answer:The factored form of the given quadratic is: -5 left(x - frac{29}{sqrt{3}}right) left(x - frac{10}{sqrt{3}}right) Alternatively, to make the factors more readable with rational coefficients, we can multiply both sides by 3 to remove the square root from the denominators: -15 (3x - 29) (x - frac{10}{3}) This expression represents the completely factored form of the quadratic.

answer:The cross product of Vector A and Vector B is: Vector C = left( begin{array}{c} (1)(-3) - (1)(-4) -(6)(-3) - (-4)(1) 6(-1) - (-4)(1) end{array} right) = left( begin{array}{c} -3 + 4 18 + 4 -6 + 4 end{array} right) = left( begin{array}{c} 1 22 -2 end{array} right) The cross product of Vector A and Vector B is left( begin{array}{c} 1 22 -2 end{array} right).

answer:The divergence of the vector field vec{F} is found by computing the sum of the partial derivatives of its components with respect to their respective variables. Applying the divergence theorem: nabla cdot vec{F} = frac{partial f}{partial x} + frac{partial g}{partial y} + frac{partial h}{partial z} Given f(x, y, z) = x, g(x, y, z) = sin(y), and h(x, y, z) = sqrt[3]{frac{x}{z}} = left(frac{x}{z}right)^{frac{1}{3}}, we have: frac{partial f}{partial x} = 1 frac{partial g}{partial y} = cos(y) frac{partial h}{partial z} = -frac{1}{3z^2}left(frac{x}{z}right)^{frac{1}{3}-1} = -frac{x}{3z^2sqrt[3]{frac{x}{z}}^2} Adding these derivatives together: nabla cdot vec{F} = 1 + cos(y) - frac{x}{3z^2sqrt[3]{frac{x}{z}}^2} Therefore, the divergence of the vector field vec{F} is: nabla cdot vec{F} = 1 + cos(y) - frac{x}{3z^2sqrt[3]{frac{x}{z}}^2}

answer:Using the formula for cost of goods sold (COGS): COGS = Beginning inventory + Net purchases - Ending inventory Where Net purchases = Purchases - Purchase returns - Purchase discounts So, Net purchases = 8,500 - 1,100 - 240 COGS = 34,000 + (8,500 - 1,100 - 240) - 28,500 COGS = 34,000 + 7,160 - 28,500 COGS = 12,660 Thus, the cost of goods sold is 12,660.