answer:First, calculate the length of one side of the square, let's choose PQ. By the Pythagorean theorem, the length of PQ is given by: [ PQ = sqrt{(-3-2)^2 + (4-3)^2} = sqrt{(-5)^2 + 1^2} = sqrt{25 + 1} = sqrt{26} ] Since all sides of a square are equal, the side length of the square is sqrt{26}. Therefore, the area of the square is: [ text{Area} = (text{side length})^2 = (sqrt{26})^2 = 26 ] Thus, the area of the square is boxed{26} square units.

answer:To find the expected value of the median of three fair six-sided dice, we can use symmetry and properties of expected values. 1. **Symmetry Argument**: Consider a roll of the dice resulting in values (a, b, c) such that (a leq b leq c). By symmetry, for each roll ((a, b, c)), there is a corresponding roll ((7-a, 7-b, 7-c)). The median of the roll ((a, b, c)) is (b), and the median of the roll ((7-a, 7-b, 7-c)) is (7-b). 2. **Expected Value Calculation**: Since each roll ((a, b, c)) has a corresponding roll ((7-a, 7-b, 7-c)), the expected value of the median (b) can be calculated as follows: [ text{Expected value of the median} = frac{b + (7-b)}{2} = frac{7}{2} ] 3. **Verification by Expected Value of Each Die**: The expected value of a single fair six-sided die is: [ E(X) = frac{1 + 2 + 3 + 4 + 5 + 6}{6} = frac{21}{6} = frac{7}{2} ] Since the dice are fair and independent, the expected value of the median of three dice should also be (frac{7}{2}). Thus, the expected value of the median of the numbers rolled is (frac{7}{2}). This can be written as (frac{m}{n}) where (m = 7) and (n = 2). Since (m) and (n) are relatively prime, we have (m+n = 7+2 = 9). The final answer is (boxed{9}).

answer:Given the set ( A = {1, 2, ldots, 366} ), we denote a binary subset ( B = {a, b} ) as having property ( P ) if ( 17 mid (a+b) ). 1. **Counting subsets with property ( P )**: We first determine the number of subsets in ( A ) where the sum ( a + b ) is divisible by 17. 1.1. **Classifying elements by their residues modulo 17**: [ {1, 2, ldots, 366} pmod{17} ] Since ( 366 = 21 times 17 + 9 ), each residue class ( [k] ) (for ( k = 0, 1, ldots, 16 )) will contain 21 elements, except ( [9] ) and ( [8] ) which will contain 22 elements each. 1.2. **Formulating subsets that sum to 0 modulo 17**: Consider ( a in [k] ) and ( b in [17-k] ): - **Case ( k = 0 )**: [ begin{aligned} &text{Elements in } [0] = 21 &|B| = binom{21}{2} = frac{21 times 20}{2} = 210 end{aligned} ] - **Other cases ( 1 leq k leq 7 )**: [ begin{aligned} &text{Pairing } [k] text{ and } [17-k]: &21 times 21 = 441 quad text{since both classes have 21 elements} end{aligned} ] There are 7 such pairs: [ 441 times 7 = 3087 ] - **Case ( k = 8 text{ and } 9 ): [ begin{aligned} &22 times 21 = 462 &|B| = binom{22}{1} times binom{21}{1} = 22 times 21 = 462 end{aligned} ] [ begin{aligned} &&text{Sum of all cases} &210 + 3087 + 484 = 3928 end{aligned} ] Thus, the total number of binary subsets ( B ) in ( A ) possessing property ( P ) is: [ boxed{3928} ] 2. **Finding non-intersecting binary subsets with property ( P )**: We select pairs of elements that do not intersect: 2.1. **Case ( a, b in [0] ) with unique pairs**: [ text{From } 21 text{ elements, we can select } binom{21}{2} = 210 quad text{subtract} frac{21 times 20}{2 times 2} text{ accounting for non-overlapping, result } 10 ] 2.2. **Case generic ( a in [i], b in [17-i]): [ begin{aligned} &21 times 21 pairs, non-overlapping 22 quad, repeat 7* 21= 147 end{aligned} ] 2.3. **Case (a=b) intersection, no overlap:** ] 22* binom{9 with 10 pairs }boxed{179} ensuring no overlap} 179 cells for with no overlap}

answer:Given the functional equation for the function ( f: mathbb{R} to mathbb{R} ): [ (x-y) f(x+y) - (x+y) f(x-y) = 4xy (x^2 - y^2) ] we aim to find all possible functions ( f ) that satisfy this equation. Let's proceed step-by-step: 1. **Substitution for simplification**: Let: [ x + y = a quad text{and} quad x - y = b ] Then, we have: [ x = frac{a + b}{2} quad text{and} quad y = frac{a - b}{2} ] Also, note: [ x^2 - y^2 = left( frac{a + b}{2} right)^2 - left( frac{a - b}{2} right)^2 = ab ] 2. **Rewrite the functional equation**: Using the substitutions ( a ) and ( b ), the original equation becomes: [ (x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2) ] Substituting ( x + y = a ), ( x - y = b ), and ( x^2 - y^2 = ab ): [ b f(a) - a f(b) = 4 left(frac{a + b}{2}right)left(frac{a - b}{2}right) ab ] Simplifying the right side: [ b f(a) - a f(b) = 4 cdot frac{a^2 - b^2}{4} cdot ab = ab(a^2 - b^2) ] 3. **Introduction of special inputs**: Let's evaluate the equation at different specific values. - For ( a = 0 ): [ b f(0) - 0 f(b) = 0 implies b f(0) = 0 implies f(0) = 0 ] 4. **Eliminate constant dependency**: Now consider ( a, b neq 0 ) and divide the equation by ( ab ): [ frac{f(a)}{a} - frac{f(b)}{b} = a^2 - b^2 ] We observe that the quantity ( frac{f(a)}{a} - a^2 ) is independent of the choice ( a ) when compared to ( b ). Hence, define: [ frac{f(a)}{a} - a^2 = c quad text{(a constant for all non-zero }, a) ] So, for ( f(a) ): [ frac{f(a)}{a} = a^2 + c implies f(a) = a^3 + c a ] 5. **Verify the solution for all ( x )**: The above form suggests: [ f(x) = x^3 + cx quad text{for all } x in mathbb{R} ] Check for ( x = 0 ), we already have ( f(0) = 0 ), which is consistent. 6. **Final Verification**: Substitute ( f(x) = x^3 + cx ) into the original equation to ensure it holds: [ (x-y) left((x+y)^3 + c(x+y)right) - (x+y) left((x-y)^3 + c(x-y)right) = 4xy (x^2 - y^2) ] Expand and simplify to confirm both sides are equal. Detailed algebra will show the left-hand side indeed equals the right-hand side, confirming our function form is correct. Thus, the general solution to the given functional equation is: [ f(x) = x^3 + cx ] # Conclusion: [ boxed{f(x) = x^3 + cx} ]