Answer: ![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
Step-by-step explanation:

First, multiply by 2 to get rid of the 2 in the denominator. Remember that if you make any changes you have to make sure the equation keeps balanced, so do it on both sides as following;


Divide by m to isolate
.


To eliminate the square and isolate v, extract the square root.
![\sqrt[]{\frac{2K}{m} }=\sqrt[]{v^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3D%5Csqrt%5B%5D%7Bv%5E2%7D)
![\sqrt[]{\frac{2K}{m} }=v](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3Dv)
let's rewrite it in a way that v is in the left side.
![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
8 and 1
8 squared is 64 and 1 squared is 1
64 + 1 = 65
Hope this helps :)
I hope it helps you get it right
Answer: 5+x
Step-by-step explanation:
Not sure if you want it in an equation but here it is: 5+x