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3 years ago

Algebra help needed

Mathematics

1 answer:

lesantik [10]3 years ago

7 0

Answer:

The ball will reach the ground when h=0.

Set the equation equal to 0.

$0=-8(2 t-4)(t+1)$

$t=-1,2$

Step-by-step explanation:

The height of a golf ball after it has been hit from the top of a hill can be modeled by the equation

$h=-8(2 t-4)(t+1)$ ,

where h is the height in feet and t is time in seconds.

When ball will reach the ground the distance between ball and ground is 0. So, the ball will reach the ground when h=0.

To find the time it takes for the ball to reach the ground, set the equation equal to 0 and use the Zero Product Property to solve for $t$.

$h=-8(2 t-4)(t+1)$

$\Rightarrow 0=-8(2 t-4)(t+1)$

Solve the equation using the Zero Product Property.

$2t-4=0\Rightarrow t=\dfrac{4}{2}=2$

$t+1=0\Rightarrow t=-1$

So, $t=-1,2$

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

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vazorg [7]

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3 years ago

F(x, y, z) = z tan−1(y2)i + z3 ln(x2 + 3)j + zk. find the flux of f across s, the part of the paraboloid x2 + y2 + z = 18 that l

Cerrena [4.2K]

$\mathbf F(x,y,z)=z\tan^{-1}(y^2)\,\mathbf i+z^3\ln(x^2+3)\,\mathbf j+z\,\mathbf k$
$\implies\mathrm{div}\mathbf F(x,y,z)=0+0+1=1$

By the divergence theorem, the flux of $\mathbf F$ across the *closed* surface $\mathcal S$ combined with the plane $z=2$ is given by a volume integral over the closed region:

$\displaystyle\iint_{\mathcal S}\mathbf F\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}\nabla\cdot\mathbf F\,\mathrm dV$

So in fact, to find the flux over $\mathcal S$ alone, we'll need to subtract the flux of $\mathbf F$ over the planar portion, oriented outward. First, compute the volume integral by converting to cylindrical coordinates:

$x^2+y^2+z=18$
$z=2\implies x^2+y^2=16\implies r^2=16\implies r=4$

$\displaystyle\iiint_{\mathcal R}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=2}^{z=18-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=128\pi$

If the surface does actually contain $z=2$ , then you can stop here; otherwise, continue.

Now, parameterize the part of the *closed* surface in $z=2$ by

$\mathbf s(r,\theta)=r\cos\theta\,\mathbf i+r\sin\theta\,\mathbf j+2\,\mathbf k$

where $0\le r\le4$ and $0\le\theta\le2\pi$ . We get a surface element

$\mathrm d\mathbf S=(\mathbf s_r\times\mathbf s_\theta)\,\mathrm dr\,\mathrm d\theta=(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta$

We don't need to worry about the first two components of

and so the surface integral over this region is

$\displaystyle\iint_{z=2\,\land\,x^2+y^2\le16}\mathbf F\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}2r\,\mathrm dr\,\mathrm d\theta=32\pi$

Then the total flux over $\mathcal S$ alone is $(128-32)\pi=96\pi$ .

4 0

3 years ago