The given equation is

hence

But x cannot be zero so x=3
So the value of x is 3
h
Step-by-step explanation:

20, 20, 20, 30, 20
Tip for rounding nearest 10th: if the 2nd number is above 5 it’s the one above, if it’s below 5 it’s the one below.
There you go!! Let me know if you have questions