💡Okay, so I have this problem where I need to find all functions ( f ) from the reals to the reals that are strictly increasing and satisfy the equation ( f(x) + g(x) = 2x ) for all real ( x ), where ( g ) is the composition inverse of ( f ). Hmm, composition inverse means that ( f(g(x)) = x ) and ( g(f(x)) = x ) for all ( x ). Alright, let's start by understanding what it means for ( g ) to be the inverse of ( f ). If ( g ) is the inverse, then ( g(f(x)) = x ) and ( f(g(x)) = x ). So, ( g ) essentially "undoes" what ( f ) does. Given that ( f(x) + g(x) = 2x ), I can think of this as a relationship between ( f ) and its inverse. Maybe I can express ( g(x) ) in terms of ( f(x) ) and substitute it into the equation. Let me try that.Since ( g ) is the inverse of ( f ), we have ( g(x) = f^{-1}(x) ). So, substituting into the equation, we get:[ f(x) + f^{-1}(x) = 2x ]This is interesting. So, the function plus its inverse equals twice the identity function. I wonder if there are standard functions that satisfy this property. For example, if ( f(x) = x + c ), then its inverse would be ( f^{-1}(x) = x - c ). Let's test this:[ f(x) + f^{-1}(x) = (x + c) + (x - c) = 2x ]Hey, that works! So, linear functions of the form ( f(x) = x + c ) satisfy the equation. But are these the only solutions?Wait, the problem says ( f ) is strictly increasing. Linear functions with a positive slope are strictly increasing, so that's good. But could there be other non-linear functions that also satisfy ( f(x) + f^{-1}(x) = 2x )?Let me think. Suppose ( f ) is non-linear. Then ( f^{-1} ) would also be non-linear. Adding them together and getting a linear function like ( 2x ) seems restrictive. Maybe only linear functions can satisfy this condition.Let me try to see if that's the case. Suppose ( f ) is differentiable. Then, maybe I can take derivatives of both sides of the equation ( f(x) + f^{-1}(x) = 2x ). Taking the derivative, we get:[ f'(x) + frac{1}{f'(f^{-1}(x))} = 2 ]Hmm, that's a bit complicated. Let me denote ( y = f^{-1}(x) ), so ( x = f(y) ). Then, the derivative equation becomes:[ f'(x) + frac{1}{f'(y)} = 2 ]But since ( y = f^{-1}(x) ), we can write ( f'(x) + frac{1}{f'(f^{-1}(x))} = 2 ). This seems a bit circular. Maybe I can assume ( f ) is linear and see if that's the only possibility. If ( f(x) = x + c ), then ( f'(x) = 1 ), and substituting back into the derivative equation:[ 1 + frac{1}{1} = 2 ]Which is true. So, the derivative condition is satisfied for linear functions. But does this mean only linear functions satisfy the original equation?Alternatively, maybe I can consider the functional equation ( f(x) + f^{-1}(x) = 2x ) and try to find ( f ) in terms of itself. Let's see.Since ( f^{-1}(x) = 2x - f(x) ), I can substitute this into the definition of inverse functions. That is, ( f(f^{-1}(x)) = x ). So:[ f(2x - f(x)) = x ]Hmm, that's an interesting equation. Let me write that down:[ f(2x - f(x)) = x ]So, applying ( f ) to ( 2x - f(x) ) gives me ( x ). This seems like a recursive equation. Maybe I can iterate this or find a pattern. Let's try plugging in ( x ) into the equation:[ f(2x - f(x)) = x ]Let me denote ( y = f(x) ). Then, ( f(2x - y) = x ). But since ( y = f(x) ), we can write:[ f(2x - y) = x ]But ( y = f(x) ), so:[ f(2x - f(x)) = x ]Wait, that's the same equation as before. Maybe I can express this in terms of ( f ) and its inverse.Alternatively, let's consider the function ( f ) and its inverse ( g ). We have ( f(x) + g(x) = 2x ), and ( f(g(x)) = x ). Maybe I can use these two equations together.From ( f(x) + g(x) = 2x ), we can express ( g(x) = 2x - f(x) ). Then, substituting into ( f(g(x)) = x ), we get:[ f(2x - f(x)) = x ]Which is the same equation as before. So, this seems to be a key equation to work with.Let