answer:Solution: For the hyperbola ( dfrac {x^{2}}{a^{2}}-y^{2}=1(a > 0)), we have (b=1), (c= sqrt {a^{2}+1}), Given that the eccentricity (e= dfrac {c}{a}= dfrac { sqrt {1+a^{2}}}{a}= sqrt {2}), Solving the equation, we get (a=1), Thus, the equation of the hyperbola becomes (x^{2}-y^{2}=1), Hence, the equation of the asymptote is (y=±x). Therefore, the correct choice is: boxed{B}. By finding the values of (b) and (c), and using the formula for eccentricity, we can calculate (a=1), and then obtain the equation of the asymptote. This question tests the method of finding the equation of the asymptote of a hyperbola, emphasizing the use of the eccentricity formula and calculation skills, and is considered a basic question.

answer:To convert the number 0.000000007 into scientific notation, we follow the steps below: 1. Identify the significant figures in the number, which in this case is 7. 2. Determine how many places the decimal point needs to be moved to get a number between 1 and 10. For 0.000000007, the decimal point moves 9 places to the right. 3. Write the number as a product of the significant figure and 10 raised to the power of the number of places the decimal point was moved. Thus, 0.000000007 becomes 7 times 10^{-9}. Therefore, the correct representation of 0.000000007 in scientific notation is boxed{7 times 10^{-9}}, which corresponds to option D.

answer:Let's denote the length of the rectangle as L meters and the breadth as B meters. According to the problem, the difference between the length and breadth is 23 meters, so we can write: L - B = 23 (Equation 1) We are also given that the area of the rectangle is 2030 m². The area of a rectangle is given by the product of its length and breadth, so we have: L * B = 2030 (Equation 2) We need to find the values of L and B to calculate the perimeter. The perimeter of a rectangle is given by the formula: Perimeter = 2 * (L + B) To find L and B, we can solve the two equations simultaneously. From Equation 1, we can express L in terms of B: L = B + 23 Now, we can substitute this expression for L into Equation 2: (B + 23) * B = 2030 Expanding the equation, we get: B^2 + 23B - 2030 = 0 This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve for B using the quadratic formula: B = [-b ± sqrt(b^2 - 4ac)] / (2a) In this case, a = 1, b = 23, and c = -2030. Plugging these values into the quadratic formula, we get: B = [-23 ± sqrt(23^2 - 4(1)(-2030))] / (2(1)) B = [-23 ± sqrt(529 + 8120)] / 2 B = [-23 ± sqrt(8649)] / 2 B = [-23 ± 93] / 2 We have two possible solutions for B: B1 = (-23 + 93) / 2 = 70 / 2 = 35 B2 = (-23 - 93) / 2 = -116 / 2 = -58 Since the breadth cannot be negative, we discard B2 and take B = 35 meters. Now we can find L using the expression L = B + 23: L = 35 + 23 = 58 meters Finally, we can calculate the perimeter: Perimeter = 2 * (L + B) = 2 * (58 + 35) = 2 * 93 = 186 meters Therefore, the perimeter of the rectangle is boxed{186} meters.

answer:To find the number of factors of 2 in the product of numbers from 2011 to 4020, we must count how many times the factor 2 appears in the prime factorization of each number in this range. 1. **Range Identification**: The range includes integers from 2011 to 4020. The total number of integers in this range is: [ 4020 - 2011 + 1 = 2010 ] Therefore, there are 2010 integers in this range. 2. **Factors of 2 Calculation**: For each number in the range, we need to determine how many times 2 (i.e., powers of 2) appears as a factor. This can be done by examining the greatest power of 2 that divides any given number. Specifically, we need to sum the divisors counted by each power of 2. We employ the formula for counting factors of a prime p in n!: [ sum_{k=1}^{infty} leftlfloor frac{n}{2^k} rightrfloor ] In our case, we adjust for the specific range. 3. **Power Increment**: Compute the number of multiples for each power of 2 in the segment from 2011 to 4020. [ sum_{k=1}^{infty} left(leftlfloor frac{4020}{2^k} rightrfloor - leftlfloor frac{2010}{2^k} rightrfloor right) ] 4. **Detailed Calculation**: - For each k from 1 to the greatest integer k such that 2^k leq 4020: begin{align*} leftlfloor frac{4020}{2} rightrfloor - leftlfloor frac{2010}{2} rightrfloor &= 2010 - 1005 = 1005 leftlfloor frac{4020}{4} rightrfloor - leftlfloor frac{2010}{4} rightrfloor &= 1005 - 502 = 503 leftlfloor frac{4020}{8} rightrfloor - leftlfloor frac{2010}{8} rightrfloor &= 502 - 251 = 251 leftlfloor frac{4020}{16} rightrfloor - leftlfloor frac{2010}{16} rightrfloor &= 251 - 125 = 126 leftlfloor frac{4020}{32} rightrfloor - leftlfloor frac{2010}{32} rightrfloor &= 125 - 62 = 63 leftlfloor frac{4020}{64} rightrfloor - leftlfloor frac{2010}{64} rightrfloor &= 62 - 31 = 31 leftlfloor frac{4020}{128} rightrfloor - leftlfloor frac{2010}{128} rightrfloor &= 31 - 15 = 16 leftlfloor frac{4020}{256} rightrfloor - leftlfloor frac{2010}{256} rightrfloor &= 15 - 7 = 8 leftlfloor frac{4020}{512} rightrfloor - leftlfloor frac{2010}{512} rightrfloor &= 7 - 3 = 4 leftlfloor frac{4020}{1024} rightrfloor - leftlfloor frac{2010}{1024} rightrfloor &= 3 - 1 = 2 leftlfloor frac{4020}{2048} rightrfloor - leftlfloor frac{2010}{2048} rightrfloor &= 1 - 0 = 1 end{align*} Summing these results: [ 1005 + 503 + 251 + 126 + 63 + 31 + 16 + 8 + 4 + 2 + 1 = 2010 ] Conclusion: [ boxed{2010} ]