3 ms·
As I recall it, the available space is divided by the sum. So in this case the width for "2" would be: 2 * (available_width / (1 * other_items + 2))
by Excavator 12y ago
As I recall it, the available space is divided by the sum.
So in this case the width for "2" would be: 2 * (available_width / (1 * other_items + 2))