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 09-01-2007 #61 Sillysidley   Join Date: Oct 2006 Posts: 822 No, it's $2^{4^x}=2^{2(2^{x})}-> 4^x=2^{x+1}-> 2^{2x}=2^{x+1}-> 2x=x+1-> x=1$ __________________ .
 09-01-2007 #62 Temperal Guest   Posts: n/a Fine, whatever. Next question.
09-01-2007   #63
pianoforte
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 Originally Posted by pianoforte Find area of triangle with vertices the intersections of lines y=3x+8, y=(1/2)x-5, 3+2x+2y=0.
*cough*

09-02-2007   #64
Temperal
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Quote:
 Originally Posted by pianoforte *cough*
Did you just make up random equations? Those numbers come out very non-whole.

 09-02-2007 #65 Sillysidley   Join Date: Oct 2006 Posts: 822 VERY non-whole. All you do is put them in slope intercept form, put two of them at a time to be equal, find the coordinates that way, and use shoelace or whatever method you want to. __________________ .
 09-02-2007 #66 Temperal Guest   Posts: n/a Do we really have to solve the problem? Come on, I mean...
 09-02-2007 #67 Sillysidley   Join Date: Oct 2006 Posts: 822 Do we have to do all that "figuring out" stuff /whine __________________ .
 09-08-2007 #68 hunter34 Guest   Posts: n/a \sigh This is a long one. Time to play with graphing programs. So you have three sets of linear equations for each vertex of the triangle. Each solution for each on is $(2\frac{1}{3},-3\frac{5}{6}),(-2\frac{3}{8},\frac{7}{8}),(-5\frac{1}{5},-7\frac{3}{5})$ So take the midpoint of $(2\frac{1}{3},-3\frac{5}{6})$ and $(-2\frac{3}{8},\frac{7}{8})$ which is $(\frac{1}{48},-\frac{71}{48})$. Now take the distance between the midpoint and two endpoints, and multiply the two distances and divide by two to get the answer. I have to go soon so someone can finish it up.
 10-29-2007 #69 Temperal Guest   Posts: n/a Eh, I decided to revive this thread. Find the volume of a tetrahedron with a surface area of 361 square meters.
 10-29-2007 #70 Sillysidley   Join Date: Oct 2006 Posts: 822 Regular or not? __________________ .

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