answer:Let's denote the side length of the first cube as ( s_1 ) and the side length of the second cube as ( s_2 ). The volume of a cube is given by the formula ( V = s^3 ), where ( s ) is the side length of the cube. Since both cubes are made of the same metal, their densities are the same. Therefore, the weight of the cube is directly proportional to its volume. This means that the ratio of their weights is equal to the ratio of their volumes. Given that the first cube weighs 6 pounds and the second cube weighs 48 pounds, the ratio of their weights is ( frac{48}{6} = 8 ). Now, let's express the volumes of the cubes in terms of their side lengths: ( V_1 = s_1^3 ) for the first cube, ( V_2 = s_2^3 ) for the second cube. The ratio of their volumes is: ( frac{V_2}{V_1} = frac{s_2^3}{s_1^3} ). Since the ratio of their weights is equal to the ratio of their volumes, we have: ( 8 = frac{s_2^3}{s_1^3} ). To find the ratio of the side lengths, we take the cube root of both sides: ( sqrt[3]{8} = frac{s_2}{s_1} ). ( 2 = frac{s_2}{s_1} ). Therefore, the ratio of the side length of the second cube to the side length of the first cube is boxed{2:1} .

answer:Given f(x)= frac {x^{2}}{2}-kln x, the domain of the function is (0,+infty). Differentiating f(x) with respect to x gives f′(x)=x- frac {k}{x}= frac {x^{2}-k}{x}. When k > 0, solving f′(x)=0 yields x= sqrt {k} or x=- sqrt {k} (the latter is discarded), For x > sqrt {k}, we have f′(x) > 0, For 0 < x < sqrt {k}, we set f′(x) < 0, Therefore, the decreasing interval of f(x) is (0, sqrt {k}), and the increasing interval is ( sqrt {k},+infty); Hence, the answer is: boxed{( sqrt {k},+infty)}. By deriving the domain and f′(x) from the analytical expression, and then discussing according to the value of k, we can determine the intervals of increase and decrease of the function based on the relationship between the derivative and the monotonicity of the function; finding the derivative of the function, solving the inequality related to the derivative function, and determining the monotonic intervals of the function accordingly.

answer:Solution: S_{10}= frac {10(a_1+a_{10})}{2}=5(a_2+a_9)=5times(3+17)=100. Hence, the answer is: boxed{100}. This can be obtained using the properties of the general term formula of an arithmetic sequence and its sum formula. This question tests the understanding of the general term formula and sum formula of an arithmetic sequence, as well as reasoning and computational skills. It is of moderate difficulty.

answer:Solution: (()Ⅰ()) From (rhocos left(theta+ dfrac{pi}{4}right)= sqrt{2} ), we get (rho(dfrac{ sqrt{2}}{2} cos theta−dfrac{ sqrt{2}}{2} sin theta)=sqrt{2} ), which means the Cartesian equation of curve (C_{2}) is (x-y-2=0), Since the parametric equation of curve (C_{1}) is (begin{cases}x=3cos alpha y=sin alphaend{cases} (alpha) is the parameter()), Therefore, according to the problem, we have (|OP|=sqrt{9{cos }^{2}alpha+{sin }^{2}alpha}= sqrt{8{cos}^{2}alpha+1} ), Thus, the maximum value of the distance (|OP|) from the moving point (P) on curve (C_{1}) to the origin (O) is (boxed{|OP|_{max}=3}); (()Ⅱ()) From (()Ⅰ()), we know the intersection point (E) of the line (x-y-2=0) with the (x)-axis has coordinates ((2,0)), The parametric equation of curve (C_{2}) is: (begin{cases}x= dfrac{ sqrt{2}}{2}t+2 y= dfrac{ sqrt{2}}{2}tend{cases} (t) is the parameter()), The Cartesian equation of curve (C_{1}) is (dfrac{{x}^{2}}{9}+{y}^{2}=1 ), Solving them together, we get (5t^{2}+2sqrt{2} t−5=0), Since (|EA|+|EB|=|t_{1}|+|t_{2}|), (t_{1}+t_{2}=-dfrac{2 sqrt{2}}{5} ), (t_{1}t_{2}=-1), Therefore, (|EA|+|EB|=|t_{1}−t_{2}|=sqrt{{left({t}_{1}+{t}_{2}right)}^{2}-4{t}_{1}{t}_{2}}= boxed{dfrac{6 sqrt{3}}{5}} ).