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[0xFFFFFF, 0xFFFFFF], [0, 100], [0, 255], mxBox); duDrawer.drawRectangle(40, 20, 32.5, 20); duDrawer.endFill(); } When the button is pressed, call drawButton() for both mLine and mPen so as to reset the visual state to the default. Then, if the instance is mEraser, call removeMovieClip() for each object nested in mBoard.mForeground. Call drawButton() with a second parameter of true to draw the button in the pressed state. Set the value of mSelected to the button MovieClip instance. mButton.onPress = function():Void { drawButton(mLine.mButton); drawButton(mPen.mButton); if(this == mEraser) { for(var sItem:String in mBoard.mForeground) { mBoard.mForeground[sItem].removeMovieClip(); } } drawButton(this.mButton, true); mSelected = this; };

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Figure 14-4: Checkboxes are one way of offering the Web page visitor any number of options to choose from. Unlike checkboxes and text fields, each radio button in the set does not have a unique name instead, each group of radio buttons has a name. If you give the entire set of radio buttons the same name, browsers can assign one value to the radio button set. That value is determined by the contents of the Checked Value text box in the Property inspector. Figure 14-5 shows two different sets of radio buttons. The figure shows the Property inspector for one of the radio buttons in the osRadio group, on the right. In this example, each button in the group is assigned the name osRadio. To designate the default selection for each radio button group, you select the particular radio button and make the Initial State option Checked instead of Unchecked. In the form shown in Figure 14-5, the default selection for the osRadio group is Macintosh.

<HTML> <HEAD> <TITLE>Reading and Displaying Browser Information</TITLE> </HEAD> <BODY> <SCRIPT language= JavaScript > <!-- Make this script invisible to old browsers. document.writeln( Your browser is: + navigator.appName + <BR> ); document.writeln( The version is: + navigator.appVersion + <BR> ); // Back to normal. --> </SCRIPT> </BODY> </HTML>

2.7 Reuse 1/3 Reuse 2/6 Reuse 3/9 Reuse 4/12 2.4 2.1 1.8 1.5 1.2 0.9 0.6 0.3 0 90th percentile delay (s)

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If you use a MultiSelect ListBox, you can t specify a LinkedCell; you need to write a macro to determine which items are selected.

Problem 2.10 By combining the equivalent circuits shown in Figs. 2-9A and 2-9B, we can determine the answer. a. 25 mV across one of the resistors and 0 V across the other resistor. b. The two noise currents add in one of the terminating resistors and cancel in the other terminating resistor. c. The terminating resistor that had 25 mV across it will now have 0 V, and the terminating resistor that had 0 V across it will now have 25 mV. Problem 2.11 From Eq. 2-2, we know that the capacitive (electric eld) pickup is a function of the terminating resistor R. If the two conductors of a twisted pair have different terminating resistances, then they will each pickup a different voltage, and a net noise voltage will appear between them. Only if the terminating resistances are equal will they pickup the same voltage hence, the net noise voltage between them will be zero. Problem 2.12 a. 100% outside the shield, and 0% inside the shield. b. 0% outside the shield, and 100% inside the shield. CHAPTER 3 Problem 3.1 Safety grounds. Problem 3.2 False, power grounding will only affect the common-mode noise. Problem 3.3 By using a ground plane or grid (ZSRP). Problem 3.4 a. The current Ig. b. The impedance Zg.

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