Calculus HELP

A particle is moving along the curve y=5 sqrt (2x+6). As the particle passes through the point (5,20 , its x-coordinate increases at a rate of 5 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

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  1. y = 5√(2x+6)
    dy/dt = 5/√(2x+6) dx/dt

    d^2 = x^2+y^2
    d dd/dt = x dx/dt + y dy/dt
    at (5,20), d=5√17, dy/dt = 5/4
    5√17 dd/dt = 5*5 + 20*5/4
    dd/dt = (25+25)/(5√17) = 10/√17

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