What is the tangent of y=x+sqrt(x),(16,20) ?

The tangent of y=x+sqrt(x),(16,20) is y= 9/8 x+2

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tangent of y=x+sqrt(x),(16,20)

tangent of

y = x + x, (16, 20)

y = \frac{9}{8} x + 2

Solution steps

Find the slope of

y = x + x : \frac{d y}{d x} = 1 + \frac{1}{2 x}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{9}{8}

Find the line with slope m=

\frac{9}{8}

and passing through

(16, 20) : y = \frac{9}{8} x + 2

y = \frac{9}{8} x + 2

\frac{d y}{d x} + e^{x} y = 1

- x^{2} + 5 x + a

h (t) = (t^{4} - 1)^{9} (t^{3} + 1)^{6}

n = 1 \sum \infty 2 + 7 n

\frac{\partial}{\partial x} (- \frac{1}{2} ln (\frac{x}{a}))

What is the tangent of y=x+sqrt(x),(16,20) ?
The tangent of y=x+sqrt(x),(16,20) is y= 9/8 x+2