HelloKitty17
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[U=0.5CV^2\] looks right?
HelloKitty17
  • HelloKitty17
@jcoury
OregonDuck
  • OregonDuck
yes

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More answers

anonymous
  • anonymous
idk on this one
anonymous
  • anonymous
multiply by 2 on both sides
anonymous
  • anonymous
can i ask you guys some questions when ur done
HelloKitty17
  • HelloKitty17
then
OregonDuck
  • OregonDuck
i will give 3 medals for the answer, sure
anonymous
  • anonymous
\[U \times 2 = 2 \times 0.5CV^2\]
anonymous
  • anonymous
what do ya get?
anonymous
  • anonymous
this one is easy, trust me
OregonDuck
  • OregonDuck
\[1.0CV ^{2}\]
HelloKitty17
  • HelloKitty17
I don't understand how to do this that is why I asked
anonymous
  • anonymous
\[2U=CV^2\]
OregonDuck
  • OregonDuck
ks that the answer?
anonymous
  • anonymous
now divide both sides by V^2
anonymous
  • anonymous
\[\frac{ 2U }{ V^2 } = \frac{ CV^2 }{ V^2 }\]
OregonDuck
  • OregonDuck
is that the answer?
HelloKitty17
  • HelloKitty17
so what is the answer?!
anonymous
  • anonymous
@jcoury just ask i have no problem helping ppl 2 or 3 at once
anonymous
  • anonymous
nah, you need to simplify it
anonymous
  • anonymous
k hold on
anonymous
  • anonymous
\[\frac{ 2U }{ V^2 }=C\]
anonymous
  • anonymous
thats your answeer
anonymous
  • anonymous
@saseal check your messages
anonymous
  • anonymous
ok
anonymous
  • anonymous
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anonymous
  • anonymous
guys its Saseal's birthday today!
OregonDuck
  • OregonDuck
Happy Birthday @saseal
anonymous
  • anonymous
ty
OregonDuck
  • OregonDuck
:)
anonymous
  • anonymous
lol
HelloKitty17
  • HelloKitty17
HAPPY B-DAY!!! How old are you?!
anonymous
  • anonymous
24
HelloKitty17
  • HelloKitty17
Sweet!!! Any plans for today?!
anonymous
  • anonymous
@saseal https://www.youtube.com/watch?v=nX6N2tgLmaQ
anonymous
  • anonymous
watch it
anonymous
  • anonymous
lol
anonymous
  • anonymous
i just felt that was appropriate
anonymous
  • anonymous
i have a few more questions can i ask them?
anonymous
  • anonymous
maybe 1 or 2 fast ones, i have some engineering problems to do also :(
anonymous
  • anonymous
okay hold on
anonymous
  • anonymous
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anonymous
  • anonymous
and if you have time
1 Attachment
anonymous
  • anonymous
i think i did the 2nd question wrong earlier
anonymous
  • anonymous
so c for 2 right

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