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mote1985 [20]
3 years ago
14

The stream of water from a fountain follows a parabolic path. The stream reaches a maximum height of 12 feet, represented by a v

ertex of (8,12)​ , and lands 16 feet from the water jet, represented by (16,0)​ . Write a function in vertex form that models the path of the stream
Mathematics
2 answers:
tia_tia [17]3 years ago
8 0

Answer:

y=-3/16(x-8)^2+12

Step-by-step explanation:

Refer to the vertex form equation for a parabola:

y=a(x-h)^2+k where (h,k) is the vertex.

Therefore, we have y=a(x-8)^2+12 as our equation so far. If we plug in (16,0) we can find a:

0=a(16-8)^2+12

0=64a+12

-12=64a

-12/64=a

-3/16=a

Therefore, your final equation is y=-3/16(x-8)^2+12

Stels [109]3 years ago
7 0

Answer:

y = (-5.33)(x - 8)^2 + 12

Step-by-step explanation:

The graph is a parabola which opens down and has its vertex at (8, 12).  We can immediately write y = a(x - 8)^2 + 12, knowing that if x = 8, y = 12.  To find the coefficient '1', substitute 16 for x and 0 for y:

0 = a(16 - 8)^2 + 12, or

-64a = 12, or a = -64/12 = 5.33

Then the desired function is:

y = (-5.33)(x - 8)^2 + 12

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anyanavicka [17]

Answer:

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liubo4ka [24]

Answer:

Step-by-step explanation:

\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a}  \qquad  f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a}  \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}}  }{\dfrac{1}{2\sqrt{x}} }\\\\

\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}  =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\

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