answer:Solution: (1) From the given conditions, we have begin{cases} a^{2}-b^{2}=3 frac{1}{a^{2}} + frac{3}{4b^{2}}=1 end{cases}, which simplifies to frac{1}{b^{2}+3} + frac{3}{4b^{2}}=1. Solving this, we get b^{2}=1, a^{2}=4, therefore The equation of the ellipse E is frac{x^{2}}{4} + y^{2} = 1; (2) Method 1: If k exists, let the equation of line l be y=k(x+ sqrt{3}). Substituting into frac{x^{2}}{4} + y^{2} = 1, we get (4k^{2}+1)x^{2} + 8sqrt{3}k^{2}x + 12k^{2} - 4 = 0, Let M(x_{1},y_{1}), N(x_{2},y_{2}), then x_{1}+x_{2}=-frac{8sqrt{3}k}{4k^{2}+1}, x_{1}x_{2}=frac{4(3k^{2}-1)}{4k^{2}+1}. therefore The distance d from F_{2} to line MN is d= frac{|2sqrt{3}k|}{sqrt{k^{2}+1}}, MN= sqrt{1+k^{2}} sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=4sqrt{1+k^{2}}sqrt{frac{1+k^{2}}{(4k^{2}+1)^{2}}}, therefore S_{triangle F_{2}MN}= frac{1}{2}|MN|cdot d=4sqrt{frac{3k^{2}(k^{2}+1)}{(4k^{2}+1)^{2}}}leqslant 4sqrt{frac{[3k^{2}+(k^{2}+1)]^{2}}{4(4k^{2}+1)^{2}}}=2, When and only when 3k^{2}=k^{2}+1, i.e., k= frac{sqrt{2}}{2}, the maximum value is achieved, At this time, the equation of line l is x-sqrt{2}y+sqrt{3}=0 or x+sqrt{2}y+sqrt{3}=0. If k does not exist, i.e., lperp x axis, then S_{triangle F_{2}MN}= frac{1}{2}|MN||F_{1}F_{2}|=sqrt{3} < 2 (discard), In summary: The equation of line l is x-sqrt{2}y+sqrt{3}=0 or x+sqrt{2}y+sqrt{3}=0. Method 2: Let the equation of line l be x=my-sqrt{3}. Substituting into frac{x^{2}}{4}+y^{2}=1, we get (m^{2}+4)y^{2}-2sqrt{3}my-1=0, Let M(x_{1},y_{1}), N(x_{2},y_{2}), then y_{1}+y_{2}=frac{2sqrt{3}m}{m^{2}+4}, y_{1}y_{2}=frac{-1}{m^{2}+4}. therefore S_{triangle F_{2}MN}= frac{1}{2}times|F_{1}F_{2}|times(|y_{1}|+|y_{2}|)= frac{1}{2}times|F_{1}F_{2}|times(|y_{1}-y_{2}|)= sqrt{3}timessqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}} = frac{4sqrt{3}timessqrt{m^{2}+1}}{m^{2}+4}= frac{4sqrt{3}}{sqrt{m^{2}+1}+frac{3}{sqrt{m^{2}+1}}}leqslant frac{4sqrt{3}}{2sqrt{3}}=2, When and only when sqrt{m^{2}+1}=frac{3}{sqrt{m^{2}+1}}, i.e., m^{2}=2, the equality holds, therefore When the area of triangle F_{2}MN reaches its maximum value, the equation of line l is x-sqrt{2}y+sqrt{3}=0 or x+sqrt{2}y+sqrt{3}=0. Thus, the final answers are boxed{frac{x^{2}}{4} + y^{2} = 1} for the equation of the ellipse, and boxed{x-sqrt{2}y+sqrt{3}=0} or boxed{x+sqrt{2}y+sqrt{3}=0} for the equation of line l when the area of triangle F_{2}MN reaches its maximum value.

