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public class AndroidLBS extends MapActivity { /** Called when the activity is first created. */ @Override public void onCreate(Bundle icicle) { super.onCreate(icicle); setContentView(R.layout.main); final MapView myMap = (MapView) findViewById(R.id.myMap); final MapController myMapController = myMap.getController(); final Button zoomIn = (Button) findViewById(R.id.buttonZoomIn); zoomIn.setOnClickListener(new Button.OnClickListener() { public void onClick(View v){ ZoomIn(myMap,myMapController); }}); final Button zoomOut = (Button) findViewById(R.id.buttonZoomOut); zoomOut.setOnClickListener(new Button.OnClickListener() { public void onClick(View v){ ZoomOut(myMap,myMapController); }}); final Button gpsButton = (Button) findViewById(R.id.gpsButton); gpsButton.setOnClickListener(new Button.OnClickListener() { public void onClick(View v){ LoadProviders(myMap,myMapController); }});

06/29/02

If you want to prove to yourself that free energies sum, you can write the equilibrium expressions for the first and second reactions and multiply them together, and you ll get the equilibrium expression for the hydrolysis of ATP. Multiplication is equivalent to the addition of logarithms, so that when you multiply equilibrium constants, you re actually adding free energies (or vice versa).

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at the point representing 2. If you start with 10 and divide by 5, you cut your distance from the number reflector by a factor of 5 and stay on the same side, so you end up at 2. Dividing by a negative number is trickier. In Fig. 5-2, suppose you start with 6 and divide it by 3. You jump to the other side of the number reflector and then reduce your distance from it by a factor of 3, finishing at 2. If you start with 10 and divide by 5, you jump to the other side of the number reflector and then cut your distance from it by a factor of 5, finishing at 2. In these situations, the integers divide each other cleanly without remainders. Remainders occur when one nonzero integer divides another integer and the result is not an integer. Messy division produces fractions, which we ll study in Chap. 6. An example is the division of 5 by 3. Another example is the division of 7 by 12. You can add, subtract, or multiply any two integers and always end up with another integer. In math jargon, the operations of addition, subtraction, and multiplication are closed over the set of integers. When you do a division problem with integers, you don t always get another integer. Therefore, the operation of division is not closed over the set of integers. When you divide a quantity by another quantity, you get a quotient. Sometimes it is called a ratio. In a division problem, the first number is sometimes called the dividend, and the second number is sometimes called the divisor.

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The lower the better A workable resistance is 50 to 500 ohms but the ideal resistance would be around 5 to 10 ohms In most cases, however, this would be unobtainable unless you used many ground rods or copper strapping at each end Keep adding rods at each probe site, spaced 20 feet apart from one another connected together with #6 copper bare wire, until the resistance gets down to between 25 and 500 ohms Keep in mind two items The further apart the probe sites, the lower the resistance, and the more rods driven into the ground, the lower the resistance We have found that on the average, five grounds at each end of the site are sufficient to obtain this reading.

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Another option rather than using ground rods, in rocky or extremely hard soil, you may use a ditch trencher and bury #6 or #4 copper wire at least 24 inches deep The copper wire should be placed in an X fashion with a length of at least 25 feet The lead cable will then be connected to the center point of the X Now obtain a length of CATV lead in wire such as RG-11 or RG-59, from your local electronic parts store A good source to try is the local Cable TV company Sometimes they have lengths of scrap lying around their warehouse and will be happy to get rid of it Connect the center conductor only (not the outside shield) to each antenna array using copper clamps Take each piece of co-ax to the location of your receiver Then install a suitable connector to the two ends.

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