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1 year ago

Please I don't understand help me

Mathematics

1 answer:

Murrr4er [49]1 year ago

6 0

Answer:

6.2 seconds

Step-by-step explanation:

To translate into math, the question is asking you "for what approximate t value does h = 0?" In order to solve, you can plug in 0 for h and solve for t.

0 = -16t^2 + 620

-620 = -16t^2

620 = 16t^2

38.75 = t^2

$\sqrt{38.75}$ = t

6.2 ~~ t (rounding to the nearest tenth)

Don't forget your units!

Let me know if you need anything explained in more detail; I'm happy to help.

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$\boxed{x_{1}=-15} \\ \\ \boxed{x_{2}=-17}$

<h2>Explanation:</h2>

In this case, we have the following equation:

$x^2+32x+256=1$

But we can write this equation as:

$x^2+32x+256=1 \\ \\ Subtract \ -1 \ from\ both \ sides: \\ \\ x^2+32x+256-1=1-1 \\ \\ x^2+32x+255=0$

So this final result is a quadratic equation written in Standard Form ( $ax^2+bx+c=0$ ). We need to find the solutions to this equations, so let's use quadratic formula:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ a=1 \\ b=32 \\ c=255 \\ \\ \\ x=\frac{-32 \pm \sqrt{(32)^2-4(1)(255)}}{2(1)} \\ \\ x=\frac{-32 \pm \sqrt{1024-1020}}{2} \\ \\ x=\frac{-32 \pm \sqrt{4}}{2} \\ \\ x=\frac{-32 \pm 2}{2} \\ \\ Finally, \ two \ solutions: \\ \\ \boxed{x_{1}=-15} \\ \\ \boxed{x_{2}=-17}$

<h2>Learn more:</h2>

Quadratic Equations: brainly.com/question/10278062

#LearnWithBrainly

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