me try to see if ( f ) is linear, does this hold? If ( f(x) = x + c ), then ( f(2x - f(x)) = f(2x - (x + c)) = f(x - c) = (x - c) + c = x ). So, yes, it holds. But what if ( f ) is non-linear? Let's suppose ( f(x) = ax + b ), a linear function. Then, ( f^{-1}(x) = frac{x - b}{a} ). Plugging into the equation ( f(x) + f^{-1}(x) = 2x ):[ ax + b + frac{x - b}{a} = 2x ]Multiplying through by ( a ):[ a^2x + ab + x - b = 2ax ]Grouping like terms:[ (a^2 + 1)x + (ab - b) = 2ax ]For this to hold for all ( x ), the coefficients must be equal:1. ( a^2 + 1 = 2a )2. ( ab - b = 0 )From equation 1:[ a^2 - 2a + 1 = 0 ][ (a - 1)^2 = 0 ]So, ( a = 1 ).From equation 2:If ( a = 1 ), then ( b - b = 0 ), which is always true. So, ( b ) can be any constant. Therefore, the only linear solutions are ( f(x) = x + c ), where ( c ) is a constant.But the problem doesn't specify that ( f ) has to be linear, only that it's strictly increasing. So, could there be non-linear solutions?Let me think about the functional equation ( f(2x - f(x)) = x ). Suppose ( f ) is non-linear. Then, ( 2x - f(x) ) would be some function, and applying ( f ) to it gives ( x ). This seems like a kind of involution. An involution is a function that is its own inverse, but here it's a bit different. Maybe I can iterate this equation or find a pattern.Let me define ( h(x) = 2x - f(x) ). Then, the equation becomes:[ f(h(x)) = x ]But ( h(x) = 2x - f(x) ), so substituting back:[ f(2x - f(x)) = x ]Which is the same as before. If I apply ( h ) again, I get:[ h(h(x)) = 2h(x) - f(h(x)) = 2(2x - f(x)) - f(2x - f(x)) ]But ( f(2x - f(x)) = x ), so:[ h(h(x)) = 4x - 2f(x) - x = 3x - 2f(x) ]Hmm, not sure if that helps.Alternatively, maybe I can express ( f ) in terms of ( h ). Since ( f(h(x)) = x ), then ( h(x) = f^{-1}(x) ). But we already have ( h(x) = 2x - f(x) ), so:[ f^{-1}(x) = 2x - f(x) ]Which is consistent with our earlier substitution.Wait, so ( f^{-1}(x) = 2x - f(x) ). Let me write that as:[ f^{-1}(x) + f(x) = 2x ]Which is the original equation. So, this is just reinforcing the relationship.Maybe I can consider the function ( f ) and its inverse ( g ) as reflections over the line ( y = x ). The equation ( f(x) + g(x) = 2x ) suggests that for each ( x ), the sum of ( f(x) ) and ( g(x) ) is twice ( x ). Graphically, this would mean that for each point ( (x, f(x)) ), the corresponding point ( (x, g(x)) ) is such that their average is ( (x, x) ). So, the points ( (x, f(x)) ) and ( (x, g(x)) ) are symmetric with respect to the line ( y = x ).Wait, but ( g(x) ) is the inverse function, so its graph is the reflection of ( f(x) ) over the line ( y = x ). So, if ( f(x) + g(x) = 2x ), then for each ( x ), the sum of the function value and its inverse function value is twice ( x ). This seems to imply that the function ( f ) is symmetric in some way with respect to the line ( y = x ). Maybe only linear functions have this property.Alternatively, let's consider specific values. Let me plug in ( x = 0 ). Suppose ( f(0) = c ). Then, ( g(0) = 2*0 - f(0) = -c ). But since ( g ) is the inverse of ( f ), ( f(g(0)) = f(-c) = 0 ). So, ( f(-c) = 0 ).Similarly, ( g(c) = 2c - f(c) ). But ( f(c) = 2c - g(c) ). Hmm, not sure if that helps.Wait, let's try to find ( f ) in terms of itself. From ( f(2x - f(x)) = x ), let me denote ( y = 2x - f(x) ). Then, ( f(y) = x ). But ( y = 2x - f(x) ), so substituting ( x = f(y) ) into this, we get:[ y = 2f(y) - f(f(y)) ]But ( f(f(y)) = f(x) ) since ( x = f(y) ). Wait, this is getting a bit tangled.Alternatively, let's consider the function ( f ) and its inverse ( g ). Since ( f ) is strictly increasing, ( g ) is also strictly increasing. Let me try to see if ( f ) must be linear. Suppose ( f ) is differentiable. Then, from the equation ( f(x) + g(x) = 2x ), taking derivatives:[ f'(x) + g'(x) cdot g'(x)^{-1} = 2 ]Wait, no. Let me correct that. If ( g(x) = 2x - f(x) ), then ( g'(x) = 2 - f'(x) ). But since ( g ) is the inverse of ( f ), we have ( f'(g(x)) cdot g'(x) = 1 ). So:[ f'(g(x)) cdot (2 - f'(x)) = 1 ]Hmm, this is a differential equation involving ( f ) and its derivative. Let me denote ( y = f(x) ). Then, ( g(y) = x ). So, ( f'(y) cdot (2 - f'(x)) = 1 ). This seems complicated. Maybe I can assume ( f ) is linear and see if that's the only solution. If ( f(x) = x + c ), then ( f'(x) = 1 ). Substituting into the differential equation:[ 1 cdot (2 - 1) = 1 ]Which is true. So, linear functions satisfy this condition. But what if ( f ) is non-linear? Suppose ( f(x) = ax + b ), but we already saw that only ( a = 1 ) works. So, maybe only linear functions satisfy this.Alternatively, suppose ( f(x) = x + c ). Then, ( g(x) = x - c ), and indeed ( f(x) + g(x) = 2x ). So, this works. But could there be a non-linear function ( f ) such that ( f(x) + f^{-1}(x) = 2x )? Let's try to see.Suppose ( f(x) = x + c ) for ( x leq k ) and ( f(x) = x + d ) for ( x > k ), where ( c neq d ). But this would make ( f ) piecewise linear. However, since ( f ) is strictly increasing, ( c ) and ( d ) must be such that the function remains increasing. But then, the inverse function would also be piecewise linear, and adding them might not give ( 2x ) everywhere.Alternatively, suppose ( f(x) ) is quadratic, say ( f(x) = ax^2 + bx + c ). Then, finding its inverse would require solving a quadratic equation, which might not lead to a simple expression. Moreover, adding ( f(x) ) and ( f^{-1}(x) ) and expecting it to be linear seems unlikely.Another approach: suppose ( f ) is its own inverse, i.e., ( f(f(x)) = x ). Then, ( f^{-1}(x) = f(x) ), so the equation becomes ( f(x) + f(x) = 2x ), which implies ( f(x) = x ). But this is a special case where ( f ) is linear with ( c = 0 ). However, the problem doesn't state that ( f ) is its own inverse, just that it has an inverse.Wait, but in our case, ( f(x) + f^{-1}(x) = 2x ). If ( f ) were its own inverse, then ( f(x) + f(x) = 2x ) implies ( f(x) = x ). So, that's a specific case, but we already have a more general solution with ( f(x) = x + c ).Let me think about the functional equation again: ( f(2x - f(x)) = x ). Suppose I define ( u(x) = 2x - f(x) ). Then, ( f(u(x)) = x ). So, ( u(x) ) is such that applying ( f ) to it gives ( x ). This suggests that ( u(x) ) is the inverse function of ( f ), i.e., ( u(x) = f^{-1}(x) ). But we also have ( u(x) = 2x - f(x) ). Therefore:[ f^{-1}(x) = 2x - f(x) ]Which is consistent with our earlier substitution.So, we have ( f^{-1}(x) = 2x - f(x) ). Let me apply ( f ) to both sides:[ f(f^{-1}(x)) = f(2x - f(x)) ]But ( f(f^{-1}(x)) = x ), so:[ x = f(2x - f(x)) ]Which is the same equation as before. This seems to be a loop, not giving me new information. Maybe I need to consider higher iterations or some kind of fixed point.Alternatively, let's consider the function ( f ) and its iterates. Suppose I define ( f_1(x) = f(x) ), ( f_2(x) = f(f(x)) ), and so on. Then, perhaps I can find a pattern or a recursive relationship.But I'm not sure if that's the right path. Maybe I can look for fixed points of ( f ). A fixed point is a value ( x ) such that ( f(x) = x ). If ( f(x) = x ), then ( g(x) = x ), and the equation ( f(x) + g(x) = 2x ) becomes ( x + x = 2x ), which is true. So, fixed points satisfy the equation, but they don't necessarily tell me about the entire function.Wait, but if ( f(x) = x + c ), then the only fixed point is when ( x + c = x ), which implies ( c = 0 ). So, unless ( c = 0 ), there are no fixed points. But the problem allows ( c ) to be any real constant, so ( f(x) = x + c ) can have no fixed points unless ( c = 0 ).Hmm, maybe that's not helpful. Let me try another approach.Suppose I consider the function ( f ) and its inverse ( g ). Since ( f ) is strictly increasing, ( g ) is also strictly increasing. The equation ( f(x) + g(x) = 2x ) suggests that ( f ) and ( g ) are symmetric around the line ( y = x ) in some way.Let me think about the graph of ( f ) and ( g ). Since ( g ) is the inverse of ( f ), their graphs are reflections over the line ( y = x ). The equation ( f(x) + g(x) = 2x ) implies that for each ( x ), the sum of the function value and its inverse function value is twice ( x ). This seems to suggest that the function ( f ) is symmetric with respect to the line ( y = x ) in a specific way. Maybe only linear functions have this property.Alternatively, let me consider specific points. Suppose ( x = a ), then ( f(a) + g(a) = 2a ). But since ( g(a) = f^{-1}(a) ), we have ( f(a) + f^{-1}(a) = 2a ). Let me denote ( b = f(a) ). Then, ( f^{-1}(b) = a ), and the equation becomes ( b + a = 2a ), which implies ( b = a ). Wait, that would mean ( f(a) = a ) for all ( a ), which would imply ( f(x) = x ). But that's only one specific solution where ( c = 0 ). However, earlier we saw that ( f(x) = x + c ) works for any constant ( c ). So, this seems contradictory.Wait, no. Let me correct that. If ( f(a) = b ), then ( f^{-1}(b) = a ). The equation ( f(a) + f^{-1}(a) = 2a ) becomes ( b + f^{-1}(a) = 2a ). But ( f^{-1}(a) ) is not necessarily ( a ). It's the value ( c ) such that ( f(c) = a ). So, ( f^{-1}(a) = c ) where ( f(c) = a ). So, the equation is ( b + c = 2a ), where ( f(a) = b ) and ( f(c) = a ). So, ( f(a) + f^{-1}(a) = 2a ) translates to ( b + c = 2a ). But since ( f(c) = a ), we can write ( f(c) = a ), and ( f(a) = b ). So, ( f(f(c)) = f(a) = b ). But ( f(f(c)) = f(a) = b ), and ( f(c) = a ). So, ( f(a) = b ), ( f(c) = a ), and ( b + c = 2a ).This seems to form a system where ( f ) swaps ( a ) and ( c ) in some way, but scaled by the factor of 2. Wait, if ( b + c = 2a ), and ( f(a) = b ), ( f(c) = a ), then ( f(f(a)) = f(b) = c ), and ( f(f(c)) = f(a) = b ). So, ( f ) swaps ( a ) and ( c ) when applied twice. But this seems to suggest that ( f ) has some periodicity or symmetry. However, since ( f ) is strictly increasing, it can't have periodicity unless it's linear. Because if ( f ) were non-linear and strictly increasing, it couldn't swap values in a way that maintains the strictly increasing property unless it's linear.Wait, let me think about that again. If ( f ) is strictly increasing, then ( f(a) > f(b) ) whenever ( a > b ). So, if ( f(a) = b ) and ( f(c) = a ), then ( a > c ) implies ( f(a) = b > f(c) = a ). So, ( b > a ). But from ( b + c = 2a ), if ( b > a ), then ( c < a ). So, ( c < a < b ). But then, ( f(c) = a ), and since ( c < a ), ( f(c) = a > f(c) ) because ( f ) is strictly increasing. Wait, that doesn't make sense. If ( c < a ), then ( f(c) < f(a) ) because ( f ) is strictly increasing. But ( f(c) = a ) and ( f(a) = b ), so ( a < b ). Therefore, ( f(c) = a < b = f(a) ), which is consistent with ( c < a ) because ( f ) is increasing.But then, from ( b + c = 2a ), we have ( c = 2a - b ). Since ( c < a ), ( 2a - b < a ), which implies ( a < b ). Which is consistent with ( f(a) = b > a ).So, for each ( a ), there's a corresponding ( c ) such that ( f(a) = b ), ( f(c) = a ), and ( b + c = 2a ). This seems to form a kind of reciprocal relationship between ( a ) and ( c ). But how does this help me find ( f )?Maybe I can consider that for all ( x ), ( f(x) + f^{-1}(x) = 2x ). So, for each ( x ), ( f(x) ) and ( f^{-1}(x) ) are symmetric around ( x ). If I think of ( f(x) ) as a reflection over the line ( y = x ), then ( f(x) ) and ( f^{-1}(x) ) are symmetric with respect to that line. The