What is the tangent of y=10sqrt(x)-x+3,(100,3) ?

The tangent of y=10sqrt(x)-x+3,(100,3) is y=-1/2 x+53

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tangent of y=10sqrt(x)-x+3,(100,3)

tangent of

y = 10 x - x + 3, (100, 3)

y = - \frac{1}{2} x + 53

Solution steps

Find the slope of

y = 10 x - x + 3 : \frac{d y}{d x} = \frac{5}{x} - 1

Compute the slope of

y = 10 x - x + 3

(100, 3) : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(100, 3) : y = - \frac{1}{2} x + 53

y = - \frac{1}{2} x + 53

\int_{0}^{10} 2 π y (100 - y^{2}) d y

\int x^{2} 0. 4^{- x} x d x

(4, 5), (1, - 3)

t^{2} y^{^{''}} - 11 t y^{^{'}} + 35 y = 0

\int \frac{x ^{4} + 8}{x ^{3}} d x

What is the tangent of y=10sqrt(x)-x+3,(100,3) ?
The tangent of y=10sqrt(x)-x+3,(100,3) is y=-1/2 x+53