answer:Let's call Diana's initial biking speed "S" in miles per hour (mph). We know that she bikes at this speed for 2 hours before she gets tired. So, the distance she covers in those 2 hours is: Distance = Speed × Time Distance = S × 2 After biking for 2 hours, Diana gets tired and bikes at a speed of 1 mph for the remaining distance. We know that the total time it takes her to get home is 6 hours, so the time she spends biking at 1 mph is: Time at 1 mph = Total time - Time at initial speed Time at 1 mph = 6 hours - 2 hours Time at 1 mph = 4 hours The distance she covers at 1 mph is: Distance at 1 mph = Speed × Time Distance at 1 mph = 1 mph × 4 hours Distance at 1 mph = 4 miles Now, we know that the total distance to get home is 10 miles. So, the distance she covers at her initial speed is: Distance at initial speed = Total distance - Distance at 1 mph Distance at initial speed = 10 miles - 4 miles Distance at initial speed = 6 miles We already have the equation for the distance at her initial speed: Distance at initial speed = S × 2 Now we can substitute the distance at initial speed with the value we found: 6 miles = S × 2 To find her initial speed, we divide both sides of the equation by 2: S = 6 miles / 2 S = 3 mph Diana's initial biking speed is boxed{3} mph.

answer:Solve 1-x^{2} > 0 to get -1 < x < 1; Therefore, |x-2|=2-x; Thus, f(x)= frac {lg (1-x^{2})}{2-x+a}; Since f(x) is an odd function; Therefore, f(-x)=-f(x); That is, frac {lg (1-x^{2})}{2+x+a}=- frac {lg (1-x^{2})}{2-x+a}; Therefore, 2+x+a=-(2-x+a); Therefore, 2+a=-2-a; Therefore, a=-2. Hence, the answer is: boxed{-2}. By solving 1-x^{2} > 0, we get -1 < x < 1, and thus |x-2|=2-x, which leads to f(x)= frac {lg (1-x^{2})}{2-x+a}. Since f(x) is an odd function, we have f(-x)=-f(x), which allows us to conclude that 2+x+a=-(2-x+a), and thus we can find the value of a. This problem examines the definition of odd functions, the solution method for quadratic inequalities, and the method for dealing with absolute value functions: removing the absolute value sign.

answer:Let the sum be S. The series is similar to the original problem, but with a base of 3 in the denominator. We proceed as follows: begin{align*} S &= frac{1}{3^1} + frac{2}{3^2} + frac{3}{3^3} + frac{4}{3^4} + cdots frac{1}{3}S &= hspace{0.9 cm} frac{1}{3^2} + frac{2}{3^3} + frac{3}{3^4} + cdots end{align*} Subtracting the second equation from the first, we have: [frac{2}{3}S = frac{1}{3^1} + frac{1}{3^2} + frac{1}{3^3} + cdots ] This is a geometric series with first term a = frac{1}{3} and common ratio r = frac{1}{3}. The sum of an infinite geometric series is given by frac{a}{1-r} = frac{frac{1}{3}}{1 - frac{1}{3}} = frac{1}{2}. So, we have frac{2}{3}S = frac{1}{2}, thus S = frac{3}{4}. Conclusion with boxed answer: The sum of the series is boxed{frac{3}{4}}.

answer:In the original problem, the last number in each row followed the pattern of being 5 times the row number. With the new number of elements per row being 8, the pattern adjusts to the last number in each row being 8 times the row number. 1. **Find the last number in the 12th row**: [ text{Last number} = 8 times 12 = 96 ] 2. **Identify the fourth number in the 12th row**: Considering the sequence in each row starts with the previous row's last number plus one, the numbers in the 12th row start at ( (8 times 11) + 1 = 89 ). Thus, the fourth number is the third increment from 89: [ text{Fourth number} = 89 + 3 = 92 ] Conclusion with boxed answer: [ boxed{92} ]