After some ancient bones were excavated, carbon was taken from one of the bones to find that there was 17% as much Superscript 14 Baseline Upper C as from current bones. The decay constant of Superscript 14 Baseline Upper C is k almost equals 0.0001216, for time in years. How many years old is the bone?

The bone is 14572 years old.

Step-by-step explanation:

There is 17% as much Carbon-14 in the ancient bones as from current bones and the decay constant is given as 0.0001216 for time in years.

The equation for radioactive decay is N = N₀ e^(-λt)

where t = time in years

          λ = decay constant

          N = final amount of carbon

          N₀ = initial amount of carbon.

There is 17% as much Carbon-14 in the ancient bones as in current bones. So,

N = 17% of N₀

N = 0.17N₀

So, N/No = 0.17.

N = N₀ e^(-λt)

N/N₀ = e^(-λt)

0.17 = e^(-0.0001216t)

ln (0.17) = -0.0001216t

-0.0001216t = -1.7719568

t =  -1.7719568/-0.0001216

t = 14572 years

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

make sure you have version 2.50

try looking in the very back of the book : )

step-by-step explanation:

you can download it here