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2 years ago

Plz help me with this

Mathematics

2 answers:

8090 [49]2 years ago

4 0

Answer:

8.216 put a line on top of the six because it's repeating

Step-by-step explanation:

gtnhenbr [62]2 years ago

3 0

That is your answer

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Sketch the curve y=x(x-2)^3 showing any stationary points and points of inflection

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3 years ago

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In the diagram, the radius of the outer circle is

JulijaS [17]

the value of x is 8 cm.

<u>Step-by-step explanation:</u>

Correct Question : In the diagram, the radius of the outer circle is 2x cm and the radius of the inside circle is 6 cm. The area of the shaded region is 220π cm2. What is the value of x? Enter your answer in the box.

We have ,

the area of a circle = πr²

the outer circle area = $\pi(2x)^2 =4\pi x^2$

the inside circle area = $\pi (6)^2= 36\pi$

According to Question,

the outer circle area - the inside circle area = he shaded region

⇒ $\pi (4x^2)-36\pi =220\pi$

⇒ $x^2-9 =55$

⇒ $x^2=64$

⇒ $x =\sqrt{64}=8$

Therefore , the value of x is 8 cm.

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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

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

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

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