condition ( f(x) + f^{-1}(x) = 2x ) adds another layer of symmetry, that their average is ( x ).This seems to suggest that the function ( f ) is symmetric in a way that its graph and the graph of its inverse are symmetric around the line ( y = x ) and also their average is the line ( y = x ). I think this kind of symmetry is only possible for linear functions. Because for non-linear functions, their inverses would not maintain this additive symmetry around ( y = x ).Let me try to formalize this intuition. Suppose ( f ) is non-linear. Then, ( f^{-1} ) is also non-linear. The sum ( f(x) + f^{-1}(x) ) would involve adding two non-linear functions, which generally results in a non-linear function. However, in our case, the sum is linear, ( 2x ). This suggests that the non-linear parts of ( f ) and ( f^{-1} ) must cancel each other out when added together. But since ( f ) and ( f^{-1} ) are inverses, their non-linear behaviors are related in a specific way. It's not clear how they could cancel each other out unless they are linear.Therefore, it's plausible that the only solutions are linear functions of the form ( f(x) = x + c ).To confirm this, let's assume ( f(x) = x + c ) and check if it satisfies all conditions.1. ( f(x) ) is strictly increasing: Yes, because the derivative is 1, which is positive.2. ( f(x) + g(x) = 2x ): Let's find ( g(x) ). Since ( f(x) = x + c ), the inverse function ( g(x) = x - c ). Then, ( f(x) + g(x) = (x + c) + (x - c) = 2x ), which satisfies the equation.Therefore, ( f(x) = x + c ) is indeed a solution for any constant ( c ).But are there any other solutions? Suppose there exists a non-linear function ( f ) satisfying the conditions. Then, as discussed earlier, the sum ( f(x) + f^{-1}(x) ) would have to be linear, which seems highly restrictive. Moreover, considering the functional equation ( f(2x - f(x)) = x ), if ( f ) is non-linear, it's unclear how this equation could hold for all ( x ). The left-hand side would involve a non-linear transformation of ( 2x - f(x) ), which would likely not simplify to ( x ) unless ( f ) is linear.Therefore, after considering various approaches and checking the properties, it seems that the only functions satisfying the given conditions are linear functions of the form ( f(x) = x + c ), where ( c ) is a real constant.

💡To find the coordinates of point ( M ) where the tangent line to the curve ( y = x^2 + 2x - 2 ) is parallel to the ( x )-axis, we need to determine where the slope of the tangent line is zero.1. **Find the derivative of the function:** [ y = x^2 + 2x - 2 ] The derivative ( y' ) is: [ y' = 2x + 2 ]2. **Set the derivative equal to zero to find the critical point:** [ 2x + 2 = 0 ] Solving for ( x ): [ 2x = -2 implies x = -1 ]3. **Substitute ( x = -1 ) back into the original equation to find ( y ):** [ y = (-1)^2 + 2(-1) - 2 = 1 - 2 - 2 = -3 ]Therefore, the coordinates of point ( M ) are ((-1, -3)).

💡First, I need to determine the total earnings from selling the ducks and chickens. The farmer sells 2 ducks at 10 each, which amounts to 20. Let ( C ) represent the number of chickens sold at 8 each, so the earnings from chickens are ( 8C ).Next, the total earnings from both ducks and chickens are ( 20 + 8C ). The farmer spends half of this amount on a new wheelbarrow, so the cost of the wheelbarrow is ( frac{20 + 8C}{2} ).After purchasing the wheelbarrow, someone agrees to pay the farmer double the amount he paid for it, resulting in an additional 60. This means the cost of the wheelbarrow is half of 60, which is 30.Setting up the equation:[frac{20 + 8C}{2} = 30]Multiplying both sides by 2 to eliminate the fraction:[20 + 8C = 60]Subtracting 20 from both sides:[8C = 40]Finally, dividing both sides by 8:[C = 5]Therefore, the farmer sold 5 chickens.

