To store a value for later use use the green memory

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To store a value for later use, use the green memory keys instead. Green letters appear above various keys on the calculator and are accessed by using the green A LPHA key. For instance, to display the letter ‘A’ type: A LPHA M ATH . To display ‘T’ type: A LPHA 4 . We will write A or T for those two-keystroke descriptions. To enter the current value into memory A , type S TO A E NTER . When A is later typed into a formula, the stored value is used. The machine’s memory ‘A’ holds that value until something else is stored there using the S TO key. For instance, if 3.45678 is displayed, store that value in T by typing S TO T E NTER . Then typing e x T ) E NTER will compute e 3.45678 , yielding the display 31.71469062 as before. 5
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4 Solving Systems of Exponential Equations As you know, systems of equations can be solved using different methods. Here’s the sample problem we will analyze today: Problem: Suppose Pa 2 = 10 and Pa 4 = 40, where P and a are positive num- bers. Find P and a . Solution: The usual method is to eliminate one unknown in order to obtain an equation in just one variable, which is easier to solve than the original system. In this particular system, we can divide to get rid of P : 40 10 = Pa 4 Pa 2 = a 2 , and we find that a 2 = 4. Since a is positive, we conclude a = 2. Put this value back into the first equation to find: P · ( 2 ) 2 = 10. Then P = 10 / 4 or P = 2.5. Check: Let’s check our answer by setting P = 2.5 and a = 2: Pa 2 = ( 2.5 ) · ( 2 ) 2 = 10 X Pa 4 = ( 2.5 ) · ( 2 ) 4 = 40 X 6
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Part IV. The Art of Estimating When working with estimated data such as values obtained by measurement, it is important to use all available calculator precision in intermediate results. But final results should be rounded appropriately. In most applications, a final result should be rounded to the least significant position. Financial applications are an exception as results in dollars are usually rounded to the nearest cent. Consider the following problem: Problem: A researcher finds that the concentration of a detergent in a lake decays from 1.27 ppm 1 to 0.86 ppm over a period of 27 hours. Use an expo- nential model to estimate the concentration 13 hours into the experiment. Solution: We will use an exponential function of the form f ( t ) = Pa - t to make our estimate. At the beginning of the experiment ( t = 0), the concentration is 1.27 ppm ( f ( 0 ) = 1.27 ppm). At the end of the experiment ( t = 27 hours), the concentration has decayed to 0.86 ppm ( f ( 27 ) = 0.86 ppm). Using the form of the decay equation, we have: 1.27 = f ( 0 ) = Pa - 0 = P 0.86 = f ( 27 ) = Pa - 27 The first equation tells us that P = 1.27. Putting this into the second equation, we have: a - 27 = 0.86 1.27 , so a = 1.27 0.86 1 / 27 1.0145.
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