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As you probably know, in programming it is sometimes easier to use a number system based on 16 instead of 10 The base 16 number system is called hexadecimal and uses the digits 0 through 9 plus the letters A through F, which stand for 10, 11, 12, 13, 14, and 15 For example, the hexadecimal number 10 is 16 in decimal Because of the frequency with which hexadecimal numbers are used, C# allows you to specify integer literals in hexadecimal format A hexadecimal literal must begin with 0x (a 0 followed by an x) Here are some examples:

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Enclosing character literals in single quotes works for most printing characters, but a few characters, such as the carriage return, pose a special problem when a text editor is used In addition, certain other characters, such as the single and double quotes, have special meaning in C#, so you cannot use them directly For these reasons, C# provides special escape sequences, sometimes referred to as backslash character constants, shown in Table 3-2 These sequences are used in place of the characters they represent For example, this assigns ch the tab character:

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(e) [x ln(cos x)] d s(s 3) e ds d cos(x2 ) e (g) dx (f) (h) [ln(e x x)] 5 Imitate the example in the text to do each of these falling body problems (a) A ball is dropped from a height of 64 feet How long will it take that ball to hit the ground (b) Suppose that the ball from part (a) is thrown straight down with an initial velocity of 5 feet per second Then how long will it take the ball to hit the ground (c) Suppose that the ball from part (a) is thrown straight up with an initial velocity of 20 feet per second Then how long will it take the ball to hit the ground 6 Use the Chain Rule to perform each of these differentiations: d cos(ln(sin x)) (a) dx

ch = '\t';

The next example assigns a single quote to ch:

3 (1) 2

ch = '\'';

Escape Sequence \a \b \f \n \r \t \v \0 \' \" \\

(b) (c) (d) (e) (f)

Here is an example that demonstrates the reference type constraint:

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// Demonstrate a reference constraint using System; class MyClass { // } // Use a reference constraint class Test<T> where T : class { T obj; public Test() { // The following statement is legal only // because T is guaranteed to be a reference // type, which can be assigned the value null obj = null; } // } class ClassConstraintDemo { static void Main() { // The following is OK because MyClass is a class Test<MyClass> x = new Test<MyClass>(); // The next line is in error because int is a value type Test<int> y = new Test<int>(); } }

First, notice how Test is declared:

d cos(sin x) e dx d ln(e cos x + x) dx d arccos(x 2 + sec x) dx d arcsin(ln x + e x /2) dx d arctan(x e x ) dx

class Test<T> where T : class {

The class constraint requires that any type argument for T be a reference type In this program, this is necessary because of what occurs inside the Test constructor:

public Test() { // The following statement is legal only // because T is guaranteed to be a reference // type, which can be assigned the value null obj = null; }

Try reading the two sentences aloud The rst sentence starts with a clich that focuses on the company: We are pleased to announce The revised sentence begins with bene ts to the customer: Your microbiological tests will be turned around more quickly Can you hear how much stronger the revised sentence is

Here, obj (which is of type T) is assigned the value null This assignment is valid only for reference types As a general rule, you cannot assign null to a value type (The exception to this rule is the nullable type, which is a special structure type that encapsulates a value type and allows the value null See 20 for details) Therefore, without the constraint, the

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