Question

standard form of the line which passes through (6, 3) and has a slope of two thirds?

Answer 1

\(\bf \begin{array}{lllll}
&x_1&y_1\\
&(6\quad ,&3)
\end{array}
\\\quad \\

slope = m= \cfrac{2}{3}
\\ \quad \\

y-y_1=m(x-x_1)\qquad \textit{plug in your values and solve for "y"}\)

Answer 2

y-y1 = m(x-x1)
m is the slope
plug in the points
then turn it into a standard form of ax+by = c

Answer 3

Alright thank you! I was just double checking my answer! :)

Answer 4

Does anyone know how to find this: Given f(x) = the quantity of 5x plus 1, divided by 2 solve for f-1(8). I'm not very sure of my process. Thanks!

Answer 5

\(\bf f(x)=y=\cfrac{5x+1}{2}\qquad inverse\implies x=\cfrac{5y+1}{2}\Leftarrow f^{-1}\)

notice, all we did was swap the 2 variables about
then solve for "y"

Answer 6

once you've solve for "y", that will be \(\bf f^{-1}\) and then just pass the "8" to it like \(\bf f^{-1}(8)\)

Answer 7

Yes, I've got that, I'm just not sure what to do with the
8

Answer 8

\(\bf f(x)=y=\cfrac{5x+1}{2}\qquad inverse\implies x=\cfrac{5y+1}{2}\\ \quad \\
f^{-1}=y=\cfrac{2x-1}{5}\qquad f^{-1}(\color{red}{8})=\cfrac{2(\color{red}{8})-1}{5}\)

Answer 9

Ohhh, i see I see! Thanks! :)

Answer 10

yw