answer:Given matrices mathbf{D} = begin{pmatrix} k & 0 0 & k end{pmatrix} and mathbf{R} = begin{pmatrix} cos phi & -sin phi sin phi & cos phi end{pmatrix}, the product mathbf{R} mathbf{D} is: [ mathbf{R} mathbf{D} = begin{pmatrix} cos phi & -sin phi sin phi & cos phi end{pmatrix} begin{pmatrix} k & 0 0 & k end{pmatrix} = begin{pmatrix} k cos phi & -k sin phi k sin phi & k cos phi end{pmatrix} ] Comparing this with the given matrix begin{pmatrix} 5 & -12 12 & 5 end{pmatrix}, we equate and solve: - k cos phi = 5 - k sin phi = 12 Now, tan phi = frac{k sin phi}{k cos phi} = frac{12}{5}. To find k, we use the Pythagorean identity sin^2 phi + cos^2 phi = 1: [ left(frac{k sin phi}{k}right)^2 + left(frac{k cos phi}{k}right)^2 = 1 implies left(frac{12}{k}right)^2 + left(frac{5}{k}right)^2 = 1 frac{144 + 25}{k^2} = 1 implies k^2 = 169 implies k = 13 ] Thus, we have k = 13 and tan phi = boxed{frac{12}{5}}.

answer:To calculate the total amount Tom spends on bricks for his shed, we can break down the calculation into smaller steps: 1. Determine the number of bricks Tom buys at half price: - Total bricks needed = 1000 - Half of 1000 = 1000 / 2 = 500 bricks 2. Calculate the cost per brick for the half-priced bricks: - Original price per brick = 0.50 - Half of 0.50 = 0.50 / 2 = 0.25 per brick 3. Calculate the total cost for the half-priced bricks: - Number of half-priced bricks = 500 - Cost per half-priced brick = 0.25 - Total cost for half-priced bricks = 500 * 0.25 = 125 4. Calculate the total cost for the full-priced bricks: - Number of full-priced bricks = 500 (the other half) - Cost per full-priced brick = 0.50 - Total cost for full-priced bricks = 500 * 0.50 = 250 5. Calculate the total amount spent: - Total cost for half-priced bricks = 125 - Total cost for full-priced bricks = 250 - Total amount spent = 125 + 250 = 375 Therefore, Tom spends boxed{375} dollars in total.

answer:**Analysis** This question examines the method of finding the eccentricity of an ellipse, paying attention to the use of the definition of an ellipse, as well as the relationship between a, b, and c, testing computational skills, and is considered a basic question. According to the definition of an ellipse, we have |PF_1| + |PF_2| = 2a, and also |PF_1| cdot |PF_2| = a^2, we can derive |PF_1| = |PF_2| = a, meaning P is at the endpoint of the short axis of the ellipse. From the condition, we can get b = c, and then calculate the eccentricity of the ellipse. **Solution** From the definition of an ellipse, we have |PF_1| + |PF_2| = 2a, and also |PF_1| cdot |PF_2| = a^2, we can derive |PF_1| = |PF_2| = a, meaning P is at the endpoint of the short axis of the ellipse, |OP| = b, and |OP| = dfrac{1}{2}|F_1F_2| = c, thus we have c = b = sqrt{a^2 - c^2}, which gives a = sqrt{2}c, e = dfrac{c}{a} = dfrac{sqrt{2}}{2}. Therefore, the correct answer is boxed{C}.

answer:Since three lines can determine three planes, there are two possible scenarios: 1. The three lines intersect pairwise and have one common point, such as the lines containing the three lateral edges SA, SB, SC of a tetrahedron S-ABC; 2. The three lines are pairwise parallel and have no common points, such as the lines containing the three lateral edges of a prism ABC-A₁B₁C₁. In summary, the number of common points of these three lines can be either 0 or 1. Therefore, the answer is: boxed{0 text{ or } 1}.