answer：First, I need to determine the total amount of work required to dig a 30-meter-deep hole in sandy soil. With 45 workers working for 8 hours, the total man-hours can be calculated by multiplying the number of workers by the hours worked.Next, since digging in clay soil takes twice as long as in sandy soil, the total man-hours needed for the clay soil task will be double the man-hours required for the sandy soil task.After calculating the total man-hours needed for the clay soil, I need to find out how many workers are required to complete this task within the available 6 hours. This can be done by dividing the total man-hours by the number of hours available.Finally, to determine the number of extra workers needed, I subtract the current number of workers (45) from the total number of workers required to complete the task in the given time frame.

answer：Okay, so I have this problem: If f(x) is an even function, is f(f(f(x))) even, odd, or neither? Hmm, let me think about this step by step.First, I remember that an even function satisfies the property f(-x) = f(x) for all x in its domain. That means if I plug in -x into the function, I get the same result as when I plug in x. Examples of even functions include f(x) = x², f(x) = cos(x), and so on.Now, the question is about composing f three times, so f(f(f(x))). I need to figure out if this composition is even, odd, or neither. Let me recall what it means for a function to be even or odd.An even function satisfies f(-x) = f(x), as I mentioned. An odd function satisfies f(-x) = -f(x). If a function doesn't satisfy either of these properties, it's neither even nor odd.So, to determine if f(f(f(x))) is even, I need to check if f(f(f(-x))) equals f(f(f(x))). If it does, then f(f(f(x))) is even. If not, maybe it's odd or neither.Let me start by evaluating f(-x). Since f is even, f(-x) = f(x). So, f(-x) is just f(x). Now, let's consider f(f(-x)). Since f(-x) = f(x), this becomes f(f(x)). So, f(f(-x)) = f(f(x)). That means f(f(x)) is also even because f(f(-x)) equals f(f(x)).Now, moving on to f(f(f(-x))). Let's break it down. Starting from the inside, f(-x) = f(x). Then, f(f(-x)) = f(f(x)). So, f(f(f(-x))) = f(f(f(x))). Therefore, f(f(f(-x))) equals f(f(f(x))). Wait a second, that means f(f(f(x))) is even because f(f(f(-x))) equals f(f(f(x))). So, it seems like the composition of an even function three times is still even.But let me double-check to make sure I'm not missing something. Suppose f(x) is even, so f(-x) = f(x). Then, f(f(f(-x))) = f(f(f(x))) because f(-x) = f(x), and so on. So, each time I apply f, the evenness is preserved.Is there a case where this might not hold? Let me think of a specific example. Let's take f(x) = x², which is even. Then f(f(f(x))) would be f(f(x²)) = f((x²)²) = f(x⁴) = (x⁴)² = x⁸. Is x⁸ even? Yes, because (-x)⁸ = x⁸. So, in this case, f(f(f(x))) is even.Another example: f(x) = |x|, which is even. Then f(f(f(x))) = f(f(|x|)) = f(| |x| |) = f(|x|) = ||x|| = |x|. Wait, |x| is even, so f(f(f(x))) is still even.What about a more complicated even function? Let's say f(x) = x² + cos(x). Then f(f(f(x))) would be f(f(x² + cos(x))). First, compute f(x² + cos(x)) = (x² + cos(x))² + cos(x² + cos(x)). Then, f of that would be [ (x² + cos(x))² + cos(x² + cos(x)) ]² + cos( [ (x² + cos(x))² + cos(x² + cos(x)) ] ). This seems complicated, but regardless of the complexity, since each application of f preserves evenness, the composition should still be even.Wait, let me test this with a specific value. Let's take x = 1 and x = -1. For f(x) = x², f(f(f(1))) = f(f(1)) = f(1) = 1. Similarly, f(f(f(-1))) = f(f(1)) = f(1) = 1. So, f(f(f(-1))) = f(f(f(1))), which holds.For f(x) = |x|, f(f(f(1))) = f(f(1)) = f(1) = 1, and f(f(f(-1))) = f(f(1)) = f(1) = 1. Again, it holds.What about a non-integer example? Let's take x = π/2. For f(x) = cos(x), which is even. Then f(f(f(π/2))) = f(f(0)) = f(1) = cos(1). Similarly, f(f(f(-π/2))) = f(f(0)) = f(1) = cos(1). So, again, f(f(f(-π/2))) = f(f(f(π/2))). It holds.Hmm, so in all these cases, f(f(f(x))) is even. Is there any case where it might not be? Let me think