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consider the point (x,y) lying on the graph of y=(x+3)^1/2. let L be the distance between the points (x,y) and (4,0). write l as a function of y
 one year ago
 one year ago
consider the point (x,y) lying on the graph of y=(x+3)^1/2. let L be the distance between the points (x,y) and (4,0). write l as a function of y
 one year ago
 one year ago

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Jemurray3Best ResponseYou've already chosen the best response.2
\[L = \sqrt{ (x4)^2 + y^2 } \] and \[ y = \sqrt{x+3} \implies x = y^2  3\] so substitute that in.
 one year ago

lynn555Best ResponseYou've already chosen the best response.0
so the answer would be y^546+x?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
erm, no. Literally substitute y^23 in for x in the equation that starts with L = ...
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
there should be no x involved.
 one year ago

LeeYeongKyuBest ResponseYou've already chosen the best response.0
dw:1347425169078:dw
 one year ago

LeeYeongKyuBest ResponseYou've already chosen the best response.0
so given function is y=(x+3)^1/2 so y^2=x+3, so, x = y^23. plugging x to the L=(x4)^2 + y^2 then, L=(y^26)^2 + y^2
 one year ago

LeeYeongKyuBest ResponseYou've already chosen the best response.0
you can reduce it easily
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
That should be y^27, not y^26, and the whole thing should be under a square root. But yeah.
 one year ago
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