the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 as does living plant material. Determine when the wood was cut

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

(Eq. 1)

Where:

- First derivative of mass in time, measured in miligrams per year.

- Time constant, measured in years.

- Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

(Eq. 2)

Where:

- Initial mass of isotope, measured in miligrams.

- Time, measured in years.

And time is cleared within the equation:

Then, time constant can be found as a function of half-life:

(Eq. 3)

If we know that and , then:

The wood was cut approximately 8679 years ago.

x = 10.1

step-by-step explanation:

all of the angles of a tirangle add up to 180º

180 - 65 = 115 (since we already know the value of one angle)

(2x + 7) + (8x + 7) = 115

(combins like terms) 10x + 14 = 115

(isolate the variable subtract 14 from both sides): 10x = 101

(find the variable divide both sides by 10): x = 101/10

x = 10.1

x

step-by-step explanation: