The radioactive isotope 14C has a half-life of approximately 5715 years. A piece of ancient charcoal contains only 81% as much of the radioactive carbon as a piece of modern charcoal. How long ago was the tree burned to make the ancient charcoal

1758 years

Step-by-step explanation:

0.693/t1/2 = 2.303/t log No/N

t1/2 = half life of the C-14

t= time taken

No= amount of radioactive material originally present

N= amount of isotope after time t.

But N= 0.81 No

0.693/5715= 2.303/t log No/0.81No

0.00012 = 0.211/t

t =0.211 /0.00012

t= 1758 years

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic