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zalisa [80]
3 years ago
6

The curved part of this figures is a semicircle.

Mathematics
2 answers:
ddd [48]3 years ago
5 0
The answer is 14 + 8.125 units. 
vovangra [49]3 years ago
3 0

First we join the two endpoints of the semicircle and that will be the diameter.

And to find the length of the diameter, we have to use distance formula, with one endpoint (3,2) and the other is (-4,-2) .

SO we get

Diameter = \sqrt{ {-4-3)^2 +(-2-2)^2 } = \sqrt{49+16} = \sqrt 65

Radius is half of diameter, so the radius is

Radius= \frac{ \sqrt{65}}{2}

Formula of area of circle is

Area = \pi r^2

So the area of semicircle is

=(1/2) \pi ( \frac{ \sqrt{65}}{2})^2 = 8.125 \pi \square \units

And the other figure is a triangle, with

Base = 3-(-4)=3+4=7 \\ height = 2-(-2) = 2+2 =4

Area = (1/2)*4*7= 14 square units

Therefore, total area is the sum of area of semicircle and area of triangle,

And we will get

Total \ area= 14 + 8.125 \pi \ units^2

Correct option is the third option .

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Answer:

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How many solutions does the equation have?
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3 years ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
Maru [420]

Parameterize S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k

with 0\le u\le1 and 0\le v\le1.

Take the normal vector to S to be

\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}

6 0
3 years ago
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