 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
qr code vb.net open source Illustration for the solution to Problem 1 in Chap 3 in .NET
3 Recognizing Code 39 Full ASCII In .NET Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. Code 39 Extended Encoder In Visual Studio .NET Using Barcode creator for .NET framework Control to generate, create Code39 image in Visual Studio .NET applications. 3
Reading Code 3/9 In VS .NET Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET framework applications. Bar Code Drawer In .NET Framework Using Barcode creator for .NET Control to generate, create bar code image in Visual Studio .NET applications. 1 Figure A3 shows the graphs of the equations q = p /4 and q = p /2 in polar coordinates, where q is the independent variable and r is the dependent variable Neither of these are functions of q In the first case, r can be any real number when q = p /4 In the second case, r can be any real number when q = p /2 2 The graph of q = p /4 is a sloping line through the origin in the Cartesian xy plane The graph of q = p /2 is a vertical line that coincides with the y axis Figure A4 shows both graphs The line representing q = p /4 portrays a function of x in the Cartesian xy plane, because there is never more than one value of y for any value of x But the line representing q = p /2 does not portray a function of x in the Cartesian xy plane, because when x = 0, y can be any real number Figure A3 Bar Code Recognizer In .NET Using Barcode recognizer for .NET Control to read, scan read, scan image in .NET applications. Code 39 Drawer In C#.NET Using Barcode generation for .NET framework Control to generate, create Code 3 of 9 image in Visual Studio .NET applications. Illustration for the solution to Problem 1 in Chap 3
Code39 Drawer In .NET Using Barcode generator for ASP.NET Control to generate, create Code39 image in ASP.NET applications. Draw Code39 In Visual Basic .NET Using Barcode printer for .NET Control to generate, create Code 3/9 image in .NET framework applications. p /2 Make Code 128C In Visual Studio .NET Using Barcode encoder for Visual Studio .NET Control to generate, create Code 128A image in VS .NET applications. Drawing GS1 DataBar14 In Visual Studio .NET Using Barcode creator for .NET Control to generate, create GS1 DataBar Expanded image in Visual Studio .NET applications. q = p /2 GS1  13 Creation In .NET Using Barcode drawer for .NET framework Control to generate, create EAN13 image in VS .NET applications. Generating Identcode In VS .NET Using Barcode drawer for Visual Studio .NET Control to generate, create Identcode image in .NET applications. q = p /4 ANSI/AIM Code 39 Generation In None Using Barcode maker for Word Control to generate, create Code39 image in Word applications. Bar Code Printer In None Using Barcode maker for Font Control to generate, create barcode image in Font applications. 3p /2 Recognize EAN 13 In VB.NET Using Barcode decoder for .NET Control to read, scan read, scan image in Visual Studio .NET applications. Generating EAN13 In .NET Framework Using Barcode creator for ASP.NET Control to generate, create EAN13 image in ASP.NET applications. Figure A4 Code 128 Code Set B Printer In None Using Barcode printer for Font Control to generate, create Code 128A image in Font applications. UPC  13 Generator In .NET Framework Using Barcode generator for Reporting Service Control to generate, create GS1  13 image in Reporting Service applications. Illustration for the solution to Problem 2 in Chap 3
Draw Code 128 In VB.NET Using Barcode generation for .NET Control to generate, create Code128 image in .NET applications. Make GS1 DataBar Limited In Java Using Barcode maker for Java Control to generate, create GS1 DataBar14 image in Java applications. y 6 4 2 q = p /2 q = p /4 2 2 4 6 476 WorkedOut Solutions to Exercises: 19 3 The equation r = a represents the same circle as the equation r = a 4 Imagine a ray that points straight to the right along the reference axis labeled 0 As the ray rotates counterclockwise so that q starts out at 0 and increases positively, the corresponding radius r starts out at 0 and increases negatively This tells us that the constant a is negative When the ray has turned through 1/2 rotation so that q = p, the radius of the solid spiral reaches the value r = 2p (Don t get this confused with the apparent radius of r = 4p on the solid spiral! The larger value is actually r = 4p, which we get when the ray has rotated through a complete circle so that q = 2p) We can solve for a by substituting the number pair (q,r) = (p, 2p) in the general spiral equation r = aq This gives us 2p = ap which solves to a = 2 Therefore, the equation of the pair of spirals is r = 2q 5 Line L runs through the origin and up to the left at an angle halfway between the p /2 axis and the p axis That direction is represented by q = 3p /4 This is the equation of L But we can also imagine that line L runs down and to the right at an angle corresponding to q = 7p /4 so this can also serve as the equation of L Theoretically, we can add or subtract any integer multiple of p from 3p /4 and get a valid equation for L By convention, we stick to the range of angles 0 q < 2p, so the above two equations are preferred over any others Circle C is centered at the origin and has a radius of 3 units, as we can see by inspecting the graph and remembering that each radial division equals 1 unit Therefore, C can be represented by r=3 We can also consider the radius to be 3 units, so r = 3 is an equally valid equation for C 3
6 Based on the solution to Problem 5, we can represent the intersection point at the upper left as either P = (3p /4,3) or P = (7p /4, 3) We can represent the intersection point at the lower right as either Q = (7p /4,3) or Q = (3p /4, 3) The more intuitive representations are the coordinates with positive radii, which are P = (3p /4,3) and Q = (7p /4,3) 7 Before we can solve the system of equations for L and C as they are shown in Fig 38, we must be certain that we ve completely identified the system For L, we have q = 3p /4 or q = 7p /4 and for C, we have r=3 or r = 3 Solving this system is deceptively simple It doesn t require algebra at all! We merely combine all the possible combinations of angles and radii we ve listed above to get the following four ordered pairs: (q,r) = (3p /4,3) (q,r) = (3p /4, 3) (q,r) = (7p /4,3) (q,r) = (7p /4, 3) 478 WorkedOut Solutions to Exercises: 19 Using plusandminus notation for the radii, we can reduce this list to two items: (q,r) = (3p /4, 3) and (q,r) = (7p /4, 3) That s redundant, but it s valid If we want to be more elegant, we can get rid of the redundancy and list the solutions as (q,r) = (3p /4,3) and (q,r) = (7p /4,3) We can tell which ordered pair represents P and which one represents Q by looking again at Fig 38 It s obvious that P = (3p /4,3) and Q = (7p /4,3) 8 Let s take away the polar grid in Fig 38 and put a Cartesian grid in its place, as shown in Fig A5 Because we ve been told that line L is equally distant from the vertical and

