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Artyom0805 [142]

3 years ago

11

A ball is thrown in the air from a ledge. It's height in feet represented by f(x)=16(x^2-6x-7), where x is the number of seconds

since the ball has been thrown. The height of the ball is 0 feet when it hits the ground. How many seconds does it take the ball to reach the ground?

2 answers:

LekaFEV [45]3 years ago

7 0

Answer:

Step-by-step explanation:

Since we know that the height is 0, we can figure out how long it took the ball to reach the ground by setting $f(x) = 0$ and solving for $x$ :

$f(x) = 16(x^{2} - 6x - 7)$

$0 = 16(x^{2} - 6x - 7)$

$0 = x^{2} - 6x - 7$

$0 = (x - 7)(x + 1)$

$x = -1, 7$

Because time can only be positive, the answer is 7 seconds.

lutik1710 [3]3 years ago

4 0

Answer:

7 seconds.

Step-by-step explanation:

height h = 16(x^2-6x-7) = 0

x^2 - 6x - 7 = 0

(x - 7)(x + 1) = 0

x = 7 seconds (we ignore the negative).

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3 years ago

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Find the complex fourth roots of 81(cos(3pi/8) + i sin(3pi/8))

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By using <span>De Moivre's theorem:
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</span>
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The angle of the given complex number = $\frac{3 \pi}{8}$
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The angle of the second root = $\frac{3\pi}{32} + \frac{\pi}{2} = \frac{19\pi}{32}$
The angle of the third root = $\frac{19\pi}{32} + \frac{\pi}{2} = \frac{35\pi}{32}$
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The fourth root = $z_{4} = 3 ( cos \ \frac{51\pi}{32} + i \ sin \ \frac{51\pi}{32})$

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