💡Okay, so I need to find the sum of the squares of the lengths of the medians of a triangle with side lengths 13, 14, and 15. Hmm, medians... I remember that a median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Each triangle has three medians, one from each vertex.I think there's a formula that relates the lengths of the medians to the lengths of the sides of the triangle. Maybe it's called Apollonius's theorem? Let me recall... Apollonius's theorem states that the sum of the squares of the sides of a triangle is equal to three times the sum of the squares of the medians. Wait, no, that doesn't sound quite right. Let me check.Actually, Apollonius's theorem says that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of the half third side. Hmm, that's a bit complicated. Let me write it down:For a triangle ABC, with median AD, Apollonius's theorem states:[ AB^2 + AC^2 = 2AD^2 + 2left(frac{BC}{2}right)^2 ]So, rearranging this, the length of the median AD can be found as:[ AD^2 = frac{2AB^2 + 2AC^2 - BC^2}{4} ]Yes, that seems right. So, the square of the median is equal to (2 times the sum of the squares of two sides minus the square of the third side) divided by 4.So, if I apply this formula for each median, I can find the squares of all three medians and then add them up.Given the sides are 13, 14, and 15, let me label them as follows for clarity:- Let’s say side a = 13, side b = 14, and side c = 15.Now, the medians corresponding to these sides will be:- Median from A to midpoint of BC: let's call this median m_a- Median from B to midpoint of AC: median m_b- Median from C to midpoint of AB: median m_cUsing Apollonius's theorem, the formula for each median squared is:[ m_a^2 = frac{2b^2 + 2c^2 - a^2}{4} ][ m_b^2 = frac{2a^2 + 2c^2 - b^2}{4} ][ m_c^2 = frac{2a^2 + 2b^2 - c^2}{4} ]So, I need to compute each of these and then add them together.Let me compute each median squared step by step.First, compute m_a^2:[ m_a^2 = frac{2(14)^2 + 2(15)^2 - (13)^2}{4} ]Calculating each term:- 2*(14)^2 = 2*196 = 392- 2*(15)^2 = 2*225 = 450- (13)^2 = 169So, plugging these in:[ m_a^2 = frac{392 + 450 - 169}{4} = frac{673}{4} = 168.25 ]Wait, 392 + 450 is 842, minus 169 is 673. So, 673 divided by 4 is indeed 168.25.Next, compute m_b^2:[ m_b^2 = frac{2(13)^2 + 2(15)^2 - (14)^2}{4} ]Calculating each term:- 2*(13)^2 = 2*169 = 338- 2*(15)^2 = 2*225 = 450- (14)^2 = 196So, plugging these in:[ m_b^2 = frac{338 + 450 - 196}{4} = frac{592}{4} = 148 ]Wait, 338 + 450 is 788, minus 196 is 592. 592 divided by 4 is 148. That seems correct.Now, compute m_c^2:[ m_c^2 = frac{2(13)^2 + 2(14)^2 - (15)^2}{4} ]Calculating each term:- 2*(13)^2 = 2*169 = 338- 2*(14)^2 = 2*196 = 392- (15)^2 = 225So, plugging these in:[ m_c^2 = frac{338 + 392 - 225}{4} = frac{505}{4} = 126.25 ]Wait, 338 + 392 is 730, minus 225 is 505. 505 divided by 4 is 126.25.So, now I have:- m_a^2 = 168.25- m_b^2 = 148- m_c^2 = 126.25Adding them together:168.25 + 148 + 126.25Let me compute that step by step:168.25 + 148 = 316.25316.25 + 126.25 = 442.5So, the sum of the squares of the medians is 442.5.Wait, that seems like a decimal. Is there a way to represent this as a fraction? 442.5 is equal to 442 and a half, which is 885/2. But maybe the question expects it in decimal form.Alternatively, I can check if I did all the calculations correctly. Let me verify each step.First, for m_a^2:2*(14)^2 = 3922*(15)^2 = 450Total: 392 + 450 = 842Minus (13)^2 = 169: 842 - 169 = 673Divide by 4: 673/4 = 168.25 Correct.For m_b^2:2*(13)^2 = 3382*(15)^2 = 450Total: 338 + 450 = 788Minus (14)^2 = 196: 788 - 196 = 592Divide by 4: 592/4 = 148 Correct.For m_c^2:2*(13)^2 = 3382*(14)^2 = 392Total: 338 + 392 = 730Minus (15)^2 = 225: 730 - 225 = 505Divide by 4: 505/4 = 126.25 Correct.Adding them: 168.25 + 148 + 126.25 = 442.5. Yes, that's correct.Alternatively, I remember another formula that the sum of the squares of the medians is equal to three-fourths the sum of the squares of the sides. Wait, is that right?Let me recall: The formula is that the sum of the squares of the medians is equal to (3/4) times the sum of the squares of the sides. So, if I compute (3/4)*(13^2 + 14^2 + 15^2), does that equal 442.5?Let me check:13^2 = 16914^2 = 19615^2 = 225Sum of squares: 169 + 196 + 225 = 590Multiply by 3/4: (3/4)*590 = (3*590)/4 = 1770/4 = 442.5Yes, that's the same result as before. So, that's a quicker way to compute it.Therefore, the sum of the squares of the medians is 442.5.I think that's the answer. It's always good to verify using different methods to ensure correctness. Both methods gave me the same result, so I'm confident that 442.5 is the correct sum of the squares of the medians.**Final Answer**The sum of the squares of the lengths of the medians is boxed{442.5}.

