Question 1145098
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You have two points on a linear relation:  *[tex \LARGE (-141.5,\,0)] and *[tex \LARGE (-24.8,\,100)].


Use the two-point form of the equation of a line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ \(\frac{y_2\,-\,y_1}{x_2\,-\,x_1}\)\(x\ -\ x_1\)]


Where the horizontal axis is *[tex \LARGE C^\circ] and the vertical axis is *[tex \LARGE Z^\circ]


So, plug in the numbers, do the arithmetic, and solve for *[tex \LARGE y] when *[tex \LARGE x\ =\ 0] (the freezing point of water in *[tex \LARGE C^\circ])


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 0\ =\ \(\frac{100\,-\,0}{-24.8\,-\,(-141.5)}\)\(x\ -\ (-141.5)\)]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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