about a function that's even but maybe has some symmetry that could cause issues.Suppose f(x) is a constant function, say f(x) = c, where c is a constant. Then f(f(f(x))) = f(f(c)) = f(c) = c. Since c is a constant, it's technically even because f(-x) = c = f(x). So, even in this case, f(f(f(x))) is even.Wait, but what if f(x) is a more complicated even function, like f(x) = x² + x⁴? Then f(f(f(x))) would be f(f(x² + x⁴)) = f( (x² + x⁴)² + (x² + x⁴)^4 ). This is still an even function because all the exponents are even, so replacing x with -x doesn't change the value.I'm starting to see a pattern here. Every time I compose an even function with itself, the resulting function remains even. So, composing it three times should still result in an even function.Let me try to formalize this. Suppose f is even, so f(-x) = f(x). Then, f(f(f(-x))) = f(f(f(x))) because f(-x) = f(x). Therefore, f(f(f(-x))) = f(f(f(x))), which means f(f(f(x))) is even.Wait, that seems too straightforward. Am I missing something? Let me think about the definition again. A function g(x) is even if g(-x) = g(x). So, if I define g(x) = f(f(f(x))), then g(-x) = f(f(f(-x))) = f(f(f(x))) = g(x). Therefore, g(x) is even.Yes, that makes sense. So, regardless of how many times I compose an even function with itself, the result will always be even. Therefore, f(f(f(x))) is even.I think I've convinced myself that f(f(f(x))) is even. All the examples I tried support this, and the formal proof using the definition of even functions also shows that f(f(f(-x))) equals f(f(f(x))), which is the definition of an even function.So, the conclusion is that f(f(f(x))) is even.

answer：Okay, so Jim sold a statue for 550, and it cost him 407.41 to make it. He made a profit, and I need to find out what percentage that profit is. Hmm, percentage profit... I think that means how much more money he made compared to his original cost, expressed as a percentage.First, I should figure out how much profit Jim actually made. Profit is usually the selling price minus the cost price, right? So, let me subtract the cost from the selling price:550 (selling price) - 407.41 (cost price) = 142.59.Okay, so Jim made a 142.59 profit. Now, to find out what percentage this is relative to his original cost, I think I need to divide the profit by the original cost and then multiply by 100 to get the percentage.So, that would be:(142.59 / 407.41) * 100.Let me do the division first. 142.59 divided by 407.41. Hmm, I'm not sure about this division. Maybe I can approximate it or use a calculator. Wait, if I divide 142.59 by 407.41, what do I get?Well, 407.41 times 0.35 is approximately 142.59 because 407.41 * 0.35 = 142.5935, which is very close to 142.59. So, that means the division result is approximately 0.35.Now, to get the percentage, I multiply by 100:0.35 * 100 = 35%.So, Jim made a 35% profit on the statue.Wait, let me double-check my calculations to make sure I didn't make a mistake. I subtracted the cost from the selling price to get the profit, which seems right. Then I divided the profit by the original cost and multiplied by 100 to get the percentage. That all makes sense.Just to be thorough, let me redo the division:142.59 divided by 407.41.Well, 407.41 goes into 142.59 about 0.35 times because 407.41 * 0.35 is roughly 142.59. So, yes, that checks out.Therefore, the percentage profit is 35%.

answer：First, I need to determine the number of cookies each person has based on Emily's cookies, denoted as ( e ).Emily has ( e ) cookies.Fiona has three times as many cookies as Emily, so Fiona has ( 3e ) cookies.Greg has twice as many cookies as Fiona, which means Greg has ( 2 times 3e = 6e ) cookies.Helen has four times the number of cookies Greg has, so Helen has ( 4 times 6e = 24e ) cookies.Next, I'll calculate the total number of cookies by adding up the cookies each person has:[e + 3e + 6e + 24e = 34e]According to the problem, the total number of cookies is 90. Therefore, I can set up the equation:[34e = 90]To find the value of ( e ), I'll solve for ( e ):[e = frac{90}{34} = frac{45}{17}]So, the value of ( e ) is ( frac{45}{17} ).

