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Continuity of the absolute value function Prove that the absolute value function |x| is continuous for all values of x. (Hint: Using the definition of the absolute value function, compute
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- • • Let > be a potential for the vector field F = (−2 y³, −6 xy² − 4 z³, −12 yz² + 4 2). Then the value of sin((-1.63, 2.06, 0.57) – (0,0,0)) is - 0.336 -0.931 -0.587 0.440 0.902 0.607 -0.609 0.146arrow_forwardThe value of cos(4M) where M is the magnitude of the vector field with potential ƒ = e² sin(лy) cos(π²) at x = 1, y = 1/4, z = 1/3 is 0.602 -0.323 0.712 -0.816 0.781 0.102 0.075 0.013arrow_forwardThere is exactly number a and one number b such that the vector field F = conservative. For those values of a and b, the value of cos(a) + sin(b) is (3ay + z, 3ayz + 3x, −by² + x) is -0.961 -0.772 -1.645 0.057 -0.961 1.764 -0.457 0.201arrow_forward
- A: Tan Latitude / Tan P A = Tan 04° 30'/ Tan 77° 50.3' A= 0.016960 803 S CA named opposite to latitude, except when hour angle between 090° and 270°) B: Tan Declination | Sin P B Tan 052° 42.1'/ Sin 77° 50.3' B = 1.34 2905601 SCB is alway named same as declination) C = A + B = 1.35 9866404 S CC correction, A+/- B: if A and B have same name - add, If different name- subtract) = Tan Azimuth 1/Ccx cos Latitude) Tan Azimuth = 0.737640253 Azimuth = S 36.4° E CAzimuth takes combined name of C correction and Hour Angle - If LHA is between 0° and 180°, it is named "west", if LHA is between 180° and 360° it is named "east" True Azimuth= 143.6° Compass Azimuth = 145.0° Compass Error = 1.4° West Variation 4.0 East Deviation: 5.4 Westarrow_forwardds 5. Find a solution to this initial value problem: 3t2, s(0) = 5. dt 6. Find a solution to this initial value problem: A' = 0.03A, A(0) = 100.arrow_forward2) Drive the frequency responses of the following rotor system with Non-Symmetric Stator. The system contains both external and internal damping. Show that the system loses the reciprocity property.arrow_forward
- 1) Show that the force response of a MDOF system with general damping can be written as: X liax) -Σ = ral iw-s, + {0} iw-s,arrow_forward3) Prove that in extracting real mode ø, from a complex measured mode o, by maximizing the function: maz | ቀÇቃ | ||.|| ||.||2 is equivalent to the solution obtained from the followings: max Real(e)||2arrow_forwardDraw the unit circle and plot the point P=(8,2). Observe there are TWO lines tangent to the circle passing through the point P. Answer the questions below with 3 decimal places of accuracy. L1 (a) The line L₁ is tangent to the unit circle at the point 0.992 (b) The tangent line 4₁ has equation: y= 0.126 x +0.992 (c) The line L₂ is tangent to the unit circle at the point ( (d) The tangent line L₂ has equation: y= 0.380 x + x × x)arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageTrigonometry (MindTap Course List)TrigonometryISBN:9781337278461Author:Ron LarsonPublisher:Cengage Learning