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Law Incorporation [45]
3 years ago
8

Part of a cheerleading routine is throwing a flyer straight up and catching her on the way down. The flyer begins this move with

her center of gravity 4 feet above the ground. She is thrown with an initial vertical velocity of 30 feet per second.
Mathematics
2 answers:
o-na [289]3 years ago
6 0
The formula would be
y=4 + 30t -16t^2
Her maximum height is when the velocity is zero.  so V = 30 -32t  0 = 30 -32t  t=15/16 seconds.


Eva8 [605]3 years ago
5 0
What is the rest of the question?
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Step-by-step explanation:

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3 years ago
(Sorry forgot to put picture)
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2 years ago
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3 years ago
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lisov135 [29]

In the inverse we replace the place of x by y .

g(x) =  \sqrt[3]{x}  - 3 \\  \\ x =  \sqrt[3]{y}  - 3 \\  \sqrt[3]{y}  = x + 3 \\ y =  {(x + 3)}^{3}  \\ y =  {(x + 3)}^{2} (x + 3) \\ y =  ({x}^{2}  + 6x + 9)(x + 3) \\ \\  y =  {x}^{3}  + 6 {x}^{2}  + 9x + 3 {x}^{2}  + 18x + 27 \\  \\ y =  {x}^{3}  + 9 {x}^{2}  + 27x + 27 \\   \\ \\ g(x)^{ - 1}  =  {x}^{3}  + 9 {x}^{2}  + 27x + 27

I hope I helped you^_^

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