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2 years ago

Is this problem convergent or divergent?

Mathematics

1 answer:

Svetllana [295]2 years ago

6 0

Answer:

Its divergent

Step-by-step explanation:

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Find the slope of a line that passes through the following points (-1,-5), (3,-3)

svet-max [94.6K]

Answer:

$m=\frac{1}{2}$

Step-by-step explanation:

defition of slope(m) of a line:

$m=\frac{Y_b-Y_a}{X_b-X_a}$

A=(-1,-5)=(Xa,Ya) then Xa= -1 and Ya= -5

B=(3,-3)=(Xb,Yb) then Xb=3 and Yb= -3

$m=\frac{-3-(-5)}{3-(-1)} \\\\m=\frac{-3+5}{3+1} \\\\m=\frac{2}{4} \\\\m=\frac{1}{2}$

8 0

2 years ago

Find the solution set.

kow [346]

Answer: 5, -5

Step-by-step explanation:

There are only 2 square roots of 25: plus and minus 5.

7 0

3 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

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