How do you add positives and negatives ?

A rectagle has a height of 5x and a width of x+2<br><br>area=???

astra-53 [7]

Answer: $Area=5x^2+10x$

Step-by-step explanation:

The area of a rectangle:

$a=h$ × $w$

$a=5x(x+2)$

$a=5x$ × $x+5x$ × $2$

$a=5x^2+10x$

-Result:

$a=5x^2+10x$

5 0

3 years ago

Combine like terms - 7 -3(4-2x)-10x

Shtirlitz [24]

Answer:

The simplified form would ve -19 - 4x

Step-by-step explanation:

Before we combine anything, we first must expand the expression by distributing the -3 to the terms in the parenthesis.

-7 - 3(4 - 2x) - 10x

-7 - 12 + 6x - 10x

Now we can combine the terms with and without variables.

-19 - 4x

8 0

3 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

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

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

In how many ways can you choose 2 side dishes out of 8

Alina [70]

The first choice can be any one of the 8 side dishes.
For each of these . . .
The 2nd choice can be any one of the remaining 7.

Total number of ways to pick 2 out of 8 = (8 x 7) = 56 ways .

BUT ...

That doesn't mean you can get 56 different sets of 2 side dishes.

For each different pair, there are 2 ways to choose them . . .
(first A then B), and (first B then A). Either way, you wind up with (A and B).

So yes, there are 56 different 'WAYS' to choose 2 out of 8.
But there are only 28 different possible results, and 2 'ways'
to get each result.

4 0

3 years ago

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Which is a better deal: five tulips for $3.00 or seven tulips for $4.00?

guajiro [1.7K]

Answer:

Step-by-step explanation:

7 divided by $4.00 = $1.75/per tulip

5 divided by $3.00 = $1.67/per tulip

Five tulips for 3.00 dollars

5 0

3 years ago

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