answer:Metabolism involves a complex network of chemical reactions, catalyzed by enzymes. Key processes include catabolism, which breaks down food molecules to release energy in the form of ATP, and anabolism, which synthesizes organic matter such as sugars and fats for storage and other cellular functions.

answer:The Jacobian matrix of mathbf{r}(x, y, z) is given by: J(mathbf{r}) = begin{bmatrix} frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z} frac{partial g}{partial x} & frac{partial g}{partial y} & frac{partial g}{partial z} frac{partial h}{partial x} & frac{partial h}{partial y} & frac{partial h}{partial z} end{bmatrix} Calculating each partial derivative, we get: frac{partial f}{partial x} = -frac{1}{x^2}, quad frac{partial f}{partial y} = 0, quad frac{partial f}{partial z} = 0 frac{partial g}{partial x} = 0, quad frac{partial g}{partial y} = e^y, quad frac{partial g}{partial z} = 0 frac{partial h}{partial x} = 0, quad frac{partial h}{partial y} = 0, quad frac{partial h}{partial z} = sec^2(z) Therefore, the Jacobian matrix of mathbf{r}(x, y, z) is: J(mathbf{r}) = begin{bmatrix} -frac{1}{x^2} & 0 & 0 0 & e^y & 0 0 & 0 & sec^2(z) end{bmatrix}

answer:** **Given:** * Mass of the block: m * Force applied to pull the block: P * Kinetic friction coefficient: μk = 0.3 * Angle of inclination of the surface: θ **Free Body Diagram:** [Image of a free body diagram of a block on an inclined surface with forces labeled] **Analysis:** 1. **Normal Force (N):** The normal force N is the force exerted by the surface on the block perpendicular to the surface. It balances the component of the block's weight perpendicular to the surface. N = mgcostheta 2. **Frictional Force (Ff):** The frictional force Ff is the force that opposes the motion of the block due to the interaction between the block and the surface. It acts in the direction opposite to the direction of motion. F_f = mu_k N = mu_k mgcostheta 3. **Equation of Motion:** In the direction parallel to the surface, the forces acting on the block are the applied force P, the component of the block's weight parallel to the surface (mg sin θ), and the frictional force Ff. sum F_x = P - mgsintheta - F_f = ma P - mgsintheta - mu_k mgcostheta = ma P = m(a + gsintheta + mu_k gcostheta) **Therefore, the equation of motion for the block in the direction parallel to the surface is:** P = m(a + gsintheta + mu_k gcostheta)

answer:To find the rank of a matrix, we can reduce it to row echelon form. Row echelon form is a matrix where each row has a leading 1 (the first nonzero entry in the row) and all other entries in the column of the leading 1 are 0. We can use elementary row operations to reduce a matrix to row echelon form. Elementary row operations are: 1. Swapping two rows 2. Multiplying a row by a nonzero scalar 3. Adding a multiple of one row to another row Using elementary row operations, we can reduce the matrix A to row echelon form as follows: left( begin{array}{cc} -10 & -10 7 & -5 8 & 0 2 & -2 end{array} right) rightarrow left( begin{array}{cc} -10 & -10 0 & -12 8 & 0 2 & -2 end{array} right) rightarrow left( begin{array}{cc} -10 & -10 0 & -12 0 & -8 2 & -2 end{array} right) rightarrow left( begin{array}{cc} -10 & -10 0 & -12 0 & -8 0 & -4 end{array} right) rightarrow left( begin{array}{cc} -10 & -10 0 & -12 0 & 0 0 & -4 end{array} right) rightarrow left( begin{array}{cc} -10 & -10 0 & -12 0 & 0 0 & 0 end{array} right) The row echelon form of the matrix A has 2 nonzero rows, so the rank of A is 2. The rank of the matrix A is 2.