The amount of water in a barrel decreased is 9 5/8 pints in 7 weeks

Use the distributive property to simplify the expression. 8(3 + 4) = 24 +

Marina CMI [18]

Using the distributive property is done by multiplying the factor (outside the parentheses) to each term within it. In this case, 8(3 + 4), we multiply the 8 by 3, and then add it to the product of 8 x 4:
8(3 + 4) = 8 x 3 + 8 x 4 = 24 + 32
If we need to simplify further, this adds up to 56.

9 0

2 years ago

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What is the horizontal asymptote of f(x)=-2x/x+1<br> A.-2<br> B.-1<br> C.0<br> D.1

Mrrafil [7]

A horizontal asymptote of a function f(x) is given by y = lim f(x) as x --> ∞ and x --> –∞. In this case,
$\lim_{x \to \infty} \frac{-2x}{x+1} = \lim_{x \to \infty} \frac{-2x}{x(1+ \frac{1}{x} )}\\
=\lim_{x \to \infty} \frac{-2}{1+ \frac{1}{x} }\\
= -2
$

Thus, the horizontal asymptote of f(x) is y = –2.

7 0

2 years ago

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$