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3 years ago

3. PENTAGON The Pentagon is a five-sided

1 answer:

5 0

The answer is 4605 feet since 921 • 5 is 4605. There are 5 sides and each side 921 so 921 • 5 = 4605 would be the answer

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Find the Two Values that x can have if X^2 -10=26

Zolol [24]

Answer:

x1= -6 x2=6

Step-by-step explanation:

x^2-10 = 26

x^2 = 26 +10

x^2 = 36

x = sqrt(36)

x = -6 or x = 6 because both 6 and -6 squared are equal to 36

3 0

3 years ago

Read 2 more answers

Use the front end addition strategy to add

ruslelena [56]

I cant see anything there needs to be more to the question

4 0

3 years ago

ANSWER AS MANY THROUGH 1 and 9 FOR 15 POINTS

Tamiku [17]

Number 9 is 1.52 because you divide 15.3 by 10 it will be 1 .52

7 0

3 years ago

Solve for x. 3(x + 2) + 4(x - 5) = 10 x = 36/7 x = 24/7 x = 16/7 x = 9/7

IRISSAK [1]

3(x + 2) + 4(x - 5) = 10

Use the distributive property:

3x + 6 + 4x - 20 = 10

Combine like terms:

7x -14 = 10

Add 14 to both sides:

7x = 24

Divide both sides by 7:

x = 24/7

5 0

3 years ago

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