answer:The curl of the vector field mathbf{F} is given by the determinant of the Jacobian matrix of mathbf{F} with the standard unit vectors as the columns: nabla times mathbf{F} = left| begin{array}{ccc} mathbf{i} & mathbf{j} & mathbf{k} frac{partial}{partial x} & frac{partial}{partial y} & frac{partial}{partial z} f & g & h end{array} right| Computing the curl, we get: nabla times mathbf{F} = left{ frac{partial h}{partial y} - frac{partial g}{partial z}, -frac{partial f}{partial z} + frac{partial h}{partial x}, frac{partial g}{partial x} - frac{partial f}{partial y} right} Now, let's compute each component: 1. frac{partial h}{partial y} - frac{partial g}{partial z} = frac{1}{y} - left(-frac{3z}{2(y-z^2)^{5/2}}right) = frac{1}{y} + frac{3z}{2(y-z^2)^{5/2}} 2. -frac{partial f}{partial z} + frac{partial h}{partial x} = -frac{2}{z} + 0 = -frac{2}{z} 3. frac{partial g}{partial x} - frac{partial f}{partial y} = 0 - 0 = 0 Therefore, the curl of the vector field is: nabla times mathbf{F} = left{ frac{1}{y} + frac{3z}{2(y-z^2)^{5/2}}, -frac{2}{z}, 0 right}

answer:The correct answer is b. flashpoint. Flashpoint is the lowest temperature at which a volatile liquid solvent forms an ignitable solvent vapor-air mixture. It is a measure of the flammability of a solvent and has no direct relation to the effectiveness of a microbial agent. Acidity (a), temperature (c), and incubation time (d) are all factors that can affect the growth and survival of microorganisms. Acidity affects the pH of the environment, which can be harmful to some microorganisms. Temperature affects the rate of microbial growth and metabolism, and incubation time is the length of time that a microbial culture is allowed to grow.

answer:To find the equation of the plane, we first calculate the vectors formed by the given points: vec{v_1} = {P_2} - {P_1} = (2, -2, -4) - (2, -4, 3) = (0, 2, -7) vec{v_2} = {P_3} - {P_1} = (-4, 2, -3) - (2, -4, 3) = (-6, 6, -6) Next, we find the cross product of these two vectors to get the normal vector of the plane: vec{n} = vec{v_1} times vec{v_2} = (0, 2, -7) times (-6, 6, -6) Computing the cross product gives: vec{n} = (42 - 0) hat{i} - (42 - 42) hat{j} + (0 - 12) hat{k} = 42 hat{i} - 12 hat{k} Now, we have the normal vector vec{n} = 42 hat{i} - 12 hat{k}. To find the equation of the plane, we use the point-normal form with point {P_1}: 42(x - 2) - 12(z - 3) = 0 Expanding and simplifying the equation, we get: 42x - 84 - 12z + 36 = 0 42x - 12z = 58 Finally, dividing through by 2 to simplify further: 21x - 6z + 29 = 0 Therefore, the equation of the plane is 21x - 6z + 29 = 0.

answer:To find the roots of the polynomial without using complex numbers, we can use the fact that the product of two polynomials is zero if and only if at least one of the polynomials is zero. In this case, we have: x(frac{1}{3}x^2-2x+4)=0 Therefore, either x=0 or frac{1}{3}x^2-2x+4=0. The first equation gives us the root x=0. The second equation is a quadratic equation, which can be solved using the quadratic formula. However, since we are not allowed to use complex numbers, the discriminant of the quadratic formula must be non-negative. The discriminant is (-2)^2-4(1/3)(4)=-16/3, which is negative. Therefore, the quadratic equation has no real roots. Hence, the only real root of the polynomial x(frac{1}{3}x^2-2x+4) is x=0.