Question

Okay so just seeing if i have this right, solving for s=4v^2 and I have v=sqrts/4

Answer 1

\[v=\sqrt{S/5}\]

Answer 2

Solving for v?
Start by isolating v^2.
v^2 is being `multiplied` by 4, so we'll undo this operation by dividing by 4,\[\large\rm \frac{s}{4}=v^2\]Then apply square root to undo the `square` on v,\[\large\rm \sqrt{\frac s4}=\pm v\]When you take the square root of a square,
you get two results, a positive and negative case.

If this is for some type of physics class though,
where v can only be a positive quantity,
then you can ignore the plus/minus symbol in front.

Otherwise, move the plus/minus to the other side,\[\large\rm \pm\sqrt{\frac s4}=v\]

Answer 3

okay i see just double checking thank you so much