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wlad13 [49]
2 years ago
10

Given; 8(x+y) = 48 and y ≠ 2 Proof: x ≠ 4

Mathematics
1 answer:
uranmaximum [27]2 years ago
8 0

Answer:

8(x+y) = 48

x+y=6

when y=2,

x=6-2=4

but y ≠ 2,

so x ≠ 4

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Find the distance between the two points in simplest radical form.<br> (1,8)(6,-4)
dezoksy [38]
Distance in simplest radical form is 13

Or root 169
8 0
3 years ago
Can someone help me with this problem?????
snow_tiger [21]

The perimeter is composed of two straight parts and two semicircles. We can use this to break down the problem.

We can find the straight parts easily. They are given in the problem.

straight parts: 82 x 2 = 164 m

The two semicircles make a circle. We just have to find the circumference of a circle with a diameter of 66 cm to get the length of the semicircles.

semicircles: 2(π)(66/2) = 66(3.14) = 207.24 m

answer: 164+207.24 = 371.24 m

7 0
3 years ago
Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of
GenaCL600 [577]

Close off the hemisphere S by attaching to it the disk D of radius 3 centered at the origin in the plane z=0. By the divergence theorem, we have

\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV

where R is the interior of the joined surfaces S\cup D.

Compute the divergence of \vec F:

\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2

Compute the integral of the divergence over R. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}

So the volume integral is

\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5

From this we need to subtract the contribution of

\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S

that is, the integral of \vec F over the disk, oriented downward. Since z=0 in D, we have

\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k

Parameterize D by

\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath

where 0\le u\le 3 and 0\le v\le2\pi. Take the normal vector to be

\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k

Then taking the dot product of \vec F with the normal vector gives

\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v

So the contribution of integrating \vec F over D is

\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==>  486π/5 - (-81π/4) = 2349π/20

5 0
3 years ago
On the coordinate plane below, Point P is located at (2,-3), and point Q is located at (-4,4).
nadezda [96]

Answer:

9

Step-by-step explanation:

We can use the distance formula

d = sqrt (   ( y2-y1)^2 + ( x2-x1) ^2)

d = sqrt (   ( 4- -3)^2 + ( -4 -2) ^2)

   = sqrt (  ( 7^2  + ( -6)^2)

   = sqrt( 49+ 36)

   = sqrt(85)

      9.219544457

Rounding to the nearest whole number

 = 9

5 0
3 years ago
(08.03)
VMariaS [17]

Answer:

rearrange them properly to get

x-y=7

-3x+y=12

( by elimination method)

x-y = 7

-3x+y=

(x+ –3x) + (–y+y) = (7+12)

-2x+0= 19

x= -9.5

from eqn(i)

x-y=7

-9.5 - y=7

-y=16.5

y= -16.5

8 0
3 years ago
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