answer:Firstly, recognize that 0.overline{36} = 36 cdot 0.overline{01} and 0.overline{12} = 12 cdot 0.overline{01}. Thus, we can write the expression as: [ frac{0.overline{36}}{0.overline{12}} = frac{36 cdot 0.overline{01}}{12 cdot 0.overline{01}} ] This further simplifies to: [ frac{36}{12} cdot frac{0.overline{01}}{0.overline{01}} = frac{36}{12} ] Reducing frac{36}{12} to its simplest form, we divide both the numerator and the denominator by their greatest common divisor, which is 12: [ frac{36}{12} = frac{3}{1} = 3 ] Thus, the solution to frac{0.overline{36}}{0.overline{12}} is boxed{3}.

answer:The sum of the elements in the set is 6. Let x = p + q + r + s, so then t + u + v + w = 6 - x. The expression simplifies as follows: [ (p+q+r+s)^{2} + (t+u+v+w)^{2} = x^{2} + (6-x)^{2} = 2x^{2} - 12x + 36 ] Complete the square: [ 2x^{2} - 12x + 36 = 2(x-3)^{2} + 18 geq 18 ] The value of 18 is attained if and only if x = 3. To check if it's possible to achieve x = 3 with distinct elements, consider: [ p+q+r+s = 3 ] Choose p = 7, q = 5, r = -4, s = -5 to satisfy p+q+r+s = 3. Then the rest of the elements are t = -8, u = -6, v = -1, w = 10 and indeed t + u + v + w = 6 - 3 = 3. Thus: [ (p+q+r+s)^2 + (t+u+v+w)^2 = 3^2 + 3^2 = 9 + 9 = boxed{18}. ] Conclusion: The minimum possible value of (p+q+r+s)^{2} + (t+u+v+w)^{2} is boxed{18}.

answer:**Analysis** This problem mainly examines the general formula and properties of a geometric sequence. First, given that the sequence left{a_nright} consists of positive real numbers, we can deduce that a_3 > 0. Then, by using the properties of a geometric sequence, we can find that a_3= sqrt{a_1a_5}, from which we can derive the answer. **Solution** Since the sequence left{a_nright} consists of positive real numbers, it follows that a_3 > 0. Given a_1=1 and a_5=9, we have a_3= sqrt{a_1a_5}= sqrt{1×9}=3. Therefore, the answer is boxed{3}.

answer:Prime factorize 420: [ 420 = 2 times 210 ] [ 210 = 2 times 105 ] [ 105 = 3 times 35 ] [ 35 = 5 times 7 ] Thus, 420 can be factorized as: [ 420 = 2 times 3 times 5 times 7 ] The distinct prime factors of 420 are 2, 3, 5, and 7. The sum of these distinct prime factors is: [ 2 + 3 + 5 + 7 = boxed{17} ]