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3 years ago

Please show me how to do this!!

Mathematics

1 answer:

Murrr4er [49]3 years ago

7 0

(2x^4 + 3x^2)(x^2-1)
you multiply and you'll get:
2x^8-2x^4+3x^4-3x^2
then add the coefficients with x^4
= 2x^8+1x^4-3x^2

the coefficient for x^4 is 1, so the answer is B

You might be interested in

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

Y=x7<br><br><br> Please help I really confused

tigry1 [53]

Answer:

Where is the rest of the question? What is it asking?

Step-by-step explanation:

Well I guess if you want to change up the equation you can do 7 * X = Y

The * is just a multiplecation symbol if you didnt know

8 0

3 years ago

How do you do this?????

Setler79 [48]

The lowest common multiple of 7 and 5 is 35, so you divide 365 by 35, the whole number is your answer,
the answer is 10

4 0

3 years ago

My questions are in the picture please help

Marrrta [24]

21.

So to solve for the value of a variable in an equation, you would need to isolate x on one side. The first thing you have to do is multiply 2 and (x + 7) together, which can be rewritten as 2(x) + 2(7). After multiplying, your equation would be: $2x+14+3x=12$ , or A.

22.

So the rule with multiplying exponents with the same base is to add the exponents together, and the rule with dividing exponents with the same base is to subtract the exponents. Your equation will be simplified as such:

$4^{(3+4)-9}\\ 4^{7-9}\\ 4^{-2}$

Now the rule with converting negative exponents into fractions is $x^{-m}=\frac{1}{x^m}$ . In this case, 4^-2 would turn into $\frac{1}{4^2}$ , or C, which is your final answer.

6 0

3 years ago

Nas funções f(x) = -3x+9; f(x) = 2x-4 e f(x) = 5x-5, caso construamos seus respectivos gráficos, informe respectivamente os pare

Likurg_2 [28]

1st option

{(3,0) e (0,9)}; {(2,0) e (0,-4)}; {(1,0) e (0,-5)}

see screenshot

sorry btw, no hablo espanol

8 0

3 years ago