answer:1. First, confirm DEF is a right triangle by checking if DF is the hypotenuse. Since DF is the longest side given, it likely serves as the hypotenuse. 2. Calculate EF using Pythagoras' theorem for a right triangle: [ EF = sqrt{DF^2 - DE^2} = sqrt{20^2 - 15^2} = sqrt{400 - 225} = sqrt{175} = 5 sqrt{7} ] 3. Determine the distance from F to the midpoint of DE, which is half the length of the hypotenuse: [ text{Distance} = frac{DF}{2} = frac{20}{2} = 10 ] Conclusion: Therefore, the distance from point F to the midpoint of segment DE is boxed{10} units.

answer:Let's call the weight of each of the remaining two statues X pounds. The total weight of the marble block was 80 pounds. According to the information given, the weight of the discarded marble is 22 pounds, and the weights of the first two statues are 10 pounds and 18 pounds, respectively. So, the weight of the marble after carving the first two statues and discarding the excess would be: 80 pounds (initial block) - 10 pounds (first statue) - 18 pounds (second statue) - 22 pounds (discarded marble) = 30 pounds Since the remaining two statues weigh the same, we can divide the remaining weight by 2 to find the weight of each statue: 30 pounds / 2 = 15 pounds Therefore, each of the remaining two statues Hammond carved weighs boxed{15} pounds.

answer:For the first line, [ begin{pmatrix} x y end{pmatrix} = begin{pmatrix} 1 1 end{pmatrix} + t begin{pmatrix} 2 -3 end{pmatrix} = begin{pmatrix} 1 + 2t 1 - 3t end{pmatrix}. ] For the second line, [ begin{pmatrix} x y end{pmatrix} = begin{pmatrix} -1 2 end{pmatrix} + u begin{pmatrix} 5 -2 end{pmatrix} = begin{pmatrix} -1 + 5u 2 - 2u end{pmatrix}. ] Setting equations equal: [ 1 + 2t = -1 + 5u quad text{(1)} ] [ 1 - 3t = 2 - 2u quad text{(2)} ] From equation (1), [ 2t - 5u = -2 quad text{(3)} ] From equation (2), [ -3t + 2u = 1 quad text{(4)} ] Solving the system of equations (3) and (4) using elimination or substitution method: Multiply equation (3) by 3 and (4) by 2, [ 6t - 15u = -6 quad text{(5)} ] [ -6t + 4u = 2 quad text{(6)} ] Adding equations (5) and (6), [ -11u = -4 implies u = frac{4}{11} ] Substituting ( u = frac{4}{11} ) into equation (3), [ 2t - 5 left(frac{4}{11}right) = -2 implies 2t = -2 + frac{20}{11} = frac{-2 times 11 + 20}{11} = frac{-22 + 20}{11} = frac{-2}{11} ] [ t = -frac{1}{11} ] Substitute (t = -frac{1}{11}) and (u = frac{4}{11}) into either line equation to find the intersection: [ begin{pmatrix} x y end{pmatrix} = begin{pmatrix} 1 + 2(-frac{1}{11}) 1 - 3(-frac{1}{11}) end{pmatrix} = begin{pmatrix} 1 - frac{2}{11} 1 + frac{3}{11} end{pmatrix} = begin{pmatrix} frac{9}{11} frac{14}{11} end{pmatrix} ] Conclusion: [ boxed{begin{pmatrix} frac{9}{11} frac{14}{11} end{pmatrix}} ]

answer:**Analysis** This problem primarily tests the application of derivatives to study the monotonicity of a function and the distribution of its roots. It also assesses the ability to analyze problems and is considered moderately difficult. First, determine the function's domain, then differentiate it. Since f(x) is not monotonic in the interval (0,+infty), we know that f′(x)=0 has two positive solutions. That is, 4x^{2}-ax+1=0 has two positive solutions. According to the relationship between the roots and coefficients of a quadratic equation, we can derive an inequality group about a. Solving this group of inequalities will yield the desired result. **Solution** The domain of the function f(x) is (0,+infty), Because f(x)=2x^{2}-ax+ln x, We have f′(x)=4x-a+frac{1}{x}=frac{1}{x}(4x^{2}-ax+1). Since f(x) is not monotonic in the interval (0,+infty), we know that f′(x)=0 has two positive solutions. That is, 4x^{2}-ax+1=0 has two positive solutions, denoted as x_1 and x_2. Hence, we can set up the following system of equations: begin{cases}begin{matrix} Delta > 0 x_1+x_2=frac{a}{4} > 0 x_1x_2=frac{1}{4} > 0 end{matrix}end{cases} Solving this system gives a>4. Therefore, the range of values for a is boxed{(4,+infty)}.