What is the slop-intercept of the line that passes through the points (7,-5) and (3,-9)?

1 answer:

4 0

$\bf (\stackrel{x_1}{7}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-9}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-9-(-5)}{3-7}\implies \cfrac{-9+5}{-4}\implies \cfrac{-4}{-4}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-5)=1(x-7) \\\\\\ y+5=x-7\implies y=x-12$

You might be interested in

Q8 Q21.) Find the amount invested at each rate.

Pavlova-9 [17]

See the attached picture:

6% = 1200
9% = 2400
12% = 3800

5 0

3 years ago

Read 2 more answers

Elizabeth attempts a field goal by kicking a football from the ground with an initial vertical

sammy [17]

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf h(t)=-16t^2+64t+0\implies h(t)=-16t^2+64t\implies \stackrel{\textit{hits the ground}~\hfill }{0=-16t^2+64t} \\\\\\ 0=-16t(t-4)\implies t= \begin{cases} 0\\ \boxed{4} \end{cases}$