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2 years ago

5

A rectangular prism has a length of 412 mm, a width of 412 mm, and a height of 6 mm. Sally has a storage container for the prism

that has a volume of 143 mm³.
What is the difference between the volume of the prism and the volume of the storage container?

Enter your answer in the box.

mm³

2 answers:

ryzh [129]2 years ago

8 0

First find the volume of the rectangle prism. Do this by multiplying all three dimensions. 412 x 412 x 6 = 1,018,464.

Subtract 1,018,464 from 143.

You should get 1,018,321 mm^3

(Remember that the units are cubed because you multipled 412 mm x 412 mm x 6 mm which is multiplying not only the numbers, but the mm three times as well.)

Sedaia [141]2 years ago

7 0

Answer:The answer

is 21 1/2

Step-by-step explanation:

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kkurt [141]

Parallel lines have equal slopes.

-x + y = 5

y = x + 5

Slope = 1

y + 5 = 1(x - 2)

y + 5 = x - 2

y = x - 2 - 5

y = x - 7

Answer: Choice A

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3 years ago

Which is greater 50 Hectoliters or 10 kiloliters

Tomtit [17]

Kiloliters are bigger so if they are bigger they have to be kiloliters right>:)
10kL<span> = 100hL, so </span>10kL<span> > </span><span>50hL</span>

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2 years ago

For f(x) = 2x+1 and g(x) = x2 - 7, find (f+ g)(x).

____ [38]

Answer:

the answer is A. it good answer

5 0

2 years 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

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

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

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

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

8 0

3 years ago

The model represents an equation. What value of x makes the equation true?

ch4aika [34]

Let's build the equation counting how many x's and 1's are there on each side.

On the left hand side we have 5x's and 8 1's, for a total of $5x+8$

On the left hand side we have 3x's and 10 1's, for a total of $3x+10$

So, the equation we want to solve is

$5x+8 = 3x+10$

Subtract 3x from both sides:

$2x+8 = 10$

Subtract 8 from both sides:

$2x=2$

Divide both sides by 2:

$x=1$

8 0

3 years ago