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Sophie [7]
3 years ago
15

Please Help                                                                                                                    

  Which is true about the volume or surface area of these prisms? A. The volume of B is greater than the volume of A. B. The surface area of A is greater than the surface area of B. C. The surface area of B is greater than the surface area of A. D. The volume of A is greater than the volume of B.

Mathematics
1 answer:
Elina [12.6K]3 years ago
3 0
They have the same volume, so it's not a or d, but the surface area of a is 104, and the surface area of b is 94, so the answer is B.
Hope i helped! i also have this question in k12.<span />
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The sum of 2112 + 2212 + 2312 when divided by 4, leaves no remainder. What will be the remainder when the sum of 2113 + 2213 + 2
trasher [3.6K]

Step-by-step explanation:

2112 + 2212 + 2312 / 4 leaves no remainder, as already the individual numbers could be divided by 4 without remainder.

2113 = 2112 + 1

2213 = 2212 + 1

so, we get a remainder of +1 and +1 for these 2 numbers, so we get in total a remainder of +2.

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What is 2xy-3yz+5+2yz-xy
melomori [17]

To simplify this expression, we are going to combine like terms.


First, we can see that we have two terms with the variables xy being multiplied together, 2xy and -xy. Together, these add to xy, so our expression is now:

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4 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
ANSWER PLZ QUICK AND FAST
kherson [118]

Half of n, is option D, n/2.

This is because if you divide something by 2, you are dividing it in half.

Hope this helps!

3 0
3 years ago
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