What is the tangent of y=sqrt(3x+1),(8,5) ?

The tangent of y=sqrt(3x+1),(8,5) is y= 3/10 x+13/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=sqrt(3x+1),(8,5)

tangent of

y = 3 x + 1, (8, 5)

y = \frac{3}{10} x + \frac{13}{5}

Solution steps

Find the slope of

y = 3 x + 1 : \frac{d y}{d x} = \frac{3}{2 3 x + 1}

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{3}{10}

Find the line with slope m=

\frac{3}{10}

and passing through

(8, 5) : y = \frac{3}{10} x + \frac{13}{5}

y = \frac{3}{10} x + \frac{13}{5}

\int 2 x^{2} - 6 x d x

\frac{d}{d x} (ln ((2 x + 3)^{4}) (x + 1)^{3})

y = - 2 e^{5 x^{2}}

4 y^{^{''}} - 25 y = 0

f (x) = 2 + x

What is the tangent of y=sqrt(3x+1),(8,5) ?
The tangent of y=sqrt(3x+1),(8,5) is y= 3/10 x+13/5