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attashe74 [19]
2 years ago
11

The length of a rectangle is 3 times the width. If

Mathematics
2 answers:
Margaret [11]2 years ago
8 0

Answer:

Length = 27 metresWidth = 9 meters

Step-by-step explanation:

VikaD [51]2 years ago
8 0
I’m not sure but yes
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Find the value of this expression if x= -6 and y=1. xy^2/-7
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How to solve this kind of question?
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<span>96 degrees Looking at the diagram, you have a regular pentagon on top and a regular hexagon on the bottom. Towards the right of those figures, a side is extended to create an irregularly shaped quadrilateral. And you want to fine the value of the congruent angle to the furthermost interior angle. So let's start. Each interior angle of the pentagon has a value of 108. The supplementary angle will be 180 - 108 = 72. So one of the interior angles of the quadrilateral will be 72. From the hexagon, each interior angle is 120 degrees. So the supplementary angle will be 180-120 = 60 degrees. That's another interior angle of the quadrilateral. The 3rd interior angle of the quadrilateral will be 360-108-120 = 132 degrees. So we now have 3 of the interior angles which are 72, 60, and 132. Since all the interior angles will add up to 360, the 4th angle will be 360 - 72 - 60 - 132 = 96 degrees. And since x is the opposite (or congruent) angle to this 4th interior angle, it too has the value of 96 degrees.</span>
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Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t
pashok25 [27]

Direct computation:

Parameterize the top part of the circle x^2+y^2=1 by

\vec r(t)=(x(t),y(t))=(\cos t,\sin t)

with 0\le t\le\pi, and the line segment by

\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)

with 0\le t\le1. Then

\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)

=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt

=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}

Using the fundamental theorem of calculus:

The integral can be written as

\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}

If there happens to be a scalar function f such that \vec F=\nabla f, then \vec F is conservative and the integral is path-independent, so we only need to worry about the value of f at the path's endpoints.

This requires

\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)

\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C

So we have

f(x,y)=-\cos x+\sin y+C

which means \vec F is indeed conservative. By the fundamental theorem, we have

\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}

3 0
3 years ago
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