Aliving specimen in equilibrium with the atmosphere contains one atom of 14c (half-life = 5730 yr) for every 7.7 ✕ 1011 stable carbon atoms. an archeological sample of wood (cellulose, c12h22o11) contains 22.2 mg of carbon. when the sample is placed inside a shielded beta counter with 88.0% counting efficiency, 811 counts are accumulated in one week. assuming that the cosmic-ray flux and the earth's atmosphere have not changed appreciably since the sample was formed, find the age of the sample.

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0