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zimovet [89]
2 years ago
14

an outfielder throws a ball to home plate. the balls path is modeled by the graph of y=x/5-x^2/1000 where x represents the horiz

ontal distance of the ball from the outfield and y represents the balls vertical distance above the ground. what are the coordinates of vertex of the ball's path?​
Mathematics
1 answer:
saul85 [17]2 years ago
4 0

The coordinates of the vertex for the quadratic function are given as follows: (100,10).

<h3>What is the vertex of a quadratic equation?</h3>

A quadratic equation is modeled by:

y = ax^2 + bx + c

The vertex is given by:

(x_v, y_v)

In which:

  • x_v = -\frac{b}{2a}
  • y_v = -\frac{b^2 - 4ac}{4a}

In this problem, the function is given as follows:

y = -x²/1000 + x/5

y = -0.001x² + 0.2x.

Hence the coefficients are a = -0.001, b = 0.2, c = 0 and the coordinates of the vertex are given as follows:

  • x_v = -\frac{0.2}{2(-0.001)} = 100
  • y_v = -\frac{(0.2)^2}{4(-0.001)} = 10

More can be learned about the vertex of a quadratic function at brainly.com/question/24737967

#SPJ1

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<u>Answer</u><u>:</u>

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Answer:

Your answer is <em>6x</em>

Step-by-step explanation:

Given,

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A = 6x    ans.

Hope its helpful!

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4 0
2 years ago
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.

\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

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