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3 years ago

Use the function rule. Find y for x = 1, 2, 3, and 4. y = x – 5

Mathematics

2 answers:

Fed [463]3 years ago

7 0

Answer:

is the fourth one

Step-by-step explanation:

just replace x with the numbers given

and then work it out.

y=x-5

y=1-5=-4

y=2-5=-3

and so on

hope this helps

Musya8 [376]3 years ago

3 0

For this case we have a function of the form $y = f (x)$ where:

$f (x) = x-5$

We must find the values of "y" when $x = 1,2,3,4$

Then, keeping in mind that different signs are subtracted and the sign of the major is placed:

$x = 1\\f (1) = 1-5 = -4\\So, y = -4$

$x = 2\\f (2) = 2-5 = -3\\So, y = -3$

$x = 3\\f (3) = 3-5 = -2\\Thus, y = -2$

$x = 4\\f (4) = 4-5\\f (4) = - 1\\Thus, y = -1$

Answer:

Option D

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2 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

Solve for e.<br><br><br>9e − 7 = 7e − 11

natka813 [3]

Answer:

e = -2

Step-by-step explanation:

9e - 7 = 7e - 11

9e - 7e = - 11 + 7

2e = -4

e = -2

6 0

3 years ago

The sum of three consecutive even numbers is 324. What are the three numbers?

Natalka [10]

Answer:

the ans is 108

Step-by-step explanation:

let the three consecutive no. be 'x'

by question:

x+x+x=324

or,3x=324

or,x =324÷3

or, x=108

checking this answer

108+108+108

=324

4 0

3 years ago