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public double Height { get { return pri_height; } set { pri_height = value < 0 -value : value; } } public void ShowDim() { ConsoleWriteLine("Width and height are " + Width + " and " + Height); } } // A derived class of TwoDShape for triangles class Triangle : TwoDShape { string Style; /* A default constructor This automatically invokes the default constructor of TwoDShape */ public Triangle() { Style = "null"; } // Constructor that takes three arguments public Triangle(string s, double w, double h) : base(w, h) { Style = s; } // Construct an isosceles triangle public Triangle(double x) : base(x) { Style = "isosceles"; } // Return area of triangle public double Area() { return Width * Height / 2; } // Display a triangle's style public void ShowStyle() { ConsoleWriteLine("Triangle is " + Style); } } class Shapes5 { static void Main() { Triangle t1 = new Triangle(); Triangle t2 = new Triangle("right", 80, 120); Triangle t3 = new Triangle(40); t1 = t2; ConsoleWriteLine("Info for t1: "); t1ShowStyle(); t1ShowDim(); ConsoleWriteLine("Area is " + t1Area());

SOLUTION We calculate: (e4 )ln 3 = e4 ln 3 = (eln 3 )4 = 34 = 81; 1 5 7 4 2; = 5 7 ( 3) 4 2 = 5 4 2 = 625 5 3 2 (32 x 3 )4 = (32 )4 (x 3 )4 = 38 x 12 = 6561 x 12 You Try It: Simplify the expression ln[e3x e y 5 24 ] EXAMPLE 618

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fails because S(1)(1) is unknown To overcome this dif culty, an iterative approach can be used The value of hc can be rst approximated by hch/(Kf + Ks)/Kf and then can be iteratively adjusted until the solution of the rst order differential Eq (518) yields the normalized value of S(1)(1) = 1 A suitable value can usually be found in around a dozen iterations The spline procedure used for the interpolation of the kinematic constraints to yield output motion programs described previously can be used here as well

Part I:

// Cast a long into a uint, no data lost l = 64000; u = (uint) l; ConsoleWriteLine("u after assigning 64000: " + u + " -- no data lost"); // Cast a long into a uint, data lost l = -12; u = (uint) l; ConsoleWriteLine("u after assigning -12: " + u + " -- data lost"); ConsoleWriteLine(); // Cast an int into a char b = 88; // ASCII code for X ch = (char) b; ConsoleWriteLine("ch after assigning 88: " + ch); } }

Cam Motion and Dynamic Behavior Spline Collocation Method for the Solution of Differential Equations As indicated earlier, it is necessary to solve ordinary differential equations to investigate the follower motion and to nd the cam motion The method of collocation has been found to work very well in this application The method has been explored by a number of authors (Ascher et al, 1979; Ascher, 1980; de Boor and Swartz, 1973; Carey and Finlayson, 1975; Cerutti, 1974, Russell and Shampine, 1972; Russell, 1974 and 1977) It is based on the simple concept that the solution of the differential equation can be represented as a linear combination of known functions In forming the linear combination, coef cients are determined such that prescribed initial or boundary conditions together with the differential equation itself are satis ed at a series of points (the collocation points) on the range of interest Methods of this type have generally been used for boundary value problems because of the effort that is normally required to solve such equations by other means The method is also applicable to initial value problems when the more usual explicit methods run into stability dif culties such as in cases involving stiff equations Various suggestions for using

The output from the program is shown here:

SOLUTION We have (x 3 5)8 = 9 x 3 5 = 91/8 x 3 = 91/8 5 1 x = (91/8 5 1 )1/3 x = 91/24 51/3

Integer outcome of x / y: 3 b after assigning 255: 255 -- no data lost b after assigning 257: 1 -- data lost s after assigning 32000: 32000 -- no data lost s after assigning 64000: -1536 -- data lost u after assigning 64000: 64000 -- no data lost u after assigning -12: 4294967284 -- data lost ch after assigning 88: X

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