Need help asap!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

zvonat [6]3 years ago

8 0

The second one would be 4 inches.

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PLEASE HELP, THIS IS REALLY IMPORTANT. PLEASE ANSWER ASAP.

Assoli18 [71]

Yes the diagonals of a parallelogram have the same midpoint since they ... of the intersection of the diagonals of parallelogram AB CD given the vertex points ... If a parallelogram is a rhombus then its diagonals are? , statement 2 is the answer

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3 years ago

What is the factored form of x^12y^18+1

Black_prince [1.1K]

Answer: The required factored form of the given expression is

$(x^4y^6+1)(x^8y^{12}-x^4y^6+1).$

Step-by-step explanation: We are given to find the factored form of the following algebraic expression:

$E=x^{12}y^{18}+1.$

We will be using the following formula:

$a^3+b^3=(a+b)(a^2+ab+b^2).$

Now, we have

$E\\\\=x^{12}y^{18}+1\\\\=(x^4y^6)^3+1^3\\\\=(x^4y^6+1)\{(x^4y^6)^2-x^4y^6\times1+1^2\}\\\\=(x^4y^6+1)(x^8y^{12}-x^4y^6+1).$

Thus, the required factored form of the given expression is

$(x^4y^6+1)(x^8y^{12}-x^4y^6+1).$

4 0

3 years ago

Read 2 more answers

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

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$