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SOLUTION First note that the answer is not we lose two grams every hundred years so The rate of decay depends on the amount of material present That is the key Instead, we let R(t) denote the amount of radioactive material at time t Equation ( ) guarantees that R has the form R(t) = P eKt Letting t = 0 denote the time at which there are 5 grams of isotope, and measuring time in years, we have R(0) = 5 and R(100) = 3

w Ni , k1 (f 2 )

do { ConsoleWrite(""); ThreadSleep(100); } while (mt1Count < 10 || mt2Count < 10 || mt3Count < 10); ConsoleWriteLine("Main thread ending"); } }

Sample output from this program is shown next:

(543) (544)

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Main thread starting Child #1 starting Child #2 starting Child #3 starting In Child #1, Count is 0 In Child #2, Count is 0 In Child #3, Count is 0 In Child #1, Count is 1 In Child #2, Count is 1 In Child #3, Count is 1 In Child #1, Count is 2 In Child #2, Count is 2 In Child #3, Count is 2 In Child #1, Count is 3 In Child #2, Count is 3 In Child #3, Count is 3 In Child #1, Count is 4 In Child #2, Count is 4 In Child #3, Count is 4 In Child #1, Count is 5 In Child #2, Count is 5 In Child #3, Count is 5 In Child #1, Count is 6 In Child #2, Count is 6 In Child #3, Count is 6 In Child #1, Count is 7 In Child #2, Count is 7 In Child #3, Count is 7 In Child #1, Count is 8 In Child #2, Count is 8 In Child #3, Count is 8 In Child #1, Count is 9

From the rst piece of information we learn that 5 = P eK 0 = P Hence P = 5 and R(t) = 5 eKt = 5 (eK )t

Part II:

(w ) (w ) k - 1 (f 2 - xi ) Ni , k1 -1 (f 2 ) ( xi + k1 - f 2 ) Ni +1, k1 -1 (f 2 ) + Ni(,w1) (f 2 ) = 1 k k1 - w - 1 xi + k1 -1 - xi xi + k1 - xi +1 ) M (j vk2 ( s2 ) ,

Child #1 terminating In Child #2, Count is 9 Child #2 terminating In Child #3, Count is 9 Child #3 terminating Main thread ending

As you can see, once started, all three child threads share the CPU Again, because of differences among system configurations, operating systems, and other environmental factors, when you run the program, the output you see may differ slightly from that shown here

Fig 118(b)

Often it is useful to know when a thread has ended In the preceding examples, this was attempted by watching the Count variable hardly a satisfactory or generalizable solution Fortunately, Thread provides two means by which you can determine whether a thread has ended First, you can interrogate the read-only IsAlive property for the thread It is defined like this: public bool IsAlive { get; } IsAlive returns true if the thread upon which it is called is still running It returns false otherwise To try IsAlive, substitute this version of MoreThreads for the one shown in the preceding program:

v M j , k2 ( s2 )

// Use IsAlive to wait for threads to end class MoreThreads { static void Main() { ConsoleWriteLine("Main thread starting"); // Construct MyThread mt1 MyThread mt2 MyThread mt3 three = new = new = new threads MyThread("Child #1"); MyThread("Child #2"); MyThread("Child #3");

do { ConsoleWrite(""); ThreadSleep(100); } while (mt1ThrdIsAlive && mt2ThrdIsAlive && mt3ThrdIsAlive); ConsoleWriteLine("Main thread ending"); } }

( - ) ( ) M j vk21-1 ( s2 ) M j v -,1k2 -1 ( s2 ) , +1 = ( k2 - 1) y j + k2 - y j +1 y j + k2 -1 - yi

The equation of a line in the plane will describe in compact form all the points that lie on that line We determine the equation of a given line by writing its slope in two different ways and then equating them Some examples best illustrate the idea EXAMPLE 110

This version produces the same output as before The only difference is that it uses IsAlive to wait for the child threads to terminate Another way to wait for a thread to finish is to call Join( ) Its simplest form is shown here: public void Join( )

23:

(545)

Part I:

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