So.. take a peek at the picture... let's get two points from it, hmm say 0,4 notice it touches the y-axis there, and say hmmm -4, 1, almost at the bottom of the line

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 0}}\quad ,&{{ 4}})\quad 
% (c,d)
&({{ -4}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-4}{-4-0}$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=4\\
x_1=0\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

once you get the slope and solve for "y", that'd be the equation of the line.

7 0

3 years ago

Erica invests $10,000 at 5% interest compounded annually.

grigory [225]

A=P(1+i $)^{n}$
A=10000(1+15% $)^{3}$
=11576.25
your answer is D

8 0

3 years ago