💡To determine Gabriella's overall score when combining all three tests into a single 60-problem test, follow these steps:1. **Calculate the number of correct answers for each test:** - **First Test:** ( 60% ) of 15 problems [ 0.60 times 15 = 9 text{ correct answers} ] - **Second Test:** ( 75% ) of 25 problems [ 0.75 times 25 = 18.75 approx 19 text{ correct answers} ] - **Third Test:** ( 85% ) of 20 problems [ 0.85 times 20 = 17 text{ correct answers} ]2. **Sum the correct answers:** [ 9 + 19 + 17 = 45 text{ correct answers} ]3. **Calculate the overall percentage:** [ frac{45}{60} = 0.75 text{ or } 75% ]4. **Round to the nearest percent:** The overall score is already a whole number, so no rounding is needed.**Final Answer:**boxed{75%}

💡Okay, so I have this problem about a regular quadrilateral pyramid ( V-ABCD ). Let me try to visualize it first. A regular quadrilateral pyramid means that the base is a square, right? So, the base is square ( ABCD ), and the apex is ( V ). The height of the pyramid is equal to half the length of ( AB ). Hmm, okay, so if I let the length of ( AB ) be, say, ( 2a ), then the height ( h ) would be ( a ). Maybe I can set ( AB = 2 ) for simplicity, so the height is ( 1 ). That might make calculations easier.Now, ( M ) is the midpoint of the side edge ( VB ). So, ( M ) divides ( VB ) into two equal parts. And ( N ) is a point on the side edge ( VD ) such that ( DN = 2VN ). That means ( N ) divides ( VD ) in a 1:2 ratio. So, ( VN:ND = 1:2 ). I need to find the cosine of the angle between the skew lines ( AM ) and ( BN ).Skew lines are lines that don't intersect and aren't parallel. In a pyramid, these lines are in different planes, so they don't intersect. To find the angle between skew lines, I remember that we can use vectors. The angle between two skew lines is the angle between their direction vectors when projected onto a common plane.So, maybe I should assign coordinates to the vertices of the pyramid to make this easier. Let me place the base ( ABCD ) on the xy-plane with point ( A ) at the origin. Let's define the coordinates:- Let’s set ( A = (0, 0, 0) ).- Since it's a square base, if ( AB = 2 ), then ( B = (2, 0, 0) ), ( C = (2, 2, 0) ), and ( D = (0, 2, 0) ).- The apex ( V ) is directly above the center of the square. The center of the square is at ( (1, 1, 0) ). Since the height is ( 1 ), ( V = (1, 1, 1) ).Okay, so now I have coordinates for all the points:- ( A = (0, 0, 0) )- ( B = (2, 0, 0) )- ( C = (2, 2, 0) )- ( D = (0, 2, 0) )- ( V = (1, 1, 1) )Now, let's find the coordinates of ( M ) and ( N ).Starting with ( M ), the midpoint of ( VB ). The coordinates of ( V ) are ( (1, 1, 1) ) and ( B ) are ( (2, 0, 0) ). The midpoint ( M ) can be found by averaging the coordinates:( M_x = frac{1 + 2}{2} = frac{3}{2} )( M_y = frac{1 + 0}{2} = frac{1}{2} )( M_z = frac{1 + 0}{2} = frac{1}{2} )So, ( M = left( frac{3}{2}, frac{1}{2}, frac{1}{2} right) ).Next, point ( N ) is on ( VD ) such that ( DN = 2VN ). So, ( N ) divides ( VD ) in a 1:2 ratio. Let me find the coordinates of ( N ).The coordinates of ( V ) are ( (1, 1, 1) ) and ( D ) are ( (0, 2, 0) ). To find ( N ), which divides ( VD ) in the ratio ( VN:ND = 1:2 ), I can use the section formula.In