answer:Given y = sqrt{x}, we substitute y^2 = x and y^4 = x^2 into the expression. Thus, the problem becomes evaluating the minimum value of [frac{x^2+x^2}{x cdot x} = frac{2x^2}{x^2} = 2.] Since the expression simplifies to a constant, 2, the minimum value of the expression is trivially boxed{2}. Conclusion: The problem does not offer a range for values but directly leads to a constant value. All steps verify that the simplifications and substitutions based on the initially imposed condition (y = sqrt{x}) are logically correct and mathematically valid.

answer:The general term of the expansion of ((x- frac {1}{x})^{4}) is given by T_{r+1}= C_{ 4 }^{ r } cdot x^{4-r} cdot (-1)^{r} cdot x^{-r} = (-1)^{r} cdot C_{ 4 }^{ r } cdot x^{4-2r} To find the constant term, we set the exponent of x to zero: 4-2r=0 Solving for r, we get r=2. Therefore, the constant term in the expansion is T_{3}= C_{ 4 }^{ 2 } = 6 The answer is boxed{A}.

answer:To find the surface area of the cube, we first need to determine the length of its sides. We are given three vertices of the cube, P=(7,12,10), Q=(8,8,1), and R=(11,3,9). We can calculate the distances between these points to understand the geometry of the cube. 1. Calculate the distance between P and Q: [PQ = sqrt{(8-7)^2 + (8-12)^2 + (1-10)^2} = sqrt{1^2 + (-4)^2 + (-9)^2} = sqrt{1 + 16 + 81} = sqrt{98}] 2. Calculate the distance between P and R: [PR = sqrt{(11-7)^2 + (3-12)^2 + (9-10)^2} = sqrt{4^2 + (-9)^2 + (-1)^2} = sqrt{16 + 81 + 1} = sqrt{98}] 3. Calculate the distance between Q and R: [QR = sqrt{(11-8)^2 + (3-8)^2 + (9-1)^2} = sqrt{3^2 + (-5)^2 + 8^2} = sqrt{9 + 25 + 64} = sqrt{98}] From these calculations, we observe that PQ = PR = QR = sqrt{98}, indicating that triangle PQR is an equilateral triangle. This implies that the sides of the cube, when projected onto a plane, form an equilateral triangle, suggesting that the edges of the cube are equal in length. Given that the diagonal of a cube is sqrt{2} times longer than its side, we can set up the equation for the side length a of the cube: [asqrt{2} = sqrt{98}] Solving for a: [a = frac{sqrt{98}}{sqrt{2}} = sqrt{49} = 7] Now that we have the side length of the cube, we can calculate its surface area. A cube has 6 faces, each of which is a square with area a^2. Therefore, the total surface area A is: [A = 6a^2 = 6 times 7^2 = 6 times 49 = 294] Thus, the surface area of the cube is boxed{294}.

answer:1. **Determine coordinates of B and D:** - B is on y = x, so B = (b, b). - D is on y = 2x, so D = (d, 2d). 2. **Determine coordinates of C:** - Since C is on y = 3x, and should also fulfill the parallelogram condition with C = (b+d, b+2d), it implies that b + 2d = 3(b + d). Solving this gives b = 2d. 3. **Check area condition using Shoelace Theorem:** - Apply Shoelace Theorem for vertices A = (0,0), B = (b, b), D = (d, 2d), and C = (3d, 3 times 2d) = (3d, 6d): [ text{Area} = frac{1}{2} left| 0 + b times 6d + 3d times 2d - (b times 2d + 3d times b) right| ] [ = frac{1}{2} left| 6bd + 6d^2 - 4bd right| = frac{1}{2} left| 2bd + 6d^2 right| = |bd + 3d^2| ] - Setting b = 2d, the area becomes: [ 2d^2 + 3d^2 = 5d^2 ] - The area condition is 2,!000,!000 = 5d^2, leading to: [ d^2 = frac{2,!000,!000}{5} = 400,!000 ] - Solve for d: d = sqrt{400,!000} = 200 times 10 = 2000 (only positive d since coordinates are in the first quadrant). 4. **Conclusion:** - The pair (b, d) where b = 2d and d = 2000 uniquely determines the vertices of the parallelogram. - Since b = 4000 and d = 2000, there is exactly one such parallelogram. 1 The final answer is boxed{textbf{(B)} 1}