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4 years ago

Simplify the expression. Show your work. 22 + (3^2 – 4^2)

Mathematics

1 answer:

irga5000 [103]4 years ago

6 0

22 + (3^2 – 4^2)
= 22 + (9 – 16)
= 22 -7
= 15

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Cos xcos y=1/2(sin(x+y)+ sin(x-y))

garri49 [273]

Taking Rhs

1/2 [sin(x+y)+ sin (x-y)]

using sinA + sin B = 2 cos (A+B)/2 cos( A - B)/2

1/2 [ 2 Cos ( x+y+x-y)/2 . Cos (x+y-x+y)/2]

=Cos 2x/2. Cos 2y/2 = Cos x . Cos y

Hence true

7 0

4 years ago

Please Answer ASAP TIMED

otez555 [7]

Answer:

It should be D, friend.

8 0

4 years ago

Read 2 more answers

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

kifflom [539]

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$ , where $R$ is the region enclosed by $S$ . Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$ ) so that $R$ has boundary $S\cup D$ .

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

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

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

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

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

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

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4$

Finally,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}$

6 0

4 years ago

. Evaluate each expression given that X = -2, Y = 8 and Z = 4

Liula [17]

Hello:-
this(*)is counted as multiplication
1. 4*2 + (-2+8)
=8+(6)
=14
2. (-2)*2 + 4
= -4+4
=0
3. 8*4/-2
= -32/2 (as -ve sign cannot be in denominator)
= -16
4. -2*4/8
= -8/8
= -1
5. -2*8/4
= -16/4
= -4
therefore it is a. -4
I hope it will help u !!!!!!!!!

8 0

3 years ago

Abcd is a square. find bc

chubhunter [2.5K]

Slope is equal to 20 2,0

3 0

3 years ago