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3 years ago

Can someone help me with my latest question

Mathematics

2 answers:

Artyom0805 [142]3 years ago

6 0

Answer:

yea

Step-by-step explanation:

Gelneren [198K]3 years ago

5 0

Answer:

Hey!

I understand you want help, but there are tons and tons of users on Brainly. Whether they come for help or <em>to </em>help, there are a lot. Like a lot! So there are so many questions to answer.

So if you're latest question doesn't get answered, whether it be difficulty or time, it probably just means that someone didn't quite get to it.

Also, yes, I will look at it!

Thank you!

Sahkfam

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How do you find point-slope form with 2 points?

kumpel [21]

y-y1=m(x-x1) where m is the slope, and (x1,y1) can be either one of the two points you are given. First, we need to find the slope m from the given points. Now put the point (2,3) in the point-slope equation along with the slope m to find the equation of the line between the two points.

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3 years ago

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Could someone help me with this

lianna [129]

The answer is:

5.7•10^-3

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3 years ago

I need help to solve this please -6 - 8x = -2(4x + 3)

OverLord2011 [107]

Answer:

infinite solutions, true for all x

Step-by-step explanation:

-6 - 8x = -2(4x + 3)

Distribute

-6-8x =-8x-6

Add 8x to each side

-6-8x+8x = -8x-6+8x

-6 =-6

Since this is always true, the equation is true for all x

8 0

3 years ago

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Surface integrals using a parametric description. evaluate the surface integral \int \int_{s} f(x,y,z)dS using a parametric desc

DiKsa [7]

You can parameterize $S$ using spherical coordinates by

$\vec s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial\vec v}\times\dfrac{\partial\vec s}{\partial\vec u}=36\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle$

(I use $\vec s_v\times\vec s_u$ to avoid negative signs. The orientation of the normal vector doesn't matter for a scalar surface integral; you could just as easily use $\vec s_u\times\vec s_v=-(\vec s_v\times\vec s_u)$ .)

Then

$f(x,y,z)=f(6\cos u\sin v,6\sin u\sin v,6\cos v)=36\sin^2v$

and the integral of $f$ over $S$ is

$\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\int_0^{\pi/2}\int_0^{2\pi}36\sin^2v\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}(36\sin^2v)(36\sin v)\,\mathrm du\,\mathrm dv$

$=\displaystyle2592\pi\int_0^{\pi/2}\sin^3v\,\mathrm dv=\boxed{1728\pi}$

7 0

3 years ago

1. What is the slope and y-intercept of y=-3x - 2?

Nadya [2.5K]

Answer: The equation is y=mx+b where m is the slope and b is the intercept, so for your problem slope is 3 and the intercept is -2

Step-by-step explanation:

6 0

3 years ago