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vlada-n [284]
3 years ago
10

In the problem 10-4 = 6 what is the correct term for the number for

Mathematics
2 answers:
JulijaS [17]3 years ago
8 0
In 10 - 4 = 6

6 is the difference
10 is the minuend
4 is the subtrahend
Damm [24]3 years ago
8 0
The number 4, in 10-4=6, is called the subtrahend.
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HELP I NEED HELP FINDING THE AREA!!!!!!!!!!!!!!!!<br> (brainlyest)
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Answer:

27 units squared?

Step-by-step explanation:

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3 years ago
What is 4 divide 120<br> Show me that<br> divide by 120
Lena [83]

Answer: 120 divided by 4 = 30

Step-by-step explanation:

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3 years ago
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What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po
emmasim [6.3K]
\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA
\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k
\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j

The cross product is

\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j

So, the flux is given by

\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA
\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du
\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv
\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv

where t=-\sin v in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi
5 0
3 years ago
Help me please, it's easy kinda
trasher [3.6K]
Karl ate 5/6 of a small bag of popcorn .
7 0
3 years ago
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John has a storage bin in the shape of a rectangular prism the storage bed is still in the and 1/2 ft long 2 ft wide and 2 ft ta
Phoenix [80]

We know, volume of rectangular prism is given by :

V = lbh\\\\V=\dfrac{1}{2}\times 2\times 2\ ft^2\\\\V=2\ ft^2

Volume of cubic box :

v=a^3\\\\v=(\dfrac{1}{2})^3\ ft^3\\\\v=\dfrac{1}{8}\ ft^3

Number of cubes can be filled :

N=\dfrac{V}{v}\\\\N=\dfrac{2}{\dfrac{1}{8}}\\\\N=16

Therefore, 16 cubes can be put in the bin.

Hence, this is the required solution.

8 0
4 years ago
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