4 ms·
> A goes as B to the power of C over D orders of magnitude. As a non-native English speaker, am I correct in assuming the above to mean this: A = B^C, A <
by msvan 11y ago
> A goes as B to the power of C over D orders of magnitude.
As a non-native English speaker, am I correct in assuming the above to mean this:
A = B^C, A < 10^D
I've never heard this grammatical construction before, and the images are not showing.
- monochromatic 11y agoClose; two minor details though. 1. A = k*B^C. We can have an arbitrary constant k in there. 2. 10^N > A > 10^(N+D). That is, A need not be small, it just has a constrained range. The "goes as" language is common, but a little loose/imprecise.
- msvan 11y agoThanks!