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3 years ago

Someone explain what happened from line 1 to line 2

Mathematics

1 answer:

ser-zykov [4K]3 years ago

7 0

Expand the 1st line ==> xlog2+3log2 = xlog6 - log6

Transpose : xlog2 - xlog6 = -log6 - 3log2

MULTIPLY BOTH SIDE BY (-1)==> -xlog2+xlog6 = log6 +3log2
FACTORIZE: x(log6-log2) = log6+3log2
Got it??

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

Write and solve and equation to find how many movies they have time to show.

GalinKa [24]

Answer:

Equation 0.5 + 1.75(x) = 4

x = 2movies

Step-by-step explanation:

Given that,

movie night will last for 4 hours

each movie take 1.75 hours

time for eating popcorn 0.5 hours

To find,

number of movies which can be watched

<h2>Equation</h2><h2>0.5 + 1.75(x) = 4</h2><h2>Solve for x</h2><h3>1.75(x) = 4 - 0.5</h3><h3>1.75(x) = 3.5</h3><h3>x = 3.5/1.75</h3><h3>x = 2 </h3>

So, total number of movies which could be watched in 4hr = 2

7 0

3 years ago

PLS ANWSER ASAP IN TI<br> ED!!

NikAS [45]

Answer:

Option (2)

Step-by-step explanation:

x = 1 is represented by a solid point on a number line.

x > 1 is represented by an arrow starting from x = 1 towards infinity

If we mix both the properties, x ≥ will be represented by an arrow starting from a solid point at x = 1 and moving towards the values greater than one.

From the options given,

Arrow mentioned in Option (2) will be the correct representation of the inequality.

7 0

2 years ago

Points C, D, and G lie on plane X. Points E and F lie on plane Y.Which statements are true? Select three options.

Slav-nsk [51]

Answer:

2nd , 3rd, & 4th

Step-by-step explanation:

4 0

3 years ago

How do I solve this?

lubasha [3.4K]

Here is your answer to the question

8 0

3 years ago