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e^(x/10) = -y^2 + 100y how would I make y the subject
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add y^2 to both sides
does that mean "solve for y'?
yes, y = ..
then subtract e^(x/10) to both sides
you could treat it like a quadratic where e^x/10 is your "c" term \[y= 50\pm \sqrt{2500 - e ^{x/10}}\]
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y^2 = -e^(x/10) + 100y
and this leads to dumbcows answer,,lol
0 = -y^2 + 100y -e^(x/10) quadratic form of equation
y^2 -100y +e^(x/10) =0
y=(-(-100)-+sqrt(100^2 -4(1)(4e^(x/10))))/2 y=(100-+sqrt(10000-4e^(x/10)))/2 y=(50-+sqrt(2500-e^(x/10))
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