answer:The eigenvectors are as follows: [ begin{align*} & v_1 = {-0.7, 0.688, 1} & v_2 = {-0.488, 2.36, 1} & v_3 = {1.181, 1.758, 1} end{align*} ]

answer:To find the distance the skier glides on the horizontal surface, we can use the work-energy theorem, which states that the work done by the net force equals the change in kinetic energy: [ W_{text{net}} = Delta K ] The work done by friction on the inclined part is: [ W_{text{fric, incline}} = -mu mgd_{text{incline}} cos{theta} ] At the bottom, the skier's initial kinetic energy will be used to overcome friction on the level surface. The work done by friction on the horizontal part is: [ W_{text{fric, horizontal}} = -mu mgd_{text{horizontal}} ] The initial potential energy at the top of the hill is: [ U_i = mgh sin{theta} ] where ( h ) is the height of the hill, ( m ) is the skier's mass, ( g ) is the acceleration due to gravity, ( mu ) is the coefficient of friction, ( d_{text{incline}} ) is the distance along the inclined surface, and ( d_{text{horizontal}} ) is the horizontal distance. Since the skier starts from rest, the final kinetic energy at the bottom of the incline is: [ K_f = frac{1}{2}mv^2 ] Applying the work-energy theorem: [ -mu mgd_{text{incline}} cos{theta} + mgh sin{theta} = frac{1}{2}mv^2 ] Solving for the final velocity ( v ): [ v = sqrt{2g(h sin{theta} - mu d_{text{incline}} cos{theta})} ] Substituting ( h = d_{text{incline}} tan{theta} ): [ v = sqrt{2g(d_{text{incline}} sin{theta} - mu d_{text{incline}}^2 cos^2{theta} tan{theta})} ] [ v = sqrt{2g(d_{text{incline}} (sin{theta} - mu d_{text{incline}} cos{theta} tan^2{theta}))} ] Given values: ( theta = 9.9^circ ), ( d_{text{incline}} = 220 , text{m} ), ( mu = 0.0750 ): [ v = 49.4 , text{m/s} ] On the horizontal surface, the skier's kinetic energy is converted to work done by friction: [ -mu mgd_{text{horizontal}} = frac{1}{2}mv^2 ] Solving for ( d_{text{horizontal}} ): [ d_{text{horizontal}} = frac{v^2}{2mu g} ] [ d_{text{horizontal}} = frac{(49.4 , text{m/s})^2}{2(0.0750)(9.81 , text{m/s}^2)} ] [ d_{text{horizontal}} approx 249 , text{m} ] So, the skier glides 249 meters horizontally before coming to a stop.

answer:Let the inverse function be denoted as h(x), such that h(f(x)) = x. Using implicit differentiation, we obtain h'(f(x)) cdot f'(x) = 1, h''(f(x)) cdot (f'(x))^2 + h'(f(x)) cdot f''(x) = 0. At x=frac{pi}{4}, f(x)=0, hence h''(0) = -frac{f''left(frac{pi}{4}right)}{left(f'left(frac{pi}{4}right)right)^3}. Now, calculate f'(x) and f''(x) using the fundamental theorem of calculus and the chain rule: For f(x), we have f'(x) = frac{d}{dx}left[int_{sin x}^{cos x}{frac{dt}{e^tsqrt{1-t^2}}}right] = frac{-cos(x)}{e^{sin(x)}sqrt{1 - sin^2(x)}} - frac{-sin(x)}{e^{cos(x)}sqrt{1 - cos^2(x)}} = frac{sin(x)e^{cos(x)} - cos(x)e^{sin(x)}}{e^{sin(x)+cos(x)}sqrt{1 - sin^2(x)cos^2(x)}}. Since sin^2(x)cos^2(x) = frac{1}{4}sin^2(2x), for x=frac{pi}{4}, we have sin^2(2x)=1 and thus sin^2(x)cos^2(x) = frac{1}{4}. Therefore, f'left(frac{pi}{4}right) = frac{frac{1}{sqrt{2}}e^{frac{1}{sqrt{2}}} - frac{1}{sqrt{2}}e^{-frac{1}{sqrt{2}}}}{esqrt{2}}. Next, we differentiate f'(x) to find f''(x): f''(x) = frac{cos(x)e^{sin(x)}sqrt{1 - cos^2(x)} + sin(x)e^{cos(x)}sqrt{1 - sin^2(x)}}{e^{sin(x)+cos(x)}sqrt{1 - sin^2(x)cos^2(x)}} + frac{cos(x)sin(x)e^{sin(x)}sqrt{1 - cos^2(x)} - sin(x)cos(x)e^{cos(x)}sqrt{1 - sin^2(x)}}{e^{sin(x)+cos(x)}sqrt{1 - sin^2(x)cos^2(x)}}. At x=frac{pi}{4}, f''left(frac{pi}{4}right) can be calculated using this expression. Finally, plug the values of f'left(frac{pi}{4}right) and f''left(frac{pi}{4}right) into the expression for h''(0) to find the desired second derivative.

answer:To determine the difference in the initial deposit needed for a 3.5% interest rate compared to 2.5%, we'll use the future value (FV) formula: [ FV = PV times (1 + r)^n ] where: - ( FV ) is the future value (30,000), - ( PV ) is the present value (to be found), - ( r ) is the annual interest rate (as a decimal), - ( n ) is the number of years (5). For a 2.5% rate: [ 30,000 = PV times (1 + 0.025)^5 ] Solving for ( PV ): [ PV_{2.5%} = 26,516 ] For a 3.5% rate: [ 30,000 = PV times (1 + 0.035)^5 ] Solving for ( PV ): [ PV_{3.5%} = 25,259 ] The difference in the initial deposit needed is: [ 26,516 - 25,259 = 1,257 ] Therefore, you would need to deposit 1,257 less today if the interest rate is 3.5% instead of 2.5%.