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Mathematics 14 Online
OpenStudy (hellokitty17):

Solve U = 0.5CV2 for C. C = U over 2V C = U over 2V squared C = 2U over V C = 2U over V squared

OpenStudy (anonymous):

\[U=0.5CV^2\] looks right?

OpenStudy (hellokitty17):

@jcoury

oregonduck (oregonduck):

yes

OpenStudy (anonymous):

idk on this one

OpenStudy (anonymous):

multiply by 2 on both sides

OpenStudy (anonymous):

can i ask you guys some questions when ur done

OpenStudy (hellokitty17):

then

oregonduck (oregonduck):

i will give 3 medals for the answer, sure

OpenStudy (anonymous):

\[U \times 2 = 2 \times 0.5CV^2\]

OpenStudy (anonymous):

what do ya get?

OpenStudy (anonymous):

this one is easy, trust me

oregonduck (oregonduck):

\[1.0CV ^{2}\]

OpenStudy (hellokitty17):

I don't understand how to do this that is why I asked

OpenStudy (anonymous):

\[2U=CV^2\]

oregonduck (oregonduck):

ks that the answer?

OpenStudy (anonymous):

now divide both sides by V^2

OpenStudy (anonymous):

\[\frac{ 2U }{ V^2 } = \frac{ CV^2 }{ V^2 }\]

oregonduck (oregonduck):

is that the answer?

OpenStudy (hellokitty17):

so what is the answer?!

OpenStudy (anonymous):

@jcoury just ask i have no problem helping ppl 2 or 3 at once

OpenStudy (anonymous):

nah, you need to simplify it

OpenStudy (anonymous):

k hold on

OpenStudy (anonymous):

\[\frac{ 2U }{ V^2 }=C\]

OpenStudy (anonymous):

thats your answeer

OpenStudy (anonymous):

@saseal check your messages

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

OpenStudy (anonymous):

guys its Saseal's birthday today!

oregonduck (oregonduck):

Happy Birthday @saseal

OpenStudy (anonymous):

ty

oregonduck (oregonduck):

:)

OpenStudy (anonymous):

lol

OpenStudy (hellokitty17):

HAPPY B-DAY!!! How old are you?!

OpenStudy (anonymous):

24

OpenStudy (hellokitty17):

Sweet!!! Any plans for today?!

OpenStudy (anonymous):

@saseal https://www.youtube.com/watch?v=nX6N2tgLmaQ

OpenStudy (anonymous):

watch it

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

i just felt that was appropriate

OpenStudy (anonymous):

i have a few more questions can i ask them?

OpenStudy (anonymous):

maybe 1 or 2 fast ones, i have some engineering problems to do also :(

OpenStudy (anonymous):

okay hold on

OpenStudy (anonymous):

OpenStudy (anonymous):

and if you have time

OpenStudy (anonymous):

i think i did the 2nd question wrong earlier

OpenStudy (anonymous):

so c for 2 right

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