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Zigmanuir [339]
3 years ago
14

Find all points on the y-axis that are a distance 6 from p(4, 7).

Mathematics
1 answer:
natali 33 [55]3 years ago
8 0

\sqrt{16 +  {(7 - y)}^{2} } = 6 \\ 16 +  {(7 - y)}^{2}   = 36 \\   {(7 - y)}^{2} = 20
7 - y =  \sqrt{20}  >  > y = 7 -  \sqrt{20}  \\ 7 - y =  -  \sqrt{20}  >  > y = 7 +  \sqrt{20}
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Multiply & Simplify.
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HELP!!!
Naddika [18.5K]

Check the picture below, so it reaches the maximum height at the vertex, let's check where that is

h(t)=64t-16t^2+0 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 64}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (64)^2}{4(-16)}\right)\implies \stackrel{maximum~height}{(2~~,~~\stackrel{\downarrow }{64})}

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