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3 years ago

For the given function, x can have what value?

Mathematics

1 answer:

Lyrx [107]3 years ago

5 0

Answer:

x cannot be -4,-3, or 13

x can be anything else

Step-by-step explanation:

There are infinitely many values x can take where the relation above will be a function.

For it to be a function, you just need to make sure each x is only assigned one y value.

So x couldn't be -4 because it would by assigned to y=2 and y=0.

x couldn't be -3 because it would be assigned to y=1 and y=0.

x couldn't be 13 because it would be assigned to y=5 and y=0.

So as long as x is not chosen to be -4,-3, or 13 your relation here is a function.

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A system of equations is given below. y=1/2x-3 and -1/2x-3. Which of the following statements best describes the two lines?

pantera1 [17]

"They have different slopes but the same y-intercept, so they have one solution" is the statement which best describes the two lines.

Answer: Option D

Step-by-step explanation:

Given equations:

$y=\left(\frac{1}{2} \times x\right)-3$

$y=\left(-\frac{1}{2} \times x\right)-3$

As we know that the slope intercept form of a line is

y = m x + c

So, from equation 1 and equation 2 we can see that

$m_{1}=\frac{1}{2} \quad \text { and } c_{1}=-3$

$m_{2}=-\frac{1}{2} \text { and } c_{2}=-3$

So, from the above expressions, we can say that both lines have different slopes but have same y – intercept with one common solution when x = 0.

4 0

3 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

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

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