answer：Okay, so I have this problem here: Given that ( f(cos x) = cos 2x ), I need to find the expression for ( f(sin x) ). Hmm, let's see. I remember there are some trigonometric identities that relate double angles to single angles, so maybe I can use those here.First, I recall that ( cos 2x ) can be expressed in terms of ( cos^2 x ) using the double-angle identity. Specifically, ( cos 2x = 2cos^2 x - 1 ). So, if I substitute ( cos x ) into the function ( f ), I get ( f(cos x) = 2cos^2 x - 1 ). That makes sense.Now, I need to find ( f(sin x) ). Since ( f ) is a function that takes ( cos x ) and gives ( cos 2x ), maybe I can express ( f ) in terms of its argument. Let me think. If I let ( y = cos x ), then ( f(y) = 2y^2 - 1 ). So, ( f ) is a function that squares its input, multiplies by 2, and subtracts 1. That seems straightforward.So, if ( f(y) = 2y^2 - 1 ), then to find ( f(sin x) ), I just need to substitute ( sin x ) into this function. That would give ( f(sin x) = 2(sin x)^2 - 1 ). Simplifying that, it's ( 2sin^2 x - 1 ).Wait a minute, I also remember another identity: ( cos 2x = 1 - 2sin^2 x ). So, ( 2sin^2 x - 1 ) is actually equal to ( -cos 2x ). Let me check that:Starting from ( cos 2x = 1 - 2sin^2 x ), if I rearrange it, I get ( 2sin^2 x = 1 - cos 2x ), so ( 2sin^2 x - 1 = -cos 2x ). Yes, that's correct.So, putting it all together, ( f(sin x) = 2sin^2 x - 1 = -cos 2x ). Therefore, the expression for ( f(sin x) ) is ( -cos 2x ).I think that makes sense. I used the double-angle identity to express ( cos 2x ) in terms of ( cos^2 x ), then figured out the form of the function ( f ), applied it to ( sin x ), and then used another identity to simplify the result. It all checks out.Just to double-check, let me plug in a specific value of ( x ) and see if both sides match. Let's take ( x = 0 ). Then, ( cos 0 = 1 ), so ( f(1) = cos 0 = 1 ). Using my expression for ( f(y) = 2y^2 - 1 ), ( f(1) = 2(1)^2 - 1 = 2 - 1 = 1 ). That matches.Now, for ( f(sin 0) ), ( sin 0 = 0 ), so ( f(0) = 2(0)^2 - 1 = -1 ). On the other hand, ( -cos 0 = -1 ). That also matches.Let's try another angle, say ( x = pi/4 ). Then, ( cos(pi/4) = sqrt{2}/2 ), so ( f(sqrt{2}/2) = cos(pi/2) = 0 ). Using ( f(y) = 2y^2 - 1 ), ( f(sqrt{2}/2) = 2*(sqrt{2}/2)^2 - 1 = 2*(0.5) - 1 = 1 - 1 = 0 ). Good.For ( f(sin(pi/4)) ), ( sin(pi/4) = sqrt{2}/2 ), so ( f(sqrt{2}/2) = 2*(sqrt{2}/2)^2 - 1 = 0 ). And ( -cos(pi/2) = -0 = 0 ). Perfect, that matches too.Alright, I feel confident that my reasoning is correct. So, the expression for ( f(sin x) ) is indeed ( -cos 2x ).

answer：Alright, so I have this problem: I need to prove that there exists a triangle whose sides are equal and parallel to the medians of a given triangle. Hmm, okay. Let me try to wrap my head around this.First off, let's recall what a median of a triangle is. A median is a line segment joining a vertex to the midpoint of the opposite side. So, in any triangle, there are three medians, one from each vertex. Now, the problem is saying that there exists another triangle whose sides are equal in length and parallel to these medians. Interesting.I think I need to visualize this. Maybe drawing a diagram would help. Let me sketch a triangle ABC, label the midpoints of each side, and then draw the medians. So, if I have triangle ABC, with midpoints D, E, and F of sides BC, AC, and AB respectively, then the medians would be AD, BE, and CF.Now, I need to construct a triangle whose sides are equal and parallel to these medians. So, the sides of this new triangle should be AD, BE, and CF in length and direction. But how do I arrange these vectors to form a triangle?Wait, maybe I can use vector addition here. If I consider the medians as vectors, then perhaps I can arrange them tip-to-tail to form a triangle. But I remember that in a triangle, the sum of the vectors representing the medians is zero. Is that true? Let me think.Actually, in any triangle, the three medians intersect at the centroid, which divides each median into a ratio of 2:1. So, if I consider the vectors from the centroid, they might sum to zero. But I'm not sure if that directly helps me here.Maybe I should consider translating the medians so that they originate from the same point. If I can arrange them such that each median starts where the previous one ends, then the figure formed should