Hello! So something I learned in OH was that the it seemed weird to me that there was no proportion for (1-x).
If you actually split up the equation, we can get
v0-xv0 + xv1, which is (1-x)v0 + xv1.
Also! Someone brought up a good way to think about this is that you start from v0, and you want a certain amount higher, and that certain amount is proportional to x and (v1-v0)
yup, one interpretation of this is that it's a weighted average between v1 and v0, where x and (1-x) are the weights, respectively