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2 years ago

An object is launched from a platform.

Mathematics

1 answer:

tankabanditka [31]2 years ago

8 0

Answer:

Step-by-step explanation:

Ground level is where h = 0, so solve the equation ...

h(x) = 0

-5(x -4)^2 +180 = 0 . . . . substitute for h(x)

(x -4)^2 = 36 . . . . . . . . . . divide by -5, add 36

x -4 = 6 . . . . . . . . . . . . . . positive square root*

x = 10 . . . . . . add 4

The object will hit the ground 10 seconds after launch.

_____

* The negative square root also gives an answer that satisfies the equation, but is not in the practical domain. That answer would be x = -2. The equation is only useful for time at and after the launch time: x ≥ 0.

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Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

$\frac{3}{x + 4} - \frac{1}{x + 3} = \frac{x + 9}{(x^2 + 7x + 12)}$

To solve this we need to remove the denominators.

Then we can first multiply both sides by (x + 4) to get:

$\frac{3*(x + 4)}{x + 4} - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

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Now we can multiply both sides by (x + 3)

$3*(x + 3) - \frac{(x + 4)*(x+3)}{x + 3} = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

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Now we can multiply both sides by (x^2 + 7*x + 12)

$(2*x + 5)*(x^2 + 7x + 12) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)$

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$2*x^3 + 19*x^2 + 59*x + 60 = (x^2 + 13*x + 3)*(x + 3)$

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Then we get:

$2*x^3 + 19*x^2 + 59*x + 60 - ( x^3 + 16*x^2 + 42*x + 9) = 0$

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So now we only need to solve this.

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But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.

This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

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Then the other two solutions are given by:

x = ±√17

And neither of these have problems in the denominators, so we can conclude that the solutions are:

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Answer:

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Step-by-step explanation:

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