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)
<span>Ok. You would set up the problem like this: 32/x=64/100 and then cross multiply; 3200=64x; divide both sides by 64; x=50. Mary has 50 customers.</span>
Answer:
Step-by-step explanation:
I gather that you need to find the area of the sector and then subtract from it the area of the segment to get the area of the triangle (although there are other ways in which to find the area of the triangle).
The area of a sector is:
where our angle is given as 34, pi is 3.1415, and the radius is 5:
and if you multiply and divide that all out you get that the area of the sector is:
A = 7.417
Now subtract from it the area of the segment, 2.209. to get the area of the triangle:
7.417 - 2.209 = 5.208
Answer:
y=45
x=9
z= 9 root 2
Step-by-step explanation:
This is a right isosceles triangle, so
y=45
x=9
z= 9 root 2
9²+9²=z²
81+81=z²
162=z²
√162=z
z=9√2