answer:The region of a sphere centered at the origin that satisfies the condition x^2+y^2+z^2 >2z is a cone-shaped region that extends infinitely in the positive z-direction. This region can be visualized by imagining a sphere centered at the origin, and then slicing the sphere with a plane that is parallel to the xy-plane and passes through the point (0, 0, 2). The resulting region is the cone-shaped region that satisfies the condition x^2+y^2+z^2 >2z.

answer:The divergence of a vector field vec{F}(x, y, z) = f(x, y, z)uvec{i} + g(x, y, z)uvec{j} + h(x, y, z)uvec{k} is given by nabla cdot vec{F} = frac{partial f}{partial x} + frac{partial g}{partial y} + frac{partial h}{partial z}. For the given vector field, we have: - frac{partial f}{partial x} = -frac{1}{x^2}, - frac{partial g}{partial y} = -frac{1}{y^2 z^5} (since g is not a function of x and z), - frac{partial h}{partial z} is zero because h is not a function of z. Hence, the divergence of the vector field vec{F} is: nabla cdot vec{F} = -frac{1}{x^2} - frac{1}{y^2 z^5} This result confirms the original answer, which is already accurate.

answer:The addition of the two matrices results in: left( begin{array}{ccc} frac{1}{10} + frac{24}{5} & frac{26}{5} - frac{31}{10} & -frac{15}{2} - frac{39}{5} frac{17}{5} - frac{47}{5} & frac{47}{5} - frac{61}{10} & frac{9}{10} - frac{13}{2} end{array} right) = left( begin{array}{ccc} frac{1}{10} + frac{48}{10} & frac{52}{10} - frac{31}{10} & -frac{75}{10} - frac{78}{10} -6 & frac{94}{10} - frac{61}{10} & frac{9}{10} - frac{65}{10} end{array} right) = left( begin{array}{ccc} frac{49}{10} & frac{21}{10} & -frac{153}{10} -6 & frac{33}{10} & -frac{28}{5} end{array} right)

answer:Using the same formula as in the original answer, we have: {eq}begin{align*} frac{partial{f(x,y)}}{partial{x}} &= 3x^2 - 3y & Rightarrow frac{partial{f(1,2)}}{partial{x}} &= 0 frac{partial{f(x,y)}}{partial{y}} &= 3y^2 - 3x & Rightarrow frac{partial{f(1,2)}}{partial{y}} &= 9 end{align*} {/eq} Thus, the equation of the tangent plane is {eq}begin{align*} z - 3 &= 0(x-1) + 9(y-2) z &= 9y - 15 end{align*} {/eq}