answer:Let's denote the actual average speed as ( v ) miles per hour and the actual time taken to drive from home to the store as ( t ) hours. The distance from home to the store, therefore, is ( d = vt ). If the driver had increased the average speed by 18 miles per hour, the new speed would be ( v + 18 ) miles per hour. The problem states that by doing so, the time taken would have been reduced by 1/3, so the new time taken would be ( frac{2}{3}t ). The distance would remain the same, so we can set up the equation: [ d = vt = (v + 18) cdot frac{2}{3}t ] Now we can solve for ( v ): [ vt = frac{2}{3}vt + frac{2}{3} cdot 18t ] Divide both sides by ( t ) to get rid of ( t ): [ v = frac{2}{3}v + frac{2}{3} cdot 18 ] Now, let's solve for ( v ): [ v - frac{2}{3}v = frac{2}{3} cdot 18 ] [ frac{1}{3}v = frac{2}{3} cdot 18 ] Multiply both sides by 3 to solve for ( v ): [ v = 2 cdot 18 ] [ v = 36 ] So, the actual average speed when the driver drove from home to the store was boxed{36} miles per hour.

answer:**Analysis** This problem tests our ability to use Horner's method to compute the value of a polynomial at a given point, focusing on computational skills. It's a fundamental question. **Step-by-step Solution** 1. First, rewrite the given polynomial in the form suitable for Horner's method: f(x) = (((7x+3)x-5)x+11. 2. Substitute the given value x=2 into the rewritten polynomial. 3. Calculate intermediate values: - v_0 = 7 - v_1 = 7 times 2 + 3 = 17 4. Continue the calculations to find the final value of the polynomial: - v_2 = 17 times 2 - 5 = 34 - 5 = 29 - f(2) = 29 times 2 + 11 = 58 + 11 = boxed{69}

answer:1. The sum of the first n terms of the sequence {a_n} is S_n, and S_n = 1 - a_n. We can derive that S_{n-1} = 1 - a_{n-1}. By subtracting the two equations, we get 2a_n = a_{n-1}, which means sequence {a_n} is a geometric sequence with a common ratio of frac{1}{2}. Since S_1 = 1 - a_1, the first term is frac{1}{2}. Therefore, the general term formula for the sequence {a_n} is: a_n = frac{1}{2^n} 2. b_n = log_4 a_1 + log_4 a_2 + dots + log_4 a_n = log_4(a_1 a_2 dots a_n) =log_4 left(frac{1}{2}right)^{1+2+3+dots+n} = -frac{n(n+1)}{4}. The general term formula for the sequence {frac{1}{a_n} + frac{1}{b_n}} is: 2^n - frac{4}{n(n+1)} The sum of the first n terms for the sequence {frac{1}{a_n} + frac{1}{b_n}} is: T_n = (2 + 2^2 + 2^3 + dots + 2^n) - 4(1 - frac{1}{2} + frac{1}{2} - frac{1}{3} + dots + frac{1}{n} - frac{1}{n+1}) =frac{2(1-2^n)}{1-2} - 4(1 - frac{1}{n+1}) =2^{n+1} + frac{4}{n+1} - 6. The final answers are: 1. boxed{a_n = frac{1}{2^n}} 2. boxed{T_n = 2^{n+1} + frac{4}{n+1} - 6}

answer:(Ⅰ) Since the equation x^2-4x+|a-3|=0 has real roots, we have Delta=16-4|a-3| geq 0, which implies |a-3| leq 4, thus -4 leq a-3 leq 4, so -1 leq a leq 7. Therefore, the set of values for the real number a is A={a|-1 leq a leq 7}. (Ⅱ) Since for all a in A, the inequality t^2-2at+12<0 always holds, let f(a)=-2at+t^2+12, then f(a)<0 always holds. Thus, f(-1)<0 and f(7)<0, which gives 2t+t^2+12<0 (1), and -14t+t^2+12<0 (2). Solving (1) gives t in emptyset, and solving (2) gives 3<t<4. In conclusion, the range of values for t is boxed{(3, 4)}.