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3 years ago

IM BEGGING YOU, IF YOU SEE THIS PLEASE TRY AND ANSWER IT!!! BRAINLIEST WOULD BE ALL YOURS!

Mathematics

1 answer:

Oksana_A [137]3 years ago

5 0

Answer:

$\displaystyle a_n=8+9(n-1)$

<em>Correct option: C]</em>

Step-by-step explanation:

<u>Arithmetic Sequences</u>

Each term in an arithmetic sequence is obtained as the previous term plus a constant number called the common difference. The general term is

$\displaystyle a_n=a_1+(n-1).r$

We are given this information

$\displaystyle a_1=8\ ,\ a_7=62$

Replacing those values in the formula

$\displaystyle 62=8+(7-1).r$

Solving for r

$\displaystyle r=\frac{62-8}{6}=\frac{54}{6}=9$

$\displaystyle r=9$

The general term is, then

$\displaystyle a_n=8+(n-1)9$

Or equivalently

$\displaystyle a_n=8+9(n-1)$

Correct option: C]

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katen-ka-za [31]

Hi there
let me help you out

The answer is 30
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3 years ago

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2 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$