5 ms·
In the Introduction, under functions: f = m*a f(20) = 20a Which is ok, but then farther down: f(a) = m*a f(20) => 20a This seems incorrect, since I
by ringmaster 13y ago
In the Introduction, under functions:
f = m*a
f(20) = 20a
Which is ok, but then farther down:
f(a) = m*a
f(20) => 20a
This seems incorrect, since I was expecting m*20. On a separate page, this seems to work fine, though.
Otherwise, quite nice.
- praeclarum 13y agoBug in the doc. :-) just ran on the device and got the expected: f(a) = m*a f(20) => 20m