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// An example that uses an implicit conversion operator using System; // A three-dimensional coordinate class class ThreeD { int x, y, z; // 3-D coordinates public ThreeD() { x = y = z = 0; } public ThreeD(int i, int j, int k) { x = i; y = j; z = k; } // Overload binary + public static ThreeD operator +(ThreeD op1, ThreeD op2) { ThreeD result = new ThreeD(); resultx = op1x + op2x; resulty = op1y + op2y; resultz = op1z + op2z; return result; } // An implicit conversion from ThreeD to int public static implicit operator int(ThreeD op1) { return op1x * op1y * op1z; } // Show X, Y, Z coordinates public void Show() { ConsoleWriteLine(x + ", " + y + ", " + z); } } class ThreeDDemo { static void Main() { ThreeD a = new ThreeD(1, 2, 3); ThreeD b = new ThreeD(10, 10, 10); ThreeD c = new ThreeD(); int i; ConsoleWrite("Here is a: "); aShow(); ConsoleWriteLine(); ConsoleWrite("Here is b: "); bShow(); ConsoleWriteLine(); c = a + b; // add a and b together ConsoleWrite("Result of a + b: "); cShow();

or r = 3 5000/ Since the problem makes sense for 0 < r < , and since it clearly has no maximum, we conclude that r = 3 5000/ , h = 3 5000/ 7 We calculate that g (x) = sin x + x cos x and g (x) = 2 cos x x sin x The roots of these transcendental functions are best estimated with a calculator or computer The provided sketch gives an idea of where the extrema and inflection points are located

9:

ConsoleWriteLine(); i = a; // convert to int ConsoleWriteLine("Result of i = a: " + i); ConsoleWriteLine(); i = a * 2 - b; // convert to int ConsoleWriteLine("result of a * 2 - b: " + i); } }

This program displays the output:

Here is a: 1, 2, 3 Here is b: 10, 10, 10 Result of a + b: 11, 12, 13 Result of i = a: 6 result of a * 2 - b: -988

8 We know that v0 = 10 and h0 = 500 Hence p(t) = 16t 2 10t + 500 The body hits the ground when 0 = p(t) = 16t 2 10t + 500 Solving, we find that t 5286 seconds 9 We see that h(x) = x x2 4 x2 + 4 (x 2 4)2

As the program illustrates, when a ThreeD object is used in an integer expression, such as i = a, the conversion is applied to the object In this specific case, the conversion returns the value 6, which is the product of coordinates stored in a However, when an expression does not require a conversion to int, the conversion operator is not called This is why c = a + b does not invoke operator int( ) Remember that you can create different conversion operators to meet different needs You could define a second conversion operator that converts ThreeD to double, for example Each conversion is applied automatically and independently An implicit conversion operator is applied automatically when a conversion is required in an expression, when passing an object to a method, in an assignment, and also when an explicit cast to the target type is used Alternatively, you can create an explicit conversion operator, which is invoked only when an explicit cast is used An explicit conversion operator is not invoked automatically For example, here is the previous program reworked to use an explicit conversion to int:

// Use an explicit conversion using System; // A three-dimensional coordinate class class ThreeD { int x, y, z; // 3-D coordinates public ThreeD() { x = y = z = 0; } public ThreeD(int i, int j, int k) { x = i; y = j; z = k; } // Overload binary + public static ThreeD operator +(ThreeD op1, ThreeD op2) { ThreeD result = new ThreeD();

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Part I:

h (x) = h (x) =