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Ymorist [56]
3 years ago
8

When an object is thrown from a height of 20 meters with an initial speed of 20 meters per second, its height ℎ above the ground

(measured in meters) at any given time (measured in seconds) can be modeled by the quadratic equation: ℎ=−52+20+20 If it is known that the object will reach a maximum height of 40 meters, find the time at which this happens. The object will reach its maximum height of 40 meters at seconds. How long will it take for the object to hit the ground? The object will hit the ground at seconds.
Mathematics
1 answer:
KatRina [158]3 years ago
6 0

Answer:

5.85 seconds

Step-by-step explanation:

The equation;

h = -5t² + vt + h

where;

v = initial upward velocity in meter per seconds

h = height above ground level when the object is thrown

t = time of flight in seconds

h = -5t² + 20t + 20

h = 0

⇒ -5t² + 20t + 20= 0

-t² + 5t + 5 = 0

t² - 5t - 5 = 0

a = 1

b = -5

c = -5

Using the quadratic formula:

\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

\dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-5)}}{2*1}

\dfrac{5 \pm \sqrt{25 +20}}{2}

\dfrac{5 \pm \sqrt{45}}{2}

\dfrac{5 \pm 6.7082}{2}

\dfrac{5 + 6.7082}{2} \ \  OR  \ \ \dfrac{5 - 6.7082}{2}

t = 5.8541    OR t =  -0.8541

Since t can only be positive; t ≅ 5.85 sec

Then;

After 5.85 seconds, the object will hit the ground.

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Find the perimeter of the shaded region. Round your answer to the nearest hundredth.
nataly862011 [7]

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59.16  units

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The breadth of the rectangle, B = 6

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What numbers are zeros of g(x) = x^2 - 2x - 4? take your time if you need to!
ipn [44]

Answer:

x = 1 + \sqrt5, x = 1 - \sqrt5

Step-by-step explanation:

Hello!

We can solve the quadratic by using the quadratic formula.

Standard form of a quadratic: ax^2 + bx + c = 0

Quadratic Formula: x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}

Given our Equation: g(x) = x^2 - 2x - 4

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  • b = -2
  • c = -4

Plug the values into the equation and solve.

<h3>Solve</h3>
  • x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}
  • x = \frac{-(-2)\pm\sqrt{(-2)^2 - 4(1)(-4)}}{2(1)}
  • x = \frac{2\pm\sqrt{4 +16}}{2}
  • x = \frac{2\pm\sqrt{20}}{2}
  • x = \frac{2\pm\sqrt{4 * 5}}{2}
  • x = \frac{2\pm(\sqrt4 * \sqrt5)}{2}
  • x = \frac{2\pm2\sqrt5}{2}
  • x = 1 + \sqrt5, x = 1 - \sqrt5
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2 years ago
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