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3 years ago

Using the number line shown, subtract 1 2/6 from 4 4/6.

Mathematics

1 answer:

Soloha48 [4]3 years ago

4 0

Hey there!

$\bold{1\frac{2}{6}-4\frac{4}{6}}$

$\bold{1\frac{2}{6}=1\frac{1}{3}=\frac{4}{3}}$
$\bold{4\frac{4}{6}=4\frac{2}{3}=\frac{14}{3}}$
Problem becomes: $\bold{\frac{4}{3}-\frac{14}{3}}$
To solve our new problem $\uparrow;\downarrow$
$\bold{4-14=-10}$
The $\bold{3's}$ stay the same because the same
$\bold{\rightarrow=\frac{-10}{3}}$
$\boxed{\boxed{\bold{Answer:\frac{-10}{3} (or)=-3\frac{1}{3}}}}$

Good luck on your assignment and enjoy your day!

~ $\bold{LoveYourselfFirst:)}$

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2 years ago

Of the following three statements, choose the best option and write a viable argument showing how it is reflected by the cost fu

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A viable argument showing how it is reflected by the Cost Function for Wichita’s factory is; The cost of production per unit of item produced is 1$ as long as total units produced is less than 2500. If more than 2500 units are produced than the cost per unit actually comes down to 0.8$per unit.

<h3>How to Interpret Cost Functions?</h3>

From the given options, the best statement is;

The plant’s production processes are performed primarily by robots that are able to work longer hours, when needed, at no additional cost.

A cost function is a formula used to predict the cost that will be experienced at a certain activity level. Thus, the cost function for the best option is;

The cost of production per unit of item produced is 1$ as long as total units produced is less than 2500. If more than 2500 units are produced than the cost per unit actually comes down to 0.8$per unit.

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___ The electricity contract with the utility company is structured so that higher daily energy usage is charged at a lower rate.

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1 year ago

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

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