answer:Let's denote the sum of the first 10 numbers as S1, the sum of the last 10 numbers as S2, and the sum of all 25 numbers as S. From the given information, we have: 1. The average of the first 10 numbers is Q, so the sum of the first 10 numbers (S1) is 10Q. 2. The average of the last 10 numbers is R, so the sum of the last 10 numbers (S2) is 10R. 3. The average of all 25 numbers is P, so the sum of all 25 numbers (S) is 25P. Now, the sum of all 25 numbers (S) can also be expressed as the sum of the first 10 numbers (S1), plus the sum of the last 10 numbers (S2), plus the sum of the 5 numbers in the middle (which includes the 13th number we are looking for). Let's denote the sum of the 5 middle numbers as S3. Then we have: S = S1 + S3 + S2 Substituting the values we have for S1 and S2, we get: 25P = 10Q + S3 + 10R Now, we need to find S3, the sum of the 5 middle numbers. We know that the 7th number is X and the 19th number is Y, but we don't have information about the 8th, 9th, 10th, 11th, and 12th numbers, nor about the 14th, 15th, 16th, 17th, and 18th numbers. However, we are interested in the 13th number, which is part of S3. Let's denote the 13th number as M (the middle number we are looking for). The sum of the 5 middle numbers (S3) includes M and four other numbers. Without loss of generality, let's denote the sum of the other four numbers as O. Then we have: S3 = M + O Substituting this into our equation for S, we get: 25P = 10Q + (M + O) + 10R Rearranging the equation to solve for M, we get: M = 25P - 10Q - 10R - O Without additional information about the other four numbers in the middle (which contribute to O), we cannot determine the exact value of M, the boxed{13th} number. We need more information about the other numbers in the sequence to find the value of M.

answer:(I) Let 2c be the focal length of the ellipse (C_{1}). From the given information, we have 2c=4 sqrt {2}, and frac{c}{a}= frac{ sqrt {6}}{3}. Solving these equations yields a=2 sqrt {3}, and b=2. Thus, the standard equation of the ellipse (C_{1}) is frac{x^{2}}{12}+ frac{y^{2}}{4}=1. The parabola (C_{2}) opens upward, so F is the upper vertex of the ellipse (C_{1}), and F(0,2). This implies p=4, and the standard equation of the parabola (C_{2}) is x^{2}=8y. (II) From the given information, we know that the slope of the line PQ exists. Let the equation of the line PQ be y=kx+m. Suppose P(x_{1},y_{1}), and Q(x_{2},y_{2}). Then overrightarrow{FP}=(x_{1},y_{1}-2), and overrightarrow{FQ}=(x_{2},y_{2}-2). Thus, overrightarrow{FP} cdot overrightarrow{FQ}=x_{1}x_{2}+y_{1}y_{2}-2(y_{1}+y_{2})+4=0. This equation can be rewritten as (1+k^{2})x_{1}x_{2}+(km-2k)(x_{1}+x_{2})+m^{2}-4m+4=0(*). Solving the system of equations: [ begin{cases} y=kx+m frac{x^{2}}{12}+ frac{y^{2}}{4}=1 end{cases} ] and eliminating y, we obtain (3k^{2}+1)x^{2}+6kmx+3m^{2}-12=0(**). According to the problem, x_{1} and x_{2} are the two roots of equation (**), and triangle =144k^{2}-12m^{2}+48 > 0. Consequently, x_{1}+x_{2}= frac{-6km}{3k^{2}+1}, and x_{1}cdot x_{2}= frac{3m^{2}-12}{3k^{2}+1}. Substituting x_{1}+x_{2} and x_{1}cdot x_{2} into (*), we get m^{2}-m-2=0. Solving this equation, we obtain m=-1 (we discard the solution m=2 as it does not meet the requirements). Solving the system of equations: [ begin{cases} y=kx-1 x^{2}=8y end{cases} ] and eliminating y, we get x^{2}-8kx+8=0. Letting triangle {{'}}=64k^{2}-32=0, we obtain k^{2}= frac {1}{2}. It can be verified that k^{2}= frac {1}{2}, and m=-1 satisfy the conditions. Now, |x_{1}-x_{2}|= sqrt {(x_{1}+x_{2})^{2}-4x_{1}x_{2}}= sqrt { frac {72}{25}-4(- frac {18}{5})}= frac {12 sqrt {3}}{5}. Therefore, S_{triangle FPQ}= frac {1}{2} times 3 times |x_{1}-x_{2}| = boxed{frac {18 sqrt {3}}{5}}.

answer:Given: P, Q, R are the midpoints of the equal sides AB, BC, CD of a convex quadrilateral ABCD. 1. **Identify the midpoints**: - Let P be the midpoint of side AB, - Let Q be the midpoint of side BC, - Let R be the midpoint of side CD. 2. **Draw the perpendicular bisectors**: - Draw the perpendicular bisector l_1 to the segment PQ, - Draw the perpendicular bisector l_2 to the segment QR. 3. **Position of points on perpendicular bisectors**: - Since AB = BC = CD, points B and C will lie on lines l_1 and l_2 respectively because P and R bisect these equal sides. 4. **Utilize the midpoint property**: - The segments PQ and QR are midsegments for triangle ABC and triangle BCD respectively. - We know from the midpoint theorem that PQ and QR are parallel to sides AC and BD, and half as long. Therefore, the line segment connecting the midpoints will be half the distance and parallel to the original segment. 5. **Construct point Q**: - Since point Q is the midpoint of BC, let's use Q to construct an intercept on both l_1 and l_2. - This ensuring PQ = QR and Q being the midpoint creates symmetry requirements that B and C must satisfy. 6. **Determine the sides**: - By ensuring symmetry and equality: - Through point Q, draw a segment such that Q is the midpoint of it, - This intersection gives two points which are B and C, as Q is a midpoint BQ = QC. 7. **Form the quadrilateral**: - Finally, once B and C are defined, connect them sequentially forming the vertices A, B, C and D such that sides AB, BC, and CD are equal and validate midpoints P, Q, and R. Thus the described steps ensure that the convex quadrilateral ABCD is constructed given the midpoints of its three equal sides. boxed{}

answer:First, we need to find the weight of the rice in each container in pounds. Since the bag of rice weighs sqrt(50) pounds and it was divided equally into 7 containers, we divide sqrt(50) by 7: Weight of rice in each container (in pounds) = sqrt(50) / 7 Now, to convert the weight from pounds to ounces, we multiply by 16 (since 1 pound = 16 ounces): Weight of rice in each container (in ounces) = (sqrt(50) / 7) * 16 Let's calculate the value: Weight of rice in each container (in ounces) = (sqrt(50) / 7) * 16 Weight of rice in each container (in ounces) = (5√2 / 7) * 16 Weight of rice in each container (in ounces) = (5 * 16√2) / 7 Weight of rice in each container (in ounces) = 80√2 / 7 Now, we can calculate the numerical value: Weight of rice in each container (in ounces) ≈ (80 * 1.414) / 7 Weight of rice in each container (in ounces) ≈ 113.12 / 7 Weight of rice in each container (in ounces) ≈ 16.16 Therefore, each container has approximately boxed{16.16} ounces of rice.