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November 29, 2014

November 29, 2014

Posted by **Mike** on Sunday, March 3, 2013 at 4:35pm.

- Calculus -
**Damon**, Sunday, March 3, 2013 at 4:45pmwhen x = 0, y = sqrt 3

y = (x+3)^.5

so

dy/dx = .5 (x+3)^-.5

slope of tangent line at x = 0 is thus

.5/sqrt 3

y = m x + b

sqrt 3 = (.5/sqrt 3)(0) + b

b = sqrt 3

y = (.5/sqrt 3) x + sqrt 3

- Calculus -
**Mike**, Sunday, March 3, 2013 at 5:00pmThank you

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