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Andrej [43]
2 years ago
14

Help!! I need to pass this class by the end of this week!!!!

Mathematics
2 answers:
timofeeve [1]2 years ago
7 0

Answer:

-3 and 1

Please Mark Brainliest If This Helped!

Kitty [74]2 years ago
6 0
It’s -3 and 1 because the horizontal line is the x-axis and there are only two interceptions
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A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft
lakkis [162]

\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}

\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}

\bf t=\cfrac{3\pm \sqrt{5}}{2}\implies t= \begin{cases} \frac{3+ \sqrt{5}}{2} \approx 2.618\\\\ \frac{3- \sqrt{5}}{2}\approx 0.382 \end{cases}

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2 years ago
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Luba_88 [7]
Refrection across the y-axis changes the function to y = |-x|.
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Shifting the graph by 4 units downward changes the graph to y = 1.5|-x| - 4

Required equation is y = 1.5|-x| - 4
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