answer:To find the locations of the vertices, we can use the Law of Cosines. Let the third vertex be (x, y). Then, we have: 9.4^2 = 2.3^2 + x^2 - 2(2.3)(x)cos(89) x = 9.12852 2.3^2 = 9.4^2 + y^2 - 2(9.4)(y)cos(89) y = 2.24281 Therefore, the vertices of the triangle are (0, 0), (9.63822, 0), and (9.12852, 2.24281). To find the measures of the three interior angles, we can use the Law of Sines. We have: sin(A) / 2.3 = sin(89) / 9.4 A = 0.240921 radians sin(B) / 9.4 = sin(89) / 2.3 B = 1.34733 radians C = 180 - A - B = 1.55334 radians Therefore, the measures of the three interior angles are 0.240921 radians, 1.34733 radians, and 1.55334 radians. Vertices: (0, 0), (9.63822, 0), (9.12852, 2.24281) Angles: (0.240921 radians, 1.34733 radians, 1.55334 radians)

answer:The Jacobian matrix of mathbf{r}(x, y, z) is given by: J(mathbf{r}(x, y, z)) = begin{bmatrix} frac{partial}{partial x} [sinh(y)] & frac{partial}{partial y} [sinh(y)] & frac{partial}{partial z} [sinh(y)] frac{partial}{partial x} [log(x+y)] & frac{partial}{partial y} [log(x+y)] & frac{partial}{partial z} [log(x+y)] frac{partial}{partial x} [e^{y+z^4}] & frac{partial}{partial y} [e^{y+z^4}] & frac{partial}{partial z} [e^{y+z^4}] end{bmatrix} = begin{bmatrix} 0 & cosh(y) & 0 frac{1}{x+y} & frac{1}{x+y} & 0 0 & e^{y+z^4} & 4z^3 e^{y+z^4} end{bmatrix}

answer:Given Data: Mass of nitrogen gas: {eq}{rm{m}} = 50.0;{rm{g}} {/eq}. Given Reaction: {eq}3;{{rm{H}}_2};left( {rm{g}} right) + {{rm{N}}_2};left( {rm{g}} right) to 2;{rm{N}}{{rm{H}}_3};left( {rm{g}} right) {/eq}. To Calculate: The number of moles of ammonia gas produced. The molecular weight of nitrogen gas is 28.014 g/mol. The moles of nitrogen gas can be calculated using the formula given below. {eq}{rm{n}} = dfrac{{rm{m}}}{{{{rm{M}}_{rm{w}}}}} {/eq} On substituting the corresponding values in the above equation, we get, {eq}begin{align*} {rm{n}} &= dfrac{{50.0;{rm{g}}}}{{28.014;dfrac{{rm{g}}}{{{rm{mol}}}}}} &= 1.78;{rm{mol}} end{align*} {/eq} From the reaction stoichiometry, 1 mole of nitrogen gas {eq}left( {{{rm{N}}_2}} right) {/eq} produces 2 moles of ammonia gas {eq}left( {{rm{N}}{{rm{H}}_3}} right) {/eq}. Thus, the moles of ammonia gas produced from 1.78 moles of nitrogen gas can be calculated as shown below. {eq}begin{align*} {rm{Moles}};{rm{of}};{{rm{N}}_2};{rm{gas}} &= 1.78;{rm{mol}};{{rm{N}}_2} times dfrac{{2;{rm{mol}};{rm{N}}{{rm{H}}_3}}}{{1;{rm{mol}};{{rm{N}}_2}}} &= 3.56;{rm{mol}} end{align*} {/eq} Hence, 3.56 moles of ammonia gas can be produced from 50.0 grams of nitrogen gas at STP.

answer:Given: - Initial velocity, ( v_i = 0 ) m/s - Acceleration, ( a = 4 ) m/s² - Time with constant acceleration, ( t_1 = 3 ) s - Time with constant speed, ( t_2 = 20 ) s The displacement during acceleration (first 3 seconds) is calculated using the kinematic equation: [ x_1 = frac{1}{2} at_1^2 ] [ x_1 = frac{1}{2} (4)(3)^2 ] [ x_1 = 18 text{ meters} ] The final velocity after 3 seconds is: [ v_f = v_i + at_1 ] [ v_f = 0 + (4)(3) ] [ v_f = 12 text{ m/s} ] The displacement during constant speed (next 20 seconds) is: [ x_2 = v_f cdot t_2 ] [ x_2 = (12)(20) ] [ x_2 = 240 text{ meters} ] Therefore, the total distance traveled by the bicyclist is the sum of ( x_1 ) and ( x_2 ): [ x = x_1 + x_2 ] [ x = 18 + 240 ] [ x = 258 text{ meters} ] So, the bicyclist travels a total of 258 meters.