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3 years ago

Simplify 2m - [n - (m - 2n)].

Mathematics

2 answers:

marta [7]3 years ago

8 0

The answer to the problem is 3m-3n

Maurinko [17]3 years ago

5 0

Answer: the correct answer is 3m-3n

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True or false: Polar equations can describe graphs as functions, even when their equations in the rectangular coordinate system

pshichka [43]

The statement above is true. Polar equations indeed can describe graphs as functions, even if when the equations in the rectangular coordinate system are not one of the functions. Polar equations can be graphed accurately using hands by using the Polar Coordinate System.

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3 years ago

Read 2 more answers

How to find the arc length (geometry)

andrew11 [14]

Answer:

94.2

Step-by-step explanation:

Arc length is basically circumference.

Circumference formula: 2πr

All we need is the radius.

Radius = r

15 x 2 = 30 (diameter)

Then we are going to multiply by pi.

Arc length = 30π OR 94.2

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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

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