Answer: x ≤ 10
Step-by-step explanation:
20 + 13x ≤ 150
13x ≤ 150 - 20
13x ≤ 130
x ≤ 130 ÷ 13
x ≤ 10
Answer:
and 
Step-by-step explanation:
The equation of the isotope decay is:

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:


The decay time is:
(There is no a year 0 in chronology).

Lastly, the relative amount is estimated by direct substitution:





Your answer would be x-4=17.
4 less than (#) is 17.
That number would be 21.
21-4=17
Hope that helps
1.7/1.9 = 0.894 (rounded)