be a triangle. Let me try to formalize this.Suppose I have three vectors representing the medians: vector m1, m2, and m3. If I place them head-to-tail, meaning the tail of m2 is at the head of m1, and the tail of m3 is at the head of m2, then the figure formed by connecting the tails and heads appropriately should be a triangle. But for this to work, the sum of these vectors should bring me back to the starting point, forming a closed shape.So, mathematically, if I have vectors m1, m2, and m3, then m1 + m2 + m3 should equal zero for them to form a triangle when placed head-to-tail. Is this the case for medians?Wait, I think there's a theorem that says the sum of the medians of a triangle, when considered as vectors from the same point, equals zero. Let me check that.Yes, in vector terms, if you take the medians from the same vertex, their vector sum is zero. But in this case, the medians are from different vertices. Hmm, maybe I need to adjust my approach.Alternatively, perhaps I can use coordinate geometry. Let me assign coordinates to the vertices of triangle ABC and compute the coordinates of the midpoints. Then, I can find the vectors representing the medians and see if they can form a triangle.Let's say triangle ABC has vertices at points A(x1, y1), B(x2, y2), and C(x3, y3). The midpoints D, E, F would then be at ((x2+x3)/2, (y2+y3)/2), ((x1+x3)/2, (y1+y3)/2), and ((x1+x2)/2, (y1+y2)/2) respectively.The vectors representing the medians AD, BE, and CF can be calculated as follows:- Vector AD: from A to D is ((x2+x3)/2 - x1, (y2+y3)/2 - y1)- Vector BE: from B to E is ((x1+x3)/2 - x2, (y1+y3)/2 - y2)- Vector CF: from C to F is ((x1+x2)/2 - x3, (y1+y2)/2 - y3)Now, if I want to form a triangle with sides equal and parallel to these medians, I need to see if these vectors can be arranged such that their sum is zero. That is, if I place them head-to-tail, the last vector should bring me back to the starting point.So, let's denote the vectors as:m1 = AD = ((x2+x3)/2 - x1, (y2+y3)/2 - y1)m2 = BE = ((x1+x3)/2 - x2, (y1+y3)/2 - y2)m3 = CF = ((x1+x2)/2 - x3, (y1+y2)/2 - y3)If I add these vectors together:m1 + m2 + m3 = [((x2+x3)/2 - x1) + ((x1+x3)/2 - x2) + ((x1+x2)/2 - x3), ((y2+y3)/2 - y1) + ((y1+y3)/2 - y2) + ((y1+y2)/2 - y3)]Let's compute the x-component:= (x2 + x3)/2 - x1 + (x1 + x3)/2 - x2 + (x1 + x2)/2 - x3= [ (x2 + x3)/2 + (x1 + x3)/2 + (x1 + x2)/2 ] - [x1 + x2 + x3]= [ (x2 + x3 + x1 + x3 + x1 + x2)/2 ] - (x1 + x2 + x3)= [ (2x1 + 2x2 + 2x3)/2 ] - (x1 + x2 + x3)= (x1 + x2 + x3) - (x1 + x2 + x3) = 0Similarly, the y-component:= (y2 + y3)/2 - y1 + (y1 + y3)/2 - y2 + (y1 + y2)/2 - y3= [ (y2 + y3)/2 + (y1 + y3)/2 + (y1 + y2)/2 ] - [y1 + y2 + y3]= [ (y2 + y3 + y1 + y3 + y1 + y2)/2 ] - (y1 + y2 + y3)= [ (2y1 + 2y2 + 2y3)/2 ] - (y1 + y2 + y3)= (y1 + y2 + y3) - (y1 + y2 + y3) = 0So, m1 + m2 + m3 = 0. That means if I place these vectors head-to-tail, they form a closed triangle. Therefore, the triangle formed by vectors m1, m2, and m3 is the required triangle whose sides are equal and parallel to the medians of triangle ABC.Wait, but does this triangle actually exist? I mean, I've shown that the vectors sum to zero, implying they can form a triangle, but is there a specific construction for it?Maybe I can construct it by translating the medians so that they originate from the same point. For example, if I take median AD and place it starting at point A, then take median BE and place it starting at the end of AD, and then take median CF and place it starting at the end of BE, the last vector should bring me back to the starting point, forming a triangle.Alternatively, perhaps using the centroid. Since the centroid divides each median into a 2:1 ratio, maybe scaling the medians appropriately could help in constructing the triangle.But I think the vector approach suffices. Since the sum of the medians as vectors is zero, they must form a triangle when placed head-to-tail. Therefore, such a triangle exists.I should probably write this up more formally to make sure I haven't missed anything. Maybe using vector notation and showing the sum equals zero explicitly.Yeah, I think that's the way to go. So, in summary, by considering the medians as vectors and showing their sum is zero, we can conclude that they can form a triangle when arranged appropriately. Therefore, such a triangle exists.