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lesya692 [45]
3 years ago
10

does this inequality sometimes, always, or never true for any value of the variable. 12s. > 10s

Mathematics
1 answer:
Vadim26 [7]3 years ago
6 0
If you take away the S variable, the inequality is
12>10
Since 10 will always be less than 12 the answer is:
The inequality is always true for any value of the variable
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satela [25.4K]

Answer:

i need this too so any body that knows it plz help

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3 years ago
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What is the solution to the system -2x + y = 6 and y= 2x - 3?
Irina18 [472]
<span>
replace y= 2x - 3 into </span>-2x + y = 6 

-2x + 2x - 3  = 6 
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3 years ago
Bill's chocolate bar is 69% cocoa. If the weight of the chocolate bar is 83 grams, how many grams of cocoa does it contain? Roun
Maurinko [17]

Answer:

Bill's bar has 57.3 grams of cocoa.

Step-by-step explanation:

The math part that your question is asking is: What is 69% of 83?

In order to use the percent 69% in a calculation, you need to change it to a decimal.

69% is 0.69

You move the decimal to the left, two places.

69% is 69.% If you don't see a decimal to start with and it is a whole number, then the decimal is starting on the right.

Move it two places to the left, you get

.69

Now MULTIPLY.

.69 × 83

= 57.27

ROUND your answer to the nearest tenth. That means only one digit to the right of the decimal.

57.27 rounds to 57.3

Bill's bar has 57.3 grams of cocoa.

8 0
2 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
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}}

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4 years ago
Classify the triangle based on side lengths 25, 15, 9.
madreJ [45]
Im pretty sure its obtuse
7 0
3 years ago
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