$\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{57})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{91}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{91}-\stackrel{y1}{57}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{4}}}\implies \cfrac{34}{2}\implies 17 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{57}=\stackrel{m}{17}(x-\stackrel{x_1}{4})\implies y-57=17x-68$

$\bf y=17x-11\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad \begin{cases} \stackrel{slope}{17}\\\\ \stackrel{y-intercept}{(0,-11)} \end{cases}$

4 0

3 years ago

A football is kicked from the ground and follows the graph of the function y = –t2 + 3t, where y represents height and t represe

vesna_86 [32]

Answer:

The height of the objects are the same after 2 seconds.

Step-by-step explanation:

In order to calculate at which time both objects have the same height we need to find the value of t that makes both equations equal. Therefore:

$-t^2 + 3t = -t + 4\\t^2 - 3t - t + 4 = 0\\t^2 - 4t + 4 = 0\\t_{1,2} = \frac{-(-4) \pm \sqrt{(-4)^2 - 4*1*4}}{2*1} = \frac{4 \pm \sqrt{16 - 16}}{2}\\t_{1,2} = \frac{4}{2} = 2$

The height of the objects are the same after 2 seconds.

8 0

3 years ago

Jaime went to the mall with $42. If he bought a T-shirt and had $18 left, how much did the T-shirt cost Jaime in dollars?

nirvana33 [79]

Answer:

$24

Step-by-step explanation:

You simply do $42-$18

=24

4 0

4 years ago

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