y = | 4x + 5z |
x2 z |
|
In Figure 1 we take the first three steps in the check of the problem. In this
example we have chosen the arbitrary value of 2.053 to store into the variable x.
We would do this via the keys
. Note that we are using the lower case
variable x. Next we have assigned .1589 to the variable Z via the
keys
. Note that we have used the upper case
variable Z. We did this because it takes fewer keystrokes to generate the upper case Z.
Finally, we use the answer of the problem,
|
||||
|
Now that we have values assigned to x, Y, and Z, we evaluate the two sides of the
original equation. Thus, we evaluate
And, we evaluate
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PRECALCULUS: College Algebra and Trigonometry
© 2000 Dennis Bila, James Egan, Roger Palay