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2 years ago

For the equation y=-3x+2, determine the value of y when x=-3

Mathematics

1 answer:

ludmilkaskok [199]2 years ago

6 0

Answer:

y=11

Step-by-step explanation:

y=-3(-3)+2

y=-3*-3+2

y=9+2

y=11

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satela [25.4K]

Answer:

Step-by-step explanation:

The anwser is C

the inverse of 5x is x/5 and the inverse of +4 is -4 so the answer would be x-4/5

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4 years ago

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2)1694 what is the answer

8_murik_8 [283]

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0.00118063754

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Step-by-step explanation:

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3 years ago

The volume of a sphere is a function of its radius. V equals 4 over 3 πr cubed . Evaluate the function for the volume of a volle

Nataly_w [17]

Answer:

The answer to your questin is: V = 5276.7 cm³

Step-by-step explanation:

Data

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3 years ago

Write an equation of the line perpendicular to the line y=-8/3x+2 and containing the point (32,6) write the answer in slope inte

miss Akunina [59]

Y = -8/3x + 2...slope here is -8/3. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal of the slope -8/3, u just flip the slope and change the sign. So our perpendicular line will have a slope of 3/8.

y = mx + b
slope(m) = 3/8
(32,6)...x = 32 and y = 6
now we sub and find b, the y int
6 = 3/8(32) + b
6 = 12 + b
6 - 12 = b
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so ur perpendicular equation is : y = 3/8x - 6

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3 years ago

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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

For the equation y=-3x+2, determine the value of y when x=-3​

For the equation y=-3x+2, determine the value of y when x=-3