Determine the two roots of the equation x3− 5x5 + e0.9x + 4(x + 1) = −2. how would I go about solving this?
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Determine the two roots of the equation x3− 5x5 + e0.9x + 4(x + 1) = −2.
how would I go about solving this?
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- the answer must be: [(13/6)m + (11/9r2)] (double dot) x + (11/9)cẋ + (10/3)kxPLEASE DO REVERSE ENGINEERING IN SOLVING THE PROBLEM THANK YOU. Determine B such that 5x + 2y – 7 = 0 is perpendicular to 2x – By + 2 = 0. a. 5 b. 6 c. 4 d. 7Solve the given exponential equation correct to 4 significant figures. e (x+1) = 3e(2x−5)
- Some iron has a coefficient of linear expansion of 12 × 10−6 K−1. A 100 mm length of iron piping is heated through 20 K. The pipe extends by: (a) 0.24 mm (b) 0.024 mm (c) 2.4 mm (d) 0.0024 mm E. No answerJust answer (i) R=31.79 h=56.68The windscreen washers of an old car don't work (W = 0). This could be for two reasons: the windscreen fluid tank is empty (E = 1) or the tube is has a blockage (B = 1). Being an old car, the light to indicate that the windscreen fluid tank is empty only comes on (L = 1) 75% of the time. It will sometimes comes on even if the tank has some fluid in it, with probability 0.25. Of course the tube could be both blocked and the tank empty. (a) Draw the Bayesian network describing the scenario just described, and express the joint probability in factorised form, defining appropriate variables.
- 5. A round aluminum sleeve with an outer diameter of 300 mm and an inner diameter of 200 mm is pressed onto a round steel sleeve with an inner diameter of 100 mm. The radial interference between the two sleeves is 6 = 0.25 mm. Let Ex = 72 GPa, VA = 0.33, E, = 200 GPa and v; = 0.29. Determine the following: (19 points) The interference pressure, p. in MPa between the two sleeves The change in the inner radius in the aluminum sleeve, u, in millimeters If the allowable tangential stress, do, in the aluminum is 180 MPa, determine the safety factor in the aluminum with respect to failure due to tangential stress. You may ignore the radial stres 7how to calculate this equation to get 18.3 kN get u write for me the step T1/sin(45) = 25 kN/sin(105) T1 = (25 kN/sin(105)) * sin(45) T1 = 18.3 kNNewton's law of cooling states that dT/dt = k(T−t0), where T(t) is the temperature of the body that is cooling, T0 is the temperature of the surrounding medium and k is constant. Solve the differential equation.