A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 mg/L of the material present and after 2 hours it is observed that the material has lost 10 percent of its original concentration, find (a) an expression for the concentration of the material remaining at any time.

Step-by-step explanation:

A certain radioactive material is known to decay at a rate proportional to the amount present.

This means that the situation can be described by the following differential equation:

In which r is the decay rate.

Solving the differential equation by separation of variables, we have that:

Integrating both sides:

In which K is the integrative constant.

Applying the exponential to both sides, to remove the ln, we get:

In which K is the initial amount present.

Initially there is 50 mg/L

This means that .

After 2 hours it is observed that the material has lost 10 percent of its original concentration

This means that . So

So

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

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step-by-step explanation:

what is the original number? you forgot to add it.