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Angelina_Jolie [31]
3 years ago
9

The graph h = −16t2 + 25t + 15 models the height and time of a ball that was thrown off a building where h is the height in feet

and t is the time in seconds. At what height was the ball initially thrown off the building?
Mathematics
1 answer:
Tamiku [17]3 years ago
6 0

Answer:

  • 15 ft

Step-by-step explanation:

<u>Given</u>

  • The graph h = −16t² + 25t + 15

<u>Initial point is the value of h at t= 0</u>

  • h = -16*0 + 25*0 + 15
  • h = 15 ft

The ball was thrown at 15 ft height

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GEOMETRY 10TH GRADE NEED HELP ASAP
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Step-by-step explanation:

Given the ratio 5:3

and Points A and B where A is located at (-6, 3), and B is located at (26, -13).

Let point A be (x_{1}, y_{1}), and let point B be (x_{2}, y_{2}).

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(26, -13) → (x_{2}, y_{2}).

Let 5 be n, and 3 be m.

5:3 → n:m

(\frac{nx_{1} + mx_{2}}{n+m}, \frac{ny_{1} + my_{2}}{n + m}).

To solve, just substitute these variables into the expressions of these coordinates to get the answer.

(\frac{nx_{1} + mx_{2}}{n+m}, \frac{ny_{1} + my_{2}}{n + m}) →

(\frac{(5)x_{1} + (3)x_{2}}{(5)+(3)}, \frac{(5)y_{1} + (3)y_{2}}{(5) + (3)}) →

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(\frac{78 – 30}{8}, \frac{15 – 39}{8}) →

(\frac{48}{8}, \frac{-24}{8})

→

(\frac{6}{1}, \frac{-3}{1})

→

(6, -3).

Thus the coordinates of B are:

\boxed{(6, -3)}

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