answer:(1) From the given equation a=bcos C+ frac{sqrt{3}}{3}csin B and applying the Law of Sines, we have: sin A=sin Bcos C+ frac{sqrt{3}}{3}sin Csin B quad text{... (1 point)} Therefore, we get: sin(B+C) = sin Bcos C + frac{sqrt{3}}{3}sin Csin B quad text{... (2 points)} This simplifies to: sin Bcos C + cos Bsin C = sin Bcos C + frac{sqrt{3}}{3}sin Csin B quad text{... (3 points)} Which further gives us: cos Bsin C = frac{sqrt{3}}{3}sin Csin B quad text{... (4 points)} Since C is an angle of a triangle, we know sin C neq 0, thus we derive: tan B = sqrt{3} quad text{... (5 points)} As the angle B in (0, pi), we conclude: B = frac{pi}{3} quad text{... (6 points)} Therefore, the value of angle boxed{B = frac{pi}{3}}. (2) Given that the area of triangle ABC is frac{3sqrt{3}}{2}, we have: frac{1}{2}acsin B = frac{3sqrt{3}}{2} Simplifying, we get: frac{1}{2}accdot frac{sqrt{3}}{2} = frac{3sqrt{3}}{2} This leads to ac = 6. By the Law of Cosines, we find b^2: b^2 = a^2 + c^2 - 2accos B = (a+c)^2 - 3ac = 36 - 18 = 18 Hence, we obtain: b = 3sqrt{2} quad text{... (7 points)} So the length of side b is boxed{b = 3sqrt{2}}.

answer:The problem involves simplifying frac{sqrt{3}cdot sqrt{5} cdot sqrt{6}}{sqrt{7} cdot sqrt{9} cdot sqrt{8}}. We can simplify sqrt{9} to 3 and sqrt{8} to 2sqrt{2}, and sqrt{6} to sqrt{2}cdotsqrt{3}, which allows some terms to be simplified: 1. Combine and simplify terms: [ frac{sqrt{3}cdot sqrt{5} cdot sqrt{2}cdotsqrt{3}}{sqrt{7} cdot 3 cdot 2sqrt{2}} = frac{sqrt{3}^2 cdot sqrt{5} cdot sqrt{2}}{sqrt{7} cdot 6 cdot sqrt{2}} = frac{3 cdot sqrt{5}}{6 cdot sqrt{7}} = frac{sqrt{5}}{2 cdot sqrt{7}} ] 2. Rationalize the denominator: [ frac{sqrt{5}}{2sqrt{7}} cdot frac{sqrt{7}}{sqrt{7}} = frac{sqrt{35}}{14} ] 3. Final simplification: [ boxed{frac{sqrt{35}}{14}} ]

answer:1. **Understanding the problem**: - We start with 30 units written on a board. - Each minute, two arbitrary numbers are erased and replaced with their sum. - Karlson eats a number of candies equal to the product of the two erased numbers. - The goal is to find the maximum number of candies Karlson could eat in 30 minutes. 2. **Visualization**: - Consider each number on the board as a point in a plane. - When two numbers, (x) and (y), are combined to form (x + y), an edge connecting (x) and (y) is effectively created with weight (xy). - This process continues until there’s only one number left on the board after 30 operations. 3. **Formulation of the problem**: - Every time (x) and (y) are combined, Karlson eats (xy) candies. 4. **Calculation of the total candies eaten**: - Each number starts as "1". - After one minute, two of these “1”s are erased, and Karlson eats (1cdot1) = 1 candy, and the board now reads "... 1 + 1 = 2". - The next minute, combining another two "1"s repeats the process, but we should strategically aim for maximizing the product terms. - Think of the process forming a complete graph incrementally: - Initially, we have 30 points. - Each minute we connect two points with an edge labeled by their product (both are "1"). - After (30 - 1 = 29) minutes, we will have connected creating a new cluster. 5. **Using Combinatorics**: - The total number of edges (products eaten), over all combinations, can be found: [ text{Total Product Sum} = sum_{text{all connections}} (product) = frac{30 cdot (30 - 1)}{2} ] - This represents a maximized strategy each step, targeting the highest number of total possible connections. - Given (n cdot (n-1)/2): Quintuples to: [ frac{30 cdot 29}{2} = 435 ] Thus, Karlson will eat 435 candies in 30 minutes. # Conclusion: [ boxed{435} ]

answer:To determine the order of infinitesimals of the given expressions as ( x rightarrow 0 ), we compare each expression to ( x ) by computing the limit of their ratio to ( x ). 1. Analyze ( 10x ): [ lim_{x rightarrow 0} frac{10x}{x} = lim_{x rightarrow 0} 10 = 10 ] Since the limit is a finite non-zero constant, ( 10x ) is an infinitesimal of the same order as ( x ). 2. Analyze ( x^3 ): [ lim_{x rightarrow 0} frac{x^3}{x} = lim_{x rightarrow 0} x^2 = 0 ] Since the limit is zero, ( x^3 ) is an infinitesimal of a higher order than ( x ). 3. Analyze ( sqrt{3x} ): [ lim_{x rightarrow 0} frac{sqrt{3x}}{x} = lim_{x rightarrow 0} sqrt{frac{3}{x}} = +infty ] Since the limit tends to infinity, ( sqrt{3x} ) is an infinitesimal of a lower order than ( x ). 4. Analyze ( tanleft(frac{x}{5}right) ): [ lim_{x rightarrow 0} frac{tanleft(frac{x}{5}right)}{x} = lim_{x rightarrow 0} frac{sinleft(frac{x}{5}right)}{x cosleft(frac{x}{5}right)} = left( lim_{x rightarrow 0} frac{sinleft(frac{x}{5}right)}{x} right) cdot left( lim_{x rightarrow 0} frac{1}{cosleft(frac{x}{5}right)} right) ] Using the small angle approximation, ( sinleft(frac{x}{5}right) approx frac{x}{5} ): [ lim_{x rightarrow 0} frac{sinleft(frac{x}{5}right)}{x} = frac{1}{5} quad text{and} quad lim_{x rightarrow 0} frac{1}{cosleft(frac{x}{5}right)} = 1 ] Therefore, [ lim_{x rightarrow 0} frac{tanleft(frac{x}{5}right)}{x} = frac{1}{5} cdot 1 = frac{1}{5} ] Since the limit is a finite non-zero constant, ( tanleft(frac{x}{5}right) ) is an infinitesimal of the same order as ( x ). 5. Analyze ( lg(1+x) ): [ lim_{x rightarrow 0} frac{log(1+x)}{x} = lim_{x rightarrow 0} log(1+x)^{1/x} ] Using the approximation ( log(1+x) approx x ) for small ( x ): [ lim_{x rightarrow 0} frac{log(1+x)}{x} = 1 ] Therefore, ( log(1+x) ) is an infinitesimal of the same order as ( x ). **Conclusion**: - Same order as ( x ): ( 10x ), ( tanleft(frac{x}{5}right) ), ( log(1+x) ) - Higher order than ( x ): ( x^3 ) - Lower order than ( x ): ( sqrt{3x} ) [ boxed{text{The functions are classified as follows:}} ]