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3 years ago

Which one of the following reasons does NOT explain why 4x+5y does not equal 9xy

Mathematics

1 answer:

gladu [14]3 years ago

4 0

Answer:

Because you can't add different variables

Step-by-step explanation:

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A company made $4 million in the second quarter. This is 1/3 more than it made in the first quarter

Paraphin [41]

It made 3 million in the first quarter

8 0

2 years ago

What is the converse of “if it is raining then we will not go the the beach

Dmitrij [34]

If we do not go to the beach, it is raining

4 0

2 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

Determine whether or not the relation is a function. Please help asap

OLga [1]

Answer:

A function can have many x values for a given y value and still be a function but it cannot have many y values for a given x value. You can easily test this by graphing the equation and doing the vertical line test by seeing if a vertical line will intersect the graph only once at all x values, if it passes it is a function. In short what you have here is a function

Step-by-step explanation:

4 0

2 years ago

Jacob earned $80 babysitting and deposited the money into his savings account. The next week he spent $85 Lon video games. Use i

nexus9112 [7]

After one week, Jacob has earned $80 from babysitting, so his change is an increase of $80. The next week, he spends $85, so he experiences a decrease of $85. Over the two weeks, his overall change is $80-$85=-$5, or a decrease of $5.

5 0

2 years ago