answer:The angle theta between two vectors can be found using the dot product formula: cos(theta) = frac{mathbf{A} cdot mathbf{B}}{|mathbf{A}| |mathbf{B}|}, where mathbf{A} cdot mathbf{B} is the dot product, and |mathbf{A}| and |mathbf{B}| are the magnitudes of vectors A and B, respectively. First, calculate the dot product of A and B: mathbf{A} cdot mathbf{B} = left(frac{1}{2}right) left(frac{7}{2}right) + left(frac{17}{2}right) (6) + (4) left(-frac{9}{2}right) + (-7)(-5) + left(frac{3}{2}right) left(frac{9}{2}right) = frac{7}{4} + frac{102}{2} - 18 + 35 + frac{27}{4} = frac{7 + 204 - 72 + 140 + 27}{4} = frac{306}{4} = 76.5 Next, calculate the magnitudes of A and B: |mathbf{A}| = sqrt{left(frac{1}{2}right)^2 + left(frac{17}{2}right)^2 + 4^2 + (-7)^2 + left(frac{3}{2}right)^2} = sqrt{frac{1}{4} + frac{289}{4} + 16 + 49 + frac{9}{4}} = sqrt{frac{1 + 289 + 64 + 196 + 9}{4}} = sqrt{frac{559}{4}} = frac{sqrt{559}}{2} |mathbf{B}| = sqrt{left(frac{7}{2}right)^2 + 6^2 + left(-frac{9}{2}right)^2 + (-5)^2 + left(frac{9}{2}right)^2} = sqrt{frac{49}{4} + 36 + frac{81}{4} + 25 + frac{81}{4}} = sqrt{frac{49 + 144 + 81 + 100 + 81}{4}} = sqrt{frac{455}{4}} = frac{sqrt{455}}{2} Now, find the angle: cos(theta) = frac{76.5}{left(frac{sqrt{559}}{2}right) left(frac{sqrt{455}}{2}right)} theta = cos^{-1}left(frac{76.5 cdot 2 cdot 2}{sqrt{559} cdot sqrt{455}}right) theta = cos^{-1}left(frac{306}{sqrt{559 cdot 455}}right) theta = cos^{-1}left(frac{306}{sqrt{252735}}right) The angle between the two vectors is theta = cos^{-1}left(frac{306}{sqrt{252735}}right).

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

answer:Net Present Value (NPV) = -Initial Investment + (Cash Flow Year 1 / (1+discount rate)^1) + (Cash Flow Year 2 / (1+discount rate)^2) + ... + (Cash Flow Year n / (1+discount rate)^n) NPV = -4,500 + (1,050 / (1+0.05)^1) + (1,250 / (1+0.05)^2) + (2,150 / (1+0.05)^3) + (1,150 / (1+0.05)^4) NPV = -4,500 + 1,000 + 1,145.03 + 1,920.73 + 997.70 NPV = 763.46

answer:To orthogonalize the vectors, we first normalize the non-zero vector. The given set contains one non-zero vector, {-1, 3}. The normalization process involves dividing the vector by its magnitude. The magnitude of the vector is calculated as: [ sqrt{(-1)^2 + 3^2} = sqrt{1 + 9} = sqrt{10} ] After normalization, the vector becomes: [ left{ begin{array}{c} -frac{1}{sqrt{10}} frac{3}{sqrt{10}} end{array} right} ] The zero vector, {0, 0}, is already orthogonal to any other vector, so it remains unchanged. Therefore, the orthogonalized set of vectors is: [ left{ begin{array}{c} left{-frac{1}{sqrt{10}}, frac{3}{sqrt{10}}right} {0, 0} end{array} right} ]