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OpenStudy (anonymous):
it still hard to do, and i looked at that and i don't understand that.... :(
OpenStudy (anonymous):
I just don't understand it at all, and I have looked at alot of examples but can never get it sadly :(
OpenStudy (jdoe0001):
hmm well using the side-splitter theorem
hmm lemme check
OpenStudy (jdoe0001):
notice that the midsegment in that link
DE is 1/2 of the longer AB
OpenStudy (anonymous):
yes
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OpenStudy (jdoe0001):
|dw:1416092940110:dw| and you have those values thus
we konw that FE = x+5
and that BC = 3x+2
thus \(\bf FE=x+5\qquad BC=3x+2\qquad FE=\cfrac{1}{2}BC\to \cfrac{BC}{2}\quad thus
\\ \quad \\
x+5=\cfrac{3x+2}{2}\)
solve for "x"
OpenStudy (anonymous):
shoot okay so you would end up with x=2
am i right??
OpenStudy (anonymous):
i don't think i did that correct....
OpenStudy (jdoe0001):
well.. is not 2
OpenStudy (anonymous):
okay so i did do that wrong..
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