## Related questions with answers

In firefighting, a good water stream can be modeled by y = -0.003x$^2$ + 0.62x + 3

where x is the water’s horizontal distance traveled (in feet) and y is its corresponding height (in feet). If a firefighter is aiming a good water stream at a building’s window 25 feet above the ground, at what two distances can the firefighter stand from the building?

Solution

VerifiedBased on the problem, to get the two distances, we just need to equate the given equation to $25$ then solve for the value of $x$.We must also manipulate the equation so that it is in the form $x^2+bx=k$ where $b$ and $k$ are any real number.

$\begin{aligned} 25&=-0.003x^2+0.62x+3\\ \dfrac{25}{-0.003}&=-\dfrac{1}{0.003}(-0.003x^2+0.62x+3)\\ -\dfrac{25000}{3}&=x^2-\dfrac{620}{3}x-1000\\ -\dfrac{25000}{3}+1000&=x^2-\dfrac{620}{3}x-1000+1000\\ -\dfrac{22000}{3}& =x^2-\dfrac{620}{3}x \end{aligned}$

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