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Let Q represent a mass (in grams) of carbon (^14C), whose half-life is 5,715 years. The quantity of carbon-14 present after t years is given by the following
Q = 11(1/2)^(t/5715)

a. Determine the initial quantity (when t=0) g
b. Determine the quantity present after 2000 years. (Round your answer to 2 decimal places).

c. What is the graph of the function over the interval t = 0 to t = 10,000



Answer :

semsee45

Answer:

(a)  11 g

(b)  8.63 g

(c)  Attached

Step-by-step explanation:

Given function:

[tex]Q = 11 \left(\dfrac{1}{2}\right)^{\dfrac{t}{5715}}[/tex]

where:

  • Q = quantity of carbon-14 present
  • t = time (in years)

Part (a)

Substitute t = 0 into the given function to determine the initial quantity:

[tex]\begin{aligned} t=0 \implies Q &= 11 \left(\dfrac{1}{2}\right)^{\dfrac{0}{5715}}\\\\&=11\left(\dfrac{1}{2}\right)^0\\\\&=11\left(1\right)\\\\&=11\; \rm g\end{aligned}[/tex]

Part (b)

Substitute t = 2000 into the given function to determine the quantity present after 2000 years:

[tex]\begin{aligned} t=2000 \implies Q &= 11 \left(\dfrac{1}{2}\right)^{\dfrac{2000}{5715}}\\\\&=11\left(0.784697887...\right)\\\\&=8.63\; \rm g\;\; \sf (2 \; d.p.)\end{aligned}[/tex]

Part (c)

[tex]\begin{aligned} t=10000 \implies Q &= 11 \left(\dfrac{1}{2}\right)^{\dfrac{10000}{5715}}\\\\&=11\left(0.297346855...\right)\\\\&=3.27\; \rm g\;\; \sf (2 \; d.p.)\end{aligned}[/tex]

The graph of the function over the interval t = 0 to t = 10000 is attached.

View image semsee45

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