answer:To evaluate the product [ (a-10) cdot (a-9) cdot dotsm cdot (a-1) cdot a, ] where a=2, we observe that one of the terms in the product is a-2. Given that a=2, we have: [ a-2 = 2-2 = 0. ] Therefore, the product can be rewritten as: [ (a -10) dotsm (a-3) cdot (a-2) cdot (a-1) cdot a = (a-10) dotsm (a-3) cdot 0 cdot (a-1) cdot a. ] Since multiplying any real number by zero results in zero, the entire product simplifies to: [ (a-10) dotsm (a-3) cdot 0 cdot (a-1) cdot a = 0. ] Thus, the evaluated product is boxed{0}.

answer:Given the conditions: [ 3a = b^3 ] [ 5a = c^2 ] and the assumption that ( a ) is divisible by ( d^6 ) where ( d = 1 ). # Part 1: Prove that ( a ) is divisible by ( 3 ) and ( 5 ). 1. From the equation ( 3a = b^3 ), we can see that ( a = frac{b^3}{3} ). For ( a ) to be an integer, ( b^3 ) must be divisible by ( 3 ). Since ( b ) is a positive integer, ( b ) itself must be divisible by ( 3 ). Let ( b = 3k ) for some integer ( k ). Then: [ b^3 = (3k)^3 = 27k^3 ] Substituting back, we get: [ a = frac{27k^3}{3} = 9k^3 ] Thus, ( a ) is divisible by ( 3 ). 2. From the equation ( 5a = c^2 ), we can see that ( a = frac{c^2}{5} ). For ( a ) to be an integer, ( c^2 ) must be divisible by ( 5 ). Since ( c ) is a positive integer, ( c ) itself must be divisible by ( 5 ). Let ( c = 5m ) for some integer ( m ). Then: [ c^2 = (5m)^2 = 25m^2 ] Substituting back, we get: [ a = frac{25m^2}{5} = 5m^2 ] Thus, ( a ) is divisible by ( 5 ). Since ( a ) is divisible by both ( 3 ) and ( 5 ), we conclude that ( a ) is divisible by ( 15 ). # Part 2: Prove that the prime factors of ( a ) are limited to ( 3 ) and ( 5 ). 1. From the previous part, we have: [ a = 9k^3 quad text{and} quad a = 5m^2 ] Since ( a ) must satisfy both conditions simultaneously, we equate the two expressions: [ 9k^3 = 5m^2 ] This implies that ( a ) must be a common multiple of ( 9 ) and ( 5 ). The least common multiple of ( 9 ) and ( 5 ) is ( 45 ). Therefore, ( a ) must be of the form: [ a = 45n ] for some integer ( n ). 2. Since ( a ) is of the form ( 45n ), the prime factors of ( a ) are ( 3 ) and ( 5 ). # Part 3: Find ( a ). 1. From the equation ( 9k^3 = 5m^2 ), we need to find integers ( k ) and ( m ) such that this equation holds. We rewrite it as: [ 9k^3 = 5m^2 ] This implies that ( k ) and ( m ) must be chosen such that both sides are equal. Since ( 9k^3 ) and ( 5m^2 ) must be equal, we need to find the smallest ( k ) and ( m ) that satisfy this condition. 2. By trial and error or by solving the equation, we find that the smallest values that satisfy this condition are ( k = 5 ) and ( m = 3 ). Substituting these values back, we get: [ a = 9k^3 = 9 cdot 5^3 = 9 cdot 125 = 1125 ] [ a = 5m^2 = 5 cdot 3^2 = 5 cdot 9 = 45 ] Since both expressions must be equal, we find that the smallest value of ( a ) that satisfies both conditions is ( 1125 ). The final answer is ( boxed{1125} ).

answer:Let the slant height (hypotenuse) be denoted as l = 17 cm, the height from the vertex to the center of the base (one of the legs) be h = 15 cm, and the radius of the base be r cm. We again use the Pythagorean theorem: [ l^2 = r^2 + h^2 ] [ 17^2 = r^2 + 15^2 ] [ 289 = r^2 + 225 ] [ r^2 = 289 - 225 ] [ r^2 = 64 ] [ r = 8 text{ cm} ] Now, we calculate the volume V of the cone using: [ V = frac{1}{3} pi r^2 h ] [ V = frac{1}{3} pi (8^2)(15) ] [ V = frac{1}{3} pi (64)(15) ] [ V = frac{960}{3} pi ] [ V = 320pi ] So, the volume of the cone is boxed{320pi} cubic centimeters.