What happens when $f(t)=rect(\frac{x-10}{2})rect(\frac{x}{2}) $ ? Which takes priority?

Question

What happens when $f(t)=rect(\frac{x-10}{2})rect(\frac{x}{2}) $ ? Which takes priority?

The first function occurs between $9<x<11$ while the second rect occurs when $-1<x<1$. What exactly happens in this function?

EDIT:

Rect is defined as:

$$-\frac12<x<\frac12: 1 $$ $$else: 0$$

Answer 1

Just multiply point by point. You will see that $f(x)=0$.