the section formula, if a point divides a line segment joining ( (x_1, y_1, z_1) ) and ( (x_2, y_2, z_2) ) in the ratio ( m:n ), then the coordinates are:( left( frac{m x_2 + n x_1}{m + n}, frac{m y_2 + n y_1}{m + n}, frac{m z_2 + n z_1}{m + n} right) )Here, ( m = 1 ) and ( n = 2 ). So,( N_x = frac{1 cdot 0 + 2 cdot 1}{1 + 2} = frac{0 + 2}{3} = frac{2}{3} )( N_y = frac{1 cdot 2 + 2 cdot 1}{3} = frac{2 + 2}{3} = frac{4}{3} )( N_z = frac{1 cdot 0 + 2 cdot 1}{3} = frac{0 + 2}{3} = frac{2}{3} )So, ( N = left( frac{2}{3}, frac{4}{3}, frac{2}{3} right) ).Now, I need to find the vectors ( overrightarrow{AM} ) and ( overrightarrow{BN} ) to compute the angle between them.First, ( overrightarrow{AM} ) is from ( A ) to ( M ). Since ( A = (0, 0, 0) ) and ( M = left( frac{3}{2}, frac{1}{2}, frac{1}{2} right) ), the vector is:( overrightarrow{AM} = left( frac{3}{2}, frac{1}{2}, frac{1}{2} right) )Next, ( overrightarrow{BN} ) is from ( B ) to ( N ). ( B = (2, 0, 0) ) and ( N = left( frac{2}{3}, frac{4}{3}, frac{2}{3} right) ). So, subtracting coordinates:( overrightarrow{BN} = left( frac{2}{3} - 2, frac{4}{3} - 0, frac{2}{3} - 0 right) = left( -frac{4}{3}, frac{4}{3}, frac{2}{3} right) )Now, to find the angle between vectors ( overrightarrow{AM} ) and ( overrightarrow{BN} ), I can use the dot product formula:( overrightarrow{AM} cdot overrightarrow{BN} = |overrightarrow{AM}| |overrightarrow{BN}| cos theta )So, solving for ( cos theta ):( cos theta = frac{overrightarrow{AM} cdot overrightarrow{BN}}{|overrightarrow{AM}| |overrightarrow{BN}|} )First, let's compute the dot product ( overrightarrow{AM} cdot overrightarrow{BN} ):( overrightarrow{AM} cdot overrightarrow{BN} = left( frac{3}{2} times -frac{4}{3} right) + left( frac{1}{2} times frac{4}{3} right) + left( frac{1}{2} times frac{2}{3} right) )Calculating each term:1. ( frac{3}{2} times -frac{4}{3} = -2 )2. ( frac{1}{2} times frac{4}{3} = frac{2}{3} )3. ( frac{1}{2} times frac{2}{3} = frac{1}{3} )Adding them up:( -2 + frac{2}{3} + frac{1}{3} = -2 + 1 = -1 )So, the dot product is ( -1 ).Next, let's find the magnitudes of ( overrightarrow{AM} ) and ( overrightarrow{BN} ).First, ( |overrightarrow{AM}| ):( |overrightarrow{AM}| = sqrt{left( frac{3}{2} right)^2 + left( frac{1}{2} right)^2 + left( frac{1}{2} right)^2} = sqrt{frac{9}{4} + frac{1}{4} + frac{1}{4}} = sqrt{frac{11}{4}} = frac{sqrt{11}}{2} )Next, ( |overrightarrow{BN}| ):( |overrightarrow{BN}| = sqrt{left( -frac{4}{3} right)^2 + left( frac{4}{3} right)^2 + left( frac{2}{3} right)^2} = sqrt{frac{16}{9} + frac{16}{9} + frac{4}{9}} = sqrt{frac{36}{9}} = sqrt{4} = 2 )So, ( |overrightarrow{AM}| = frac{sqrt{11}}{2} ) and ( |overrightarrow{BN}| = 2 ).Now, plugging into the formula:( cos theta = frac{-1}{left( frac{sqrt{11}}{2} times 2 right)} = frac{-1}{sqrt{11}} )But cosine is positive in the first quadrant, and since we're talking about the angle between two lines, we take the absolute value. So,( cos theta = frac{1}{sqrt{11}} )Rationalizing the denominator,( cos theta = frac{sqrt{11}}{11} )So, the cosine of the angle between the skew lines ( AM ) and ( BN ) is ( frac{sqrt{11}}{11} ).**Final Answer**The cosine of the angle between the skew lines ( AM ) and ( BN ) is boxed{dfrac{sqrt{11}}{11}}.