answer:Since the given hyperbola is frac {x^{2}}{a^{2}}- frac {y^{2}}{3}=1 and its eccentricity is 2, we have the following relationship for a hyperbola: e = frac{sqrt{a^2 + b^2}}{a} Given that the eccentricity e = 2, we plug into the equation: 2 = frac{sqrt{a^2 + b^2}}{a} Since b^2 = 3 (from the coefficient of y^2 in the hyperbola equation), we have: 2 = frac{sqrt{a^2 + 3}}{a} Squaring both sides, we get: 4 = frac{a^2 + 3}{a^2} 4a^2 = a^2 + 3 3a^2 = 3 a^2 = 1 a = pm 1 Because a is a length, we take a = 1. The coordinates of the vertices of the hyperbola are (pm a, 0), which are (pm 1, 0). The equations of the asymptotes are y = pm sqrt{3}x. Now, the distance d from the vertex (a, 0) to the line y = sqrt{3}x can be found using the point-to-line distance formula: d = frac{|ax_{1} - by_{1} + c|}{sqrt{a^2 + b^2}} The line y = sqrt{3}x can be rewritten as -sqrt{3}x + y = 0, so a = -sqrt{3}, b = 1, and c = 0. Plugging in the vertex coordinates (1, 0), we get: d = frac{|-sqrt{3} cdot 1 - 1 cdot 0 + 0|}{sqrt{(-sqrt{3})^2 + 1^2}} d = frac{sqrt{3}}{sqrt{3 + 1}} d = frac{sqrt{3}}{2} So the distance from the vertex of the hyperbola to the asymptote is boxed{frac {sqrt {3}}{2}}.

answer:To solve this problem, we consider two cases based on the given information that we have an isosceles triangle with two sides of lengths 2 and 4. **Case 1:** The equal sides are of length 2. - In this scenario, the sides of the triangle would be 2, 2, and 4. - To check if a triangle can be formed, we use the triangle inequality theorem which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. - For our sides, we see that 2 + 2 = 4, which is exactly equal to the length of the third side. - This means that the sides would lie in a straight line and thus, a triangle cannot be formed in this case. **Case 2:** The length of the base is 2, and the equal sides are of length 4. - Here, the sides of the triangle are 2, 4, and 4. - Applying the triangle inequality theorem again, we find that 4 + 4 > 2, which satisfies the condition for a triangle to be formed. - The perimeter of a triangle is the sum of its sides, so for this triangle, the perimeter is 2 + 4 + 4. - Calculating the sum gives us a perimeter of 10. Therefore, considering both cases, we conclude that the only valid scenario for forming an isosceles triangle with the given side lengths results in a perimeter of 10. Hence, the perimeter of the triangle is boxed{10}.

answer:To analyze the condition "overrightarrow{a}cdot overrightarrow{c}=overrightarrow{b}cdot overrightarrow{c}" and its relation to "overrightarrow{a}=overrightarrow{b}", let's consider two scenarios: 1. **Scenario 1: overrightarrow{a} perp overrightarrow{c} and overrightarrow{b} perp overrightarrow{c}** - If overrightarrow{a} perp overrightarrow{c}, then overrightarrow{a} cdot overrightarrow{c} = 0. - Similarly, if overrightarrow{b} perp overrightarrow{c}, then overrightarrow{b} cdot overrightarrow{c} = 0. - Therefore, in this case, overrightarrow{a} cdot overrightarrow{c} = overrightarrow{b} cdot overrightarrow{c} = 0. - However, overrightarrow{a} and overrightarrow{b} are not necessarily equal just because they both are perpendicular to overrightarrow{c}. This shows that overrightarrow{a} cdot overrightarrow{c} = overrightarrow{b} cdot overrightarrow{c} does not imply overrightarrow{a} = overrightarrow{b}. 2. **Scenario 2: overrightarrow{a} = overrightarrow{b}** - Starting from overrightarrow{a} = overrightarrow{b}, we can subtract overrightarrow{b} from both sides to get overrightarrow{a} - overrightarrow{b} = overrightarrow{0}. - Taking the dot product of both sides with overrightarrow{c}, we get (overrightarrow{a} - overrightarrow{b}) cdot overrightarrow{c} = overrightarrow{0} cdot overrightarrow{c} = 0. - Expanding the left side gives overrightarrow{a} cdot overrightarrow{c} - overrightarrow{b} cdot overrightarrow{c} = 0, which simplifies to overrightarrow{a} cdot overrightarrow{c} = overrightarrow{b} cdot overrightarrow{c}. - This shows that if overrightarrow{a} = overrightarrow{b}, then overrightarrow{a} cdot overrightarrow{c} = overrightarrow{b} cdot overrightarrow{c} must also be true. From the analysis above, we can conclude that "overrightarrow{a}cdot overrightarrow{c}=overrightarrow{b}cdot overrightarrow{c}" is a necessary condition for "overrightarrow{a}=overrightarrow{b}" because if overrightarrow{a} = overrightarrow{b}, then their dot products with any other vector overrightarrow{c} will be equal. However, it is not a sufficient condition because overrightarrow{a} cdot overrightarrow{c} = overrightarrow{b} cdot overrightarrow{c} does not guarantee that overrightarrow{a} = overrightarrow{b}, as demonstrated in Scenario 1. Therefore, the correct answer is: boxed{text{B: Necessary but not sufficient condition}}

answer:**Analysis** This problem mainly examines the application of the binomial theorem and is considered a basic question. According to the binomial theorem, the expression in the problem can be rewritten as ({(lg 2+lg 5)}^{20}), which leads to the solution. **Solution** Given: ({{(lg 2)}^{20}}+{C}_{20}^{1}{{(lg 2)}^{19}}lg 5+ldots +{C}_{20}^{r-1}{{(lg 2)}^{21-r}}{{(lg 5)}^{r-1}}+ldots +{{(lg 5)}^{20}}=(lg 2+) (lg 5)^{20}={(lg 10)}^{20}=1). Therefore, the answer is (boxed{1}).