If 14 g of a radioactive substance are present initially and 2 yr later only 7 g remain, how much of the substance will be present after 5 yr?

2.47g

Step-by-step explanation:

Given that Ao=14

A(t)= Aoe^kt

k =2yrs

Considering the function where Ao is the initial substance, k is the rate of decay and t is the time.

At the 2nd year A(t) is 7

That is,

7 = 14e^2k

Solve for k

7/14 = e^2k

In(7/14) = 2k

k = In(7/14)÷2

k = In(0.5)/2

k = -0.34657

Hence, A(t)= 14e^-0.34657

So, when t is 5

A(5) = 14e^-0.34657×5

= 2.4749

Approx = 2.47g

i need more details sorry

step-by-step explanation:

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb