3 + sqrt{5}][x - 3 - sqrt{5}]]This is again a difference of squares:[(x - 3)^2 - (sqrt{5})^2 = (x^2 - 6x + 9) - 5 = x^2 - 6x + 4]So, the polynomial becomes:[(x^2 - 6x + 4)(x - r) = 0]Let's expand this:[x^3 - rx^2 - 6x^2 + 6rx + 4x - 4r = 0]Combine like terms:[x^3 + (-r - 6)x^2 + (6r + 4)x - 4r = 0]Comparing this with the original polynomial (x^3 + ax^2 + bx + c = 0), we can equate coefficients:1. Coefficient of (x^3): 1 = 1 (which is consistent)2. Coefficient of (x^2): (-r - 6 = a)3. Coefficient of (x): (6r + 4 = b)4. Constant term: (-4r = c)Since (a), (b), and (c) are rational, and (r) is an integer, we need to find an integer (r) such that all these coefficients are rational. But since (r) is an integer, all these expressions will automatically be rational. So, we need another condition to find (r).Wait, maybe I can use the fact that the polynomial must have integer coefficients if (r) is an integer. But the problem only states that (a), (b), and (c) are rational, not necessarily integers. Hmm, that complicates things a bit.Alternatively, perhaps I can use the Rational Root Theorem. The Rational Root Theorem states that any possible rational root, expressed in lowest terms (p/q), must have (p) dividing the constant term and (q) dividing the leading coefficient. In our case, the leading coefficient is 1, so any rational root must be an integer that divides the constant term (c).But in our polynomial, the constant term is (-4r). So, the possible rational roots are the divisors of (-4r). However, since (r) is an integer, the possible integer roots are the divisors of (-4r). But since (r) itself is a root, it must divide (-4r). That seems a bit circular.Wait, maybe I can think of it differently. Since (r) is an integer root, and the polynomial has rational coefficients, then (r) must satisfy the equation (x^3 + ax^2 + bx + c = 0). But without knowing (a), (b), or (c), it's hard to plug in values.Alternatively, maybe I can use the fact that the sum of the roots is (-a). We have:[(3 - sqrt{5}) + (3 + sqrt{5}) + r = 6 + r = -a]So, (6 + r = -a). Since (a) is rational, (r) must be such that (6 + r) is rational, which it is because (r) is an integer. But this doesn't give me a specific value for (r).Wait, maybe I can consider that the minimal polynomial of (3 - sqrt{5}) is quadratic, which is (x^2 - 6x + 4), as we found earlier. So, the cubic polynomial must be a multiple of this quadratic. Therefore, the cubic polynomial can be written as ((x^2 - 6x + 4)(x - r)), which we already did.Since the cubic polynomial has rational coefficients, and (r) is an integer, perhaps I can find (r) by considering that the product must result in integer coefficients. But since (a), (b), and (c) are only required to be rational, not necessarily integers, this might not help directly.Wait, but if I consider that the coefficients (a), (b), and (c) are rational, and (r) is an integer, then the expressions for (a), (b), and (c) in terms of (r) must all be rational. But since (r) is an integer, these expressions are already rational. So, this doesn't give me new information.Maybe I need to think about the possible integer roots. Since (r) is an integer, and the polynomial is cubic, perhaps I can test possible integer values for (r) that would make the polynomial have rational coefficients.But without knowing (a), (b), or (c), it's hard to test specific values. However, perhaps I can use the fact that the product of the roots is (-c). We have:[(3 - sqrt{5})(3 + sqrt{5})r = 4r = -c]So, (c = -4r). Since (c) is rational, and (r) is an integer, this is fine. But again, this doesn't help me find (r).Wait, maybe I can use the sum of the products of the roots taken two at a time, which is (b). We have:[(3 - sqrt{5})(3 + sqrt{5}) + (3 - sqrt{5})r + (3 + sqrt{5})r = 4 + 6r = b]So, (b = 4 + 6r). Since (b) is rational, and (r) is an integer, this is consistent.But I still don't have enough information to find (r). Maybe I need to consider that the polynomial must have integer coefficients if (r) is an integer. Wait, no, the problem only states that (a), (b), and (c) are rational, not necessarily integers. So, that approach might not work.Alternatively, perhaps I can consider that the minimal polynomial of (3 - sqrt{5}) is quadratic, and since the cubic polynomial has this quadratic as a factor, the third root (r) must be such that the cubic polynomial is the product of the quadratic and a linear term with rational coefficients. But since we already have that, and (r) is an integer, maybe I can find (r) by considering that the cubic polynomial must have integer coefficients if (r) is an integer. Wait, but the problem doesn't specify that (a), (b), and (c) are integers, just rational.Hmm, this is tricky. Maybe I need to think differently. Let's consider that the sum of the roots is (6 + r = -a). If I can find (a), I can find (r). But I don't know (a).Wait, but perhaps I can use the fact that the polynomial must have integer coefficients if (r) is an integer. Let me check that.If (r) is an integer, then the coefficients of the polynomial would be:- (a = - (6 + r))- (b = 4 + 6r)- (c = -4r)Since (a), (b), and (c) are rational, and (r) is an integer, these expressions are all rational. But the problem doesn't specify that (a), (b), and (c) are integers, just rational. So, (r) could be any integer, and (a), (b), and (c) would adjust accordingly.But the problem states that there is an integer root, so (r) must be an integer. However, without additional constraints, there are infinitely many possible integer values for (r). But the problem must have a unique solution, so perhaps I'm missing something.Wait, maybe I can consider that the polynomial must have integer coefficients if (r) is an integer. Let me think about that. If (r) is an integer, then the coefficients (a), (b), and (c) would be rational, but not necessarily integers. However, if the polynomial is monic (which it is, since the coefficient of (x^3) is 1), and if all roots are algebraic integers, then the coefficients would be integers. But (3 - sqrt{5}) is not an algebraic integer because its minimal polynomial is (x^2 - 6x + 4), which has a leading coefficient of 1, so it is an algebraic integer. Similarly, (r) is an integer, so it's also an algebraic integer. Therefore, the sum and products would be algebraic integers, implying that (a), (b), and (c) are integers.Wait, is that correct? Let me recall that the sum and products of algebraic integers are algebraic integers. Since (3 - sqrt{5}) and (3 + sqrt{5}) are algebraic integers, and (r) is an integer (hence an algebraic integer), then the coefficients (a), (b), and (c) must be algebraic integers. But since they are rational, they must be rational integers, i.e., integers.So, this implies that (a), (b), and (c) are integers. Therefore, the coefficients must be integers. So, now I can say that (a), (b), and (c) are integers.Given that, let's revisit the expressions:1. (a = - (6 + r))2. (b = 4 + 6r)3. (c = -4r)Since (a), (b), and (c) are integers, and (r) is an integer, this is consistent.Now, perhaps I can find (r) by considering that the polynomial must have integer coefficients, and the minimal polynomial of (3 - sqrt{5}) is (x^2 - 6x + 4). So, the cubic polynomial is ((x^2 - 6x + 4)(x - r)). Expanding this, as we did earlier, gives:[x^3 - (r + 6)x^2 + (6r + 4)x - 4r]Since (a), (b), and (c) are integers, and (r) is an integer, this is fine. But how does this help me find (r)?Wait, perhaps I can consider that the cubic polynomial must have integer coefficients, and since (3 - sqrt{5}) is a root, the minimal polynomial must divide the cubic polynomial. But since the minimal polynomial is quadratic, and the cubic is of degree three, the third root (r) must be such that the cubic polynomial is the product of the minimal polynomial and a linear term with integer coefficients.But we already have that, so perhaps I need to find (r) such that the cubic polynomial has integer coefficients, which it does for any integer (r). So, this approach might not help.Wait, maybe I can consider that the cubic polynomial must have integer coefficients, and since (r) is an integer, the coefficients (a), (b), and (c) are integers. Therefore, the expressions for (a), (b), and (c) in terms of (r) must be integers, which they are.But I still don't have a specific value for (r). Maybe I need to consider that the polynomial must have integer coefficients, and the minimal polynomial of (3 - sqrt{5}) is (x^2 - 6x + 4), so the cubic polynomial is ((x^2 - 6x + 4)(x - r)). Therefore, the coefficients of the cubic polynomial are:- (x^3): 1- (x^2): (- (r + 6))- (x): (6r + 4)- Constant term: (-4r)Since all coefficients are integers, (r) must be an integer. But without additional constraints, (r) can be any integer. However, the problem states that there is an integer root, so (r) must be such that the cubic polynomial has integer coefficients, which it does for any integer (r).But the problem must have a unique solution, so perhaps I'm missing something. Maybe I can consider that the polynomial must have integer coefficients, and the minimal polynomial of (3 - sqrt{5}) is (x^2 - 6x + 4), so the cubic polynomial is ((x^2 - 6x + 4)(x - r)). Therefore, the coefficients of the cubic polynomial are:- (x^3): 1- (x^2): (- (r + 6))- (x): (6r + 4)- Constant term: (-4r)Since all coefficients are integers, (r) must be an integer. But without additional constraints, (r) can be any integer. However, the problem states that there is an integer root, so (r) must be such that the cubic polynomial has integer coefficients, which it does for any integer (r).Wait, but the problem says that (a), (b), and (c) are rational, not necessarily integers. So, maybe my earlier assumption that they must be integers is incorrect. Let me reconsider.If (a), (b), and (c) are only required to be rational, then (r) can be any integer, and (a), (b), and (c) would adjust accordingly to be rational. Therefore, without additional constraints, there are infinitely many possible integer roots (r). But the problem states that there is an integer root, implying that there is a unique solution.This suggests that perhaps there is a specific value of (r) that makes the polynomial have rational coefficients, but I'm not sure how to find it without more information.Wait, maybe I can consider that the minimal polynomial of (3 - sqrt{5}) is (x^2 - 6x + 4), and since the cubic polynomial has this as a factor, the third root (r) must be such that the cubic polynomial is the product of the minimal polynomial and a linear term with rational coefficients. But since (r) is an integer, the linear term must have integer coefficients, implying that (r) is an integer.But again, this doesn't give me a specific value for (r).Wait, perhaps I can use the fact that the sum of the roots is (6 + r = -a). If I assume that (a) is an integer, then (r) must be such that (6 + r) is an integer, which it is because (r) is an integer. But the problem doesn't specify that (a) is an integer, only that it's rational.I'm stuck. Maybe I need to think differently. Let me consider specific integer values for (r) and see if they lead to rational coefficients.Let's try (r = -6):Then,- (a = - (6 + (-6)) = 0)- (b = 4 + 6*(-6) = 4 - 36 = -32)- (c = -4*(-6) = 24)So, the polynomial would be (x^3 + 0x^2 - 32x + 24 = x^3 - 32x + 24). Let's check if (3 - sqrt{5}) is a root:Plug (x = 3 - sqrt{5}) into the polynomial:[(3 - sqrt{5})^3 - 32(3 - sqrt{5}) + 24]First, calculate ((3 - sqrt{5})^3):[(3 - sqrt{5})^3 = (3 - sqrt{5})(3 - sqrt{5})(3 - sqrt{5})]First, compute ((3 - sqrt{5})^2):[(3 - sqrt{5})^2 = 9 - 6sqrt{5} + 5 = 14 - 6sqrt{5}]Then multiply by (3 - sqrt{5}):[(14 - 6sqrt{5})(3 - sqrt{5}) = 14*3 - 14sqrt{5} - 18sqrt{5} + 6*5 = 42 - 14sqrt{5} - 18sqrt{5} + 30 = 72 - 32sqrt{5}]Now, compute (-32(3 - sqrt{5})):[-32*3 + 32sqrt{5} = -96 + 32sqrt{5}]Add all terms together:[72 - 32sqrt{5} - 96 + 32sqrt{5} + 24 = (72 - 96 + 24) + (-32sqrt{5} + 32sqrt{5}) = 0 + 0 = 0]So, (3 - sqrt{5}) is indeed a root when (r = -6). Therefore, the integer root is (-6).I think this is the correct answer because when I tested (r = -6), it satisfied the polynomial equation with (3 - sqrt{5}) as a root, and all coefficients (a), (b), and (c) are rational (in fact, integers in this case). Since the problem states that there is an integer root, and we've found that (-6) works, I believe that's the answer.

💡First, I need to determine the length of the goods train. The train is moving at a speed of 72 kilometers per hour and crosses a 290-meter-long platform in 26 seconds.To make the units consistent, I'll convert the speed from kilometers per hour to meters per second. This is done by multiplying the speed by 1000 meters per kilometer and dividing by 3600 seconds per hour.Next, I'll calculate the total distance the train travels while crossing the platform by multiplying the speed in meters per second by the time taken in seconds.Finally, the length of the train can be found by subtracting the length of the platform from the total distance traveled.