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question:The distance from the vertex of the hyperbola dfrac{x^2}{4} y^2 = 1 to its asymptote is ______.
answer:Solution: The coordinates of the vertex of the hyperbola dfrac{x^2}{4} - y^2 = 1 are (2,0), and the equations of its asymptotes are y = pm dfrac{1}{2}x. Therefore, the distance we are looking for is dfrac{1}{sqrt{dfrac{1}{4} + 1}} = dfrac{2sqrt{5}}{5}. Hence, the answer is boxed{dfrac{2sqrt{5}}{5}}. To solve this problem, we find the equations of the asymptotes of the hyperbola and the coordinates of the vertex, and then use the formula for the distance from a point to a line. This question tests the application of simple properties of hyperbolas and computational skills.
question:w many right interior angles are in a right triangle?
answer:1. Recognize that by definition, a right triangle contains one interior angle that is exactly 90 degrees. 2. The remaining two angles must be acute to sum to 90 degrees, satisfying the total of 180 degrees for all interior angles of a triangle. 3. Thus, there is exactly one right angle in a right triangle. Conclusion: There is boxed{1} right angle in a right triangle.
question:There are two distinguishable flagpoles, and there are 19 flags, of which 10 are identical blue flags, and 9 are identical green flags. Let N be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find the remainder when N is divided by 1000.
answer:To solve this problem, we follow a step-by-step approach based on the given solution: 1. **Understanding the General Problem**: For a string of y elements, where we want to arrange x elements such that none of the x elements are adjacent, the number of such arrangements is given by {y-x+1choose x}. This is our foundational formula. 2. **Generalizing for Blue and Green Flags**: Given a blue flags and b green flags, we aim to arrange them such that no two green flags are adjacent. Additionally, we want to split these flags between two distinguishable flagpoles. 3. **Addressing Consecutive Green Flags Across Poles**: To ensure no two green flags are adjacent across the two poles, we introduce an extra blue flag. This allows us to divide the arrangement into two parts (for the two poles) without having a green flag at the end of one pole and the start of the next. 4. **Calculating the Arrangements**: With the extra blue flag, we have {a+2choose b} ways to arrange the flags such that no greens are adjacent. To decide the division between the two poles, we have a+1 choices for the extra blue flag's position. This gives us a total of (a+1){a+2choose b} arrangements. 5. **Correcting for Overcounting**: The above method overcounts cases where one pole might not have any flags. To correct this, we calculate the cases where the extra blue flag is at either end, giving 2{a+1choose b} such arrangements. 6. **Final Calculation**: The total number of valid arrangements is (a+1){a+2choose b} - 2{a+1choose b}. Substituting a=10 and b=9, we calculate the arrangements as follows: - First, calculate {10+2choose 9} = {12choose 9} = 220. - Then, calculate 2{10+1choose 9} = 2{11choose 9} = 2 times 55 = 110. - The total arrangements are (10+1) times 220 - 2 times 110 = 11 times 220 - 220 = 2200. - Subtracting the overcounted arrangements, we get 2200 - 220 = 1980. However, this calculation seems to have deviated from the original solution's final steps. Let's correct this and follow the original solution more closely: - For a=10 and b=9, we directly apply the formula: (a+1){a+2choose b} - 2{a+1choose b} = (10+1){10+2choose 9} - 2{10+1choose 9} = 11 times 66 - 2 times 55 = 726 - 110 = 616. Realizing the mistake in the calculation, we should adhere strictly to the original solution's final step: - The correct final step is to calculate (10+1){10+2choose 9} - 2{10+1choose 9} = 11 times 220 - 2 times 55 = 2310. Since we are looking for the remainder when N is divided by 1000, we find: [N equiv 2310 pmod{1000} equiv boxed{310}.] This corrects the calculation error and aligns with the original solution's conclusion.
question:the Cartesian coordinate system xoy, the parametric equation of curve C_1 is begin{cases} x=2sqrt{2}frac{sqrt{2}}{2}t y=sqrt{2}+frac{sqrt{2}}{2}t end{cases} (t text{ is the parameter}). In the polar coordinate system with the origin as the pole and the positive xaxis as the polar axis, the equation of curve C_2 is rho=4sqrt{2}sin theta. (Ⅰ) Convert the equation of C_2 into a Cartesian coordinate equation; (Ⅱ) Suppose C_1 and C_2 intersect at points A and B, and the coordinates of point P are (sqrt{2},2sqrt{2}), find |PA|+|PB|.
answer:Solution: (Ⅰ) Since the equation of curve C_2 is rho=4sqrt{2}sin theta, we have rho^2=4sqrt{2}rhosin theta. Therefore, the Cartesian coordinate equation of C_2 is x^2+y^2=4sqrt{2}y, which simplifies to x^2+(y-2sqrt{2})^2=8. (Ⅱ) Converting begin{cases} x=2sqrt{2}-frac{sqrt{2}}{2}t y=sqrt{2}+frac{sqrt{2}}{2}t end{cases} to begin{cases} x=sqrt{2}-frac{sqrt{2}}{2}t y=2sqrt{2}+frac{sqrt{2}}{2}t end{cases} (t text{ is the parameter}). Substituting begin{cases} x=sqrt{2}-frac{sqrt{2}}{2}t y=2sqrt{2}+frac{sqrt{2}}{2}t end{cases} into x^2+(y-2sqrt{2})^2=8, we get t^2-2t-6=0. Then, t_1+t_2=2 and t_1t_2=-6. Therefore, |PA|+|PB|=sqrt{(t_1+t_2)^2-4t_1t_2}=sqrt{4+24}=2sqrt{7}. Thus, the final answer is boxed{2sqrt{7}}.