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- Thread starter jhanson58
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- #2

- Mar 5, 2012

- 8,876

Welcome to MHB, jhanson58!I am trying to understand my math problem. I know what an exponent is. But what does it mean when X 2 you have a little 2 in the lower right hand corner.

V1=V2-u/1-uV2/C2 Solve for V2 How do I go about doing this.

A little 2 in the lower right hand corner distinguishes variables.

So $x_1$ is a different variable from $x_2$.

We might also call them simply $x$ and $y$.

As for your equation, can you clarify it?

I read it as:

$$V_1=V_2-{u \over 1}-{uV_2 \over C_2}$$

But I suspect that is not what you intended.

What did you intend?

Btw, as you can see I moved your post to a new thread, since it's a new topic.

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- Jan 30, 2012

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On our toolbar, you will see a button with the \(\displaystyle \sum\) character on it. Clicking this button will generate the MATH tags, between which you can put your $\LaTeX$ code. To see the code used by others, right click the expression, and from the pop-up menu choose Show Math As ► Tex Commands. You can then copy/past the code for your own use/modification.The problem is

V1=V2-u

-----

1-uV2

----

C2

Solve for V2

How do I put symbols or equation on this page.

There is a small learning curve to get familiar with the characters and commands, but once you get some practice it becomes second nature.

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- #6

- Mar 5, 2012

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Is it like this?The problem is

V1=V2-u

-----

1-uV2

----

C2

Solve for V2

How do I put symbols or equation on this page.

$$V_1 = \frac {V_2 - u} {1 - \frac{uV_2}{C_2}}$$

Note that when you click

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- #7

- Jan 30, 2012

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Well, you should right-click aI do not understand how to put the problem in correctly. when I right click on it.

\[

V_1 = \frac {V_2 - u} {1 - \frac{uV_2}{C_2}}

\]

You may need to turn Javascript on in your browser. If this does not work, use other suggestions above.

Last edited:

- Aug 30, 2012

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[tex]\large v_1 = \frac {v_2 - u} {1 - \frac{uv_2}{c^2}}[/tex]

-Dan

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- #10

- Mar 5, 2012

- 8,876

Nice one!

[tex]\large v_1 = \frac {v_2 - u} {1 - \frac{uv_2}{c^2}}[/tex]

-Dan

Then, to solve $v_2$ from it, we can observe that in SR from the perspective of the other observer, the velocities are added instead of subtracted. Therefore:

$$v_2 = \frac {v_1 + u} {1 + \frac{uv_1}{c^2}}$$

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- #12

Where does "it" say that?i need help! it says 0.999999.... is 1! how can That be?