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2 years ago

(-8X^8)(8x^8) What is the answer to this?

Mathematics

1 answer:

jonny [76]2 years ago

4 0

Answer:

If you are trying to simplify this, the answer would be: $-64x^{10}$

Hope this helps! If not, feel free to comment below and I'll see what else I can do to help. Thanks and good luck!

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A teenager who is 5 feet tall throws an object into the air. The quadratic function LaTeX: f\left(x\right)=-16x^2+64x+5f ( x ) =

tia_tia [17]

Answer:

At approximately x = 0.08 and x = 3.92.

Step-by-step explanation:

The height of the ball is modeled by the function:

$f(x)=-16x^2+64x+5$

Where f(x) is the height after x seconds.

We want to determine the time(s) when the ball is 10 feet in the air.

Therefore, we will set the function equal to 10 and solve for x:

$10=-16x^2+64x+5$

Subtracting 10 from both sides:

$-16x^2+64x-5=0$

For simplicity, divide both sides by -1:

$16x^2-64x+5=0$

We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute:

$\displaystyle x=\frac{-(-64)\pm\sqrt{(-64)^2-4(16)(5)}}{2(16)}$

Evaluate:

$\displaystyle x=\frac{64\pm\sqrt{3776}}{32}$

Simplify the square root:

$\sqrt{3776}=\sqrt{64\cdot 59}=8\sqrt{59}$

Therefore:

$\displaystyle x=\frac{64\pm8\sqrt{59}}{32}$

Simplify:

$\displaystyle x=\frac{8\pm\sqrt{59}}{4}$

Approximate:

$\displaystyle x=\frac{8+\sqrt{59}}{4}\approx 3.92\text{ and } x=\frac{8-\sqrt{59}}{4}\approx0.08$

Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.

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