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ad-work [718]
3 years ago
6

Make u the subject of formula in M=m(v-u).u​

Mathematics
1 answer:
Naily [24]3 years ago
8 0

Answer:

u = v - 1

Step-by-step explanation:

I'm guessing the formula you've given me is m = m(v - u), and I'm going to solve it that way. If it's actually different, let me know so I can redo the answer! :)

So, we have the equation m = m(v - u) and we're trying to make u the subject, or isolate it on one side of the equation. Right now, it's being subtracted from v and then multiplied by m. Let's start by dividing both sides of the equation by m so we can stop multiplying (v - u) by m.

m = m(v - u)

m/m = m(v-u)/m

1 = v - u

Now, our equation is pretty simple, but our u is still being subtracted from v, so we should add a u to both sides.

1 = v - u

1 + u = v - u + u

1 + u = v

Now, we're adding 1 to u, so let's go ahead and subtract 1 from both sides.

1 + u = v

1 - 1 + u = v - 1

u = v - 1

And we now we have u as the subject of our formula.

Hopefully that was helpful! :)

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State the domain and range of the function:<br>domain:<br>range:​
agasfer [191]

Answer:

Domain: [-5, 1)

Range: (5, 1]

General Formulas and Concepts:

<u>Algebra I</u>

  • Domain is the set of x-values that can be inputted into function f(x)
  • Range is the set of y-values that are outputted by function f(x)

Step-by-step explanation:

According to the graph, our x-values span from -5 to 1. Since -5 is a closed dot, it is inclusive in the domain. Since -1 is an open dot, it is exclusive in the domain:

[-5, 1)

According to the graph, our x-values span from -5 to 1. Since -5 is an open dot, it is exclusive in the range. Since 1 is a closed dot dot, it is exclusive in the range:

(5, 1]

7 0
3 years ago
Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an expl
BlackZzzverrR [31]

The tetrahedron passes through the intercepts (5, 0, 0), (0, 2, 0), and (0, 0, 10). Parameterize the surface (call it \Sigma) by

\vec r(u,v)=(1-v)\langle5,0,0\rangle+v\left((1-u)\langle0,2,0\rangle+u\langle0,0,10\rangle\right)

\vec r(u,v)=\langle5(1-v),2(1-u)v,10uv\rangle

with 0\le u\le1 and 0\le v\le1. Take the normal vector to \Sigma to be

\vec r_v\times\vec r_u=\langle20v,50v,10v\rangle

Then the flux of \vec F(x,y,z)=\langle x,y,z\rangle across \Sigma is

\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\langle5(1-v),2(1-u)v,10uv\rangle\cdot\langle20v,50v,10v\rangle\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1100v\,\mathrm du\,\mathrm dv=\boxed{50}

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What fraction of 2 minutes is 6<br>seconds?​
Inga [223]

Answer:

1/20 of 2 minutes = 6 seconds

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3 years ago
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Flauer [41]
Sorry but what are you asking
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3 years ago
quadrilateral WXYZ is translated (-5,3). if point W is located at (5,8), where will its image be located?
xeze [42]

The translation simply tells you how to operate on the coordinates of the points. So, translating (-5,3) means that you have to subtract 5 to the x coordinate and 3 to the y coordinate.

So, point (5,8) is mapped onto

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