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Find the equation of the tangent line to the curve `y=sqrt(5x-3)-2` which is parallel to the line `4x-2y+3=0`

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slope = `m= -4/(-2) = 2`
`dy/dx|_p = 2`
`=> 1/(2sqrt(5x-3)) xx 5 = 2`
`x = 73/80`
putting it in the curve eqn
`y= -3/4 p(75/80, -3/4)`
`m=2`
eqn of tangent
...
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