$\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{\frac{1}{3}})
\qquad \qquad \qquad \qquad 
% slope = m
slope = m\implies \cfrac{3}{4}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-\cfrac{1}{3}=\cfrac{3}{4}(x-4)\implies y-\cfrac{1}{3}=\cfrac{3}{4}x-3
\\\\\\
y=\cfrac{3}{4}x-3+\cfrac{1}{3}\implies y=\cfrac{3}{4}x-\cfrac{8}{3}$

4 0

3 years ago

Outline the process for inscribing a regular hexagon in a circle. Perform the construction in GeoGebra

mr_godi [17]

The following are the general steps to draw a hexagon inscribed in a circle:

First draw a circle with center A through B:

Then the intersection points will be vertex C and vertex D of the hexagon. Doing this for each vertex (from left to right) we will get the following:

4 0

1 year ago

What is the square root 55 explanation plzz​

What is the square root 55 explanation plzz