answer:**Analysis** This question mainly examines the transformation rules of the graph of the function y=Asin (omega x+varphi), which is a basic problem. By applying the transformation rules of the function y=Asin (omega x+varphi) based on the given conditions, we can draw a conclusion. **Solution** Solution: For the function y=sin left(x+ frac{pi}{6}right) (x in mathbb{R}), reducing the x-coordinates of all points on the graph to half of their original values gives the graph of y=sin left(2x+ frac{pi}{6}right). Then, shifting the graph of these points to the left by frac{pi}{4} units, the equation of the resulting graph is y=sin left[2left(x+ frac{pi}{4}right)+ frac{pi}{6}right]=sin left(2x+ frac{pi}{6}+ frac{pi}{2}right)=sin left(2x+ frac{2pi}{3}right). Therefore, the correct option is boxed{C}.

answer:1. **Define the Problem:** Let (a_n) represent the expected number of rolls starting with an (n)-sided die. 2. **Base Case:** For (n = 1), we immediately stop, so: [ a_1 = 0 ] 3. **Expected Number of Rolls for (n > 1):** For (n > 1), the expected number of rolls (a_{n}) can be derived as: [ a_{n} = 1 + frac{1}{n} sum_{i=1}^{n} a_{i} ] This represents the one initial roll plus the average expected rolls from the subsequent configurable (k)-sided dice. 4. **Simplify Recurrence:** To simplify for (n = 2): [ a_2 = 1 + frac{1}{2} (a_1 + a_2) ] Since (a_1 = 0): [ a_2 = 1 + frac{1}{2} a_2 ] Solving for (a_2): [ a_2 - frac{1}{2} a_2 = 1 implies frac{1}{2} a_2 = 1 implies a_2 = 2 ] 5. **General Recurrence Relation:** For (n geq 3), rearrange the original equation: [ a_{n} = 1 + frac{1}{n} sum_{i=1}^{n}a_{i} ] Since (a_{n}) is included in the sum, we can express the sum as: [ sum_{i=1}^{n} a_{i} = a_{n} + sum_{i=1}^{n-1} a_{i} ] Substituting: [ a_{n} = 1 + frac{1}{n}(a_{n} + sum_{i=1}^{n-1} a_{i}) ] Simplify: [ n a_{n} = n + a_{n} + sum_{i=1}^{n-1} a_{i} ] [ n a_{n} - a_{n} = n + sum_{i=1}^{n-1} a_{i} ] [ (n - 1) a_{n} = n + sum_{i=1}^{n-1} a_{i} ] [ a_{n} = frac{n}{n - 1} + frac{1}{n - 1} sum_{i=1}^{n-1} a_{i} ] 6. **Find the Summation Explicitly:** Noticing a pattern when calculating individual cases, derive: [ a_{n} = 1 + sum_{i=1}^{n-1} frac{1}{i} ] 7. **Calculate Specifically for (n = 6):** [ a_{6} = 1 + sum_{i=1}^{5} frac{1}{i} ] The Harmonic Series Sum for (i = 1) to (5): [ sum_{i=1}^{5} frac{1}{i} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} ] 8. **Calculate the Exact Sum:** Convert to common denominator: [ 1 = frac{60}{60}, quad frac{1}{2} = frac{30}{60}, quad frac{1}{3} = frac{20}{60}, quad frac{1}{4} = frac{15}{60}, quad frac{1}{5} = frac{12}{60} ] Summing these fractions: [ frac{60 + 30 + 20 + 15 + 12}{60} = frac{137}{60} ] 9. **Add Initial 1:** [ a_{6} = 1 + frac{137}{60} = frac{60 + 137}{60} = frac{197}{60} ] 10. **Conclusion:** [ boxed{frac{197}{60}} ]

answer:Solution: ((1)) From (ρsin^2θ=2acos θ (a > 0)), we get (ρ^2sin^2θ=2aρcos θ (a > 0)), (therefore) The Cartesian coordinate equation of curve (C) is (y^2=2ax (a > 0)), The general equation of line (l) is (y=x-2); ((2)) Substituting the parametric equation of line (l) into the Cartesian coordinate equation of curve (C) (y^2=2ax), We get (t^2-2 sqrt{2}(4+a)t+8(4+a)=0), Let the parameters corresponding to points (A) and (B) be (t_1) and (t_2) respectively. Then we have (t_1+t_2=2 sqrt{2} (4+a)) and (t_1t_2=8(4+a)), Thus, (|AB|=|t_1-t_2|= sqrt{(t_1+t_2)^2-4t_1t_2}= sqrt{[2 sqrt{2}(4+a)]^2-4×8(4+a)}=2 sqrt{10}), Simplifying, we get (a^2+4a-5=0), Solving this, we find: (a=1) or (a=-5) (discard this solution) (therefore) The value of (a) is boxed{1}.

answer:The problem states that three friends receive one extra gold coin each. Therefore, if these three coins were not extra, the number of gold coins would be a multiple of 15. We can express the number of gold coins you have as 15k + 3. Given that you have fewer than 150 gold coins, the condition is 15k + 3 < 150, which simplifies to 15k < 147. Solving for k, we have: [ k < frac{147}{15} = 9.8 ] Since k must be an integer (as you can't have a fraction of a gold coin for equal distribution), the highest integer value for k that satisfies the inequality is k = 9. Thus, the largest number of gold coins that fits the problem's setup is: [ 15 times 9 + 3 = 135 + 3 = boxed{138} ]