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Shtirlitz [24]
3 years ago
12

The angle is 1/12 of the circle how many degrees is it

Mathematics
1 answer:
Ket [755]3 years ago
7 0
A circle is 360 degrees, so 360/12 = 30 degrees.
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Which underfined term is used to define an angle?
AleksAgata [21]

From the term line the term angle can be defined because when two lines intersect at a point an angle is formed. An angle is defined as a figure which is formed when two lines intersect each other at a point. ... Therefore, the undefined term which is used to define the term an angle is line.

7 0
3 years ago
What is the solution to 7 × p = -56? A. -49 B. -8 C. 8 D. 49
Serga [27]

Answer:

-8

Step-by-step explanation:

Hello!

What we do to one side of the equation we do to the other side

7 * p = -56

Divide both sides by 7

p = -8

The answer is -8

Hope this helps!

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3 years ago
How many times dose 3 go in to 81
puteri [66]
3 goes into 81, 27 times.
6 0
3 years ago
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Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi<span />
4 0
3 years ago
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SIZIF [17.4K]
It is C  parallelogram and rumbus

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3 years ago
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