In this scenario we know that carbon14 at a given time A(t) = A0e^(-kt), where A0 is the original carbon present, t is time in years and k is the constant.
As we are working with the half life of carbon being 5730 years, we assume original carbon-14 content, A0 = 1, and carbon-14 at half life 5730 years, A(t) = 0.5.
i.e.
0.5 = 1e^(-5730k)
apply Ln to both sides of equation to cancel e
ln(0.5) = -5730k
k = ln(0.5) / -5730
k = -0.69315 / - 5730 = 1.20968 x 10^-4
Answer:
f(5) = 16.5
Step-by-step explanation:
Substitute x = 2.5 into f(x) , that is
f(2.5) = 6(2.5) + 1.5 = 15 + 1.5 = 16.5
Answer:
b
Step-by-step explanation:
Answer:
Step-by-step explanation:
Let a = - 1 2/5 and b = 2 1/3
<u>We have a + b, lets compare with the options:</u>
- A) a + (- b) = a - b, no
- B) a - b, no
- C) b + a = a + b, yes
- D) b - a, no
