All categories

4 years ago

What is u = kx + ух, for x

Mathematics

1 answer:

Ivahew [28]4 years ago

6 0

Answer:

Step-by-step explanation:

u = kx + ух

First of all factorize x out at the right side of the equation

That's

u = x(k + y)

Divide both sides by ( k + y) to make x stand alone

That's

We have the final answer as

Hope this helps you

You might be interested in

What is 15.37×5 show work

just olya [345]

This should be quite easy to explain. So the first thing you want to do is multiply them. This isn't to hard if you know your times tables. Don't worry about the decimal till the end.

15.37
× 5
---------
76.85

15.37 times 5 is 76.85

Now on to the decimal. All you need to do is move it two times to the left. You see how it's in between the 5 and 3? The same goes for your answer. So it is 76.85

I hope this helped!! ^^ I can make this a little more easier to explain if you want me to just in case, ok?

3 0

3 years ago

Read 2 more answers

Evalute x(-y+z) for x =-1 y=2, and z=5

Taya2010 [7]

Answer:

-3

Step-by-step explanation:

x(-y+z)

-1(-2+5)

-1(3)

-3

8 0

4 years ago

Find the missing angles in the diagram below (pls help no links)

damaskus [11]

Step-by-step explanation:

here you have the answers

3 0

3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$