Question 806271
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No, no, no:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{1}{3}\ =\ \frac{-1}{3}\ =\ \frac{1}{-3}\ \not =\ \frac{-1}{-3}\ =\ \frac{1}{3}]


See?


Furthermore, there is no reason to change the fraction to a decimal.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v\ =\ -\frac{1}{3}t\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v\ =\ -\frac{1}{3}(6)\ +\ 3\ =\ -2\ +\ 3\ =\ 1]


Which makes sense because if the value is 3 at time 0, 6 years later, the value needs to be smaller because the value of things like software depreciate over time.


Mark your coordinate axes with a *[tex \Large t] on the horizontal and a *[tex \Large v] on the vertical.  Then whatever value you select for *[tex \Large t] is the first coordinate in your ordered pair, and the *[tex \Large v] that results from that selection is the second coordinate in your ordered pair.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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